Edition 09.99
SIMOVERT MASTERDRIVES
Application Manual
SIMOVERT VC
Load
D
IM
Gearbox Cable drum
09.99
2Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
No guarant ee can be accept ed for the general validity of the calculation techniques described and
presented here. As pr eviously, the configuring engineer is fully responsible for the results and the
subsequently selected motors and drive converters.
09.99 Contents
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Contents
1 FOREWORD............................................................................................................7
2 FORMULAS AND EQUATIONS ..............................................................................9
3 VARIOUS SPECIAL DRIVE TASKS......................................................................25
3.1 Traction- and hoisting drives..................................................................................25
3.1.1 General information................................................................................................25
3.1.2 Traction drive with brake motor..............................................................................38
3.1.3 Hoisting drive with brake motor..............................................................................49
3.1.4 High-bay racking vehicle........................................................................................60
3.1.5 Hoisting drive for a 20 t gantry crane......................................................................77
3.1.6 Elevator (lift) drive..................................................................................................83
3.1.7 Traction drive along an incline................................................................................92
3.2 Winder drives.......................................................................................................103
3.2.1 General information..............................................................................................103
3.2.2 Unwind stand with closed-loop tension control using a tension transducer...........110
3.2.3 Winder with closed-loop tension control using a tension transducer.....................117
3.2.4 Unwinder with 1FT6 motor...................................................................................123
3.2.5 Unwind stand with intermittent operation..............................................................131
3.3 Positioning drives................................................................................................. 139
3.3.1 General information..............................................................................................139
3.3.2 Traction drive with open-loop controlled positioning using Beros .........................147
3.3.3 Elevator drive with closed-loop positioning control (direct approach)....................149
3.4 Drives with periodic load changes and surge loads..............................................151
3.4.1 General information..............................................................................................151
3.4.2 Single-cylinder reciprocating compressor.............................................................161
3.4.3 Three-cylinder pump ............................................................................................167
3.4.4 Transport system for freight wagons.................................................................... 171
3.4.5 Drive for an eccentric press..................................................................................184
3.5 Load distribution for mechanically-coupled drives................................................197
3.5.1 General information..............................................................................................197
3.5.2 Group drive for a traction unit...............................................................................204
3.5.3 Drive for an extraction tower with 10 mechanically-coupled motors......................207
3.6 Crank drive...........................................................................................................208
3.7 Rotary table drive.................................................................................................221
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3.8 Pivot drive ............................................................................................................227
3.9 Spindle drive with leadscrew................................................................................238
3.10 Cross-cutter drive.................................................................................................254
3.11 Centrifugal drive...................................................................................................261
3.12 Cross-cutter with variable cutting length...............................................................272
3.13 Saw drive with crank ............................................................................................293
3.14 Saw drive as four-jointed system..........................................................................306
3.15 Mesh welding machine.........................................................................................323
3.16 Drive for a pantograph .........................................................................................357
3.17 Drive for a foil f eed with sin2 rounding-of f.............................................................377
4 INFORMATION FOR SPECIFIC APPLICATIONS...............................................387
4.1 Accelerating time for loads with square-law torque characteristics.......................387
4.1.1 General information ..............................................................................................387
4.1.2 Accelerating time for a fan drive...........................................................................389
4.1.3 Accelerating time for a blower drive (using the field-weakening range)................392
4.1.4 Re-accelerating time for a f an drive after a brief supply failure.............................398
4.1.5 Calculating the minimum braking time for a fan drive...........................................406
4.2 Connecting higher output motors to a drive converter than is normally permissible....418
4.2.1 General information ..............................................................................................418
4.2.2 Operating a 600 kW motor under no-load conditions ...........................................422
4.3 Using a transformer at the drive converter output ................................................425
4.3.1 General information ..............................................................................................425
4.3.2 Operating a 660 V pump motor through an isolating transformer.........................427
4.3.3 O perating 135V/ 200 Hz handheld gr inding machines through an isolating
transformer...........................................................................................................428
4.4 Emergency off......................................................................................................430
4.4.1 General information ..............................................................................................430
4.5 Accelerat ing - and decelerating time with a constant load torque in the
field-weakening range..........................................................................................436
4.5.1 General information ..............................................................................................436
4.5.2 Braking time for a grinding wheel drive ................................................................438
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4.6 Influence of a rounding-off f unction...................................................................... 442
4.7 Possible braking torques due to system losses....................................................453
4.7.1 General information..............................................................................................453
4.7.2 Braking time for a fan drive.................................................................................. 460
4.8 Minimum acceleration t im e for fan drives with a linear acceleration
characteristic........................................................................................................465
4.8.1 General information..............................................................................................465
4.8.2 Example............................................................................................................... 471
4.9 Buffering multi-motor drives with kinetic energy...................................................474
4.9.1 General information..............................................................................................474
4.9.2 Example for buffering with an additional buf fer drive............................................479
4.10 Harmonics fed back into the supply in regenerative operation ............................. 483
4.11 Calculating the braking energy of a mechanical brake.........................................491
4.11.1 General information..............................................................................................491
4.11.2 Hoisting drive with counterweight.........................................................................491
4.11.3 Traversing drive ...................................................................................................495
4.12 Criteria for selecting motors for clocked drives.....................................................499
4.12.1 General information..............................................................................................499
4.12.2 Example 1, rotary table drive with i=4 and short no-load interval..........................507
4.12.3 Example 2, rotary table drive with i=4 and long no-load interval...........................510
4.12.4 Example 3, rotary table drive with a selectable ratio and short no-load interval.... 512
4.12.5 Example 4, rotary table drive with selectable ratio and long no-load interval........515
4.12.6 Example 5, traversing drive with selectable ratio and longer no-load time............517
4.13 Optimum traversing characteristics regarding the maximum motor torque
and RMS torque...................................................................................................520
4.13.1 Relationships for pure flywheel drives (high-inertia drives)................................... 520
4.13.2 Relat ionships for flywheel drives (high-iner tia drives) with a constant load torq ue 524
4.13.3 Summary..............................................................................................................526
4.13.4 Example with 1PA6 motor....................................................................................527
5 INDEX..................................................................................................................533
6 LITERATURE....................................................................................................... 536
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09.99 1 Foreword
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1 Foreword
This Application Manual f or the new SIMOVERT MASTERDRIVES drive converter ser ies is
conceived as supplement to the Engineering Manual and the PFAD configuring/ engineering
progr am . Using numerous examples, it provides support when engineering com plex drive tasks,
such as for example t r action- and hoisting dr ives, winders etc. I t is shown how to calculate the
torq ue- and motor output s from the mechanical drive data and how to select the motor, drive
converter, cont r ol type and additional components. As result of the diversity of the various drive
tasks, it is not possible to pr ovide detailed inst ructions with wiring diagrams, parameter set t ings
etc. T his would also go far beyond the scope of this particular docum ent. In some cases, the
examples are based on applications using previous drive converter system s and actual inquiries
with the appropriate dr ive data.
Further, this Application Manual provides informat ion for specif ic applications, f or example, using a
transformer at the converter output, accelerating time for square-law load torq ue char acteristics
etc. T his is supplemented by a summary of formulas and equations with units, formulas and
equations for induction motors, converting linear into rotat ional m ot ion, the use of gearboxes and
spindles as well specifying and converting moments of inertia.
Your support is r equired in updating the Application Manual, which means improvements,
corrections, and the inclusion of new examples. This docum ent can only become a real support
tool when we receive the appropriate feedback as far as its contents ar e concer ned and inquiries
regarding the applicat ion in conjunction with the new SIMOVERT MASTERDRIVES dr ive
converter series.
This Application Manual is conceived as a working document for sales/m ar keting per sonnel. At the
present tim e, it is not intended to be passed on to cust omers.
1 Foreword 09.99
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09.99 2 Formulas and equations
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2 Formulas and equations
Units, conversion
1 N = 1 kg m / s 2 = 0. 102 kp force
1 kg mass
1 W power
1 Ws = 1 Nm = 1 J work, energ y
1 kg m 2 = 1 Ws3 = 1 Nms2 moment of inertia
g = 9.81 m/ s 2 acceleration due to t he force of gravity
Induction mot or
nf
p
060=⋅ synchronous speed
f in Hz, n in RPM
p: Pole pair number , f: Line supply f r equency
snn
n
=−
0
0slip
fnn
nf
s=−⋅
0
0slip fr equency
PUI
M
n
=⋅⋅⋅ ⋅=⋅
39550
cos
ϕη
shaft out put
MP
n
=⋅9550 torque
M in Nm, P in kW, n in RPM
Mnn
nn
M
nn
≈−
−⋅
0
0torq ue in t he r ange n0 to nn
Mn: Rated torque, nn: Rat ed speed
MV
VM
stall nstall n
≈⋅()
2stall torque at reduced voltag e
2 Formulas and equations 09.99
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MffM
stall nstall n
≈⋅()
2stall torque in the f ield- weakening range
Vn: Rated voltag e, fn: Rat ed freq uency
Mstall n: Rated stall torque
RMS torque
The RMS torque can be calculat ed as follows if the load duty cycle is small with respect t o t he
motor thermal time constant:
MMt
tkt
RMS
ii
i
efp
=⋅
+⋅
∑2
The following must be true: MRMS ≤ Mpermissible for S1 duty
titime segment s with constant t or que Mi
tepower-on duration
tpno load time
k
freduction f act or ( de- r at ing factor)
Factor kf for the no- load time is a derating factor for self-ventilated mot or s . For 1LA5/1LA6 motors
it is 0.33. For 1PQ6 motors, 1PA6 motors and 1FT 6/ 1FK6 m otors, this factor should be set to 1.
Example
tt
21
M
t
MRMS
tp
te
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Comparison of the paramet er s for linear- and r otational motion
Linear mot ion Rotational mo t ion
adv
dt
=[m/s2]acceleration
α
ω
=d
dt [s-2]angular acceler.
vds
dt adt==⋅
∫[m/s]velocity
ω
ϕ
α
==⋅
∫
d
dt dt [s-1]angular velocity
svdt=⋅
∫
[m]distance
ϕω
=⋅
∫
dt [rad]angle
m[kg]mass J[kgm2]moment of inertia
F[N]force
M
[Nm]torque
Fma=⋅ [N]accelerating force
M
J=⋅
α
[Nm]acceler. torque
PFv=⋅ [W]power P
M
=⋅
ω
[W]power
WFs=⋅ [Ws]work WM
=⋅
ϕ
[Ws]work
Wmv=⋅⋅
1
22[Ws]kinetic energy WJ=⋅⋅
1
22
ω
[Ws]rotat ional ener gy
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Example : Relationships between acceleration, velocity and distance for linear mot ion
Area corresponds to velocity v
Area corresponds to distance s
t
t
t
Acceleration a
Velocity v
Distance s
09.99 2 Formulas and equations
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Converting linear - into rotat ional m otion
Example: Hoisting a load with constant velocity
sr
m
F
v
ϕ
ω
, n
vvelocity [m/s]
r
roll radius [m]
ttime [s]
ω
=v
r
angular velocity [s-1]
nv
r
=⋅⋅=
⋅⋅⋅
ω
π
π
260 260 speed [RPM]
ω
=⋅tangle [rad]
svt
r
=⋅=⋅
distance [m]
mmass moved [kg]
Fmg=⋅ hoisting force [N]
M
F
r
mg
r
=⋅=⋅⋅ torque [Nm]
P
M
Fv mgv=⋅=⋅=⋅⋅
ω
power [W]
WFsM mgs=⋅= ⋅=⋅⋅
ϕ
hoisting work [Ws]
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Example: Accelerating a m ass on a conveyor belt with constant acceler ation
sF
a, v
r
m
ϕ
αω
,
aacceleration [m/s2]
r
roll radius [m]
taccelerating time [s]
α
=a
r
angular acceler at ion [s-2]
vat
r
t=⋅=⋅⋅
α
velocity [m/s]
ωα
==⋅
v
r
tangular velocity [s-1]
nv
r
=⋅⋅=
⋅⋅⋅
ω
π
π
260 260 speed [RPM]
ϕα
=⋅
t2
2angle [rad]
sa
tvt
=⋅ =⋅
2
22 distance [m]
mmass moved [kg]
Fma=⋅ accelerating force [N]
M
F
r
ma
r
=⋅=⋅⋅ accelerating torque [Nm]
P
M
Fv mav=⋅=⋅=⋅⋅
ω
accelerating power [W]
WFsM mv mr=⋅= ⋅=⋅⋅ =⋅⋅⋅
ϕω
1
21
2
222
accelerating work [Ws]
09.99 2 Formulas and equations
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Converting linearly moved masses into m om ents of inertia
The conversion is realized by equating t he kinetic energ y to t he r otational energy, i. e.
1
21
2
22
⋅⋅ =⋅⋅mv J
ω
and Jmv
=⋅()
ω
2
Example: Conveyor belt
Jm
r
=⋅
2moment of inertia referred to the roll [kgm2]
mmass [kg]
r
radius [m]
Example: Hoisting dr ive
Jmr=⋅⋅
1
42moment of inertia referred to the fixed roll [kgm2]
mmass [kg]
r
fixed roll r adius [m]
r
v
=
ω
m
r
v
ω
r
mv
2v
ω
r
v2
=
ω
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Using gearboxes
in
nMotor
load
=gearbox ratio
gearbox ef ficiency
Referring moments of inert ia t o t he motor shaft
JJ
Motor Load
Gearbox
nn
Motor Load
i,
η
JJi
load load
∗=2load moment of inertia referred to the motor shaft [kgm2]
JJ Ji
motor load
∗=+
2total mom ent of inertia refer r ed t o the motor shaft (without
gearbox and coupling)
[kgm2]
Referring load t or ques t o the motor shaf t
Example: Hoisting dr ive, hoist ing and lowering with constant velocity
η
m
Motor Gearbox
r
i,
09.99 2 Formulas and equations
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r
roll radius [m]
Fmg
H=⋅ hoisting force [N]
Mmgr
load =⋅⋅ load torque at the roll [Nm]
MM
i
load load
∗=⋅
η
load torq ue r eferred t o the motor shaft when hoisting
(energ y f low from the mot or t o the load) [Nm]
MMi
load load
∗=⋅
η
load torq ue r eferred t o the motor shaft when lowering
(energ y f low from the load to t he m otor) [Nm]
Gearbox rat io f or the shortest acceler at ing t im e w it h a constant motor t orque
a) Mload = 0, i. e., pure high-inertia drive without friction
iJ
Jload
motor
=and JiJ
load motor
2=
this means, t hat the load moment of inertia, r eferred t o the motor shaft, m ust be the same as the
motor m om ent of inertia
b) Mload ≠ 0 , i. e., t he dr ive accelerat es against f r ict ional forces
iM
MM
MJ
J
load
motor
load
motor
load
motor
=+ +()
2
2 Formulas and equations 09.99
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Using spindle drives
m
hSp
Motor
l
D
Mmotor
nmotor
Guide v, a
vfeed velocity [m/s]
hSp pitch [m]
D
spindle diameter [m]
lspindle length [m]
Jl
D
spindle ≈⋅⋅ ⋅ ⋅
π
2785 21000
4
.() spindle moment of inert ia ( steel) [kgm2]
mmoved masses [kg]
nvh
motor Sp
=⋅60 motor speed [RPM]
α
π
SW Sp
h
D
=⋅
arctan spindle pit c h angle [rad]
ρπη
α
=⋅⋅−arctan h
D
Sp SW fr ict ion angle of t he spindle [rad]
spindle efficiency
α
π
bvSp bv Sp
ah
,,
=⋅
⋅2angular acceleration - deceleration of the spindle [s-2]
abv,acceleration- deceleration of the load [m/s2]
MJ
bvmotor mot bvSp,,
=⋅
α
accelerating- and decelerating t or que for t he
motor
[Nm]
MJ
bvSP Sp bvSp,,
=⋅
α
accelerating- and decelerating t or que for t he
spindle
[Nm]
Fma
bv bv,,
=⋅ accelerating- and deceleration force for m ass m [N]
09.99 2 Formulas and equations
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Horizontal spindle drive
Fmgw
RF
=⋅⋅ frictional for c e in t he guide [N]
wFspecific t r aversing resistance
Motor torque when accelerating
MM MFF
D
motor b motor b Sp b R SW
=+++⋅⋅+()tan()
2
αρ
[Nm]
Motor torque when decelerating
MMMFF
Dsign F F
motor v motor v Sp v R SW v R
=− − + − + ⋅ ⋅ + ⋅ − +()tan( ())
2
αρ
[Nm]
Vertical spindle drive
Fmg
H=⋅ hoisting force [N]
Motor torque when accelerating upwards
MM MFF
D
motor b motor b Sp b H SW
=+++⋅⋅+()tan()
2
αρ
[Nm]
Motor torque when decelerating downwards
MMMFF
Dsign F F
motor v motor v Sp v H SW v R
=− − + − + ⋅ ⋅ + ⋅ − +()tan( ())
2
αρ
[Nm]
Motor torque when accelerating upwards
MMMFF
Dsign F F
motor b motor b Sp b H SW b H
=− − +−+ ⋅⋅ −⋅ −+()tan( ())
2
αρ
[Nm]
Motor torque when decelerating downwards
MM MFF
D
motor v motor v Sp v H SW
=+++⋅⋅−()tan()
2
αρ
[Nm]
Blocking occur red for a downwards movement for
ρ
α
>SW
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Accelerating- and braking time at constant m ot or torque and const ant load t orque
()
tnJ
MM
a
motor load
=⋅⋅ ⋅
⋅−
∗
∗
2
60
π
max accelerating time from 0 to nmax [s]
()
tnJ
MM
br motor load
=⋅⋅ ⋅
⋅+
∗
∗
2
60
π
max braking t ime from nmax to 0 [s]
J in kgm2, M in Nm, nmax in RPM (final mot or speed)
(moment of inertia and load t orques converted to t he motor shaft)
Accelerating time at constant motor tor que and for a square-law load torq ue
Example: Fan
MM n
n
load load
=⋅
max max
()
2
tnJ
MM
M
M
M
M
amotor load
motor
load
motor
load
=⋅⋅
⋅⋅ ⋅+
−
π
max
max
max
max
ln
60
1
1accelerating time from 0 to nmax [s]
J in kgm2, M in Nm, nmax in RPM (final mot or speed)
Mmotor > Mload max when accelerating, otherwise the final speed will not be reached.
09.99 2 Formulas and equations
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Moments of inert ia of various bodies (shapes)
Solid cylinder
r
lJmr lr=⋅⋅ =⋅⋅⋅⋅
1
2210
243
π
ρ
Hollow cylinder
rR
lJmRr lRr=⋅⋅ + =⋅⋅⋅ − ⋅
1
2210
22 44 3
() ()
π
ρ
Thin-walled hollow cylinder
r
l
m
δ
Jmr l r
mm
= ⋅ =⋅⋅⋅⋅⋅ ⋅
233
210
π
δ
ρ
()
r
R
r
m
≈≈
J in kgm2
l, r, R, rm, δ in m
m in kg
ρ in kg/dm3 (e. g. steel: 7.85 kg/dm3)
2 Formulas and equations 09.99
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Calculating t he m om ent of inertia of a body
If t he m oment of inertia of a body around t he cent er of gr avity S is known, t hen t he m om ent of
inertia to a parallel axis A is:
S
A
m
s
JJ ms
S
=+⋅
2
Example: Disk, t hickness d with 4 holes
J
=−⋅+⋅JJms
disk without hole S hole hole
42
()
=⋅⋅⋅⋅−⋅⋅⋅+⋅⋅⋅
ρπππ
10 242
34 422
dR dr r ds()
J in kgm2
d, r, R, s in m
m in kg
ρ in kg/dm3
s
R
A
S
r
09.99 2 Formulas and equations
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Tract ion drives - friction components
For tract ion dr ives, t he friction consist s of the component s for rolling fr ict ion, bear ing fr iction and
wheel flange friction. T he following is valid for t he frictional force (r esist ance force) :
F
mgw
WF
=⋅⋅ resistance force [N]
mmass of t he t raction drive [kg]
w
F
specific t r act ion r esist ance
If factor w
F
is not known, it can be calculated as follows:
w
D
D
fc
FWr
=⋅ ⋅++
22
()
µ
D
wheel diameter [m]
D
Wshaft diam et er for bear ing fr iction [m]
µ
r
bearing friction coefficient
f
lever arm of the rolling fr ict ion [m]
ccoefficient f or wheel f lange friction
Values for w
F
can be taken from t he appr opriate equat ions/ formulas ( e. g. /5/ , /6/).
D
D
mg
W
.
f
2 Formulas and equations 09.99
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09.99 3 Various special drive tasks
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3 Various special drive tasks
3.1 Traction- and hoisting drives
3.1.1 General information
Tract ion drives
Motor
D
v
Basic traction dr ive
Travelling at constant velocity
Fmgw
WF
=⋅⋅ drag force (this always acts against the direction of
motion, i. e. it has a braking effect) [N]
mmass being m oved [kg]
wFspecific t r act ion r esist ance (this tak es int o account
the inf luence of the rolling - and bearing friction)
MF
D
load W
=⋅
2load torq ue at the drive wheel [Nm]
Ddrive wheel diameter [m]
PFv
motor W
=⋅
⋅max
η
103motor output (power) at maximum velocity
full- load, st eady-state output [kW]
MM
i
motor load
=⋅
η
motor torque [Nm]
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vma
x
ma x. velocity [m/s]
mech. efficiency
in
nmotor
wheel
=gearbox ratio
ni
vD
motor max max
=⋅ ⋅
⋅60
π
motor speed at vmax [RPM]
Note:
For tract ion unit s used outside, in addition to r esist ance force FW, t her e is also the wind force
(wind resistance) FWi.
FAp
W
i
W
i
W
i
=⋅ wind resistance (wind force) [N]
AW
i
surf ace exposed to t he force of the wind [m2]
pW
i
wind pressure [N/m2]
Accelerating and braking
α
bvmotor bv
ia D
,,max
=⋅ ⋅
2angular m ot or acceler ation and -braking of t he m ot or [s-2]
abv, max max. acceleration and braking of the traction unit [m/s2]
MJ
b v motor motor b v motor,,
=⋅
α
accelerating- and braking torque for the motor [Nm]
Jmotor motor m om ent of inertia [kgm2]
MJi
b v load load bvmotor
,,
=⋅
α
accelerating- and braking torque for the load,
ref er r ed t o the drive wheel [Nm]
Jm
D
load =⋅()
22moment of inertia of the linearly-moved masses,
ref er r ed t o the drive wheel [kgm2]
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The component of other m om ents of inertia such as couplings, g ear boxes, dr ive wheels etc., is
generally low, and can be neglected with respect t o the load moment of inertia and the m otor
moment of inertia. O ften, the rotating m asses are taken int o account by adding bet ween 10 and
20 % to the linearly-m oving m asses.
Motor torque when accelerating (start ing torq ue):
MM MM
i
motor b motor b load load
=++⋅
⋅
()
1
η
[Nm]
Motor torque when decelerating (brak ing):
MM MM
i
motor v motor v load load
=− + − + ⋅()
)1
(reg ener at ive operat ion) [Nm]
1) If t he expression in br ackets should be > 0 for very low deceleration values, then the factor η
must be chang ed t o 1/η (so that t he deceleration component is not pr edom inant).
When acceler at ing (start ing) the drive wheels should not spin.
Force which can be transf er r ed t hrough a drive wheel:
FR
Fwheel
FF
Rwheel
=⋅
µ
fr ict ional force due to st atic friction (stiction) [ N]
Fwheel for ce due to the weight acting on t he drive wheel [N]
µ
coefficient of friction for the st at ic friction, e. g. 0.15
for st eel on st eel
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The following must be valid for t he rotating m asses if the accelerating f or ces ar e neglected:
FF
Rb
∑>
Fma
bb
=⋅max accelerating f or c e for linear m ot ion [N]
The maximum per m issible acceleration is obtained as follows if all of the wheels are driven:
ag
bmax <⋅
µ
[m/s2]
Example of a tr action cycle
v
vmax
vmax
-
t
Forwards Interval Reverse
Mmotor
t
Pmotor
Regenerative operation
t
tbtktvtp
Interval
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Hoisting dr ives
Motor
Gearbox
D
Load
Cable dr um
Principle of oper ation of a hoist ing drive (sing le- cable)
Hoisting a load with const ant velocity
Fmg
H=⋅ hoisting force [N]
mload [kg]
MF
D
load H
=⋅
2load torq ue at the cable drum (it always acts in the
same direction)
[Nm]
Dcable drum diamet er [m]
PFv
motor H
=⋅
⋅max
η
103motor output at the maximum hoisting velocity
(steady-st a t e full-load out put )
[kW]
MM
i
motor load
=⋅
η
motor torque [Nm]
vma
x
ma x. ho i sting veloc i ty [m/s]
mechanical ef ficiency
in
nmotor
drum
=gearbox ratio
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ni
vD
motor max max
=⋅ ⋅
⋅60
π
motor speed at vmax [RPM]
Lowering the load at constant velocity
PFv
motor H
=− ⋅⋅
max
103
η
motor out put (regener ative operation) [kW]
MMi
motor load
=⋅
η
motor torque [Nm]
Acceleration and decelerat ion
α
bvmotor bv
ia D
,,max
=⋅ ⋅
2angular acceler at ion and angular deceleration of the
motor [s-2]
abv, max max. acceleration or deceler ation of the load [m/s2]
MJ
b v motor motor b v motor,,
=⋅
α
accelerating- or decelerating torque for the motor [Nm]
Jmotor motor m om ent of inertia [kgm2]
MJi
b v load load bvmotor
,,
=⋅
α
accelerating- or decelerating torque for the load
ref er r ed t o the cable drum [Nm]
Jm
D
load =⋅()
22load moment of inertia referred to the cable drum [kgm2]
Hoisting the load, motor t or que when accelerating (st ar ting torque):
MM MM
i
motor b motor b load load
=++⋅
⋅
()
1
η
[Nm]
Hoisting the load, motor t or que when decelerating:
MM MM
i
motor v motor v load load
=− + − + ⋅ ⋅
()
1
η
[Nm]
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Lowering the load, m ot or torque when accelerating :
MM MM
i
motor b motor b load load
=− + − + ⋅()
(reg ener at ive operat ion) [Nm]
Lowering the load, m ot or torque when decelerating :
MM MM
i
motor v motor v load load
=++⋅()
(reg ener at ive operat ion) [Nm]
For hoisting dr ives, t he influence of the masses to be accelerat ed ( load and r ot at ing masses) is
generally low with respect to the load tor que. Often, t hese ar e t aken into account by adding 10%
to the load tor que.
Example of a hoisting and lowering cycle
v
vmax
vmax
-
t
Hoisting Interval Lowering
MMotor
t
PMotor
t
Regen. op.
tbtktvtp
Interval
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Selecting an induct ion m ot or
For tract ion- and hoist ing drives, when selecting the induct ion m ot or , intermittent duty S3 can be
applied (power-on duration ≈ 20 to 60%) . Depending on the relative power-on durat ion and m otor
type, the ther m ally perm issible motor output is incr eased accor ding to the following form ula:
PP th
kt
th permissible r
r
≤⋅+
−⋅
−⋅
11
10
()
()
(f or load dut y cycles < 10 min; h, k0, refer to Catalog
M11)
[kW]
Ppermissible permissible output at t he dr ive converter for S1 duty
(control r ange 1:2)
[kW]
tPON
r=100% PON = power-on duration
Thus, for example, for a 4-pole 1LA6 mot or, frame size 180 and 40% power-on duration, the
output can be increased by approx. 25% referred t o Ppermissible. T his value is used as basis for
the steady-stat e full-load output. When utilizing motors according to t em per at ure rise class F, the
rated output can generally be used for Ppermissible.
For tract ion dr ives with high accelerat ing torques, the maximum tor que is decisive when selecting
the motor . It must have an adequate safet y margin to the motor stall tor que, and at least :
Mstall > 13.max
⋅M
Thus, even f or line supply undervoltage conditions, saf e, reliable operation is guaranteed. I n or der
to guar ant ee the correct functioning of the closed-loop cont r ol, 200% of t he r at ed m otor torque
should not be exceeded, i. e.:
M
M
nmax ≤⋅2(for SI MOVERT VC)
If acceler at ion and br aking have a significant inf luence on t he load duty cycle, the RMS motor
torq ue should also be calculat ed. The f ollowing must be true:
MMt
tkt M
RMS ii
efp perm
=⋅
+⋅≤
∑2
.(for load dut y cycles < 10 min) [Nm]
Mperm.permissible tor que demanded f r om the drive
converter f or S1 duty (e.g. control range 1:2)
[Nm]
titime segment s with constant t or que Mi[s]
tepower-on duration [s]
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k
fderating factor
tpno-load time [s]
Factor kf for the no- load time is a derating factor for self-ventilated mot or s . For 1LA5/1LA6 motors
it is 0.33. This fact or should be set to 1 f or 1PQ 6 m otors, 1PA6 mot ors and 1FT6/1FK6 m ot or s .
If t he pr oportion of low speed duty within the load duty cycle cannot be neglected (e. g. a high
proportion of time is taken for acceler ating and decelerating or longer per iods of duty at low
speeds), then t his must be appropriat ely taken into account. In this case, the ar it hmetic average of
the speeds within the cycle time T is generated, so t hat the appropriate r educt ion factor can be
taken from t he r eduction characteristic for S1 duty.
n
nn
t
tkt
mot average
mot i A mot i E
i
i
efP
=
+⋅
+⋅
∑2[RPM]
nn
mot i A mot i E
+
2
average motor speed in time sector t i ( A: Initial value,
I: Final value)
[RPM]
It is possible to also calculate the motor RMS current for a load duty cycle instead of MRMS for a
more accurat e analysis and for operation in t he field-weakening range. T he following must be t r ue:
I
II t
tkt I
RMS
mot i A mot i E
i
i
efP perm
=
+⋅
+⋅ ≤
∑()
.
22
[A]
II
mot i A mot i E
+
2average motor current in time sector ti (A: Initial
value, I: Final value)
[A]
Iperm.permissible current at the drive converter for S1 duty
at naverage (this is derived from the Mper S1 curve;
for example, for 1PA6 m otors, Iper=Imot n)
[A]
Refer t o t he next Sect ion „Selecting the dr ive converter“ to calculate the mot or cur rent.
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Selecting t he dr ive converter
Traction drives
For acceleration and braking, t he dr ive converter overload capabilit y can be used for 60 s, as the
accelerating- and braking times are gener ally low. The induct ion motor current at t he maximum
motor t or que is approximately given by:
IM
MIIkI
k
motor motor
motor n motor n n n n n
max max
()( )≈⋅−⋅+⋅
22 2 22 2
1
µµ
[A]
I
n
µ
rated mot or magnetization curr ent [A]
kn=1for nn
mot mot n
≤constant flux range
kn
n
nmot
mot n
=for nn
mot mot n
>field-weakening range
From approx. Mmot ≥ 1.5 Mmot n, saturation effect s occur in the motor cur r ent which can no
longer be neglected. T his can be t aken into account by linearly increasing the motor cur r ent in the
range from Mmot n to 2 Mmot n (0% at Mmot n, approx. 10% at 2 Mmot n).
When select ing the drive converter, in addition to the maximum motor current , t he RMS motor
current, r eferred to the load duty cycle, must also be taken into account ( in t his case, kf must
always be set to 1). The r at ed dr ive converter cur r ent must be greater than or equal to the RMS
motor cur r ent within a 300 s interval. For a motor m atched to the drive converter, this condition is
generally f ulfilled. It may be necessary to make a check, f or example, for induction m ot or s with
high magnetizing current s and shor t no-load intervals.
For tract ion dr ives, vector control is pref erable due to the impr oved contr ol char act er ist ics ( e. g.
accelerating along the curr ent lim it) and also due to t he im pr oved imm unit y to stalling. G enerally,
vector control is also possible, even when several motors are connected to a single drive
converter, as nor m ally, t he m ot ors are similarly loaded.
Hoisting drives
For hoisting dr ives, t he m aximum induction motor curr ent is calculated the same way as for
traction dr ives. T he proportion of acceler ating- and decelerat ing torq ues in t he load torque is
generally low. Generally, the m aximum m otor current is only approximately 100 to 120 % of the
rated mot or current. As hoist ing drives must be able to acceler at e suspended loads, it is
recommended t hat a higher saf ety margin is used between the calculated m aximum m otor current
and the maximum drive converter cur r ent. Depending on t he acceler at ing time, t he st ar ting current
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should be approximately 150 to 200 % of t he required motor current . Vect or control is also
pref er able for hoisting drives due to the impr oved control characteristics, and as it is possible to
impress a defined current when start ing. A speed encoder impr oves the dynamic charact er ist ics.
Further, the tachometer can be used t o m onitor the setpoint - actual values (to identif y if the load
has dropped) .
Generally, holding br akes are used to hold a suspended load. In or der t hat the load doesn’t drop
when the brake is released, the motor m ust first develop an opposing torque to the brake. This
means:
•An appropriately high m otor current set point is entered for closed-loop f requency control
(depending on t he st ar ting torque)
•The mot or flux has been established (approx. 0.1 t o 1 s depending on the motor size)
•A slip fr equency is specified f or closed- loop freq uency contr ol
•Speed controller pr e- control for closed-loop speed control if the load is to be held at zero
speed
For closed-loop frequency control, after the signal to r elease t he br ake has been output, a shor t
delay should be inserted befor e starting. This prevents the drive star t ing, in the open-loop
controlled range, against a br ake which still hasn’t been completely released. After the load has
been hoisted, the signal to close the brake is output at fU=0. T he inverter is only inhibited
somewhat later when it is ensured that the brak e has been applied. During this delay time, the
motor develops a holding torque as result of the DC current braking as fU=0.
fslip
fU
t
Enable inverter,
establis h motor flux
Brak e release signal
Start of acceleration
U
f = 0, b rake
application s ignal
Inverter inhibit
~
~
Example for invert er enable, brake contr ol and acceler at ion enable for a hoisting drive with closed-
loop fr equency control.
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Dimensioning t he br aking resist or
When dim ensioning the brak e r esistor, the m ot or output in the regenerative mode as well as the
load duty cycle are important. The maximum output in regenerative operation occurs when the
drive is braked at m aximum speed. Thus, the max. br aking power of t he brake resist or is given by:
PP Mn
br W br motor motor motor v motor motormax max max
=⋅=
⋅⋅
ηη
9550 [kW]
motor motor efficiency
The maximum br aking power of the brake r esist or may not exceed the permissible peak braking
power. Further, the following must also be t r ue:
W
TP
br brcont
≤.
WPdt
br brW
T
=⋅
∫
0
braking energy f or one cycle [kWs]
Tcycle time (≤ 90 s) [s]
Pbr cont.perm issible continuous braking power [kW]
For cycle times gr eat er than 90 s, f or t he condition
W
TP
br br cont
≤.
a time segm ent T = 90 s is selected from t he load dut y cycle so that t he highest average braking
power occurs in that seg m ent.
For tract ion dr ives, t he following is obtained, for example, when braking from vmax to 0 within
time tv:
WPt
br br W v
=⋅
max
2[kWs]
For hoisting dr ives, t he br aking power is constant when the load is lowered at const ant velocity
vmax:
PFv
br W const Hmotor
=⋅⋅⋅
max
103
ηη
[kW]
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Thus, t he br aking power during lowering is appr oximat ely g iven by:
WP tt t
br br W const b k v
≈⋅⋅++⋅()
1
21
2[kWs]
tbaccelerating time [s]
tklowering time at const ant velocity vmax [s]
tvdecelerating t im e [s]
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3.1.2 Traction drive with brake motor
This application involves a traction drive with brake motor (sliding-rotor motor ) . The mot o r is
operated at a cr awl speed f or positioning. When the limit switch is reached, the motor is shutdown,
and is mechanically braked down to standst ill using the integ ral brake.
Drive data
Mass to be moved m = 4200 kg
Load wheel diameter D = 0. 27 m
Specific t r act ion r esist ance wF= 0.03
Gearbox ratio i = 18.2
Mech. efficiency η= 0.85
Max. traversing speed vmax = 1. 1 m / s
Min. traversing velocity (crawl velocity) vmin = 0.1 m/ s
Max. acceleration ab max = 0.35 m/s2
Max. deceleration av max = 0.5 m/s2
Rated motor output Pn= 2.3 kW
Rated motor current Imotor n = 6.2 A
Rated motor torque Mn= 15.4 Nm
Rated motor speed nn= 1425 RPM
Motor efficiency ηmotor = 0.79
Motor moment of inertia Jmotor = 0.012 kgm2
Coupling mom ent of iner tia J K = 0.00028 kgm2
Max. distance moved smax = 3.377 m
Cycle time tcycle = 60 s
Crawl operation tim e ts= 1.2 s
Brake application time tE= 0.1 s
Mechanical brake torq ue Mmech = 38 Nm
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Constant velocity m ot ion
Resistance f or ce, load torq ue:
FW=⋅⋅mgw
F
=⋅⋅=4200 981 003 12361.. .N
Mload =⋅FD
W2
=⋅=12361 027
21669...Nm
Motor output, motor torque:
PFv kW
motor W
=⋅
⋅=⋅
⋅=
max ..
..
η
10 12361 11
085 10 16
33
MM
iNm
motor load
=⋅=⋅=
η
1669
182 085 108
.
.. .
Motor speed at vmax:
nini
vDRPM
motor loadmax max max ...
=⋅ =⋅ ⋅
⋅=⋅
⋅
⋅=
60 182 11 60
027 1416
π
π
Motor speed at vmin:
nini
vDRPM
motor loadmin min min ....=⋅ =⋅ ⋅
⋅=⋅
⋅
⋅=
60 182 01 60
027 1287
π
π
Calculating the mot or torques when acceler ating and decelerating
Angular acceler at ion and angular deceleration of the mot or :
α
bmotor b
ia Ds=⋅ ⋅ = ⋅ ⋅ = −
max .. ..
2182 035 2
027 47 2 2
α
vmotor v
ia Ds=⋅ ⋅ = ⋅ ⋅ = −
max ....
2182 05 2
027 67 4 2
Accelerating torque and decelerating torque for m otor and coupling:
Mb motor K+=+⋅()JJ
motor K b motor
α
=+ ⋅=(. . ) . .0012 000028 472 058 Nm
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Mv motor K+=+⋅()JJ
motor K v motor
α
=+ ⋅=(. . ) . .0012 000028 674 0828 Nm
Load moment of inertia referred to the load wheel:
Jload =⋅mD
()
22
=⋅ =4200 027
27655
22
(,).kgm
Accelerating torque and decelerating torque for the load:
Mb load =⋅Ji
load bmotor
α
=⋅=7655 472
182 1985..
..Nm
Mv load =⋅Ji
load vmotor
α
=⋅=7655 674
182 2835..
..Nm
Motor torque when accelerating:
Mmotor =++⋅
⋅
+
MMM
i
b motor K b load load
()
1
η
=+ + ⋅⋅=058 1985 1669 1
182 085 24 2.(. .)
.. .Nm
Motor torque when decelerating:
Mmotor =− + − + ⋅
+
MMM
i
v motor K v load load
()
=− + − + ⋅ =−0828 2835 1669 085
182 627.(. .)
...Nm (reg ener at ive operat ion)
The highest motor t or que is requir ed when accelerating, namely 157 % of t he m otor rated t or que.
The mom ent s of inertia for the gearbox and other rotating masses have been neglected.
Accelerating time:
tv
as
bb
===
max
max
.
..
11
035 314
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Calculating the maxim um braking power
The maximum m ot or output in the regenerative mode occurs when the drive starts to decelerate
fr om t he m aximum motor speed.
PMn kW
br motor motor v motor
max max ..=⋅=− ⋅=−
9550 627 1416
9550 093
Drive converter select ion
For the hig hest r equired motor torque, a motor cur r ent of approximately:
IM
MIII
motor motor
motor n motor n n nmax max
()( )≈⋅−+
22 2 2
µµ
is obtained.
The following is obtained with II
n motor n
µ
=⋅07. (assumption):
IA
motor max (.
.)(. . .) . . .≈⋅−⋅+⋅=
242
154 62 07 62 07 62 82
2222 22 for tb=3.14 s
Selected drive converter:
6SE7018-0EA61
PV n=3 kW; IV n=8 A, IV max=11 A
Closed-loop f r equency control type
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Dimensioning the brake r esist or
The brake resistor is used dur ing deceleration (br aking). As t he braking characteristics are the
same f or forwards as well as reverse motion, only forwards motion must be analyzed.
Velocity-time diagram f or f orwards motion
tk
tvts
tE
v
t
T
Forwards motion No-load interval
Mo tor pow ered-down
Brake applied
vmax
vmin
tb
tv me ch
~
~
Maximum distance m oved:
sm
max .=3377
Cycle time for forwards- and reverse motion:
Tts
cycle
==
230
Time for operation at the crawl speed:
ts
s=12.(specified)
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Accelerating time from 0 to vmax:
ts
b=314.
Decelerating time from vmax to vmin:
tvv
as
vv
=−=−=
max min
max
..
.
11 01
05 2
When neglecting t he dist ance t ravelled during the brake application time tE and t he deceler at ion
time as a result of the mechanical br ake tv mech, smax is given by the area in the v-t diagram:
svt
vtvv
tv t
bkvsmax max max max min min
≈⋅+⋅++⋅+ ⋅
22
The tr avelling time at constant velocity vmax is given by:
tk≈−⋅−+⋅− ⋅svtv vtv t
v
bvsmax max max min min
max
22
≈−⋅−+⋅− ⋅ =
3377 11 3143
211 01
220112
11 03
... . . ..
..s
No-load time:
tp=− =− +++
Tt T t t t t
total b k v s
()
=− +++ =30 314 03 2 12 234(. . .) . s
Max. brak ing power of the br ake resistor when decelerating from vmax to vmin:
PP kW
br W br motor motormax max .. .=⋅=⋅=
093 079 073
Min. braking power of the brak e r esist or when decelerating f r om vmax to vmin:
Pbr W min =⋅=
⋅⋅PMn
br motor motor motor v motor motormin min
ηη
9550
=⋅⋅=
627 1287
9550 079 0067
..
..kW
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Braking diagra m
t
Forwards motion No-load interv al
Deceleration
0. 73 kW
0.067 kW
br W
P
3. 14 s
0.3 s
2 s
30 s
~
~
Braking energy f or one cycle (cor r esponds to the area in the br aking diag r am ):
WPP
tkWs
br br W br W v
=+⋅= +⋅=
max min .. .
2073 0067
2208
The following must be valid for t he brake resist or :
W
TkW P
br br cont
== ≤
08
30 0027
...
With
PP
br cont.=20
36 (with an internal brake resistor)
the f ollowing is obt ained
36 0027 0972 20
⋅= ≤..kW P
Thus, t he smallest brak ing unit is selected with P20 = 5 kW (6SE7018-0ES87-2DA0). The internal
braking resistor is adequate.
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Thermal mo t or analysis
Torq ue char act er istics for f or wards motion
t
Forwards motion No-load interval
3.14 s
0.3 s
30 s
2 s
1.2 s
24.2 Nm
10.8 Nm
-6.27 N m
10.8 Nm
Mmotor
23.4 s
~
~
The RMS torque is obt ained from t he torque characteristics:
MRMS =⋅+ ⋅+ ⋅+ ⋅
++++ ⋅
242 314 108 03 627 2 108 12
314 03 2 12 033 234
2222
.. ... ..
.. ...
=20925
144.
.
=1205. Nm
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Calculating t he average speed
nmotor average,Pfe
i
E i motA i mot
tkt
t
nn
⋅+
∑⋅
+
=2=⋅⋅+⋅+ +⋅+ ⋅
++++ ⋅
1
22
ntntnn
tnt
ttttkt
bk vs
bkvs fP
max max max min min
=⋅⋅+⋅++⋅+ ⋅
++++ ⋅ =
1
21416 314 1416 03 1416 1287
22 1287 12
314 03 2 12 033 234 301
.. ..,
.. ... RPM
For a reduction factor of 0.8 (cont r ol r ange 1:5), a per m issible t or que is obtained as f ollows:
08 08 154 1232....⋅=⋅=MNm
mot n
Operat ion is t hermally permissible.
Calculating the distance travelled after the brake motor has been powered-down
When t he traction unit reaches the limit switch at the cr awl speed, the brake mot or is powered-
down by inhibiting the drive converter pulses. T he brake application delay time tE now starts
befor e t he m echanical br ake start s to become effective. T he t raction unit continues t o m ove
without any motor torque and is only braked by the ef fect of drag and t he mechanical efficiency.
Aft er the brake is applied, it takes time tv mech to reach st andst ill.
~
~
v
v
v
min
E
t
ss
Ev
E
t
tv mech
Motor shutdown
Brake appl ie d
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To calculate t he dist ance travelled after the brake m ot or has been shutdown, a sub-division is
made into range 1 (deceleration due t o dr ag and the mechanical efficiency) and range 2
(deceleration as a r esult of the mechanical brake).
Calculating t he deceler ation after shutdown (range 1)
The following is obtained with Mmotor=0:
MMMM
i
motor v motor K v load load
==− +− + ⋅
+
0()
or
0121
=− + ⋅ − ⋅ ⋅ + ⋅()JJ J
iMi
motor K v motor load v motor load
α
α
The mot or angular deceleration is t hen given by:
α
η
vmotor
load
motor K load
Mi
JJJ
i
s
1
22
2
1669 085
182
0012 000028 7655 085
182
3735=⋅
++ ⋅=⋅
++⋅
=−
...
.. .
..
.
and for t he deceler ation of the traction unit:
aiDms
vv motor
112
23735
182 027
20277=⋅=⋅=
α
..../
Distance tr avelled during tE:
s
E
=⋅−⋅⋅vt at
EvEmin 1
212
=⋅−⋅ ⋅ =01 01 1
20277 01 00086
2
.. . . . m
Velocity after tE:
vvat ms
EvEmax min .. .. /=−⋅=− ⋅=
101 0277 01 00723
Calculating t he deceler ation as result of the mechanical brake (rang e 2)
The following is obtained with Mmotor=-Mmech:
MMM MM
i
motor mech v motor K v load load
=− =− + − + ⋅
+()
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and
−=−+⋅ −⋅⋅ +⋅MJJ J
iMi
mech motor K v motor load v motor load
()
α
α
222
The mot or angular deceleration is t hen given by:
α
η
v motor
mech load
motor K load
MM
i
JJJ
i
s
2
22
2
38 1669 085
182
0012 000028 7655 085
182
2194=+⋅
++ ⋅=+⋅
++⋅
=−
...
.. .
..
.
and for t he t raction unit deceleration:
aiDms
vv motor
222
22194
182 027
21627=⋅=⋅=
α
.
.../
Deceleration time as t he result of the mechanical brak e:
tv
as
vmech E
v
== =
2
00723
1627 0044
...
Distance travelled during deceleration caused by the mechanical brake:
svt m
vEvmech
=⋅=⋅=
200723 0044
200016
.. .
As the mechanical brake is only applied after a delay time tE, t he traction unit travels a total
distance of
sss mmm
ges E v
=+= + = =00086 00016 001 10...
This load-dependent dist ance m ust be taken int o account by the positioning contr ol.
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3.1.3 Hoisting drive with brake motor
This application involves a hoisting drive with brake motor (shift -rotor m otor). Posit ioning is
realized at the crawl speed.
Drive data
Mass to be moved m L= 146 kg
Intrinsic mass mE= 66 kg
Cable drum diamet er D = 0. 1 m
Gearbox ratio i = 20.5
Mech. efficiency η= 0.85
Max. hoisting velocity vmax = 0.348 m/s
Min. hoisting velocity (crawl velocity) vmin = 0.025 m/s
Max. acceleration amax = 0.58 m/s2
Rated motor output Pn= 1.05 kW
Rated motor current Imotor n = 3.8 A
Rated motor torque Mn= 7.35 Nm
Rated motor speed nn= 1365 RPM
Motor efficiency ηmotor = 0.75
Motor moment of inertia Jmotor = 0.0035 kgm2
Coupling mom ent of iner tia J K = 0.00028 kgm2
Max. hoisting height hmax = 0.8 m
Cycle time tcycle = 90 s
Crawl time ts= 0.4 s
Brake application time tE= 0.04 s
Mechanical brake torq ue Mmech = 14.5 Nm
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Hoisting the load wit h const ant velocity
Hoisting force, load t or que:
FH=+⋅()mmg
L
E
=+⋅=(). .146 66 9 81 2079 7 N
Mload =⋅FD
H2
=⋅=2079 7 01
2104..Nm
Motor output, motor torque:
PFv kW
motor H
=⋅
⋅=⋅
⋅=
max ..
..
η
10 20797 0348
085 10 0852
33
MM
iNm
motor load
=⋅=⋅=
η
104
205 085 597
.. .
Motor speed at vmax:
nini
vDRPM
motor loadmax max max ...
=⋅ =⋅ ⋅
⋅=⋅ ⋅
⋅=
60 205 0348 60
01 1362
π
π
Motor speed at vmin:
nini
vDRPM
motor loadmin min min ....=⋅ =⋅ ⋅
⋅=⋅ ⋅
⋅=
60 205 0025 60
01 97 9
π
π
Lowering the load at constant velocity
Motor output, motor torque:
PFv kW
motor H
=− ⋅⋅=− ⋅⋅=−
max .. ..
10 20797 0348
10 085 0615
33
η
(reg ener at ive operat ion)
MMiNm
motor load
=⋅=⋅=
η
104
205 085 431..
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Calculating the mot or torques when acceler ating and decelerating
Motor angular accelerat ion:
α
motor ia Ds=⋅ ⋅ = ⋅ ⋅ = −
max .. ..
2205 058 2
01 2378 2
Accelerating torque for motor and coupling:
Mbmotor K+=+⋅
()JJ
motor K motor
α
=+ ⋅=(. . ) . .00035 000028 2378 0899 Nm
Load moment of inertia referred to the cable drum:
Jload =+⋅()()mm D
LE
22
=+⋅ =()(
.).146 66 01
2053
22
kgm
Load accelerating torque:
Mb load =⋅Ji
load motor
α
=⋅ =053 2378
205 615..
..Nm
As deceleration is the sam e as acceler ation, the f ollowing is t r ue:
Mvmotor K+=+
Mb motor K
Mv load =Mb load
Hoisting the load, motor t orque when accelerating:
Mmotor =++⋅
⋅
+
MMM
i
b motor K b load load
()
1
η
=++⋅
⋅=0899 615 104 1
205 085 722.(. )
.. .Nm
Hoisting the load, motor t or que when decelerating:
Mmotor =− + − + ⋅ ⋅
+
MMM
i
v motor K v load load
()
1
η
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Mmotor =− + − + ⋅ ⋅=0899 615 104 1
205 085 472.(. )
.. .Nm
Lowering the load, m ot or torque when accelerating :
Mmotor =− + − + ⋅
+
MMM
i
b motor K b load load
()
=− + − + ⋅ =0899 615 104 085
205 316.(. )
...Nm (reg ener at ive operat ion)
Lowering the load, m ot or torque when decelerating :
Mmotor =++⋅
+
MMM
i
v motor K v load load
()
=++⋅=0899 615 104 085
205 547.(. )
...Nm (reg ener at ive operat ion)
The highest motor t or que is requir ed when accelerating and hoisting t he load, nam ely 98 % of the
rated mot or torque. The highest m ot or torque in regenerative operation is r equired when
decelerating the drive while the load is being lowered.
Calculating the maxim um braking power
The maximum m ot or output in regenerative operation occurs when the drive starts to decelerate
fr om t he m aximum motor speed while the load is being lowered.
PMn kW
br motor motor v motor
max max ..=⋅=− ⋅=−
9550 547 1362
9550 078
Selecting the drive conver ter
The highest required motor tor que approximately corresponds to t he rated motor t or que. An
appropriate overload capabilit y is required to st ar t the hoisting dr ive, i. e. the drive converter
should be able to provide approximately double the rat ed m otor current as IU max.
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Selected drive converter:
6SE7016-1EA61
PV n=2.2 kW; IV n= 6.1 A; I V max=8.3 A
Closed-loop f r equency control type
Dimensioning the brake r esist or
The brake resistor is used when the load is lowered.
Velocity-time diagram for hoist ing and lowering
v
vmax
vmin
vmin
vmax
t
ttotal
tbtv1
tk
ts
tv2
-
-
T
Hoisting No-load interval Lowering
~
~
~
~
No-load in terval
Maximum distance t r avelled when hoisting and lowering:
hm
max .=08
Cycle time f or hoisting and lowering:
Tt s
cycle
==90
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Time when the drive operates at t he crawl velocity:
ts
s=04.(specified)
Accelerating time from 0 to vmax:
tv
as
b== =
max
max
...
0348
058 06
Decelerating time from vmax to vmin:
tvv
as
v10348 0025
058 0557=−=−=
max min
max
..
..
Deceleration time from vmin to 0:
tv
as
v20025
058 0043== =
min
max
...
From the ar ea in t he v-t diagram, hmax is given by:
hvt
vtvv
tvt
vt
bkvs
v
max max max max min min min
=⋅+⋅++⋅+ ⋅+ ⋅
22 2
12
The unit t r avels at const ant velocity vmax for:
tk=−⋅−+⋅− ⋅− ⋅
hvtv vtvt
vt
v
bvs
v
max max max min min min
max
22 2
12
=−⋅−+⋅−⋅−⋅
=
08 0348 06
20348 0025
20557 0025 04 0025 0043
2
0348 167
.... ....
..
..s
Thus, the total t r avelling time fo r h max is given by:
ttotal =++ ++ttt tt
bkv sv12
=+ + ++ =06 167 0557 04 0043 327.. . .. .s
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Max. brak ing power rating while accelerating from 0 t o vmax:
Pbr W vmax ( )
max
0→=⋅=
⋅⋅
→
PMn
br motor v motor motor b motor motormax ( ) max
max
0
9550
ηη
=⋅⋅=
316 1362
9550 075 0338
...kW
Braking power while travelling at constant velocity, vmax:
Pbr W const v v()
max
==⋅=
⋅⋅
=
PMn
br motor const v v motor motor const motor motor() max
max
ηη
9550
=⋅⋅=
431 1362
9550 075 046
...kW
Max. braking power while decelerating from vmax to vmin:
Pbr W v vmax ( )
max min
→=⋅=
⋅⋅
→
PMn
br motor v v motor motor v motor motormax( ) max
max min
ηη
9550
=⋅⋅=
547 1362
9550 075 0585
...kW
Min. braking power while decelerating from vmax to vmin:
Pbr W v vmin ( )
max min
→=⋅=
⋅⋅
→
PMn
br motor v v motor motor v motor motormin ( ) min
max min
ηη
9550
=⋅⋅=
547 979
9550 075 0042
..
..kW
Braking power while travelling at constant velocity vmin:
Pbr W const v v()
min
==⋅=
⋅⋅
=
PMn
br motor const v v motor motor const motor motor() min
min
ηη
9550
=⋅⋅=
431 979
9550 075 0033
..
..kW
Max. braking power while decelerating from vmin to 0:
Pbr W vmax ( )
min →0=⋅=
⋅⋅
→
PMn
br motor v motor motor v motor motormax ( ) min
min 0
9550
ηη
=⋅=
→
PkW
br motor v v motormin ( )
max min .
0042
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Braking diagra m
~
~
90 s
0.585 KW
0.338 kW
0.042 kW
0.033 kW
0.460 kW
t
Lowering
Pbr W Interval IntervalHoisting
1.67 s
0.557 s
0.6 s
0.4 s 0.043 s
3.27 s
Constant velocity (v=vmax)
Deceleration
Constant vel. (v=v
Acceleration
min)
~
~
Deceleration
Braking energy f or one cycle (cor r esponds to the area in the br aking diag r am )
Precise calculation:
Wbr =⋅ ⋅+ ⋅+ +⋅
→= →→
1
22
0 1
PtPt
PP
t
br W v b br W const v v k br W v v br W v v vmax ( ) ( ) max ( ) min ( )
max max
max min max min
+⋅+⋅⋅
=→
PtPt
br W const v v s br W v v() max( )
min min
1
202
=⋅ ⋅ + ⋅ + +⋅
1
20338 06 046 167 0585 0 042
20557......
.
+⋅+⋅ ⋅ =0033 04 1
20042 0 043 1058.. .. .kWs
In this particular case, the following estimation can be made:
Wbr ≈⋅++⋅
==
PttPt
br W const v v b k br W const v v s() ()
max min
()
≈⋅+ + ⋅=046 06 167 0033 0 4 1057.(. .). . . kWs
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The following must be valid for t he brake resist or :
W
TkW P
br br cont
== ≤
1057
90 0012
...
With
PP
br cont.=20
36 (with internal br aking r esistor)
the f ollowing is obtained
36 0012 0432 20
⋅= ≤
..kW P
Thus, t he sm allest braking unit is selected with P20 = 5 kW (6SE7018-0ES87-2DA0). The internal
brake r esist or is adequate.
Thermally checking the m ot or
As the motor is only subject to a maximum load of approx. rated torque, and is only powered-up
for appr ox. 7 s from a t ot al of 90 s, a ther mal analysis is not required.
Positioning error as result of the br ake application time t
E
Aft er t he brake mot or is powered-down by inhibiting the dr ive converter pulses, there is first a
brake applicat ion delay time t E before the m echanical br ake becomes ef fective. T he dr ive is
shutdown at fU=0 and n≈0 ( t he slip speed is neglected). The load drops, accelerat ing due to its
own weight, as t here is no opposing motor torque. Aft er t he br ake has been applied, standst ill is
reached af t er time tv mech.
tE
tv mech
v
s
sE
v
t
vmax E
-
Motor
powered-down
Brake applied
~
~
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The positioning error calculat ion is sub- divided into range 1 (acceleration due t o t he load itself) and
range 2 ( deceler at ion due to the mechanical brake).
Calculating t he acceler ation caused by the load itself ( r ange 1)
For Mmotor=0:
MMMM
i
motor b motor K b load load
==− +− + ⋅
+
0()
and
02
=− + ⋅ − ⋅ ⋅ + ⋅()JJ J
iMi
motor K b motor load b motor load
α
α
The mot or angular acceleration is t hen given by:
α
η
bmotor
load
motor K load
Mi
JJJ
i
s=⋅
++ ⋅=⋅
++⋅
=−
22
2
104 085
205
00035 000028 053 085
205
889
..
.. .
..
and for t he load acceler ation:
aiDms
bb motor
=⋅=⋅=
α
2889
205 01
2217 2
.../
The distance m oved while the system accelerates due to the load:
sat m
EbE
=⋅⋅=⋅ ⋅ =
1
21
2217 004 00017
22
.. .
Velocity after the brak e applicat ion time has expired:
vat ms
EbEmax .. . /=⋅= ⋅ =217 004 0087
Calculating t he deceler ation as result of the mechanical brake (rang e 2)
For Mmotor=Mmech:
MMM MM
i
motor mech v motor K v load load
== + +⋅
+()
and
MJJ J
iMi
mech motor K v motor load v motor load
=+⋅+⋅⋅+⋅()
α
α
2
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The angular deceleration of the motor is t hen given by:
α
η
v motor
mech load
motor K load
MM
i
JJJ
i
s=−⋅
++ ⋅=−⋅
++⋅
=−
22
2
145 104 085
205
00035 0 00028 053 085
205
2100
...
.. .
..
and the deceleration of the load itself:
aiDms
vv motor
=⋅=⋅=
α
22100
205 01
2512 2
.../
Delay time as result of the mechanical brake:
tv
as
vmech E
v
===
max ...
0087
512 0017
Distance moved during the delay caused by the mechanical brake:
svt m
vEvmech
=⋅=⋅
max ..
.
20087 0017
2000074
Due to the delayed application of the mechanical brake aft er the brake m ot or has been powered
down - time tE - the load m oves a tot al dist ance of
sss m mm
total E v
=+= + = =
00017 000074 0 00244 2 44.. . .
downwards. T his load- dependent positioning error must be taken into account by the positioning
system.
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3.1.4 High-bay racking vehicle
This application involves a high- bay racking vehicle with traversing drive, elevating dr ive and
telescopic drive. In t his case, only the traversing- and elevating drives are investigated. 1PA6
compact induction m ot ors are to be used.
Traversing drive
Drive data
Mass to be moved m = 5500 kg
Load wheel diameter D = 0. 34 m
Specific t r act ion r esist ance wF= 0.02
Gearbox ratio i = 16
Mech. efficiency η= 0.75
Max. traversing velocity vmax = 2.66 m /s
Max. acceleration amax = 0.44 m/s2
Max. traversing distance smax = 42.6 m
No-load interval af t er a traversing seq uence tP= 10 s
Traversing at const ant velocity
Resistance f or ce, load torq ue:
FW=⋅⋅mgw
F
=⋅⋅=5500 981 002 10791.. .N
Mload =⋅FD
W2
=⋅=10791 034
218345...Nm
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Motor output, motor torque:
PFv kW
motor W
=⋅
⋅=⋅
⋅=
max ..
..
η
10 10791 2 66
075 10 383
33
MM
iNm
motor load
=⋅=⋅=
η
18345
16 075 1529
...
Motor speed at vmax:
nini
vDRPM
motor loadmax max max ..
=⋅ =⋅ ⋅
⋅=⋅ ⋅
⋅=
60 16 266 60
034 2391
π
π
Calculating the load tor ques when accelerating and decelerat ing
Angular acceler at ion of the load wheel:
α
load aDs=⋅=⋅= −
max ...
2044 2
034 259 2
Load moment of inertia referred to the load wheel:
Jload =⋅mD
()
22
=⋅ =5500 034
215895
22
(.).kgm
Load accelerating torque:
Mb load =⋅Jload load
α
=⋅=15895 259 4114.. .Nm
As deceleration is the sam e as acceler ation, the f ollowing is t r ue:
Mv load =Mb load
Motor selection
The following motor is selected ( r efer to Calculating the motor torques):
1PA6 103-4HG., Pn=7.5 kW, Mn=31 Nm, nn=2300 RPM, In=17 A, Iµ=8.2 A
Jmotor=0. 017 kgm2, ηmotor=0.866
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Calculating the mot or torques when acceler ating and decelerating
Motor accelerating- and deceler at ing torq ues:
MM
b motor v motor
==⋅⋅
Ji
motor load
α
=⋅⋅=0017 259 16 0704.. .Nm
Motor torque when accelerating:
Mmotor =++⋅
⋅
MMM
i
b motor b load load
()
1
η
=+ + ⋅
⋅=0704 41138 18345 1
16 075 5027.(. .)..Nm
Motor torque when decelerating:
Mmotor =− + − + ⋅MMM
i
v motor v load load
()
=− + − + ⋅ =−0704 41138 18345 075
16 1139.(. .)
..Nm (reg ener at ive operat ion)
The highest motor t or que is requir ed when accelerating, namely 162 % of t he r ated motor t or que.
The mom ent s of inertia of the gearbox, coupling, brake and other rotating masses have been
neglected.
Accelerating- and decelerating t ime:
ttv
as
bv
== = =
max
max
.
.
266
044 6
Investigating the condition so that t he w heels do not spin when starting
The following must be true so t hat the drive wheels don’t spin when starting:
FR
∑ >
F
b
Whereby:
Fma N
bb
=⋅ = ⋅ =
max .5500 044 2420 (accelerating force)
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For two axes where one is driven, and when assuming that the driven axis carries half of the
vehicle weight:
FR
∑=⋅⋅⋅
1
2mg
µ
=⋅ ⋅ ⋅ =
1
25500 981 015 4047.. N(sum of the f r ictional forces)
Thus, t he above condition is fulf illed.
Calculating the maxim um braking power
The maximum m ot or output in regenerative operation occurs when the drive starts to decelerate
fr om t he m aximum motor speed.
PMn kW
br motor motor v motor
max max ..=⋅=− ⋅=−
9550 1139 2391
9550 285
Selecting the drive conver ter
For the hig hest r equired motor torque, the approximate motor current is given by:
IM
MIIkI
k
motor motor
motor n motor n n n n n
max max
()( )≈⋅−⋅+⋅
22 2 22 2
1
µµ
kn
n
n
mot
mot n
===
max ,
2391
2300 10396 field weakening factor
With IA
n
µ
=82., the f ollowing is obtained:
IA
motor max (.)( .). . ..≈⋅−⋅+⋅=
5027
31 17 82 10396 82 1
10396 263
22 2 2 2 2 for tb=6 s
Selected drive converter:
6SE7022-6EC61
PV n=11 kW; IV n=25.5 A; IV max=34.8 A
Closed-loop speed control
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Dimensioning the brake r esist or
The brake resistor is used dur ing deceleration. As the braking characteristics ar e the same f or
both f or wards- and r everse m ot ion, only forwards motion has to be investig at ed.
Velocity-time diagram f or f orwards motion
tbtv
ttotal
vmax
vmax
tp
T
No-load in terval
t
vForwards m otion
tk
ts
p=10
tt s
vb
==6
sm
max .
=42 6
Thus, t he t otal time f or the maximum travel (distance) is obtained as follows:
ttotal =⋅+ −⋅
2tsvt
v
bbmax max
max
=⋅+ −⋅
=26 42 6 2 66 6
266 22
..
.s
Time for traversing at constant velocity:
tt t s
ktotal b
=−⋅=−⋅=2222610
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Cycle time for forwards- and reverse t r avel:
Tt t s
total p
=+=+=22 10 32
Maximum braking power for the brake r esist or :
Pbr W max =⋅⋅=⋅⋅=PkW
br motor motor Invmax .. , .
η
η
285 0866 0 98 242
Braking diagra m
No-load in terval
t
Pbr W F or wards m ot ion
22 s
6 s10 s
32 s
2.42 kW
Deceleration
Braking energy f or a cycle (cor r esponds to the area in the br aking diag r am ):
Wbr =⋅ ⋅=⋅ ⋅=
1
21
2242 6 726Pt kWs
brW vmax ..
The following must be valid for t he brake resist or :
W
TkW P
br br cont
== ≤
726
32 0227
...
With
PP
brcont.=20
36 (with an internal brake resistor)
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the f ollowing is obtained
36 0227 817 20
⋅= ≤
..kW P
Thus, a br aking unit is select ed with P20 = 10 kW (6SE7021-6ES87- 2DA0) . The inter nal br ake
resistor is adequate.
Thermally checking the m ot or
Torque characteristic for forwards motion
t
50.27 Nm
15.29 Nm
-11.39 Nm
6 s 10 s
32 s
6 s
Forwa rd mo tion No-load interva l
10 s
Mmotor
The RMS torque is obt ained from t he torque characteristic as f ollows:
MRMS =⋅+ ⋅ + ⋅
+++
5027 6 1529 10 1139 6
610610
22 2
.. .
=182787
23 .
=239. Nm
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The calculated RMS torque is lower than the ra t ed motor t or que with 30 Nm. T hus, operation is
therma lly perm issible. When calculating the RMS value, the slight field-weakening r ange is
neglected.
Elevating drive
Drive data
Mass to be moved m = 1350 kg
Load wheel diameter D = 0. 315 m
Gearbox ratio i = 63.3
Mech. efficiency η= 0.7
Max. elevating velocity vmax = 0.6 m/s
Max. acceleration amax = 0.6 m/s2
Max. height hmax = 6 m
No-load interval after elevating or lowering tP= 8 s
Elevating the load at constant velocity
Elevating force, load t orque:
FH=⋅mg
=⋅=1350 981 132435..N
Mload =⋅FD
H2
=⋅=132435 0315
220859...Nm
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Motor output, motor torque:
PFv kW
motor H
=⋅
⋅=⋅
⋅=
max ..
..
η
10 132435 0 6
07 10 1135
33
MM
iNm
motor load
=⋅=⋅=
η
20859
633 07 471
.
.. .
Motor speed at vmax:
nini
vDRPM
motor loadmax max max ...
=⋅ =⋅ ⋅
⋅=⋅⋅
⋅=
60 633 06 60
0315 2303
π
π
Lowering the load at constant velocity
Motor output, motor torque:
PFv kW
motor H
=− ⋅⋅=− ⋅⋅=−
max ....
10 132435 0 6
10 07 556
33
η
(reg ener at ive operat ion)
MMiNm
motor load
=⋅= ⋅=
η
20859
633 07 2307
.
...
Calculating the load tor ques when accelerating and decelerat ing
Angular acceler at ion of the load wheel:
α
load aDs=⋅=⋅ = −
max ...
206 2
0315 381 2
Load moment of inertia referred to the load wheel:
Jload =⋅mD
()
22
=⋅ =1350 0315
23349
22
(.).kgm
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Load accelerating torque:
Mb load =⋅
Jload load
α
=⋅=3349 381 12758.. .Nm
As deceleration is the sam e as acceler ation, the f ollowing is t r ue:
Mv load =Mb load
Motor selection
The following motor is selected ( r efer to Calculating the motor torques):
1PA6 103-4HG., Pn=7.5 kW, Mn=31 Nm, nn=2300 RPM, In=17 A, Iµ=8.2 A
Jmotor=0. 017 kgm2, ηmotor=0.866
Calculating the mot or torques when acceler ating and decelerating
Accelerating- and decelerating t or que for t he m otor:
MM
b motor v motor
==⋅⋅
Ji
motor load
α
=⋅⋅=
0017 381 633 41....Nm
Elevating the load, m ot or torque when accelerating :
Mmotor =++⋅
⋅
MMM
i
b motor b load load
()
1
η
=++⋅
⋅=41 12758 20859 1
633 07 5405.(. .) .. .Nm
Elevating the load, m ot or torque when decelerating :
Mmotor =− + − + ⋅ ⋅
MMM
i
v motor v load load
()
1
η
=− + − + ⋅ ⋅=41 12758 20859 1
633 07 401.( . .) .. .Nm
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Lowering the load, m ot or torque when accelerating :
Mmotor =− + − + ⋅MMM
i
b motor b load load
()
=− + − + ⋅ =41 12758 20859 07
633 1755.( . .)...Nm (reg ener ative operation)
Lowering the load, m ot or torque when decelerating :
Mmotor =++⋅MMM
i
v motor v load load
()
=++⋅=41 12758 20859 07
623 2858.(. .)...Nm (reg enerative operation)
The highest motor t or que is requir ed when accelerating, while elevating the load, namely 174 % of
the rated m otor torque. The hig hest motor t or que in regener at ive operat ion is r equired when
decelerating while the load is being lowered. The accelerating- or decelerating component in this
case is low with respect t o t he load torque. The moments of inertia for g ear box, br ake and
coupling have been neglected.
Accelerating and deceler ating time:
ttv
as
bv
== = =
max
max
.
.
06
06 1
Calculating the maxim um braking power
The maximum m ot or output in regenerative operation occurs when the drive starts to decelerate
fr om t he m aximum motor speed while the load is being lowered.
PMn kW
br motor
motor v motor
max
max ..=⋅=− ⋅=−
9550 2858 2303
9550 689
Drive converter select ion
For the hig hest m otor torque, the approximat e m ot or current is given by:
IM
MIII
motor motor
motor n motor n n nmax max
()( )≈⋅−+
22 2 2
µµ
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With IA
n
µ
=82., the f ollowing is obtained:
IA
motor max (.)( .) . .≈⋅−+=
5405
31 17 82 82 272
22 2 2 for tb=1 s
5% is added for sat ur ation effects, so that a max. motor curr ent of 28. 5 A is obt ained.
Selected drive converter:
6SE7022-6EC61
PV n=11 kW; IV n=25.5 A; IV max=34.8 A
Closed-loop speed control
Dimensioning the brake r esist or
The brake resistor is used when lowering the load.
Velocity-time diagram for elevating and lowering
tbtv
ttotal
vmax
-
vmax
tp
ttotal
tp
tbtv
T
No-load
interval
t
vLowering
tk
Elevating
tk
No-load
interval
ts
p=8
tt s
vb
==1
hm
ma
x
=6
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Thus, a t ot al time is obtained f or hmax:
ttotal =⋅+ −⋅
2thvt
v
bbmax max
max
=⋅+−⋅
=21 6061
06 11
.
.s
Time for travel at const ant velocity:
tt t s
ktotal b
=−⋅=−⋅=
211219
Cycle time f or elevating and lowering:
Ttt s
total p
=⋅ + =⋅ + =2 2 11 8 38()()
Max. brak ing power for t he br ake resistor when decelerating:
Pbr W vmax =⋅⋅=⋅⋅=PkW
br motor v motor Invmax .. . .
η
η
689 0866 098 585
Max. brak ing power for t he br ake resistor when accelerating :
Pbr W bmax =⋅⋅=
⋅⋅⋅PMn
br motor b motor Inv
motor b motor
motor Invmax
max
ηη ηη
9550
=⋅⋅⋅=
1755 2303
9550 0866 098 359
....kW
Braking power for t he br ake resistor when travelling at const ant velocity:
Pbr W const =⋅⋅=⋅⋅=PkW
br motor const motor Inv
η
η
556 0866 098 472.. . .
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Braking diagra m
Interval Low ering Interval
t
Elevating
Pbr W
Constant velocity
Acceleration
Deceleration
11 s 8 s 9 s
11 s
8 s
5.85 kW
4.72 kW
3.59 kW
38 s
1
s1
s
Braking energy f or a cycle (cor r esponds to the area in the br ake diagram ) :
Wbr =⋅ ⋅+ ⋅+⋅ ⋅
1
21
2
PtPtPt
brW b b brW const k brW v vmax max
=⋅ ⋅+ ⋅+⋅ ⋅=
1
2359 1 472 9 1
2585 1 472.. . .kWs
The following must be valid for the brake resistor:
W
TkW P
br br cont
== ≤
472
38 124
...
With
PP
br cont..
=20
45 (f or an exter nal br ake resistor)
the f ollowing is obt ained
45 124 558 20
.. .⋅= ≤kW P
Thus, a br aking unit is selected with P 20 = 10 kW (6SE7021-6ES87-2DA0), with an external brake
resistor ( 6SE7021-6ES87-2DC0).
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Thermally checking the m ot or
Motor characteristic when hoisting and lowering
No load
interval
t
LoweringElevating
54.05 Nm 47.1 Nm 40.1 Nm
17.55 Nm 23.07 Nm 28.58 Nm
1 9 s 1s
11 s
8 s 1 9 s
11 s
1 8 s
38 s
Mmotor
sss
No load
interval
The RMS torque is obt ained from t he torque characteristic:
MRMS =⋅+ ⋅+ ⋅+ ⋅+ ⋅+ ⋅
⋅+++
5405 1 471 9 401 1 1755 1 2307 9 2858 1
21918
22 2 2 2 2
..... .
()
=30410
38
=283.Nm
The calculated RMS torq ue is less than the rated m ot or torque with 31 Nm. Thus, operation is
therma lly perm issible.
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Using a rectifier / regenerat ive feedback unit with inverters
As an alternative, inst ead of using 3 dr ive converters with braking units, a multi-motor drive system
can also be configured using a rectifier/regenerative feedback unit , DC bus and 3 inverters.
Selecting the rectifier/r egener ative feedback unit
The low output of the telescopic drive with respect to the traversing- and elevating drive, can be
neglected when dimensioning the rectifier/regenerative feedback unit, as t he telescopic drive is
always operational just by itself . On the other hand, the traversing- and elevating drive can operate
simultaneously. When selecting the rectifier/regenerative feedback unit, it is assumed, that under
worst case conditions, when accelerating, the maximum power and therefore the m aximum DC
link curr ent of both drives occur simultaneously.
Maximum m otor power and maximum DC link current for t he t raversing drive:
PMn kW
motor traverse motor traverse traverse
max max max ..=⋅=⋅=
9550 5027 2391
9550 1259
IP
UA
DClink traverse motor traverse
ZK motor traverse inv
max max .
...
.=⋅⋅
=⋅
⋅⋅ ⋅ =
ηη
1259 10
135 400 0866 098 2747
3
Max. motor power and maximum DC link current for t he elevating drive:
PMn kW
motor hoist motor hoist hoist
max max max ..=⋅=⋅=
9550 5405 2303
9550 1303
IP
UA
DClink hoist motor hoist
DClink motor hoist inv
max max .
...
.=⋅⋅
=⋅
⋅⋅ ⋅ =
ηη
1303 10
135 400 0866 098 2843
3
Thus, the maximum DC link cur rent is given by:
II I A
DClink total DClink traverse DClink hoistmax max max ...=+=+=2747 2843 559
Selected rectifier/r egenerative f eedback unit:
6SE7024-1EC85-1AA0
Pn=15 kW; IDC link n=41 A; IDC link max=56 A
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Selecting the regener ative feedback transf or mer
When select ing the reg ener at ive feedback tr ansformer , the RMS DC link current for the
rectifier/r egenerative f eedback unit is calculated in the regener ative mode. Previously, to
dimension the braking resist or s, the brak ing powers Pbr W=Pbr DC link in t he DC link, were already
calculated f or the traversing- and elevating drive. In this case, the inverter ef ficiency was
neglected. Thus, the calculat ion is on t he safe side. Using the equation
IP
V
DClink gen br DClink
DClink
=
the RMS value in regenerative operation is obtained:
IP
V
DClink gen RMS br DClink RMS
DClink
=
The RMS value of the DC link braking power is obtained with the braking powers P br W, specified
in the brak e diagrams for the t r aversing- and elevating drive:
PkW
br DClink RMS =⋅⋅+⋅⋅+⋅+⋅⋅
=
1
3242 6 1
3359147291
3585 1
38 245
2222
....
.
To calculate t he RMS value of the non- const ant segment s, t he following equation is used.
Pdt P P P P t
brW
i
i
brW i brW i br W i brW i i
2
121
21
1
3
⋅=⋅ + + ⋅ ⋅
+
++
∫()∆
The cycle time for the elevating dr ive, 38 s, was used as time int er val.
The following is now obtained for the RMS value of the DC link curr ent:
IA
DClink gen RMS =⋅
⋅=
245 10
135 400 454
3
.
..
The perm issible RMS value is as follows for a r egenerative feedback transformer with 25% duty
ratio:
II I A
DClink RMS permissible DClink n DClink n
=⋅⋅=⋅=⋅=092 25
100 046 41 046 1886....
Thus, a r egenerative feedback transformer 4AP2795-0UA01-8A with a 25% duty ratio is sufficient.
In addition, a 4% uk line r eactor 4EP3900-5US is re quired.
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3.1.5 Hoisting drive for a 20 t gantry crane
This application involves a four - line hoist ing unit with 2 cable runs. T he t est load is 25 t.
Previously, a 6-pole wound-rotor mot or (US: slipring motor) was used, directly connected t o t he
line supply with Pn=21.5 kW; this has now been replaced by a 6-pole induction motor with
Pn=22 kW.
i, G
Cable drum
Load
D
IM
η
Hoisting unit data
Cable drum diamet er D = 0.6 m
Gearbox ratio i = 177
Gearbox efficiency ηG= 0.9
Permissible load Q = 20 t
Test load Qmax = 25 t
Accelerating time, decelerating time tb, tv= 10 s
Rated motor output Pn= 22 kW
Rated motor torque Mn= 215 Nm
Rated motor speed nn= 975 RPM
Motor moment of inertia JMotor = 0.33 kg m 2
Maximum hoist ing height hmax = 5 m
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Hoisting the load at const ant velocity
Hoisting force and load tor que at the cable drum :
Fmg N
Hload
=⋅=⋅⋅ =
220 10 9 81
298100
3.
MF
DNm
load H
=⋅= ⋅=
298100 06
229430
.
Circumferential speed at t he cable dr um:
vdrum max =⋅=
⋅⋅
⋅⋅
ω
π
n motor n motor
iDn
iD
2
2
60 2
=⋅⋅
⋅⋅=
2 975
177 60 06
20173
π
../ms
Motor output, motor torque:
PFv kW
motor Hdrum
G
=⋅⋅=⋅
⋅=
max .
..
η
10 98100 0173
09 10 1886
33
MM
iNm
motor load
G
=⋅=⋅=
η
29430
177 09 184 7
..
For a 25 t test load, a motor out put of 23.57 kW and a motor torque of 231 Nm ar e obt ained. The
selected 22 kW motor is suff icient , as it involves a drive with intermittent duty.
Lowering the load at constant velocity
Motor output, motor torque:
PFv kW
motor Hdrum G
=− ⋅⋅=− ⋅⋅=−
max ...
10 98100 0173
10 09 1527
33
η
(reg ener at ive operat ion)
MMiNm
motor load G
=⋅= ⋅=
η
29430
177 09 149 6..
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Calculating the mot or torques when acceler ating
Motor angular accelerat ion:
α
motor ==
⋅⋅
⋅
ω
π
n motor
b
nmotor
b
t
n
t
2
60
=⋅⋅
⋅=−
2 975
60 10 102 2
π
.s
Accelerating torque for motor + gearbox:
Mb motor gearbox+=+ ⋅()JJ
motor gearbox motor
α
=+⋅=(. ) . .033 0 102 337 Nm
(Jgearbox neglected)
Moment of inert ia of the load referred t o t he cable drum:
Jload =⋅ ⋅
1
42
2
mD
load ()
=⋅⋅ ⋅ =
1
420 10 06
2450
32 2
(.)kgm
Moment of inert ia of the cable drum :
JTr=20 kgm2 (estim ated)
Accelerating torque for the load+wire drum:
Mb load drum+=+⋅=+⋅()()JJ JJ i
load drum drum load drum motor
α
α
=+⋅=()
..450 20 102
177 27 08 Nm
Hoisting the load, motor t or que when accelerating:
Mmotor =++⋅
⋅
++
MMM
i
b motor gearbox b load drum load G
()
1
η
=+ + ⋅
⋅
337 27 08 29430 1
177 09
.(. ) .
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Due to the relatively long accelerating time, the accelerating torques are low with respect to the
load torq ue.
Selecting the mot or and drive converter
Selected motor:
1LA5 207-6
Pn=22 kW; Imotor n=42.5 A; Mn=215 Nm; ηmotor=90.8%; Jmotor=0.33 k gm2
Selected drive converter:
6SE7027-2ED61
PV n=37 kW; IV n=72 A; IV max=98 A
Closed-loop f r equency control
The maximum cur r ent of the 37 kW drive converter is 98 A. Thus, t he 22 kW m otor (Imo tor n = 42. 5
A) can also easily start even with a suspended test load of 25 t. T he st arting curr ent must be set
high enough, so that the load doesn’t slip when the brake is released and ther e ar e no pr oblem s at
the transit ion from open- loop controlled operation t o closed- loop controlled operation. I t is practical
to set the slip freq uency for the brake release somewhat higher t han t he rated slip frequency.
Rated motor slip freq uency:
fnn
nHz Hz
slip n Sn
S
=−⋅=
−⋅=50 1000 975
1000 50 125.
As a result of the relatively long accelerat ing time fr om 10 s t o 50 Hz, t he brake enable signal can
be simply realized using the signal fV ≥ fx, with fx = 2 Hz. I n the still available time from 0. 4 s to
when the brake is released, the flux will have had time to establish itself.
Dimensioning the brake r esistor
The brake resistor is used when the load is lowered. As result of the low influence of the
accelerating- and decelerating torques, only the br aking power due to t he const ant load torq ue
when the load is being lowered is taken int o account. The worst case sit uation is investigated,
when the nominal load is lowered fr om the full height.
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Velocity-time diagram when lowering the load from hmax=5 m
tbtv
ttotal
t
vload max
vload
-
vvms
load drum
max max ./==
200865
tt s
vb
==10
Total time f or the load to be lowered:
ttotal ==⋅+ −⋅
tt
hv t
v
br b load b
load
2max max
max
=⋅ +−⋅
=210 5 0 0865 10
00865 678
.
..s
Max. brak ing power of the br aking resistor:
Pbr W max =⋅⋅⋅
Fv
HTr Gmotor
max
103
ηη
=⋅⋅⋅ =
98100 0173
10 09 0 908 1387
3... .kW
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Braking diagra m
t
Pbr W
10 s 67.8 s 10 s
13.87 kW
Braking energy f or a 90 s cycle (cor r esponds to the area in the br aking diag r am ):
WP t t kWs
br br W total b
=⋅−=⋅−=
max ().(.).1387 678 10 8017
The following must be true for the br ake resistor:
W
TkW P
br br cont
== ≤
8017
90 891
...
With
PP
br cont..
=20
45 (f or an exter nal br ake resistor)
the f ollowing is obt ained
891 45 401 20
.. .⋅= ≤kW P
Thus, a br aking unit is selected with P 20 = 50 kW (6SE7028-0EA87-2DA0) with an external brake
resistor ( 6SE7028-0ES87-2DC0).
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3.1.6 Elevator (lift) drive
This drive involves an elevator with a 1:1 suspension system, and a load capability of 1080 kg for
14 people. The elevator moves through a tot al height of 23 m with 8 floors. T he elevator
approaches the floors directly without crawl using a closed-loop position control function.
i, G
Drive pulley
D
Counter-weight
Cabin +
load
Cable Suspended
cable
IM
η
Elevator data
Driving pulley diameter D = 0. 64 m
Gearbox ratio i = 35
Gearbox efficiency ηG= 0.69
Load capability Q = 1080 kg
Max. weight of the elevator cabin Fmax = 1200 kg
Max. counter- weight (Fmax+0.5Q) Gmax = 1740 kg
1/2 suspended cable weight 1/2 mHK = 28. 75 kg
Cable weight mcable = 48.3 k g
Operat ional velocity vmax = 1.25 m/s
Acceleration amax = 0.9 m/s2
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Rated motor output Pn= 13.5 kW
Rated motor current IMotor n = 33.5 A
Rated motor torque Mn= 94 Nm
Rated motor speed nn= 1455 RPM
Motor efficiency ηMotor = 0. 89
Motor moment of inertia JMotor = 0.21 k gm2
Moment of inert ia of the hoist ( r eferred to the drive pulley) JW= 105.7 kgm2
Moment of inert ia of the handwheel (referred to t he dr ive pulley)JH= 27 kg m2
Elevator height range FH = 23 m
Ascending with a full load and at constant velocity
Hoisting force:
FH=⋅+−⋅+ ⋅(. ( ) . )102 097mm mm g
F Q G cable
=⋅ + −⋅+⋅(. ( ) . .) .102 1200 1080 097 1740 483 981
=67306. N
(1,02; 0, 97: additional fact ors)
Load torq ue, m otor torque:
MF
DNm
load H
=⋅= ⋅ =
267306 064
221538...
MM
iNm
motor load
G
=⋅=⋅=
η
21538
35 069 8918
.
..
Motor output:
PFv kW
motor H
G
=⋅⋅=⋅
⋅=
max ..
..
η
10 67306 125
069 10 12 2
33
The existing 13. 5 kW m ot or is sufficient, as the dr ive operat ion is int ermittent ( r efer to t he thermal
check later in the text).
Motor speed at vmax:
nini
vDRPM
motor drummax max max ..
=⋅ =⋅ ⋅
⋅=⋅ ⋅
⋅=
60 35 125 60
064 1305
π
π
D
G
F+Q
Cable
FH
Mload
nTr
v
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Descending with full load and const ant velocity
Motor torque, motor out put :
MMiNm
motor load G
=⋅= ⋅=
η
21538
35 069 4246
...
PFv kW
motor HG
=− ⋅⋅=− ⋅⋅=−
max .. ..
10 67306 125
10 069 58
33
η
(reg ener at ive operat ion)
Calculating the mot or torques when acceler ating and decelerating
Motor angular accelerat ion:
α
motor ia Ds=⋅ ⋅ = ⋅ ⋅ = −
max ...
235 09 2
064 9844 2
Accelerating torque for the motor:
Mbmotor =⋅
Jmotor motor
α
=⋅ =021 9844 2067.. .Nm
Load moment of inertia referred to the drive pulley:
mlinear
∑=⋅++++102 1
2
.( )mmmm m
FQGSeil HK
=⋅ ++++ =102 1200 1080 1740 483 2875 4180.( . .) kg
Jload =⋅
∑mD
linear ()
22
=⋅ =4180 064
2428
22
(.)kgm
Accelerating torque for load+hoist+handwheel:
Mb load W H++ =++⋅()JJJ i
load W H motor
α
=++⋅ =(.)
.
428 1057 27 9844
35 1577 Nm
D
G
F+Q
Cable
FH
Mload
nTr
v
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As deceleration is the sam e as acceler ation, the f ollowing is valid:
Mvmotor =Mbmotor
Mv load W H++ =++
Mbload W H
Ascending with full load, m otor torque when accelerating:
Mmotor =+ +⋅
⋅
++
MM M
i
b motor b load W H load G
()
1
η
=++ ⋅
⋅=2067 1577 21538 1
35 069 1752.( .) ..Nm
Ascending with full load, m otor torque when decelerating:
Mmotor =− + − + ⋅ ⋅
++
MM M
i
v motor v load W H load G
()
1
η
=− + − + ⋅ ⋅=2067 1577 21538 1
35 069 32.( .) ..Nm
Descending with full load, m otor torque when accelerating:
Mmotor =− + − + ⋅
++
MM M
i
b motor b load W H load G
()
=− + − + ⋅ =−2067 1577 21538 069
35 93.( .)
..Nm
Descending with f ull load, motor t orque when decelerating :
Mmotor =+ +⋅
++
MM M
i
v motor v load W H load G
()
=++ ⋅=2067 1577 21538 069
35 94 22.( .)
..Nm (reg ener at ive operat ion)
The highest motor t or que is requir ed when accelerating while ascending with a full load. The
highest m ot or torque in t he r egenerative mode is when decelerating while descending with a f ull
load.
Accelerating- and decelerating t imes:
ttv
as
bv
== = =
max
max
...
125
09 139
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Calculating the maxim um braking power
The maximum m ot or output in regenerative operation occurs when the drive starts to decelerate
fr om t he m aximum motor speed while descending with full load.
PMn kW
br motor motor v motor
max max ..=⋅=− ⋅=−
9550 94 22 1305
9550 1288
Selecting the drive conver ter
Selected drive converter:
6SE7026-0ED61
PV n=30 kW; IV n=59 A; IV max=80.5 A
T300 technology board with closed-loop elevator control
The drive converter is adequately dimensioned, as the r at ed convert er cur r ent corresponds to 176
% of t he r at ed motor curr ent . For the highest r equired motor t orque, t he appr ox. m ot or cur rent is
given by:
IM
MIII
motor motor
motor n motor n n nmax max
()( )≈⋅−+
22 2 2
µµ
With II
nmotorn
µ
=⋅035. (assumpt ion) , t he following is obtained:
IA
motor max (.)(. . .) . . .≈ ⋅ −⋅ +⋅=
1752
94 335 035 335 035 335 59 7
2 2 22 22
Dimensioning the brake r esist or
When dim ensioning the brak e r esistor, the worst case sit uation is investigated which is when the
elevator is descending with a f ull load t hr ough all the f loor s. An 8 s no- load interval is assumed
between ending a descent and star t ing to mak e an ascent.
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Velocity-time diagram f or a descent and an ascent thr ough all of t he floors (max. height)
tbtv
ttotal
vmax
-
vmax
tp
tp
tbtv
T
Descent No-load
interval Ascent
t
v
ttotal
tk
tk
No-load
interval
ts
p=8
tt s
vb
==
139.
hm
ma
x
=23
Thus, the following t ot a l t im e is obtained for travelling thr ough the full height:
ttotal =⋅+ −⋅
=⋅ + −⋅=22139
23 125 139
125 1979thvt
vs
bbmax max
max
...
..
Constant t r avel velocity time:
tt t s
ktotal b
=−⋅= −⋅=21979213917..
Cycle time for t he descent and ascent:
Ttt s
total p
=⋅ + =⋅ + =2 2 1979 8 5558()(.).
Max. brak ing power of the br ake resistor :
Pbr W max =⋅=⋅=PkW
br motor motormax .. .
1288 089 1146
Brake r esist or power when the elevator is moving at a constant velocity:
Pbr W const =⋅=⋅=PkW
br motor const mot
58 089 516.. .
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Braking diagra m
Descent No-load
interval Ascent
t
Pbr W
17 s
19.8 s
19.8 s
8 s 8 s
1.39 s 1.39 s
11.46 kW
5.16 kW
55.58 s
No-load
interval
Braking energy f or a cycle (cor r esponds to the area in the br aking diag r am ):
Wbr =⋅+⋅⋅PtPt
br W const k br W v
1
2max
=⋅+⋅ ⋅516 17 1
21146 139...
=+=
877 7 96 957.. .kWs
The following must be valid for the brake resistor:
W
TkW P
br br cont
== ≤
957
5558 172
.
...
With
PP
br cont..
=20
45 (with an external brak e r esist or )
the f ollowing is obt ained
45 172 774 20
.. .⋅= ≤kW P
Thus, a br aking unit is selected with P 20 = 10 kW (6SE7021-6ES87-2DA0) with an external brake
resistor ( 6SE7021-6ES87-2DC0).
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Thermally checking the m ot or
This elevator makes 180 ascents/descent s per hour , i. e. 20 s per ascent /descent including 8 s no-
load interval. Thus, when thermally checking t he m ot or, a duty cycle at full load t hr ough 5 floors
has been assumed.
Height difference between the bottom and t op floors (5 floors):
hm
5523
8144=⋅ = .
Total t im e t o t ravel from t he first t o the fifth f loor ( through 5 floors):
ttotal =⋅+ −⋅
25
thv t
v
bbmax
max =⋅ + −⋅=2139 144 125 139
125 12 91....
..s
Cycle time for a descent and an ascent:
Ttt
total p
=⋅ + =⋅ + =2 2 12 91 8 4182()(.).
Time for which the elevator moves with continuous velocity:
tt t s
k total b
=−⋅= −⋅=
2 1291 2 139 1013...
Torque characteristics when descending and ascending
Mmotor 175.2 Nm
89.18 Nm
3.2 Nm
-9.3 Nm
42.46 Nm
94.22 Nm
1.39 s 1.39 s
10.13 s 10.13 s
1.39 s
1.39 s
Ascending DescendingNo-load
interval
8 s 8 s
41.82 s
t
No-load
interval
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The RMS torque is obt ained from t he torque characteristic:
MRMS =⋅+ ⋅ +⋅+⋅+ ⋅ + ⋅
⋅+⋅
1752 139 8918 1013 32 139 9 3 139 42 46 1013 9422 139
212910338
2222 2 2
.. . . .. .. . . . .
(. . )
=1539677
311 .
.
=704. Nm
Calculating t he average speed:
nmot average =
+⋅
+⋅
∑nn
t
tkt
mot i A mot i E
i
efP
2=⋅⋅⋅+⋅
⋅++ ⋅
1
22
2
nt nt
ttkt
bk
bk fP
max max
=⋅⋅⋅+⋅
⋅+ +⋅ =
1
21305 139 2 1305 1013
2 139 1013 033 8 967
..
... RPM
The average m ot or speed is greater than half the rated mot or speed ( c ont r ol range 1:2) and the
calculated RMS torque is less than the motor r at ed torque with 94 Nm. Thus, operation is
therma lly perm issible.
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3.1.7 Traction drive along an incline
In this case, it involves a rack and pinion drive. The g r adient is 3 degrees. At standstill, t he vehicle
is held at standstill by the m ot or holding brake.
Motor
Pinion Rack
Gearbox
Drive data
Mass to be moved (fully laden) m = 3500 kg
Mass to be moved (empty) m = 3000 kg
Pinion diameter D = 0.12 m
Angle of inclination (g r adient) α= 3 degrees
Specific t r act ion r esist ance wF= 0.05
Mech. efficiency η= 0.9
Max. ve loci ty vmax = 0.2 m/s
Max. acceleration ab max = 0.8 m/s2
Max. deceleration av max = 0.8 m/s2
Max. distance moved smax = 2.4 m
Cycle time tcycle = 210 s
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Specified m ot ion char act er ist ic
tp1
tp2
tp3 tp4
tp5
vmax
vmax
_
v
t
T
1 2
3456
Descending
Ascending
tp6
1
~
~
Distances travelled (as specified):
Range 1: s1= 1.4 m with m=3500 kg (ascending )
Range 2: s2= 1 m with m=3000 kg (ascending)
Total 2.4 m
Range 3: s3= 0.15 m with m=3000 kg (descending )
Range 4: s4= 0.65 m with m=3000 kg (descending )
Range 5: s5= 0.2 m with m=3000 kg (descending )
Range 6: s6= 1.4 m with m=3500 kg (descending )
Total 2.4 m
No-load intervals (as specified):
tp1=tp5=2 s
tp2=tp3=tp4=8 s
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Accelerating time, decelerating time:
tt v
as
bv b
== = =
max
max
.
..
02
08 025
With
tsvtt
v
ki ibv
=−⋅ ⋅+05.()
max
max
the f ollowing tim es ar e obt ained when moving at a constant velocity:
ts
k1675
=.
ts
k2475
=.
ts
k305
=.
t
s
k43=
ts
k5075
=.
ts
k6675
=.
Thus, t he t im e for the com plete distance moved is given by:
ttttt s
ascent descent pi
i
i
ki
i
i
bv+=
=
=
=
=++⋅+=++=
∑∑
1
5
1
66282253535() . .
With the cycle time (specif ied)
Tt s
cycle
==210
the rem aining no-load inter val is g iven by:
tTt s
p ascent descent6210 535 1565=− = − =
+..
Precise positioning is only possible using a ser vo drive due to t he shor t accelerating- and
decelerating t im es.
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Ascending at constant velocity (loaded)
v
FH
m g
FW
Fload
FN
α
Component of gravity down the incline, normal com ponent of f or c e, dr ag:
F
H
=⋅⋅mgsin
α
=⋅⋅ =3500 981 3 1797
0
.sin N
F
N=⋅⋅mgcos
α
=⋅⋅ =3500 981 3 34288
0
.cos N
FW=⋅
F
w
NF
=⋅=34288 0 05 1714 4..N
Force due to the load, load torque:
FFF N
load ascend H W,..=+= + =1797 1714 4 35114
Mload ascend,=⋅FD
load 2
=⋅=35114 012
22107...Nm
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Descending at constant velocity (loaded)
v
FH
m g
FW
Fload
FN
α
Force due to the load, load torque:
FFF N
load descend H W
=−= − =1797 1714 4 82 6..
Mload descend =⋅FD
load 2
=⋅=82 6 012
2495...Nm
Calculating the load tor ques when accelerating and decelerat ing ( loaded)
Angular acceler at ion of the pinion:
α
bpinion b
aDs=⋅=⋅= −
max ...
208 2
012 1333 2
Load moment of inertia:
Jm
Dkgm
load =⋅ = ⋅ =() (
.).
23500 012
212 6
222
As acceleration is the sam e as deceler at ion, the following is valid f or the load torques:
MMJ Nm
b load v load load b pinion
==⋅=⋅=
α
12 6 1333 168..
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Selecting the mot or
The following motor is empir ically selected (r efer to calculating the motor torques):
1FT6 041-4AF7 with gear box i=81, nn=3000 RPM, Mn(100)=2.2 Nm, In(100)=1.7 A, ηmotor=0.86
Jmotor+brake=0.00039 kg m 2, Jgearbox=0.0008 kgm 2
Motor speed at vmax:
ni
vDRPM
motor max max ...=⋅ ⋅
⋅=⋅ ⋅
⋅=
60 81 02 60
012 25783
π
π
Calculating the mot or torques while t he vehicle is moving at constant velocit y ( loaded)
Ascending:
MM iNm
motor load ascend
=⋅
⋅=⋅
⋅=
12107 1
81 09 289
η
...
Descending:
MM iNm
motor load descend
=⋅=⋅=
495 09
81 0055...(regenerative operation) 1)
1) Regenerat ive operat ion is obt ained from t he various sig ns for tor que and speed
Calculating the mot or torques when acceler ating and decelerating (loaded)
Motor accelerating- and deceler at ing torq ues:
MM
b motor v motor
==⋅⋅Ji
motor b pinion
α
=⋅⋅=000039 81 1333 0 42...Nm
The accelerat ing - and decelerating torques for the g earbox, refer r ed to the pinion:
MM
b gearbox v gearbox
==⋅⋅Ji
gearbox b pinion
2
α
=⋅⋅=00008 81 1333 70
2
..Nm
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Ascending, m ot or torque when accelerating :
Mmotor =++ + ⋅
⋅
MMM M i
b motor b load b gearbox load ascend
()
1
η
=+++ ⋅
⋅=042 168 70 2107 1
81 09 658.( .) ..Nm
Ascending, m ot or torque when decelerating :
Mmotor =− + − − + ⋅ ⋅
MMMM i
v motor v load v gearbox load ascend sign
()
)(...)
11
η
=− + − − + ⋅ ⋅=−
−
042 168 70 2107 1
81 09 072
1
.( .) ..Nm (r egenerative operation)
Descending, m ot or torque when accelerating :
Mmotor =− + − − + ⋅MMMM i
b motor b load b gearbox load descend
sign
()
)(...)
1
η
=− + − − + ⋅ =−
−
042 168 70 4 95 09
81 362
1
.( .)
..Nm
Descending, m ot or torque when decelerating :
Mmotor =++ + ⋅MMMM i
v motor v load v gearbox load descend
()
=+++ ⋅=042 168 70 4 95 09
81 312.( .)
..Nm (reg ener at ive operat ion)
1) For a neg at ive expression in brack ets, η changes to η-1
The highest motor t or que is requir ed when accelerating as the vehicle ascends. The hig hest motor
torq ue in r egenerative operation is r equired when decelerating as t he vehicle descends. The
selected motor is adequately dimensioned, as it can be overloaded up t o appr ox. 7 Nm at
nmax=2578.3 RPM and 400 V supply voltage.
The steady-st at e holding brake torque corr esponds to the motor t orque when descending at
constant velocity:
MNm
hold =0055.
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Selecting the drive conver ter
For the hig hest r equired motor torque, the motor cur r ent is given by:
II
M
MA
motor motor n motor
motor n
max max ....=⋅ =⋅=17 658
22 508 ( har dly any saturat ion effect for this motor)
Selected drive converter:
6SE7013-0EP50 (Compakt Plus type)
PV n=1.1 kW; IV n=3 A, IV max=9 A (300% overload capability)
Dimensioning the brake r esistor
The brake resistor is used when the vehicle decelerates as it ascends and when it descends at
constant velocity and when decelerating. In the following calculat ion, the slight decrease in the
motor t orque during motion without load is neglected ( thus the calculation is on the safe side).
Max. braking power for the brake resistor when decelerating from vmax to 0:
Pbr W vmax ( )
max→0=⋅=
⋅⋅
→→
PMn
br motor v motor motor v v motor motormax ( ) () max
max
max
00
9550
ηη
=⋅⋅=
072 25783
9550 086 0168
..
..kW
Braking power for t he br ake resistor while moving at const ant velocity with -vmax:
Pbr W const v v()
max
=− =⋅=⋅⋅
=− =−
PMn
br motor const v v motor motor const v v motor motor() () max
max
max
ηη
9550
=⋅⋅=
0055 25783
9550 086 0013
..
..kW
Max. brak ing power for t he br ake resistor when decelerating fr om - vmax to 0:
Pbr W vmax ( )
max
−→0=⋅=
⋅⋅
−→ −→
PMn
br motor v motor motor v v motor motormax ( ) () max
max
max
00
9550
ηη
=⋅⋅=
312 2578 3
9550 086 0724
..
..kW
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Brake diagram
Pbr W
tvtvt
Ascending Descending
(2x) (4x)
T
tk3,4,5,6
0.724 kW
0.168 kW
0.013 kW
~
~
~
~
Braking energy f or a cycle (cor r esponds to the area in the br aking diag r am ):
W
br =⋅⋅ ⋅+ ⋅ + + + +⋅⋅ ⋅
→=− −
→
21
241
2
03456
0
PtP ttttPt
brW v v brWconst v v k k k k brW v vmax ( ) ( ) max ( )
max max max
()
=⋅⋅ ⋅ + ⋅ ++ + +⋅⋅ ⋅ =21
20168 025 0013 05 3 075 675 4 1
20724 025 0545...(. ..) ...kWs
The following must be valid for the brake resistor:
W
TkW P
br br cont
== ≤
0545
90 00061
...
As the cycle time with 121 s is great er t han 90 s, T is set t o 90 s.
With
PP
br cont.,
=20
45 (f or Com pact Plus, only an external brake resist or )
the f ollowing is obt ained
45 00061 0027 20
.. .⋅= ≤kW P
Thus, t he smallest brak ing resistor is select ed with P20 = 5 kW (6SE7018-0ES87- 2DC0) .
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Thermally checking the m ot or
In the following calculation, the slight decrease in the m ot or torque when the vehicle is not carrying
a load is neglected ( t hus, the calculation is on the safe side).
Torq ue char act er istics
tv
tb
tb
tv
Mmotor
6.58 Nm
2.89 Nm
-0.72 N m
-3.62 N m
0.055 Nm
3.12 Nm
T
t
Ascending
(2x) Descending
(4x)
tk3,4,5,6
tk1,2
~
~
~
~
The RMS torque is obt ained from t he torque characteristic as f ollows
tt s
kk12
675 4 75 115+= + =.. .
tttt s
kkkk3456
05 3 075 675 11+++=++ + =...
Thus,
MRMS =⋅+⋅+⋅+⋅+⋅+⋅
⋅ ⋅+ + +⋅ +++++ =
2 658 072 025 289 115 4 362 312 025 0055 11
6 025 2 115 11 1 2 8 8 8 2 1565 082
22 2 22 2
(. . ) . . . (. . ) . .
.. ( .) .Nm
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For the 1FT 6 m ot or, factor kf is set to 1 for the no-load intervals. T he calculated RMS torque is
less than the rat ed motor tor que with 2.2 Nm. Thus, operation is ther m ally perm issible.
Inform ation regarding the calculation
Due to the tedious calculations required when trying- out various motors and g ear box rat ios, it is
recommended t hat a spreadsheet program such as Excel is used.
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3.2 Winder drives
3.2.1 General information
DD
max
FZ=const.
V
i, G
Reel
nW
nmot, Mmot
core
D
Motor
η
master=const.
Mode of operation of an axial winder
A material web, with width b, is wound or unwound at constant velocity and tension.
nvD
W=⋅⋅60
π
reel speed [RPM]
vweb velocity [m/s]
Dactual reel diamet er [m]
nvD
Wmin max
=⋅
⋅60
π
minimum speed of the reel at D=Dmax [RPM]
nvD
WKern
max =⋅
⋅60
π
max. speed of the reel at D=Dcore [RPM]
Fzb
Z
=⋅ web tension [N]
zweb tension/m [N/m]
bweb width [m]
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MF
DFvn
WZ Z W
=⋅=⋅
⋅⋅⋅
260
21
π
winding torque ( pr opor tional to 1/nW)[Nm]
MF
D
WZmax max
=⋅
2max. winding torq ue at D=Dmax [Nm]
MF
D
WZ
core
min =⋅
2min. winding torque at D=Dcore [Nm]
PFv
WZ
=⋅
103winding power (constant) [kW]
MP
WW,
MW max
MW min
PW= cons t.
MW~ 1/nW
Winding
Unwinding
Dmax Dcore
nW min nW max nW
Winding torque and winding power at v = const. and z = const.
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Influence of t he f riction
Further, a speed- dependent frictional torque is pr esent due to the bearing friction and other
effects. Thus, the winding torque is increased when winding and reduced when unwinding.
M
Rspeed-dependent frictional tor que [Nm]
M
M
M
W res W R
=+ resulting winding torq ue when winding [Nm]
M
M
M
W res W R
=− resulting winding torq ue when unwinding [Nm]
Motor tor que, motor out put (steady-st at e oper ation)
With
inn
Dn
v
motor
W
core motor
==
⋅⋅
⋅
π
max
60 gearbox ratio
Ggearbox ef ficiency
the f ollowing is obtained for the m otor:
MMM
i
motor WR
G
=+
⋅
η
motor t or que when winding [Nm]
PMM
n
motor WR
GW
=+⋅⋅
9550
η
motor out put when winding
(energ y f low from the mot or t o the load)
[kW]
MMM
i
motor WR
G
=− −⋅
η
motor t or que when unwinding [Nm]
PMM
n
motor WR
WG
=− −⋅⋅
9550
η
motor out put when unwinding
(energ y f low from the load to t he m otor)
[kW]
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Acceleration, deceler at ion
When acceler at ing and decelerating the reel at constant web tension, additional accelerating- and
deceleration tor ques are present. The ef fective moment of inertia consist s of a f ixed component
(motor , gearbox/coupling = supplem entary, winding core) and a variable component (reel as
hollow cylinder).
JJ J Ji
F motor plementary core
=+ +
sup 2fixed moment of inertia referred to the motor
shaft
[kgm2]
JbDD
VWcore
=⋅⋅ ⋅⋅ −
π
ρ
32 1034 4
()
variable reel moment of iner t ia
(as a f unct ion of reel diamet er D)
[kgm2]
and
JmDD
VW core
=⋅ ⋅ +
1
822
() [kgm2]
JbDD
VWcoremax max
()=⋅⋅ ⋅⋅ −
π
ρ
32 1034 4
maximum reel m o m ent of iner t ia [kgm2]
and
JmDD
VW coremax max
()=⋅ ⋅ +
1
822 [kgm2]
JforDD
Vcoremin ==0min. r eel m om ent of inertia
b, D in m; ρ in kg/dm3; m in kg
Accelerating- and decelerating t or que for t he m otor + gearbox/coupling ref er r ed to the motor shaft:
MJJi
Dv
t
b v motor plement motor plement bv
,sup sup ,
()
+=+ ⋅⋅⋅
2[Nm]
Accelerating- and decelerating t or que for cor e+r eel r eferred t o the reel:
MJJ
Dv
t
b v core reel core V bv
,,
()
+=+⋅⋅
2[Nm]
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tbaccelerating time from 0 to v [s]
t
v
decelerating t im e from v to 0 [s]
When acceler at ing and decelerating, the variable moment of inertia is assumed to be constant.
For constant t ension during acceleration and deceler at ion, the motor t orque is given by:
Winding , motor tor que when accelerating f r om 0 to v:
MM M MM
i
motor b motor plement b core reel W R G
=+++⋅
⋅
++sup ()
1
η
[Nm]
Winding , motor tor que when decelerating from v to 0:
MM M MM
i
motor v motor plement v core reel W R G
=− + − + + ⋅ ⋅
++sup )
()
11
η
[Nm]
1) If t he expression in t he br ackets is < 0, t he factor 1/ηG must be changed to ηG (the
deceleration component is pr edominant)
Unwinding, motor torque when accelerating from 0 t o v:
MM M MMi
motor b motor plement b core reel W R G
=+−+⋅
++sup )
()
2
[Nm]
Unwinding, motor torque when decelerating from v to 0:
MM M MM
i
motor v motor plement v core reel W R G
=− + − − + ⋅
++sup ()
[Nm]
2) If t he expression in br ackets is > 0, factor ηG must be changed to 1/ηG (acceleration
component is predom inant )
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The mot or out put is obtained f r om t he motor tor que as follows
PMn
motor motor motor
=⋅
9550 [kW]
Motor selection
The mot or is select ed according to t he m aximum t or que at nmin and t he m aximum speed. For
additional accelerating - and deceler ating tor ques, it may be possible to ut ilize the overload
capability of the drive. As the winding torque is proportional to 1/n, then it is possible to use t he
field-weakening range, with Mperm. ∼1/n. Further , there must be sufficient safety margin to the
stall torque in the f ield- weakening range.
The winding power PW is constant over the speed. Thus, the motor must pr ovide this out put at
nmin.
M
M
nn n
min max
perm.
Mmotor stat.
stall
1,3
motor
M
nmotor
Fiel d we akening range
´
n
Motor torque under steady-state operat ing conditions and permissible tor que for S1 duty
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Closed-loop winder cont r ol
There ar e several ways of providing const ant tension:
•indirect closed-loop t ension control (without tension t r ansducer )
In this case, the constant tension is im plem ented using a torque ref er ence value input adapt ed
by the diameter. I t is important t o accur ately compensate the f rictional- and accelerating
torq ues, as these parameter s cannot be com pensated without using a tension tr ansducer.
•direct closed-loop t ension control with dancer roll
The tension is ent er ed using a position-controlled dancer roll.
•direct closed-loop t ension control with tension transducer
The tension act ual value is sensed using a tension transducer and is appr opr iat ely contr olled.
Note:
An Excel program is available to dim ension winders: "McWin - Motorcalculation f or Winder " . Also
ref er to ASI Information E20125-J3001-J409-X-7400 from Sept em ber 94.
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3.2.2 Unwind stand with closed-loop tension control using a tension transducer
This application involves an unwind stand with "flying " roll change. Before t he reel is changed, t he
new reel is prepared for splicing. When the old reel falls below a specific diameter , the new reel is
accelerated until the circumferential speed corr esponds t o the web velocity of t he paper. The reel
support mechanism then rotates int o the splice position. T he new reel is autom atically spliced, and
the old reel ret racted.
Unwind stand data
Max. reel diameter Dmax = 1.27 m
Core diamet er Dcore = 0.11 m
Web width b = 1.7 m
Material density ρW= 0.93 kg/dm3
Max. web veloci ty vmax = 15 m/s
Web t ension z = 100 N/m
Acceleration tb= 40 s
Stop (coast down) tv= 40 s
Friction MR= 20 Nm at Dmax
Gearbox ratio i = 1 to 3
Steady-state operation
nv
DRPM
Wmin max
max .
=⋅
⋅=⋅
⋅=
60 15 60
127 226
ππ
nvDRPM
Wcore
max max .
=⋅
⋅=⋅
⋅=
60 15 60
011 2604
ππ
Fzb N
Z=⋅= ⋅ =100 17 170.
MF
DNm
WZmax max .
=⋅ = ⋅ =
2170 127
2108
MF
DNm
WZ
core
min ..=⋅ = ⋅ =
2170 011
294
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PFv kW
WZ
=⋅=⋅=
max .
10 170 15
10 255
33
Gearbox ratio i = 2 is selected
The gearbox eff iciency ηG is assumed to be ≈ 1 ( belt-driven gear box, pinion wheel)
n i n RPM
motor Wmin min
=⋅ = ⋅ =2 226 452
n i n RPM
motor Wmax max
=⋅ = ⋅ =2 2604 5208
MMM
iNm
motor WR
max max
=− −=− −=−
108 20
244
MMM
iNm
motor WR
min min ..=− −=− −=−
94 0
247
The frictional torque at nW min was assumed to be 0. T he motor tor ques are negat ive, as t his
involves an unwind st and ( regenerative operat ion) .
Calculating the maxim um motor tor ques when accelerating and decelerating
JVmax =⋅⋅ ⋅⋅ −
π
ρ
bDD
Wcore
32 1034 4
()
max
=⋅⋅ ⋅⋅ − =
π
17 093
32 10 127 011 40376
344 2
.. (. . ) . kgm
Jcore ≈0 (cardboard core)
JF=+ ≈JJ kgm
motor plementsup .008 2 (estimat ed)
Relationships at D=Dmax
Mb motor plement+sup =+ ⋅⋅⋅()
sup max
max
JJ i
Dvt
motor plement b
2
=⋅⋅⋅=008 2 2
127 15
40 0094...Nm
Mbcore reel+=+⋅⋅()
max
max
JJDvt
core roll b
2
=+ ⋅ ⋅ =(.)
..0 40376 2
127 15
40 2384 Nm
MM
v
b
= (as tb = tv)
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Accelerating from 0 to vmax (without tension, with f r ict ion) :
Mmotor =+−+⋅
++
MMMM
i
b motor plement b core reel W Rsup max
()
1
=+ −+⋅0094 2384 0 20 1
2
.(. )
=+=
0094 1292 1293...Nm
Decelerating from Vmax to 0 (with tension, with frict ion):
Mmotor =− + − − + ⋅
++
MMMM
i
v motor plement v core reel W Rsup max
()
1
=− + − − + ⋅0094 2384 108 20 1
2
.(. )
=− − =−
0094 1632 1633...Nm
The highest motor t or que is requir ed when decelerating with tension applied. At Dmax the
accelerating- or decelerating torque for the motor+supplement is essent ially of no signif icance.
Relationships at D=Dcore:
Mb motor plement+sup =+ ⋅⋅⋅()
sup max
JJ i
Dvt
motor plement core b
2
=⋅⋅⋅=008 2 2
011 15
40 109...Nm
Mbcore reel+≈=≈000(, )JJ
Vcore
MM
v
b
= (as tb = tv)
Accelerating from 0 to vmax (without tension, without friction):
Mmotor =+−+⋅
++
MMMM
i
b motor plement b core reel W Rsup min
()
1
=+−+⋅=109 0 0 0 1
2109.( ) .Nm
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Decelerating from vmax to 0 (with tension, without friction):
Mmotor =− + − − + ⋅
++
MMMM
i
v motor plement v core reel W Rsup min
()
1
=− + − + ⋅ =−109 0 94 0 1
2579.(.) .Nm
The highest motor t or que is requir ed when decelerating with tension.
Motor selection
The mot or select ed as result of the calculated maximum mot or torques:
1PA6 133-4HD.
Pn=13.5 kW; Mn=112 Nm; nn=1150 RPM; n1=2500 RPM; In=29 A; Iµ=13 A; Jmotor=0. 076 kgm2
When deceler at ing fr om vmax to 0 (with tension, with fr ict ion) , at nmin, the motor m ust pr ovide
1633
112 146
..= x
rated mot or torque for 40 s. T he maximum motor out put is then given by:
PMn kW
motor motor motor
=⋅=⋅=
min ..
9550 1633 452
9550 773
The 44 Nm st eady-st at e winding t or que is permissible f or the motor at nmin.
To check t he m otor in the field-weakening range, the perm issible m ot or output at nmax, with 130
% safety margin, is given by:
PP
n
nkW
perm n n.max
max ..=⋅ = ⋅ =
1135 2500
5208 648
Permissible motor torque at nmax and 130 % safet y margin fr om the stall torque:
MP
nNm
perm n
perm n
.
.
max
max
max ..=⋅=⋅=
9550 648 9550
5208 119
Thus, t he m otor can also provide the req uir ed torque of 5.79 Nm, at nmax, when decelerating with
tension. T he m aximum m otor output is t hen given by:
PMn kW
motor motor motor
=⋅=⋅=
max ..
9550 579 5208
9550 316
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Selecting the invert er and the rectif ier /regenerative f eedback unit
The maximum m ot or cur rent in dynamic operation is g iven by:
IM
MIII
motor dyn motor dyn
motor n motor n n nmax max
()()≈⋅−+
22 2 2
µµ
With IA
n
µ
=13 , the following is obtained:
IA
motor dynmax (.)( )≈⋅−+=
1633
112 29 13 13 40
222 2
The maximum m ot or cur rent for winding operation is:
IM
MIII
motor stat motor stat
motor n Motor n n nmax max
()()≈⋅−+
22 2 2
µµ
IA
motor statmax ()( ) ,≈⋅−+=
44
112 29 13 13 16 5
222 2
Thus, an inverter is selected (2x):
6SE7023-4TC20
PV n=15 kW; IV n=34 A; IV max=46.4 A
T300 technology board with closed-loop winder control
Inf eed/ r egenerative f eedback unit:
6SE7024-1EC85-1AA0
Pn=15 kW; IDC link n=41 A
Autotransformer with 25% duty ratio:
4AP2795-0UA01-8A (25% is sufficient , as the power when winding is only 2.55 kW)
4% uk line reactor:
4EP3900-5UK
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-120
-100
-80
-60
-40
-20
001000 2000 3000 4000 5000 6000
Speed in RPM
Torque in Nm
M mot stat
M S1 charac.
Limited by M stall
P = const
M mot = Mn
Diagram for the winding tor que (M mot stat) and t he MS1 character istic
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-250
-200
-150
-100
-50
0
50
100
150
200
250
0 1000 2000 3000 4000 5000 6000
Speed in RPM
Torque in Nm
M mot b
M mot v+tens.
M max perm.
M mot = 2 Mn
Limited by Mstall
Diagram for the acceler at ing torq ue without t ension ( M mot b), the deceleration t or que with tension
(M mot v+tension) and the Mperm max characteristic
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3.2.3 Winder with closed-loop tension control using a tension transducer
This involves a winder with accumulator. The material web, running at max. 50 m/ m in, is wound-
up, tension contr olled. Just bef or e t he final diameter is r eached, the reel support assem bly is
rotated, the winder is stopped, the paper web cut, and t he second r eel spliced t o the empty core.
The mat er ial accum ulat or m ust accept the web coming from the m achine dur ing this time ( appr ox.
15-20 s).
Aft er t he empty reel starts , this runs, speed-cont r olled, 17.8 m/min faster than the web speed of
the machine. Thus, the material accumulator is em ptied in approx. 60 s. Af ter this, t he closed- loop
tension controlled m ode is selected.
Winder dat a
Max. reel diameter Dmax = 1 m
Core diamet er Dcore = 0.12 m
Max. reel weight mW= 870 kg
Max. web velocity in operation vmax B= 50 m/min = 0. 833 m / s
Max. web velocity when r etrieving paper vmax L= 67.8 m/m in = 1.13 m / s
fr om t he accum ulator
Web t ension FZ= 600 N
Acceleration tb= 2 s
Fast stop tv= 2 s
Material thickness d = 2 mm
Steady-state operation
nv
DRPM
WB
min max
max
..=⋅
⋅=⋅
⋅=
60 0833 60
1159
ππ
nv
DRPM
WL
core
max max ..
=⋅
⋅=⋅
⋅=
60 113 60
012 180
ππ
MF
D
Nm
WZ
max max
=⋅ = ⋅=
2600 1
2300
MF
DNm
WZ
core
min .
=⋅ = ⋅ =
2600 012
236
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PFv kW
WZB
=⋅=⋅=
max ..
10 600 0833
10 05
33
Selected gear box rat io, i = 20
The gearbox eff iciency ηG is assumed to be 0.9
n i n RPM
motor Wmin min .=⋅ = ⋅ =20 159 318
n i n RPM
motor Wmax max
=⋅ = ⋅ =20 180 3600
MM
iNm
motor W
G
max max
..=⋅=⋅=
η
300
20 09 16 7
MM
iNm
motor W
G
min min
.
=⋅=⋅=
η
36
20 09 2
Calculating the maxim um motor tor ques when accelerating and decelerating
JVmax =⋅ ⋅ +
1
822
mD D
Wcore
()
max
=⋅ ⋅ + =
1
8870 1 012 1103
22 2
(.) .kgm
Jcore ≈0 (cardboard cor e)
JF=+ ≈JJ kgm
motor plementsup .003 2 (estim ated)
Relationships at D=Dmax
Mb motor plement+sup =+ ⋅⋅⋅()
sup max
max
JJ i
D
v
t
motor plement B
b
2
=⋅⋅⋅ =003 20 2
10833
205...Nm
Mbcore reel+=+⋅⋅()
max
max
JJD
v
t
core reel B
b
2
=+ ⋅⋅ =(.)
..0 1103 2
10833
29192 Nm
MM
v
b
= (as tb = tv)
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Accelerating from 0 to vmax B (with tension):
Mmotor =++⋅
⋅
++
MMM
i
b motor plement b core reel W G
sup max
()
1
η
=+ + ⋅⋅
05 9192 300 1
20 09
.(. ) .
=+ =05 2177 22 27.. .Nm
Decelerating from vmax B to 0 (with tension):
Mmotor =− + − + ⋅ ⋅
++
MMM
i
v motor plement v core reel W G
sup max
()
1
η
=− + − + ⋅ ⋅
05 9192 300 1
20 09
.( . ) .
=− + =05 1156 1106.. .Nm
Decelerating from vmax B to 0 (without tension, e. g. mater ial web brok en) :
Mmotor =− + − ⋅05 9192 09
20
.( .)
.(the factor 1/ηG is chang ed int o ηG, as
the expression in brack et s is < 0)
=− − =−05 414 464.. .Nm
The highest motor t or que is requir ed when the winder is accelerating with tension. The highest
regener at ive mot or t orque is req uir ed when decelerating without tension. T he acceler ating- and
decelerating t or que for m otor+supplement pr act ically plays no role at Dmax.
Relationships at D=Dcore:
Mb motor plement+sup =+ ⋅⋅⋅()
sup max
JJ i
D
v
t
motor plement core
L
b
2
=⋅⋅⋅=003 20 2
012 113
2565....Nm
Mbcore reel+≈=≈000(, )JJ
Vcore
MM
v
b
= (as tb = tv)
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Accelerating from 0 to vmax L (with tension) :
Mmotor =++⋅
⋅
++
MMM
i
b motor plement b core reel W G
sup min
()
1
η
=++⋅
⋅=565 0 36 1
20 09 765.( ) ..Nm
Decelerating from vmax L to 0 (with tension):
Mmotor =− + − + ⋅ ⋅
++
MMM
i
v motor plement v core reel W G
sup min
()
1
η
=− + + ⋅ ⋅
565 0 36 1
20 09
.( ) .
=− + =−565 2 365..Nm
Decelerating from vmax L to 0 (without tension):
Mmotor =−565.Nm
The highest motor t or que is requir ed when accelerating with tension. The hig hest regenerative
motor t or que is requir ed when decelerating without tension.
Calculating the maxim um braking power
The following is valid for t he m otor output :
PMn
motor motor motor
=⋅
9550
The maximum m ot or output in the regenerative mode occurs when decelerating without tension at
maximum speed (i. e. at D=Dcore).
PkW
br motor max ..=−⋅ =−
565 3600
9550 213
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Motor selection
The mot or was selected as a result of the calculat ed m aximum m ot or torques:
1PQ5 113-4AA20- Z
Pn=4 kW; Mn=27 Nm; nn=1435 RPM; In=9.2 A;
Mstall=3 Mn; ηmotor=0.83; Jmotor=0.011 kgm2
Checking the motor in t he field-weakening r ange at nmax
Motor stall torque in the field- weakening range:
MM n
n
stall stall n n
=⋅()
2
Permissible motor torque at nmax and 130 % safet y margin fr om the stall torque:
Mperm. == ⋅ =
⋅⋅=
MMn
nNm
stall stall n n
13 13 327
13 1435
3600 99
22
..
() .().
max
The mot or can also pr ovide the dem anded t orque of 7.65 Nm when accelerating with tension, at
nmax.
Selecting the drive conver ter
Selected drive converter:
6SE7021-0EA61
PV n=4 kW; In=10.2 A
T300 technology board with closed-loop winder control
Dimensioning the brake r esist or
The max. braking power f or the brake r esist or is:
PP kW
br W br motor motormax max .. .=⋅=⋅=
213 083 177
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Braking energy:
WPt kWs
br br W v
=⋅=⋅=
max ..
2177 2
2177
Winding time:
tDD
dv
Wcore
=−
⋅⋅ ⋅= −
⋅⋅
⋅=
max
max
.
..min
22 22
41012
4 0002 50 774
ππ
> 90 s
Thus, t he cycle time for braking is set to 90 s. The following must be valid for t he brake resist or :
W
TkW P
br br cont
== ≤
177
90 002
...
With
PP
br cont.=20
36 (with an internal brake resistor)
the f ollowing is obt ained
36 002 072 20
⋅= ≤
..kW P
The lowest brak ing unit is selected with P20 = 5 k W (6SE7018-0ES87- 2DA0). The int er nal brake
resistor is sufficient.
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3.2.4 Unwinder with 1FT6 motor
A 1FT6086-8SF71- 1AA0 m ot or has been selected for a winder with flying roll change. The
dimensioning and cor r ect m otor selection are t o be checked. The unwinder inverter is t o be
integr ated into a mult i- m otor gr oup.
Unwind stand data
Max.reel diameter Dmax = 0.8 m
Core diamet er Dcore = 0.09 m
Reel weight m = 100 kg
Max.web veloci ty vmax = 200 m/min
Steady-state t orque referred to the motor speed at Dmax Msteady-st. max = 31 Nm
Required acceler a t ion t im e tb= 0.5 s
Required deceler ation time tv= 0.5 s
Gearbox rat io ( belt driven) i = 4
Moment of inertia of the m andr el Jmandrel = 0.08 kgm2
Diameter of the large gear wheel on the mandr el ( 112 t eeth) DZ mandrel = 290 mm
Diameter of the small gear wheel at the motor ( 28 teeth) DZ mot = 72.5 mm
Width of the g ear wheels b = 38 mm
Gear wheel mater ial: Steel
Motor data
Rated output Pn= 9.7 kW
Rated speed nmot n = 3000 RPM
Moment of inert ia Jmot = 0.00665 kgm 2
Efficiency ηmot = 0.91
Torque constant kT0= 1. 33 Nm / A
Stall torque M0= 35 Nm
Stall current I0= 26.4 A
Max. permissible t or que Mmax = 90 Nm
Max. permissible cur r ent Imax = 84 A
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Max. web velocity in m/s:
vms
max ./==
200
60 333
Web t ension:
FM i
DN
Zstat
=⋅⋅=⋅⋅=
max max .
231 4 2
08 310
Steady-state operation
PFv kW
WZ
=− ⋅=− ⋅=−
max ..
1000 310 333
1000 103
ni
vDRPM
mot min max
max
...=⋅ ⋅
⋅=⋅ ⋅
⋅=
60 4333 60
08 3183
ππ
ni
vDRPM
mot core
max max ...=⋅ ⋅
⋅=⋅ ⋅
⋅=
60 4333 60
009 28294
ππ
Steady-state m otor torque as a function of the mot or speed:
mot
mot
statesteadystatesteadymot n
n
MM min
max ⋅−= −−
Determining t he m ot or torques when accelerat ing and deceler at ing as a function of the m ot or
speed
Max. moment of iner tia of t he full roll:
JmDD kgm
vcore
max max
(( ) ( ) ) (( .)(
.)) .=⋅ + = ⋅ + =
22 2 100
208
2009
281
22 22 2
Estimating t he additional moments of inertia:
23434 066.010)
2
29.0
(85.7038.0
2
1
1085.7
2
1kgmrbJ mandrelZ =⋅⋅⋅⋅=⋅⋅⋅⋅≈
Jbr kgm
Zmot ≈⋅⋅ ⋅⋅ =⋅ ⋅ ⋅ ⋅ =
1
2785 10 1
20038 785 008
210 000026
43 43 2
...(
.).
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Thus, t he supplem entary moment of inertia, r eferred t o t he motor is given by:
2
22
sup 01.000026.0
408.0066.0 kgmJ
i
JJ
JmotZ
mandrelZmandrel
plement ≈+
+
=+
+
=
Accelerating from 0 to vmax without tension (with tension)
MJJ i
Dvtn
n
mot b mot plement b
mot
mot
=+ ⋅⋅⋅()
sup max
max
min
2component, m otor+supplement
+⋅⋅⋅
⋅−
−⋅JDvt
Dn
nD
DD i
vb
mot
mot core
core
max max
max max
min
max
()
21
444
44 component, winder
)( min
max mot
mot
statesteady n
n
M⋅− −(component tension)
The maximum m ot or torques when accelerating ar e obt ained without t ension. As this unwind stand
has a flying roll change, the accelerat ing curve is calculated without tension. At a r oll change, the
full or par tially full roll m ust first be acceler at ed in preparation for the flying roll chang e ( where t he
two material webs are glued together). I n operation, after standst ill, acceleration only with tension.
Decelerating from vmax to 0 with tension
MJJ i
Dvtn
n
mot v mot plement v
mot
mot
=− + ⋅ ⋅ ⋅()
sup max
max
min
2component, m otor+supplement
−⋅⋅⋅
⋅−
−⋅JDvt
Dn
nD
DD i
vv
mot
mot core
core
max max
max max
min
max
()
21
444
44 component, winder
mot
mot
statesteady n
n
Mmin
max ⋅− −component t ension
Checking the selected mot or
The steady-st at e and dynamic t or que characterist ics, together with the motor limiting
characterist ics ar e plot ted to check t he select ed motor. As can be seen from t he following
characterist ics, oper ation with the selected motor is permissible.
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-35
-30
-25
-20
-15
-10
-5
0500 1000 1500 2000 2500 3000
Motor speed in RPM
Torque in Nm
M mot steady-state
M S1 charac.
Steady-state t orque characteristic and Mperm S1 character ist ic
-100
-80
-60
-40
-20
0
20
40
60
80
100
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Torque in Nm
M mot b
M mot v+tens.
M max per.
Acceleration charact er ist ic without t ension, deceler ation characteristic with tension and Mperm max
characteristic
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Selecting the invert er
The inverter is selected according to the peak curr ent in dynamic oper at ion and accor ding to the
highest m ot or current when winding. The highest motor cur r ent when winding must be less t han
the inverter base load cur r ent so that the overload capability of t he invert er can be used. Thus, t he
following m ust be valid:
nINVdynmot II ⋅≤ 6.1
max
nINVstatesteadymot II ⋅≤
−91.0
max
The mot or cur r ent is given by, taking int o account saturation:
IM
kT b
mot mot
=⋅
01 for MM
mot ≤0
IM
kT b MM
MM MI
MI
mot mot
mot
=⋅⋅− −
−⋅− ⋅
⋅
01 0
0
20
0
11(( )( ))
max
max
max
for MM
mot >0
with
bb
nmot
115
16000
=−⋅()
,(b=0.1 for f r am e size <100, ot her wise 0.15)
The max. motor current when winding is 23.34 A. T he m ax. m ot or current in dynamic operat ion is
52.77 A. T hus, the following inverter is selected:
6SE7023-4TC51 SI MOVERT MASTERDRI VES Motion Contr ol
PINV n=15 kW
IINV n=34 A
IINV max=54.5 A
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0
5
10
15
20
25
30
35
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Motor current stat. in A
I mot ste ady -st a.
I WR base load
Motor current when winding, inverter base load curr ent
0
10
20
30
40
50
60
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Motor current dyn. in A
I mot b
I mot v+tens.
I WR max
Motor currents when accelerating without t ension and when decelerating with tension, maximum
inverter current
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DC link currents
In order t o integrate the inverter into a multi- m otor drive gr oup, the maximum DC link current s
must be calculated in dynamic oper at ion as well as the const ant DC link current when winding.
The DC link power is obt ained from:
VZ
INVmot
mot
motlinkDC n
MP )( 1
60
2
ηη
π
⋅
⋅
⋅⋅
⋅=
with
)( mot
MsignVZ =
Thus, the DC link curr ent is:
ly
linkDC
linkDC U
P
I
sup
35.1 ⋅
=
The maximum DC link powers, and therefore also the maximum DC link currents occur at the
maximum speed. As accelerat ion without t ension at m aximum speed does not occur (empt y roll),
acceleration with tension is used when dimensioning (acceler at ing aft er a st andstill). The following
values are obtained with Usupply=400 V:
AI blinkDC 92.3
max =acceleration with tension
AI vlinkDC 53.6
max −= decelerat ion with tension
AI statesteadylinkDC 71.1−=
−winding operation
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-8
-6
-4
-2
0
2
4
6
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
DC link current in A
I DC link v+tens.
I DC link b+tens.
DC link curr ent when accelerating with tension and when decelerating with tension
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3.2.5 Unwind stand with intermittent operation
An unwind stand is to be operated using t wo mechanically-coupled 1PA6 mot ors. The inverters for
the two motors are supplied from a rectifier/regenerative feedback unit. The web velocity is not
constant, but it is specified by the following characterist ic.
max
v
k
t
t
v
b
tv
t
T
Unwind stand data
Max. roll diameter Dmax = 1, 25 m
Core diamet er Dcore = 0,125 m
Roll weight m = 2000 kg
Max. web veloci ty vmax = 600 m/min
Web t ension Fz= 500 N
Required acceler a t ing time tb= 7 s
Required br aking t im e tv= 7 s
Constant- velocity tim e tk= 1 s
Cycle time T = 25 s
Gearbox ratio i = 3
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Max. web velocity in m/s:
smv /10
60
600
max ==
Each motor should accept 50 % of the load. Thus, f or the calculation, t he weight of the roll and the
web tension is always divided by two.
Steady-state operation
kW
v
F
P
Z
W5.2
1000
10
2
500
1000
2max −=
⋅
−=
⋅
−=
RPM
D
v
inmot 4.458
25.16010
3
60
max
max
min =
⋅⋅
⋅=
⋅⋅
⋅=
ππ
RPM
D
v
in
core
mot 7.4583
125.0 6010
3
60
max
max =
⋅⋅
⋅=
⋅⋅
⋅=
ππ
Steady-state m otor torque as a function of the mot or speed:
mot
mot
statmotstatmot n
n
MM min
max ⋅−=
with
Nm
i
DF
Mz
statmot 08.52
3
1
2
25.1
2
5001
22 max
max −=⋅⋅−=⋅⋅−=
Determining t he m ot or torques when accelerat ing and deceler at ing as a function of the m ot or
speed
Max. moment of iner tia of t he full roll:
22222
max
max 27.197))
2
125.0
()
2
25.1
((
2
2
2000
))
2
()
2
((
2
2kgm
DD
m
Jcore
v=+⋅=+⋅=
Estimating t he additional moments of inertia (r eferred t o t he motor):
2
sup 1.0 kgmJ plementary ≈
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Accelerating from 0 to vmax with tension
min
max
max
sup 2
)(
mot
mot
b
plementarymotbmot nn
t
v
D
iJJM ⋅⋅⋅+= component, motor+supplement ar y
iDD
D
n
n
D
t
v
D
J
core
core
mot
mot
b
v1
)(
244
max
44
min
4
max
max
max
max ⋅
−
−⋅
⋅⋅⋅+ winding component
mot
mot
stat n
n
Mmin
max ⋅− tension component
Deceleration f rom vmax to 0 with tension
min
max
max
.sup 2
)(
mot
mot
v
plmotvmot nn
t
v
D
iJJM ⋅⋅⋅+−= component, motor+ supplem ent ar y
iDD
D
n
n
D
t
v
D
J
core
core
mot
mot
v
v1
)(
244
max
44
min
4
max
max
max
max ⋅
−
−⋅
⋅⋅⋅− winding component
mot
mot
stat n
n
Mmin
max ⋅− tension component
Selecting the mot or
The following motor is selected:
1PA6 137-4HF.
Pn=25 kW; Mn=136 Nm; nn=1750 RPM; Mstall=485 Nm ; In=56 A; Iµ=23 A; Jmotor=0. 109 kgm2
ηmotor=0.902
When select ing the motor , it must be t aken into account t hat the motor acceler at ing torq ue and t he
motor deceler at ing torq ue at each speed must be below the dynamic limiting curve of the mot or . In
addition, the RMS current, obtained from the tr aversing characteristic must be less or equal to the
rated mot or current at each speed. The highest t or ques, and ther efore also the highest RMS
current is obt ained at nmot=nmot min as follows:
NmMbmot 65.99
max =max. accelerating torque
NmMvmot 81.203
max −= max. decelerating torque
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With
2222 )()( nnnmot
nmot
mot
mot III
M
M
I
µµ
+−⋅≈ (in t he constant flux rang e)
the maximum mot or currents when accelerating, when decelerating and when unwinding with
constant velocity are given by:
AI bmot 9.43
max =
AI vmot 9.79
max =
AI statmot 2.30
max =
v
t
T
b
t
k
t
t
mot
I
p
t
maxbmot
I
maxstatmot
I
maxvmot
I
n
I
µ
Thus, the following RMS value is obtained:
A
T
tItItItI
Ipnvvmotkstatmotbbmot
RMS 75.50
22 max
2max
2max =
⋅+⋅+⋅+⋅
=
µ
This value lies below the rated motor current with 56 A.
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To check t he dynamic r elat ionships, t he dynamic lim it ing curve of t he m ot or toget her with the
accelerating and deceler ating tor que will be investigated.
-300
-200
-100
0
100
200
300
0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000
Motor speed in RPM
Torque in Nm
M mot b+tens .
M mot v+tens.
M per m. max
Limited by M stall
2 M n
Selecting the invert er
The inverter is selected according to the peak curr ent in dynamic oper at ion and accor ding to the
highest RMS current. The f ollowing must therefore be valid:
nINVdynmot II ⋅≤ 36.1
max or nINVdynmot II ⋅≤ 6.1
max (f or 160 % overload capability)
nINVrmsmot II ≤
max
The max. RMS motor cur r ent is 50.75 A. The m ax. motor current in dynamic operation is 79.9 A.
Thus, t he following inverters are select ed:
6SE7026-0TD61 SI MOVERT MASTERDRI VES Vect or Control
PINV n=30 kW; IINV n=59 A; IINV max=94.4 A ( 160 % overload capability)
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The 160 % overload capability can be used here, as t he m otor torque and therefore also the mot or
current decr ease as t he speeds increase, and as the maximum m ot or current is required in the
constant flux range. T hus, the fact that f ield weakening is entered somewhat earlier (at 90 % rated
speed) is of no im por t ance.
Selecting the rect ifier/regener at ive feedback unit
The rect ifier/r egenerative feedback unit is dimensioned accor ding to the maximum DC link current
in dynamic operation and according to the highest RMS DC link curr ent . The DC link cur r ent s for
each inverter is calculat ed using the DC link power.
VZ
INVmot
mot
motDClink n
MP )( 1
60
2
ηη
π
⋅
⋅
⋅⋅
⋅=
with
)( mot
MsignVZ =
Thus, the DC link curr ent is given by:
ply
DClink
DClink V
P
I
sup
35.1 ⋅
=
-20
-15
-10
-5
0
5
10
15
0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000
Motor speed in RPM
DC link current in A
I DC link b+tens.
I DC link v+tens.
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The maximum DC link powers when accelerating and when decelerating, and t her efore also the
maximum DC link cur r ents, occur at t he m inim um speed. Thus, with Vsupply=400 V, the following
values are obtained for one motor:
AI bDClink 02.10
max =max. DC link current when accelerating
AI vDClink 01.16
max −= max. DC link current when decelerating
AI statDClink 09.4−= DC link current at constant velocity
b
t
v
t
k
t
T
t
motoring
regenerating
DC link b max
I
DC link v max
I
DC link stat
I
The following is valid for t he RMS value when motoring, for one m o t or:
A
T
tI
I
bbDClink
motRMSDClink 06.3
3
2max
=
⋅
=
The following is valid for t he RMS value when generating for one m otor:
A
T
tI
tI
I
vvDClink
kstatDClink
genRMSDClink 96.4
3
2max
2
=
⋅
+⋅
=
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The rect ifier/r egenerative feedback unit is selected accor ding to the following criteria for the
summed curr ent s of both motors:
nRDClinkbDClink II Re/max 36.104.2002.102 ⋅≤=⋅=
∑maximum value, motoring
nRDClinkvDClink II Re/max 92.036.102.3201.162 ⋅⋅≤=⋅=
∑maximum value, g enerating
nRDClinkmotrmsDClink II Re/
12.606.32 ≤=⋅=
∑RMS value, motoring
nRDClinkgenrmsDClink II Re/
92.092.996.42 ⋅≤=⋅=
∑RMS value, g ener ating
This result s in a 15 kW r ectifier/regenerative feedback unit with IDC link R/Re n=41 A. However, as the
output of the rectifier/r egenerative f eedback unit should be at least 30% of the power rating of the
connected inverter, a 37 kW r ect ifier/regenerative feedback unit is selected.
6SE7028-6EC85-1AA0
Pn=37 kW; IDC link n=86 A
The perm issible RMS value for a regenerative feedback transformer with 25% duty cycle is given
by:
AAIII nRDClinknRDClinkpermrmsDClink 56.3946.08646.0
100
25
92.0 Re/Re/. =⋅=⋅=⋅⋅=
As this value is greater than the regenerative RMS value, regenerative f eedback transformer with
a 25% duty cycle is suff icient. In addition, a 4% u k line reactor 4EU2451-4UA00 is required.
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3.3 Positioning drives
3.3.1 General information
Positioning procedur e
Positioning is possible in t wo ways:
•open-loop controlled
•closed-loop controlled
Open-loop cont r olled posit ioning
With open- loop controlled positioning , dur ing positioning, there is no feedback reg ar ding the actual
position. T hus, it is not guar ant eed t hat the actual end position coincides with the desir ed end
position. A reference point t ransmitter with a time-dependent velocity signal or a position sensor
(e.g . Bero, opt o- bar r ier ) is r e quired f or open- loop controlled positioning .
Effects on the accuracy
•changing deadtimes (reading-in, processing ) in the setpoint processing and when processing
the control com m ands
•ref er ence point transmitter and position sensor resolution
•play in the mechanical system (g ear box, j oints etc.)
•dependency on the load condition (e. g . motor slip)
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Example showing the influence of changing deadt imes when reading-in/processing control
commands
t
v
vmax
smax
Shutdown command (e.g. via BERO)
tmax
∆
~
~
The maximum r esponse t im e t o t he shutdown command is, in the most unfavorable case, ∆tmax.
Thus, t he m aximum positioning error is:
∆∆sv t
ma
x
ma
x
ma
x
=⋅
Positioning err ors can be reduced by approaching the end position at crawl speed (rapid tr averse,
crawl speed positioning).
t
v
vmax
smax
Shutdown command
tmax
Crawl spee d st arts
vmin
~
~
∆
∆
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The dwell time at vmin m ust be at least ∆tmax, ot herwise, positioning could be completed
prematurely.
The maximum posit ioning error is now only:
∆∆sv t
ma
x
mi
n
ma
x
=⋅
The disadvantage when using a cr awl speed is the long er posit ioning time.
Closed-loop contro lled posit ioning
For closed-loop contr olled posit ioning, the cur r ent position is compared to t he r eference value fr om
the reference value transmitt er. Deviations are equalized through the closed-loop position
controller.
Position
controller
Ref. point
generator
Motor
Speed
controller Actuator
vset
sact
sset
xd
Block diag ram f or closed-loop controlled posit ioning
xd: Position difference
The reference point generator and closed- loop posit ion cont roller can be externally set-up in the
PLC or can also be integ r at ed into a technology board in the dr ive converter . I nstead of ent er ing a
motion charact er ist ic via the reference point t r ansmitter, the complete position diff er ence can be
fed t o t he closed- loop position controller. I n this case, quant it ies vmax and amax during positioning
are kept constant elsewhere (ramp- function generator, lim it er, square-root function).
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The accuracy is inf luenced by the following factor s
•position encoder resolut ion
•play in the mechanical system
•deadtimes in the r eference value processing
•drive speed control rang e
For high r equirements r egarding fast and precise posit ioning without overshoot, the drive must
have the highest possible limiting f requency (> 40 Hz) and a high speed cont r ol range (> 1: 1000).
For applications, for example, feed drives, packing machines etc., ser vo drives are preferably used
(SIMOVERT MC). If the req uir em ents regarding dynamic perf or m ance ar e not as high, e. g. f or
high-bay rack ing vehicles, elevators etc., even standar d AC dr ives can be used (SI MOVERT VC) .
The somewhat rest r ict ed speed cont r ol range can be compensated for using an appr opriate
gearbox ratio and a longer positioning t im e.
Example for t he influence of t he encoder r esolution when using a double-pulse encoder
Motor
Gearbox
i
s
D/2
Encoder
Block diag ram
Distance s moved during one motor revolution is:
sDi
=⋅
π
[mm]
Ddrive wheel diameter [mm]
igearbox ratio
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Thus, sm allest detectable position difference xdi at an edge change is given by:
xD
iz
di =⋅
⋅⋅
π
4[mm]
zpulses per encoder revolution
The lowest detectable posit ion diff er ence xdi should be at least sm aller than the required accuracy
by a factor of 4.
If t he dr ive is still t o respond to a specific position difference xd , then it must still be possible to
adjust a minim um velocity of
v
x
k
dvmi
n
=⋅ [mm/s]
kvposition controller gain fact or [s-1]
The position cont r oller gain must not be t oo high for st ability reasons. It is dependent on t he dr ive
configuration (m om ent of inertia, gearbox play, deadtimes etc. ) and t he r equired characteristics.
Values which can be implemented are:
kvmax = 10 to 20 for st andar d AC dr ives (SI MOVERT VC)
kvmax = 50 to 100 for servo drives (SIMOVERT MC)
With the minimum velocity, the minim um m otor speed is given by:
niv
D
min min
=⋅⋅
⋅
60
π
[RPM]
The required speed control r ange can be determined via the minimum m otor speed.
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Fast positioning
In order t o achieve fast positioning , the drive must accelerate and brake at the maximum possible
values, and the constant velocity rang e must be traversed at the maximum velocity.
Example for fast posit ioning with tb=tv
tk
tbtv
ttotal
a
amax
amax
-
v
vmax
t
t
stotal
With
ttv
a
bv
==
max
max (acceleration=deceleration) [s]
ama
x
max. acceleration [m/s2]
vma
x
ma x. velocity [m/s]
results in the dist ance travelled stotal:
svtv
a
total total
=⋅−
max max
max
2[m]
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or f or positioning t im e ttotal f or the specif ied dist ance travelled:
tv
as
v
total total
=+
max
max max [s]
If the positioning t im e ttotal and the distance travelled stotal are specified, then vmax is given by:
vat at as
total total totalmax max max max
()=⋅−⋅−⋅
22
2[m/s]
The constant velocity time t k is obtained from :
tt v
a
k total
=−
⋅2max
max [s]
For the special case with tk=0, t he trapezoidal velocity characteristic changes int o a triangular
characteristic.
Example for fastest positioning with tb=tv, special case, tk=0
tbttotal
a
amax
amax
-
v
vmax
t
t
stotal
v
t
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With
ttv
a
bv
==
max
max (acceleration=deceleration) [s]
the f ollowing is obtained:
svt v
a
total total
=⋅=
max max
max
2
2[m]
ts
vv
a
total total
=⋅=⋅22
max
max
max [s]
vat as
total totalmax max max
=⋅=⋅
2[m/s]
Other acceler ation characterisics
In addition to fast positioning with constant acceleration, other acceleration characteristics are
possible, especially for ser vo drives, e. g.:
•Sawtooth acceleration
This characteristic results in positioning with minimum motor losses. However, for the same
positioning t im e, t he maximum acceleration increases with respect t o fastest positioning and
the end position is rapidly approached ( m echanical st r essing).
•Sinusoidal acceleration
In this case, acceler ation steps, which occur for fastest positioning or for sawtooth accelerat ion
characterist ics, ar e elim inated as well as the associated high mechanical st r esses. However,
disadvantages are the higher maximum acceleration for t he sam e positioning time with respect
to f ast est positioning and t he com plex implementation.
•Jerk- free accelerat ion char act er ist ic
The accelerat ion can be smoothly entered (jer k-f r ee) to keep the st r essing on the mechanical
system low and to prevent oscillations being excited, i. e., no steps and t ransitions in the
accelerating char acteristic. However, disadvantages are the higher maximum acceleration for
the same positioning time with respect to fastest posit ioning and the somewhat complex
implementation.
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3.3.2 Traction drive with open-loop controlled positioning using Beros
The maximum posit ioning error due t o t he changing deadtim e when reading - in/ pr ocessing control
commands at t he SIMOVERT VC drive converter are to be det ermined.
Drive data
Max. traversing velocity vmax = 36 m/min
Min. traversing velocity (crawl velocity) vmin = 2 m/min
Max. deadtime when processing t he control commands ∆tmax = 4xT0 = 4.8 ms
Velocity-time diagram
Bero 1
v
vmax
vmin
t
~
~
Bero 2
Bero 1: Chang eover from vmax to vmin via fixed reference values (t er m inal st r ip)
Bero 2: Off 1 com m and via the t er m inal st r ip ( t he br ake is energized at fU= 0)
The maximum posit ioning error as r esult of the chang ing deadtime is:
∆sma
x
=⋅vt
mi
n
ma
x
∆
=⋅=2 48 0016mmscm/min . .
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Without using a crawl velocity, a maximum positioning error, given by the following, would be
obtained:
∆sma
x
=⋅vt
ma
x
ma
x
∆
=⋅=
36 48 029mmscm/min . . !
Closed-loop f r equency control is used as result of the enhanced contr ol char act eristics. Using
closed-loop f r equency control and the possibility of im pressing the current at low frequencies, it is
possible to obtain defined characterist ics when ramping down from the crawl speed to 0. Closed-
loop speed control can also be used to pr event an excessive dependency on the load.
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3.3.3 Elevator drive with closed-loop positioning control (d irect approach)
A SIMOVERT VC drive converter with T300 technology board for posit ioning is used. The r e quired
accuracy when stopping at f loor s is ± 2.5 mm .
Drive data
Gearbox ratio i = 35
Drive wheel diameter D = 640 mm
Max. ve loci ty vmax = 1.25 m/s
Max. acceleration amax = 0.9 m/s2
Checking the accuracy
A double-pulse encoder with 1024 pulses per revolution is used. Thus, the smallest det ectable
position dif ference is g iven by:
xD
iz mm
di =⋅
⋅⋅=⋅
⋅⋅ =
π
π
4640
35 4 1024 0014.
This value is more than 400 % less than the required accuracy of ± 2. 5 mm. T hus, t he selected
encoder is adequat e as far as the r esolut ion is concer ned. The maximum motor speed is given by:
nvi
DRPM
max max ..
=⋅⋅
⋅=⋅⋅
⋅=
60 125 35 60
064 1305
π
π
For a speed control r ange of 1: 1000 and a r ated motor speed of nn=1500 RPM, the minimum
velocity is given by:
vnD
imm s
n
min ./=⋅⋅
⋅⋅ =⋅⋅
⋅⋅ =
π
π
1000 60 1500 640
1000 35 60 144
If t he closed- loop control should still respond t o a posit ion diff er ence of
xmm mm
d==
25
40625
..
the f ollowing kv fact or m ust be able to be set for the closed-loop posit ion controller:
kvxs
vd
== =−
min .
..
144
0625 23 1
This value can be realized for VC drive converters using T 300.
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Velocity characteristic when moving bet w een floors
With
sh m
total ==
12875.
this results in a posit ioning time for a f loor of:
tv
as
vs
total total
=+=+ =
max
max max
.....
125
09 2875
125 396
Accelerating- and decelerating t ime:
ttv
as
bv
== = =
max
max
...
125
09 139
The constant velocity time is given by:
tt v
as
ktotal
=−
⋅=−
⋅=
2369 2125
09 091
max
max
..
..
Velocity characteristic when moving between f loors
1.39 s
3.69 s
0.91 s 1.39 s t
t
a
-
v
1.25 m/s
0.9 m/s2
0.9 m/s2
stotal = 2.875 m
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3.4 Drives with periodic load changes and surge loads
3.4.1 General information
Periodic load changes and surge loads can occur, for example, for reciprocat ing compressor or
press drives.
For drives with periodic load changes, t he following must be observed:
•In order t hat the drive stability is guar ant eed, a drive converter with vector control should be
used.
•If t he excitat ion frequency of the load changes is higher than the lim it ing fr equency of the
closed-loop control ( nom inal values for closed-loop frequency control: Appr ox. 5 Hz, closed-
loop speed control: Appr ox. 15 Hz), t hen t he closed-loop frequency/speed controller m ust be
set as slow as possible, so that it does not attempt to compensate the speed fluctuations.
•The mot or t orque to be g enerated is reduced by the drive moment of inertia, and for the
extreme case, t o t he ar ithmetic average value of the load torque. This dam ping effect
decreases as the speed decreases.
•For drives with a low moment of iner t ia and high load torque changes (e. g . a single-cylinder
reciprocating compressor) , the motor cannot be sim ply dimensioned accor ding to the RMS
value of the load torque and the speed cont r ol r ange. In this case, the rated m ot or torque
should be at least 80% of the maximum load tor que.
•When st ar t ing loaded (open-loop contr olled r ange), under cer tain circumstances, the rated
motor t or que must be 130 t o 150 % of the maximum load t or que when starting. Vect or control
with tachometer is pr eferably used. In t his case, it is suff icient when the rat ed motor tor que
approximately corresponds to t he maximum load torque.
•If t he m otor and load are elastically coupled (e. g. through a belt drive), t hen t orsional
oscillations can be generated. The excitation frequency of the load changes m ay not coincide
with the resonant frequency.
Drives with surge/shock loading:
•If the drive moment of inert ia is appr opr iately high, the surge tor que can briefly be a multiple of
the rated m otor torque.
•If t her e are only low moments of inertia, t he motor can be braked down to standstill when load
surges occur . Under cer tain circumstances, the drive can even briefly accelerate in the
opposing direct ion of rotat ion ( e. g. a driven vehicle collides with a stationary vehicle). In order
to handle cases such as this, vector cont rol with tachometer must be used in order to pr event
the mot or st alling.
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Supplementary moment of inertia
By using a supplementary mom ent of inertia, it is possible to reduce the mot or peak torques.
When using a gearbox, it m akes the most sense t o locate this supplementar y moment of iner t ia at
the gear box side with the higher speed. This results in smaller dimensions.
Calculating t he motor t or que
Under the assumpt ion t hat the motor and load ar e r igidly coupled (no two-mass system with elastic
coupling) and when neglecting the electr ical m ot or time constants, t he following differential
equation is valid:
TdM
dt MMt
mech motor motor load
⋅+=
()
with
TJns
M
mech total n
n
=⋅⋅⋅
⋅
2600
π
mech. time constant [s]
Jtotal total moment of inertia [kgm2]
M
nrated motor torque [Nm]
snn
n
nn
=−
0
0rated motor slip
n0motor synchronous speed [RPM]
nnmotor r at ed speed [RPM]
If t he m oment of inertia is known, f or simple cases, the mot or torque can be calculated.
However, this calculation assumes that t he drive is not in the closed-loop speed controlled mode.
When using closed-loop speed control, depending on the setting, higher m ot or torques ar e
obtained. In the extreme case, if the speed control could keep the speed precisely constant , t he
motor t or que would precisely correspond to the load tor que.
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Case 1
The load tor que is specified in t he form of an arithm et ic average value component with
superimposed oscillation:
MtM M t
load load load load
() sin( )=+ ⋅ ⋅∆
ω
Thus, t he m otor torque is given by:
MtM M
Tt
motor load load
load mech
load
() ()
sin( )=+
+⋅ ⋅⋅−
∆
12
ωωϕ
with
ω
=⋅
arctan( )
load mech
T
The rat io of the amplit udes is t hen given by:
∆
∆M
MT
motor
load load mech
=+⋅
1
12
()
ω
Thus, load t or que oscillations are bett er damped the higher the mechanical time constant, and the
higher t he freq uency of the load torque oscillation.
t
Mload, Mmot
Mload
T= _
2
load
load
π
ω
∆
∆
ω
ϕ
Mload Mmot
Mload
Mmot
Example of load t orque oscillation with
ω
load mech
T⋅=3
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Case 2
The load tor que is a surge t or que in the f or m of a step function. T hus, the motor t or que is given
by:
MtM e
motor load
t
Tmech
() ( )
max
=⋅−
−
100
≤≤
t
t
and
MtM e e
motor load
t
T
tt
T
mech mech
() ( )
max
=⋅−⋅
−−
−
100
t
t
≥0
Mload, Mmot
Mload max
t0
Mload
Mmot
Mmot max
Example for a load sur ge with
T
t
mech =⋅30
If Tmech is significantly long er than t0, the motor only has to provide a fraction of the maximum
load torq ue. In this case, the m aximum m otor torque can be approximately calculated as f ollows:
MM
t
T
motor load mech
max max
≈⋅
0
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Case 3
The load tor que is a periodic sequence of torque steps. T hus, t he motor tor que is given by:
MtM Ce
motor load
t
Tmech
() max
=+⋅
−
100
≤≤
t
t
and
MtCe
motor
tt
Tmech
()=⋅
−−
2
0
t
t
T
0≤≤
With
CM e
e
load
t
T
T
T
p
mech
mech
11
1
=− ⋅ −
−
−
−
max
CM Ce
load
t
Tmech
21
0
=+⋅
−
max
t
Mload max
t0T
tp
Mload
Mmot max
Mmot m in
Mload, Mmot
Mload
Mmot
Example of a periodic sequence of load st eps with Tt
mech =⋅25 0
.
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If Tmech is significantly long er than to, the maximum and minimum m otor torques can be
approximately calculated as follows:
MMMt
T
motor load load mech
max max
≈+ ⋅
⋅0
2
MMM t
T
motor load load mech
min max
≈− ⋅
⋅0
2
with
MM t
T
load load
=⋅
max 0(arithm etic average value component of t he load torque)
Case 4
The load tor que is specified as diagram. The following calculation pr ocedur es can be used for the
motor torque:
•The load tor que characterist ic can be appr oximat ed using simple functions, which can be
solved using different ial equations.
•Fourier analysis of the load torq ue char act er ist ic, and therefore effective as f or case 1. In this
case, in addition to t he basic fundamental, only the har m onics have to be t aken into account,
which represent a significant propor t ion of the mot or t orque.
•Numerical integration, e. g., using the Runge Kutta t echnique
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Example
The load tor que characterist ic of a single- cylinder r eciprocating compressor can be approximately
represented as:
2load t
Mload (approximation)
Mload
Mload
Mload max
ω
π
π
2π corresponds to a cr ank revolution
The following is valid for t he appr oximat ion:
MtM t
load load load
( ) sin( )
max
=⋅⋅
ω
0≤≤
ω
π
load t
Mt
load ()=0
π
ω
π
≤≤
load t2
with
ω
π
π
load motor load
n
if=⋅⋅=⋅
260 2load torq ue oscillat ion frequency
i gearbox ratio
Thus, t he m otor torque is obtained from the following diff erential equat ion:
MtCe M
Tt
mot
t
Tload
load mech
load
load
load mech
() ()
sin( )
max
≈⋅ + +⋅ ⋅⋅−
−⋅
⋅
ω
ω
ωωϕ
120≤≤
ω
π
load t
MtCe
mot
t
T
load
load mech
()≈⋅
−⋅−
⋅
ωπ
ω
π
ω
π
≤≤
load t2
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with
CM
eT
load
Tload mech
load mech
=⋅
−⋅+⋅
−⋅
max sin
()()
ω
π
ω
11
2
ω
=⋅
arctan( )
load mech
T
The mot or t orque is now to be represented for a control range of 1:4.
ω
load mech
T
⋅=
15. for
nn
motor motor n
=. For nn
motor motor n
=/4, a value of 0. 375 is obt ained.
Mload max m
Mmot
Mload,Mmot
Mload (approximation)
Mload
2load t
ω
π
π
Motor torque at nn
motor motor n
= and
ω
load mech
T⋅=15.
Using the Mload characteristic,the arithmetic average value component of the load torque is
approximately given by:
MM
load load
≈max
π
The calculation indicat es t hat the maximum mot or torque is only approximately 65 % of t he
maximum load torque.
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2load t
ω
π
π
Mload max
Mmot
Mload,Mmot Mload (approximation)
Mload
Motor torque at nn
motor motor n
=4 and
ω
load mech
T
⋅=
0375.
As a result of the low influence of t he moment of inertia at low speeds, the maximum motor tor que
is almost as hig h as t he m aximum load torque.
This approximation does not take int o account that the load tor que is also brief ly negat ive. For low
moments of inertia, or low speeds, the m otor torque can also be negative (regenerative operation) .
Instead of directly solving a dif ferential equation, the m otor torque can also be calculated using
Fourier analysis. T he load t or que is approximated by:
Mt
load ()
=⋅⋅⋅+⋅−⋅ ⋅⋅
−
=
∞
∑
Mt t
load load load
max sin( ) ( cos( ))
1
2112 2
41
2
1
ωπνω
ν
ν
≈ ⋅ + ⋅ ⋅ − ⋅+ − ⋅ ⋅+
Mttt
load load load loadmax sin( ) sin( ) sin( )
11
22
3222
15 42
πωπωππωπ
(term inat ed after 4 t er ms of t he pr ogression)
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Thus, t he m otor torque is given by:
MtM Tt
motor load
load mech
load
() ()
sin( )
max
≈⋅+
⋅+ ⋅ ⋅−
11
21 21
πωωϕ
−⋅⋅ + ⋅ ⋅⋅−+
2
312 22
22
πω ωϕ
π
()
sin( )
load mech
load
Tt
−⋅⋅ + ⋅ ⋅⋅−+
2
15 1 4 42
24
πω ωϕ
π
()
sin( )
load mech
load
Tt
with
ω
1=⋅
arctan( )
load mech
T
ω
22
=⋅
arctan( )
load mech
T
ω
44
=⋅
arctan( )
load mech
T
The calculation m et hod using 4 terms of the progression practically provides the same results as
the previous direct calculat ion.
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3.4.2 Single-cylinder reciprocating compressor
A single-cylinder reciprocating compressor is to be operat ed in a speed cont r ol range f r om 94 to
281 RPM. A gearbox with i = 5.33 is used. The load torque characteristic is known, and the load
moment of inertia referred to the motor shaft.
Crankshaft angle / degr ees
Mload / Nm
*
-20
0
20
40
60
80
100
120
140
160
90 180 270 360
Load torq ue char act eristic of a single-cylinder reciprocat ing compressor r eferred t o t he motor shaft
Drive data
Max. load torque ref er r ed to the motor shaft Mload max
∗
Min. load torque referred t o t he m otor shaft Mload min
∗= -15 Nm
Load moment of inertia referred to the motor shaft Jload
∗= 0.0213 kgm2
Max. motor speed nmotor max = 1500 RPM
Min. motor speed nmotor min = 500 RPM
As there is no sig nificant mom ent of inertia, a m otor is required with a rated torq ue of at least 80%
of t he m aximum load t orque referred to the motor shaft.
MM Nm
motor n load
≥⋅ =⋅ =
∗
08 08 153 122 4...
max
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A 22 kW, 4-pole mot or is selected (1LA5 186-4AA.. )
Mn =144 Nm, nn =1460 RPM, Imotor n = 41 A, Jmotor = 0. 15 kgm2, ηmotor = 0.912
Estimating the influence of the moment of inertia on t he m axim um motor tor que
Total m om ent of inertia r eferred t o the motor shaft:
JJ J kgm
total motor load
=+=+ =
∗015 00213 01713 2
.. .
Rated mot or slip:
snn
n
nn
=−=−=
0
0
1500 1460
1500 00267.
Mechanical time constant:
TJns
Ms
mech total n
n
=⋅⋅⋅
⋅=⋅⋅ ⋅
⋅=
260 01713 2 1500 00267
144 60 0005
0
π
π
..
.
Excitation fr equency at nmotor max = 1500 RPM:
ω
π
π
load motor
n
is=⋅⋅=⋅⋅=−
2
60 2 1500
533 60 29 47 1
max ..
or
fHz
load load
== =
ω
π
π
22947
2469
..
Thus, t he following is obtained f or
ω
load mech
T⋅ at nmotor max = 1500 RPM:
ω
load mech
T⋅= ⋅ =29 47 0005 0147.. .
The basic fundamental amplit ude r at io is t hen given by:
∆
∆M
MT
motor
load load mech
∗=+⋅ =+=
1
1
1
1 0147 099
22
().
.
ω
The damping effect of the moment of inertia can in this case, even at maximum speed, be
neglected for the basic fundamental. This means, that the maximum mot or t orque, m ainly
determined by the basic fundamental, is pr actically the same as the maximum load tor que ref er r ed
to the motor shaft.
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Estimating the br aking energy
-15 Nm
180 210
Mload
*
27.5
Crankshaft angle/degrees
~
~
Section of t he load torque characteristic
The mot or t racks the posit ive section of the load tor que characterist ic pr actically 1:1, with the
exception of a phase shift. I n t he negative section, the peak of t he load torque is bet t er dam ped as
a result of the higher freq uency. I n or der t o estimate the maximum negative motor t orque, a
sinusoidal oscillation with 180/27.5 = 654 % frequency is assumed. The approximate mot or torque
at the maximum m otor speed is given by:
∆∆
MM
TNm
motor load
load mech
≈+⋅⋅ =+⋅ =
1654
15
1 654 0147 108
22
(. ) (. . ) .
ω
The maximum br aking power is given by:
PMn kW
br motor motor motor
max max ..
.≈⋅⋅
=⋅⋅ =
∆
9550 108 1500 0912
9550 155
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Braking diagra m
Pbr max
Pbr
180 210 Crankshaft angle/degrees
~
~
The braking energy of a cycle corresponds to t he ar ea in t he br aking diag ram. The braking
characterist ic is given by:
PP t
br br load
≈⋅ ⋅⋅
max sin( . )654
ω
Thus, t he br aking energy at nmotor max = 1500 RPM is given by:
WPkWs Ws
br br
load
≈⋅⋅=⋅⋅==
2
654 2155
654 2947 00161 161
max
..
....
ω
It should now be checked, as to whether the drive converter DC link can accept this energy. T he
maximum amount of energ y which can be accepted by the DC link for a 22 kW drive converter on
the 400 V line supply is as follows:
∆WCVV Ws
ddnmax max
( ) (820 ) .=⋅⋅ − =⋅ ⋅ ⋅ − =
−
1
21
22700 10 540 5141
22 622
Thus, t he br aking energy, which occurs during the br ief reg ener at ive phase, can be t emporarily
accepted by the DC link. I t is then reduced to zero in t he subsequent motor oper ation. Thus, a
pulsed resistor is not necessar y.
Drive converter select ion
6SE7024-7ED61
PV n = 22 kW, I V n = 47 A
Closed-loop f r equency control
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Thermally checking the m ot or
As the motor torque practically corresponds to the load tor que ref er r ed to the motor shaft, when
calculating t he RMS motor t or que, the RMS load torq ue can be used as basis. The load torq ue is
approximated using a sinusoidal half wave (as described previously under "General information") .
0 90 180 270 360 Crankshaft angle/degrees
Mload m ax
*
Approximation of the load torq ue char act er ist ic
The RMS motor torque is appr oximat ely g iven by:
MMNm
motor RMS load
≈==
∗max .
2153
2765
In a speed control r ange 1:3 f or the 22 kW motor, t he per missible torque for S1 duty is
approximately 111 Nm. Thus, there is sufficient reser ve.
Starting under load
If a single-cylinder reciprocat ing compressor is t o st art under load, for closed-loop frequency
control, a 30 kW m otor and a 30 kW drive converter are required (6SE7026-0ED61) .
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Motor tor que calculated using the Runge Kutta t echnique
Reciprocating compressor (n=1500 RPM , i=5.33,
Tm ech=0.005 s)
C rankshaft angle in deg rees
Torque in Nm
-20
0
20
40
60
80
100
120
140
160
0 30 60 90 120 150 180 210 240 270 300 330 360
Load torqu e
Mo to r torq ue
Av erage v alue
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3.4.3 Three-cylinder pump
A three-cylinder pump is to be operated in a contr ol r ange 102 to 238 RPM. A gearbox with i = 6.11
and η = 0.97 is used. The load t orque characteristic is specified and t he torque at t he pum p shaft is
796 Nm.
1,235
0,686
Crankshaft angle/degrees
Mload / Mload
0
0,2
0,4
0,6
0,8
1
1,2
1,4
0 30 60 90 120 150 180 210 240 270 300 330 360
Load torq ue char act eristic of t he three-cylinder pump
With the average torque at the pump shaft and a g ear box efficiency of 0.97, the average torque at
the motor shaft is obtained as follows:
MM
iNm
motor load
=⋅=⋅=
η
796
611 097 134 3
.. .
The load mom ent of inertia is not known, but can probably be neglected with respect to the motor
moment of inertia. Thus, it is assumed t hat the motor t or que has the same ripple as t he load
torq ue. The maximum and minimum m ot or torques are given by:
MM Nm
motor motormax ....=⋅ =⋅=1235 1235 134 3 1659
MM Nm
motor motormin ....=⋅ =⋅=0686 0686 134 3 921
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The maximum and m inimum motor speeds are given by:
n n i RPM
motor loadmax max ..=⋅=⋅=238 611 1454 2
n n i RPM
motor loadmin min ..=⋅=⋅=102 611 6232
As result of the relatively low load t o r que oscillations, t he RMS motor torq ue and the RMS load
torq ue can be used to dimension and select the m ot or. The load t or que characterist ic is
approximated using individual sinusoidal half waves to estimate the RMS value.
1
120
/
Mload Mload
in degre es
123 4
a4
a1a2a3
ϕϕϕ ϕ ϕ
The load charact er ist ic is appr oximat ed using sinusoidal half waves
The approximat e RMS value is obtained as f ollows:
MM
aa
load RMS load
iii
i
i
=⋅ ⋅+⋅+
=
=
∑
ϕπ
()142
120
2
1
4
With
a1 = 0.235, a2 = -0.061, a3 = 0.093, a4 = -0.3
ϕ1 = 44, ϕ2 = 11, ϕ3 = 24, ϕ4 = 41
1.0111 is obtained as the square root. Thus, t he RMS motor torque is approximat ed as follows:
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MM Nm
motor RMS motor
≈⋅=⋅=10111 134 3 10111 1358....
If t he ar ithmetic average value is high with respect to the oscillating t orque amplitudes, then the
RMS value practically corresponds to the average value.
A 4-pole, 30 k W motor is selected due to the speed contr ol r ange 1:2.33 (1LA5 207-4AA..)
Mn = 196 Nm, nn = 1465 RPM, Imotor n = 55 A, Jmotor = 0.24 kgm2
Estimating the influence of the moment of inertia on t he m axim um motor tor que
Total m om ent of inertia r eferred t o the motor shaft:
JJ kgm
total motor
≈=024 2
.
Rated mot or slip:
snn
n
nn
=−=−=
0
0
1500 1465
1500 00233.
Mechanical time constant:
TJns
Ms
mech total n
n
=⋅⋅⋅
⋅=⋅⋅ ⋅
⋅=
260 024 2 1500 00233
196 60 00045
0
π
π
..
.
Excitation fr equency at nmotor max = 1454.2 RPM:
ω
π
π
load motor
n
is=⋅ ⋅⋅=⋅ ⋅⋅=−
32
60 32 1454 2
611 60 748 1
max .
..(3 periods for a crank shaft revolution)
and
fHz
load load
===
ω
π
π
2748
2119
..
Thus,
ω
load mech
T⋅ at nmotor max = 1454.2 RPM is given by:
ω
load mech
T⋅=⋅ =748 00045 0336.. .
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The basic fundamental amplit ude r at io is now given by:
∆
∆M
MT
motor
load load mech
∗=+⋅ =+=
1
1
1
1 0336 095
22
().
.
ω
Thus, t he dam ping effect of the moment of inertia, even at m aximum speed can be neglected f or
the basic f undam ent al.
Selected drive convert er
6SE7027-2ED61
PV n = 37 kW, IV n = 72 A
Closed-loop f r equency control type
Motor tor que num erically calculated using the Runge Kutt a technique
T hree-cylinder pump (n=1454 RPM , i=6.11, Tm ech=0.0045 s)
C ran ksh aft an gle in d egrees
Torque referred to the average value
0
0,2
0,4
0,6
0,8
1
1,2
1,4
0 30 60 90 120 150 180 210 240 270 300 330 360
Load torqu e
Mo to r torq ue
Av erage v alue
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3.4.4 Transport system for freight wagons
This drive is f or a trolley that pushes f r eight wagons to form tr ain units on a 500 m rail section. The
vehicle is driven using a fixed 37 kW AC motor with g ear box.
IM
Wagon
Trolley Cable
Driving pulle y v
Approx. 500 m rail section
Gearbox
~~ ~~
Schematic of the trolley system!
For a 50 Hz motor frequency, t he t rolley runs at a constant velocity of 1.25 m/s towards the
wagons. The cable t ension is est ablished after a delay due to t he spr ing effect of the steel cable. If
the wagons are braked, or if a group of several wagons doesn’t accelerate quickly enough due to
the high m ass, then a slip coupling between the mot or and gearbox is activated when a maximum
motor t or que is reached. T he pulley wheel speed decreases. If the wagon g r oup doesn’t m ove or
the pulley wheel speed does not increase after a specific tim e, the trolley is retracted.
In order t o elim inate the slip coupling, t he m otor is now to be fed from a drive converter. Further, in
order to save time, the trolley can be withdrawn at twice the velocity. The slip coupling is now
replaced by a closed-loop speed/tor que control. I f the motor torque, transferred via the cable,
reaches the set t orque limit, the closed-loop speed contr ol can no longer maintain t he reference
speed, and the drive system automatically goes over into the closed- loop t or que controlled mode.
The mot or m aintains the torque, the speed decreases, and is now def ined by the load. However,
without the damping eff ect of the slip coupling, problems can occur as the motor - cable-wagon
combination repr esent s a system which can oscillate. This will now be shown using the f ollowing
simplif ied diagram. The friction and gearbox efficiency have, in this case, been neglect ed.
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IM ir
c
x1 x2
v1 v2
m
Equivalent diagr am for system behavior after t he t r olley has „hit “ the wagon
x1cable motion over the drive wheel
x2wagon mot ion
v1velocity of the cable over the drive wheel
v2velocity of the wagon(s)
r drive wheel radius
c spring constant of t he cable ( dependent on the cable lengt h)
m mass of the wagon(s)
i gearbox rat io
Jmotor motor moment of inertia
n0motor no- load speed at r ated frequency
The following is valid for t he dr ive wheel:
rFload
i M motor
ω
load
iM M i J ddt
motor load motor load
⋅−=⋅⋅
2
ω
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with
MFrcxxr
load load
=⋅=⋅−⋅
()
12
ω
load v
rr
dx
dt
==⋅
11
1
The following is valid for the wagon(s) :
v2
m
Ftension
Fm
dv
dt mdx
dt
tension =⋅ =⋅
222
2
with
Fcxx
tension =⋅ −
()
12
Behavior when a wagon is braked
To star t of with, the behavior is investigat ed when the trolley hits a braked wagon. In this case,
v2=0 and x2=0. It is now assumed, t hat the speed remains constant due to the closed-loop speed
control, unt il Mmotor=Mmax. Thus, the following is true f or this range:
ω
π
load const v
r
n
i
r
dx
dt
===
⋅
⋅=⋅
10 0 1
260 1
Solution:
x
v
t
110
=⋅
Mcrv
it
motor =⋅⋅ ⋅
10
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Mmotor=Mmax is reached at
ti
M
crv
110
=⋅⋅⋅
max
Aft er the tor que limit has been reached, Mmotor=Mmax remains constant. The following is valid for
this rang e:
dx
dt cr
iJ xiM r
iJ
motor motor
21
2
2
212
+⋅
⋅⋅=
⋅⋅
⋅max
Solution:
xvtt i
M
cr
mm110
111
=⋅ ⋅−+
⋅⋅
ωω
sin( ( )) max
d
x
d
t
vv tt
m
1110 1 1
== ⋅ ⋅−cos( ( ))
ω
with
ω
mmotor
cr
iJ
1
2
2
=⋅
⋅mechanical oscillation f requency
For operation at the torque limit, undamped oscillat ion occur s. However, in pract ice t her e is always
some damping due t o t he cable friction and gearbox losses, which in this particular analysis, have
been neglected. Addit ional dam ping can be achieved by compensating the speed to the torque
limit. This compensation has, in this case, the following form :
MMkM
n
nMkM
v
v
motor D D
=−⋅⋅=−⋅⋅
max max max max
0
1
10
For example, f or kD=0.1, a 10 % change in the torque limit is achieved at n=n0. With this
compensation, t he diff er ential equation now has a damping com ponent:
dx
dt ik M r
viJ dx
dt cr
iJ xiM r
iJ
D
motor motor motor
21
210 212
212
+⋅⋅ ⋅
⋅⋅ ⋅+⋅
⋅⋅=
⋅⋅
⋅
max max
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If , for example, t he st art of compensation is set to Mmotor=0.2 Mmax, then, for non-periodic
damping, t he solution is as follows:
xCe Ce i
M
c
r
ptt ptt
11 2
12 22
=⋅ +⋅ +
⋅⋅
⋅− ⋅−() () max
d
x
d
t
vCpe Cpe
ptt ptt
1111 22
12 22
==⋅⋅ +⋅⋅
⋅− ⋅−() ()
with
paa m12 11
21
2
22
,()=− ± −
ω
aik M r
viJ
D
motor
110 2
=⋅⋅ ⋅
⋅⋅
max
Cpvti
M
cr v
pp
1
2102 10
21
=⋅⋅−
⋅⋅−
−
()
max
Cvpvt
i
M
cr
pp
2
10 1 10 2
21
=−⋅ ⋅−
⋅⋅
−
()
max
t
M
i
crv
210
02
=⋅⋅
⋅⋅
,max
Wit h the following values
r = 0.36 m
c = 20000 N/m
i = 22.5
Jmotor = 1.2 kgm2
Mn= 483 Nm
n0= 750 RPM
when the trolley hits the braked wagons and for Mmax=Mn, t he following characterist ics ar e
obtained f or t he cable motion along the dr ive wheel x1, velocity of the cable along driving wheel v1,
the motor torque Mmotor and tension Ftension. I n order to k eep t he equation tr anspar ent , the
quantit ies ar e r eferred to the relevant nominal values:
xi
M
c
r
n
10 =⋅⋅and
F
c
x
n=⋅
10
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Cable velocity across the drive wheel (v1), Cable route along the drive
wheel (x1)
Time in s
-1
-0,5
0
0,5
1
1,5
01234567
x1/x10
v1/v10
kD=0
Motor torque (M Mot), tension (F tension)
Time in s
0
0,2
0,4
0,6
0,8
1
1,2
1,4
1,6
01234567
M Mot/Mn
F tension/Fn
kD=0
Behavior when the trolley hits the wagon(s) which is(are) br aked. Without compensation, the
system continuously oscillates (kD=0).
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Cable velocity across the drive wheel (v1), Cable route along the drive
wheel (x1)
Time in s
0
0,5
1
1,5
01234567
x1/x10
v1/v10
kD=1,4
M otor torque (M Mot), tension (F tension)
Time in s
-0,4
-0,2
0
0,2
0,4
0,6
0,8
1
1,2
1,4
1,6
01234567
F tension/Fn
M Mot/Mn
kD=1,4
Behavior when the vehicle impacts wagon(s) which is(are) braked. With the appropriat e
compensation (e. g. k D= 1. 4) a non-periodic damped oscillation is obtained. Compensation star t s
at Mmotor=0.2 Mmax.
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Behavior when accelerat ing t he wagon(s)
To star t of with, the accelerat ion of an unbrak ed group of wagons will be investigated. As v2 and
x2 now also change, t he analytical calculat ion is significant ly mor e com plex. The equations ar e
now changed-over as follows:
d
x
d
t
v
11
=
d
x
d
t
v
22
=
dv
d
t
c
mxx
212
=⋅−()
as well as
dv
d
t
10=without compensation and for vv
110
=
dv
dt ri
iJ MkM
v
vcr
iJ xx
motor Dmotor
121
10
2
212
=⋅
⋅⋅−⋅⋅−
⋅
⋅⋅−()()
max max with com pensat ion
This equation system can be solved using the Rung e Kut t a technique.
Using the sam e num er ical values as before for braked wagons, Mmax=Mn, and mass m of 400 t or
10 t, the following characterist ics ar e obtained for the velocity of the cable over drive wheel v1, the
velocity of the wagon(s) v2, motor torque Mmotor and tension Ftension. Motor limit compensation is
again withdrawn after appr ox. 1. 5 s using a PT1 element with a time constant of appr ox. 5 s;
otherwise, the mot or t orque during acceler ation would be permanently reduced by factor kD.
The overshoot of velocity v2 of the wagon(s) is certainly lower as f r iction has been neglected. This
effect can be significantly reduced, for example, by decreasing t he setpoint to approx. 80% after
the vehicle has hit the wagon. I f the mot or speed again reaches this 80%, t he set point is slowly
increased to 100% along a slow ramp ( r am p- function g ener at or). Thus, the cable tension (and also
overshoot) is reduced before the final velocity is reached.
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Cable velocity across the drive wheel (v1), velocity of the wagons (v2)
Time in s
0
0,2
0,4
0,6
0,8
1
1,2
1,4
0 5 10 15 20 25 30
v1/v10
v2/v10
m=400 t, kD=1,4
Motor torque (M Mot), tension (F tension)
Time in s
-0,4
-0,2
0
0,2
0,4
0,6
0,8
1
0 5 10 15 20 25 30
M Mot/Mn
F tension/Fn
m= 4 00 t, kD =1,4
Behavior when accelerating the wagons with m=400 t and a com pensat ion of k D=1.4.
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Cable velocity across the drive wheel (v1), velocity of the wagons (v2)
Tim e in s
0
0,2
0,4
0,6
0,8
1
1,2
00,511,522,53
v1/v10
v2/v10
m=10 t, kD=1,4
Motor torque (M Mot), tension (F tension)
Time in s
-0,4
-0,3
-0,2
-0,1
0
0,1
0,2
0,3
0,4
00,511,522,53
M Mot/Mn
F tension/Fn
m= 10 t, kD= 1 ,4
Behavior when accelerating a wagon with m=10 t and a compensat ion of k D=1.4.
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Implement ing the torque compensat ion f unction
n
n
*
Speed controller
n
kDMnn0
n
_
Mnm*
S
S
1
2
= 5 s
kDMn
n0
T
Switch S2 is closed before impact. Switch S1 closes af t er im pact at Mmotor=0.2 Mmax. S2 opens
aft er 1. 5 s and ensures that compensat ion is slowly withdrawn. This can be realized, for example,
using the T100 technology board.
Selecting the mot or and drive converter
The mot or alr eady exists.
Type: ABB with In=81 A; nn=735 RPM; Jmotor=1.2 kgm2; η=0.93; Mn=483 Nm; Mstall/Mn=2.2
Drive converter selected:
6SE7031-0EE60
PV n=45 kW; IV n=92 A; I V max=125.6 A
Closed-loop speed control
T100 technology board
Dimensioning the brake r esist or
Braking energy is gener at ed when:
a) the trolley brakes to 0 after it moves backwards at 2.5 m / s ( twice the velocity)
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b) the trolley brakes from 1.25 m /s to 0 while moving forwards
c) during impact
Case a)
When br aking from twice the velocity in the speed rang e bet ween 1500 RPM and 750 RPM, the
motor is oper at ed in the field-weakening range. Per m issible m ot or torque at nmax and 130 %
safet y margin fr om the stall torque:
MMMn
nNm
perm stall stall n
.max
..
() ..() .== ⋅ =
⋅⋅=
13 13 2 2 483
13 750
1500 204 4
022
For Mmotor=Mperm=constant while braking, then the maximum braking power is given by:
PMn kW
br perm motormax .max ...=⋅⋅=⋅⋅=
9550 204 4 1500
9550 093 299
η
The load mom ent of inertia is r equired to deter m ine t he br aking tim e. The linearly-moved masses
(trolley and cable) are assumed to be 1000 kg. Then, t he following is true:
Jmr kgm
load =⋅= ⋅ =
222
1000 036 129 6..
or, referred t o t he motor
JJikgm
load load
*.
..== =
22 2
1296
225 0256
Thus, when neglect ing the f r ict ion and gearbox losses, the braking time is given by:
tJJ n
Ms
br motor load
perm
=+⋅⋅
⋅=+⋅⋅
⋅=
()(. . ) ..
*max
.
2
60 12 0256 2 1500
60 204 4 112
ππ
Case b)
Braking fr om 750 RPM to 0 in the const ant - flux range is t o be r ealized in, for example, 0.5 s. Thus,
the motor torque is g iven by:
MJJ n
tNm
motor motor load
br
=+⋅⋅
⋅=+⋅⋅
⋅=
()(..)
..
*2
60 12 0256 2 750
60 05 2287
0
ππ
Thus, t he m aximum br aking power is given by:
PMn kW
br motor motormax ...=⋅⋅=⋅⋅=
0
9550 2287 750
9550 093 167
η
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Case c)
As can be seen fr om t he diagrams, t he m ost unfavorable case is when the wagon(s) is(are)
braked without t or que compensation. The theoretical maximum braking power for Mmotor=Mn and
n=-n0 is obtained as f ollows:
PMn kW
br nmotormax ..=⋅⋅=
⋅⋅=
0
9550 483 750
9550 093 353
η
However, this value does not actually occur in practice as the cable friction and g ear box losses
were neglected. The behavior with a damping compensation of kD=1.4 is more realistic. In this
case, the maximum br aking power af t er impact with Mmotor=-0.4 Mn and n=n0 is given by:
PMn kW
br nmotormax ..
..=⋅⋅=
⋅⋅⋅=
04
9550 04 483 750
9550 093 141
0
η
Thus, a m aximum br aking power f or case a) of 29.9 kW is obtained. A braking unit with P20=20
kW (6SE7023-2EA87-2DA0) and external brake resist or ( 6SE7023- 2ES87-2DC0) would be
sufficient. T his allows a peak br aking power of 1. 5 P20=30 k W to be achieved f or 3 s.
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3.4.5 Drive for an eccentric press
A press drive with a 160 kW DC motor and nn = 2400 RPM is to be replaced by a three-phase
drive. The DC motor speed control r ange is 436 RPM (setting-up speed) up t o 2400 RPM
(maximum speed).
D
D
S
M
r
l
z=19
Flywheel
α
z=31
z=64
z=170
Simplif ied function schematic
Drive data
Max. pressing force 25 mm before lower end stop Fload max = 10000 kN
Max. def or m ing work W max = 250 kWs
Diameter of the mo t or pulley wheel DM= 300 mm
Diameter of the drive wheel DS= 1670 mm
Moment of inert ia of the flywheel JS= 2845 kg m 2
Crank radius r = 470 mm
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Effective crank angle during im pact αSt = 18.8 degrees
Number of strok es at 436 RPM (crawl speed nmin)z
min = 4/min
Number of strok es at 873 RPM (base speed nG)z
G= 8/min
Number of strok es at 2400 RPM (max. speed nmax)z
max = 22/min
Required acceler at ing time f r om 0 to nmax th< 60 s
Required br aking ( deceler ation) time from nmax to 0 tbr < 60 s
Using this dr ive data, it has to be checked whether a 4-pole 160 kW induct ion motor is suitable for
the press drive.
Motor data
1LA6 316-4AA; nn=1485 RPM; Mn=1030 Nm; Jmotor=3.2 kgm2; ηmotor=0.958;
Mstall=2.7 Mn; Im o tor n =275 A
The average m ot or power at maximum speed and an assumed press efficiency of 0. 8:
PW z kWs RPM kW
motor average press press
=⋅=⋅=⋅
⋅=
max max ..
ηη
250 22 250 22
08 60 114 6
As far as t he average power is concerned, this mot or is adequate. However, the behavior during
impact as well as the RMS motor loading should now be investigated.
Calculating the r at ios
The rat io bet ween the mot or and gearbox is obtained as f ollows:
iD
D
SS
M
== =
1670
300 557.
The tot al r atio between the motor and the crank wheel is obtained from the number of teeth as
follows:
itotal =⋅⋅=
1670
300 64
19 170
31 10283.
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Calculating the load t or que
In the following calculation, the pr ess eff iciency, additional frictional forces and the m om ent of
inertia of the gear are neglected. Further, it is assumed that the load force is constant during
impact.
α
r
l
FT
Fload max
load
h
β
ω
F : Tangential force
h: Ram disp l ace m ent during pressing
T
MFrF r
load T load
=⋅= ⋅⋅ +
max sin( )
cos
α
β
β
h
r
St
=⋅−( cos )1
α
For l/r >> 1, or for a small angle α, t he equation f or Mload is simplified as follows:
MFrF r
load T load
=⋅≈ ⋅⋅
max sin
α
For α=αSt=18.8 deg r ees, the maximum load torque refer red to the mot or is obtained as follows:
kNm
kN
i
rF
Mtotal
St load
load 7314
83102 180
81847010000 .
.
).sin(.
sin
max
*max =
π
⋅⋅⋅
=
α⋅⋅
=
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The mot or cannot produce this torque. It m ust be provided by the flywheel.
Calculating t he m echanical time constant:
TJns
M
mech nn
n
=⋅⋅⋅
⋅
*260
π
With
JJ
iJkgm
S
Smotor
*...=+ = +=
22 2
2845
557 32 949 ( t ot al moment of inertia referred to t he motor)
snn
n
nsyn n
syn
=−=−=
1500 1485
1500 001.(motor slip)
the f ollowing is obtained:
Ts
mech =⋅⋅ ⋅
⋅=
94 9 2 1485 001
1030 60 0143
..
.
π
The worst case condition, with impact at t he lowest operating speed, is now investigated i. e. at the
base speed. The t im e for the im pact at nG =873 RPM, but neglecting the speed dip, is obtained as
follows:
tins
St load
total
G
≈=
⋅⋅
⋅=⋅⋅ ⋅
⋅=
α
ωαπ
π
π
60
2
188 180 102 83 60
2 873 0 369
..
.
As the f ollowing is not valid for the mechanical t im e const ant : Tmech >> tSt, the motor quickly
reaches the t or que limit. The motor speed is then mainly defined by the load. In order to sim plify
the calculation, it is assumed that t he torque is limited at the star t of impact. Then, the following is
true:
iM F r Jddt
total motor load load load
⋅−⋅⋅=⋅
max max sin
α
ω
ω
α
load d
dt
=−
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With
JJi
load total
=⋅
*2
Mmotor max = torque lim it
This differential equation system can be solved, for example, using the Runge Kutt a t echnique.
For a small ang le α, with sin
α
α
≈ , a dif ferent ial equation can be analytically solved as follows:
d
dt
Fr
J
iM
J
load
load
total motor
load
2
2
αα
−⋅⋅=− ⋅
max max
Solution:
α
=⋅ +⋅ + ⋅⋅
⋅⋅
Ce Ce iM
Fr
pt pt total motor
load
12
12 max
max
ω
α
load d
dt
=− =−Cpe Cpe
pt pt
11 22
12
⋅⋅ −⋅⋅
⋅⋅
ni
motor total load
=⋅ ⋅
ω
π
60
2
with
p
12,=±Fr
J
load
load
max ⋅
CiM
Fr
n
pi
St total motor
load
motor
total
11
1
1
2
2
60
=⋅ − ⋅⋅−⋅
⋅⋅
()
max
max
α
π
CiM
Fr
n
pi
St total motor
load
motor
total
21
1
1
2
2
60
=⋅ − ⋅⋅+⋅
⋅⋅
()
max
max
α
π
With the specified data, and the specif icat ion MM
motor nmax .=⋅15 , the characteristic of
ω
load and
nmotor can be calculat ed dur ing the impact . For nm o tor 1 =nG (motor speed before impact) , by
iterat ing for α=0, the following values are obtained:
n RPM
motor 2613=(motor speed after t he im pact )
ts
St =047.(impact time)
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Towards the end of the impact, if the mot or torque exceeds the load t orque, t he flywheel is again
accelerated with MM
motor motor
=max . T he following is valid:
iM Jddt
total motor load load
⋅=⋅
max
ω
Thus, t he m otor speed is given by:
nn iM
Jtt
motor motor total motor
load St
=+
⋅⋅
⋅⋅−
2
260
2max ()
π
or f or t he acceler ating time from nm o tor 2 to nmotor 1:
t
b=⋅⋅⋅−
J
Mnn
motor motor motor
*
max
()
260 12
π
=⋅
⋅⋅
⋅−=
949 2
15 1030 60 613 167
.
.(873 ) .
π
s
The time from t he start of impact until t he or iginal speed has been reached again, is given by:
tt s
St b
+= + =
047 167 214.. .
The maximum available tim e at nG, with 8 strok es/ min is obtained as follows:
tzss
G
max min .== = =
1860
875
The load tor que and the motor speed at the lowest operating speed nm o tor 1=nG=873 RPM is
illustrat ed in t he following diagrams.
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t / s
t / s
M / kN m
load
n / RPM
motor
tt
St b
nmot or 1
mot or 2
n
0
200
400
600
800
1000
0 0.5 1 1.5 2 2.5
0
400
800
1200
1600
0 0.5 1 1.5 2 2.5
Load torq ue and m ot or speed at nmotor 1=nG=873 RPM
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Simplified calculat ion
Using the conser vation of energy equation, the magnitude of t he speed dip as well as the impact
time can be approximately calculated. The energy supplied from the flywheel is:
∆WJ n n WW
motor motor motor
=⋅⋅ ⋅ − = −
1
22
60 21
22
2* max
()( )
π
Wmotor is the energy supplied from the m ot or during impact:
WMidMi
motor motor total motor total St
St
=⋅⋅= ⋅⋅
∫
0
ααα
max
Thus, t he m otor speed after the impact is given by:
nn
WM i
J
motor motor motor total St
21
2
2
1
22
60
=−
−⋅⋅
⋅⋅
max max
*()
α
π
The approximate im pact time is given by:
ti
St total St
motor
≈⋅
α
ω
with the average motor speed:
ωπ
motor motor motor
nn
=⋅ +
2
60 2
12
Using these formulas, t he speed dip and the impact time for nmotor 1=873 RPM are ag ain checked.
nRPM
motor 22
3
2
873 250 10 15 1030 10283 188 180
1
2949 2
60
618=−
⋅−⋅ ⋅ ⋅⋅
⋅⋅ =
...
.( )
π
π
ts
St ≈⋅⋅
⋅+=
10283 188 180
2
60 873 618
2
043
.. .
π
π
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Behavior in the field- w eakening r ange
The drive is to be operated with constant power in the field- weakening range. Fur t her , there
should be a 130 % safet y margin to the st all lim it . Thus, t he per m issible m ot or torque is
represented by the following characteristics.
nnnmax
nmotor
Mmotor perm
P = const
Mstall
1,3
´
If a t or que of 1. 5 Mn is assumed in the const ant -flux rang e, t hen the permissible torque at constant
power for nmax is obtained as follows:
Mn
nMNm
perm n nn.( ) max
max ..=⋅⋅= ⋅⋅=15 1485
2400 15 1030 966
On the other hand, with Mstall=2.7 Mn, for nmax, the permissible torque f or a 130 % safety margin
is obtained as f ollows:
MMn
nMNm
perm n stall n n
.( ) max
max .()
..()
..
== ⋅
⋅=⋅
⋅=
′
13 27
13 1485
2400 2 7 1030
13 819
22
Thus, the limit for Mperm.(nmax) is defined by the stall torque.
With Mperm.(nmax), the speed dip and the im pact time can be calculated for nmotor=nmax=2400
RPM.
n RPM
motor 22
3
2
2400 250 10 819 10283 188 180
1
2949 2
60
2309=−
⋅−⋅ ⋅⋅
⋅⋅ =
..
.( )
π
π
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ts
St ≈⋅⋅
⋅+=
10283 188 180
2
60 2400 2309
2
0137
.. .
π
π
The following is valid for the accelerat ing time f rom nmotor 2 to nmotor 1:
tJ
Mnn s
bmotor perm n motor motor
=⋅⋅⋅− =
⋅
⋅⋅−=
*
.( )
max
()
.().
260 94 9 2
819 60 2400 2309 11
12
ππ
The condition
t
t
St b
+ < tz
max max
=1
124.s < 60
22 273
ss=.
is also fulfilled.
Calculating the accelerating- and decelerating tim es
If the drive is accelerated f r om 0 to nmax, with the t or que permissible at nmotor = nmax in the field
weakening range, or is braked fr om nmax to 0, the accelerating- and decelerating times are g iven
by:
tt Jn
Ms
hbr motor perm n
== ⋅⋅
⋅=⋅⋅
⋅=
*max
.( )
max
..
2
60 94 9 2 2400
60 819 2912
ππ
This time is less than the specified 60 s.
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Thermally checking the m ot or
Torque characteristic
t +tbSt
T
Mmotor m ax
Mmotor
t
The RMS torque is obt ained from t he torque characteristic:
MM tt
T
RMS motor St b
=⋅
+
max
With
Tz
=1
In this case, factor kf is 1, as it involves continuous operation. The most unfavorable case is
obtained at the lowest operating speed nG, as in this case, t he speed dip, and thus also the rat io
tSt+tb/T is its highest (at nmax, t he RMS torque is always less than Mn due to Mmotor max < Mn).
For nmotor=nG, the following value is obtained for MRMS:
MM
tt
z
Nm
RMS n St b
G
=⋅ ⋅ +=⋅ ⋅ =15 115 1030 214
75 825..
..
Thus, t he calculat ed RMS torque is less than the perm issible m ot or torque in t he speed cont r ol
range 1: 2. Thus, operation is thermally permissible.
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Selecting the drive conver ter
At 150 % motor torque, the maximum motor cur r ent is given by:
IIII
motor motor n n nmax .( )≈⋅ −+1522 2 2
µµ
With II
n motor n
µ
=⋅027., the following is obt ained:
IA
motor max .( . ).≈⋅−⋅+⋅=15 275 027 275 027 275 404
22 22 22
Thus, t he RMS motor cur r ent can be calculated for the most unfavorable case at nmotor=nG.
t + t
St b T
Imotor
t
Imotor max
Imag n
From the diagram above, the following is obtained f or Im otor RMS:
Imotor RMS =⋅++⋅−+IttITtt
T
motor St b n St bmax () (())
22
µ
=⋅+ ⋅⋅− =
404 214 027 275 75 214
75 225
222
.. (..)
.A
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Selected drive converter:
6SE7033-2EG60
PV n=160 kW; IV n=315 A; IV max=430 A
Closed-loop f r equency control mode
The rat ed dr ive converter cur rent is greater than the RMS motor current. The maximum motor
current t her efore does not exceed the m aximum dr ive converter cur r ent .
Dimensioning the brake r esist or
If t he dr ive is also braked with the permissible torque at nmax in the field-weakening range, t he
maximum brak ing power for t he br ake resistor is obtained as follows:
PMn kW
br W motor perm n motormax .( ) max
max ..=⋅⋅=
⋅⋅=
9550 819 2400
9550 0958 197 2
η
The braking energy when braking fr om nmax to 0 at tbr=th is then as follows:
WPt kWs
br br W br
=⋅ ⋅ =⋅ ⋅ =
1
21
21972 2912 2871
max ..
The following must then be valid for the brake r esist or :
W
TkW P
br br cont
== ≤
2871
90 319..
With
PP
br cont..
=20
45 (with an external brak e r esist or )
the f ollowing is obt ained
319 45 1436 20
.. .⋅= ≤kW P
Thus, a br aking unit is select ed with P20 = 170 kW (6SE7032-7EB87- 2DA0) with an external
brake r esist or (6SE7032-7ES87-2DC0).
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3.5 Load distribution for mechanically-coupled drives
3.5.1 General information
For mechanically-coupled dr ives, for example traction drives with several driven axes, or roller
table drives which are coupling through the material, it is important to uniformly distribute t he load
over the individual drive motors. Depending on the particular application case, this can be
implemented by using a group (sect ional) drive with one drive converter, or using individual drives,
each with its own dedicated drive converter.
IM IM
Ring gear
Pinion Pinion
Example of a m echanically-coupled drive
Multi-motor dr ive configuration
The simplest way of achieving a uniform load dist r ibut ion for mechanically-coupled dr ives with
motor s having the same output is by using a multi-mo t or drive configuration. In this case, all of the
motors are connected to a drive converter, and are therefore supplied with the same stator
fr equency. As a result of the mechanical coupling, the motor s have the sam e speed with
symmetrical relationships (identical gear box rat io, ident ical roll diameter et c.). Deviations in the
load distribution can t her efore only come if the individual motors have dif ferent slip char acteristics.
In accordance with DIN VDE 0530, deviations of up to 20% are perm issible. A specially-selected
motor can also help. Problems can also occur for tract ion unit s with slig ht ly differ ent drive wheel
diameters. The thus obtained speed differences increasingly influence t he load dist ribution the
lower the rated motor slip is.
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MM
M
21
total
n
M
nn1
2
Example for t he load distribution of 2 m echanically-coupled m otors with different slip characteristics
(the mot or with the flatter char act eristic accepts the hig her load component)
MM
M
21
total
n
M
nn1
2n
∆
Example for t he load distribution of 2 m echanically-coupled m otors with different speeds, e. g. as a
result of different wheel diameter s ( t he m otor with the lowest speed accepts the higher load
component)
Group drives cannot be used with dif ferent m otor outputs t o pr ovide a uniform load distr ibut ion if
technological r equirements oppose t his ( e. g. r edundancy), or if the speed diff er ences are too
high.
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Single-m otor drive conf iguration
For single- m otor drives with motors having the same output, when the same set point is entered at
all of t he dr ive converter s, characteristics can be achieved which are similar t o those for a group
drive. The setpoints should be digitally entered so t hat the stator freq uencies, are, as far as
possible, identical (e. g. via a peer-to-peer connection of the individual drive converters). Further ,
the same ram p-up and ramp-down times must be selected for all of the drive converters . When
using the V/ f characteristic or closed-loop frequency control, slip compensation must be disabled.
Otherwise, load distr ibut ion cannot be realized through t he motor slip. I t is also not possible to use
closed-loop speed control for the sam e reason (i.e. without "dr oop" ).
Further, it is possible to inf luence t he load dist ribution by activating the so-called " dr oop"
characteristic.
Droop character ist ic
SIMOVERT VC drive converters m ust be used when implement ing the droop function. T he dr oop
funct ion can be used for closed-loop frequency- or for closed-loop speed control. For closed- loop
fr equency control, it is only ef fective in the contr olled r ange (f r om approx. 5 Hz).
K
Speed/frequency controller
m*
n /f
**
n /f
act act
p
Block diag ram of " droop" f or SIMOVERT VC
The droop function causes the motor slip to increase when a load is applied to the dr ive due to t he
negative f eedback of the torq ue set point to the speed / freq uency setpoint . Thus, under st eady-
state conditions, uniform load distribution can be achieved by setting the sam e slip char acteristic.
"Load equalization control" and "mast er /slave drive" techniques provide a f lexible way of ensuring
unified load distribution, even at various speeds.
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Load equalizat ion cont rol
SIMOVERT VC drive converter s m ust be used when implem ent ing load equalization control.
Closed-loop f r equency control must be set . All of the drive converters r eceive the sam e freq uency
setpoint, and the same ramp-up and ramp-down times must be set . A uniform load distribution is
achieved by comparing the tor que setpoint of the f ir st drive with the torque set point s of all other
drives, and then a frequency correct ion set point for these drives is generated by the PI cont roller,
provided as standard in SIMOVERT VC dr ive converters . This correction set point must be limit ed
to approx. 300 to 400 % of the mot or r ated slip frequency. The load eq ualization cont r ol m ay only
be enabled, if t he closed-loop fr equency control is f ully operat ional (fr om appr ox. 5 Hz). Below this
range, t he open- loop controlled mode is used.
Drive 1
frequency controlled
Drive 2
frequency controlled
with higher -level
load equalization
control
f*
RFG F requenc y c ontroller
Load equalization
controller
m*
RFG
f*
∆
fact
m*
fact
Frequency controller
Block diag ram of a load equalization control function
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The frequency setpoint, torque set point and t he control commands from t he first converter are
transferred to all of the other dr ive converter s, for example, t hr ough a peer-to-peer coupling. I n t his
case, the drive converters r equire an SCB2 board.
IM
SIMOVER T VC
SC B 2
SIMOVER T VC
Drive 1
frequency
controlled
SCB 2
f*
RS485
Drive 2
frequency
controlled
IM
Example of a load eq ualization cont r ol function with peer-to-peer connection (RS485)
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Master/slave drive
SIMOVERT VC drive converter s ar e required to implement master/s lave drives. Each dr ive
requir es a speed t achometer (pr eferably a pulse encoder due to t he enhanced accur acy). For
master/ s lave drives, the load is uniformly distribut ed over t he complete speed range by
transferring the torque setpoint of the speed-controlled m aster drive to the torque-controlled slave
drive.
Speed contr oller
nist
n*
m*
m*
Mas ter drive
spe ed contr olled
Slav e drive 1
torque controlled
Block diag ram of a m aster/slave drive
In order t hat the slave drives don’t accelerate up to excessive speeds when the load is removed
(e. g . this could happen if the material web breaks) , the closed-loop tor que control of the slave
drive can be implemented using a speed control, overdriven with ∆n* with an appropriate torque
limit. I n this case, each slave drive requires, in addit ion t o t he torque set point of the mast er drive,
the speed setpoint of the master drive.
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n*m*
Speed controller
nact
m*
∆n*
From the master drive
Block diag r am of a slave drive with over-driven closed-loop speed control
The tor que setpoint, speed setpoint and the contr ol com m ands can be transferred to the slave
drives, for example, via a peer-to-peer connection.
IM
SIMOVERT VC
Master drive
speed controll ed
SCB 1/2
n*
IM
SIMOVERT VC
Slave drive 1
torque-controlled
SCB 1/2
Fiber optic /
RS485
Example of a mast er/slave drive with peer-to-peer connect ion (SCB1: Fiber-optic cable, SCB2:
RS485)
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3.5.2 Group drive for a traction unit
The four wheels of a traction unit are each driven by a motor. The two motors on one side of the
traction unit ar e fed f r om one convert er . As result of the different wheel loads, t he wheels are
subject to diff er ent wear factors. The load distr ibution of the motors on one side of t he traction unit
at maximum differ ent wheel diameter s and m aximum velocity are t o be det er mined.
Dmax Dmin
IM
SIMOVERT
v
Motor 1 Motor 2
ii
IM
Block diag r am of a traction unit side
Drive data
Rated motor output (8-pole) Pn= 40 kW
Motor rated speed nn= 740 RPM
Max. ve loci ty vmax = 49.73 m /min
Wheel diam et er , new
D
max = 1.2 m
Wheel diam et er , after wear
D
min = 1.19 m
Gearbox ratio i= 55.345
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With v = vmax = 49.73/60 = 0.829 m/s, the m ot or speeds ar e given by:
nvi
DRPM
160 0829 60 55345
12 730=⋅⋅
⋅=⋅⋅
⋅=
max
max
..
.
ππ
nvi
DRPM
260 0829 60 55345
119 736=⋅⋅
⋅=⋅⋅
⋅=
max
min
..
.
ππ
The following assumption is made to calculat e t he m otor torques:
nnn M
M
slip n n
10 1
=− ⋅
nnn M
M
slip n n
20 2
=− ⋅
MMM
load12
+=
With
n n n RPM
slip n n n
=−=− =
0750 740 10
Thus, t he m otor torques are given by:
MMnnM
n
load n
slip n
121
22
=+
−⋅
⋅
()
MMnnM
n
load n
slip n
221
22
=−
−⋅
⋅
()
Wit h the differential torque:
∆MnnM
nMM
n
slip n
nn
=−⋅
⋅=−⋅
⋅=⋅
() ( ) .
21
2736 730
210 03
The mot or with the lar ger wheel diameter is loaded with 30% higher r ated motor t or que and the
motor with the lower wheel diameter, with 30% less rated motor tor que. When low velocities are
approached, the diff er ence bet ween the t or ques becomes less. T he diff er ent loading must be
taken int o account when dimensioning the mot or s and defining t he closed- loop control type. A
vector control can no longer be optimally set for such dif ferent t orque loads. I f vector contr ol is
requir ed, e. g. as a angular synchronous cont r ol is being used for both traction unit sides, then a
configuration with 4 individual drives could be considered. The two single-motor drives on a
traction unit side ar e designed as master/ s lave drives, and the m aster drives coupled throug h the
angular synchronous contr ol function.
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730
736
743
n / RP M
M = 2 M
load n
M
nn
12
M =1.3 Mn
1M = 0.7 Mn
2
Example with Mload = 2 Mn
For the above example with Mload = 2 Mn, n0 is obtained as:
nnn MMM
MRPM
slip n load
nn
01 2730 10 1 03 743=+ ⋅⋅+=+⋅+=()(.)
∆
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3.5.3 Drive for an extraction tower with 10 mechanically-coupled motors
An extraction tower is to be driven thr ough a ring g ear , pinion and gearbox with 10 motors, each
37 kW/ 6-pole, nn = 980 RPM. For redundancy reasons, each motor is provided with its own drive
converter.
In order t o r educe gear stressing, the m ot or s m ay only generat e t orque in one direction (motoring ) .
Braking is not req uir ed. As result of the symmetrical configurat ion, regenerative operation of t he
individual motors can practically only occur under no-load condit ions without material, as even the
smallest differ ences her e influence the stat or freq uency. The rated slip frequency of t he m otors is
given by:
fHzHz
slip n =−⋅=
1000 980
1000 50 1
Thus, for a 0.01 Hz resolution st ator fr equency refer ence, t he torque error lies below 1%. This can
be achieved with a digit al set point input. A peer-t o- peer connection cannot be considered due to
the req u ir ed r edundancy. Eit her a SINEC L2 DP bus connection via the CB1 board or an USS bus
connection via the SCB2 board can be used. A V/f character ist ic without slip com pensat ion should
be used. The r am p- up and r amp-down times must be the sam e for all drive converters .
IM
SIM O VER T VC
Dr iv e 2
CB1
From the mas ter
to the other drives
Rack
Pinion
SIM O VER T VC
Dr iv e 1
CB1
SINEC L2-DP
Pinion
IM
Implementing a bus connection via SINEC L2- DP
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3.6 Crank drive
Mass m is to be raised by 2 r, within 0.25 s, to t he upper deadpoint (ϕ = π) using a crank drive
configuration, st ar ting f r om t he lower deadpoint ( ϕ = 0). The return strok e is r ealized after a 0.25 s
delay time. The st r oke cycle time is 5 s.
r
l
m
ϕ
αω
,
Drive data
Mass to be moved m = 300 kg
Crank radius r = 0.06 m
Connecting rod length l = 0.3 m
Positioning t im e ttotal = 0.25 s
Cycle time T = 5 s
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Angular velocity
ω
and angular acceleration
α
Positioning should be as fast as possible, i. e. a t riangular characteristic with
tb = tv = 0.125 s is used.
t
max
0.125 s 0 .2 5 s 0.5 s 0. 625 s 0. 75 s 5 s
t
max
max
-
Upwards Interval Downwards
ω
ω
α
α
α
~
~
~
~
The maximum angular velocity during the upwards mot ion occurs at
t = ttotal/2 = 0.125 s or at an angle of ϕ = π/2.
t
ttotal
max _
2
t
2
ω
ω
ϕ
π
=
total
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Using the following equat ion
ϕωπ
=⋅=
max ttotal
2
22
the maximum ang ular velocity is given by:
ω
π
π
max ..=⋅=⋅=−
22
025 2513 1
ts
total
With
ωα
max max
=⋅
ttotal
2
the maximum ang ular acceler ation is given by:
α
ω
max max .
..=== −
ts
total
2
2513
0125 20106 2
The maximum speed at
ω
ω
=ma
x
is given by:
n RPM
max max .
=⋅
⋅=⋅
⋅=
ω
π
π
60
22513 60
2240
Due to the short accelerating- and deceler at ing times, pr ecise positioning is only possible using a
servo drive.
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Converting the cir cular motion into a linear m ovement
r
l
r cos r sin
x
x, vxax
,
ϕϕ
ϕ
αω
,
From the diagram, x is obtained as follows:
x=−⋅ −⋅lr r
22 2
sin cos
ϕϕ
≈−⋅⋅−⋅lrlr
22
2sin cos
ϕϕ
(approximation)
Thus, t he following is obtained for velocity vx and acceleration ax:
vd
x
dd
dt r
r
l
x=⋅≈⋅⋅ −
⋅⋅
ϕ
ωϕ ϕ
(sin sin )
22
adv
dd
dt rrl
xx
=⋅≈⋅⋅ +⋅ −⋅⋅ +⋅
ϕϕωϕαϕ ω ϕ
αϕ
( cos sin ) ( cos sin )
222222
To evaluate these eq uations for vx and ax, angular velocity ω, angular acceleration α and angle ϕ
should be determined as a function of t im e t . To realize this, m otion is sub-divided into four
ranges.
x
l
r
mi
n
=− (lower deadpoint)
x
l
r
ma
x
=+ (upper deadpoint)
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Range 1 (. )0 0125 0 2
≤≤ ≤ ≤t s and
ϕ
π
upwards motion, accelerat ing
α
α
1=ma
x
ω
α
1=⋅
ma
x
t
ϕα
12
1
2
=⋅ ⋅
max t
Range 2 (. . )0125 025 2
s t s and≤≤ ≤ ≤
π
ϕπ
upwards motion, decelerat ing
α
α
2=− ma
x
ω
α
2025
=⋅ −
max (. )st
ϕπ α
22
1
2025=−⋅ ⋅ −
max (. )st
Range 3 (. . )05 0625 3 2
s s and≤≤ ≤≤
ϕπϕ
π
downwards motion, acceler at ing
α
α
3=ma
x
ω
α
305
=⋅−
max (.)ts
ϕπ α
32
1
205=+⋅ ⋅−
max (.)ts
Range 4 (. . )0625 075 3
22s s and≤≤ ≤≤
ϕπϕπ
downwards motion, deceler at ing
α
α
4=− ma
x
ω
α
4075=⋅ −
max (. )st
ϕπα
42
21
2075=−⋅ ⋅ −
max (. )st
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Calculating the load t or que
δ = π - ϕ
δ + β + γ = π
r
r
l⋅=⋅=⋅sin sin sin
δ
β
With linear acceleration ax, the following force acts on mass m:
Fmgma
x
=⋅+⋅ linear force
Using the angular relationships, which can be derived f r om the diagram above, the connecting r od
for ce, the tang ent ial force and t he load force are obtained as follows:
FF
S=cos
β
connecting rod force
FF F
TS S
=⋅ +=⋅ −+sin( ) sin( )
δ
β
π
β
tangential for ce
MFrmga r
load T x
=⋅=⋅+⋅ −+ ⋅()
sin( )
cos
π
β
β
load torq ue
With
βϕ
=⋅arcsin( sin )
r
l
ω, α
ϕ
r
l
ax
F
F
T
S
γ
δ+β
β
F
δ
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As result of the irr egular ang ular velocity and the complex angular relat ionships, it is relatively
complex to calculate the load t orque. In order to evaluate the equations, it is r ecom mended that a
progr am such as Excel is used.
If l is m uch greater than r (approx. from r / l < 0.1), t he following approximation can be used as a
funct ion of ϕ to calculate the load torque:
Mmrg r r
load () max max
( sin sin sin )
ϕ
ϕα ϕ ϕα ϕ
≈⋅⋅⋅+⋅⋅⋅ +⋅⋅22for 02
≤≤
ϕ
π
Mmrg r r
load () max max
( sin ( ) sin sin )
ϕ
ϕα πϕ ϕα ϕ
≈⋅⋅⋅ + ⋅⋅ − ⋅ − ⋅⋅22for
π
ϕπ
2≤≤
The range
π
π
≤≤2 is a mir r or image of r ange 0≤≤
π
.
The maximum load t or que can be determined using this approximation. The maximum lies within
the rang e π/4 < ϕ < π/2. I f the approximat ion is used for this par ticular case, the maximum for
Mload is obtained at ϕ = 1.185 to 530 Nm. Ang le ϕ, at which the maximum occurs, can also be
determined by tr ial and er ror or by iteration.
Without this approximation, for the particular case, the following load torque characteristic is
obtained
-600
-400
-200
0
200
400
600
0.125 0.25 0.375 0.5 0.625 0.75
Upwards motion Downwards motionInterval
M/
Nm
Load
t / s
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The maximum load t or que when accelerating upwards is 572 Nm and when decelerating
downwards, -572 Nm.
Motor selection
Selected motor:
1FK6 103-8AF71 with gearbox i=10 (η =0.95), nn=3000 RPM, Jmotor=0. 01215 kgm2
Mn(100)=17 Nm, M0(100)=36 Nm, I 0(100)=23.8 A, ηmotor=0.94, JGearbox=0.00174 kg m 2
Maximum mot or speed:
n i n RPM
motor max max
=⋅ = ⋅ =10 240 2400
Calculating the m ot o r torques
Accelerating- and decelerating torques for the motor and the gearbox:
MMJi Nm
b motor gearbox v motor gearbox motor gearbox+++
==⋅⋅=+⋅⋅=
α
max (. . ) . .001215 000174 10 20106 2793
Motor torque when accelerating:
MM
M
i
motor t b motor load t
() ()
=+
⋅
η
for 0 0125≤≤ts. and 05 0625..st s≤≤ 1)
1) If t he load torque is neg ative in these ranges, t hen at the appropriat e locat ions, factor 1/η must
be changed to η.
Motor torque when decelerating:
MM
M
i
motor t v motor load t
() ()
=− + ⋅
η
for 0125 025..st s≤≤ and 0625 075..st s≤≤ 2)
2) If t he load torque is positive in these r anges, at t he appr opr iat e locations, fact or η must be
changed into 1/ η.
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The highest motor t or que is requir ed when accelerating upwards at nmotor=2220 RPM:
MNm
motor bmax ...=+
⋅=2793 572
10 095 881
The highest motor t or que in regener at ive operat ion is r equired when decelerating dur ing
downwards motion:
MNm
motor vmax ...=− +−⋅=−2793 572
10 095 823
Frictional f orces and for ces due t o other moving masses have been neglect ed. The selected mot or
is adequately dimensioned, as at nmotor=2220 RPM and 400 V supply voltage, it can be
overloaded up to approx. 102 Nm.
Motor torque characteristic for this particular case
-100
-80
-60
-40
-20
0
20
40
60
80
100
0.125 0.25 0.375 0.5 0.625 0.75
Upwards motion Downwards motionInterval
M/
Nm
motor
t / s
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Calculating the maxim um motor output
The mot or out put is obtained f r om t he motor tor que and the angular velocity:
PMi
motor t motor t t() () ()
=⋅⋅
ω
From the calculat ion, the maximum mot or out put is 21.2 kW when accelerating during upwards
motion. The maximum motor out put in regenerative operat ion is - 19.8 kW. It occur s when
decelerating dur ing downwards motion.
Motor output charact er ist ic for this par t icular case
-25
-20
-15
-10
-5
0
5
10
15
20
25
0.125 0.25 0.375 0.5 0.625 0.75
Upwards motion Downwards motion
Interval
motor / kWP
t / s
Regenerative operation
Thermally checking the m ot or
The RMS torque for 1FT 6 m ot or s is generally obtained as follows f or any torque charact er ist ics:
MMdt
T
RMS =∫2
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The integral expression can be approximately solved as follows by sub-dividing it into m seg ments:
Mdt MM
tt
ii
ii
i
im
21
22
1
12
≈+⋅−
−−
=
=
∑
∫()
To evaluate this approximation, Mmotor must be available in tabular form . The following is obtained
(refer to the table later on in the text):
Mdt Mdt Nms
ts
ts
ts
ts
2
0
025 2
05
075 22
1009
=
=
=
=
∫∫
+≈
.
.
.
Thus, t he following is obtained:
MRMS ≈1009
5=142.Nm
The average m ot or speed is given by:
n
nt
TRPM
average
mot
b
=⋅⋅ =⋅⋅ =
max .
242400 2 0125
5120
The calculated RMS torque lies, at naverage, below the MS1 charact er ist ic. Thus, operat ion is
permissible.
Dimensioning the brake r esist or
Maximum braking power for the brake r esist or :
PP kW
br W br motor motormax max .. .=⋅=⋅=
η
198 094 186
Using the m ot or output, the br aking energy for a cycle can be calculated as follows:
WP dtP dt
br motor
ts
ts
motor motor motor
ts
ts
=⋅+⋅
=
=
=
=
∫∫
0 125
025
0 625
075
.
.
.
.
ηη
The integral expansion can be solved as follows by sub-dividing into m segment s:
Pdt PP
tt
motor motor motor
i
im motor i motor i ii
∫∑
⋅≈⋅+⋅−
=
=−−
ηη
1
11
2()
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The calculation r esult s in ( refer to the table later in t he text):
Wbr ≈14.kWs
The following must be true for the br ake resistor:
W
TkW P
br br cont
== ≤
14
5028
...
With
PP
br cont.=20
36 (with an internal brake resistor)
the f ollowing is obt ained
36 028 1008 20
⋅= ≤
..kW P
Further, the following must also be true:
PkWP
br W max ..=≤⋅186 15 20
Thus, a br aking unit is selected with P 20 = 20 kW (6SE7023-3EA87-2DA0). The internal brake
resistor is adequate.
Selecting the drive conver ter
The highest required motor tor que is given by:
M
M
motor max
(100)
..
0
881
36 245==
The following value is obtained from t he torque/ cur r ent characteristic:
I
I
motor max
(100) .
027=
and
II A
motor max (100) ....=⋅=⋅=
027 238 27 643
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Drive converter selected:
6SE7024-7ED51
PV n=22 kW; IV n=47 A; I V max=75.2 A
Table for the approximate com putation of t he int egral expression
tM
load Mmotor Pmotor ≈⋅
∫
Mdt
2≈⋅⋅
∫
Pdt
η
0 0.00 27.93 0.00 0.00 0.00
0.025 10.52 29.03 1.46 20.29 0.00
0.05 61.48 34.40 3.46 45.62 0.00
0.072 180.63 46.94 6.80 82.87 0.00
0.094 394.58 69.46 13.13 160.19 0.00
0.116 572.09 88.15 20.56 298.73 0.00
0.1205 564.89 87.39 21.17 333.39 0.00
0.125 530.16 83.73 21.04 366.35 0.00
0.125 95.87 -17.84 -4.48 366.35 0.00
0.1375 -115.74 -38.92 -8.80 377.81 -0.08
0.15 -208.92 -47.77 -9.61 401.54 -0.19
0.15625 -210.25 -47.90 -9.03 415.85 -0.24
0.18125 -79.76 -35.50 -4.91 460.29 -0.40
0.20625 6.72 -27.22 -2.39 485.31 -0.49
0.25 0.00 -27.93 0.00 518.57 -0.54
0.5 0.00 27.93 0.00 518.57 -0.54
0.54375 -6.72 27.29 2.40 551.93 -0.54
0.56875 79.76 36.32 5.02 577.73 -0.54
0.59375 210.25 50.06 9.44 625.54 -0.54
0.6 208.92 49.92 10.04 641.16 -0.54
0.6125 115.74 40.11 9.073 666.79 -0.54
0.625 -95.87 18.82 4.73 679.06 -0.54
0.625 -530.16 -78.29 -19.68 679.06 -0.54
0.6295 -564.89 -81.59 -19.77 707.83 -0.62
0.634 -572.09 -82.28 -19.19 738.04 -0.71
0.656 -394.58 -65.41 -12.36 859.57 -1.03
0.678 -180.63 -45.09 -6.53 929.00 -1.23
0.7 -61.48 -33.77 -3.39 963.90 -1.33
0.725 -10.52 -28.93 -1.45 988.62 -1.39
0.75 0.00 -27.93 0.00 1008.83 -1.40
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3.7 Rotary table drive
This involves a so-called applicator drive with a maximum of 36000 applicat ions per hour. An
application means a rot ation through 18 degrees within 50 ms and a subsequent int er val of 50 ms.
Rotary table (application rol l)
Gearbox
Motor
Drive data
Angle for an application (i. e . attaching a label) ∆
= 18 degrees
Rotary table m om ent of iner t ia Jload = 0. 05 kgm2
Positioning t im e ttotal = 50 ms
Cycle time T = 100 m s
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Positioning must be as fast as possible, i. e., a t riangular characteristic with
tb=tv=ttotal/2=25 ms is used.
t
load
load max
load
load max
load max
t
T
tbttotal tv
ω
ω
α
α
α
-
ϕ
∆
Wit h the following ang le
∆
ϕ
ω
π
=⋅=
load total
t
max
210
the maximum ang ular velocity is given by:
ωϕ
π
load total
ts
max ..=⋅=⋅=−
2210
005 1257 1
∆
With
ωα
load load total
t
max max
=⋅
2
the f ollowing is obtained for the m aximum angular acceleration:
α
ω
load load
total
ts
max max .
..=⋅=⋅=−
221257
005 502 7 2
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The maximum load speed at
ω
ω
=load max is obtained as follows:
nRPM
load load
max max
=⋅=
ω
π
60
2120
As a result of the short acceler ating- and decelerat ing times, pr ecise posit ioning is only possible
using a servo drive.
Calculating the load tor ques when accelerating and decelerat ing
As acceleration is the sam e as deceler at ion, the load torques are given by:
MMJ Nm
b load v load load load
==⋅ =⋅=
α
max ...005 502 7 2513
Selecting the mot or
The following motor is selected ( r efer to Calculating the motor torques):
1FT6 062-6AC7 with gear box i=5 (η=0.95), nn=2000 RPM, Jmotor=0.00085 kgm2
Jgearbox=0.0003 kgm2, Mn(100)=5.2 Nm, In(100)=2.6 A, ηmotor=0.84
The gearbox ratio was selected close to the optimum :
iJ
JJ
opt load
motor gearbox
=+=+=
005
000085 00003 66
.
..
.
Thus, t he m aximum m otor speed is given by:
n i n RPM
motor loadmax max
=⋅ =⋅ =5 120 600
Calculating the mot or torques when acceler ating and decelerating
Accelerating- and decelerating t or ques for t he m otor:
MMJi Nm
b motor v motor motor load
==⋅⋅=⋅⋅=
α
max ...000085 5 5027 214
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Accelerating- and decelerating t or que for t he gearbox:
MMJi Nm
b gearbox v gearbox gearbox load
==⋅⋅=⋅⋅=
α
max ...00003 5 5027 0754
Motor torque when accelerating:
Mmotor =+ +⋅
⋅
MM M
i
b motor b gearbox b load 1
η
=+ + ⋅
⋅=214 0754 2513 1
5095 818.. . ..Nm
Motor torque when decelerating:
Mmotor =− − − ⋅MM M
i
v motor v gearbox v load
η
=− − − ⋅ =−214 0754 2513 095
5767.. .
..Nm (reg ener at ive operat ion)
The highest motor t or que is requir ed when accelerating. The select ed m ot or is adequately
dimensioned, as it can be overloaded up t o appr ox. 24 Nm at nmax=600 RPM and a 400 V supply
voltage.
Selecting the drive conver ter
A motor cur r ent, for the highest r equired motor t orque, is obt ained as follows:
II
M
MA
motor motor n motor
motor n
max max ....=⋅ =⋅=26 818
52 41 (f or t his m otor there is har dly any saturation
effect)
Selected drive converter:
6SE7013-0EP50 (Compact Plus)
PV n=1.1 kW; IV n=3 A, IV max=4.8 A (160% overload capabilit y)
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Dimensioning the brake r esist or
Max. brak ing power for t he br ake resistor dur ing deceleration from ωload max to 0:
Pbr W max =⋅=
⋅⋅PMn
br motor motor motor v motor motormax max
ηη
9550
=⋅⋅=
767 600
9550 084 0405
...kW
Braking diagra m
Pbr W
0.405 kW
t
T
tv
tb
Braking energy f or a cycle (cor r esponds to the area in the br aking diag r am ):
W
br =⋅ ⋅
1
2Pt
br W vmax
=⋅ ⋅ =
1
20405 0025 0005.. .kWs
The following must be true for the br ake resistor :
W
TkW P
br br cont
== ≤
0005
01 005
....
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With
PP
br cont..
=20
45 (f or an exter nal br ake resistor)
the f ollowing is obt ained
45 005 0225 20
.. .
⋅= ≤
kW P
Thus, t he smallest brak ing resistor is select ed with P20 = 5 kW (6SE7018-0ES87- 2DC0) .
Thermally checking the m otor
Torque characteristic
8.18 N m
-7 .67 Nm
0.1 s
t
0.025 s
0.025 s
Mmotor
The RMS torque and t he average motor speed is given by:
MRMS =⋅+⋅=
818 0025 767 0025
01 561
22
.. ..
..Nm
n
nt
TRPM
average
mot
b
=⋅⋅ =⋅=
max ,
.
22600 0 025
01 150
The calculated RMS torque lies, at naverage, below the MS1 charact er ist ic. Thus, operat ion is
therma lly perm issible.
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3.8 Pivot drive
This involves a pivot drive with an 180 degree rotational movement forwards and backwards
throug h a horizontal axis. A servo drive is to be used. After t he dr ive has rotated, it is held in t he
initial position by a mechanical br ake.
r = 650 r = 650
21
m1
m2
Fz
m1g
m2g
max= 180 degrees
ϕ
∆
Initial posit ion of the pivot dr ive
Drive data
Angle of rotation ∆ϕmax = 180 degrees
Weight m1= 170 kg
Opposing weight m2= 130 k g
Supplementary f orce Fz= 500 N
Pivot time when moving throug h 180 degrees t total = 3 s
Delay time befor e t he start (dr ive on) tw= 3 s
Interval between forwards and return mot ion (dr ive on) tp= 1 s
Cycle time T = 45 s
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Positioning should be as fast as possible, i. e. , a triangular accelerating charact er ist ic with
tb = tv = t total/2 = 1. 5 s is used. Thus, t he following motion diagram is obtained.
T
Drive on Brak e on
twttotal tp
t
max load
-
t
-
t
tt
bv
load Delay time Forwards Interval Backwards
ω
ω
ω
αα
α
~
~
~
~
max load
load
max load
max load
total
The maximum angular velocity and the maximum angular acceler at ion ar e given by:
ω
π
load total
ts
max max .=⋅=⋅=−
22
32094 1
α
ω
load load
total
ts
max max ..=⋅=⋅=−
22094 2
31396 2
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The maximum load speed at
ω
ω
load load
=max is given by:
n RPM
load loadt
max max .
=⋅
⋅=⋅
⋅=
ω
π
π
60
22094 60
220
As the geom et r ical dim ensions of the weights ar e not known, it is assumed that their centers of
gravity act at single points. T hus, the load moment of inertia is calculat ed as follows:
Jmrmr kgm
load =⋅+⋅= ⋅ + ⋅ =
11
222
222 2
170 065 130 065 12675...
Calculating the load t or ques
F
mg
z
1
mg
2
1
2
r
r
ϕ
Mload =⋅+⋅⋅ −⋅⋅⋅( ) cos cosmg
F
r
mg
r
z11 22
=⋅+⋅−⋅⋅⋅= ⋅(( . ) . . . )
cos
.cos170 981 500 065 130 981 065 58006
Nm
The accelerat ing - and decelerating torque of the load is given by:
MMJ Nm
b load v load load load
==⋅ =⋅=
α
max ..12675 1396 177
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Selecting the mot or
The following motor is selected ( r efer to Calculating the motor torques):
1FT6 062-1AF71 with gear box i=150 (η =0.95); nn=3000 RPM; Jmotor+brake=0.00105 k gm2
Mn(100)=4.6 Nm; In(100)=3.4 A; ηmotor=0.88
Maximum mot or speed:
n i n RPM
motor loadmax max
=⋅ = ⋅ =150 20 3000
Calculating the m ot o r torques
Accelerating- and decelerating torque for the motor:
MMJi Nm
b motor v motor motor load
==⋅⋅=⋅⋅=
α
max ...000105 150 1396 022
Holding torque for the motor during delay tim e at ϕ = 0 and during the no- load int erval at ϕ =180
degrees:
MMiNm
H motor load
== =
58006
150 387
..
Motor torque when accelerating forwards:
Mmotor t()=+ +⋅
⋅
MMM
i
b motor load t b load
()
() 1
η
for 345st s≤≤. and 02
≤≤
ϕ
π
=+ ⋅ +⋅
⋅
022 06 177 1
150 095
. (580. cos ) .
ϕ
=+⋅146 407..cosNm Nm
With
ϕα
=⋅ ⋅−
1
22
load w
tt
max ()
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Motor torque when decelerating for wards:
Mmotor t()=− + − ⋅MMM
i
v motor load t v load
()
()
for 45 6.st s≤≤ and
π
ϕπ
2≤≤
=− + ⋅ − ⋅022 06 177 095
150
. (580. cos ) .
ϕ
=− + ⋅134 367..cosNm Nm
(reg ener at ive operat ion)
With
ϕπ α
=−⋅ ⋅ + −
1
22
load w total
tt t
max ()
Motor torque when accelerating backwards:
Mmotor t()=− + − ⋅ ⋅
MMM
i
b motor load t b load
()
() 1
η
for 785st s≤≤. and
πϕ
π
≥≥
2
=− + ⋅ − ⋅ ⋅
022 06 177 1
150 095
. (580. cos ) .
ϕ
=− + ⋅146 407..cosNm Nm
With
ϕπ α
=−⋅ ⋅− −−
1
22
load total p w
tt t t
max ()
Motor torque when decelerating backwards:
Mmotor t()=+ +⋅MMM
i
v motor load t v load
()
()
for 85 10.st s≤≤ and
π
ϕ
20≥≥
=+ ⋅ +⋅022 06 177 095
150
. (580. cos ) .
ϕ
=+⋅134 367..cosNm Nm
(reg ener at ive operat ion)
With
2
max )2(
2
1tttt wptotalload −++⋅⋅⋅=
αϕ
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The highest motor t or que is requir ed when accelerating forwards at ϕ = 0 and when accelerating
backwards at ϕ = π:
MNm
motor bmax ..cos.=+⋅ =146 407 0 553
The highest motor t or que in regener at ive operat ion is r equired when decelerating forwards at ϕ =
π and when decelerating back wards at ϕ = 0:
MNm
motor vmax ..cos .=− + ⋅ =−134 367 501
π
The gearbox moment of iner tia was neglected.
Motor torque characteristic for this particular case
-6
-4
-2
0
2
4
6
2 4 6 8 10 12 14 45
~
~
t / s
Delay time Forwa rds In terval Backwards
M/ Nm
motor
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Selecting the drive conver ter
For the hig hest r equired motor torque, the motor cur r ent is given by:
II
M
MA
motor motor n motor
motor n
max max ....=⋅ =⋅=34 553
46 409 (for this m otor there is har dly any saturation
effect)
Drive converter selected:
6SE7013-0EP50 (Compact Plus)
PV n=1.1 kW; IV n=3 A; IV max=4.8 A ( 160% overload capabilit y)
Dimensioning the brake r esistor
The mot or output must be det ermined when dimensioning the brake resistor.
PM i
motor motor load
=⋅⋅
ω
With
ω
load =0for 03≤≤
t
s and
=0
ω
α
load load w
tt=⋅−
max () for 345st s≤≤. and 02
≤≤
ϕ
π
ω
α
load load w total
tt t=⋅+−
max () for 45 6.st s≤≤ and
π
ϕπ
2≤≤
ω
load =0for 67s
t
s≤≤ and
ϕ
π
=
ω
α
load load total p w
tt t t=− ⋅ − − −
max () for 785st s≤≤. and
πϕ
π
≥≥
2
ω
α
load load total p w
tttt=− ⋅ ⋅ + + −
max ()2for 85 10.st s≤≤ and
π
ϕ
20≥≥
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Motor output charact er ist ic for this par t icular case
-1000
-800
-600
-400
-200
0
200
400
600
800
1000
2 4 6 8 10 12 14 45
~
~
t / s
Delay time F orwards Interval B ackwards
Regenerative operation
P / W
motor
Regenerat ive operat ion occur s when decelerating in the rang es 45 6.st s≤≤ and 85 10.st s≤≤ .
From the calculat ion, the maximum mot or out put in regenerative operation is -875 W. Thus, t he
maximum brak ing power for t he br ake resistor is given by:
PP W
br W br motor motormax max .=⋅=−⋅=−
875 088 770
The braking energy for a cycle can be calculated as f ollows from the m ot or output:
WP dtP dt
br motor
ts
ts
motor motor motor
ts
ts
=⋅+⋅
=
=
=
=
∫∫
45
6
85
10
..
ηη
The integrals can be approximately calculated as follows by sub-dividing into m segments:
Pdt PP
tt
motor motor motor
i
im motor i motor i ii
∫∑
⋅≈⋅+⋅−
=
=−−
ηη
1
11
2()
To evaluate this approximation f or mula, Pmotor must be available in tabular for m .
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The calculation pr ovides the following results (r efer to t he table later in the t ext) :
Wbr ≈− −75767 75767..
≈−151534. Ws
The following must be true for the br ake resistor:
W
TWP
br br cont
==≤
151534
45 3367
...
With
PP
br cont..
=20
45 (with an external brak e r esist or )
the f ollowing is obt ained
45 3367 1515 20
.. .
⋅= ≤
WP
Further, the following must be t r ue:
PWP
br W max .=≤⋅770 15 20
Thus, t he sm allest braking resistor is select ed with P20 = 5 kW (6SE7018-0ES87- 2DC0) .
Thermally checking the m ot or
For the RMS torque for 1FT 6 m otors, g ener ally the following torq ue char act eristics are valid:
MMdt
T
RMS =∫2
The integral expression can be approximately calculated by sub-dividing into m segments:
Mdt MM
tt
ii
ii
i
im
21
22
1
12
≈+⋅−
−−
=
=
∑
∫()
To evaluate this approximation f or mula, Mmotor must be available in tabular form. The calculation
provides the following results (refer to the t able lat er in the text):
Mdt Nms
ts
ts
2
0
10 22
18354
=
=∫≈.
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Thus, t he following is obtained:
MRMS ≈18354
45
.≈202.Nm
The calculated RMS torq ue is less than the torque permissible in S1 duty. T hus, oper ation is
permissible.
Table for approximately calculating t he int egral expressions
tM
motor Pmotor ≈⋅
∫
Mdt
2≈⋅⋅
∫
Pdt
η
0.00 3.87 0.00 0.00 0.00
3.00 3.87 0.00 44.86 0.00
3.00 5.53 0.00 44.86 0.00
3.15 5.53 173.79 49.45 0.00
3.30 5.52 347.11 54.03 0.00
3.45 5.49 517.59 58.55 0.00
3.60 5.40 679.16 62.94 0.00
3.75 5.22 820.36 67.03 0.00
3.90 4.90 923.40 70.63 0.00
4.05 4.39 964.32 73.51 0.00
4.20 3.64 915.58 75.50 0.00
4.35 2.66 751.75 76.56 0.00
4.50 1.46 459.25 76.88 0.00
4.50 -1.34 -421.21 76.88 0.00
4.65 -2.42 -684.52 77.76 -72.98
4.80 -3.31 -831.70 79.41 -173.05
4.95 -3.98 -875.02 81.78 -285.69
5.10 -4.44 -837.41 84.74 -398.71
5.25 -4.73 -743.75 88.10 -503.07
5.40 -4.90 -615.63 91.70 -592.79
5.55 -4.98 -469.15 95.42 -664.38
5.70 -5.01 -314.61 99.18 -716.11
5.85 -5.01 -157.52 102.95 -747.27
6.00 -5.01 0.00 106.72 -757.67
Continued on the next pag e
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tM
motor Pmotor ≈⋅
∫
Mdt
2≈⋅⋅
∫
Pdt
η
6.00 -3.87 0.00 106.72 -757.67
7.00 -3.87 0.00 121.68 -757.67
7.00 -5.53 0.00 121.68 -757.67
7.15 -5.53 173.79 126.27 -757.67
7.30 -5.52 347.11 130.85 -757.67
7.45 -5.49 517.59 135.37 -757.67
7.60 -5.40 679.16 139.75 -757.67
7.75 -5.22 820.36 143.84 -757.67
7.90 -4.90 923.40 147.44 -757.67
8.05 -4.39 964.32 150.33 -757.67
8.20 -3.64 915.58 152.32 -757.67
8.35 -2.66 751.75 153.38 -757.67
8.50 -1.46 459.25 153.70 -757.67
8.50 1.34 -421.21 153.70 -757.67
8.65 2.42 -684.52 154.58 -830.65
8.80 3.31 -831.70 156.22 -930.72
8.95 3.98 -875.02 158.60 -1043.36
9.10 4.44 -837.41 161.56 -1156.38
9.25 4.73 -743.75 164.92 -1260.74
9.40 4.90 -615.63 168.52 -1350.46
9.55 4.98 -469.15 172.24 -1422.05
9.70 5.01 -314.61 176.00 -1473.78
9.85 5.01 -157.52 179.77 -1504.94
10.00 5.01 0.00 183.54 -1515.34
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3.9 Spindle drive with leadscrew
This drive involves a plastic machining unit , with vertical- and horizontal spindle dr ives without
gearbox. Servo drives with Compact Plus as 2-axis system should be used.
M o to r for the
horizontal drive
M o to r for the
vertical drive
Principle of oper ation
Vertical drive
Drive data
Total mass to be raised mtotal = 116 k g
Spindle diameter D = 0.04 m
Spindle pitch h = 0.01 m
Spindle length l = 1.2 m
Spindle efficiency ηSp = 0.9
Moment of inert ia of the coupling ( acc. to the manufacturer) JK= 0.00039 kg m 2
Max. elevating vmax = 0.33 m/s
Accelerating time tb= 0.1 s
Decelerating time tv= 0.1 s
Max. distance raise hmax = 0.651 m
No-load interval between elevating and lowering tP= 10 s
Cycle time T = 20 s
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Elevating f or c e, acceler ation force:
F
H
=⋅
mg
total =⋅=116 981 1138.N
FFm am vtN
b v total total b
== ⋅= ⋅ = ⋅ =
max .,.116 033
01 3828
Spindle pitch ang le:
α
π
π
SW hDrad=⋅=⋅=arctan arctan ...
001
004 007941
Friction angle of the spindle:
ρπη
απ
=⋅⋅ −= ⋅⋅−=arctan arctan .
.. ..
h
Drad
Sp SW 001
004 095 007941 000416
Calculating the tor ques for spindle and coupling when accelerat ing and deceler at ing
Angular velocity of the spindle at vmax:
ω
π
π
max max .
..
Sp v
hs=⋅⋅ =⋅⋅ =−
22033
001 207 3 1
Angular acceler at ion of the spindle:
α
ω
bsp sp
b
ts===
−
max .
.
207 3
01 2073 2
According t o t he m anufacturer, the spindle mom ent of inertia 40x10 is as f ollows:
Jl kgm
sp =⋅+⋅⋅=⋅+⋅⋅=
−− −−
119 27 10 1607 10 119 27 10 1607 10 12 000205
63 63 2
.. ....
The first component is a const ant value and the second com ponent represents t he formula for the
moment of inertia of the solid cylinder.
Accelerating- and decelerating t or que for spindle+coupling:
MM JJ Nm
b sp c v sp c sp c b sp++
==+⋅= + ⋅=() (. .) .
α
000205 0 00039 2073 506
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Motor selection
The following motor is selected ( r efer to Calculating the motor torques):
1FK6 042-6AF71, nn=3000 RPM, Mn(100)=2.6 Nm, M0(100)=3.2 Nm, I 0(100)=2.7 A, ηmotor=0.89
Jmotor+brake=0.000368 kg m2, Mmotor perm.=10.3 Nm
Motor speed at vmax:
n RPM
motor sp
max max .
=⋅=⋅=
ω
π
π
60
2207 3 60
21980
Calculating the mot or torques during t he const ant-velocity phase
Upwards:
MF
DNm
motor H SW
=⋅⋅ += ⋅ ⋅ + =
21138 004
2007941 000416 191tan( ) .tan( . . ) .
αρ
Downwards:
MF
DNm
motor H SW
=⋅⋅ −= ⋅ ⋅ − =
21138 004
2007941 000416 172tan( ) .tan( . . ) .
αρ
(reg. op.)
Calculating the mot or torques when acceler ating and decelerating
Motor accelerating- and deceler at ing torq ue:
MM
b motor v motor
==⋅Jmotor b sp
α
=⋅=0000368 2073 0763..Nm
Upwards, motor t or que when accelerating:
Mmotor =+++⋅⋅+
+
MMFF
D
b motor b sp c b H SW
()tan()
2
αρ
=++ +⋅⋅ + =0763 506 3828 1138 004
2007941 000416 837..(. )
.tan( . . ) . Nm
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Upwards, motor t or que when decelerating:
Mmotor =− − + − + ⋅ ⋅ + ⋅
+
MM FF
Dsign
vmotor vsp c v H SW
( ) tan( (...))
)1 2
αρ
=− − + − + ⋅ ⋅ + =−0763 506 3828 1138 004
2007941 000416 456..(. )
.tan( . . ) . Nm (reg. op.)
Downwards, mot or torque when accelerating :
Mmotor =− − + − + ⋅ ⋅ − ⋅
+
MM FF
Dsign
b Motor b sp c b H SW
( ) tan( (...))
)1 2
αρ
=− − + − + ⋅ ⋅ − =−0763 506 3828 1138 004
2007941 000416 468..(. )
.tan( . . ) . Nm
Downwards, mot or torque when decelerating :
Mmotor =+++⋅⋅−
+
MMFF
D
v Motor v sp c v H SW
()tan()
2
αρ
=++ +⋅⋅ − =0763 506 3828 1138 004
2007941 000416 812..(. )
.tan( . . ) . Nm (reg. op.)
1) If the expression in brack et s has a negative sign, t he sign of
ρ
changes
The highest motor t or que is requir ed when accelerating upwards. The highest motor tor que in
regener at ive operat ion is r equired f or deceler ation when moving downwards. The selected m otor
is adequately dimensioned, as it can be overloaded up to approx. 10.3 Nm at nmax=1980 RPM
and for a 400 V line supply voltag e.
Selecting the drive conver ter
The following is valid for t he highest required mot or t o r que:
M
M
motor max
(100)
...
0
837
32 262==
Thus, t he following value is obtained fr om t he torque/current charact er ist ic:
I
I
motor max
(100) .
027=and II A
motor max (100) ....=⋅=⋅=
027 27 27 73
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Drive converter selected:
6SE7013-0EP50 (Compact Plus)
PV n=1.1 kW; IV n=3 A; IV max=9 A (300% overload capabilit y)
Calculating of the braking power and the br aking energy
The brake resistor is used when the drive decelerat es when descending or ascending.
Velocity-time diagram
tbtv
ttotal
vmax
-
vmax
tp
ttotal
tbtv
T
Interval
t
vLowering
tk
Elevating
tk
~
~
~
~
The tr aversing time is given by:
thvt
vs
total b
=+⋅
=+⋅
=
max max
max
...
..
0651 033 01
033 207
Time for constant- velocity mot ion:
tt t s
k total b
=−⋅=−⋅=2 207 2 01 187...
Max. brak ing power for t he br ake resistor when decelerating fr om vmax to 0:
Pbr W vmax ( )
max→0=⋅=
⋅⋅
→→
PMn
br motor v motor motor v v motor motormax ( ) () max
max
max
00
9550
ηη
=⋅⋅=
456 1980
9550 089 084
...kW
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Max. brak ing power for t he br ake resistor when decelerating fr om - vmax to 0:
Pbr W vmax ( )
max
−→0=⋅=
⋅⋅
−→ −→
PMn
br motor v motor motor v v motor motormax ( ) () max
max
max
00
9550
ηη
=⋅⋅=
812 1980
9550 089 15
...kW
Braking power while travelling at constant velocity downwards:
Pbr W konst v v()
max
=− =⋅=
⋅⋅
=− =−
PMn
br Motor konst v v Motor Motor v v v Motor Motor() () max
max
max
ηη
9550
=⋅⋅=
172 1980
9550 089 0317
...kW
Braking diagra m
Interval Lowering
Elevating
Pbr W
1.5 kW
0.84 kW
tt
vv
T
~
~
~
~
tk
0.317 kW
Braking energy f or a cycle (cor r esponds to the area in the br aking diag r am ):
W
br =⋅ ⋅+ ⋅+⋅ ⋅
→=− −
→
1
21
2
00
PtP tPt
br W v v br W konst v v k br W v vmax ( ) ( ) max ( )
max max max
=⋅ ⋅ + ⋅ +⋅⋅ =
1
2084 01 0317 187 1
215 01 071... . ...kWs
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Thermally checking the m ot or
Torque characteristic
Interval
t
LoweringElevating
8.37 Nm
1.91 Nm
-4, 56 Nm -4.68 Nm
1.72 Nm
8.12 Nm
1.87 s
2.07 s
1.87 s
2.07 s
20 s
Mmotor
0.1 s
0.1 s
0.1 s
0.1 s
10 s
~
~
~
~
The RMS torque is obt ained from t he torque characteristic as f ollows:
MRMS =⋅+ ⋅ + ⋅+ ⋅+ ⋅ + ⋅837 01 191 187 456 01 468 01 172 187 812 01
20
22 2 22 2
............
=123.Nm
The average m ot or speed is given by:
n
ntn t
TRPM
average
motor
b motor k
=⋅⋅⋅+⋅
=⋅⋅⋅+⋅
=
2222
1980
201 2 1980 187
20 390
()(..)
max
max
The calculated RMS torque lies, at naverage, below the MS1 charact er ist ic. Thus, operat ion is
permissible.
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0
2
4
6
8
10
12
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
M mot, M per. in Nm
Absolute mot or torq ue and dynamic lim iting charact e r ist ic
0
0,5
1
1,5
2
2,5
3
3,5
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Motor torque i n Nm
M per . S1
M RMS / n avera ge
RMS motor torque at naverage and Mperm S1 characteristic
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Horizontal drive
Drive data
Mass to be moved m total = 100 kg
Spindle diameter D = 0.032 m
Spindle pitch h = 0.01 m
Spindle length l = 1 m
Spindle efficiency ηSp = 0.95
Slide fr iction wF= 0,02
Coupling mom ent of inertia ( acc. t o the manufacturer) JK= 0.0002 kgm2
Max. traversing velocity vmax = 0.4 m / s
Accelerating time tb= 0.16 s
Decelerating time tv= 0.16 s
Max. distance moved hmax = 0.427 m
Interval between feed and retraction tP= 5 s
Cycle time T = 20 s
Frictional force, accelerat ing forc e:
FR=⋅⋅
mgw
total F =⋅⋅ =
100 981 002 1962.. .N
FFm am vtN
b v total total b
== ⋅= ⋅ = ⋅ =
max .
.
100 04
016 250
Spindle pitch ang le:
α
π
π
SW hDrad=⋅=⋅=arctan arctan ...
001
0032 009915
Friction angle of the spindle:
ρπη
απ
=⋅⋅ −= ⋅⋅
−=arctan arctan .
....
h
Drad
Sp SW 001
0032 095 009915 000518
Calculating the tor ques for spindle and coupling when accelerat ing and deceler at ing
Angular velocity of the spindle at vmax:
ω
π
π
max max .
..
sp v
hs=⋅⋅ =⋅⋅ =−
2204
001 2513 1
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Angular acceler at ion of the spindle:
α
ω
bsp sp
b
ts===
−
max .
.
2513
016 1571 2
According t o t he m anufacturer, the spindle mom ent of inertia 32x10 is calculated as follows:
Jl kgm
sp =⋅+⋅⋅=⋅+⋅⋅=
−− −−
33573 10 0712 10 33573 10 0712 10 1 0000746
63 63 2
.. .. .
The first component is a const ant value and the second com ponent represents t he formula for the
moment of inertia of a solid cylinder.
Accelerating torque and decelerating torque for spindle+coupling:
MM JJ Nm
bspc vspc sp c bsp++
==+⋅= +⋅=()(. .) .
α
0000746 00002 1571 149
Selecting the mot or
The following motor is selected ( r efer to calculating the motor torques):
1FK6 042-6AF71, nn=3000 RPM, Mn(100)=2.6 Nm, M0(100)=3.2 Nm, I 0(100)=2.7 A, ηmotor=0.89
Jmotor+brake=0.000368 kg m2, Mmotor perm.=10.3 Nm
Motor speed at vmax:
nRPM
motor sp
max max .
=⋅=⋅=
ω
π
π
60
22513 60
22400
Calculating the mot or torques during t he const ant-velocity phase
MF
DNm
motor R SW
=⋅⋅ += ⋅ ⋅ + =
21962 0032
2009915 000518 00329tan( ) . .tan( . . ) .
αρ
Calculating the mot or torques when acceler ating and decelerating
Accelerating- and decelerating t or que for t he m otor:
MM
b motor v motor
==⋅Jmotor b sp
α
=⋅=0000368 1571 0578..Nm
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Motor torque when accelerating:
Mmotor =+++⋅⋅+
+
MMFF
D
b motor b sp c b R SW
()tan( )
2
αρ
=+++⋅⋅ + =0578 149 250 1962 0032
2009915 000518 252..( .)
.tan( . . ) . Nm
Motor torque when decelerating:
Mmotor =− − + − + ⋅ ⋅ + ⋅
+
MM FF
Dsign
v Motor v sp c v R SW
( ) tan( (...))
)1 2
αρ
=− − + − + ⋅ ⋅ − =−0578 149 250 1962 0032
2009915 000518 242..( .)
.tan( . . ) . Nm (reg. op.)
1) If the expression in brack et s has a negative sign, t he sign of
ρ
changes
The highest motor t or que is requir ed when accelerating. The hig hest motor t or que in the
regener at ive mode is r equired when decelerating. The selected motor is adequately dimensioned,
as it can be overloaded up to approx. 10.3 Nm for nmax=2400 RPM and a 400 V line supply
voltage.
Selecting the drive conver ter
The highest required motor tor que is given by:
M
M
motor max
(100)
...
0
252
32 0788==
The following value is obtained from t he torque/ cur r ent characteristic:
I
I
motor max
(100) .
00788=
and
II A
motor max (100) ....=⋅=⋅=
00788 27 0788 213
Drive converter selected:
6SE7012-0TP50 ( Compact Plus)
PV n=0.75 kW; IV n=2 A; IV max=6 A (300% overload capability)
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This inverter is connected to the previously selected drive converter for the horizontal dr ive (2-axis
operation)
Calculating of the br aking power and the braking ener gy
The brake resistor is used when the drive decelerat es.
Velocity-time diagram
tbtv
ttotal
vmax
-
vmax
tp
ttotal
tbtv
T
Interval
t
vBackwards
tk
Forwards
tk
~
~
The tr aversing time is obtained as f ollows:
tsvt
vs
total b
=+⋅
=+⋅ =
max max
max
...
..
0427 04 016
04 123
Time where the drive moves with a constant velocity:
tt t s
k total b
=−⋅=−⋅=2 123 2 016 091...
Max. brak ing power for t he br ake resistor when decelerating fr om vmax to 0:
Pbr W vmax ( )
max→0=⋅=
⋅⋅
→→
PMn
br motor v motor motor v v motor motormax ( ) () max
max
max
00
9550
ηη
=⋅⋅=
242 2400
9550 089 0541
...kW
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Braking diagra m
Interval
t
BackwardsForwards
Pbr W
0.541 kW
tt
vv
T
~
~
~
~
Braking energy f or a cycle (cor r esponds to the area in the br aking diag r am ):
W
br =⋅⋅ ⋅
→
21
20
Pt
br W v vmax ( )
max
=⋅⋅ ⋅ =21
20541 016 0087...kWs
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Thermally checking the m ot or
Torque characteristic
Interval
t
BackwardsForwards
2.52 Nm
0.0329 Nm
-2. 42 Nm -2. 52 Nm
-0.0329 Nm
2.42 Nm
0.91 s
1.23 s
0.91 s
20 s
Mmotor
0.16 s
0.16 s
0.16 s
0.16 s
5 s
~
~
~
~
1.23 s
The RMS torque is obt ained from t he torque characteristic as f ollows:
MRMS =⋅⋅+ ⋅+⋅2 232 016 00329 091 242 016
20
222
(. . . . . . )=0442.Nm
The average m ot or speed is given by:
n
ntn t
TRPM
average
motor
b motor k
=⋅⋅⋅+⋅
=⋅⋅⋅+⋅
=
2222
2400
2016 2 2400 091
20 257
()(..)
max
max
The calculated RMS torque lies, at naverage, below the MS1 charact er ist ic. Thus, operat ion is
permissible.
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0
2
4
6
8
10
12
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
M m ot , M per. in Nm
Absolute mot or torq ue and dynamic lim iting charact e r ist ic
0
0,5
1
1,5
2
2,5
3
3,5
0500 1000 1500 2000 2500 3000
Motor speed in RPM
Motor torque in Nm
M per. S1
M RPM / n averag e
RMS motor torque at naverage and Mperm S1 characteristic
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Dimensioning the brake r esist or
When dim ensioning the common br ake resistor, it is assumed that t he max. brake power of the
two drives can occur simultaneously. Thus, the following is tr ue:
PPP kWP
br W tot br W br Wmax . max max .. . .=+=+= ≤⋅
12 20
15 0541 2041 15
Further, the following must be t r ue for the br ake resistor :
WW
TP
br br br cont
12
071 0087
20 00399
+=+=≤
.. .
With
PP
br cont =20
45.(f or Com pact Plus, only an external brake resist or )
the f ollowing is obt ained
45 0 0399 018 20
.. .
⋅= ≤
kW P
Thus, t he smallest brak e r esistor is selected with P20 = 5 kW (6SE7018-0ES87-2DC0).
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3.10 Cross-cutter drive
This drive involves a cross-cutt e r dr ive with a direct ly driven k nife roll. A ser vo drive is to be used.
vweb
DKn ife ro ll
load
s
ω
Drive data
Reel diameter D = 150 m m
Reel weight mW= 1 k g
Web velocity vweb = 100 m/m in
Cut length s = 250 mm
The feed time f or the 250 mm cut is calculat ed from the web velocity as follows:
ts
vmm
ms
sweb
== =⋅=
250
100 025 60
100 015
/min ..
When cut t ing, the k nife roll must rotate with the same circ um ferential velocity as the web velocity.
If t his velocity is maint ained, t hen a feed time per cut is obtained as follows:
tD
vmm
ms
sweb
′=⋅=⋅=⋅⋅ =
π
π
π
150
100 015 60
100 0283
/min ..
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Aft er each cut , the knife roll must be acceler ated so that it has t he sam e velocity as the web, and it
must then be deceler ated to a circumferential velocity which is the same as the feed velocity.
This should be realized as f ast as possible, e.g. a t r iangular characteristic is used with
tt
t
ms ms ms ms
bvs
==−= −=
210 150
210 65
The 10 ms t akes into account deadt im es and controller stabilization.
load
tt
ts
bv
-
t
t
=2
^
Area
Cut
ω
ω
α
α
α
π
load max
ω
load min
load
load max
load max
The minim um angular velocity of the knife roll is obtained from the web velocity:
ω
load web
vDmmm s
min /min ..=⋅= ⋅ =⋅= −
2100 2
150 100
60 2
015 2222 1
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The maximum angular velocity is derived fr om the angle t hrough which the kn ife roll has moved
per revolution. T his angle corresponds to 2π. T hus, the following is tr ue:
22
πω
ω
ω
=⋅+−⋅+
load s load load bv
ttt
min max min ()(corr esponds to the area under
ω
load )
and
ωω πω π
load load bv load s
tt ts
max min min
().
.(..).=+
+⋅− ⋅= + ⋅− ⋅ = −
222222
2
013 2 22 22 015 67 6 1
The maximum angular acceleration of the knife roll is g iven by:
α
ω
ω
load load load
b
ts
max max min ..
..=−=−=−
676 22 22
0065 6982 2
The maximum speed of the knife r oll is:
nRPM
load load
max max ..=⋅=⋅=
ω
π
π
60
267 6 60
26456
Calculating the load tor ques when acelerating and decelerat ing
Moment of inert ia of the load:
Jm
Dkgm
load W
=⋅ ⋅ =⋅⋅ =
1
22
1
21015
2000281
22 2
() (
.).
As acceleration is the sam e as deceler at ion, the load torques are given by:
MMJ Nm
b load v load load load
==⋅ = ⋅=
α
max ...000281 6982 196
Selecting the mot or
The following motor is selected ( r efer to calculating the motor torques):
1FT6 061-6AC7, nn=2000 RPM, Jmotor=0.0006 k gm2
Mn(100)=3.7 Nm, M0(100)=4 Nm, I0(100)=2 A, ηmotor=0.82
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Calculating the mot or torques when acceler ating and decelerating
Accelerating- and decelerating t or que for t he m otor:
MMJ Nm
b motor v motor motor load
==⋅=⋅=
α
max ...00006 6982 042
Motor torque when accelerating:
Mmotor =+MM
bmotor bload
=+=042 196 2 38.. .Nm
Motor torque when decelerating:
Mmotor =− −MM
v motor v load
=− − =−042 196 2 38.. .Nm (regener at ive operat ion)
The highest motor t or que is requir ed when accelerating and when decelerating. T he selected
motor is adequately dimensioned, as it has sufficient r eserves at nmax=645. 6 RPM and a 400 V
supply voltage. The coupling moment of inertia was neglected. Further, no frictional forces were
taken int o account.
Selecting the drive conver ter
The following is valid for t he highest required mot or t o r que:
M
M
motor max
()
..
0100
238
406==
The following value is obtained from t he torque/ cur r ent characteristic:
I
I
motor max
()
.
0100
06=
and
II A
motor max ( ) ...=⋅=⋅=
0 100 06 2 06 12
Selected drive converter:
6SE7011-5EP50 (Compact Plus)
PV n=0.55 kW; IV n=1.5 A, IV max=2.4 A (160% overload capabilit y)
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Dimensioning the brake r esist or
The maximum br aking power for the brak e r esist or when decelerating f r om ωload max to
ωload min:
Pbr W max =⋅⋅Mmotor v load motor
ω
max
=⋅⋅=238 676 082 132... W
Braking power for t he br ake resistor at ωload min:
Pbr W load
()
min
ω
=⋅⋅Mmotor v load motor
ω
min
=⋅ ⋅=2 38 22 22 082 434... .W
Braking diagra m
Pbr W
132 W
t
tv
43.4 W
ts
Braking energy f or a cycle (cor r esponds to the area in the br aking diag r am ):
W
br =+⋅
PP t
br W br W
v
load
max ( )
min
ω
2
=+⋅=
132 434
20065 57
...Ws
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The following must be valid for t he brake resist or :
W
tkW P
br
sbr cont
== ≤
00057
015 0038
....
with
PP
br cont..
=20
45 (with an external brak e r esist or )
the f ollowing is obt ained
45 0038 0171 20
.. .
⋅= ≤
kW P
Thus, t he smallest brak ing resistor is select ed with P20 = 5 kW (6SE7018-0ES87- 2DC0) .
Thermally checking the m otor
Torque characteristic
Mmotor
2.38 N m
-2 .38 Nm
0.15 s
t
0.065 s
0.065 s
The RMS torque is obt ained from t he torque characteristic:
MRMS =⋅⋅ =
2 238 0065
015 222
2
..
..Nm
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Wit h the following
n RPM
load
load
min
min ..=⋅
⋅=⋅
⋅=
ω
π
π
60
22222 60
22122
the average motor speed is given by:
nntt
nn t
t
average
load s b
load load
b
s
=⋅−⋅+ +⋅⋅
min
min max
()222
=⋅−⋅ + + ⋅ =
2122 015 2 0065 2122 6456 0065
015 400
.(. . )( . .).
.RPM
The calculated RMS torque is, at naverage, below the S1 curve. Thus, operation is thermally
permissible.
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3.11 Centrifugal drive
This involves a centrifugal drive with a specified load duty cycle. A 6-pole 250 kW induction mot or
is to be used (force-ventilated) .
Drive data
Moment of inert ia of the empt y drum JL= 565 kgm 2
Moment of iner tia of t he full drum JV= 1165 k gm2
Moment of inert ia of the drum , water removed JE= 945 kgm2
Filling speed nF= 180 RPM
Centrif uging speed nS= 1240 RPM
Discharge speed nR= 70 RPM
Filling time tF= 34 s
Centrifuging t im e tS= 35 s
Discharge t ime tR= 31. 83 s
Accelerating time from nF to nStbS = 45 s
Braking time from nS to nRtv= 39 s
Discharge t or que MR= 500 Nm
Frictional tor que Mfriction = 120 Nm
Rated motor output Pn= 250 kW
Rated motor current In= 430 A
Rated motor torque Mn= 2410 Nm
Rated motor speed nn= 989 RPM
Motor efficiency ηmotor = 0.96
Motor moment of inertia Jmotor = 7.3 kgm2
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Speed characterist ic r equired f or a cycle
t
n
1
2
345
6
T
nS
nF
nR
Range 1: Accelerating from the discharge- to the f illing speed ∆ t1=tbF
Range 2: Filling at the filling speed ∆ t2=tF
Range 3: Accelerating from the filling - to the centr ifuging speed ∆ t3=tbS
Range 4: Centrifuging speed ∆ t4=tS
Range 5: Brak ing fr om t he centrifuging- to the discharge speed ∆ t5=tv
Range 6: Discharg ing at the discharge speed ∆ t6=tR
Times tF, tbS, tS, tv and tR are specified. Acceler ation from nR to nF should be realized with the
same increase is speed as when accelerating to the centrifugal speed. Thus, the following is
obtained f or tbF:
tt
nn
nn s
bF bS FR
SF
=⋅−
−=⋅ −
−=45 180 70
1240 180 467.
The cycle time is g iven by:
Tt t t t t t s
bF F bS S v R
=+++++= +++++ =4673445353931831895,..
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It is assumed t hat the moment of inertia linear ly increases and decreases dur ing the f illing phase
(rang e 2) , the accelerating phase t o the centrif uging speed (range 3) and the discharge phase
(rang e 6) . Thus, the following moment of inertia charact eristic is obtained over a cycle.
Tt
J
JV
JE
JL
1
23456
The following is generally true for the drive tor que for changing speed and changing moment of
inertia:
MJ dn
d
t
nd
J
d
t
=⋅ ⋅ +⋅⋅
2
60 60
π
π
[Nm]
J in kgm2, n in RPM
Using the linear char acteristics of curves n(t) and J(t), the following are obtained f or the
derivatives:
dn
dt n
t
=∆
∆
d
J
dt
J
t
=∆
∆
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Calculating the m ot o r torques
The mot or torques in each range are calculated, at the start ( index 0) and the end (index 1). The
characterist ic is linear between these two points.
Range 1, acceler ating from the discharge- to the filling speed (∆n≠0, ∆J=0)
MM
motor motor10 11
==+ ⋅⋅
−+()JJ nn
tM
L motor FR
bF friction
2
60
π
=+⋅⋅
−+=(565 . ) .
73 2
60 180 70
467 120 1532
π
Nm
Range 2, filling at the f illing speed (∆n=0, ∆J≠0)
MM
motor motor20 21
==⋅⋅−+
π
nJ J
tM
FV L
Ffriction
60
=⋅⋅−+=
π
180
60 1165 565
34 120 286 Nm
Range 3, acceler at ing, f r om t he filling- to the centrifuging speed ( ∆n≠0, ∆J≠0)
Mmotor 30 =+ ⋅⋅
−+⋅⋅−+()JJ nn
t
nJ J
tM
V motor SF
bS
FE V
bS friction
2
60 60
π
π
=+⋅⋅
−+⋅⋅−+=(.)1165 7 3 2
60 1240 180
45 180
60 945 1165
45 120 2966
π
π
Nm
Mmotor 31 =+ ⋅⋅
−+⋅⋅−+()JJ nn
t
nJ J
tM
E motor SF
bS
SE V
bS friction
2
60 60
π
π
=+⋅⋅−+⋅⋅−+=(.)945 7 3 2
60 1240 180
45 1240
60 945 1165
45 120 2152
π
π
Nm
Range 4, cent r ifuging at the centrif uging speed ( ∆n=0, ∆J=0)
MMM Nm
motor motor friction40 41 120===
Range 5, br aking from the cent r ifuging - to the discharge speed (∆n≠0, ∆J=0)
MM
motor motor50 51
==+ ⋅⋅
−+()JJ nn
tM
Emotor RS
vfriction
2
60
π
=+⋅⋅
−+=−(.)945 7 3 2
60 70 1240
39 120 2872
π
Nm
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Range 6, dischar ging at the discharge speed (∆n=0, ∆J≠0)
MM
motor motor60 61
==⋅⋅−++
π
nJ J
tMM
RL E
Rfriction R
60
=⋅⋅−++=
π
70
60 565 945
3183 120 500 576
.Nm
Torq ue char act er istic for a cycle
-3000
-2000
-1000
0
1000
2000
3000
20 40 60 80 100 120 140 160 180 200 t / s
M / Nm
motor
1
2
3
4
5
6
Thermally checking the m ot or
The mot or ent ers the field- weakening range above the rated speed. In this r ange, a higher cur rent
for t he sam e torque is required than in the constant-flux range. In or der to therm ally check t he
motor, the RMS motor torq ue is not used, but instead, t he RMS motor cur r ent.
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The mot or cur r ent is calculated in the constant -flux rang e as follows:
IM
MIII
motor const motor
motor n motor n n n()
()( )
Φ= =⋅−+
22 2 2
µµ
for n ≤ nn
The following is obtained in the f ield-weakening range:
IM
MII
n
nIn
n
motor field motor
motor n motor n n motor
motor n nmotor n
motor
() ()( )() ()=⋅−⋅+⋅
22 2 22 2
µµ
for n ≥ nn
Thus, t he m otor current s in each r ange, are calculated at the start ( index 0) and at the end (index
1) and in rang es 3 and 5 in addit ion t o t he start of the field- weakening range ( index FW). In
ranges 3 and 5, a sim plification has been made by assuming t hat the motor cur r ent has a linear
characterist ic bet ween the two points. The field- weakening range is obt ained from t he following
speed diagram .
t
n
35
nS
nF
nR
nn
tbS tv
t3 FW 5 F W
t
Field weakening
range
tS
4
∆∆
∆tt
nn
nn s
FW bS Sn
SF
345 1240 989
1240 180 1066=⋅−
−=⋅ −
−=.
∆tt
nn
nn s
FW v Sn
SR
539 1240 989
1240 70 837=⋅ −
−=⋅ −
−=.
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Range 1, acceler at ing from the discharge- to the filling speed
MM Nm
motor motor10 11 1532==
II A
motor motor10 11 22 2 2
1532
2410 430 1333 1333 292== ⋅−+=()( .) .
Range 2, filling at the f illing speed
MM Nm
motor motor20 21 286==
II A
motor motor20 21 22 2 2
286
2410 430 1333 1333 142== ⋅−+=()( .) .
Range 3, acceler at ing from the filling- to cent r ifuging speed (field-weakening range from nn to nS)
MNm
motor 30 2966=
IA
motor 30 22 2 2
2966
2410 430 1333 1333 520=⋅−+=()( .) .
Mmotor FW3=−−⋅=−
−⋅=MMM
ttNm
motor motor motor
bS FW31 31 30 32152 2152 2966
45 1066 23448∆..
IA
motor FW322 2 2
23448
2410 430 1333 1333 4195=⋅−+=(.)( .) . .
MNm
motor 31 2152=
IA
motor 31 22 2 2 2 2
2152
2410 430 1333 1240
989 1333 989
1240 469 9=⋅−⋅+⋅=()( .)() .() .
Range 4, cent r ifuging at the centrif uging speed (field-weakening r ange)
MMM Nm
motor motor friction40 41 120===
II A
motor motor40 41 22 2 2 2 2
120
2410 430 1333 1240
989 1333 989
1240 109 3== ⋅−⋅ +⋅ =()( .)() .() .
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Range 5, br aking from the cent r ifuging - to the discharge speed (field-weakening range from nS to
nn)
MM M Nm
motor motor FW motor50 5 51 2872===−
IA
motor 50 22 2 2 2 2
2872
2410 430 1333 1240
989 1333 989
1240 620=⋅−⋅+⋅=()( .)() .()
II A
motor FW motor551 22 2 2
2872
2410 430 1333 1333 505== ⋅−+=()( .) .
Range 6, dischar ge at the discharge speed
MM Nm
motor motor60 61 576==
II A
motor motor60 61 22 2 2
576
2410 430 1333 1333 1653== ⋅−+=()( .) . .
Current char act eristic for a cycle
0
100
200
300
400
500
600
700
0 20 40 60 80 100 120 140 160 180 200
Field weakening
range
I / A
motor
t / s
1
2
3
4
5
6
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The RMS motor curr ent is calculated from the curr ent char acteristic. For non- constant current s in
ranges 3 and 5, t he following is obtained f or an interval:
iiiii
i
i
tIIIIdtI ∆⋅⋅++⋅=⋅ ++
+
∫)(
3
11
21
2
12
Thus, t he following is true:
IT
motor RMS
2⋅=⋅+⋅+⋅ + +⋅ ⋅ −It It I I II t t
bF F FW FW bS FW10
220
230
23
230 3 3
1
3()()∆
+⋅ ++ ⋅ ⋅ +⋅
1
33
231
23313 40
2
()IIIItIt
FW FW FW S
∆
+⋅ + + ⋅ ⋅
1
350
25
250 5 5
()II II t
FW FW FW
∆+⋅− +⋅Itt It
FW vFW R
5
2560
2
()∆
=⋅+⋅+⋅ + +⋅ ⋅−292 4 67 142 34 1
34195 520 4195 45 1066
22 22
. (520 . . ) ( . )
+⋅ + + ⋅ ⋅ + ⋅
1
34195 469 9 4195 469 9 1066 109 3 35
22 2
(. . . .). .
+⋅ + + ⋅ ⋅ + ⋅ − + ⋅
1
3620 505 620 505 837 505 39 837 1653 3183
22 2 2
().(.)..
=22557472 2
As
and
IA
motor RMS ==
22557472
1895 345
.
This RMS motor curr ent is permissible as the rat ed motor curr ent is 430 A.
Calculating the braking power
Braking occurs when decelerating from t he cent rifug ing speed to the discharge speed. In this
case, the drive is braked with a constant motor t or que of 2872 Nm .
Max. braking power in the DC link at nS:
PMn kW
br DClink motor v S motor WRmax .. .=⋅⋅⋅=⋅⋅⋅=
9550 2872 1240
9550 096 098 3508
ηη
Minimum braking power in the DC link at n R:
PMn kW
br DClink motor v R motor WRmin .. .=⋅⋅⋅=⋅⋅⋅=
9550 2872 70
9550 096 098 198
ηη
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Braking power characteristic for a cycle
5
tv
Pbr DC link
Pbr D C link max
Pbr D C link min
t
Braking energy f or a cycle (cor r esponds to the area in the br aking diag r am ):
WPP
tkWs
br br DClink br DClink v
=+⋅= +⋅=
max min .. .
23508 198
239 72267
The following is obtained for t he average braking power:
PW
TkW
br DClink average br
== =
72267
1895 38
.
.
As a result of the high averag e br aking power, a brake resistor is not used, but instead, a dr ive
converter with regener at ive feedback into t he line supply.
Selecting the drive conver ter
The maximum m ot or cur rent is 620 A (braking mode) and t he RMS motor cur r ent is 345 A. Thus,
the f ollowing drive converter is selected with regenerative f eedback into the line supply.
6SE7135-1EF62-4AB0
PV n=250 kW; IV n=510 A, IV max=694 A, IDC link n=605 A
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It must still be checked whether the m aximum DC link current when braking, in regenerative
operation, is per m issible.
IP
U
P
VA
DClink gen br DClink
ZK
br DClink
ply
max max max
sup
..
.
==
⋅=⋅
⋅=
135 3508 10
135 400 650
3
The perm issible DC link current in r egenerative operation is 92% of the value permissible f or
motor oper at ion:
II A
DClink gen perm DClink n... ..=⋅⋅=⋅⋅=136 092 605 136 092 757
Thus, t he infeed/regenerative feedback unit is adeq uat ely dimensioned.
Selecting the regener ative feedback transf or mer
When select ing the regenerative f eedback transforme r , t he RMS DC link cur rent in reg ener ative
feedback operation is first calculated.
IP
V
DClink gen br DClink
DClink
=
The RMS value in regenerat ive feedback operation is obtained using t he following equation:
IP
V
DClink gen RMS br DClink RMS
DClink
==⋅
⋅++⋅⋅
11
322
V
PPPPt
T
DClink
br DClink br DClink br DClink br DClink v
()
max min max min
=⋅
⋅++⋅⋅
=
1
540
1
33508 198 3508 198 39
1895 175
22
(. . . .)
.A
The perm issible RMS value for a regenerative feedback transformer with 25% duty rat io factor is
given by:
II I A
DClink RMS perm DClink n DClink n.....= ⋅⋅ = ⋅=⋅=092 25
100 046 605 046 2783
Thus, a r egenerative feedback transformer 4BU5195-0UA31-8A, with 25% duty ratio f actor is
adequate.
For several centrifugal drives, a multi-motor dr ive can be conf igured, consist ing of a
rectifier/r egenerative f eedback unit, a DC link bus and an inverter for each cent r ifuge. Energy is
transferred thr ough the DC link bus by appropr iat ely harm onizing t he individual cycles. Thus, a
lower-rating rectifier/regenerative feedback unit can be selected.
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3.12 Cross-cutter with variable cutting length
For a f or ce- ventilat ed induction motor, the possible web speed should be defined as a funct ion of
the cut leng t h. The mechanical configurat ion is as follows.
IM
Gearbox knife rolls
vweb
Drive data
Diameter of the knife r olls D = 284 mm
Moment of iner tia of t he knife rolls Jload = 6.43 kgm2
Mechanical efficiency ηmech = 0.95
Overlap angle ϕü= 60 degrees
Max. permissible web velocity vB per. = 349 m/min
Max. permissible circumferential velocity of the kn ife rolls vU W per. = 363 m / m in
Min. cutting length sL min = 372 mm
Max. cutting length sL max = 1600 mm
Gearbox ratio i = 2.029
Supplementary moment of inertia, motor side Jmot suppl. = 0. 18 kgm2
Cutting torque MS= 115 Nm
Rated motor output Pmot n = 90 kW
Rated motor torque Mmot n = 1150 Nm
Motor stall torque Mstall = 3105 Nm
Rated motor speed nmot n = 750 RPM
Motor moment of inertia Jmot = 0.67 k gm2
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Rated motor current Imot n = 200 A
Magnetizing curr ent Imag = 97 A
Efficiency ηmot = 0.92
Rated mot or volt age Umot n = 340 V
When defining the possible web velocity as a function of the cut length, the following limits m ust
be observed.
• Permissible limiting torque for the motor (Mmot max ≤ Mper.)
• Thermal limit for the motor (Irms ≤ Imot n)
• Max. per m issible web velocity
• Max. permissible circumferent ial velocity of the kn ife rolls
The maximum m ot or torque when accelerating is obtained f r om
MMJJiJi
Mot Mot b Mot Mot pl b load b mech
max sup .
()==+ ⋅⋅+⋅⋅
⋅
αα
η
1
α
bangular acceler ation of t he knife rolls
When deceler at ing, the m ot or torque is given by
α
α
vb
=−
MJJiJ i
Mot v Mot Mot pl b load b mech
=− + ⋅ ⋅ − ⋅ ⋅()
sup .
αα
Precisely during the overlap phase ( constant motor speed) , the motor should pr ovide the cut t ing
torque.
MM
i
Mot k S
mech
=⋅
η
The mot or cur r ent is obtained f r om t he motor tor que:
II M
M
I
Ikn I
Ikn
Mot Mot n Mot
Mot n
mag
Mot n
mag
Mot n
≈⋅ ⋅− ⋅+ ⋅()(())()
2222
2
11
kn =1for nn
Mot Mot n
≤constant flux range
kn n
nMot
Mot n
=for nn
Mot Mot n
>field weakening
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In order t o calculate the character ist ic of the mot or torque and motor current over tim e as well as
the RMS motor current, the mot ion sequences of t he knife r olls m ust first be defined as a f unction
of t he cut length.
Motion sequences of t he knife r olls
For the specified overlap angle ϕü , the web speed and the circum ferent ial velocity of the k nife rolls
should be in synchronism.
vB
ϕü
ω
D
B
ωB
ϕ
üoverlap angle
vBconstant web velocity
ω
BB
vD
=⋅
2angular velocity of the kn ife rolls during the overlap
Ddiameter of the knife rolls
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Depending on the cut length sL, for t he r e m aining angular range, the following different situat ions
are relevant for the angular velocity characteristic of the knife rolls.
1. Sub-synchronous lengths ( sD
L<⋅
π
)
For sD
L<⋅
π
, the cut length is less than t he cir cum ference of the knife rolls. This means, it m ust
be appropriately accelerat ed t o " c at ch- up" and then decelerated. The following diag r am is
obtained.
tt
t
bv
s
k
_2
tk
_2
B
d
ω
ω
ω
cut
t
ω
dangular velocity of the kn ife rolls at the reversal point
tkü
B
=
ϕ
ω
time inter val with constant ang ular velocity
ω
B
tt
bv dB
b
== −
ω
ω
α
accelerating time, decelerating tim e of the knife r olls
α
bangular acceler ation of t he knife rolls
ts
vsDtt
sL
B
L
Bkb
==
⋅⋅=+⋅
22
ω
time between 2 cuts
For one revolution of t he knife r olls, the area under t he cur ve in the diagram is g iven by:
2⋅= ⋅+ − ⋅
π
ω
ω
ω
Bs d B b
tt()
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By inserting this, the f ollowing eq uat ions ar e obtained for the quantit ies which are required.
t
s
D
b
L
b
=⋅−⋅
22
π
α
ω
ϕ
B
Lü
b
s
Dt
=−2
ωω π
dB
L
b
s
D
t
=+
⋅−⋅
22
If
ω
B is known, then tim es tk and ts can be calculated with the pr eviously specified formulas. The
angular acceler at ion
α
b is then obtained as follows from t he m aximum motor tor que when
accelerating,
α
η
bMot
Mot Mot pl load
mech
M
iJ J J
i
=⋅+ +
⋅
max
sup .
()
The maximum m ot or speed is given by:
ni
Mot d
max =⋅⋅
⋅
ω
π
60
2
The following condition must be maint ained for sL min :
sD
Lü
min >⋅
ϕ
2
2. Synchronous length ( sD
L=⋅
π
)
For sD
L=⋅
π
(the cut length corresponds t o the circumference of t he knife r olls) ,
ω
ω
dB
=. This
means, that there is neither acceler at ion nor deceler at ion. The web velocity could theoretically be
set as req u ir ed. However, in pr act ice it is lim ited by vBper..
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3. Above-synchronous lengths ( sD
L>⋅
π
)
For sD
L>⋅
π
, the cut length is gr eater than the cir cum ference of the knife rolls. Therefore, after
the cut, the system must be braked and then re- acceler at ed. This results in the following diagram.
tt
t
tk
2
tk
2
bv
s
d
B
ω
ω
ω
t
cut
In this case,
tt
bv Bd
b
== −
ω
ω
α
Thus, t he acceler ation time is given by:
t
s
D
b
L
b
=
⋅−⋅
22
π
α
The maximum m ot or speed is now def ined by
ω
B:
ni
Mot B
max =⋅⋅
⋅
ω
π
60
2
All other eq uat ions cor respond to those of the sub-synchronous lengt h.
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In the lim it ing case,
ω
d=0. The following is valid for the associated cut lengt h:
sD
D
Lit ü
lim =⋅⋅− ⋅
22
π
If t he cut length is increased even further, an interval must be inserted with
ω
d=0. Thus, the
following diagram is obtained
tk
2tk
2
tbtv
ts
ω
B
ω
t
tp
cut
For the area under t he curve in the diagram, the following is obtained for one revolution of the
knife rolls:
2⋅= + ⋅
π
ϕ
ω
üBb
t
The following is still true
α
ω
bB
b
t
=
Thus, t he following is obtained:
tbü
b
=⋅−2
πϕ
α
ω
π
ϕ
Bü
b
t
=⋅−2
ttt t
psk b
=−−⋅2interval for
ω
d=0
All of t he ot her equations correspond to those of t he sub- synchronous lengths.
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Making the calculation
The calculation is m ade using an Excel/VBA program for a specific cut lengt h according to t he
following schemat ic:
1) Enter Mmot max and calculat e the motor RMS current
2) Interrogate Irms ≤ Imot n , when required, iterate up t o Irms = Imot n
3) Interrogate Mmot max ≤ Mstall (nmot max), when req uired, iterate up to Mmot max = Mstall(nmot max)
4) Interrogate vB ≤ vB per. , when req u ir ed limit
5) Interrogate vU W max ≤ vU W per. , when requir ed limit
To Point 1)
Init ially, Mmot max is set to Mper. (e.g. 2 Mmot n). T he angular acceleration αb is obtained with Mmot max.
Depending on the r ange (sub-synchronous, above-synchronous, above-synchronous with sL>sL
limit), t he quantities vB, nmot max, tk, tb, ts and tp quantities are defined. If nmot ma x > nmot n , the instant
where the field weakening becom es active must be calculated f or t he cur rent calculation.
Naturally, nmot max may not exceed the permissible limiting speed of the motor. If an interval tp
occurs f or above-synchronous leng th, then the m ot or torque is set to zero in this rang e. The
characterist ic of the mot or torque with respect t o time, and therefore also the m otor RMS current,
can be calculated.
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tnn
nn
t
FW
Mot Mot n
Mot Mot B b
=−
−⋅
max
max
Example for m otor speed, mot or t orque and motor current for sub-synchronous leng t hs
As a result of the ef ficiency when accelerating, the motor torque is great er t han when decelerating,
and the motor cur rent when accelerating is corr espondingly higher. T he current increases in t he
field-weakening range. The RMS value is obtained f or the motor t or que and the motor current with
the f or m ulas specified previously:
I
II tt
t
rms
Mot i Mot i ii
i
i
s
=
+⋅−
−−
=
=
∑()()
121
1
6
2
For above-synchronous lengths with ss
LLit
>lim , there is also an int erval with Mmot=0.
nMot
nMot max
nMot n
nMot B
Field weakening range
2 tFW
tk
2tk
2
tbtv
MMot
MMot b
MMot v
MMot k
t
t
t
IMot
IMot max
0123456
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To Point 2)
If Irms ≤ Imot n , then continue with Point 3). For Irms > Imot n , Mmot max is reduced until Irms = Imot n. This
is an iterative procedur e. After t his, the calculation continues with Point 3) .
To Point 3)
First, a check is made whether nmot max ≤ nlimit (nlimit is the applicat ion point for t he stall limit,
decreasing with 12
/nMot ).
nn M
M
it Mot n stall
per
lim .,
=⋅ ⋅13
Mper.
limit
nMot
n
Mn
n
stall Mot n
Mot
13 2
,()
⋅
If nmot max ≤ nlimit, the calculat ion cont inues with Point 4) . Otherwise, the st all lim it is checked:
MMn
n
Mot stall Mot n
Mot
max max
.()≤⋅
13 2
If this condition is f ulfilled, cont inue with Point 4). Ot her wise, Mmot max is r educed until the stall lim it
is just reached. This is an iterative procedure. Then continue with Point 4).
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To Point 4)
If vB ≤ vB per., then continue with Point 5). Otherwise, vB is limited to vB per.. The remaining values
(ωd, tk, tb, ts, tp) must be appropriat ely converted according to the reduced vB.
With
ω
Bper Bper
vD
..
=⋅
2
the remaining values are obtained:
ωω π
ϕ
dBper
L
Lü
s
D
s
D
=⋅+
⋅−⋅
−
.()122
2
or
ω
d=0for ss
LLit
>lim
t
s
D
b
Lü
Bper
=−
ω
2
.or tbü
Bper
=⋅−2
π
ω
.for ss
LLit
>lim
tkü
Bper
=
ω
.
tsD
sL
Bper
=⋅⋅
2
ω
.
tp=0or ttt t
psk b
=−−⋅2for ss
LLit
>lim
Due to the fact, that when vB is limited, the maximum circumfer ent ial velocit y vU W max of t he knif e
rolls can no longer incr ease, the interr ogation operations ar e therefor e completed.
To Point 5)
For above-synchronous lengths, t he interrogation operations have been completed, as, in this
case, the m aximum cir cum ferent ial velocity of the k nife rolls is defined by vB. For sub- synchronous
lengths, the maximum cir c um ferent ial velocity of the knife rolls is defined by ωd:
vD
UW dmax =⋅
ω
2
For vU W max ≤ vU W per., t he interrogation operations have been completed. O therwise, vU W max is
limited t o vU W per.. The remaining values (ωB, tk, tb, ts) must be converted corresponding t o the
reduced vU W max.
With
ω
dper UWper
vD
..
=⋅
2
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the remaining values are obtained:
ωω π
ϕ
Bdper L
Lü
s
D
s
D
=⋅
+⋅−⋅
−
.1
122
2
t
s
D
b
Lü
B
=−
ϕ
ω
2
tkü
B
=
ϕ
ω
tsD
sL
B
=⋅⋅
2
ω
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Dimensioning a capacit or batter y
A capacitor battery can be used for buffering, to pr event power oscillations bet ween the mot or and
line supply. This capacitor battery is dimensioned, so t hat in operation, energy is neither fed back
into the line supply nor converted into heat in the pulsed resistor . Without taking into account the
losses, the regenerative feedback energy from the kinetic energ y is obtained as follows:
WW Jww
gen kin tot d B
==⋅⋅−
1
222
.sub-synchronous and above-synchronous
with
JJ JJ i
tot load Mot Mot pl.sup.
()=+ + ⋅
2
Using the capacit or ener gy
WCUU
Cddn
=⋅⋅ −
1
222
()
max
the f ollowing is obtained for the capacitor:
CW
UU
kin
ddn
=⋅−
2
22
max
max
The maximum kinetic energy g ener ally occurs at s L min. When dimensioning the capacit or battery,
the existing capacitance in t he drive converter DC link or inverter, must be t aken into account.
If one t akes into account t he eff iciencies ηmech, ηmot, ηWR, the regenerative feedback ener gy must
be calculated using t he negative area under the cur ve of the DC link power.
WP dtM it
gen DClink gen Mot v d B Mot WR v
=⋅=⋅⋅+⋅⋅⋅⋅
∫1
2()
ωωηη
Aft er several r ear r angements, the following is obtained:
WJ
gen tot d B Mot WR
=⋅ ⋅ − ⋅ ⋅
′
1
222
.
ωωηη
with
JJ JJ i
tot load mech Mot Mot pl.sup.
()
′=⋅+ + ⋅
η
2
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Results as characteristics
The following characteristics are obtained with the specified numer ical values.
0
50
100
150
200
250
300
350
300 500 700 900 1100 1300 1500 1700
Cut length in mm
web velocity in m/min
Web velocity vB
0
5
10
15
20
25
30
35
40
45
300 500 700 900 1100 1300 1500 1700
Cut length in mm
omega in 1/s
w
B
w d
Angular velocities ωB and ωd with limits vB per. and vU W per.
(intersect ion at t he synchronous cut length sDmm
Lsyn =⋅=
π
892 2,)
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0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
300 500 700 900 1100 1300 1500 1700
Cut length in mm
Irms / I mot n
Motor RMS current referred to the rated motor current
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
300 500 700 900 1100 1300 1500 1700
Cut length in mm
M mot max / M mot n
Max. motor torque ref erred to the rated motor torque
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0
0.2
0.4
0.6
0.8
1
1.2
1.4
300 500 700 900 1100 1300 1500 1700
Cut length in mm
I mot max / I mot n
Max. motor current referred to the rated motor current
-8
-6
-4
-2
0
2
4
6
8
10
300 500 700 900 1100 1300 1500 1700
Cut length in mm
Energy in kWs
W mot
W gen
Energy when motoring and generating
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The following characteristics (over t im e) ar e obtained for 2 various cut lengths (sub- synchronous
and above-synchronous).
0
50
100
150
200
250
300
350
400
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
Time in s
v in m/min
Circumferential velocity of the knife rolls for s L=716 mm ( sub- synchronous)
The inter vention point for field-weakening is v=329.8 m/min
-1500
-1000
-500
0
500
1000
1500
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
Time in s
Motor torque i n Nm
Motor torque for sL=716 mm (sub-synchronous)
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0
50
100
150
200
250
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
Time in s
Current in A
Motor current for sL=716 mm (sub-synchronous)
Current incr ease in t he field-weakening r ange
-80
-60
-40
-20
0
20
40
60
80
100
120
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
Time in s
DC link power in kW
DC link power f o r sL=716 m m ( sub- synchronous)
The negative area under the curve corresponds t o t he regenerative energy
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0
50
100
150
200
250
300
350
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
Time in s
v in m/min
Circumferential velocity of the knife rolls for s L=1016 mm ( above-synchronous)
The inter vention for f ield- weakening is v=329.8 m/m in
-800
-600
-400
-200
0
200
400
600
800
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
Time in s
Motor torque in Nm
Motor torque for sL=1016 mm (above-synchronous)
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0
20
40
60
80
100
120
140
160
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
Time in s
Current in A
Motor current for sL=1016 mm (above-synchronous)
Current incr ease in t he field-weakening r ange
-60
-40
-20
0
20
40
60
80
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
Time in s
DC link power in kW
DC link power f o r sL=1016 m m ( above synchronous)
The negative area under the curve corresponds t o t he regenerative energy
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Drive converter selection
The drive converter is select ed according to the maximum motor cur r ent and the maximum RMS
value of the motor current for all cut lengths. T he maximum RMS value is exactly the same as the
rated mot or current ( 200 A) . The maximum motor current is obt ained when accelerating and at the
maximum motor speed (in the field-weakening range) . The maximum value is at precisely the
minimum cut length with Imo t max=259.2 A. T hus, the following SI MOVERT MASTERDRI VES
MOTION CO NTROL drive converter is obtained:
6SE7032-6EG50
PV n=110 kW
IV n=218 A
IV max=345 A
Selecting t he addit ional capacit or battery
The maximum r egenerative energy occurs in t his case, at the minimum cut length.
WkWs
gen max ,=677
Thus, t he r equired capacitance is g iven by
CW
UU mF
gen
ddn
=⋅−=⋅−=
22 6770
675 520 73
22 22
max
max
The DC link capacit ance of the drive converter is 14.1 mF. T his m eans t hat an additional
capacitance of 58. 9 m F is required. The pre-charging of the external capacitor bat t er y is realized
aft er int ernal charging has been com plet ed, using a separate pre-charging circuit.
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3.13 Saw drive with crank
Stacks of paper napkins should be cut using a series of circular saw blades, connected to a crank
pinion. The saw blades, driven using a t oothed belt, are m oved in a vertical direction via the crank
drive.
M
Gearbox
Saw blades
Be lt d riv e
Motor
Mode of operation
Drive data
Weight of the saw blades msaw = 80 kg
Weight of the crank pinion mpinion = 13.5 k g
Weight of the connecting rod mrod = 6.5 k g
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Weight of the crank disk mdisk = 10 kg
Length of the crank pinion r2= 1.25 m
Length of the connecting r od l = 0.6 m
Radius of t he cr ank disk r = 0. 12 m
Clearance between the center of the saw blades and the f ixed
point of r ot ation r1= 0.8 m
Cycle time T = 0.625 s
Motion sequence
Starting from the upper dead center of the cam disk, it accelerates throug h 180 degrees to the
lower dead center, followed by deceleration through 180 deg rees back to t he upper dead center
followed by a no-load interval.
Upper de ad cent er
Lower de ad cent er
ab
ϕ
The 180 degree angular range between points a and b should be traveled through within 160 ms.
The cycle time is 625 ms. The characterist ics for the angular velocity and the angular acceler ation
of t he cr ank disk are as follows.
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160 m s
ttotal 625 ms
t
max
ω
ωπ
π
2
π
2
t
α
α
α
max
max
-
ωmax and t tot ar e given by:
ωπ
max ,
=⋅⋅+=−
2
016 2
12
23 1
s
ts
tot..( ) .=⋅+=016 2 2 0546
Thus, t he angular acceleration of the crank disk is g iven by:
α
ω
max max
...==
⋅=−
ts
tot
2
23 2
0546 842 2
The maximum cr ank disk speed is g iven by:
nRPM
max max .=⋅
⋅=⋅
⋅=
ω
π
π
60
223 60
22196
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The angular acceleration α, t he angular velocity ω and the ang le ϕ of the crank disk can now be
represented as a function of time. The zero point of the ang le is at the lower dead center.
Acceleration rang e: 02
≤≤tttot. or −≤≤
π
ϕ
0
α
α
=max
ω
α
=⋅
max t
ϕπ α
=− + ⋅ ⋅
1
22
max t
Deceleration rang e: ttt
tot tot
..
2≤≤ or 0≤≤
ϕ
π
α
α
=− max
ω
α
=⋅−
max .
()tt
tot
ϕπ α
=−⋅ ⋅ −
1
22
max .
()tt
tot
The following simplificat ions were made when calculating t he forces and t or ques.
0
A
B
msaw
mrod
2
mrod
2
l
r1
r2
Crank pinion
Connecting rod
Crank disk
The weight of the saw blades including the mounting, are assumed to act at a single point at a
distance r1 from the fixed point of r ot at ion 0. The weight of t he connecting rod mrod is dist r ibut ed
evenly between points A and B. T he crank pinion is assumed t o be a t hin bar with mass mpinion.
The cutt ing torque of the saw blades is neglected.
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
Forces at the cr ank pinion
Wit h t he pr eviously specif ied sim plifications, t he force Fx in the x axis, acting at point A is obtained.
This force consists of the accelerating component Fbx and a weight com ponent FGx. The
accelerating force Fbx acts in t he opposite direction to acceler at ion ax.
0
A
msaw
mrod
2
r1
r2
ax
Fx
The mom ent of inertia of this arr angement referred t o t he fixed point of r otation 0:
Jmrmr m r
tot rod saw pinion.=⋅+⋅+⋅ ⋅
21
3
221222
With acceleration ax, the accelerating force at point A is obt ained as follows:
Fr J a
r
bx tot x
⋅= ⋅
22
.
or
FJ
a
rmmr
rma
bx tot xSt
saw pinion x
=⋅= +⋅ +⋅ ⋅
.(() )
221
2
2
21
3
The accelerat ion ax , initiated by the crank movement is, with ω and α (also refer to Example 3.6),
given by:
ar rl
x≈⋅ ⋅ +⋅ − ⋅ ⋅ + ⋅( cos sin ) ( cos sin )
ωϕαϕ ω ϕ
αϕ
222222
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The following is valid for t he force due to the weight at point A:
Fmmr
rmg
Gx rod saw pinion
=+⋅+⋅⋅()
21
2
1
2
Torques at the cr ank disk
The tor que at the crank disk due t o force Fx (also refer to Example 3.6) is given by:
MFF r
Fx bx Gx
=+⋅ −+ ⋅()
sin( )
cos
π
ϕ
β
β
with
βϕ
=⋅arcsin( sin )
r
l
In addition, a t or que is effective due to the weight of half of the rod mass at Point B.
ϕ
ϕ
mrod
2B
r
Mmgr
rod rod
/sin
22
=⋅⋅⋅
ϕ
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
The mom ent of inertia of the crank disk is calculated as follows when assuming a full cylinder:
Jmr
disk disk
=⋅ ⋅
1
22
For half of the rod m ass at point B, the mom ent of inertia is given by:
Jmr
rod rod/2 2
1
2
=⋅ ⋅
The tot al t orque which must be provided at the cr ank disk:
MMM JJ
crank Fx rod disk rod
=+ ++ ⋅
//
()
22
α
With the specified numer ical values, t he torque required at t he cr ank disk can be calculat ed, e. g.
using an Excel prog r am .
-150
-100
-50
0
50
100
150
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Time in s
Crank torque in Nm
Selecting the mot or
Selected motor:
1FT6044-4AF7 with gear box i=10 (ηgearbox=0.95), nn=3000 RPM, Jmot=0.00062 kgm 2 ( with br ake),
ηmot=0.89, Jgearbox=0.000133 k gm2, kT0=1.67 Nm/A, M0=5 Nm, I0=3 A, Mmax=20 Nm , Imax=11 A
3 Various special drive tasks 09.99
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SIMOV E RT MASTERDRIVES - Application Manual
Maximum mot or speed:
RPMninMot 21966.21910
maxmax =⋅=⋅=
Motor torque
The following is valid for t he motor t or que:
MJ iJ iMi
mot Mot gearbox crank gearbox
VZ
=⋅⋅+ ⋅⋅+ ⋅
⋅
αα η
1
with
)( crank
MsignVZ =
-15
-10
-5
0
5
10
15
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Time in s
Motor torque in Nm
The highest motor t or que is requir ed dur ing the upwards motion when motoring . This torque is
11.91 Nm. The motor t orque during the downwards motion, in regenerative operation, is less as a
result of the gear box ef ficiency.
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
0
2
4
6
8
10
12
14
16
18
20
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
M mot, M per. in Nm
The mot or t orque lies below the limiting characteristic. Thus, the m ot or is suitable as far as the
dynamic limits are concerned.
Thermally checking the m ot or
The mot o r RMS torque is approximately given by:
MMdt
T
MMMM
tt Nm
rms
Mot
Moti Moti Moti Moti ii
=⋅≈
++ ⋅⋅− =
∫∑−−
−
21
22 11
30625 45
()
..
The average m ot or speed is obtained from:
n
nn
tt
T
nt
TRPM
average
Mot i Mot i
ii Mot tot
=
+⋅− =⋅⋅
=⋅
⋅=
−−
∑1
1
222
22196 0546
2 0625 959 2
() .
..
max .
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0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Motor torque i n Nm
M per . S1
M rms/n average
The effective mot or torque lies, at naverage, below the MS1 characteristic. This means that t he motor
is also suitable as f ar as t he thermal limit s ar e concer ned.
Dimensioning the brake r esist or
The mot or out put is obtained f r om t he motor tor que and the angular velocity as f ollows:
PMi
Mot Mot
=⋅⋅
ω
-2500
-2000
-1500
-1000
-500
0
500
1000
1500
2000
2500
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Time in s
Motor output in W
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SIMOV E RT MASTERDRIVES - Application Manual
The maximum br aking power for the brak e r esist or is obtained from the maximum negative motor
output (= m otor brak ing power):
PP W
brW br Mot Mot WRmax max ,,=⋅⋅=−⋅⋅=−
η
η
2100 0 89 0 98 1832
The braking energy for a cycle can be calculated using the negative area under t he cur ve within
the motor output character ist ic.
WP dt
br Mot Mot WR
=⋅⋅⋅
∫
ηη
for PMot ≤0
≈⋅⋅
+⋅−
∑−−
ηη
Mot WR
Mot i Mot i
ii
PP
tt
1
1
2() for PP
Mot i Mot i−≤
10,
The zero positions of the motor output must be calculat ed using linear interpolation. A zero
position between two points is obtained with the condition
PP
Moti Moti−⋅<
10
to
tt tt
PP
P
xi ii
Mot i Mot i Mot i
=+ −−⋅
−−
−−11
11
The calculation r esult s in:
WWs
br =−
1238.
The following is valid for the brake resistor:
W
TWP
br br duration
== ≤
1238
0625 1981
.
..
With
PP
br dur..
=20
45 (ext. brake resistor)
the f ollowing is obt ained
45 1981 8914 20
.. .⋅= ≤WP
3 Various special drive tasks 09.99
304 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
The following must also be valid:
PWP
br W max .=≤⋅1832 15 20
Thus, when selecting a com pact type of drive converter, a brake unit with P20=5 kW (6SE7018-
0ES87-2DA0) and external brake resistor 5 kW ( 6SE7018- 0ES87- 2DC0) are requir ed. When
selecting a Com pact Plus dr ive converter , only the external brake resist or is r equired (the chopper
is integr at ed in the drive converter).
Selecting the drive conver ter
The drive converter is select ed accor ding to the peak value and the motor RMS current. The motor
current is given by, taking into account saturation:
IM
kT b
Mot Mot
=⋅
01 for MM
Mot ≤0
IM
kT b MM
MM MI
MI
Mot Mot
Mot
=⋅⋅− −
−⋅− ⋅
⋅
01 0
0
20
0
11(( )( ))
max
max
max
for MM
Mot >0
with
bb
nMot
115
16000
=−⋅()
,(b=0.1 for f r am e size <100, ot her wise 0.15)
7.13 A is obtained for the peak current. The RMS value is obtained f r om :
I
IIII
tt
TA
rms
Mot i Mot i Mot i Mot i ii
≈
++ ⋅⋅− =
−−
−
∑1
22 11
3273
()
.
Selected SIMOVERT MASTERDRI VES MOT ION CONT ROL drive converter:
6SE7015-0EP50 (Compact Plus type)
PU n=1.5 kW; IU n=5 A; IU max=8 A
with external 5 kW brake resist or (6SE7018-0ES87-2DC0)
09.99 3 Various special drive tasks
Siem ens AG 305
SIMOV E RT MASTERDRIVES - Application Manual
0.00
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Time in s
Motor current in A
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306 Siem ens AG
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3.14 Saw drive as four-jointed system
The saw drive, investigated in 3. 13, will now be re-calculated with modified data. T he geometrical
arrang em ent is such that the approximation with the cr ank disk, can no longer be used. T hus, this
arrang em ent is known as the so-called f our - jointed system.
xr1
lpinion
bpinion
xr2
yr2
yr1 2
3
l
r
yA
xA
4
Gearbox
Motor
M
Saw blades
Crank disk 1
Mode of operation
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
Drive data
Weight of the saw blades m saw = 21. 5 kg
Weight of the crank pinion mpinion = 26.5 k g
Weight of the connecting rod mrod = 7.5 k g
Moment of inert ia of the crank disk Jdisk = 0.21 kgm2
Length of the crank pinion lpinion = 0.59 m
Width of the crank pinion bpinion = 0.18 m
Length of the connecting r od l = 0.4 m
Radius of t he cr ank disk r = 0. 1 m
Clearance, center of the saw blades to the
fixed center of rotat ion 2 ( x axis) xr 1 = 0.54 m
Clearance, center of the saw blades to the
fixed center of rotat ion 2 ( y axis) yr 1 = 0.0325 m
Clearance, point 3 t o the fixed center of rotat ion 2 (x axis) xr 2 = 0.43 m
Clearance, point 3 t o the fixed center of rotat ion 2 (y axis) yr 2 = 0.035 m
Clearance, point 1 t o point 2 (x axis) xA= 0.35 m
Clearance, point 1 t o point 2 (y axis) yA= 0.5285 m
Cycle time T = 1 s
Motion sequence
Starting from the upper dead center ( i. e. connecting rod and cr ank pinion in one line), t he cr ank
disk rot ates clockwise throug h 360 degrees within 0.5 s. This is then followed by an interval of
0.5 s. T he maximum speed of the crank disk should be 192 RPM. Thus, the angular velocity and
angular acceler at ion char acteristics f or the crank disk are as f ollows.
3 Various special drive tasks 09.99
308 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
ttotal
t
π
tbtt
kv
2
T
t
1
ω
max1
ω
max1
α
max1
α
−
1
α
For ω1 max, tb, tv and tk, the following is obt ained:
ω
π
π
1
11
2
60 2 192
60 20106
max
max .=⋅⋅ =⋅⋅ =−
ns
tt ts
bv
tot
== ⋅−⋅
=⋅−⋅
=
ω
π
ωπ
1
1
220106 05 2
20106 01875
max
max
..
..
tt t s
ktot b
=−⋅=−⋅ =2 05 2 01875 0125.. .
Thus, t he angular acceleration of the crank disk is g iven by:
α
ω
1
12
20106
01875 107232
max
max .
..== = −
ts
b
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
xr2
yr2 2
3
l
r
yA
xA
4
1
ϕ
10
r2
Diagram t o det ermine the starting angle at t he upper dead center
The star t ing angle at t he upper dead cent er at t=0 is given by:
ϕ
10
2
2
05285 0035
043 035 14101=−
−=−
−=arctan arctan ..
.. .
yy
xx rad
Ar
rA (corresponds t o 80.79 degrees)
The angular acceleration α1, the angular velocity ω1 and the ang le ϕ1 of the cam disk can now be
represented as a function of time.
Range: 0≤≤tt
b(acceleration)
α
α
11
=max
ω
α
11
=⋅
max t
ϕϕ α
110 1 2
1
2
=+⋅ ⋅
max t
3 Various special drive tasks 09.99
310 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
Range: tttt
bbk
≤≤ + (constant angular velocity)
α
10
=
ω
ω
11
=max
ϕϕ ω ω
110 1 1
1
2
=+⋅ ⋅+ ⋅−
max max ()ttt
bb
Range: tttt
bk tot
+≤≤ (deceleration)
α
α
11
=− max
ω
ω
α
11 1
=−⋅−−
max max ()tt t
bk
ϕϕ ω ω ω α
110 1 1 1 1 2
1
21
2
=+⋅⋅+⋅+⋅−−−⋅⋅−−
max max max max
() ()tt ttt ttt
b k bk bk
Angular relationships in a four-jointed syst em
2
3
l
r
yA
xA
4
1
r2
ϕ
1
ϕ
2
ϕ
3
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
The following relationships are valid for angles ϕ1, ϕ2 and ϕ3:
lxr r yr r
AA
212 2
212 2
2
=+⋅ −⋅ +−⋅ +⋅(cos cos)(sin sin)
ϕϕ ϕϕ
Equation 1
tan cos cos
sin sin
ϕ
ϕ
ϕ
ϕϕ
312 2
12 2
=+⋅ − ⋅
−⋅ + ⋅
xr r
yr r
A
AEquation 2
with
rxy
rr22
22
2
=+
Angle ϕ1 is valid over the motion sequence. Angle ϕ2 can be determ ined using Equation 1
(however, only iteratively). Ang le ϕ3 is obtained with ϕ1 and ϕ2 using Equation 2.
Torques at the cr ank pinion
In order t o calculate the accelerating torques at the crank pinion, the moment s of inertia with
ref er ence to the fixed center of rotat ion 2 m ust be known. The following simplifications are made.
lpinion
Weight of the saw blades
mrod
2
r1
r22
3
bpinion
Mb2
α
2
The weight of the saw blades including their m ounting, is assum ed t o act at a single point at a
distance r1 from the fixed center of r ot ation 2. The weight of the connecting rod mrod is dist r ibut ed
3 Various special drive tasks 09.99
312 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
evenly between points 3 and 4. T he crank pinion is assumed t o be a square with length lpinion, width
bpinion and with mass mpinion. T he m om ent of inertia with respect to the fixed center of r ot ation 2 is
then given by:
Jmrmrm lb
saw rod pinion pinion pinion
21
22222
2312
=⋅+⋅+ ⋅ +()
with
rxy
rr11
21
2
=+
The accelerat ing torque with the ang ular acceler ation of the cr ank pinion α2 is given by:
MJ
b222
=⋅
α
The ang ular velocity ω2 of the crank pinion is obtained by implicitly differ entiating Equation 1.
ω
ω
ϕ
ϕ
ϕϕ
211 1
22 2
=⋅⋅⋅ +⋅
⋅⋅ +⋅
rA B
rA B
(sin cos)
(sin cos)
with
Ax r r
A
=+⋅ −⋅
cos cos
ϕ
ϕ
12 2
By r r
A
=−⋅ +⋅
sin sin
ϕ
ϕ
12 2
The required angular acceler ation α2 is obtained by diff er entiating ω2:
ααω ω
21122
2
=⋅⋅+⋅−⋅⋅
⋅
rZZrN
rN
(&)&
with
ZA B=⋅ +⋅sin cos
ϕ
ϕ
11
NA B=⋅ +⋅sin cos
ϕ
ϕ
22
&sin ( &)cos (
&)ZAB BA=⋅−⋅+⋅+⋅
ϕωϕω
1111
&sin ( &)cos (
&)NAB BA=⋅−⋅+⋅+⋅
ϕωϕω
2222
&sin sinAr r=− ⋅ ⋅ + ⋅ ⋅
ωϕ ωϕ
1122 2
&cos cosBr r=− ⋅ ⋅ + ⋅ ⋅
ωϕωϕ
1122 2
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
In order t o calculate the torques due to the weight applied at t he crank pinion, t he angular
difference to r 2 must be known. The cut t ing torq ue of the saw blades is neglected.
Weight of the saw blades
r1
r2
mrod
22
3
ϕ
2∆
ϕ
r1
∆
ϕ
S
mpinion
MG2
∆
ϕ
r
r
r
r
r
y
x
y
x
1
2
2
1
1
=+arctan arctan angle between r1 and r2
∆
ϕ
S
r
r
y
x
=arctan 2
2angle between the center of gr avity axis and r2
Thus, t he t orque due to t he weight is given by:
Mmgr mg
lmgr
Gsaw r pinion pinion Srod
2121 2 22
22
=⋅⋅⋅ + + ⋅⋅ ⋅ + +⋅⋅⋅cos( ) cos( ) cos
ϕϕ ϕϕ ϕ
∆∆
Calculating the connecting r od f or c e
The connecting r od force can be calculat ed using torq ues M2b and M2G.
3 Various special drive tasks 09.99
314 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
2
3r2
ϕ
2
ϕ
3
Fpinion T
F
rod
M2
MM M
bG22 2
=+ torq ue ar ound point 2
2
22 r
MM
FGb
Tpinion +
=tangential for ce vertical to r2
FFMM
r
rod pinion T b G
=−=+
⋅−cos( ) cos( )
ϕϕ ϕϕ
32
22
232 connecting rod force in direct ion l
Torques at the cr ank disk
The connecting r od force can be br oken down into a tangential com ponent vertical to r.
09.99 3 Various special drive tasks
Siem ens AG 315
SIMOV E RT MASTERDRIVES - Application Manual
mg
rod
2⋅
ϕ
1
Fcran k T
F
rod
mrod
2
r
4
1
ϕ
3
FF
crank T rod
=⋅ −cos( )
ϕ
ϕ
13
Thus, t he t orque as result of the connect ing r od force is given by:
MFrFr
F crank T rod
rod =⋅=⋅⋅−cos( )
ϕ
ϕ
13
In addition, a t or que is effective due to the weight of half of the connect ing r od mass in Point 4.
Mmgr
mrod
rod/cos
221
=⋅⋅⋅
ϕ
3 Various special drive tasks 09.99
316 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
The mom ent of inertia for half of the connect ing r od mass in Point 4 is given by:
Jmr
mrod
rod/2
1
22
=⋅ ⋅
Thus, t he t orque req uir ed at the crank disk is given by:
MMM JJ
crank F m disk m
rod rod rod
=+ ++ ⋅
//
()
22
1
α
With the specified numer ical values, t he torque required at t he cr ank disk can be calculat ed, e. g.
using an Excel prog r am .
-150
-100
-50
0
50
100
150
200
0 0.2 0.4 0.6 0.8 1
Time in s
Crank torque in Nm
Selecting the mot or
Selected motor:
1FK6063-6AF71 with gearbox i=10 (ηgearbox=0.95), nn=3000RPM, Jmot=0.00167 kgm2 ( with brake),
ηmot=0.89, Jgearbox=0.000542 k gm2, kT0=1.33 Nm/A, M0=11 Nm, I0=8.3 A, Mmax=37 Nm , Imax=28 A
Maximum mot or speed:
n i n RPM
mot max max
=⋅ = ⋅ =
110 192 1920
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
Motor torque
The following is valid for t he motor t or que:
MJ iJ iMi
mot mot gearbox crank gearbox
VZ
=⋅⋅+ ⋅⋅+ ⋅
⋅
αα η
11 1
with
)( crank
MsignVZ =
-15
-10
-5
0
5
10
15
20
0 0.2 0.4 0.6 0.8 1
Time in s
Motor torque in Nm
The highest motor t or que is requir ed when moving upwards when motoring. This is 18.19 Nm. The
motor t or que when moving downwar ds when gener at ing is lower due to the gear box efficiency.
During the pause interval at ϕ1= ϕ1 0, a holding torque is obtained due to ϕ1 0 ≠ 90 degrees. O nly
the weight of the connecting r od can participate in the holding torque, as t he connect ing rod and
crank disk are in one line. T his is:
Mmgr iNm
mot hold rod
=⋅⋅⋅ ⋅=⋅⋅⋅ ⋅=
2175
2981 01 14101 1
10 00589
10
cos ...cos. .
ϕ
3 Various special drive tasks 09.99
318 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
0
5
10
15
20
25
30
35
40
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
M mot, M per. in Nm
The mot or t orque lies below the limiting characteristic. Thus, the m ot or is suitable as far as the
dynamic limits are concerned
Thermally checking the m ot or
The mot o r RMS torque is approximately given by:
MMdt
T
MMMM
tt Nm
RMS
mot
moti moti moti moti ii
=⋅≈
++ ⋅⋅− =
∫∑−−
−
21
22 11
31571
()
.
The average m ot or speed is obtained from:
n
nn
tt
T
ntn t
TRPM
average
mot i mot i
ii
mot
bmotk
=
+⋅− =⋅⋅+ ⋅ =⋅+
=
−−
∑1
1
22
21920 01875 0125
1600
() (. . )
max
max
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
0
2
4
6
8
10
12
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Motor torque in Nm
M per. S1
M RMS/n average
The effective mot or torque lies, at naverage, below the MS1 characteristic. This means that t he motor
is also suitable as f ar as t he thermal limit s ar e concer ned.
Dimensioning the brake r esist or
The mot or out put is obtained f r om t he motor tor que and the angular velocity as f ollows:
PMi
mot mot
=⋅⋅
ω
1
-3000
-2000
-1000
0
1000
2000
3000
4000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time in s
Motor power in W
3 Various special drive tasks 09.99
320 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
The maximum br aking power for the brak e r esist or is obtained from the maximum negative motor
output (= m otor brak ing power):
PP W
brW br mot mot WRmax max ..=⋅⋅=−⋅⋅=−
η
η
2379 089 098 2075
The braking energy for a cycle can be calculated using the negative area under t he cur ve within
the motor output character ist ic.
WP dt
br mot mot WR
=⋅⋅⋅
∫
ηη
für Pmot ≤0
≈⋅⋅
+⋅−
∑−−
ηη
mot WR
mot i mot i
ii
PP
tt
1
1
2() für PP
mot i mot i−≤
10,
The zero positions of the motor output must be calculat ed using linear interpolation. A zero
position between two points is obtained with the condition
PP
moti moti−⋅<
10
to
tt tt
PP
P
xi ii
mot i mot i mot i
=+ −−⋅
−−
−−11
11
The calculation r esult s in:
WWs
br =−
2047.
The following is valid for the brake resistor:
W
TWP
br br duration
== ≤
2047
12047
..
With
PP
br dur..
=20
45 (ext. Brake resistor)
the f ollowing is obt ained
45 2047 921 20
..⋅= ≤WP
09.99 3 Various special drive tasks
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The following must also be valid:
PWP
brW max .=≤⋅2075 15 20
Thus, when selecting a com pact type of drive converter, a brake unit with P20=5 kW (6SE7018-
0ES87-2DA0) and external brake resistor 5 kW ( 6SE7018- 0ES87- 2DC0) are requir ed. When
selecting a Com pact Plus dr ive converter , only the external brake resist or is r equired (the chopper
is integr at ed in the drive converter).
Selecting the drive conver ter
The drive converter is select ed accor ding to the peak value and the motor RMS current. The motor
current is given by, taking into account saturation:
IM
kT b
mot mot
=⋅
01 for MM
mot ≤0
IM
kT b MM
MM MI
MI
mot mot
mot
=⋅⋅− −
−⋅− ⋅
⋅
01 0
0
20
0
11(( )( ))
max
max
max
for MM
mot >0
with
bb
nmot
115
16000
=−⋅()
,(b=0.1 for f r am e size <100, ot her wise 0.15)
13.93 A is obtained for the peak current. The RMS value is obtained f r om :
I
IIII
tt
TA
eff
mot i mot i mot i mot i
ii
≈
++ ⋅⋅− =
−−−
∑1
22 1
1
3435
()
.
Selected SIMOVERT MASTERDRI VES MOT ION CONT ROL drive converter:
6SE7021-0EP50 (Compact Plus type)
PU n=4 kW; IU n=10 A; IU max=16 A (160% overload capabilit y)
with external 5 kW brake resist or (6SE7018-0ES87-2DC0)
3 Various special drive tasks 09.99
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0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
0 0.2 0.4 0.6 0.8 1
Time in s
Motor current in A
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
3.15 Mesh welding machine
The drives f or a m esh welding m achine ar e to be dimensioned. In t his case it involves the 4 drives:
Trolley, roller feed, t r ansverse wire f eed and m esh transport. Synchronous servo mot or s with
inverters, r ect ifier unit and cent ral brake r esist or ar e to be used.
Trolley
The tr olley drive is a traversing drive to pull- in t he longitudinal wires. A breakaway force is
assumed at the start of pull-in for 200 ms, and is then further calculated with a steady-state pull-in
for ce. When dimensioning the drive, ther e should be 100% safety margin regar ding the breakaway
for c e and pull- in force.
Drive data
Trolley mass mtrolley = 350 kg
Load mass (wires) mload = 100 k g
Breakaway force (for 200 m s ) FLB = 2750 N
Pull-in f orce, steady-st a t e FZ= 600 N
Frictional force FR= 400 N
Travel s = 6 m
Max. traversing velocity vmax = 2.5 m/s
Traversing t im e, forwards ttot, forwards = 4 s
Traversing t im e, backwards ttot, backwards = 3.25 s
Pinion diameter D = 0.15152 m
Clock cycle time T = 16 s
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Traversing charact er ist ic
vmax
vmax
-
t
v
tk for.
tb for. tv for.
tb back. tk bac k. tv back.
a
amax for.
amax back.
amax for.
amax back.
-
-
t
ttot for.
ttot back.
T
forwards backwards
s
vs
ttt fortotforvforb 6.1
5.26
4
max
... =−=−==
sttt forbfortotfork 8.06.1242 ... =⋅−=⋅−=
2
.
max
.max /5625.1
6.1 5.2 sm
t
v
a
forb
for ===
s
vs
ttt backtotbackvbackb 85.0
5.26
25.3
max
... =−=−==
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SIMOV E RT MASTERDRIVES - Application Manual
sttt backbbacktotbackk 55.185.0225.32 ... =⋅−=⋅−=
2
.
max
.max /94.2
85.0 5.2 sm
tv
a
backb
back ===
Load-force char acteristic
Fload
v
vmax
LB
- FR
2 FLB
FR+ 2 Ftension
0.2 s t
T
forwards backwards
The safety margin of 100% f or t he br eakaway force and the steady-state pull-in force are taken
into account in the load- force char act er ist ic. The traversing velocity until the breakaway force
becomes ef fective, is given by:
sm
ts
vv
forb
LB /3125.0
6.1 2.0
5.2
2.0
.
max =⋅=⋅=
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The following is valid for t he load torque:
2
.. D
FJM loadloadforloadforload ⋅+⋅=
α
2
.. D
FJM loadloadbackloadbackload ⋅+⋅=
α
with
D
a
load 2
⋅=
α
2
.)
2
()( D
mmJ loadtrolleyforload ⋅+=
2
.)
2
(D
mJ trolleybackload ⋅=
afr om t he char acteristic a(t)
load
Ffr om t he char acteristic Fload(t)
-200
-100
0
100
200
300
400
500
0246810121416
Time in s
Load torque in Nm
09.99 3 Various special drive tasks
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Selecting the mot or
Selected motor:
1FT6082-8AK7 with gear box i=16 (ηgearbox=0,95), nn=6000 RPM, Jmot=0,003 kgm 2, ηmot=0,88,
Jgearbox=0,00082 kgm2, k T 0=0,71 Nm/A, M0=13 Nm, I0=18,3 A, Mmax=42 Nm , Imax=73 A
Maximum mot or speed:
nvi
DRPM
mot max max ..
=⋅⋅
⋅=⋅⋅
⋅=
60 25 60 16
015152 5042
π
π
Motor torque
The following is valid for t he motor t or que:
VZ
gear
forloadloadgearloadmotformot i
MiJiJM
η
αα
⋅
⋅+⋅⋅+⋅⋅= 1
..
i
MiJiJM VZ
gear
backloadloadgearloadmotbackmot
η
αα
⋅+⋅⋅+⋅⋅= ..
with
)( ..,backforload
MsignVZ =
-10
-5
0
5
10
15
20
25
30
35
0246810121416
Time in s
Motor torque i n Nm
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0
5
10
15
20
25
30
35
40
45
0 1000 2000 3000 4000 5000 6000
Motor speed in RPM
M mot, M per. in Nm
The mot or t orque lies below the limiting characteristic. Thus, the m ot or is suitable as far as the
dynamic limits are concerned.
Thermally checking the m ot or
The RMS motor tor que is calculated f rom the mot or t orque characteristics
Mmt
TNm
RMS
Mot i i
=⋅=
∑2616
∆.
The average m ot or speed is obtained from:
T
tnt
n
tnt
n
T
t
nn
nbackkmotbackb
mot
forkmotforb
mot
i
imotimot
average
⋅+⋅⋅+⋅+⋅⋅
=
∆⋅
+
=∑−max
max
max
max
12
2
2
22
=⋅++ + =
5042 16 08 085 155
16 15126
(. . . . ) .RPM
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
0
2
4
6
8
10
12
14
0 1000 2000 3000 4000 5000 6000
Motor speed in RPM
Motor torque i n Nm
M per . S1
M RMS / n avera ge
The effective mot or torque lies, at naverage, below the MS1 characteristic. This means that t he motor
is also suitable as f ar as t he thermal limit s ar e concer ned.
Selecting the invert er
The inverter is selected according to the peak value and RMS value of the motor cur rent. Refer to
example 3.13 for the current calculat ion.
49.6 A is obtained for the peak current. The RMS current is 9.23 A. I n t his par ticular case, the 300
% overload capability of t he Com pact Plus type of construct ion can be used, as t he overload t im e
is less than 250 ms; after the overload, the current returns t o below 0.91 x of the rated curr ent ,
and the recovery time is greater than 750 m s.
Selected SIMOVERT MASTERDRIVES MOTIO N CONTROL inver t er:
6SE7022-0TP50 ( Compact Plus type of constr uction)
Pinv n=11 kW; Iinv n=25.4 A; I DC link n=30.4 A; Iinv max=76.5 A (300 % overload capability)
A 15 kW inverter would be considered if t he 300 % overload capabilit y was not to be used
Iinv max=54.4 A (160 % overload capability).
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0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
50.00
0 2 4 6 8 10 12 14 16
Time in s
M o to r c u rren t in A
Calculating the braking power and braking energy
The mot or out put is obtained f r om t he motor tor que and the traversing velocity as follows:
PMi
v
D
mot mot
=⋅⋅
⋅2
-3000
-2000
-1000
0
1000
2000
3000
4000
5000
6000
7000
0 2 4 6 8 10 12 14 16
Time in s
Motor power in W
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
The max. braking power f or the brake r esist or is obtained from the max. negative mot or output
(=motor brak ing power):
WPP invmotmotbrWbr 237098.088.07.2747
maxmax −=⋅⋅−=⋅⋅=
The braking energy for one cycle is obtained f rom the negative surface area under t he motor
output charact eristic
WstPW invmotdownvmotbrbr 100798.088.085.07.2747
2
1
2
1max −=⋅⋅⋅⋅−=⋅⋅⋅⋅=
ηη
The average br aking power f or one cycle is
PW
TW
br average br
==− =−
1007
16 6294.
Calculating the DC link curr ents when motor ing
The maximum DC link current and t he RMS DC link cur r ent when motoring are requir ed to
dimension the rect ifier unit. These currents are obtained from the posit ive mot or out put. The
following is valid f o r t he m ax. DC link current :
A
U
P
I
linkDCinvmot
mot
linkDC 44.14
40035.198.088.0 4.6726
max
max =
⋅⋅⋅
=
⋅⋅
=
ηη
The RMS DC link curr ent is obtained as f ollows:
T
t
PPPP
U
I
i
imotimotimotimot
linkDCinvmot
RMSlinkDC ∑∆
⋅⋅++
⋅
⋅⋅
=−− 3
)(
11
22 1
ηη
for PP
mot i mot i−≥
10,
A value of 3.76 A is obt ained
3 Various special drive tasks 09.99
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Roller f eed
For a roller feed drive, it involves a rotating drive to posit ion the mesh to weld the transverse wires.
A 50% safet y mar gin for t he feed torque should be tak en int o account.
Drive data
Load moment of inertia J load = 1.48 kg m 2
Intrinsic m om ent of inertia JE= 0.35 kgm2
Feed torq ue Mv= 450 Nm
Traversing angle per transverse wire ϕ= 5 rad
Max. angular velocity ωmax = 20 s-1
Traversing t im e per transverse wire ttot. = 0.45 s
Cycle time per transverse wire T = 0. 75 s
Traversing charact er ist ic
tbtv
tk
ttot.
t
T
ω
max
ω
α
load
α
load max
α
-load max
09.99 3 Various special drive tasks
Siem ens AG 333
SIMOV E RT MASTERDRIVES - Application Manual
sttt totvb 2.0
20
5
45.0
max
.=−=−==
ω
sttt btotk 05.02.0245.02
.=⋅−=⋅−=
1
max
max 100
2.0
20 −
=== s
tb
load
ω
α
Load torq ue
The load tor que is given by:
vloadEloadload MJJM ⋅+⋅+= 5.1)(
α
load
α
fr om t he char acteristic αload(t)
0
100
200
300
400
500
600
700
800
900
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Time in s
Load torque in Nm
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Selecting the mot or
selected motor:
1FT6134-6SB. with gear box i=7 (ηgearbox=0.95), nn=1500 RPM, Jmot=0.0547 kgm 2, ηmot=0.94,
Jgearbox=0.00471 kgm2, k T 0=2.92 Nm/A, M0=140 Nm, I0=48 A, Mmax=316 Nm, Imax=139 A
Maximum mot or speed:
niRPM
mot max max
=⋅⋅
⋅=⋅⋅
⋅=
ω
π
π
60
220 60 7
21337
Motor torque
The following is valid for t he motor t or que:
MJ iJ iM
i
mot mot load gearbox load load gearbox
VZ
=⋅⋅+ ⋅⋅+ ⋅
⋅
αα η
1
with
VZ sign Mload
=()
0
20
40
60
80
100
120
140
160
180
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
T ime in
s
M o to r to rq ue in N m
09.99 3 Various special drive tasks
Siem ens AG 335
SIMOV E RT MASTERDRIVES - Application Manual
0
50
100
150
200
250
300
350
0 200 400 600 800 1000 1200 1400 1600
Motor speed in RPM
M mot, M per. in Nm
The mot or t orque lies below the limiting characteristic. Thus, the m ot or is suitable as far as the
dynamic limits are concerned.
Thermally checking the m ot or
The RMS motor tor que is obtained f rom the mot or t orque characteristic as follows:
MMt
TNm
RMS
mot i i
=⋅=
∑29343
∆.
The average m ot or speed is obtained as follows:
n
nnt
T
ntn t
TRPM
average
mot i mot i
i
mot
bmotk
=
+⋅=⋅⋅+ ⋅=⋅+ =
−
∑1
22
21337 02 005
075 4457
∆max
max (. . )
..
3 Various special drive tasks 09.99
336 Siem ens AG
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0
20
40
60
80
100
120
140
0 200 400 600 800 1000 1200 1400 1600
Motor speed in RPM
Motor torque i n Nm
M per . S1
M RMS / n avera ge
The effective mot or torque lies, at naverage, below the MS1 characteristic. This means that t he motor
is also suitable as f ar as t he thermal limit s ar e concer ned.
Selecting the invert er
59.76 A is obtained for the peak current. The RMS value is 32.47 A.
Selected SIMOVERT MASTERDRIVES MOTIO N CONTROL inver t er:
6SE7023-8TP50 ( Compact Plus type of constr uction)
Pinv n=18.5 kW; Iinv n=37.5 A; IDC link n=44.6 A; I inv max=60 A (160 % overload capability)
09.99 3 Various special drive tasks
Siem ens AG 337
SIMOV E RT MASTERDRIVES - Application Manual
0.00
10.00
20.00
30.00
40.00
50.00
60.00
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Time in s
Motor current in A
Calculating the braking power and the braking ener gy
As can be seen fr om t he m otor torque characteristic, no braking power is obtained in normal
operation. However, as braking without load and f eed torque must also be taken int o account , the
braking power and braking ener gy are also calculated for this particular situation. The m ot or
torq ue dur ing the deceleration phase is given by:
MJJ iJ i
mot v mot gearbox load E load gearbox
=− + ⋅ ⋅ − ⋅ ⋅()
max max
αα
η
=− + ⋅ ⋅ − ⋅ ⋅ =−(. . ) . ..00547 000471 100 7 035 100 095
74634 Nm
The max. braking power f or the brake r esist or is given by:
PM i kW
brW mot v mot WRmax max ....=⋅⋅⋅⋅=−⋅⋅⋅⋅=−
ω
η
η
4634 20 7 094 098 598
Braking energy and average braking power f or one cycle:
WPt kWs
br br W v
=⋅ ⋅=−⋅ ⋅ =−
1
21
2598 02 06
max .. .
PW
TW
br average br
==−=−
600
075 800
.
3 Various special drive tasks 09.99
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Calculating the DC link curr ents when motor ing
The mot or power is calculated from t he m otor torque and angular velocity as follows:
PMi
mot mot
=⋅⋅
ω
0
5000
10000
15000
20000
25000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Time in s
Motor power in W
The following is valid for t he max. DC link curr ent:
A
U
P
I
linkDCinvmot
mot
linkDC 02.48
40035.198.094.0 23885
max
max =
⋅⋅⋅
=
⋅⋅
=
ηη
The DC link RMS current is given by:
T
t
PPPP
U
I
i
imotimotimotimot
linkDCinvmot
RMSlinkDC ∑∆
⋅⋅++
⋅
⋅⋅
=−− 3
)(
11
22 1
ηη
for PP
mot i mot i−≥
10,
A value of 16.33 A is obt ained
09.99 3 Various special drive tasks
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SIMOV E RT MASTERDRIVES - Application Manual
Transverse wire f eed
The tr ansverse wire f eed dr ive compr ises t wo disks, which clamp the t ransverse wire. These disks
are each moved fr om a motor in opposing directions to tr anspor t the transverse wire.
Drive data
Load moment of inertia per motor Jload = 0.0053 k gm2
Intrinsic m om ent of inertia per motor JE= 0.0463 kg m 2
Feed torq ue per m otor Mv= 180 Nm
Traversing angle per transverse wire ϕ= 10.05 rad
Max. angular velocity ωmax = 29 s-1
Traversing t im e per transverse wire ttot. = 0.5 s
Clock cycle time per t ransverse wire T = 0. 75 s
Traversing charact er ist ic
tbtv
tk
ttot.
t
T
ω
max
ω
α
load
α
load max
α
load max
-
3 Various special drive tasks 09.99
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SIMOV E RT MASTERDRIVES - Application Manual
ttt s
bvtot
== − = − =
ϕ
ω
max ...05 1005
29 01534
tt t s
ktot b
=−⋅=−⋅ =
2 05 2 01534 01931.. .
α
ω
load b
ts
max max .
== =−
29
01534 189 1
Load torq ue
The load tor que is given by:
MJJ M
load load E load v
=+⋅+
()
α
α
load f rom the charact er ist ic αload(t)
0
20
40
60
80
100
120
140
160
180
200
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Time in s
Load torque in Nm
09.99 3 Various special drive tasks
Siem ens AG 341
SIMOV E RT MASTERDRIVES - Application Manual
Selecting the mot or
Selected motor:
1FT6086-8AF7. with gear box i=7 (ηgearbox=0.95), nn=3000RPM, Jmot=0.00665 kg m2, ηmot=0.91,
Jgearbox=0.00174 kgm2, k T 0=1.56 Nm/A, M0=18.5 Nm, I0=17.3 A, Mmax=90 Nm , Imax=72 A
Maximum mot or speed:
niRPM
mot max max .=⋅⋅
⋅=⋅⋅
⋅=
ω
π
π
60
229 60 7
219385
Motor torque
The mot or t orque is given by:
MJ iJ iM
i
mot mot load gearbox load load gearbox
VZ
=⋅⋅+ ⋅⋅+ ⋅
⋅
αα η
1
with
VZ sign Mload
=()
0
5
10
15
20
25
30
35
40
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Time in s
Motor torque i n Nm
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0
10
20
30
40
50
60
70
80
90
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
M mot, M per. in Nm
The mot or t orque lies below the limiting characteristic. Thus, the m ot or is suitable as far as the
dynamic limits are concerned.
Thermally checking the m ot or
The mot or RMS torque is calculated from the mot or t orque characteristic as follows:
MMt
TNm
RMS
mot i i
=⋅=
∑22352
∆.
The average m ot or speed is given by:
n
nnt
T
ntn t
TRPM
average
mot i mot i
i
mot
bmotk
=
+⋅=⋅⋅+ ⋅ =⋅+=
−
∑1
22
219385 01534 01931
075 8956
∆max
max .(. . )
..
09.99 3 Various special drive tasks
Siem ens AG 343
SIMOV E RT MASTERDRIVES - Application Manual
0
5
10
15
20
25
30
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Motor torque i n Nm
M per . S1
M RMS / n avera ge
The effective mot or torque lies, at naverage, below the MS1 characteristic. This means that t he motor
is also suitable as f ar as t he thermal limit s ar e concer ned.
Selecting the invert er
26.1 A is obtained for the peak current. The RMS value is 15.34 A.
Selected SIMOVERT MASTERDRIVES MOTIO N CONTROL inver t er:
6SE7021-8TP50 ( Compact Plus type)
Pinv n=7.5 kW; Iinv n=17.5 A; I DC link n=20.8 A; Iinv max=28 A (160 % overload capability)
3 Various special drive tasks 09.99
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0.00
5.00
10.00
15.00
20.00
25.00
30.00
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
T ime in s
M o to r cu rren t in A
Calculating the braking power and the braking ener gy
As can be seen fr om t he m otor torque characteristic, there is no braking power under norm al
operating condit ions. However, as braking without load and f eed torque must be taken int o
account, the br aking power and braking energ y are also calculat ed. In this case, the m otor torque
is obtained during t he deceler ation phase:
MJJ iJ i
mot v mot gearbox load E load gearbox
=− + ⋅ ⋅ − ⋅ ⋅()
max max
αα
η
=− + ⋅ ⋅ − ⋅ ⋅ =−(. . ) . ..000665 000174 189 7 00463 189 095
71229 Nm
The max. braking power f or the brake r esist or is:
PM i kW
brW mot v mot WRmax max ....=⋅⋅⋅⋅=−⋅⋅⋅⋅=−
ω
η
η
1229 29 7 091 098 222
Braking energy and average braking power f or one cycle:
WPt kWs
br br W v
=⋅ ⋅=−⋅ ⋅ =−
1
21
2222 01534 017
max .. .
PW
TW
br average br
==−=−
170
075 2267
..
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Calculating the DC link curr ents when motor ing
The mot or out put is calculated f r om t he motor tor que and angular velocity:
PMi
mot mot
=⋅⋅
ω
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
T ime in s
Motor power in W
The following is valid for t he max. DC link curr ent:
A
U
P
I
linkDCINVmot
mot
linkDC 71.16
40035.198.091.0 8046
max
max =
⋅⋅⋅
=
⋅⋅
=
ηη
The RMS value of the DC link current is given by:
T
t
PPPP
U
I
i
imotimotimotimot
linkDCINVmot
RMSlinkDC ∑∆
⋅⋅++
⋅
⋅⋅
=−− 3
)(
11
22 1
ηη
for PP
mot i mot i−≥
10,
A value of 7.42 A is obt ained
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Mesh removal drive
The mesh r em oval is a tr aversing dr ive which transport s t he m eshes away. The t r anspor t is
realized in 3 stages over a distance of 4.85 m. Initially, two movements are made, each with 0.5 m
with intermediate pauses and t hen t he r emaining 3.85 m is traversed in one step. T he drive
traverses back t o t he initial position without any pause.
Drive data
Trolley weight mtrolley = 200 kg
Load mass (mesh) mload = 120 kg
Tension force FZ= 1900 N
Friction f orce FR= 100 N
Distance, tot al s = 4.85 m
Max. traversing velocity vmax = 2.5 m/s
Partial tr aversing distance s1= 0.5 m/s
Traversing t im e for the par tial traversing dist ance ttot. 1 = 0.5 s
Pause times tP= 0.375 s
Pinion diameter D = 0.15152 m
Clock cycle time T = 16 s
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Traversing charact er ist ics
vmax
vmax
-
t
v
a
amax
amax
-
t
T
forwards backwards
vmax 1
11
2
3
ttot. 1
tPtP
tP
ttot. 1
0.5 m 0.5 m 3.85 m
4.85 m
A triang ular velocity characteristic is used for t he 2 partial traversing dist ances. Thus, the
maximum velocity and the maximum acceleration ar e obt ained as follows:
vs
tms
tot
max .
./
11
1
2205
05 2=⋅=⋅=
av
tms
tot
max
max ./=⋅=⋅=
222
05 8
1
1
2
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The same acceler at ion is assum ed for the r em aining distances. T he m issing times are obt ained as
follows:
ttv
as
bv22
225
803125== ==
max
max
..
tsv t
vs
k
b
2
222
2
385 25 03125
25 12275=−⋅
=−⋅ =
max
max
...
..
ttv
as
bv33 325
803125== ==
max
max
..
tsv t
vs
k
b
3
333
3
485 25 03125
25 16275=−⋅
=−⋅ =
max
max
...
..
Load force characteristics
FR
-t
T
forwards backwards
Fz+ FR
Fload
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Load torque
The load tor que is given by:
2
D
FJM loadloadforwardsloadforwardsload ⋅+⋅=
α
2
D
FJM loadloadbackwardsloadbackwardsload ⋅+⋅=
α
with
D
a
load 2
⋅=
α
2
)
2
()( D
mmJ loadtrolleyforwardsload ⋅+=
2
)
2
(D
mJ trolleybackwardsload ⋅=
afr om char act eristic a(t)
load
Ffr om t he char acteristic Fload(t)
-150
-100
-50
0
50
100
150
200
250
300
350
0 2 4 6 8 10 12 14 16
Time in s
Load torque in Nm
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Selecting the mot or
Selected motor:
1FK6103-8AF71. with gearbox i=7 (ηgearbox=0.95), nn=3000 RPM, Jmot=0.01215 kgm 2, ηmot=0.94,
Jgearbox=0.00185 kgm2, k T 0=1.51 Nm/A, M0=36 Nm, I0=23.8 A, Mmax=102 Nm , Imax=78 A
Maximum mot or speed:
nvi
DRPM
mot max max ..
=⋅⋅
⋅=⋅⋅
⋅=
60 25 60 7
015152 2206
π
π
Motor torque
The mot or t orque is given by:
VZ
gearbox
forwardsloadloadgearboxloadmotforwardsmot i
MiJiJM
η
αα
⋅
⋅+⋅⋅+⋅⋅= 1
i
MiJiJM VZ
gearbox
backwardsloadloadgearboxloadmotbackwardsmot
η
αα
⋅+⋅⋅+⋅⋅=
with
)( ,backwardsforwardsload
MsignVZ =
-30
-20
-10
0
10
20
30
40
50
60
70
0 2 4 6 8 10121416
Time in s
Motor torq ue in Nm
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0
20
40
60
80
100
120
0 500 1000 1500 2000 2500 3000
M otor speed in RP M
M mot, M per. in Nm
The mot or t orque lies below the limiting characteristic. Thus, the m ot or is suitable as far as the
dynamic limits are concerned.
Thermally checking the m ot or
The RMS motor tor que is calculated f rom the mot or t orque characteristic:
MMt
TNm
RMS
mot i i
=⋅=
∑21675
∆.
The average m ot or speed is given by:
n
nnt
T
ntn ttt
T
average
mot i mot i
i
mot
bmotbkk
=
+⋅=⋅⋅+ ⋅ ⋅+ +
−
∑11
1223
22
42∆max
max ()
=
⋅⋅
⋅⋅ ⋅⋅+ ⋅ ⋅+ + =
2607
2 015152 025 4 2206 03125 2 12275 16275
16 5349
π
..(...)
.RPM
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0
5
10
15
20
25
30
35
40
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Motor torque in Nm
M per. S1
M RMS / n avera ge
The effective mot or torque lies, at naverage, below the MS1 characteristic. This means that t he motor
is also suitable as f ar as t he thermal limit s ar e concer ned.
Selecting the invert er
43.6 A is obtained for the peak current. The RMS value is 11.44 A.
Selected SIMOVERT MASTERDRIVES MOTIO N CONTROL inver t er:
6SE7023-0TP50 ( Compact Plus type of constr uction)
Pinv n=15 kW; Iinv n=34 A; IDC link n=40.5 A; Iinv max=54.4 A (160 % overload capability)
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0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
0 2 4 6 8 10121416
T ime in s
Mo to r cu rrent in A
Calculating the braking power and the braking ener gy
The mot or out put is obtained f r om t he motor tor que and the traversing velocity:
PMi
v
D
mot mot
=⋅⋅
⋅2
-10000
-5000
0
5000
10000
15000
0 2 4 6 8 10 12 14 16
T ime in s
Motor power in W
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The max. braking power f or the braking resistor is obt ained t hr ough the maximum neg ative motor
output (= m otor brak ing power):
WPP invmotmotbrWbr 8.548398.094.09.5952
maxmax −=⋅⋅−=⋅⋅=
The braking energy for one cycle is obtained thr ough the negat ive surface area under the motor
output charact eristic
WstPW ivimotbrinvmotbr 2078
2
1max −=⋅⋅⋅⋅= ∑
ηη
The average br aking power f or one cycle is
PW
TW
br average br
==− =−
2078
16 1299.
Calculating the DC link curr ents when motor ing
The following is valid for t he m ax. DC link current :
A
U
P
I
linkDCinvmot
mot
linkDC 93.28
40035.198.094.0 14390
max
max =
⋅⋅⋅
=
⋅⋅
=
ηη
The DC link RMS current is obtained as f ollows:
T
t
PPPP
U
I
i
imotimotimotimot
linkDCinvmot
RMSlinkDC ∑∆
⋅⋅++
⋅
⋅⋅
=−− 3
)(
11
22 1
ηη
for PP
mot i mot i−≥
10,
A value of 4.75 A is obt ained
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Selecting the rectifier unit
The rectifier unit is selected according t o t he peak DC link cur rent and the DC link RMS current.
maxmax EElinkDClinkDC II ≤
nEElinkDCRMSlinkDC II ≤
Further, the rated DC link cur r ent of the rectifier unit should be at least 30% of the total of the
rated DC link cur rents of the connected inverters.
nEElinkDCnINVlinkDC II ≤⋅∑
3.0
For the 15 k W rectifier unit Com pact Plus, the sum of the rat ed DC link currents of the connect ed
inverters may not be g r eater than 80 A due to t he r esistor pre-charging.
As the f our dr ives can operat e sim ultaneously, the sum of the peak values of t he individual
inverters is used to est imate IDC link max (this m eans t hat the calculation is on the safe side).
AII INVlinkDClinkDC 8.12493.2871.16202.4844.14
maxmax =+⋅++=≈ ∑
The RMS values of the individual inverters are added to determine IDC link RMS. In principle, the RMS
values of the total DC link curr ents must be generated, but t o save time this has not been done
and the error neglected.
AII RMSINVlinkDCRMSlinkDC 68.3975.442.7233.1676.3 =+⋅++=≈ ∑
When t he r at ed DC link current s ar e added for the individual inverters, t he following is obtained:
AI nINVlinkDC 1.1575.408.2026.444.30 =+⋅++=
∑
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Selected SIMOVERT MASTERDRIVES MOTIO N CONTROL rectifier unit
6SE7031-2EP85-0AA0 (Compact Plus type)
Pn=50 kW; IDC link n=120 A; IDC link max=192 A (160 % overload capability)
Selecting t he brake resist or
When select ing the brake resistor, it is assumed that the four drives must br ake simultaneously.
Thus, t he sum of the individual peak braking powers is formed (in t his case you will always be on
the saf e side) .
WPP INVbrbr 1827454842220259802370
maxmax =+⋅++=≈∑
The average br aking powers of t he individual inverters ar e also added.
WPP averageINVbraveragebr 14479.1297.226280094.62 =+⋅++=≈ ∑
With the conditions
PP
br max .≤⋅15 20
PP
br average ≤20
45.
a braking resistor with P20=20 kW (6SE7023-2ES87- 2DC0) is obt ained. The chopper for this
inverter is, for Compact Plus, integr ated in the rect ifier unit.
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3.16 Drive for a pantograph
For the systems which have been implemented up until now, the pantograph is brought into t he up
(operating position) using a spr ing which is tensioned in the quiescent ( down) position. This spring
also provides the contact force between the pantograph bar and the overhead wire. T he
disadvantage with this arrangement is that the spring is continuously tensioned in t he quiescent
position, which means that the components are subj ect to increased mechanical str essing. The
pantogr aph is br ought back into the quiescent st at e, and the spring is eit her pneumatically
tensioned or tensioned using an auxiliary electric drive.
Overhead wire
Operating position
(up position)
Bar
Quiescent position
Lower joints
Principle mode of operation
When t he spring is replaced by an electric drive, the drive must fulfill the f ollowing r e quirement s:
• The pant ograph must be retracted from t he overhead wire within approx. 1 s
• Af ter the bar com es int o cont act with the overhead wire, it must be pressed ont o t he overhead
wire with a constant force
• When the overhead wire height changes, t he drive must appropriately corr ect the pantograph
so that the contact force is, as f ar as possible, constant.
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Initially, the load t or que is investigated at one of the two lower joints of the pantograph as a
funct ion of angle ϕ. The motor t or que can, f or example, be applied through a g ear box, or an
additional lever arm with rack and pinion. Posit ioning can be implemented using a suitable
traversing cam . Aft er t he pantograph bar m akes contact with the overhead wire, the m ot or must
be closed-loop torque controlled using a t or que limiting function as a function of the height in or der
to guar ant ee a constant contact force.
Pantogr aph dat a
Weight of the upper pantog r aph assem bly (left and righthand section) mo= 64 kg
Weight of the lower pantograph assembly (left and rig ht hand section) m u= 80 kg
Weight of the bar mW= 60 k g
Length of the upper bar s lo= 1.751 m
Length of the lower bars lu= 1.25 m
Contact force FA= 160 Nm
Height in t he lower posit ion yu= 0.416 m
Height in the upper position yo= 2. 481 m
Distance between the lower joints 2a = 1 m
Positioning t im e ttot = 1 s
Angular relat ionships
ϕ
β
γ
ϕ
a
y
o
l
u
l1
2
3
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The height y of the pantograph is a f unct ion of angle ϕ:
22 )cos(sin
ϕϕ
⋅+−+⋅= uou lally Equation 1
The following is valid for t he angle:
o
uo l
lal 22 )cos(
arcsin
ϕ
β
⋅+−
=,
β
π
γ
−= 2
Using equation 1, and the g iven height s yu and yo, the associated ang le ϕu and ϕo can be
iterat ively determ ined. The following is obt ained:
rad
u15.0=
(lower quiescent position)
rad
o085.1=
(working ( up) position)
0
0.5
1
1.5
2
2.5
3
0 0.2 0.4 0.6 0.8 1 1.2 1.4
Anl ge phi in rad
y in m
y
yu
yo
Diagram to determine the initial values for iteration
Movement
Starting from the lower quiescent position, the pantog r aph should be brought int o it s final position
within 1 s. It must also be retracted t o its final position in t he lower q uiescent position also within
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1 s. The angular velocity at the lower joint 1 is specified as symmetrical t r iangular traversing
characterist ic. Thus, the following characteristic is obt ained for the angular velocity and the
angular acceler at ion.
maxload
ω
maxload
ω
−
maxload
α
maxload
α
−
load
α
tot.
t
t
t
load
ω
ϕ
∆
p
t
The following is obtained for t he m ax. angular velocity, max. angular acceler at ion and t he m ax.
load speed:
1
max 87.1
1)15,0085.1(2
)(2
2−
=
−⋅
=
−⋅
=
∆⋅
=s
tt ges
uo
ges
load
ω
2
max
max 74.3
187.12
2−
=
⋅
=
⋅
=s
tges
load
load
ω
α
1
max
max min86.17
26087.1
2
60 −
=
⋅⋅
=
⋅⋅
=
π
π
ω
load
load
n
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The angular acceleration αload, the angular velocity ωload and t he angle ϕ can now be represented
as a funct ion of time.
Range: 0≤≤
tt
b(acceleration upwards)
maxloadload
α
α
=
t
loadload ⋅= max
α
ω
2
max
2
1t
loadu ⋅⋅+=
αϕϕ
Range: totalb ttt ≤≤ (acceleration downwards)
maxloadload
α
α
−=
)(
max totalloadload tt −⋅−=
α
ω
2
max )(
2
1totalloadu tt −⋅⋅−∆+=
αϕϕϕ
Range: bptotalptotal tttttt ++≤≤+ ( acceler at ion upwards)
maxloadload
α
α
−=
)(
max ptotalloadload ttt −−⋅−=
α
ω
2
max )(
2
1ptotalloadu ttt −−⋅⋅−∆+=
αϕϕϕ
Range: totalptotalbptotal ttttttt ++≤≤++ (acceleration downwards)
maxloadload
α
α
−=
)2(
max ptotalloadload ttt −⋅−⋅=
α
ω
2
max )2(
2
1ptotalloadu ttt −⋅−⋅⋅+=
αϕϕ
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Calculating the load tor que when raising and lowering t he pant ograph
When r aising and lowering the pantog r aph, the load torque is calculated, without taking into
account the contact forc e.
Torque as result of the forces due to the various weights
24 Wo FF +
4o
F
2u
F2G
M
4o
m
24 Wo mm +
u
l
o
l
ϕ
1
2
3
The forces associated with only one half of the pantog r aph ar e investigated due to the symm et r y.
Half of the weight of the upper assem bly Fo/2 is distributed, extending from t he center of gravity,
50% to point 2 and 3 in order to simplify the calculat ion. The torque associated with point 1 does
not change. Half of the weight of the lower assembly Fu/2 act s at the center of gravity. Only half
the weight of the counterpoise FW/ 2 acts at point 3. The for ce Fo/4+FW/2, acting vertically at point
3, is brok en down into a bar stress for bar lo, and t hen into a normal component force for bar lu.
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GSt
F
GSt
F
GN
F
24 Wo FF +
ϕ
δ
ϕ
β
γ
1
o
l
u
l
4
F
2u
F
2G
M
The following is obtained for t he bar stress and the norm al com ponent of the force:
γγ
cos
)
24
(
cos 24 g
mmFF
F
WoWo
GSt
⋅+
=
+
=
δ
cos⋅= GStGN FF
with
βϕ
π
δ
−−= 2
Thus, t he t orque as result of the various weights is g iven by:
uGNu
ouuG lFl
FlFM ⋅+⋅⋅+⋅⋅=
ϕϕ
cos
4
cos
222
uGNu
ouu lFlg
ml
g
m⋅+⋅⋅⋅+⋅⋅⋅=
ϕϕ
cos
4
cos
22
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Torque due to the accelerat ion of the various masses
4o
m
24 Wo mm +
u
l
o
l
ϕ
1
2
3
yy avy ,,
2b
M
2
y
F
2u
m
Also here, due to t he symm et ry, only the forces associat ed with half of t he pant ograph are
investigated. The forc e Fy/2 is obtained f r om the acceleration in the y axis:
y
Wo
ya
mm
F⋅+= )
24
(
2
The velocity vy and acceleration ay is obt ained using equation 1 by differentiation.
ϕ
ω
ϕω
ϕ
ϕ
sin sin)cos(
cos ⋅− ⋅⋅⋅⋅+
+⋅⋅=⋅= u
loaduu
loaduy ly lla
l
dt
d
d
dy
v
ϕ
ϕ
ϕ
sin⋅−
=⋅= u
y
yly T
dt
d
d
dv
a
with
222 ))2(cossin()sincos( yloadyloadloaduloadloadu vvyyllaT −⋅⋅+⋅⋅+⋅⋅−⋅+⋅+⋅⋅⋅=
ωαϕϕωϕαϕω
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Also here the force Fy/2, acting vertically in point 3, is br oken down into a bar for ce regarding bar lo
and then into a force to t he vertical regarding bar lu.
4o
m
u
l
ϕ
1
2b
M
2u
m
bN
F
The bar st r ess and t he normal component of the force as result of the accelerat ion ar e given by:
γ
cos
2y
bSt
F
F=
δ
cos⋅= bStbN FF
Thus, t he t orque due to acceleration is given by:
ubNload
blFJ
M⋅+⋅=
α
1
2
In this case, J1 is the moment of inertia of bar lu with weight mu/2 and the weight mo/4, acting at
point 2, referred t o point 1.
22
1423
1u
o
u
ul
m
l
m
J⋅+⋅⋅=
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The load tor que at point 1, when taking into account bot h halves of the pantog r aph is given by:
)
22
(2 bG
load MM
M+⋅=
Using the num er ical values, t he load t orque can be calculated, for example, using an Excel
program.
-1000
0
1000
2000
3000
4000
5000
6000
7000
0 0.5 1 1.5 2 2.5 3 3.5 4
Time in s
Load torque in Nm
Upwards Downwards
Load torq ue when the pantograph is being r aised and lowered (pant ograph moving up and down)
Calculating the steady- s t ate load torque
When calculat ing the steady-stat e load torque during operation, t he cont act for ce FA is taken into
account. T he following is obtained:
)cos
4
cos
22
(2 ustatNu
ouu
statload lFlg
ml
g
m
M⋅+⋅⋅⋅+⋅⋅⋅⋅=
ϕϕ
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with
δ
γ
cos
cos 2
)
24
(⋅
+⋅+
=
A
Wo
statN
F
g
mm
F
0
500
1000
1500
2000
2500
3000
3500
4000
0 0.2 0.4 0.6 0.8 1 1.2
Angle phi in rad
M lo a d s tat in N m
Steady-state load t orque with contact force as a f unction of ang le ϕ
Selecting the mot or
When select ing the motor a sim plification is made in so much that it is assum ed t hat the motor
torq ue is convert ed t o t he load torque linearly thr ough a gearbox.
Selected motor:
1PA6105-4HF. with gearbox i=110 (ηgear=0.9), Mn=44 Nm, nn=17500 RPM, Mstall=114.4 Nm,
In=17.5 A, Iµ=9.3 A, JMot=0.029 kgm2, Jgear=0.02 kgm2, ηMot=0.875
Maximum mot or speed:
RPMnin loadmot 3.196486.17110
maxmax =⋅=⋅=
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Motor torque
The following is valid for t he motor t or que:
VZ
gear
loadloadgearloadmotupmot i
MiJiJM
η
αα
⋅
⋅+⋅⋅+⋅⋅= 1
i
MiJiJM VZ
gear
loadloadgearloadmotdownmot
η
αα
⋅+⋅⋅+⋅⋅=
i
MM statloadstatmot 1
⋅=
with
)( load
MsignVZ =
-40
-20
0
20
40
60
80
100
0 0.5 1 1.5 2 2.5 3 3.5 4
T ime in s
M o to r to rq u e in N m
Upwards Downwards
Motor torque when the pantograph is being r aised and lowered (pant ograph is moved up and
down)
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0
5
10
15
20
25
30
35
40
0 0.2 0.4 0.6 0.8 1 1.2
Angle phi in rad
M Mot stat in Nm
M Mot G
M Mot stat
Steady-state m otor torque with contact for c e ( Mmot stat) and without contact force (Mmot G) as a
funct ion of angle ϕ
The highest motor t or que in dynamic operation is required when the pantogr aph is being raised
(moved upwards). This t orque is 81.13 Nm. The mot or t or que when the pantograph is being
lowered (moved downwar ds) is less due to the ef ficiency of the gear.
In order t o achieve a constant contact force in steady-state oper ation, the mot or torque limit must
be entered as a function of angle ϕ according t o the characteristic Mmot stat.
This can be realized, for example, by sensing the angle using an absolute value encoder and
entering t he torque limit via a characteristic block. In order t o com pensate overhead wire height
changes mor e quickly, a torque increase can be used which is dependent on the dif ferentiat ed
angle ϕ. Mmot stat is great er than torque Mmot G, caused by the weight, by the com ponent of the
contact force FA. The maximum steady-stat e motor tor que is 33.55 Nm. However, this value is not
reached, as the overhead wire height only varies by 0.5 m. T hus, using equation 1, an angle of 0.8
rad and a steady-stat e m otor torque of approximat ely 25 Nm is obt ained. Thus, the m ot or is
therma lly suitable, as t he r a t ed t orque is 44 Nm.
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0
10
20
30
40
50
60
70
80
90
100
0 500 1000 1500 2000 2500
Motor speed in 1/min
M mot, M perm. in Nm
M Mot dyn
M perm.
The dynamic mot o r torque lies below the limiting curve. Thus, the m ot o r is also suitable regar ding
the dynamic limits.
Dimensioning the brake r esistor
The mot or output is obtained from the m otor torque and the angular velocity as follows:
loadmotmot iMP
ω
⋅⋅=
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-15000
-10000
-5000
0
5000
10000
15000
0 0.5 1 1.5 2 2.5 3 3.5 4
Time in s
Motor power in W
DownwardsUpwards
Motor output while the pantog r aph is being raised and lowered (pantog r aph m oving up and down)
The max. braking power f or the brake r esist or is obtained from the max. negative mot or output
(=motor brak ing power):
WkPP INVmotmotbrWbr 17.998.0875.07.10
maxmax −=⋅⋅−=⋅⋅=
As a result of the condition
20max 5.1 PP Wbr ⋅≤
a braking unit with minimum P 20=10 kW must be used.
The braking energy when rasing and lowering t he pantograph can be calculat ed using the area
under the neg at ive mot or power curve.
dtPW INVmotmotbr ⋅⋅⋅=
∫
ηη
for 0≤
mot
P
)(
21
1−
−−⋅
+
⋅⋅≈ ∑ii
imotimot
INVmot tt
PP
ηη
for 0,
1≤
−imotimot PP
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The calculation r esult s in:
WskWbr 58.3−=
The following must be valid for a br aking unit with inter nal br ake resistor :
36
20
P
T
Wbr ≤
or
s
P
W
Tbr 9.12
103658.3
36
20 =
⋅
=
⋅
≥(at P20=10 kW)
When a 10 kW br aking unit with inter nal br ake resistor is used, the cycle time f or upwards and
downwards motion must therefore be a minimum of 12.9 s. Otherwise, an external brak e r esistor
must be used or possibly a larg er braking unit.
Selecting the drive conver ter
The drive converter is select ed accor ding to the max. m otor current when accelerating and the
max. motor cur r ent under steady-state oper at ing conditions. T he m ot or current is calculat ed from
the motor torque as follows:
2
2222 1
)())(1()( kn
I
I
kn
I
I
M
M
II
nmot
mag
nmot
mag
nmot
mot
nmotmot ⋅+⋅−⋅⋅≈
kn =1for nmotmot nn ≤constant flux range
nmot
mot
n
n
kn =for nmotmot nn >field weakening
Thus, t he following is obtained:
AI bmot 2.31
max =max. motor cur r ent when accelerating with 81.13 Nm
AI statmot 6.12
max =max. motor cur r ent under steady-state oper at ing conditions
with 25 Nm
8 % is added for Imot b max f or saturation effects.
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Selected drive converter SIMOVERT MASTERDRIVES VECTOR CONTROL:
6SE7022-6EC61
PU n=11 kW; IU n=25.5 A; IU max=34.8 A (136 % overload capability)
with 10 kW braking unit (6SE7021-6ES87-2DA0)
0
5
10
15
20
25
30
35
0 0.5 1 1.5 2 2.5 3 3.5 4
Time in s
Motor current in A
Upwards Downwards
Motor current when the pantograph is being raised and lowered.
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Estimating the accuracy of t he cont act force
If t he steady-state tor que characterist ics at an oper ating point are investigated, then the most
unfavorable case at the lowest required motor torque is obtained, i. e. at the maximum height:
NmMstatmot 8.17=steady-stat e m ot or torque
NmMGmot 7.15=motor t orque component for the forces
due to the weight
and therefore
AGmotstatmotAmot FNmMMM ~1.2=−= m otor torque component for the contact force
If a 1% t or que accuracy is used for t he dr ive, t hen the following is obtained:
Nm
M
Mn
mot 44.0
100
44
%1
%100 ==⋅=∆
Thus, t he accur acy of the contact force is g iven by:
%21%100
1.244.0 =⋅=
∆
=
∆
=
∆
A
mot
A
A
Amot
Amot
M
M
F
F
M
M
Due to the fact that the motor has been dim ensioned accor ding to the dynamic requirements, t he
requir ed st eady-st ate motor t or que is relatively small with respect to the r at ed m otor torque and
because the component r esponsible for the contact force is only approximately 12 % of the
steady-state m otor torque, a larger error is obt ained r egarding t he accur acy of the contact force.
Better r esults would be achieved if t he com ponent of t he weight could be kept lower, e. g . using an
equalization weight, or a spring, which would compensate the weight component in t he operating
range.
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Checking the calculation using the potential energy
The calculation can be checked using the pot ential energy to move the m asses from t he quiescent
position into t he oper at ing position. To realize this, t he center of gravity travel of weight s mo, mu
and mW is calculated.
m
lyly
mh uuuouo
o492.1
215.0sin25.1416.0
2085.1sin25.1481.2
2sin
2sin =
⋅+
−
⋅+
=
⋅+
−
⋅+
=∆
m
ll
mh uuou
u459.0
215.0sin25.1
2085.1sin25.1
2
sin
2
sin =
⋅
−
⋅
=
⋅
−
⋅
=∆
myymh uoW 065.2416.0481.2 =−=−=∆
Thus, t he following is obtained f or the potential energy:
WWuuoopot mhgmmhgmmhgmW ∆⋅⋅+∆⋅⋅+∆⋅⋅=
Ws251281.9)065.260459.080492.164( =⋅⋅+⋅+⋅=
The following is obtained for t he average power:
W
t
W
P
total
pot
average 2512
1
2512 ===
On the other hand, the energ y f or the upwards motion is calculated as f ollows:
)(
21
11
00 −
−− −⋅
⋅+⋅
≈⋅⋅=⋅= ∑
∫∫ ii
iloadiloadiloadiload
load
t
load
t
loadup tt
MM
dtMdtPW totaltotal
ωω
ω
When t his r elat ionship is evaluated, for appropr iat ely smaller st eps, the same value is determined
as for Wpot. T he kinetic energ y, required when accelerating, is com plet ely fed back when
decelerating.
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-2000
-1000
0
1000
2000
3000
4000
5000
6000
7000
8000
0 0.2 0.4 0.6 0.8 1 1.2
T im e in s
P in W
P load
P average
Load power Pload and average power Paverage when moving upwards (pantograph being r aised)
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3.17 Drive for a foil feed with sin2 rounding-off
The drive will be investigated, simplif ied as a pur e flywheel drive. An 1FK6 motor is t o be used. In
order to avoid tor que steps, the speed setpoint is rounded-off according to a sin2 character ist ic
(also refer under 4.6) .
Drive data
Load moment of inertia Jload = 0.75 k gm2
Gearbox ratio i = 15.57
Gearbox moment of inertia JG= 0.000151
Gearbox efficiency ηG= 0.95
Max. motor speed nMot max = 3000 RPM
Accelerating time tb= 70 ms
Deceleration time tv= 70 ms
Constant speed tim e tk= 70 ms
No-load time tp= 0.8 s
The maximum angular velocity of the load is given by:
1
max
max 1772.20
57.1560 30002
60
2−
=
⋅⋅⋅
=
⋅
⋅⋅
=s
i
nmot
load
π
π
ω
For a sin2 rounding- off function, t he m aximum angular acceleration of the load is g iven by:
2
max
max 8.452
207.01772.20
2−
=⋅=⋅= s
tb
load
load
ππ
ω
α
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Traversing curve
maxload
ω
maxload
α
α
−
load
ω
load
α
b
tk
tv
tp
t
t
t
maxload
The tr aversing curve is symmetrical due to tb=tv.
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The ang ular velocity ωload and the angular acceler at ion αload can be r epresented as a function of
the time as follows:
Range: b
tt ≤≤0(acceleration)
)
2
(sin2
max b
loadload tt
⋅⋅
⋅=
π
ωω
)sin(
max b
loadload tt⋅
⋅=
π
αα
Range: kbb tttt +≤≤ (c onst ant speed)
maxloadload
ω
ω
=
0=
load
α
Range: vkbkb tttttt ++≤≤+ (deceleration)
)
2)(
(cos2
max v
kb
loadload tttt ⋅−−⋅
⋅=
π
ωω
)
)(
sin(
max v
kb
loadload tttt −−⋅
⋅−=
π
αα
Load torque
The following is valid for t he load t o r que:
loadloadload JM
α
⋅=
The maximum load t or que is given by:
NmJM loadloadload 6.3398.45275.0
maxmax =⋅=⋅=
α
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Selecting the mot or
Selected motor:
1FK6063-6AF71, nn=3000 RPM, JMot=0.00161 kg m 2, ηMot=0.89, kT0=1.33 Nm/A, M0=11 Nm,
I0=8.3 A, Mmax=37 Nm, I max=28 A
Motor torque
The following is valid for t he motor t or que:
)sin()
1
)(( max b
load
G
loadGmotbmot tt
i
JiJJM ⋅
⋅⋅
⋅
⋅+⋅+=
π
α
η
(acceleration, motoring)
)
)(
sin())(( max v
kb
load
G
loadGmotvmot tttt
i
JiJJM −−⋅
⋅⋅⋅+⋅+−=
π
α
(deceleration, generating )
-40
-30
-20
-10
0
10
20
30
40
0 0.2 0.4 0.6 0.8 1 1.2
Time in s
Motor torque in Nm
The highest motor t or que is achieved when accelerating, with:
maxmax )
1
)(( load
G
loadGmotbmot i
JiJJM
α
η
⋅
⋅
⋅+⋅+=
Nm37.358.452)
95.057.15 1
75.057.15)000151.000161.0(( =⋅
⋅
⋅+⋅+=
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The max. motor torque when decelerating is lower as a result of the gear box efficiency:
maxmax ))(( load
G
loadGmotvmot i
JiJJM
α
η
⋅⋅+⋅+−=
Nm14.338.452)
57.1595.0
75.057.15)000151.000161.0(( −=⋅⋅+⋅+−=
0
5
10
15
20
25
30
35
40
0 500 1000 1500 2000 2500 3000 3500
Motor speed in RPM
M mot, M perm. in Nm
The mot or t orque lies below the limiting curve. Thus, the m otor is suitable as far as the dynamic
limits are concer ned.
Thermally checking the m ot or
The RMS motor torque is given by:
pvkb
tttt
tt vmotbmot
RMS tttt
dtMdtM
M
bvkb
kb +++
⋅+⋅
=∫∫
++
+0
22
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Aft er inserting , the f ollowing is obt ained:
pvkb
tttt
tt v
kb
vmot
b
bmot
RMS tttt
dt
tttt
Mdt
tt
M
M
bvkb
kb +++
⋅
−−⋅
⋅+⋅
⋅
⋅
=∫∫
++
+0
22 max
22 max )
)(
(sin)(sin
π
π
Nm
tttt
t
M
t
M
pvkb
v
vmot
b
bmot 02.9
8.007.007.007.0 2
07.0
14.33
2
07.0
37.35
22 222 max
2max =
+++
⋅+⋅
=
+++
⋅+⋅
=
The average m ot or speed is given by, with nmot ~ ωload:
pvkb
ttt
t
ttt
tt v
kb
b
mot
pvkb
ttt
mot
average tttt
dt
tttt
dtdt
tt
n
tttt
dtn
n
bkb
b
vkb
kb
vkb
+++
⋅
⋅−−⋅
++⋅
⋅⋅
⋅
=
+++
⋅
=∫∫∫
∫+++
+
++
0
22
max
0
))
2)(
(cos)
2
(sin(
π
π
pvkb
v
k
b
mot
tttt
t
t
t
n
+++
++⋅
=)
22
(
max RPM8.415
8.007.007.007.0
)
2
07.0
07.0
2
07.0
(3000 =
+++
++⋅
=
0
2
4
6
8
10
12
0 500 1000 1500 2000 2500 3000 3500
Motor speed in RPM
Torque in Nm
M perm. S1
M RMS / n avera ge
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The RMS motor torque is, at naverage below the MS1 characterist ic. Thus, the motor is suitable as far
as the therm al lim its are concerned.
Dimensioning the brake r esist or
The mot or out put is obtained f r om t he motor tor que and the angular velocity as f ollows:
loadmotmot iMP
ω
⋅⋅=
-8000
-6000
-4000
-2000
0
2000
4000
6000
8000
0 0.2 0.4 0.6 0.8 1 1.2
T ime in s
Motor power in W
The following is valid when generating:
)
2)(
(cos)
)(
sin( 2
maxmax v
kb
load
v
kb
vmotvmot tttt
i
tttt
MP ⋅−−⋅
⋅⋅⋅
−−⋅
⋅−=
π
ω
π
The max. braking power of t he brake resist or is obt ained from the max. motor output when
generating:
INVmotvmotWbr PP
⋅⋅= maxmax
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If t he following is used
ϕ
π
=
⋅−−⋅
v
kb
tttt2)(
then the maximum of Pmot v is obtained as follows, with
))sin()cos()2sin(2)(cos)2cos(2(0 2
maxmax
ϕϕϕϕϕω
ϕ
⋅⋅⋅⋅−⋅⋅⋅⋅⋅⋅−== loadvmot
vmot iM
d
dP
for
rad5236.0)
2
1
arcsin( ==
ϕ
By inserting ϕ, t he m aximum br aking power f or the brake r esist or is given by:
)5236.0(cos)5236.02sin( 2
maxmaxmax ⋅⋅⋅⋅⋅⋅⋅−= INVmotloadvmotWbr iMP
ηηω
W5898)5236.0(cos)5236.02sin(98.089.01772.2057.1514.33 2−=⋅⋅⋅⋅⋅⋅⋅−=
The braking energy for a cycle is calculated as f ollows:
dtPW INVmot
ttt
tt vmotbr
vkb
kb
⋅⋅⋅= ∫++
+
ηη
dt
tttt
tttt
iM
v
kb
ttt
tt v
kb
INVmotloadvmot
vkb
kb
⋅
⋅−−⋅
⋅
−−⋅
⋅⋅⋅⋅⋅−= ∫++
+
)
2)(
(cos)
)(
sin( 2
maxmax
ππ
ηηω
Ws
t
iM v
INVmotloadvmot 3.202
07.0
98.089.01772.2057.1514.33
maxmax −=⋅⋅⋅⋅⋅−=⋅⋅⋅⋅⋅−=
π
π
ηηω
The following is valid for the brake resistor:
20max 5.15898 PWP Wbr ⋅≤=
5.4
3.200
8.007.007.007.0 3.202 20
P
W
T
Wbr ≤=
+++
=(ext. brake resistor)
Thus, when selecting a dr ive converter, type of construct ion Com pact Plus, an external 5 kW
brake r esist or is required ( 6SE7018-0ES87-2DC0). The chopper is integrated into the dr ive
converter.
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Selecting the drive conver ter
The drive converter is select ed accor ding to the peak- and the RMS value of the mot or cur r ent.
Refer t o example 3. 13 for a precise calculat ion of the curr ent.
The peak cur r ent is 27.1 A. T he RMS value is 6.86 A. In this case, 300 % overload capability can
be utilized for type of construct ion Com pact Plus, as the overload time is less than 250 m s, aft er
the overload current dr ops back down again to below 0.91 x of the rated current, and t he recovery
time is gr eater than 750 ms.
An approximate calculation of the mot or cur r ent results in:
A
kT
M
Imot
mot 6.26
33.1 37.35
0
max
max ==≈
A
kT
M
IRMS
RMS 78.6
33.1 02.9
0==≈
Selected SIMOVERT MASTERDRI VES MOT ION CONT ROL drive converter:
6SE7021-0EP50 (Compact Plus type of construction)
PINV n=4 kW; IU n=10 A; IU max=30 A (300 % overload capability)
A 7.5 kW drive converter could be used without utilizing the 300 % overload capability
IU max=32.8 A (160 % overload capabilit y).
0.00
5.00
10.00
15.00
20.00
25.00
30.00
0 0.2 0.4 0.6 0.8 1 1.2
Time in s
Motor current in A
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09.99 4 Information for specific applications
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SIMOV E RT MASTERDRIVES - Application Manual
4 Information for specific applications
4.1 Accelerating time for loads with square-law torque characteristics
4.1.1 General information
If t he load torque is specified in the f ollowing form for f an and pum p dr ives
MM n
n
load load
=⋅
max max
()
2
at constant m otor torque (e.g . when accelerating along the current limit in t he const ant-f lux rang e) ,
the accelerating time can be calculated using the following f ormula:
tnJ
MM
M
M
M
M
a
motor load
motor
load
motor
load
=⋅⋅
⋅⋅ ⋅+
−
π
max
max
max
max
ln
60
1
1
accelerating time from 0 to nmax [s]
J in kgm2 (moment of inertia of the motor, load and coupling ), M in Nm, n in RPM
M
Mmotor
n
nmax
Mload
Accelerating torque
Mload max
Example of the m otor- and load tor que when starting
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The difference between the motor- and load t or que acts as accelerat ing torque. After acceler at ion
to nmax t he m ot or torque and t he load t or que are the same. Acceleration is f aster, the higher the
ratio Mmotor / Mload max and the lower the total moment of inertia.
If t he load torque is not a square-load torque, or if the motor torque is not constant when
accelerating, then the accelerat ing time must be num erically calculated using the following
equation:
ta=⋅⋅−
∫
Jdn
MM
motor load
n
nn
π
30 0() ()
max
By sub-dividing the funct ion int o m segments, t he following approximation is obtained:
ta≈⋅⋅⋅ −+−⋅−
=
=
−− −
∑
JMMMM
nn
i
im
motor i load i motor i load i ii
π
30 1
211
111 1
()()
Mmotor and Mload must be available in tabular form in or der to evaluate this approximation.
M
n
nmax
ni -1 ni
Mmotor i -1
Mmotor i
Mload i -1
Mload i
Mmotor
Mload
Example for t he motor tor que- and load torque characterist ics when sub-divided into segm ent s
09.99 4 Information for specific applications
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4.1.2 Accelerating time for a fan drive
A fan is to accelerate up to nmax=4450 RPM. A gearbox with i = 1/3 is used. The power
requir ement at nmax is 67 kW. The load torq ue char acteristic and moment of inert ia of the f an ar e
known.
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
MM
load load max
/
max
n / n
Load torq ue char acteristic of the fan
The maximum load t or que is:
MP
nNm
load max max
max
.=⋅=⋅=
9550 67 9550
4450 1438
Thus, t he load t orque characteristic is approximately obtained f rom the following:
MNm
n
load ≈⋅1438 4450 2
.()
Moment of inert ia of the fan:
Jkgm
fan =25 2
A 4-pole 75 KW motor is selected ( 1LA6 280- 4AA. .).
Mn=484 Nm, nn=1480 RPM, Imotor n=134 A, Jmotor=1.4 kgm2
4 Information for specific applications 09.99
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Total m om ent of inertia r eferred t o the motor shaft:
J∗=+
J
iJ
fan motor
2
=⋅+= +=3 25 14 225 14 2264
22
...kgm
The load tor que characterist ic, r eferred t o the motor shaft, is appr oximat ely obtained as follows:
Mload
∗≈⋅⋅⋅
M
i
n
i
load
G
motor
max ()
η
4450 2
When assum ing that the g ear box ef ficiency
G is 1, the following is obtained:
Mload
∗≈⋅ ⋅ = ⋅
31438 4450 3 4314 1483
22
.(
/
).()Nm nNm n
motor motor
The following drive converter is to be used:
6SE7031-8EF60
PV n=90 kW; IV n= IV continuous=186 A
Closed-loop f r equency control type
The mot or is t o be accelerated with a current corr esponding to the cont inuous dr ive converter
current. Thus, the possible motor torque is approximated as f ollows:
MM
II
II
nmotor n
motor n n
max max
≈⋅ −
−
22
22
µ
µ
The following is obtained with IAII A
motor n motor V continuous
===134 186,max and II
n motor n
µ
=⋅035.:
MNm
max .
.
≈⋅ −⋅
−⋅=484 186 035 134
134 035 134 694
222
222
The accelerat ing time fr om n max=4450 RPM t o nmotor max=1483 RPM is given by:
ta=⋅⋅
⋅⋅ ⋅
+
−
∗
∗
∗
∗
π
nJ
MM
M
M
M
M
motor
motor load
motor
load
motor
load
max
max
max
max
ln
60
1
1
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ta=⋅⋅
⋅⋅⋅+
−=
π
1483 226 4
60 694 4314
694
4314 1
694
4314 1
686
,,ln .
.
.s
As the accelerating time is gr eat er than 60 s, the m aximum dr ive converter current cannot be used
for acceleration.
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4.1.3 Accelerating time for a blower drive (using the field-weakening range)
A blower is to be accelerated as fast as possible to nmax=3445 RPM using a dr ive converter . The
load torq ue char act eristic and the moment of inertia of the blower and coupling ar e known.
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
MM
load load max
/
max
n / n
Load torq ue char acteristic of the blower with Mload max=116.7 Nm and nmax=3445 RPM
The load tor que characterist ic is appr oximat ed as follows:
MNm
n
load ≈⋅1167 3445 2
.()
Moment of inert ia of the blower and coupling:
J J kgm kgm kgm
blower coupling
+= + =9 275 01 9 375
22 2
...
A 2-pole 90 KW motor is selected ( 1LA6 283- 2AC. .).
Mn=289 Nm, nn=2970 RPM, Imotor n=152 A, Jmotor=0.92 kgm2
Total moment of inertia:
J=+ +JJ J
blower coupling motor
=+=9 375 092 10295
22 2
...kgm kgm kgm
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The mot or should acceler ate, in the constant - flux range, up t o n n=2970 RPM with 200% of the r at ed
torq ue along the current lim it, and in the field-weakening range fr om nn=2970 RPM up to
nmax=3445 RPM, with a current limit corresponding to the torque. Acceleration is sub- divided into
two segments for the calculat ion.
Range 1 (constant-flux range)
The load tor que is obtained as f ollows at the rated motor speed:
MNm
load 2970 2
1167 2970
3445 8674=⋅ =.( ) .
M / Nm
n / RPM
578
2970
Mload
86.74
Mmotor = 2 M n
Relationships/behavior in the constant-flux range
In the constant-flux range, with Mmotor=2 Mn the accelerating time to nn=2970 RPM can be
calculated as f ollows:
ta1=⋅⋅
⋅⋅ ⋅ ⋅
⋅+
⋅−
π
2970
60 2
21
21
2970
2970
2970
J
MM
M
M
M
M
n load
n
load
n
load
ln
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ta1=⋅⋅
⋅⋅ ⋅ ⋅
⋅+
⋅−=
π
2970 10 295
60 2 289 8674
2289
8674 1
2289
8674 1
584
,
.ln .
.
.s
Range 2 (field-w eakening range)
For a constant m otor current Imotor , the mot or torque ratios are approximated by:
M
M
f
f
II
ff
II
field
const
nmotor n n
motor n
φ
µ
µ
=≈−⋅
−
22 2
22
() for f ≥ fn
As Imotor
2» In
µ
2 when starting, the expression can be simplified as follows:
M
Mffn
n
field
const
nn
φ
=≈≈ for f ≥ fn (and n ≥ nn)
Thus, t he m otor torque at nmax=3445 RPM is given by:
MMn
nMn
nNm
nnn
3445 2970 22289
2970
3445 4983≈⋅=⋅⋅=⋅⋅=
max max
.
2970 3445
Mload
Mmotor
578
n / RPM
M / N m
498.3
116.7
86.74
Relationships in the field-weakening range
The accelerat ing time in the f ield- weakening range can now be calculated as follows in one step:
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ta2=⋅⋅⋅ −+−⋅−
JMMMM
motor load motor load
π
30 1
211
3445 2970
2970 2970 3445 3445
()()
=⋅⋅⋅ −+−⋅=
10295
30 1
21
578 8674 1
4983 116 7 475 119
.(...
).
π
s
Total accelerating time:
tt t s
aa a
=+= + =
12
584 119 7 03...
Comparing the acceler at ing t im e f or direct online start ing ( w ithout drive convert er )
For comparison pur poses, the accelerating t im e t o nn=2970 RPM is to be calculated when
connected directly online to t he 50 Hz line supply. The following torq ue char acteristic is used as
basis for t he 90 KW motor with rot or class 10, MA/Mn=2 and Mstall/Mn=2.7. The starting
characterist ic is sub- divided into 15 seg m ents.
M / Nm
n / RP M
0
200
400
600
0 1000 2000 3000
800
Mmotor
Mload
Motor- and load torq ue when starting
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inM
load Mmotor ta
0 0 8.17 578 0
1 148 6 550 0.287
2 297 4.67 535 0.586
3 594 5.5 520 1.2
4 891 9 520 1.82
5 1188 13.9 530 2.45
6 1485 21.7 560 3.05
7 1782 31.2 600 3.63
8 2079 42.5 650 4.18
9 2376 55.5 710 4.69
10 2600 66.5 770 5.04
11 2673 70.3 780 5.15
12 2750 74.4 770 5.27
13 2800 77.1 740 5.35
14 2900 82.7 580 5.54
15 2970 86.7 289 5.8
Table to num er ically calculate t he st arting (acceler at ing) time
The accelerat ing time in the last column of the t able, is calculat ed using the values entered for n,
Mload and Mmotor by summing them according to the following appr oximation:
ta=⋅⋅⋅ −+−⋅−
=
=
−− −
∑
JMMMM
nn
i
i
motor i load i motor i load i ii
π
30 1
211
1
15
11 1
()()
=58.s
It is preferable to use a pr ogram such as Excel to evaluate this appr oximation formula. The direct
online starting time is somewhat shorter , but the maximum speed of 3445 RPM is not reached.
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0
500
1000
1500
2000
2500
3000
0123456
t / s
n / RPM
Speed characterist ics for direct online st ar ting up to nn=2970 RPM
Selecting the drive conver ter
The drive converter m ust be able to provide the necessary motor cur r ent to accelerate the 90 KW
motor with 200% rat ed t or que. The m otor current , as a function of the motor t or que is
approximately given by:
IM
MIII
motor nmotor n n n
≈⋅−+()( )
22 2 2
µµ
The following is obtained with IA
motor n =152 and II
n motor n
µ
=⋅025. f or t he 2-pole 90 KW motor:
IA
motor ≈⋅ − ⋅ + ⋅ =2 152 025 152 025 152 2968
2 2 22 22
(. ). .
Drive converter selected:
6SE7032-6EG60
PV n=132 kW; IV n=260 A; IV max=355 A
Closed-loop f r equency control type
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4.1.4 Re-accelerating time for a fan drive after a brief supply failure
A fan dr ive is to accelerate to its previous (old) speed after a brief supply f ailur e. The maximum
power failure duration is 2 s. As a re sult of the re lat ively high drive moment of iner t ia, the kinetic
buffering function is used to buffer the supply failure. Thus, the dead-times, which would occur if a
restart - on-the-fly circuit was to be used, are eliminated: They include the motor de- excitat ion/
excitation times, t he time to establish the power supply voltages and initialize the m icr opr ocessor
system as well as the search time (t he search time is eliminated when a tachometer/encoder is
used).
Drive data
Rated motor output Pmotor n = 75 kW
Rated motor cur rent (at 500 V) Imotor n = 102, 4 A
Rated motor torque Mmotor n = 241 Nm
Maximum f an speed nmax = 2970 RPM
Power requir em ent at nmax P
max = 44.9 kW
Moment of inert ia of the fan Jfan = 4.75 kgm 2
Moment of inert ia of the mot or Jmotor = 0.79 kgm2
The maximum load t or que is given by:
MPnNm
load max max
max
..=⋅=⋅=
9550 449 9550
2970 1444
Thus, t he load t orque characteristic is approximated by:
MNm
n
load ≈⋅1444 2970 2
.()
Total moment of inertia:
JJJ kgm
total fan motor
=+ =+=475 079 554 2
...
To star t off with, t he m otor speed must be defined at the inst ant that the line supply retur ns. The
worst case condition is assumed with a maximum 2 s line supply failur e and m aximum speed
when the power actually failed. The speed dip is t he highest at m aximum speed, as in t his case,
the highest load t orque is ef fective.
The k inet ic buffering is active during t he line supply failure. The motor speed is further decr eased,
in addition to the effect of t he load t or que, to cover the losses in t he motor, inverter and in the
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power supply. The additional brak ing torq ue as a r esult of the losses can be neglected here, as
the speed decrease during the line supply failure is low due to the relatively high moment of inertia
and in the vicinity of nmax, the load t or que component is the m ost signif icant.
With this assumption, the following equation is obtained for the speed characterist ic dur ing line
supply failure, starting from nmax, in the vicinity of nmax:
n
n
M
Jn
t
load
total A
min
max
max
max
≈+⋅⋅⋅ ⋅
1
130
2
π
The following is obtained f or tA=2 s (line supply failure time)
n RPM
min .
.
≈+⋅
⋅⋅ ⋅=
1
1
2970 30 144 4
554 2970 22544
2
π
The following drive converter is to be used:
6SE7031-3FG60
PV n=90 kW; IV n=128 A; IV max=174 A
Closed-loop f r equency control type
The mot or is t o be accelerated with a current corr esponding to the m aximum dr ive converter
current. The possible motor torque is approximat ed by:
MM
II
II
nmotor n
motor n n
max max
≈⋅ −
−
22
22
µ
µ
The following is obtained with IAIIA
motor n motor V
===102 4 174., max max and II
n motor n
µ
=⋅028.:
MNm
max ..
.. . .≈⋅ −⋅
−⋅ =241 174 028 1024
1024 028 1024 4207
222
222
The mot or can t herefore develop 420.7/ 241=175% r ated torq ue when re-acceler at ing up to n max.
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The re-accelerating time from nmin=2544 RPM to nmax=2970 RPM is given by:
t
H=⋅⋅
⋅⋅ ⋅+
−−+
−
π
nJ
MM
M
M
M
M
M
Mn
n
M
Mn
n
total
motor load
motor
load
motor
load
motor
load
motor
load
max
max
max
max
max
min
max
max
min
max
(ln ln )
60
1
1
=⋅⋅
⋅⋅
⋅+
−−+
−=
π
2970 554
60 4207 1444
4207
1444 1
4207
1444 1
4207
1444 2544
2970
4207
1444 2544
2970
084
.
..
(ln
.
.
.
.
ln
.
.
.
.
).s
Thus, t he overall t ime from the start of the line supply failur e up to when nmax is ag ain reached is
given by:
ttt s
total A H
=+=+ =
2084284..
0
500
1000
1500
2000
2500
3000
0 0.5 1 1.5 2 2.5 3
n / RPM
t / s
Speed characterist ic for line supply failur e ( 0 to 2 s) and re-st ar ting (2 to 2.84 s)
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Taking into account the losses dur ing kinetic buffering
At low speeds, the brak ing torq ue due to the losses can no longer be neglected with respect to t he
load torq ue. Thus, the speed characteristic should now be determined during the kinet ic buffer
period by estimating the losses.
Motor losses at the rat ed oper at ing point:
PPPkW
V motor n motor n
motor motor n
=−=−=
η
75
0944 75 4 45
..
Under the assumpt ion, t hat the motor is in a no- load condition during the kinetic buf fering phase,
and that the losses ar e then constant and approximat ely only half as high as they are at the rated
operating point , then the following is obtained:
PPkW
V no load V motor n
−≈==
2445
2223
..
If t he losses in the inverter and in the power supply are neglect ed, t hen the additional braking
torq ue is given by:
Mbrake ≈⋅=⋅⋅= ⋅
−−
P
n
P
nnnMnn
V no load V no load brake n
9550 9550
max
max max
max
≈⋅⋅= ⋅
223 9550
2970 72
..
max max
nnNm nn
The braking tor que becomes higher at lower speeds.
With this basis (assumption) for Mbrake, t he speed char acteristic during line supply failure, star t ing
from nset, is given by:
tc aan n
an na
a
set set
set
set
≈⋅⋅ −⋅ +
++⋅ ⋅−
(ln( ()
) arctan( )
1
61
3
2
3
22
2
−⋅ −⋅+
+−⋅ ⋅−1
61
3
2
3
22
2
ln( ()
) arctan( ))
aann
an na
a
with
an M
Mbrake n
load
=max max
max
3cnJ
MM
total
load brake n
=⋅⋅
⋅⋅
π
max
max max
30 2
3
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This formula can no longer be used for speeds less than approximately 25 to 30% of t he m aximum
speed, as the load tor que characterist ic appr oximat ion is too inaccurate (influence of t he friction) .
As the f or m ula cannot be changed-over for n, nmin must be found by insert ing various values for
n, with subsequent int er polat ion.
Example for nset=750 RPM
With
aRPM==2970 72
1444 1093
3..cs=⋅⋅
⋅⋅
=
π
2970 554
30 144 4 7 2 3242
2
3
.
.. .
and ∆n=25 RPM, then t is shown in the following table
int
0 750 0.00
1 725 0.38
2 700 0.76
3 675 1.13
4 650 1.50
5 625 1.86
6 600 2.22
The following value is obtained for nmin by linearly interpolating the values above and below tA=2 s
(1.86 s and 2.22 s) :
nn
nn
ttt t RPM
Amin () ..
(. ) .≈+−
−⋅− = + −
−⋅−=
656
65 6600 625 600
222 186 2 22 2 6153
The re-accelerating time from nmin=615.3 RPM to nset=750 RPM is then given by:
t
H=⋅⋅
⋅⋅ ⋅+
−−+
−
π
nJ
MM
M
Mn
n
M
Mn
n
M
Mn
n
M
Mn
n
total
motor load
motor
load
set
motor
load
set
motor
load
motor
load
max
max
max max
max max
max
min
max
max
min
max
(ln ln )
60
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t
H=⋅⋅
⋅⋅
⋅+
−−+
−=
π
2970 554
60 4207 1444
4207
1444 750
2970
4207
1444 750
2970
4207
1444 6153
2970
4207
1444 6153
2970
019
,
..
(ln
.
.
.
.
ln
.
..
.
..).s
Thus, the total t im e from the start of the line supply f ailur e unt il nset is again reached is obtained
from:
ttt s
total A H
=+=+ =
2019219..
0
250
500
750
00.5 11.522.5
n / RPM
t / s
without taking into account the losses
taking into account the losses
Speed characterist ic for a line power failur e (0 to 2 s) and re- acceler at ion ( r e- s t ar ting) (2 to 2.19 s)
taking into account the losses dur ing kinetic buffering (for comparison pur poses, a cur ve is also
shown where the losses were not taken into account)
4 Information for specific applications 09.99
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Estimating n
min
and t
H
nmin and tH can be estimated for shor t line supply failure tim es and high moments of inertia.
Characterist ics when coasting down
With the gradient of the speed characteristic at t he point n=nset
dn
dt
Mn
nMn
n
J
nn
load set brake n set
total
set
==− ⋅+ ⋅
⋅
max max
max
() max
2
30
π
then nmin is g iven by:
nnt
dn
dt
set A nn
set
min ≈+⋅=
If nset is close to nmax, then t he appr oximat ion used ear lier can be applied:
n
n
M
Jn
t
set
load
total A
min max
max
≈+⋅⋅⋅ ⋅
1
130
2
π
Characterist ics when re-acceler at ing
With the gradient of the speed characteristic at t he point n=nmin
dn
dt
MM n
n
J
nn
motor load
total
==−⋅
⋅
min
max min
max
()
2
30
π
then tH is g iven by:
tnn
dn
dt
Hset
nn
≈−
=
min
min
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Example for n set=750 RPM and tA=2 s (as previously calculated)
dn
dt RPM s
nn
set
==− ⋅+⋅
⋅=−
1444 750
2970 72 2970
750
554
30
65
2
.( ) .
./
π
nRPM
min ()
≈+⋅−=
750 2 65 620
dn
dt RPM s
nn==−⋅
⋅=
min
..()
../
4207 1444 620
2970
554
30
714 3
2
π
ts
H≈−=
750 620
714 3 0182
..
Thus, t he t otal time f r om the start of the line supply failure up to when nset is again reached is
given by:
ttt s
total A H
=+=+ =
2 0182 2182..
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4.1.5 Calculating the minimum braking time for a fan drive
A fan dr ive is to be braked from nmax=1500 RPM to standst ill using the existing brake resistor . I t is
assumed that t he load has a square-law characterist ic.
Drive data
Rated motor output (4 pole) Pmotor n = 250 kW
Rated motor current Imotor n = 430 A
Rated motor torque Mmotor n = 1600 Nm
Maximum f an speed nmax = 1500 RPM
Power requir em ent at nmax P
max = 198 kW
Moment of inert ia of the fan Jfan = 223 kgm2
Moment of inert ia of the mot or Jmotor = 3.6 kgm2
Motor efficiency ηmotor = 0.962
Rated converter out put s PU n = 250 kW
Brake r esist or power using an ext. brak e r esistor P20 = 100 kW
Peak braking power 1.5 P20 = 150 kW
The fan load torque helps to brake the drive - especially in the hig h speed r ange. T o optimally
utilize the brak e r esistor, it is t herefore more favorable t o br ake with constant decelerat ion and not
with constant brak ing torq ue (operation at t he cur rent limit) . When decelerating with a constant
braking torq ue, a higher peak braking power would be generated for the same brak ing time (refer
to the compar ison later on in the text) .
To optim ally utilize the brake resistor , the peak br aking power and the braking energy are
important. Both values must remain below the permissible limits.
Braking with constant deceleration
Calculating the peak braking power for t he brake resistor
The mot or torque when braking is given by:
MMJ
d
dt
motor br load
=+⋅
ω
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Whereby
MM n
n
load load
≈⋅
max max
()
2
MPnNm
load max max
max
.=⋅
⋅=⋅
⋅=
60
2198000 60
2 1500 12605
ππ
d
dt n
tbr
ω
π
=− ⋅⋅
260 max (const ant deceleration)
Thus, t he m otor torque is given by:
MM n
nJn
t
motor br load br
=⋅−
⋅⋅
⋅
max max
max
()
22
60
π
and the motor output by:
Pmotor br =⋅
⋅
Mn
motor br 260
π
=⋅ ⋅ −
⋅⋅
⋅⋅
2
60 2
60
2
π
π
(() )
max max
max
Mn
nJn
tn
load br
The maximum m ot or output when braking, for
nJn
Mt
load br
=⋅⋅
⋅⋅ ⋅
2
360 1
3
π
max
max
is given by
PJn Jn
Mt
motor br load br
max max max
max
=− ⋅⋅ ⋅⋅⋅
⋅⋅
ππ
22
3
4050 10 1
With a specified peak br aking power Pbr W max for the brake r esist or , and the f ollowing
relationship
PP
motor br br W
motor
max max
=
η
Then the m inimum permissible brak ing time can be calculated as follows:
t
br min =⋅⋅ ⋅ ⋅⋅
⋅⋅ ⋅
Jn n
MP
motor
load br W
ππη
max max
max max
22 2
22
310 4050
=+⋅⋅⋅ ⋅⋅
⋅⋅⋅ =(.) .
..223 36 1500 1500 0962
10 4050 12605 150000 1752
22 2
22
3
ππ
s
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Calculating the braking ener gy when braking
It must now be checked as to whether the permissible braking energy is exceeded. For a constant
deceleration, t he speed is given by:
nn
t
tbr
=⋅−
max ()1
and thus the mot or output when braking
PMn
t
tJn
tt
t
motor br load br br br
=⋅ ⋅⋅− −
⋅⋅
⋅⋅−
2
60 12
60 1
32
ππ
(() ())
max max max
Pmotor br
Pmotor br max
t
tt
0br
Negative area corresponds
to the braking ener gy
Example for t he motor output when brak ing with a square-law load torq ue characteristic
The braking energy for the br ake resistor is then given by:
W
br =⋅ ⋅
∫
η
motor motor br
t
tPdt
br
0
=⋅⋅⋅⋅⋅− −⋅⋅
⋅⋅ ⋅−
ηπ π
motor br load
br br br
tMn t
tJn
tt
t
2
60 4 12
260 1
04202
(() ())
max max max
Time t0 is obtained by setting Pmotor br to zero:
tJn
Mt
t
load br br012
60
=− ⋅⋅
⋅⋅
⋅()
max
max
π
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If t0 is negative, a zero should be entered f or t 0 in t he br aking energy equation. I n this case, even
at the star t of braking, t he deceler ation torque exceeds the load t orque and t0 is g iven by:
ts
012266 2 1500
60 12605 1752 1752 4 72=− ⋅⋅
⋅⋅⋅=−(...
). .
π
t0 is therefore set to zero when calculating t he braking ener gy, and the following is obt ained:
WkWs
br =⋅⋅ ⋅⋅−⋅⋅
⋅⋅ =−
0962 2 1752
60 12605 1500
42266 2 1500
260
17
52 1855
2
..
(..
.)
ππ
The perm issible braking energy for t he external brake resist or is, for the 90 s interval:
WP s kWs
br perm br W time.=⋅=⋅=90 25 90 2250
Thus, t he second condit ion is also fulf illed for inter vals, which are not less t han
TW
Wss
br
br perm
=⋅=⋅=
.
.90 1855
2250 90 742
Diagrams for m otor output and m ot or torque when br aking with constant deceler at ion
Motor breaking power for tbr=17.52 s
Time in s
Motor braking power in kw
-160
-140
-120
-100
-80
-60
-40
-20
00 2 4 6 8 10 12 14 16 18
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Motor torque for tbr=17.52 s
Time in s
Motor torque in Nm
-2500
-2000
-1500
-1000
-500
002468101214
16 18
Checking the drive convert er dimensioning
The maximum m ot or torque when braking occurs at n = 0, and is given by:
MJn
tNm
motor br br
max max ..
=− ⋅⋅
⋅=− ⋅⋅
⋅=−
2
60 2266 2 1500
60 1752 2032
π
π
For the hig hest r equired motor torque when braking, an approximat e m otor current is given by:
IM
MIII
motor motor
motor n motor n n nmax max
()( )≈⋅−+
22 2 2
µµ
With II
n motor n
µ
=⋅031. (assumpt ion) , the following is obtained:
IA
motor max ()( . ).≈⋅−⋅+⋅=
2032
1600 430 031 430 031 430 536
2 2 22 22
Drive converter selected:
6SE7035-1EK60
PV n=250 kW; IV n=510 A; IV max=694 A
This drive converter is adequately dimensioned for br aking.
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Comparison, br aking with a constant m ot or torque
The maximum br aking power is then always at n=nmax. The fo llowing is obt ained for t he r equired
motor t orque using the specified peak rating of the brake resistor :
MP
nNm
motor br br W
motor
=− ⋅
⋅⋅ =− ⋅
⋅⋅ =−
max
max .
60
2150000 60
2 1500 0962 993
πη π
Thus, t he m inim um br aking tim e is obt ained as follows:
t
br min =⋅⋅
⋅⋅
⋅
Jn
MM
M
M
load motor br
load
motor br
2
60
π
max
max
max
arctan
=⋅⋅
⋅⋅
⋅=
2266 2 1500
60 12605 993
12605
993 269
.
.arctan ..
π
s
Thus, with the same peak brak ing power, the brak ing time is long er , than when braking with
constant mot or torque. The mot or out put during braking is g iven by:
PM n
motor br motor br
=⋅
⋅260
π
with
nn
M
M
M
M
MM
Jn t
load
motor br
load
motor br
load motor br
=⋅ −
⋅⋅
⋅⋅ ⋅
max
max
max max
max
tan(arctan( ) )
60
2
π
In this case, the braking energy can only be numerically calculated. When sub-dividing into m
segment s, for example, t he following approximation can be used:
WPP tt
br motor motor br i motor br i
i
im
ii
≈⋅ +⋅−
+
=
=
+
∑
η
1
11
2()
The following is obtained by sub-dividing into 10 segment s:
W
Ws
br ≈−1739
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itP
motor ≈⋅
∫
η
motor motor
Pdt
0 0 -155.93 0
1 2.69 -131.64 -371.91
2 5.38 -110.98 -685.69
3 8.07 -92.94 -949.42
4 10.76 -76.85 -1169.02
5 13.44 -62.21 -1348.87
6 16.13 -48.63 -1492.21
7 18.82 -35.85 -1601.47
8 21.51 -23.61 -1678.37
9 24.20 -11.72 -1724.06
10 26.89 0.00 -1739.22
Table to num er ically calculate t he br aking energy
Diagrams for m otor output and m ot or speed when braking wit h a const ant motor t or que
Motor output for tbr=26.9 s
Time in s
-160
-140
-120
-100
-80
-60
-40
-20
00 5 10 15 20 25 30
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Motor s peed for tbr=26.9 s
Time in s
0
200
400
600
800
1000
1200
1400
1600
0 5 10 15 20 25 30
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Braking with constant deceleration and subsequent torque limiting
When br aking with constant deceler at ion, the motor t or que increases as the motor speed
approaches zero. Under certain circumstances, it is practical to limit the torque, f or inst ance, to the
rated mot or torque. Nat ur ally the br aking tim e is t hen extended. In order that t he br ake resistor is
optimally utilized, t or que limiting should only be used, if the m aximum m otor output was reached
when braking.
The maximum m ot or output when braking occurs at :
nJn
Mt TPM
load br
=⋅⋅
⋅⋅ ⋅ =⋅⋅
⋅⋅=
2
360 2266 2 1500
180 12605 1752 10995
33
ππ
max
max
... .
The value, previously calculated when braking with constant deceleration without torque limiting , is
inserted for tbr . The associated mot or t or que is given by:
Mmotor br =⋅−
⋅⋅
⋅
Mn
nJn
t
load br
max max
max
()
22
60
π
=⋅ −
⋅⋅
⋅=−12605 10995
1500 2266 2 1500
60 1752 1354 4
2
.( .)...
π
Nm
The mot or t orque at n=0, without t or que limiting , is given by:
MJn
tNm
motor br br
=− ⋅⋅
⋅=− ⋅⋅
⋅=−
2
60 2266 2 1500
60 1752 2032
π
π
max ..
Thus, t or que limiting can lie between -1354.4 Nm and -2032 Nm. I n t his particular case, the motor
torq ue should be lim it ed to the rated t or que (-1600 Nm) when braking.
Mmotor
Mn
-
t
tt
1br
´
Example for torque lim iting when braking (in this case t o t he rated tor que)
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Initially, t he speed n 1 at t 1 m ust be calculated.
n1=⋅ ⋅
⋅⋅
⋅−nMJn
tM
load br nmax max
max
()
12
60
π
=⋅ ⋅ ⋅⋅
⋅−=1500 1
12605 2266 2 1500
60 1752 1600 8778
.(..).
π
RPM
Thus, t1 is given by:
tt n
ns
br11
117521
8778
1500 727=⋅− = ⋅− =().(
.).
max
The following is obtained for t he „ extended“ br aking tim e:
tbr
′=+ ⋅⋅
⋅⋅
⋅⋅tJn
MM
M
Mn
n
load n
load
n
11
2
60
π
max
max
max
max
arctan( )
=+ ⋅⋅
⋅⋅
⋅⋅=727 2266 2 1500
60 12605 1600
12605
1600 8778
1500 19 28. .
.arctan( ..
).
π
s
For t0=0, up to time t1, t he br aking energy is given by:
W
br 1=⋅⋅⋅⋅⋅−− −⋅⋅
⋅⋅ ⋅−−
ηπ π
motor br mload
br br br
tMn t
tJn
tt
t
2
60 4 11 2
260 11
14212
( (()) (()))
max max max
=⋅⋅ ⋅⋅⋅−− − ⋅⋅
⋅⋅ ⋅−−
0962 2 1752
60 12605 1500
411727
15
52 2266 2 1500
2601752 11727
1552
422
..
(.(( ..)) ..(( ..)))
ππ
=−1032 kWs
To calculate t he br aking energy in the range t1 to t´br the motor output in t his range must first be
determined. The following is tr ue:
PM
n
motor br n
=− ⋅ ⋅260
π
with
nn
M
M
M
Mn
n
MM
Jn tt
load
n
load
n
load n
=⋅ ⋅−
⋅⋅
⋅⋅ ⋅−
max
max
max
max
max
max
tan(arctan( ) ( ))
11
60
2
π
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In this case, the braking energy can only be numerically determined. The following is obtained
using the pr evious approximation and by sub-dividing int o 10 segments:
W
kWs
br 2815≈−
Thus, t he t otal braking energy is given by:
W
kWs
br ≈− − =−1032 815 1847
itP
motor ≈⋅
∫
η
motor motor
Pdt
0 7.27 -147.05 0
1 8.47 -130.23 -160.11
2 9.67 -114.14 -301.22
3 10.87 -98.66 -424.10
4 12.07 -83.69 -529.40
5 13.27 -69.14 -617.65
6 14.47 -54.92 -689.28
7 15.67 -40.97 -744.66
8 16.87 -27.21 -784.02
9 18.07 -13.57 -807.57
10 19.27 0.00 -815.41
Table to num er ically calculate t he br aking energy
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Motor output and motor tor que diagr am s
Motor output for t´br=19.28 s
Time in s
-160
-140
-120
-100
-80
-60
-40
-20
00 2 4 6 8 10 12 14 16 18 20
Motor tor que for t´br=19.28 s
Time in s
-1800
-1500
-1200
-900
-600
-300
0024681012141618
20
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4.2 Connecting higher output motors to a drive converter than is
normally permissible
4.2.1 General information
Motor and drive converter ar e opt imally utilized when the induction motor is assigned to the drive
converter, as specified in the Catalog and Engineering Manual. However, in special cases, it may
be necessary to use a higher-rating motor than is generally permissible with the particular drive
converter, t aking into account a poor er utilization level. Reasons could include:
•A large m ot or is only to be oper at ed under no-load conditions f or test purposes
•The mot or is only operat ed at par tial load, e.g. derating factors up to 0.5 are obtained for
Mpermissible/Mn for a speed cont r ol r ange of 1: 100 and utilization to temper at ur e r ise class B.
In cases such as these, a standard motor / dr ive converter assignment could result in a drive
converter with an excessive rating and thus far higher costs.
The f ollowing must be observed when connecting a hig her-rating induction motor to the drive
converter than is per m issible according to the Cat alog or Engineering Manual:
•Higher current peaks are obtained due to the lower leakage inductance of higher-rating
motors, which under certain circumst ances, could cause the drive converter to be tr ipped by
overcurrent. Thus, the maximum dr ive converter cur rent must be r educed cor r esponding to the
higher r at ed m otor current . Anot her possibility is to compensate t he low leakage inductance
using an additional output reactor, or to increase the leak age inductance using ∆/Υ
changeover.
•At constant flux, the mot or power factor deter ior ates in partial load operation. The transferrable
drive converter active power is then decreased, especially f or m ot ors with a high magnet ization
current com ponent . Thus, in t he par t ial load r ange, a larg er m otor cannot g ener at e the same
shaft out put as the adapted smaller m ot or.
•The mot or dat a, required when vector contr ol is used, may, under certain cir c um st ances, no
longer be able t o be set when a larg er m otor is used. T hus, a sm aller motor must be
parameter ized. The motor m odel values are correctly set via the motor ident ification function.
•If t he m otor is only to be essentially operated under no- load conditions, it mainly draws reactive
current. This reactive current is not supplied from the line supply, but f rom the DC link
capacitors. For r easons of stabilit y, t he r eact ive current component should not be m or e t han
approx. 60% of the rated drive converter cur r ent .
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Reduction factor if the leakage inductance of the m otor is too low
As long as the m ot or leakage reactance does not f all below approximately 10% of t he r ated drive
converter impedance, oper at ion is uncr it ical. This is the case f or the standard induction motors
listed in the Catalog and Engineering Manual. When connecting a higher-rating motor without
taking any measures (refer below), the following rated motor current limit should not be exceeded
in order to pr event higher peak cur r ent s:
Imotor n ≤ 136.⋅IUn
Further, the maximum drive converter curr ent should be r educed as follows:
III
I
I
VVnVn
motor n
Vn
max ..()=⋅−⋅⋅ −136 06
21for II I
Vn motorn Vn
≤≤⋅136.
Compensating the leakage induct ance using an additional output react or
If t he m otor leakage inductance is to be raised t o at least 10% of the drive converter rated
impedance using an addit ional out put r eactor, then t he following must be true:
XX Z U
I
motor reactor V n Vn
Vn
σ
+=⋅=⋅
⋅
01 01 3
..
If t he m otor leakage is assumed to be 10%, t hen the reactor induct ance is given by:
LU
II
reactor Vn
Vn motorn
≈⋅⋅
⋅⋅−
01 10
3 314
11
3
.() [mH]
As the standard out put reactors ar e designed for appr ox. 1. 2 to 0.6% of the rated drive converter
impedance, then, for example when connecting a mot or with twice the rating as norm ally
permissible, approximat ely 4 to 8 st andar d output reactor s would be req uir ed.
Increasing the leakage induct ance using
∆
/
Υ
motor w inding changeover
The leak age inductance is increased by a factor of 3 using ∆/Υ changeover of t he m otor winding.
Thus, a m ot or with 3 tim es t he out put normally permissible can be fed from the converter under
partial load conditions. However, the per m issible m ot or output is then only one third of the rat ed
motor out put.
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Permissible motor torque
The following approximation is valid for t he per m issible m ot or torque referred to the rated m ot or
torque for a higher-rating motor:
M
M
I
I
I
I
I
I
I
I
perm
nn
Vperm
mot n mot n
n
mot n
.
.
()()
()
≈⋅ −
−
µ
µ
µ
µ
22
2
1
IVperm.: permisible drive converter current at the oper at ing point
Imot n : r ated motor cur r ent
I
µ
: m otor magnetizing current at the operating point
I
n
µ
: r at ed motor magnetizing current
Example 1
A motor with 300% output of that nor m ally perm issible in a Υ circuit configur ation (660 V) is to be
fed from a 380 V drive converter . The rated motor current in the ∆ circuit configurat ion should be
3⋅
I
Un and in the Υ circuit conf iguration 3⋅IVn . Under the assumption t hat :
II
Vperm Vn.= → I
I
Vperm
mot n
.=1
3
and with
I
I
n
µµ
≈3 (field weakening as a r esult fof the Υ circuit)
the permissible motor torque is given by:
M
M
I
I
I
I
perm
n
n
mot n
n
mot n
.
()
()
≈⋅−⋅
−=
1
3
1
31
3
1
1
3
2
2
µ
µ
The possible output at rated speed is therefore 1/3 of the rat ed m otor output.
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Example 2
A 4-pole 7.5 k W motor is to be fed from a 5.5 kW drive converter as fan/pump drive. The
continuous drive converter curr ent (rated dr ive converter current) is fully utilized.
Rated motor current: 15. 4 A
Rated magnetizing curr ent : 53%
Rated converter cur r ent: 13.2 A
The condition II
motor n V n
≤⋅136. is in this case fulfilled.
With
I
I
n
µµ
= the permissible motor torque at rated speed is given by:
M
M
perm
n
.(.
.).
..≈−
−=
132
154 053
1053 079
22
2
The output at rated speed is then 079 75 59.. .
⋅=
kW kW . 5.5 kW can be achieved as fan/pump
drive using the adapted 4-pole 5.5 kW motor connect ed to the 5.5 k W drive converter.
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4.2.2 Operating a 600 kW motor under no-load conditions
Aft er ser vice work, t raction motors ar e t o be checked for symm et rical (balanced) phase currents
under no-load conditions and mechanical running characteristics. A motor is accelerated up to
maximum speed using a PWM drive converter connect ed to the 400 V line supply.
Motor data
Rated output Pn= 600 kW
Rated voltage Vn= 1638 V
Rated current In= 266 A
Rated speed nn= 1670 RPM
Rated f r equency f n= 85 Hz
Max. speed nmax = 3145 RPM
Max. f r equency fmax = 160 Hz
Leakag e induct ance Lσ= 2.04 mH
Rotor resistance R2´ = 0.0691 Ω
No-load current at the rated oper at ing point Iµ n = 85 A
Estimating the r equired drive converter out put according to the no- load cur r ent
The mot o r voltage at 50 Hz is:
VV
Hz50 50
85 1638 963≈⋅ =
As the drive converter can only supply 400 V, the mag netizing current is given by:
II A
n
µµ
≈⋅=⋅=
400
963 400
963 85 35
The magnetizing current com ponent should not exceed 60% of the r ated drive converter current .
Thus, t he dr ive converter current is given by:
IA
Vn =⋅=
100
60 35 583.
According t o t his cr iterium, t hen at least a 30 kW drive converter with
I
Un=59 A is required.
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Estimating the stall torque at 160 Hz
The mot or st all torque can be approximat ed as follows:
MMs
s
s
s
stall n nn
K
K
n
≈⋅+
2()
With
snn
n
nn
=−=−=
0
0
1700 1670
1700 00176.
sR
XX
R
fL
Kn
=+=⋅⋅ ⋅ =⋅⋅ ⋅ ⋅ =
′
′
′
−
2
12
23
200691
2 85 2 04 10 00634
σσ σ
ππ
...
The following is obtained
MNm
stall n ≈⋅ + =
3430
200176
00634 0 0634
00176 6654(.
..
.)
The mot or is fed f r om a dr ive converter with constant voltag e from 50 Hz to 160 Hz. The following
is obtained f or t he stall torq ue at 160 Hz:
MM
V
V
f
fNm
stall Hz stall n Hz
transition
() max
max
()( ) ()()
160 50
22 22
6654 400
963 50
160 113≈⋅ ⋅ =⋅⋅=
Thus, with a 1.3 safety marg in from t he st all torque, only 87 Nm is available f or acceler at ion. This
represents 2. 54% of the rated motor t or que. The associat ed acceler ating power is thus given by:
PMn kW
bb
=⋅=⋅=
max .
9550 87 3145
9550 286
A 37 kW drive converter is to be used (6SE7027- 2ED61 with IV n=72 A).
Permissible motor leakage inductance
The perm issible motor leakage inductance is given by:
LU
ImH
Vn
Vn
σ
≥⋅⋅
⋅⋅=⋅⋅
⋅⋅
=
01 10
3314
01 400 10
3 314 72 102
33
...
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The mot or leakage inductance of 2.04 mH is t her efore sufficiently hig h ( t his is a function of the
high motor rated motor voltage).
Closed-loop control t y pe and set t ing
Closed-loop f r equency control is pref erable due to the impr oved stabilit y. When parameterizing the
system, a 6-pole 30 kW m ot or ( 1LA6 223-6) can be used as basis.
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4.3 Using a transformer at the drive converter output
4.3.1 General information
Normally, t he r ated motor voltage and the maximum drive converter output volt age are
harmonized with one another, thus ensuring optimum drive converter utilization. O ccasionally, t he
rated mot or voltage and maximum drive converter output voltage are not the same. In t his case, it
may be necessary to adapt the voltag e using an output transformer, e.g.:
•Operating processing m achines with a lower line supply voltage, e.g. grinding machines, which
up until now, were fed from a mot or - generator set .
•To step- up t he m otor voltage to keep the cable losses low if long m otor feeder cables are
used.
The following must be observed when using a transformer at the drive converter output:
•If t her e is no sinusoidal filter between the drive converter and transformer, t he transformer
must be desig ned for operat ion at the converter output . Cont rary to a transformer connect ed to
the line supply, the frequencies are higher (pulse frequency), and t he cur r ent spikes. I n or der
to prevent the effects of sat ur at ion due t o t he high current spikes, and as a result of possible
current dissymmetries (i.e unbalance), the inductance of the transformer m ust be utilized to a
lower level than otherwise.
•DC current br aking is no long er possible as t he t r ansformer can’t conduct DC current .
•The achievable motor starting t or que is generally lower than if a transformer was not used, as
the transformer can only effectively transform above a specif ic m inim um freq uency. For rated
torq ue st ar ting, the transfor mer must be designed to operate, for example, at t he rated motor
slip fr equency.
•The drive converter m ust additionally supply the losses as well as the transformer magnetizing
current.
•When using an autotransformer , it is not permissible to g r ound the transformer neut r al ( s t ar )
point, as this would represent a ground f ault for the drive converter. An isolating transformer is
pref er able if long m ot or feeder cables are used, as otherwise, the influence of the cable
capacitance with respect to ground would be significant ly reduced ( lower charg ing currents due
to the almost sinusoidal voltages, even with respect to gr ound on the motor side) .
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•When a sinusoidal filter is used, t he m aximum output voltage is only approx. 85% of the line
supply voltage. The t r ansformer r atio must take this into account. Depending on the drive
converter output , it may even be necessary to de-rate it due t o t he high pulse fr equency
requir ed (6 kHz at 380 to 460 V) .
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4.3.2 Operating a 660 V pump motor through an isolating transf ormer
A 23 kW pump motor , located in a well, is connected to a 30 kW drive converter installed 1 km
away in an equipment room. In order to keep the motor cable losses as low as possible, the motor
is connected in a Υ circuit configuration at 660 V. I n order to adapt t he maximum 380 V drive
converter output voltage, an isolating transfor mer is used. In t his case, it is not cost- eff ective to
use a 660 V drive converter due to the low motor output. Fur ther, a sinusoidal filter is required
between the drive converter and isolating transformer due to t he 1000 m feeder cable and as a
result of the high 660 V m otor voltage (a t hird-party motor is being used). 1000 m feeder cables
are permissible as the isolating tr ansformer de- couples the motor cable ground capacitances. The
isolating t ransform er ratio was selected to compensate the 15% lower output voltage when using
the sinusoidal f ilter. De-rating is not required f or the 30 kW drive converter at a 6 k Hz pulse
frequency.
IM
Isolating transformer
323/660 V, 33 kVA
380 V
660 V, 23 kW
pump moto r
Approx. 1 km
cabl e length
6SE7026-
0ED61
30 kW drive
converter
Sinusoidal
filter
C, D
~
~
Block diag ram
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4.3.3 Operating 135V/200 Hz handheld grinding machines through an isolating
transformer
Handheld grinding machines equipped with three- phase m otors are to be fed f r om a dr ive
converter which is connected to a 380 V supply. In t his case, t he convert er acts as three-phase
supply with a f ixed voltage and frequency. A m aximum of 9 motor s can be connect ed t hrough
sockets. The motors can eit her run individually or all tog ether. An isolating transformer is used t o
match the voltage. Further , a sinusoidal filt er is used as t he m otors are not designed for convert er
operation. I n addition, EMC disturbances/noise via the motor cables ar e significant ly reduced.
Ground curr ent s flowing through cable capacitances are essentially eliminated using the isolating
transformer.
Motor data
Rated voltage Vn= 135 V
Rated f r equency f n= 200 Hz
Rated speed nn= 6000 RPM
Rated output/rated curr ent of the mot ors to be connected:
2 motors with 260 W / 1.41 A
6 motors with 1000 W / 5.6 A
1 motor with 1800 W / 8. 6 A
It is assumed, that an 1800 W motor will be switched-in, in addition to 2 x 260 W motors and
6x1000 W motor s r unning at rated load. The starting current ( inr ush cur r ent) is 600% In. Thus, the
maximum current is given by:
IAAAA
max .. .≈⋅ +⋅ +⋅⋅ =2141 656 6186 88
If all of the mot or s ar e oper ated at rated load, t hen the following is tr ue:
IAAAA
n total ≈⋅ +⋅ +⋅ =2141 656 186 45...
As the current s ar e geometrically summed, the algebr aic addit ion m ade her e m eans t hat the
calculation is on the safe side. The 150% overload capability of the drive converter can be utilized
for t he m aximum cur rent when the highest- r at ing motor st ar ts.
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Thus, t he basic dr ive converter load current must be at least
IIüA
UG ≥⋅=⋅=
max
..
/
.
15 88
15 323 135 245
A 15 kW drive converter, 6SE7023-4EC61 with 30.9 A base load cur r ent is selected. Thus, t her e is
sufficient reserve for the magnetizing current and the transformer losses. The transformer r at io
takes int o account that only 85 % of the maximum drive converter output voltage can be reached
due to the sinusoidal filter. De- rating is not r equired f or the 15 kW drive converter and 6 kHz pulse
frequency.
6SE7023-
4EC61
15 kW
dr iv e co nv er ter
Sinusoidal
filter
Transformer
323/135 V, 12 kVA
380 V
Motor sockets
2 x 260 W, 6 x 1000 W,
1 x 1800 W
C, D
Block diag ram
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4.4 Emergency off
4.4.1 General information
According t o VDE 0113, stop functions fall under the following categor ies:
•Category 0
Uncontrolled shutdown by immediately disconnecting t he power feed (the mot or coasts down).
•Category 1
Controlled shutdown, whereby power is still fed to the drive so that it can shut down in a
controlled fashion. The power is only disconnected once the drive comes t o a standstill.
•Category 2
Controlled shutdown, whereby the power is still connected even aft er the drive comes to a
standstill.
Further, the following additional requirements ar e valid for emergency off :
•Emerg ency of f has priorit y over all other functions
•The power should be disconnected as quickly as possible
•It is not per missible that a reset initiates a restar t
Emerg ency of f must either be effective as a categ or y 0- or cat egory 1 stop function. The actual
categor y must be defined as a result of the machine r isk evaluation. For an emergency off
funct ion, st op cat egory 0, only permanently-wired (hardwired) elect r om agnetic component s m ay
be used (e.g . neither the drive converter elect r onics nor PLC controls can be used). For an
emerg ency of f f unct ion, stop category 1, t he power supply must be r eliably disconnected, and as
for cat egory 0, by just electr om echanical com ponents.
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Emergency off via a category 1 stop funct ion ( f ast stop)
In this case, shutdown is initiated with an emergency off command using the „off 3“ com m and ( fast
stop with a suitably selected ramp function). The line contactor is opened t o disconnect the power
aft er t he drive comes to a standstill. I n addit ion, a time relay is provided to positively disconnect
the power so that the line contactor is reliably opened, even if t he dr ive converter electronics
develops a fault. When the delay expires, the „ external fault“ command is also output, in or der to
prevent restar t ing without first acknowledging the fault. Fast shutdown is only possible if the drive
is equipped with a pulsed resistor or regenerative feedback.
Fast shutdown does not f unct ion if the drive converter or line supply develops a fault . Thus,
normally safet y-cr it ical dr ives (e. g. cranes) always have a mechanical brake, which can brak e the
drive to standstill within a specif ic time. Further , using this br ake, the drive is held at st andstill
when the electronics fails even if the DC link voltage is st ill available (t he DC link voltage only
decays slowly) .
Emergency off via a category 0 stop funct ion ( elect r ical shut down)
In this case, the system must be shutdown (into a no- voltage condition) as quickly as possible (the
motor coast s down or is mechanically brak ed). There are two methods of achieving t his:
a) Disconnecting t he line supply voltage in front of the drive converter
For saf et y reasons, it is not permissible to open the line contactor using the internal drive
converter electr onics ( t he dr ive converter elect r onics could fail). Thus, the line cont act or is dir ectly
opened via the emergency of f button/switch. In order t o pr event switching under load, an „external
fault “ com m and is simultaneously output to inhibit t he dr ive pulses (pr ocessing time: 4T0). This
also prevents the drive starting without fir st acknowledging t he fault.
b) Disconnecting t he line supply voltage in front of the drive converter and disconnecting the
motor
If only the line supply voltage is disconnect ed in front of the converter ( feeding t he dr ive
converter), t he DC link still remains charged f or a specific time. Thus, voltage is st ill pr esent at the
motor t er minals. In order to ensure that t he voltage at the mot or terminals is zero, a mot or
contactor locat ed between the motor and drive converter m ust open. The mot or cont actor must be
closed befor e t he dr ive converter is powered-up, and it may only be opened if ther e is no cur r ent
flowing thr ough the motor . This can be realized using t he m ain contactor check back signal.
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Contactor opening when the line supply fails
When t he line supply f ails, the contactor s can, under cer tain circumstances, drop-out (open) ,
befor e t he dr ive converter monitoring function can output a pulse inhibit com mand. This is t he
case, f or example, in t he par tial load range, dur ing short line supply dips, or when kinet ic buffering
becomes active. Thus, in this case, the contact or opens under load. Especially for m ot or
contactors, switching under load must be pr evented. I f the contactor switches under load at low
output frequencies, a quasi DC current could be int errupted which could result in arcing. To
prevent this happening, the contactor s ( control circuit) m ust be supplied from an uninterruptable
power supply.
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M
3~
24 V DC external auxiliary voltage
Auxiliary voltage
e.g. 23 0 V AC
On/off
K2 (off 3)
-x9: 1 2
-x9: 4
-x9: 5
-x101: 16
-x101: 17
-x101: 13
-x101: 13
-x1 : U1/
L1 V1/
L2 W1/
L3
V2/
T2 W2/
T3
-x2 : U2/
T1
-K1
6SE70
K3
K4
K2
De la y ed dr op - out,
time somewhat longer
than the fast stop ramp
K3
K2
-K4
-K4
-x101: 18
-x101: 13 K3 (ext. fault)
Emergency off
Example of an emer gency off circuit using a cat egory 1 stop (fast stop and then t he power is
disconnected)
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M
3~
24 V DC ext. auxi liary pow er supply
Auxiliary voltage
e.g. 230 V AC
ON/off
K2 (e xt. fault)
-x9: 1 2
-x9: 4
-x9: 5
-x 101 : 16
-x 101 : 17
-x 101 : 13
-x 101 : 13
-x1: U1/
L1 V1/
L2 W1/
L3
V2/
T2 W2/
T3
-x2: U2/
T1
-K1
6SE70
Emergency off
K2
K2
K3
-K3
-K3
Example of an emer gency off circuit using a cat egory 0 stop (the line supply voltage is
disconnected)
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M
3~
24 V DC ext. auxiliary power supply
Auxiliary voltage
e. g. 23 0 V AC
On/off
K3 (ext. fault)
-x9: 1 2
-x9: 4
-x9: 5
-x101: 16
-x101: 17
-x101: 13
-x101: 13
-x1: U1/
L1 V1/
L2 W1/
L3
V2/
T2 W2/
T3
-x2: U2/
T1
-K1
6SE70
-K2
K2 checkback signal
-x101: 13
-x101: 18
-K4
Emergency off
K3
K3
K4
-K4
An example of an emer gency off circuit using a cat egory 0 stop. (t he line supply voltage is
disconnected and the mot or is disconnected). Caut ion: Observe the maximum switching capability
of cont act X9: 4.5.
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4.5 Accelerating- and decelerating time with a constant load torque in
the field-weakening range
4.5.1 General information
The prer equisite f or t he calculation is a constant mot or torque in the constant-flux range and a
constant mot or output in the field-weakening r ange (e.g . accelerating along the current limit).
Thus, t he m otor torque is given by:
M M const
motor motor
==
max (constant flux range)
MM n
n
motor motor n
=⋅
max (field-weakening r ange, n ≥ nn)
Mmotor
motor
load
P
MAccelerating
torque
nn
nmax
n
Example for st arting in the field-weakening r ange
Under these assumpt ions, t he accelerating tim e ( ta) and the deceler at ing time (tbr) can be
calculated as f ollows.
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Case 1 ( Mload
*=0, i.e. pur e high-inert ia dr ive (flywheel))
In the constant-flux range, with 0≤≤nn
n the f ollowing is true:
tJn
M
abr total n
motor
,
*
max
=⋅⋅
⋅2
60
π
In the field-weakening range, with nnn
n≤≤max the following is t rue:
tJnn
Mn
abr total n
motor n
,
*max
max
()
=⋅⋅ −
⋅⋅
π
22
60
Case 2 ( Mload
*≠0, e.g . friction pr esent )
In the constant-flux range, with 0≤≤nn
n the f ollowing is true:
tJn
MM
abr total n
motor load
,
*
max *
()
=⋅⋅
⋅2
60
π
m
In the field-weakening range with nnn
n≤≤max the following is t rue:
tJnn
M
Mn
M
MM
Mn
nM
abr total n
load
motor n
load
motor load
motor nload
,
*max
*
max
*2
max *
max max *
(ln)=⋅⋅± −+⋅⋅⋅
2
60
π
m
m
The lower sign in the last two formulas is true when the drive brakes
Mload
*Constant load tor que ref er red to the mot or [Nm]
Jtotal
*Total m om ent of inertia r eferred t o the motor [kgm2]
Mmotor max Motor torque in t he const ant -flux rang e [Nm]
nnRated motor speed ( s t ar t of field weakening) [RPM]
nmax Max. motor speed in t he field-weakening r ange [RPM]
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4.5.2 Braking time for a grinding wheel drive
The motor of a grinding wheel drive is operated in the field- weakening range from 50 Hz to 60 Hz.
The minimum braking time is to be determined when a drive conver ter with brake resistor is used.
The load tor que when braking is assum ed to be zero (fr ict ion has been neglected). There is no
gearbox.
Drive data
Rated motor output Pmotor n = 11 kW
Rated motor current Imotor n = 21.4 A
Rated motor torque Mmotor n = 36 Nm
Rated motor speed ( 2 pole) nn= 2915 RPM
Motor efficiency ηmotor =0.87
Motor moment of inertia Jmotor = 0.034 kgm2
Ratio of the stall torque to the rat ed m otor torque Mstall/Mn= 2.6
Maximum speed of the grinding wheel nmax = 3600 RPM
Moment of inert ia of the g r inding wheel Jwheel = 0.34 kgm 2
Rated drive converter output PV n = 11 kW
Max. drive converter curr ent IU max = 34.8 A
External brak e r esist or rating P20 = 20 kW
Peak braking power 1.5 P20 = 30 kW
Determining the maximum motor torque
The maximum possible m ot or torque at the maximum drive converter curr ent is:
MM
II
II
motor motor n Vn
motor n n
max max
≈⋅−
−
22
22
µ
µ
With II
n motor n
µ
=⋅041. t he following is obtained:
MNm
motor max .. .
.. .
≈⋅ −⋅
−⋅=36 348 0 41 214
214 041 214 62
222
222
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In the field-weakening range at nmax and at constant mot or output, the m ot or torq ue is t hen only
given by:
MM
n
nNm
motor motor n
() max max
.
3600 62 2915
3600 502=⋅=⋅=
On the other hand, the permissible motor torque at nmax and 130% m ar gin for the stall torque is
given by:
MMn
nNm
motor perm motor n n
.( ) max
.
.() ..() .
3600 22
26
13 26 36
13 2915
3600 47 2=⋅⋅=
⋅⋅=
Thus, t he m aximum per missible motor tor que is reduced to:
MM n
nNm
motor motor perm n
max . ( ) max ..=⋅=⋅=
3600 47 2 3600
2915 583
Thus, the drive converter is not fully utilized when braking.
Determining t he br aking times
Total moment of inertia:
JJ J kgm
total motor wheel
=+=+=0034 034 0374 2
...
Braking time in t he field-weakening range:
tJnn
Mn s
br total n
motor n
1
22 22
60 0374 3600 2915
60 583 2915 051=⋅⋅ −
⋅⋅
=⋅⋅ −
⋅⋅ =
ππ
()
.( )
..
max
max
Braking time in the const ant - flux range:
tJn
Ms
br total n
motor
22
60 0374 2 2915
60 583 196=⋅⋅
⋅=⋅⋅⋅
⋅=
π
π
max
...
Thus, t he t otal braking time is given by:
ttt s
br total br br
=+=+=
12
051 196 247.. .
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Determining t he m axim um br aking power and braking energy
The maximum br aking power is given by:
PMn kW
br motor n motormax max ...=⋅⋅=
⋅⋅=
9550 583 2915
9550 087 1548
η
The brake resistor with P20=20 kW and a peak br aking power of 30 kW is more than adequately
dimensioned.
The braking energy corresponds to the area in the braking diagram.
Braking diagra m
tbr 1 tbr 2
Pbr max
Pbr
t
WP t Pt kWs
br br br br br
=⋅+⋅=⋅+⋅=
max max .. ., .
12
21548 051 1548 196
22307
With
W
TPP
br br cont
≤=
..
20
45 (with an external brak e r esist or )
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The perm issible period is given by:
TW
Ps
br
≥⋅=⋅=
45 2307 45
20 52
20
....
Thus, br aking can be realized every 5.2 s.
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4.6 Influence of a rounding-off function
Acceleration steps ar e elim inat ed, and therefore softer positioning achieved, using a rounding- off
funct ion for a traversing characterist ic. We will now investigate t he influence of such a r ounding-of f
funct ion on t he m ot or - and drive converter dimensioning. These investigat ions ar e based on the
square- law rounding-off, the sin2 rounding-of f, as well as rounding-off for jerk-free mot ion. These
various rounding-off functions are com par ed with the behavior without rounding-of f, i.e.
acceleration as a step function.
Square-law r ounding- off function
For a squar e- law rounding - off function, acceleration is entered as a trapezoid. This results in a
square- law character ist ic of the velocity during linear ly increasing or linearly decreasing
acceleration.
with 0 ≤ k ≤ 1
a
amax
t
t
kb
2t
kb
2
v
vmax
tb
123
t
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Acceleration in the ar eas 1, 2, 3:
aat
kt
b
12=⋅ ⋅⋅
max
aa
2=max
aa
tt
kt
b
b
32=⋅ ⋅ −
⋅
max
The following is true f or t he r atio between maximum acceleration and maximum velocity:
av
tk
b
max max
()
=⋅−12
Velocity in areas1, 2, 3:
va
kt t
b
12
=⋅⋅
max square-law
vakt
at
kt
bb
242
=⋅⋅ +⋅−
⋅
max max () linear
va t ka
kt tt
bbb32
12
=⋅⋅−−
⋅⋅−
max max
() () square-law
For k= 0 ( 0% r ounding-of f), ar eas 1 and 2 ar e omitted. Acceler at ion is a st ep function. For k=1
(100% rounding - off ) , ar ea 2 is omitted. Acceler at ion is t r iangular.
The relat ionships without rounding-of f (step- function accelerat ion) will now be compared with
100% rounding- off (triangular acceler ation). This assum es, traversing mot ion with the same travel,
same travel time and t he same maximum velocity.
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a
amax
v
vmax
tb
t
k=1
k=0
ttot.
tv
t
k=0
k=1
Comparison of the traversing curve with and without rounding- off
The area under t he t raversing curve v corresponds to the t ravel stotal. In this case, tb=tv.
With
av
tk
b
max max
()
=⋅−12
the f ollowing is obtained if k=1:
avtb
max max
=⋅2
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With the same travel stotal, the same travel time ttotal and the same maximum velocity vmax, the
maximum acceleration for 100% rounding - off is exactly double. For a pur e acceler ating drive, the
motor t or que would also then double (the mot or m ust be appropriately dimensioned) and,
corresponding t o t his torque, also t he m otor current ( dr ive converter dim ension) . However, the
maximum motor t orque only occurs at half the maximum motor speed.
In the following, for the previously described traversing mot ion, t he motor tor que characterist ics
are shown as a function of the speed using a pur e acceler ating drive. The m otor torque must lie
below the permissible limit s ( in t his case, the limiting cur ve for an 1FK6 mot or ) . The mot o r t or que
for 100% r ounding-of f (k =1) is com par ed with the m otor torque without rounding-off ( k=0).
0
5
10
15
20
25
30
35
40
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
M in Nm
Mper.
k=1
k=0
Comparison of the motor t orque with and without rounding-off
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Sin
2
rounding-off
For a sin2 rounding- off , the velocity is entered according t o a sin2 function. For t he acceler ation, a
sinusoidal funct ion is obt ained with twice the f requency.
v
vmax
t
a
amax
tb
t
Velocity:
vv t=⋅ ⋅
max sin ( )
2
ω
for 0 ≤ t ≤ tb
with
ω
π
=⋅2tb
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Acceleration:
aa t
=⋅ ⋅⋅
max sin( )2
ω
for 0 ≤ t ≤ tb
with
av
max max
=⋅
ω
The relat ionships without rounding-of f (step- function accelerat ion) will now be compared with sin2
rounding- off (sinusoidal acceleration). I t is assumed that the t r aversing operation has the sam e
travel, the sam e t r avel tim e and t he sam e m aximum velocity.
a
amax
v
vmax
tbttotal
tv
t
t
sin rounding-off
2
without rounding-off
sin rounding-off
2
without rounding-off
Comparison of the traversing curve, with and without rounding-off
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The area under t he t raversing curve v corresponds to the t ravel stotal. In this case, tb=tv.
With
av vtb
max max max
=⋅=⋅
ω
π
2
a fact or of π/2 is obt ained with respect t o t he acceleration value without r ounding-off. For the same
travel stotal, the same travel time ttotal and the same m aximum velocity vmax, the maximum
acceleration for sin2 rounding-of f is therefore g r eater by a factor π/ 2. For a pure accelerating dr ive,
the motor torque would thus be increased by this factor (t he m otor must be appr opr iat ely
dimensioned), and, cor responding to t his t or que, also the motor current ( t he drive converter must
be appropriately dimensioned). However, t he m aximum m ot or torque only occurs at half the
maximum motor speed.
The previously described traversing m otion is illustrated in t he following using a pure acceler at ing
drive, showing the motor torque as a function of the speed. T he m otor torque must lie within the
permissible limit s ( in t h is case, t he lim iting curve for an 1FK6 mot or ) . The motor torque for sin2
rounding- off is compared with the motor torque without rounding-of f.
0
5
10
15
20
25
30
35
40
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
M in Nm
M
2
with sin -rounding-off
without rounding-off
per.
Comparison with and without rounding- off
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Rounding-off f or jerk-f r ee motion
For jerk-f ree motion, the acceleration must be specified so that t he der ivative at t he transitions is
0. For example, t his is achieved with a sin2 f unct ion for the acceler at ion.
a
amax
v
vmax
tbt
t
Acceleration:
aa t=⋅ ⋅
max sin ( )
2
ω
for 0 ≤ t ≤ tb
with
ω
π
=tb
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Velocity:
vv t
tt t
bb
=⋅−
⋅⋅⋅⋅⋅
max (sin())
1
22
ωω
for 0 ≤ t ≤ tb
with
vat
bmax max
=⋅ ⋅
1
2
The relat ionships without rounding-of f (step- function accelerat ion) with rounding -off for jerk- free
movement (sin2- function accelerat ion) will now be compared. It is assumed that t he t raversing
operation has the sam e travel, the same travel time and the same maximum velocity.
a
amax
v
vmax
tbttotal
tv
t
t
jerk-free rounding-off
without rounding-off
jerk-free rounding-off
without rounding-off
Comparison of the traversing curve with and without rounding- off
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The area under t he t raversing curve v corresponds to the t ravel stotal. In this case, tb=tv.
With
avtb
max max
=⋅2
the same tr avel stotal, the same travel time ttotal and the same maximum velocity vmax results in twice
the acceleration value with respect to t he traversing curve without rounding - off . For a pure
accelerating dr ive, t he motor tor que would thus be increased by this factor (the mot or m ust be
appropriately dimensioned), and, corresponding to this tor que, also the motor current ( t he drive
converter must be appr opr iat ely dimensioned) . However, the m aximum m otor torque only occurs
at half t he m aximum motor speed.
The previously described traversing m otion is illustrated in t he following using a pure acceler at ing
drive, as an example for t he m otor torque characteristic as a function of the speed. T he m otor
torq ue must lie within the perm issible lim it s ( in t his case, the limit ing curve for an 1FK6 m ot o r ) . The
motor torque for sin2 rounding-off is compar ed with the m otor torque without rounding-off .
0
5
10
15
20
25
30
35
40
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
M in Nm
Mper.
jerk-free rounding-off
without rounding-off
Comparison of the motor t orque, with and without rounding-off
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Summary
If , for a positioning drive, the same t r aversing sequence is investigat ed with and without rounding-
off, then, for the sam e positioning time, and the same maximum velocity, the tr aversing curve with
rounding- off, requires a higher acceleration and deceler at ion. Thus, t he m ot or torq ue and
therefore the mot or current increases. Due t o the higher motor torque, the RMS torque also
increases. Under cer t ain circumstances, a lar ger motor and a larger drive converter must be used.
By appropriately increasing t he acceler ation- and deceleration tim es, acceler ation and deceleration
can be reduced. Natur ally, t he positioning time t hen incr eases.
For squar e- law rounding - off and sin2 rounding-of f, the acceleration steps are avoided, but
completely jerk-f r ee m otion is not achieved. This is only achieved when the sin2-type acceleration
funct ion is applied. This type of accelerat ion set point doubles the value for t he acceler ation, and
also results in the highest increase in the RMS value.
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4.7 Possible braking torques due to system losses
4.7.1 General information
Normally, when braking , the brak ing energy is either dissipated in a brake r esist or , fed back into
the line supply, or, when using DC current brak ing, converted in the m ot or itself. There are only a
very limited number of ways of stor ing the braking energy in the DC link. Below is an example of a
4-pole 55 kW motor connect ed to a 55 kW drive converter with a 400 V supply voltage.
With
Jkgm
Mot =079 2
.motor m om ent of inertia
CmF
DClink =9drive converter DC link capacitance
the kinet ic ener gy Wkin, to shutdown the motor and the ener gy WC which can be stored in the DC
link capacitor is given by:
WJnWs
kin Mot Mot Mot
=⋅ ⋅⋅⋅ =⋅ ⋅⋅⋅ =
1
2260 1
2079 2 1475
60 9424
22
().( )
π
π
WCU U Ws
C DClink DClink DClink n
=⋅ ⋅ − =⋅⋅ ⋅ − =
−
1
21
29 10 530 1616
22 322
( ) (800 )
max
Using the ener gy stored in the DC link capacit or , not even the motor alone can be shut down
without taking into account the losses. Thus, in the following, the possible braking tor que is
determined, t aking into account t he loss equation. The following diagr am makes t he sit uation
clear.
Rectifier Inverter
Line supply
PV Inverter
M
PV react.
PV motor
Pbr
PV const.
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The mot or br aking power may not exceed the sum of the losses in the m ot or , in the output r eact or,
in the inverter and in t he power supply (constant losses) . Thus, t he following must be tr ue:
PPPP P
V Mot V react V inv V const br
+++≥
.
With
PMn
br br Mot
=⋅
9550
the f ollowing is obtained:
MPP PP
n
br V Mot V react r V inv V const
Mot
≤+++⋅()
.. 9550
There is som ewhat of a problem in so much t hat the inverter losses are load-dependent , and the
motor- and r eactor losses, load- and speed- dependent . As the speed decreases, the ir on losses of
the motor and r eactor decrease while the current - dependent losses of the motor, react or and
inverter increase due t o t he incr easing braking torq ue. As a first approximation, it is assumed that
the sum of the losses is constant. As the expected braking torq ues at rated speed are low, an
operating point close to the no-load mot or is assum ed. In this case, t he m agnetizing curr ent can
be considered to be the m otor current . Thus, useable results can be obtained at rated speed using
the f or m ula above for the braking tor que. The calculat ion becom es increasingly more inaccurate
as the speed decreases.
For II
Mot mag
≈, the losses at r ated speed can be estimated as follows:
PP P
V Mot V Mot n Mot n Mot n
≈⋅=⋅⋅−04 05 04 05 11..... ..... ( )
η
PP I
IP
VDr VCun mag
Dr n VFen
≈⋅ +()
2
PPP I
I
Vinv VDn VSchn mag
Un
.(.())≈+⋅⋅ ⋅
093
3500
400 2
The constant losses com pr ise of the losses of the power supply, switching losses in the DC link
capacitors, and for small unit s , addit ionally the losses due t o t he DC fan.
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Examples for t he losses and braking torques at r at ed speed
For a 4-pole 5.5 kW m ot or with output reactor connected to a 5.5 k W drive converter with 400 V
supply voltage, the losses are given by:
PkW
V
∑≈058.
or f or t he r ated torq ue at r ated speed:
MP
nNm
br V
n
≈⋅=⋅=
∑9550 058 9550
1450 382
..
This corr esponds t o 10. 6% of the rated torq ue.
For a 4-pole 55 k W motor with output r eactor connected to a 55 kW dr ive converter with 400 V
supply voltage, the losses are given by:
PkW
V
∑≈207.
and, at rat ed speed the torq ue is:
MP
nNm
br V
n
≈⋅=⋅=
∑9550 207 9550
1475 134
..
This corr esponds t o 3. 8% of the rated torq ue.
This means t hat the possible braking torque decreases with increasing power.
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Calculating the braking time for a pure high-inertia dr ive (flywheel drive)
In this case
MJ dn
dt
br tot
=− ⋅ ⋅
.2
60
π
For a constant br aking torque, a linear braking ramp is obtained with
tJn
M
br tot
br
=⋅⋅
⋅
.max
π
30
If t he equation, derived from the losses, is now used f or Mbr
MP
n
P
nnnMnn
br VV br n
=⋅=⋅⋅= ⋅
∑∑
9550 9550
max
max max
max
then t, as a function of n is given by:
tJ
Mn ndn J
Mn
nn
tot
br n n
ntot
br n
=− ⋅
⋅⋅
⋅= ⋅
⋅⋅
⋅−
∫
.
max
.
max max
max max max
()
ππ
30 60 22
or
tJn
M
br tot
br n
=⋅⋅
⋅
π
max
max
60
The mot or speed, as a function of time is:
nn Mn
Jt
br n
tot
=−
⋅⋅
⋅⋅
max max
.
max
260
π
The braking time, with the braking torque assumed as ∼1/ n, is precisely half as long as for a
constant braking tor que. It is assum ed, that the Ud max control automatically ensures a down ramp
with a square-root char act er istic.
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n
nmax
t
M = const.
br
M ~1/n
br
Example of a braking charact eristic for a pur e high-inertia drive (flywheel drive)
Calculating t he braking t im e for a fan drive
In this case
MM J dn
dt
br load tot
+=−⋅⋅
.2
60
π
The load tor que of t he fan is given by:
MM n
n
load load
=⋅
max max
()
2
With
MM nn
br br n
=⋅
max
max
t, as a function of n, is given by:
tJdn
MnnMn
n
tot
br n load
n
n
=− ⋅⋅⋅+ ⋅
∫
.
max max max
max
max ()
π
30 2
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=⋅⋅
⋅⋅⋅⋅−⋅ +
+−−⋅+
+
Jn
Ma
aan n
an
aann
an
tot
load
.max
max
max max
max
(ln ()ln ())
π
222
2
22
2
30 1
6
+⋅⋅⋅−
⋅−⋅−
⋅
1
3
2
3
2
3a
na
a
na
a
(arctan( ) arctan( ))
max
or
tJn
Ma
aan n
an a
na
a
br tot
load
=⋅⋅
⋅⋅⋅⋅−⋅ +
++⋅⋅⋅−
⋅+
.max
max
max max
max
max
ln () (arctan( ) )
ππ
222
2
30 1
61
3
2
36
With
an M
M
br n
load
=⋅
max max
max
3
Without taking into account the mot or br aking t or que, the speed, as a function of time, is given by:
n
n
Mt
Jn
load
tot
=+⋅⋅
⋅⋅
1
130
2
max
max
.max
π
Zero speed is in this case, only reached for t → ∞.
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n
n
t
without motor braking torque (only M )
with motor braking torque
load
max
Example of a braking charact eristic for a fan drive
Summary
When br aking using t he losses, only low braking torq ues can be achieved at t he r at ed operating
point (approximat ely 2 to 10% Mn depending on the mot or size). This technique can only be used,
where, on one hand short braking tim es ar e not im portant, and where on the ot her hand, due to
large m om ent s of inertia and low friction losses, run-down takes minutes. T his occur s , e.g. for
large fan drives. In this case, at high speed, t he load torque initially has a hig h br aking effect , but
which decreases according to an n 2 function.
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4.7.2 Braking time for a fan drive
The braking time is to be estimated for a 200 kW f an dr ive with 4-pole 1LA6 317-4.. motor and 200
kW drive converter connected to 400 V. There is no brake resistor.
Drive data
Rated motor current Imotor n = 345 A
Rated motor torque Mmot n = 1280 Nm
Motor efficiency for drive converter oper at ion ηmotor = 0. 95
Motor magnetizing current Imag = 107 A
Motor moment of inertia Jmot = 4. 2 kgm2
Rated drive converter curr ent IU n = 370 A
Power requir em ent at nmax Pmax = 196 kW
Maximum f an speed nmax = 1485 RPM
Moment of inert ia of the fan Jload = 75 kgm 2
Determining the losses at nmax with Imot=Imag (values for the dr ive converter from t he Engineering
Manual for Drive Converter s)
PP kW
VMot Motn Mot n
≈⋅ ⋅ −=⋅⋅ −=04 11 04 200 1
095 142.().(
.).
η
PPI
IPkW
V react V Cu n mag
react n VFen..
() .(). .≈⋅ +=⋅ +=
22
0138 107
370 0088 01
PPP I
IkW
Vinv VDn VSchn mag
Un
.(.()) (. . .())≈ + ⋅⋅ ⋅= +⋅ ⋅=
093
3500
400 302 089 093
3500
400 107
370 1
22
PkW
V const ≈015.(fan losses are subtracted)
Thus, t he sum of the losses is
PkW
V
∑=545.
The braking time is now calculated with these losses which are assumed constant . It is assumed,
that the closed- loop Ud max control automat ically def ines the downramp.
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Maximum load tor que:
MPnNm
load max max
max
=⋅=⋅=
9550 196 9550
1485 1260
Motor braking torque at nmax:
MP
nNm
br n V
max max
.
=⋅=⋅=
∑9550 545 9550
1485 35
Thus, t he r atio of the br aking tor que to the r at ed m otor torque at nmax is:
M
M
br n
Mot n
max ..== =
35
1280 00273 273%
At nmax, this braking torque is small with respect t o the braking effect of the fan load t or que. The
braking effect due to losses only becomes effect ive at a lower speed.
Calculating t he braking t im e
an M
M
br n
load
=⋅
max max
max
3=⋅ =1485 35
1260 44974
3.RPM
tJn
Ma
aan n
an a
na
a
br tot
load
=⋅⋅
⋅⋅⋅⋅−⋅ +
++⋅⋅⋅−
⋅+
.max
max
max max
max
max
ln () (arctan( ) )
ππ
222
2
30 1
61
3
2
36
=⋅⋅
⋅⋅
792 1485
30 1260
2
.
π
1
6 44974
44974 449 74 1485 1485
44974 1485
22
2
⋅
⋅−⋅+
+.ln ..
(. )
+⋅⋅⋅−
⋅+
1
44974 3
2 1485 449 74
44974 3 6
.(arctan( .
.))
π
=29 32. s
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0
200
400
600
800
1000
1200
1400
1600
0 5 10 15 20 25 30 35 40
Time in s
Motor speed in RPM
Braking characterist ic for the fan drive
Estimating t he influence of t he ener gy which can be stored in the DC link capacitor
The closed-loop Udmax control limits t he DC link voltage t o the following value in generator
operation:
UU V
DClink linemax ..=⋅⋅ =⋅⋅=119 2 119 2 400 675
The following is obtained for t he ener gy which can be stored in the DC link capacitor with
CDClink=18.8 mF:
WCU U kWs
C DClink DClink DClink n
=⋅ ⋅ − =⋅ ⋅ ⋅ − =
−
1
21
2188 10 675 530 164
22 322
().().
max
The average power which can be absorbed in the capacitor , during braking is given by:
PW
tkW
Cm C
br
== =
164
2932 0056
...
This component is low with respect to t he t otal losses of 5.45 kW, and can therefore be neglect ed
in the calculation.
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Comparison with DC braking
As a comparison, t he possible braking torques using DC current braking will be estimated for this
fan dr ive.
The DC curr ent braking torque can be given by:
MM
I
IsR
Xs
br a a
h
≈⋅ ⋅
+⋅
() ’
12
22
2
1
with
sn
ns
=slip when braking
nssynchronous speed
Iastarting current ( inrush cur rent)
Mastarting torque
I1equivalent three- phase cur r ent to gener at e the braking torque
I1 can be converted as follows into a DC braking curr ent Ig:
II
g12
3
=⋅
Mbr
nsn
Example of a braking tor que curve for DC cur r ent braking
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The braking tor que, f or speeds which are not t oo low, can be approximat ed as:
MM
I
In
nMI
In
nn
br a a
sag
a
s
≈⋅ ⋅=⋅⋅ ⋅() () ~
122
2
31
Thus, also when brak ing using the losses, a br aking torque is obtained which is proportional t o
1/n. T his allows the braking t or que to be calculated for the 200 k W motor at rated speed. With
II A
an
=⋅ =⋅ =
7 7 345 2415
MM Nm
an
=⋅ =⋅ =
27 2 7 1280 3456..
II A
gUn
==370 (br aking with the rat ed dr ive converter cur rent)
the f ollowing is obtained at rated speed (nn
ns
≈)
MNm
br ≈⋅⋅ =3456 2
3370
2415 54
2
()
The braking tor que which can be achieved with t he r ated drive converter current and DC cur r ent
braking is therefore somewhat gr eat er than the brak ing torq ue calculat ed using the losses (35
Nm).
Estimating t he braking time
an M
M
br n
load
=⋅
max max
max
3=⋅ =1485 54
1260 51968
3.RPM
tJn
Ma
aan n
an a
na
a
br tot
load
≈⋅⋅
⋅⋅⋅⋅−⋅ +
++⋅⋅⋅−
⋅+
.max
max
max max
max
max
ln () (arctan( ) )
ππ
222
2
30 1
61
3
2
36
=⋅⋅
⋅⋅
792 1485
30 1260
2
.
π
1
6 51968
519 68 519 68 1485 1485
68 1485
22
2
⋅
⋅−⋅+
+.ln ..
(519. )
+⋅⋅⋅−
⋅+
1
51968 3
2 1485 519 68
51968 3 6
.(arctan( ,
.))
π
=241. s
A somewhat shorter br aking tim e is obt ained due to the higher braking torque. I n com par ison to
braking using the losses, for DC curr ent braking, the total m otor brak ing power is converted in the
rotor. In addition, stat or copper losses occur due to the DC current supply.
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4.8 Minimum acceleration time for fan drives with a linear acceleration
characteristic
4.8.1 General information
The linear acceler at ion of a fan dr ive with square- law load torque characterist ic is to be
investigated. At st arting, a br eakaway torque, decreasing as square- law, must be addit ionally
taken int o account. The following is obtained f or t he load torque as a function of speed n:
MM n
nMn
n
load load LB LB
=⋅+⋅−
max max
() ( )
22
10≤≤
nn
LB
MM n
n
load load
=⋅
max max
()
2nnn
LB ≤≤max
M
M
load
load max
MLB
nLB n
nmax
Breakaway compo nent
Example of a load tor que characterist ic with breakaway torque
The maximum m ot or torque during acceleration is obtained with MLB<Mload max
MJJM
Mot Mot load loadmax max
()=+⋅+
α
with
α
π
=⋅⋅
⋅
260 n
tH
max
Thus, t he m inim um acceler ation time is given by:
tnJJ
MM
HMot load
Mot load
=⋅⋅ ⋅ +
⋅−
2
60
π
max
max max
()
()
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For the g iven drive converter, the maximum motor t or que is obtained f r om the maximum
permissible drive converter current IU max=Imot max:
MM
kn
II
kn
II
Mot
Mot n Umag
Mot n mag
max
max ()
=⋅−
−
22
22
kn =1for nn
nmax ≤constant flux range
kn nnn
=max for nn
nmax >field-weakening r ange
The maximum m ot or torque should not exceed 2 Mn for reasons of stability. For operation in t he
field-weakening range at nmax, sufficient clearance must st ill be kept to t he st all limit:
MMn
n
Mot stall n
max max
.()≤⋅
13 2
Thus, t he m aximum m otor torque must lie below the f ollowing charact er ist ic:
M
limit
nn
Mn
n
stall n
13 2
.()
⋅
2n
Where:
nn M
M
it n stall
Mot n
lim .
=⋅⋅⋅213
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We st ill have to check whether the motor RMS current ( corresponds to the dr ive converter RMS
current) is less t han or equal to t he r at ed dr ive converter cur r ent when accelerating. The m ot or
RMS current is calculated in a 300 s int er val. The following is valid for tH ≤ 300 s:
I
Idt
s
IdtI st
s
rms
Mot
s
Mot Mot k H
tH
=⋅=⋅+ ⋅ −
∫∫
2
0
300 22
0
300
300
300
()
IMot k : const ant motor curr ent at the operating point Mmot=Mload max
Imot
max
Imot
tH300 s t
Imot k
Example of the m otor current char act eristic (tH ≤ 300 s)
If the acceleration time tH is to be greater than 300 s, the interval f rom tH -300s to tH is used in the
calculation f or the maximum RMS value during accelerat ion. The RMS value is then given by:
I
Idt
s
rms
Mot
ts
t
H
H
max =⋅
−∫2
300
300
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Imot
max
Imot
mot k
tH
300 s
t
I
Example for t he motor curr ent char acteristic (t H > 300 s)
In order t o calculate the RMS value, the motor curr ent must be determ ined as a function of t he
time. The motor cur r ent is obtained as follows fr om the motor t or que:
II M
M
I
Ikn I
Ikn
Mot Mot n Mot
Mot n
mag
Mot n
mag
Mot n
=⋅ ⋅− ⋅+ ⋅()(())()
2222
2
11
kn =1for nn
n
≤
kn n
nn
=for nn
n
>
The mot or t orque during acceler ation is given by:
MJJ Mn
Mot Mot load load
=+⋅+()()
α
Aft er acceler ation, the f ollowing is t r ue
MM
Mot load
=max
For Mload (n) and α, the eq uations shown at the beginning of this t ext are valid. The following is now
inserted for n:
nn t
tH
=⋅
max (linear acceler at ion)
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The mot or cur r ent is then known as a f unct ion of time. The RMS value is calculated for tH ≤ 300 s
in 4 ranges:
0≤≤ ⋅tn
nt
LB H
max break away, constant flux range
n
nttn
nt
LB HnH
max max
⋅<≤ ⋅ constant flux range
n
nttt
nHH
max ⋅<≤ field-weak ening range ( if available)
tt s
H<≤
300 steady-state with Mmot=Mload max
For tH >300 s, the last r ange is eliminated, and t he st arting point for the calculation is within one of
the f irs t 3 r anges.
The integral to calculat e the RMS value is numerically evaluated using an Excel table. Further, the
table of t he cur ve output is used as Excel g r aphics.
it
iniMload i Mmot i kniImot i ≈⋅
∫
Idt
Mot
2
0
1
2
..
0
..
t(nLB)
0
..
nLB
MLB
..
.. 1
.. .. 0
..
t(nn)n
n1
tHnmax Mload max Mmot max
ni /nnImot max
m-1
mtH
300 s nmax
nmax
Mload max
Mload max
Mload max
Mload max
Imot k
Imot k Is
rms
2300⋅
Calculation table for t H ≤ 300 s
The following is formed in the last column:
()()
II tt
Mot i Mot i
i
im
ii
−
=
=
−
+⋅−
∑12
11
2
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This approximately corr esponds t o the expression:
Idt
Mot
s2
0
300
∫⋅ or IdtI st
Mot
t
Mot k H
H2
0
2300
∫⋅+ ⋅ −()
A zero is in the first line (i=0).
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4.8.2 Example
The minimum acceler ation time f or linear acceler ation for a fan drive with breakaway torque,
decreasing accor ding to a squar e- law character ist ic, is to be calculated. Furt her , the RMS value of
the drive converter curr ent when accelerating, should be check ed. The output of the 4-pole fan
motor is 90 kW for a 400 V volt supply voltage. A 90 kW drive converter is used.
Drive data
Rated motor output Pmot n = 90 kW
Rated motor torque Mmot n = 581 Nm
Rated motor current Imot n = 160 A
Magnetizing curr ent Imag = 52.8 A
Rated motor speed nn= 1480 RPM
Max. speed nmax = 1600 RPM
Motor moment of inertia Jmot = 1.6 kgm2
Stall torque Mstall = 1568.7 Nm
Load moment of inertia Jload = 200 kg m2
Max. load torque Mload max = 500 Nm
Fan breakaway torque MLB = 100 Nm
Fan breakaway speed nLB = 200 RPM
Rated drive converter curr ent I U n = 186 A
Max. drive converter curr ent IU max = 253 A
The max. motor torque is given by
kn nnn
== =
max .
1600
1480 1081
for
MM
kn
II
kn
II
Mot
Mot n Umag
Mot n mag
max
max ()
=⋅−
−
22
22
=⋅ −
−=
581
1081
253 528
1081
160 528 8833
22
22
.
(.
.)
..Nm <2⋅MMot n
Thus, t he m inim um acceler ation time is given by:
tnJJ
MM s
HMot load
Mot load
=⋅⋅ ⋅ +
⋅− =⋅⋅ ⋅ +
⋅− =
2
60 2 1600 16 200
60 3 500 8813
π
π
max
max max
()
()
(. )
(883. ) . <300 s
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Checking t he stall limit:
Mn
n
stall n
13 2
.()
max
⋅=⋅ = >
15687
13 1480
1600 10325
2
.
.() . max
Nm MMot
In order t o calculate the RMS value, the speed range from 0 to nLB is sub- divided into 5 identical
sections, the speed r ange fr om nLB up to nn, into 10 identical sections, and the speed range fr om
nn to nmax, into 5 ident ical sect ions. Time ti is calculated over the speed ni as follows:
tn
nt
iiH
=⋅
max
Thus, t he following calculation table is obt ained.
i t i n i M load i M mot i kn i I mot i ≈⋅
∫
Idt
Mot
2
0 0 0 100 483,298127 1 136,282178 0
1 2,20314175 40 64,3125014 447,610629 1 127,779981 38405,6206
2 4,40628349 80 37,2500055 420,548133 1 121,408307 72606,5335
3 6,60942524 120 18,8125123 402,11064 1 117,11084 103941,479
4 8,81256698 160 9,00002188 392,298149 1 114,83971 133574,318
5 11,0157087 200 7,81253418 391,110662 1 114,565661 162560,407
6 18,0657623 328 21,0125919 404,310719 1 117,621629 257579,157
7 25,1158159 456 40,6126777 423,910805 1 122,196064 358945,754
8 32,1658695 584 66,6127914 449,910919 1 128,324764 469562,303
9 39,2159231 712 99,0129332 482,311061 1 136,045657 592747,395
10 46,2659766 840 137,813103 521,11123 1 145,394241 732353,29
11 53,3160302 968 183,013301 566,311428 1 156,400536 892883,165
12 60,3660838 1096 234,613526 617,911654 1 169,087684 1079608,38
13 67,4161374 1224 292,61378 675,911908 1 183,471912 1298685,75
14 74,466191 1352 357,014062 740,312189 1 199,563359 1557274,69
15 81,5162446 1480 427,814372 811,112499 1 217,367301 1863654,47
16 82,8381296 1504 441,801933 825,10006 1,01621622 224,078497 1928054,86
17 84,1600147 1528 456,014495 839,312622 1,03243243 230,99668 1996493,28
18 85,4818997 1552 470,452058 853,750186 1,04864865 238,122442 2069221
19 86,8037848 1576 485,114622 868,41275 1,06486486 245,456565 2146501,26
20 88,1256698 1600 500,000000 883,300315 1,08108108 253 2228609,8
21 88,1256698 1600 500,000000 500,000000 1,08108108 148,765367 2228609,8
22 300 1600 500,000000 500,000000 1,08108108 148,765367 6917629,08
The RMS value is then obtained using the last value from the colum n ≈⋅
∫
Idt
Mot
2:
IAI
rms U n
==<
6917629
300 1518.
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0
200
400
600
800
1000
1200
0 200 400 600 800 1000 1200 1400 1600
Speed in RPM
Torque in Nm
M per .
M mot
M load
Load torq ue, m otor torque and permissible torque as a funct ion of the speed
0
50
100
150
200
250
300
0 50 100 150 200 250 300
Time in s
Motor current in A
Motor current in a 300 s int er val during acceleration and aft er acceleration
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4.9 Buffering multi-motor drives with kinetic energy
4.9.1 General information
In addition t o buff ering the system at power failure using additional capacitors in the DC link, it is
also possible to buf fer the system using the kinet ic energy of t he dr ives. For a multi-mot or drive
with common DC link bus, for stabilit y reasons, only one drive can be used for the k inet ic
buffering. This drive must have sufficient k inet ic ener gy, to cover the power requirement of the
multi-m otor drive during t he power failure (power outag e) time. During t he buff ering phase, the
speed of t he buff er drive drops, depending on the m oment of inertia and the power requir ement. If
a specific speed r at io is r equired f or the individual drives, then this must be ensur ed using the
buffer drive as master drive.
If , within the multi-motor group, there is no drive with sufficient kinet ic ener gy, then an additional
motor with coupled flywheel can be used as buff er drive. This mot or is connected to the com mon
DC link bus via an appropriate invert er . With this arrangem ent, the speed of the driven loads can
be maintained during buffering, only the speed of the buffer dr ive decreases.
The k inet ic buff er ing funct ion m ust be activated for all inverters. For the buffer drive inverter, the
threshold when the kinet ic buff er ing kick s- in, must be set as high as possible (e.g . 85%); f or all of
the other invert er s, it should be set as low as possible (e.g . 65%). This prevents the drives from
mutually influencing one another. The kinetic buffering is activated in the inverter s of the remaining
drives to prevent shut down with DC link undervolt age.
When dim ensioning the system, it must be tak en into account, that during buf fering , the required
power of the r em aining drives including the losses, flows through the motor and the inverter of the
buffer drive. For r easons of stabilit y, t he m otor torque of t he buff er drive should not exceed the
rated tor que.
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INV
EE
buffer drive
INV INV
MM M
12 P
Pload 1 Pload 2 Pload P
()PP
kin load P mot P
−⋅
η
Power relationships during buff er ing
Buffering using a dr ive in t he m ulti-motor dr ive gr oup
For simplification, all drives are considered to be constant-t or que drives. Only the mom ent of
inertia of the buffer drive is tak en into account. During buff er ing, the speed rat ios should be
maintained bet ween the drives. The req uir ed power in t he DC link during buff er ing is then obtained
from:
PP
DClink load i
Mot i WR
i
Mot P
Mot P
=⋅⋅
∑max
max
ηη
ω
ω
motoring
The buffer drive feeds back into the DC link :
PP
DClink MotP MotP WR
=⋅⋅
generating
The following must be true:
Pload i
Mot i WR
i
Mot P
Mot P
max
max
ηη
ω
ω
⋅⋅
∑=− ⋅ ⋅PMot P Mot P WR
η
η
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During buffer ing , the buf fer drive must pr ovide a variable-speed power of
PPP
Mot P load i
Mot i Mot P WR
i
Mot P
Mot P erf Mot P
Mot P
=− ⋅⋅
⋅=−⋅
∑max
max max max
ηη η
ω
ω
ω
ω
2
The following is true:
PM J
d
dt MPP
Mot P Mot P Mot P P Mot P load P Mot P kin load P
=⋅=⋅ + ⋅=−+
ω
ω
ω
()
A linear speed reduction is obt ained with the simplifications/ appr oximat ions which have been
made. It is:
d
dt J
PM
Mot P
P
erf
Mot P load P
ω
ω
=− +
1()
max
max
with
ω
π
Mot P
Mot P
n
=⋅⋅2
60
nMot P : buffer drive speed
nMot P max : buffer drive speed at the st ar t of buffer ing
JP: moment of inertia of the buf fer drive including m ot or
For the charact er ist ics of the motor speed of the buffer drive with respect to t ime, during buffer ing,
the f ollowing expression is obtained:
nn J
PMt
Mot P Mot P P
erf
Mot P Last P
=−
⋅⋅ ⋅+⋅
max
max
max
()
60
2
πω
The mot or t orque of the buffer drive should not be g r eat er than the rated t or que during buff er ing,
i.e.:
MM
Mot P Mot P n
≤
With
MPP
Mot P
Mot P
Mot P
erf
Mot P
== =
ωω
max
max constant
the max. required power is given by the following:
PM
erf Mot P n Mot Pmax max
≤⋅
ω
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Buffering via an addit ional buf f er dr ive
The speed of the driven load should, during buff er ing, be maintained for a specific time tP. The
following is t her efore t r ue for t he power r equired in the DC link:
PP
DClink load i
Mot i WR
i
=⋅
∑
ηη
motoring
The buffer drive feeds back into the DC link ( r e generative power):
PP
DClink Mot P Mot P WR
=⋅⋅
generating
Thus, dur ing buffering, t he buff er drive must provide a constant power of
PPP
Mot P load i
Mot i Mot P WR
ierf
=− ⋅⋅
=−
∑
ηη η
2
The energy is given by:
WW J Pt
P kin P Mot P Mot P erf P
==⋅⋅ − =⋅
1
222
()
max min
ωω
For the charact er ist ics of the motor speed of the buffer drive over time, dur ing buffering, the
following expression is obtained:
nP
Jt
Mot P Mot P n erf
P
=⋅⋅−
⋅⋅
60
2
2
2
πω
The mot or t orque of the buffer drive is given by
MP
Mot P erf
Mot P
=
ω
increases with decreasing speed
At ωmot P=ωmot P min, it should be the sam e as t he rated torque. If the buf fer drive is only used up to
the rated speed, i.e. ωmot P max=ωmot P n, t hen the required rated motor out put is given by:
PM P
Mot P n Mot P n Mot P n erf
Mot P n
Mot P
=⋅=⋅
ω
ω
ω
min
The required rated m ot or output therefore increases with ωmot P n /ωmot P min. Thus, for example,
ωmot P n/ωmot P min is set to 1.1. The requir ed quantities ar e obt ained as follows:
PP
Mot P n erf
=⋅11.
JJ J Pt
PMotP pl erf P
Mot P n
=+=
⋅⋅
⋅−
sup (.)
2
11
11
22
ω
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with
ω
π
Mot P n
Mot P n
n
=⋅⋅2
60
As the ηmot P and nmot P n values, req uir ed to calculate Perf and JP, are st ill not known, a motor m ust
be assumed.
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4.9.2 Example for buffering with an additional buffer drive
A multi- m otor drive, comprising 5 mot ors 30 kW/4-pole, 5 inverters 30 kW and a rect ifier unit
160 kW, should continue to oper ate for a maximum of 1 second with the same m otor speeds when
power failures occur. The power requir ement is 150 kW. The required energy during buffering,
including the losses, should be covered by the kinetic energy of a high-inert ia flywheel load. In this
case an additional inverter with induction motor and coupled flywheel should be connected. Wit h
this arrangement, the speed of the driven loads can be maintained during buffering; only the buffer
drive speed decreases (f lywheel drive). The following configuration is obtained.
INV
EE 160 kW
30 kW
buffer drive
INV INV30 kW
30 kW 30 kW
IM IM IM
5 x
When t he system is powered-up, at first, the induction motor with the coupled flywheel is
accelerated. The driven loads are then powered-up. This prevents the rectifier unit from being
overloaded. In operat ion, the buffer drive practically only draws the reactive power. The DC link
current of the associated invert er is t herefore ≈0, as only the losses of the m otor, running under
no-load conditions, and t he invert er losses have to be covered. When a power failur e occur s, the
buffer drive supplies the r e quired power to the DC link.
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Dimensioning t he buffer drive
For the g iven multi-motor drive, with the f o llowing dat a
PkW
load =30 (t he sam e load for every motor) ,
Mot =0918.
the f ollowing dimensioning data are obtained for the buff er dr ive. I n this case, a 2-pole mot or is to
be used. The following assumption is made: A 200 kW m ot or with ηmot P=0.959 and nmot P n=2980
RPM. The inverter efficiency is assumed to be ηINV=0.98. The following is ther efore obtained:
PP
erf load i
Mot i Mot P WR
i
=⋅⋅
∑
ηη η
2=⋅ ⋅⋅=530
0918 0959 098 1774
2
... .kW
PP kW
Mot P n erf
=⋅= ⋅=11 177 4 12 1952....
ω
π
Mot P n
Mot P n
n
=⋅⋅2
60 =⋅⋅ =−
22980
60 31206 1
π
.s
JPt
Perf P
Mot P n
=⋅⋅
⋅−
2
11
11
22
ω
(.)=⋅⋅⋅
⋅− =
2 1774 10 1
31206 1 1
11
21
3
22
2
.
.(
.)kgm
The minimum m otor speed of the buffer drive is given by:
nnRPM
Mot P Mot P n
min .
===
11 2980
11 2709
Thus, a 200 kW m ot or 1LA6 317-2AC.. and a 200 k W inverter 6SE7033-7TG60 with vector control
is therefore used f or t he buff er dr ive. The additional moment of inertia is t hen given by:
JJJ kgm
pl P Motsup ..=− =−=21 23 187 2
The calculated addit ional m om ent of inertia can, for example, be im plem ent ed using a steel
flywheel. The following is valid:
Jd r=⋅⋅ ⋅⋅
π
2785 1000
4
.[kgm2]d, r in m
The following is obtained for r if d is set to 0.1 m:
rJ
dm=⋅
⋅⋅⋅ =⋅
⋅⋅ ⋅ =
210 2187
01 785 10 0351
3
43
4
πρ π
.
.. .
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The diamet er of the 10 cm t hick flywheel is theref ore 70.2 cm.
The additional 200 kW inverter in conjunction with the 160 kW rectifier unit is perm issible, as, in
operation, t he buff er dr ive practically requires no act ive power, and, additionally, the f ollowing
condition is f ulfilled:
PP kW
EE inv
≥⋅ =⋅⋅+ =
∑
03 03 30 200 105..(5)
.
In order not to overload the rectifier unit while the f lywheel drive is accelerating, the acceleration
time is, for example, adjusted, so that when the rat ed speed is reached, precisely 160 kW motor
output is required (=r at ed output of the rectifier unit). The motor output is given by:
PM J
tJn
t
Mot P b Mot P n P
Mot P n
HMot P n P
Mot P n
H
max ()=⋅ =⋅ ⋅ =⋅
⋅⋅ ⋅
ω
ω
ω
π
2
60 1
2
or f or t he acceler ation time:
tJ n
Ps
HP Mot P n
Mot P
=⋅
⋅⋅ ⋅=⋅
⋅⋅ ⋅⋅=() () .
max
2
60 121 2 2980
60 1
160 10 128
22
3
π
π
As the acceleration t ime in this case, is less than 60 s, a shor ter acceleration tim e can be achieved
by utilizing the overload capability of the rect ifier unit.
Checking the inverter rating
For the mult i- motor drive inverters, it must be observed, that t he connect ed motors must go into
the f ield-weakening mode earlier, due to the fact that t he DC link voltage is reduced t o 85% in
buffer operat ion. This means that a higher current is drawn. Depending on the pre-loading
condition and duration of the buffer time, this can result in an inverter overload. T he m aximum
motor cur r ent in field-weakening operation is g iven by:
IIM
M
I
Ikn I
Ikn
Mot Mot n load
Mot n
mag
Mot n
mag
Mot n
max ()(())()=⋅ ⋅− ⋅+ ⋅
2222
2
11
kn n
n
Mot
Mot n
=⋅max
.085
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In this part icular case, the 30 kW motors ar e oper ated with rated torq ue and r ated speed. Then
the f ollowing is t r ue:
M
Mload
Mot n =1
kn n
n
Mot n
Mot n
=⋅=
085 1
085..
With
I
Imag
Mot n =041.
IA
Mot n =55
the maximum motor current is given by:
IA
Mot max =62
The RMS value in a 300 s interval is then calculated as follows:
IItI t A
eff Mot P Mot n P
=⋅+ ⋅ − =⋅+ ⋅ − =
max () ()
.
22 22
300
300 62 1 55 300 1
300 5502
The 30 k W inverter, with a rat ed cur rent of 59 A and a maximum current of 80.2 A, ar e t her efore
adequately dimensioned. For the inverter of the buf fer drive, t her e ar e practically no restrictions
due to the reduced DC link voltage, as in t his case, there is no pre- loading, and because, at nmot P
max, the motor cur r ent increase as a result of the field-weakening factor k n=1/0.85, is part ially
compensated by the low motor t or que (Mmot P=Mmot P n/1.1).
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4.10 Harmonics fed back into the supply in regenerative operation
Presently, harm onics fed back int o t he supply can only be calculated using the PATH conf iguring
progr am , for 1Q oper ation. Thus, t he basic diff er ences between motoring using an uncont r olled
rectifier and gener at ing using a rect ifier/r egenerative f eedback unit are to be investigated here.
The following equivalent circuit diagrams are used as basis for these investig ations.
~
~
~
Iload
6-p ul s e uncontro lled
rectifier
Lsupply
CDC lin k
Usupply
3
u
iiDC lin k
supply u
DC lin k
Equivalent circ uit diagram when moto r ing with an uncontrolled rect ifier
~
~
~
Iload
L
C
Autotransformer
1:1,2
u
supply u
i
Usupply
3
i
6-p ul s e control led
rectifier
supply
DC lin k
DC lin k
DC lin k
Equivalent circ uit diagram when gener ating with a rectifier/ r egenerative feedback unit
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The load curr ent I load is assumed to be constant. It is calculated as follows:
INVMotlinkDC
Mot
load UP
I
ηη
⋅⋅
=motoring
linkDC
INVMotMot
load U
P
I
⋅⋅
=generating
The approximat e DC link voltage is obt ained, when mot or ing, without taking into account t he
voltage drops as follows:
plylinkDC UU sup
35.1 ⋅≈
When generating with t he r e ct ifier/ r egenerative feedback unit, the DC link volt age is specified
throug h the closed-loop control.
The DC link capacitor CDC link is provided thr ough the inverter or dr ive converter. The inductance
Lsupply is calculated from the fault level on the high voltage side SK HS, the nominal transformer
rating STr n, the uk of t he t ransform er, the uk of the commutating react or and the rated apparent
power of the inverter or inverter SU n. For regenerative operat ion with t he r ectifier / regenerat ive
feedback unit, the autotransformer uk is also included. The total inductance Lsupply is given by:
)(
sup SpTrKDrKTrKHSKply LLLLL +++= ′
with
ω
⋅
=
′
HSK
ply
HSK S
U
L2
sup component of the high voltag e network
ω
⋅
⋅
=nTr
Trkply
TrK S
uU
L2
sup component of the converter t r ansformer
ω
⋅
⋅
=nU
Drkply
DrK S
uU
L2
sup component of the reactor
ω
⋅
⋅
=nU
SpTrkply
SpTrK S
uU
L2
sup component of the autotr ansformer
(f or r ect ifier/regenerative feedback unit)
ω
π
=⋅⋅2f
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When calculat ing using an Excel/VBA prog r am, the charact er ist ics of the line curr ent over 3
periods of t he line supply voltage ar e calculated, start ing from practical initial values. The st eady-
state condition is t hen achieved. The 4th period is ther efore output , and Fourier analysis applied.
Depending on the load and line supply inductance, oper at ion with continuous cur r ent or
discontinuous current is obtained. When motoring using an uncontrolled rect ifier, the DC link
voltage is obtained according to the line supply voltage and load. The load current m ust be
determined it eratively. When generating with a re ct ifier/regenerat ive feedback unit , t he DC link
voltage is controlled (closed-loop). The firing angle αSt is therefore iteratively determined
corresponding to the contr ol input for t he DC link voltage.
Calculation example
For a drive with a 90 kW load, the relationships r egarding t he har monics fed back into the line
supply are to be investigated, when motoring/generating . When motoring, a 90 k W drive converter
with 2% uk line reactor is assumed, and for generating , a 90 kW inverter with 90 kW
rectifier/regenerative feedback unit and 4% uk line reactor. A uk of 2% is assumed f or t he
autotransformer . The data reg arding supply voltage (2.2 kV/400 V), t r ansformer (STr=150 kVA,
uk Tr=4%) and line supply fault level (SK HS=10 MVA) are the sam e. The motor eff iciency is 94.9%
and the inverter efficiency is assumed to be 98%. The DC link capacit ance of the drive converter
and inverter is 12 mF.
The inductance Lsupply is given by
LH
KHS
′=⋅⋅ =
400
10 10 314 51
2
6
µ
LH
KTr =⋅
⋅⋅ =
400 004
150 10 314 136
2
3.
µ
LH
KDr2%
2
400 002
3 400 186 314 79=⋅
⋅⋅⋅=
.
µ
(wit h IU n=186 A)
LH
KDr4% 158=
µ
LH
K SpTr 2% 79=
µ
to
HLLLL DrKTrKHSKmotply
µ
266
%2sup =++=
HLLLLL SpTrKDrKTrKHSKgenply
µ
424
%2%4sup =+++=
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The calculation r esult s in a 521 V DC link voltage when motoring . The load current is then given by
INVMotlinkDC
Mot
load UP
I
ηη
⋅⋅
==⋅
⋅⋅
=
90 10
521 0949 098 186
3
.. A
The same DC link voltage is assumed when generat ing. The load cur r ent is then obtained as
follows
linkDC
INVMotMot
load U
P
I
⋅⋅
==⋅⋅ ⋅ =
90 10 0949 098
521 161
3.. A
Additional results
When m ot oring:
AI ply 5.145
)1(sup =basic line cur r ent fundamental, secondary side
cos .
(1)
ϕ
=0963 power fact or of the basic fundamental
AI RMSply 2.153
sup =RMS line current, secondary side
%2.7
sec =DF distortion factor on t he secondar y side
DFprim =2% distortion factor on the pr im ary side
Generating operation:
AI ply 152
)1(sup =basic fundament al of the line curr ent, secondary side
cos .
(1)
ϕ
=0796 power factor of the basic fundamental
AI RMSply 5.162
sup =RMS line current, secondary side
%6.8
sec =DF distortion f actor on the secondary side
DFprim =24%.distor tion factor on t he primary side
α
St =14130
.rectifier firing angle
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-400
-300
-200
-100
0
100
200
300
400
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02
Time in s
Current in A, voltage in V
i supply u
u supply u
Line current and associat ed phase voltage when motoring
-400
-300
-200
-100
0
100
200
300
400
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02
Time in s
Current in A, voltage in V
i supply u
u supply u
Line current and associat ed phase voltage when generating
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0
100
200
300
400
500
600
700
800
900
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02
Time in s
Voltage in V, current in A
u DC link
i DC link
DC link voltag e and DC link current when motoring
-600
-500
-400
-300
-200
-100
0
100
200
300
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02
Time in s
Voltage in V, current in A
u DC link
i DC link
DC link voltag e and DC link current when generating
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0
10
20
30
40
50
60
70
80
90
100
1 5 7 111317192325
Line harmonic curr ent s as a % of the basic fundamental when motoring
0
10
20
30
40
50
60
70
80
90
100
1 5 7 111317192325
Line harmonic curr ent s as a % of the basic fundamental when generat ing
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0
100
200
300
400
500
600
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07
Time in s
Current in A, voltage in V
u DC link
i DC link
DC link current and DC link voltage stabilizing over th r ee line supply periods when mot or ing
Summary
When generating, the load current in t he DC link is lower than when motoring, due to the inf luence
of t he eff iciencies. However, a hig her line cur r ent Isupply (1) is obtained than when motoring due to
the f act that when generating , the cos ϕ(1) is poorer ( t he firing angle is not 0), and due t o the ratio
of t he aut otransfor mer of 1.2 (the curr ent is t ransform ed up).
Inspite of the higher line induct ance, higher harmonic currents ar e obt ained due t o the unfavorable
firing angle of the rectifier when generating .
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4.11 Calculating the braking energy of a mechanical brake
4.11.1 General information
Many drives are equipped with a mechanical holding br ake. At st andst ill with the m ot or shutdown,
this brak e is intended to prevent any motion. This is extremely important for drives with a
significant holding t or que, f or example, hoist ing and elevator drives. When the electrical braking
funct ion fails, the m echanical holding brake m ust also assum e an emergency stop function. Two
examples to calculate the br aking energy are now shown. As a result of supplementary torques
(hoisting t or que and fr ictional torque) , with these examples, it is not possible to simply apply the
condition "max. external m om ent of inertia = m ot or intrinsic mom ent of inertia“ .
4.11.2 Hoisting drive with counterweight
The braking energy is to be calculated for the most unfavorable case for a m echanical br ake for a
hoisting dr ive with counterweight. It has been assumed that t he electrical brak ing has failed.
M
Brake
Gearbox D
F+Q
Counterweight
Chain
~
~~
~
vdown
Block diag ram of t he hoisting drive
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Drive data
Weight of the elevator cabin mF= 2000 k g
Weight of the load mQ= 5100 kg
Weight of the counterweight mG= 4500 kg
Weight of the chain mchain = 200 kg
Moment of inert ia, m otor JMot = 0.36 kgm2
Moment of inert ia, gearbox Jgearbox = 0.012 kgm2
Moment of inert ia, br ake Jbrake = 0.02 kgm2
Gearbox ratio i = 16.79
Pulley diameter D = 0.26 m
Max. traversing velocity vmax = 1.3 m/s
Braking torq ue of the mechanical br ake Mbr = 400 Nm
Permissible braking energ y of the mechanical brake Wbr max = 100 kJ
Mechanical hoisting drive efficiency ηmech = 0.9
Gearbox efficiency ηgearbox = 0.95
The worst case is mot ion downwards at full velocity and with a full load. In this case, ther e is t he
highest kinetic energy and the m ass diff er ence acts to accelerate t he cabin downwards. The
braking energy can be calculated using the energy balance equat ion. The following is tr ue:
)( loadpotloadkingearboxmechZmotkinbr WWWW +⋅⋅+= +
with
2
max
2
max )
2
(
2
1
)()(
2
1D
v
iJJJJW ZmotmotgearbrakemotZmotkin ⋅
⋅⋅⋅=⋅++⋅= ∑++
ω
2
max
2
1vmW loadkin ⋅⋅= ∑
hgmW loadpot ⋅⋅∆=
Further,
hv tbr
=⋅
max 2
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2
2
2max
max br
br
br
motbrbr t
D
v
iM
t
MW ⋅
⋅
⋅⋅=⋅⋅=
ω
chainGQF mmmmm +++=
∑
∆mm m m
FQG
=+−
Appropriate ly combined, the following is obt ained:
iM
D
gm
vm
D
v
iJ
W
br
gearmech
gearmechZmot
br
⋅
⋅⋅∆⋅⋅
−
⋅⋅⋅⋅+
⋅
⋅⋅⋅
=∑∑
+
2
1
2
1
)
2
(
2
12
max
2
max
ηη
ηη
From this formula it can be seen t hat braking can only function, if:
i
D
gm
Mgearmech
br 2
⋅⋅∆⋅⋅
>
ηη
The steady-st at e load torque, referr ed to the motor speed m ust therefore be lower than the
braking torq ue of the mechanical br ake which is available.
The following is obtained with the specified number s:
mkg
∑=+++=2000 5100 4500 200 11800
∆mkg
=+−=
2000 5100 4500 2600
Jkgm
Mot Z+
∑=++ =036 002 0012 0392 2
... .
WWskJ
br =⋅⋅ ⋅
⋅+⋅ ⋅ ⋅
−⋅⋅ ⋅⋅
⋅
==
1
20392 1679 213
026 09 095 1
211800 13
109 095 2600 981 026
2
400 1679
24314 24314
22
.(. .
.).. .
.. . .
.
.
tWD
Miv s
br br
br
=⋅
⋅⋅ =⋅
⋅⋅
=
max
.
.. .
24314 026
400 1679 13 0724
Thus, t he br aking energy which occurs, is less than the permissible br aking energy of the
mechanical brak e.
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The braking energy can also be calculated by analyzing the torq ues.
The braking tor que when moving downwards is as follows:
i
MJiJM gearboxmech
loadloadloadloadZmotbr
αα
⋅
⋅+⋅+⋅⋅= ∑+)(
with
brbr
load
load tDv
t⋅
⋅
== max
max 2
ω
α
2
)
2
(
∑⋅= D
mJload
2
D
gmMload ⋅⋅∆=
Further,
2
2
2max
max br
br
br
motbrbr t
D
v
iM
t
MW ⋅
⋅
⋅⋅=⋅⋅=
ω
gearbrakemotZmot JJJJ ++=
∑+
chainGQF mmmmm +++=
∑
∆mm m m
FQG
=+−
The same formula as for Wbr is obtained by appropriately inserting.
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4.11.3 Traversing drive
For a traversing drive at maximum velocity, the brak ing energy occurring at the mechanical br ake
is to be calculated. It is assumed that t he electrical brak ing has failed.
M
Brake
Gearbox
D
v
m
Guide
Block diag ram of t he traversing dr ive
Drive data
Weight of the load m = 5500 kg
Moment of inertia, motor Jmot = 0.017 kgm2
Moment of inert ia, gearbox Jgearbox = 0.00164 kgm2
Moment of inert ia, br ake Jbrake = 0.00063 kgm2
Gear ratio i = 16
Wheel diam et er D = 0.34 m
Max. traversing velocity vmax = 2.66 m/s
Braking torq ue of the mechanical br ake Mbr = 60 Nm
Permissible braking energ y of the mechanical brake Wbr max = 25 kJ
Traversing r esistance wF= 0.02
Gearbox efficiency ηgearbox = 0.95
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The braking energy is calculated using the energy equation. The following is valid:
VZ
gearfrictionloadkinZmotkinbr WWWW
η
⋅−+= +)(
with
)( frictionloadkin WWsignVZ −= (norm ally positive)
2
max
2
max )
2
(
2
1
)()(
2
1D
v
iJJJJW ZmotmotgearbrakemotZmotkin ⋅
⋅⋅⋅=⋅++⋅= ∑++
ω
2
max
2
1vmW loadkin ⋅⋅=
swgmW Ffriction ⋅⋅⋅=
Further,
sv tbr
=⋅
max 2
2
2
2max
max br
br
br
motbrbr t
D
v
iM
t
MW ⋅
⋅
⋅⋅=⋅⋅=
ω
Re-arranging, t he following is obtained:
iM
D
wgm
vm
D
v
iJ
W
br
F
VZ
gear
VZ
gearZmot
br
⋅
⋅⋅⋅⋅
+
⋅⋅⋅+
⋅
⋅⋅⋅
=∑+
2
1
2
1
)
2
(
2
12
max
2
max
η
η
To star t off with, t he sign of t he gearbox efficiency must be determined. The limiting case with
Wkin load=Wfriction is investigat ed. The following is obt ained for Wbr:
2
2
)
2
(
2
1max
2
max br
brZmotbr t
D
v
iM
D
v
iJW ⋅
⋅
⋅⋅=
⋅
⋅⋅⋅= ∑+
Further, with Wkin load=Wfriction, the following is obtained
1
22
2
⋅⋅ =⋅⋅ ⋅ ⋅mv mgw v t
Fbr
max max
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After eliminat ing tbr, an equation is obtained for wF:
∑⋅⋅⋅
⋅
=+giJ MD
w
Zmot
br
F2
If the following is now valid
∑⋅⋅⋅
⋅
≤+giJ MD
w
Zmot
br
F2
the sign should be set posit ive. O therwise, set the sign negative (η-1=1/η).
The following is obtained with the specified numer ical values:
2
01927.000164.000063.0017.0 kgmJ Zmot =++=
∑+
02.0372.3
81.921601927.0 6034.0
2>=
⋅⋅⋅ ⋅
=
⋅⋅⋅
⋅
∑+giJ MD
Zmot
br (i. e. the sign is positive)
WWskJ
br =⋅⋅⋅
⋅+⋅ ⋅
+⋅⋅⋅⋅
⋅
==
1
2001927 16 2266
034 095 1
25500 266
1095 5500 981 002 034
2
60 16
16156 16156
22
.(.
.). .
...
..
tWD
Miv s
br br
br
=⋅
⋅⋅ =⋅
⋅⋅ =
max
.
..
16156 034
60 16 266 2151
The braking energy obtained is theref or e less than the permissible braking energy of the
mechanical brak e.
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The braking energy can also be calculated by investigating the torques. The following is valid f or
the brak ing torq ue:
i
MJiJM VZ
grearbox
frictionloadloadloadZmotbr
η
αα
⋅−⋅+⋅⋅= ∑+)(
with
)( frictionloadload MJsignVZ −⋅=
α
brbr
load
load tDv
t⋅
⋅
== max
max 2
ω
α
2
)
2
(D
mJload ⋅=
gearboxbrakemotZmot JJJJ ++=
∑+
2
D
wgmM Ffriction ⋅⋅⋅=
Further,
2
2
2max
max br
br
br
motbrbr t
D
v
iM
t
MW ⋅
⋅
⋅⋅=⋅⋅=
ω
The same formula is obt ained for Wbr by appropriately applying.
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4.12 Criteria for selecting motors for clocked drives
4.12.1 General information
Clocked drives with 1FK6/1FT6 synchronous servomot or s or 1PA6 induction servomotors ar e t o
be investigated. These can include, f or example, t r aversing drives, rotary table dr ives and spindle
drives. The motors must fulf ill, on one hand t he dynamic r equirements when accelerating and
decelerating , and on the other hand, oper at ion m ust be permissible f r om a thermal perspect ive.
A drive task can generally be solved using several motors with various rated speeds. The rated
speed is selected according t o the following crit er ia:
• minimum motor size
• minimum motor current (drive converter selection)
• preferred m otor series, e. g. 1FK6 with nn=3000 RPM and nn=6000 RPM
• existing m ot or outputs at a specific rated speed
A gearbox is gener ally used to adapt t he load torque and load speed. I n t his case, a differentiat ion
must be made bet ween wheels:
• a gearbox ratio is specified ( e. g. t oot hed belt ratio)
• the gearbox ratio can be selected ( e. g. planetary g ear box f or motor m ount ing)
For the latt er, an additional degr ee of f r eedom is obtained when selecting the mot or . For example,
the optimum gearbox ratio f or a minimum motor torque can be defined. However, the gear box rat io
which can be selected is restricted by the maximum per missible motor speed, as well as the
possible ratios f or certain gearboxes. Other restrictions are obtained due to the size, efficiency and
costs. T hese m ust be taken int o account when f inally selecting the drive package, comprising
motor, gearbox and drive converter.
Basic considerations re garding m otor utilization
Optimum m ot or utilizat ion accor ding to the dynamic requir em ent s
In order t o det ermine the optim um dynamic utilization of a m ot or, the dynamic limiting curve of the
motor is investigated with the resulting permissible motor out put. This is given by:
60
2mot
dynpermmotdynpermmot n
MP ⋅⋅
⋅=
π
(nmot in RPM)
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0
0.5
1
1.5
2
2.5
0 500 1000 1500 2000 2500 3000 3500 4000 4500
Motor speed in RPM
M perm. / M perm. max
P perm. / P rated
n limit
n rated
Example for the dynamic limiting cur ve and the dynamic permissible motor output for
1FT6082-8AF7
0
0.5
1
1.5
2
2.5
0 500 1000 1500 2000 2500 3000 3500 4000
Motor speed in RPM
M perm. / M perm. max
P perm. / P rated
n limit
n rated
Example for t he dynamic lim it ing curve and the dynamic permissible mot or output f or
1PA6103-4HG
nmotpermperm MMM ⋅== 2
max.. for itmot nn lim
≤,2
.)(
3.1 mot
nmot
stall
perm n
n
M
M⋅= for itmot nn lim
>
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A motor is dynamically suitable for a particular application if all of t he torque/ speed points lie below
the dynamic limiting curve. T he m aximum permissible motor torque M perm. max can only be utilized
up to the lim it ing speed nlimit. After this, ther e is a decr ease due to the voltage limiting cur ve and
the stall torque limiting curve. However, for synchronous ser vomot or s, the maximum possible
motor out put is generally achieved for a motor speed, which lies somewhat higher than nlimit, but is
less than nrated. For induction servomotors, t he maximum possible motor out put is always reached
at nlimit, as t he st all torque limit characteristic is proportional to 1/nmot2.
A pure f lywheel drive (high-inert ia dr ive) with constant acceler ation is now to be investigated. For
the specif ied m ax. load speed nload max and when using a gearbox with moment of inertia JG, the
maximum motor t orque and the maximum motor output when accelerating are given by:
i
JiJJM loadloadloadGmotmot 1
)( maxmaxmax ⋅⋅+⋅⋅+=
αα
60
2max
maxmax mot
motmot n
MP ⋅⋅
⋅=
π
(nmot max in RPM)
60
2
)( max
maxmax mot
loadGmotload n
iJJP ⋅⋅
⋅⋅⋅++=
π
α
with
max
max
load
mot
n
n
i=
60
2max
maxmaxmax load
loadloadload n
JP ⋅⋅
⋅⋅=
π
α
(nload max in RPM)
i
JiJJ
Mn
J
loadGmot
motload
load 1
)(
60
2maxmax
⋅+⋅+
⋅
⋅⋅
⋅=
π
The following curves are obtained, if the points Pmot max/nmot max and Pload max/nmot max are entered for a
constant Mmot max for various values of i.
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n mot max
P load max
P mot max
i
Example for t he max. motor output and the max. load power as a funct ion of the g ear box rat io i
The optim um gearbox ratio f or the highest power output t o t he load, and therefore for the shortest
accelerating time, is given by
)
1
(
)
1
)((
60
2
02
2
maxmaxmax i
JJJ
i
JiJJ
Mn
J
di
dP
loadGmot
loadGmot
motload
load
load ⋅−+⋅
⋅+⋅+
⋅
⋅⋅
⋅−==
π
to
Gmot
load
opt JJ J
i+
=
If t he acceler ation αload max is specified, with the opt imum gearbox rat io, t he lowest accelerating
torq ue Mmot max can be achieved, and therefore also the lowest motor cur rent.
The following is obtained for t he m ax. m ot or output with i=iopt:
maxmax 2loadmot PP ⋅= (i. e. power adaption)
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From the out put perspective, a motor is best utilized when it comes t o t he dynamic lim it ing
characterist ic, if the required max. mot or out put lies close to the max. possible m ot or output. T he
max. load power is reached at a constant m ax. m ot or torque using the optimum gearbox ratio. If
the max. mot or speed, achieved with the optim um gearbox ratio, is hig her than nlimit, then t he
maximum of t he load out put is obtained at nlimit.
Depending on the r at io of the mot or m oment of inertia to the load moment of inertia, a
differentiat ion can be made between 2 limiting cases.
a) The m ot or m oment of inertia has no significant inf luence on t he acceler at ion
For all of t he gearbox ratios coming into question the following is valid:
2
/iJJJ loadGmot <<+
As far as t he Pload max/nmot max characteristic is concerned, nmot max lies to the far left away from the
optimum position.
The following is obtained f or Pload max:
maxmax motload PP ≈
The required output Pload max can be achieved with a high torq ue at a low speed, or with a low
torq ue at a high speed. As f ar as the dimensions are concerned, generally, motor s with a higher
rated speed are m or e favorable, as the m aximum possible m otor output incr eases, for the sam e
dimensions, with higher rated speeds. The utilization is at an opt im um , if the m ot or is being
precisely operated at t he m aximum possible m otor output.
b) The m ot or m oment of inertia has a significant inf luence on t he acceler at ion
In this case, it is favorable to set the optimum gearbox ratio. For opt
ii =, the following is obtained
for Pload max:
2max
max mot
load P
P=
Also here, as far as the dimensions ar e concer ned, generally motor s with hig her r ated speeds are
more favorable. T he m otor utilization is at an opt im um , if t he m otor is operated at nlimit with the
maximum motor out put possible at that point .
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Optimu m utilizat ion of a motor accor ding to the ther m a l r equir e m ents
In order t o determine the optimum ther mal utilization of a motor, t he S1 limiting char acteristic of
the motor is investigated and the resulting permissible m otor output. This is given by:
60
2
.. mot
statpermmotstatpermmot n
MP ⋅⋅
⋅=
π
(nmot in RPM)
0
0.2
0.4
0.6
0.8
1
1.2
0 500 1000 1500 2000 2500 3000 3500 4000 4500
Motor speed in RPM
M perm. S1 / M 0
P perm. / P rated
n rated
Example for t he S1 limiting curve and the st eady-state permissible mot or output for
1FT6086-8AH7
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0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 500 1000 1500 2000 2500 3000 3500 4000
Motor speed in RPM
M perm. S1 / M rated
P perm. / P rated
n nenn
Example for t he S1 limiting curve and the per m issible st eady-st ate motor out put for 1PA6103-4HG
A motor is t her m a lly suitable for a part icular dr ive situation, if t he RMS mot or t orque at naverage lies
below the S1 limiting char act er ist ic. For synchronous servomotors, t he thermally permissible
torq ue decr eases with increasing motor speed. The maximum possible motor out put is however
reached f or a m ot or speed which generally lies above the rated speed. For synchronous
servomotors with higher r at ed speeds and higher rated outputs, this value can also lie below the
rated speed. For induction servomotors, t he t hermally permissible torque for m otor speeds greater
than nrated linearly decreases with 1/nmot. Thus, the m aximum possible motor output is achieved at
nrated and then remains const ant .
Thus, a m otor would be well utilized, as far as t he S1 lim iting charact eristic is concerned, if the
average motor speed is close to the power optimum. However, for clocked dr ives, t his is generally
not possible due to a reduct ion in t he per missible dynamic motor tor que at nmot max due to t he
voltage limit ing character ist ic and the stall tor que limiting char acteristic. Thus, if a drive with a
constant accelerat ing torq ue is investigated, the averag e motor speed may only be so high, t hat at
nmot max, the accelerat ing torq ue is j ust below the dynamic limiting charact eristic. The opt imum
utilization is achieved, if , in addition, the RMS motor t or que is close to the S1 lim iting charact er ist ic
and if t he averag e motor speed is just r eached with the opt imum gearbox rat io for the lowest RMS
motor t or que (for a pure flywheel drive [high-inertia dr ive]) , the minimum of the RMS motor torque
coincides with the minimum of the accelerating torque).
Of cour se, these conditions can only seldomly be simultaneously fulfilled.
4 Information for specific applications 09.99
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Procedure when selecting t he r at ed speed for a specific drive task
a) Gearbox ratio is specified
• motors with various rated speeds are deter m ined using the PATH conf iguring pr ogram
• a suitable motor/dr ive converter com bination is selected
b) The g ear box rat io can be select ed
• a motor with the smallest frame size for a specific rated speed is determined using t he PATH
configuring pr ogram, by varying t he gearbox ratio within the possible limits
• this technique is then applied for other r at ed speeds
• a suitable motor/gearbox/drive converter combination is select ed
Depending on whether the m ot or is selected according t o t he RMS value (e. g. shor t no-load
intervals), accor ding to the maximum m ot or torque (e. g. long no-load intervals) or additionally
according t o t he optimum gear box rat io, for a specific drive task, a m ot or with lower or hig her r ated
speed might be m ore favorable. The following examples will clearly illustrate this.
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4.12.2 Example 1, rotary table drive with i=4 and short no-load interval
Drive data
Load moment of inertia J load = 0.4 kgm2
Moment of inert ia of the g ear box (t oot hed belt pulleys) JG= 0.002 kg m 2
Angle ϕtot. = 180 degrees
Positioning t im e ttot. = 0.2 s
No-load interval tp= 0.2 s
A 1FT6 mot or is t o be used.
Traversing char acteristic
ttot. t
t
p
ω
load max
ω
load
ϕ
tot.
1
.
.
max 42.31
2.0
22 −
=
⋅
=
⋅
=s
ttot
tot
load
π
ω
2
.
max
max 2.314
2.0 42.312
2
−
=
⋅
== s
ttot
load
load
ω
α
RPMn load
load 300
26042.31
2
60
max
max =
⋅⋅
=
⋅⋅
=
π
π
ω
RPMnin loadmot 12003004
maxmax =⋅=⋅=
4 Information for specific applications 09.99
508 Siem ens AG
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Thus, t he m aximum m otor torque when accelerating or decelerat ing is given by:
i
JiJJM loadloadloadGmotmot 1
)( maxmaxmax ⋅⋅+⋅⋅+=
αα
The RMS motor tor que and the average m ot or speed are given by:
707.0
2.02.0 2.0 maxmax
.
.
max
.
.
2max ⋅=
+
⋅=
+
⋅=
+⋅
=motmot
ptot
tot
mot
ptot
totmot
rms MM
tt t
M
tt
tM
M
RPM
tt
t
n
n
ptot
tot
mot
average 300
2.02.0
2.0
2
1200
2
22
.
.
max
=
+
⋅
=
+
⋅⋅
=
For the specified gearbox, the maximum motor t or que and the RMS motor tor que only depend on
Jmot.
When select ing an 1FT6 mot or , the following 3 mot or s are obtained
Rated speed RMS motor current Motor type / rated out put Drive converter / rated out put
1500 RPM 13.09 A 1FT6105-8AB7 / 6.6 kW 6SE7021-4EP50 / 5.5 kW
2000 RPM 17.45 A 1FT6105-8AC7 / 8 kW 6SE7022-1EP50 / 7. 5 kW
3000 RPM 26.18 A 1FT6105-8AF7 / 9.7 kW 6SE7022-7EP50 / 11 kW
Due to the short no-load interval, the motor is selected according to the RMS value. The 3 motor s
have the same fr ame size and length, have different rated speeds and rated out put s, but the same
motor m om ent of inertia and the same S1 character ist ic. In this case, t he m ost favorable mot or is
that with the lowest motor cur r ent, i. e. in t his case, the motor whose rated speed is locat ed close
to the average motor speed naverage=300 RPM. T he motor t ype with 1500 RPM fulfills this condition
the best, and the smallest converter, with 5.5 kW is obtained.
09.99 4 Information for specific applications
Siem ens AG 509
SIMOV E RT MASTERDRIVES - Application Manual
0
5
10
15
20
25
30
35
40
45
50
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Torque in Nm
M perm. S1
M RMS / n avera ge
MS1 characteristics and MRMS at naverage for motors with nn=1500 RPM, nn=2000 RPM and
nn=3000 RPM
0
20
40
60
80
100
120
140
0 200 400 600 800 1000 1200 1400 1600
Motor speed in RPM
M mot, M perm. in Nm
Dynamic limiting cur ve and mot or torque characteristic f or the motor with nn=1500 RPM
4 Information for specific applications 09.99
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4.12.3 Example 2, rotary table drive with i=4 and long no-load interval
The rot ar y table dr ive f rom example 1, with otherwise identical data, but a no- load int erval of 4 s is
now to be investigated. The average motor speed is:
RPM
tt
t
n
n
ptot
tot
mot
average 57.28
42.0
2.0
2
1200
2
22
.
.
max
=
+
⋅
=
+
⋅⋅
=
The rat io of the RMS torque t o t he maximum torque is given by:
22.0
42.0 2.0 maxmax ⋅=
+
⋅= motmotrms MMM
The following 4 1FT6 motor s ar e selected
Rated speed Max. motor cur r ent Motor t ype / r at ed out put Drive converter / rated
output
2000 RPM 19.52 A 1FT6084-8AC7 / 3. 5 kW 6SE7018- 0EP50 / 3 kW
3000 RPM 29.42 A 1FT6064-6AF7 / 2. 2 kW 6SE7021-0EP50 / 4 kW
4500 RPM 44.15 A 1FT6064-6AH7 / 2. 3 kW 6SE7022- 1EP50 / 7. 5 kW
6000 RPM 62.15 A 1FT6082-8AK7 / 3. 5 kW 6SE7022-7EP50 / 11 kW
As a result of the longer no- load t im e, in this case the motor is select ed according to t he dynamic
limiting char act eristic. The select ed drive converter, type of construction Compact Plus, is utilized
for a 300 % r at ed cur rent. T his is possible, as t he current flows for less than 250 ms and t he no-
load time is gr eater than 750 ms. From the two smallest m ot or s with nn=3000 RPM and nn=4500
RPM, that motor with the lowest rat ed speed is m or e favorable, as in this case t he m otor torque
characterist ic can be bet t er adapted to the dynamic limiting charact er ist ic, t hus resulting in a lower
motor cur r ent. Althoug h t he m otor with nn=2000 RPM is larger , it has a lower motor current . The
reason f or t his is that the maximum speed is closer t o the rated speed. The optimum com binat ion
is the motor with nn=3000 RPM t ogether with the 4 k W drive converter.
09.99 4 Information for specific applications
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SIMOV E RT MASTERDRIVES - Application Manual
0
5
10
15
20
25
30
35
40
0 500 1000 1500 2000 2500 3000 3500 4000 4500
Motor speed in RPM
M mot, M perm. in Nm
M perm. (3000 RPM)
M perm. (4500 RPM)
M max
Dynamic limiting char acteristic and motor torque characteristic for motor s with nn=3000 RPM
and nn=4500 RPM
0
1
2
3
4
5
6
7
8
9
10
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
Torque in Nm
M perm. S1
M RMS / n avera ge
MS1 characterist ic and Mrms at naverage for the motor with nn=3000 RPM
4 Information for specific applications 09.99
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4.12.4 Example 3, rotary table drive with a selectable ratio and short no-load interval
Drive data
Load moment of inertia J load = 0.075 kg m 2
Moment of inert ia, gearbox JG= 0.0003 kgm2
Angle ϕtot. = 18 degrees
Positioning t im e ttot. = 0.05 s
No-load time tp= 0.05 s
A 1FT6 mot or is t o be used.
1
.
.
max 566.12
05.0 10
2
2−
=
⋅
=
⋅
=s
ttot
tot
load
π
ϕ
ω
2
.
max
max 65.502
05.0 566.122
2
−
=
⋅
== s
ttot
load
load
ω
α
RPMn load
load 120
260566.12
2
60
max
max =
⋅⋅
=
⋅⋅
=
π
π
ω
maxmax loadmot nin ⋅=
The maximum m ot or torque when accelerating and when decelerating as well as the RMS motor
torq ue is obt ained, as in example 1, as follows:
i
JiJJM loadloadloadGmotmot 1
)( maxmaxmax ⋅⋅+⋅⋅+=
αα
707.0
05.005.0 05.0 maxmax
.
.
max
.
.
2max ⋅=
+
⋅=
+
⋅=
+⋅
=motmot
ptot
tot
mot
ptot
totmot
RMS MM
tt t
M
tt
tM
M
For a selectable g ear box rat io, t he m aximum m otor torque and the RMS motor torque are
dependent on Jmot, JG and i. The average mot or speed is given by:
ptot
tot
load
average tt
t
ni
n+
⋅⋅
⋅
=.
.
max 2
22
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3 1FT6 mot or s are selected, taking into account t he optimum gear box rat io.
Rated speed RMS motor cur r ent Motor type / rated out put Drive converter / rated
output
2000 RPM 3.44 A 1FT6064-6AC7 / 1.7 kW 6SE7015-0EP50 / 1.5 k W
3000 RPM 4.98 A 1FT6064-6AF7 / 2.2 kW 6SE7015-0EP50 / 1.5 kW
4500 RPM 7.51 A 1FT6064-6AH7 / 2.3 kW 6SE7018-0EP50 / 3 kW
As a result of the short no- load time, the motor is selected according to the RMS value. The 3
motors have the sam e frame size and length, but have different r ated speeds and rated output s,
and the same mot or moment of inertia and the same S1 char acteristic. T hus, when the gearbox is
optimized, t he sam e value is obtained for all of the 3 mot o r s
85.6
0003.00013.0 075.0 =
+
=
+
=Gmot
load
opt JJ J
i
In the table above, 7 was used as the basis. The most favorable motor is t hat m otor with the
lowest motor curr ent , i. e. the m otor whose rated speed is the closest t o t he average motor speed.
The mot o r t ype with 2000 RPM fulfills this condition. The motor with nn=1500 RPM cannot be
used, as the smallest motor with this rated speed is the 1FT6102-8AB7 with a rated out put of
3.8 kW. T he m aximum speed and t he average motor speed ar e given by:
RPMnin loadmot 8401207
maxmax =⋅=⋅=
RPM
tt
t
ni
n
ptot
tot
load
average 210
05.005.0
05.0
2
1207
2
22
.
.
max
=
+
⋅
⋅
=
+
⋅⋅
⋅
=
The following diagram is obtained if t he points Mrms, naverage are enter ed int o diagram Mperm. S1 for
the various gearbox ratios.
4 Information for specific applications 09.99
514 Siem ens AG
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0
2
4
6
8
10
12
0 200 400 600 800 1000 1200 1400 1600 1800 2000
n mittel in 1/min
M RMS, M perm. S1 in Nm
M perm. S1
M RMS / n avera ge
i=7
i=13
i=3
Optimizing MRMS for the motor with nn=2000 RPM
If t he transition points Mmot max, nmot max ar e entered into the diagram Mperm. max, f or t he various
gearbox ratios, t he following diagram is obtained.
0
5
10
15
20
25
30
35
40
0 200 400 600 800 1000 1200 1400 1600 1800 2000
n mot max in RPM
M mot max, M perm. in Nm
i=7
i=13
i=3
Optimizing Mmot max for a motor with nn=2000 RPM
09.99 4 Information for specific applications
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SIMOV E RT MASTERDRIVES - Application Manual
4.12.5 Example 4, rotary table drive with selectable ratio and long no-load interval
The rot ar y table dr ive f rom example 1, with otherwise the same data but a no- load t im e of 4 s is
now to be investigated with a selectable ratio ( planet ar y g ear ) . The appropriate gearbox moment
of iner t ia is used depending on the selected mot or type.
The following 4 1FT6 motor s were selected
Rated speed Max. motor cur r ent Motor t ype / r at ed out put Drive converter / rated
output
2000 RPM 16.33 A 1FT6084-8AC7 / 3. 5 kW 6SE7018- 0EP50 / 3 kW
3000 RPM 14.55 A 1FT6064-6AF7 / 2. 2 kW 6SE7015-0EP50 / 1.5 kW
4500 RPM 17.56 A 1FT6064-6AH7 / 2. 3 kW 6SE7018- 0EP50 / 3 kW
6000 RPM 14.23 A 1FT6044-4AK7 / 1. 9 kW 6SE7015-0EP50 / 1.5 kW
In this case, depending on the motor type, the f ollowing values are used for i and JG.
Rated speed i iopt Nmot max JG
2000 RPM 5 8.48 1500 RPM 0.00084 kgm2
3000 RPM 7 16.12 2100 RPM 0.00024 k gm2
4500 RPM 10 16.33 3000 RPM 0.0002 k gm2
6000 RPM 16 (2stage) 24.08 4800 RPM 0.00018 kg m2
In this case, iopt cannot be used as gearbox ratio due to t he dynamic lim it ing characterist ic. The
selected drive converters, t ype of construction Compact Plus were utilized for a 300 % rated
current. This is possible, as the current flows f or less t han 250 m s , and the no-load time is g r eater
than 750 ms. Contrary to example 2 with a fixed gearbox rat io, by optim izing t he gearbox ratio, the
smallest mot or is obtained at the hig hest r ated speed nn=6000 RPM.
4 Information for specific applications 09.99
516 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
0
10
20
30
40
50
60
70
0 1000 2000 3000 4000 5000 6000
Motor speed in RPM
M mot, M perm. in Nm
M perm. (6000 RPM)
M perm. (2000 RPM)
M perm. (6000 RPM)
M perm. (2000 RPM)
Dynamic limiting char acteristics and motor torque curves for mot or s with nn=2000 RPM (fram e size
80) and nn=6000 RPM (fram e size 48)
09.99 4 Information for specific applications
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SIMOV E RT MASTERDRIVES - Application Manual
4.12.6 Example 5, traversing drive with selectable ratio and longer no-load time
Drive data
Weight to be moved (transported) m = 5500 kg
Load wheel diameter D = 0. 34 m
Specific t r aversing resistance wF= 0. 02
Gearbox ratio i = 16
Mech. efficiency η= 0.75
Max. traversing velocity vmax = 2.66 m /s
Max. acceleration amax = 0.44 m/s2
Max. traversing travel smax = 42.6 m
No-load time after a tr aversing sequence tP= 10 s
In this case, it involves the traversing drive for the high-bay racking vehicle fr om Section 3.1.4. In
this case, a 1PA6 motor is to be used.
The following 3 1PA6 motors are select ed.
Rated speed Max. motor cur r ent Motor t ype / r at ed out put Inverter / rat ed output
1150 RPM 32.82 A 1PA6107-4HD / 7. 2 kW 6SE7022-6TC61 / 11 kW
1750 RPM 30.66 A 1PA6105-4HF / 8 kW 6SE7022-6TC61 / 11 kW
2300 RPM 26.97 A 1PA6103-4HG / 7. 5 kW 6SE7022- 6TC61 / 11 kW
In this case, depending on the motor type, the f ollowing values were used for i.
Rated speed i iopt Nmot max Nlimit
1150 RPM 7 103 1046 RPM 1313 RPM
1750 RPM 10 103 1494 RPM 1750 RPM
2300 RPM 16 (2stage) 134 2391 RPM 2492 RPM
The gearbox ratios were selected, so that the maximum motor speed is as close as possible to
speed nlimit. In this case, iopt cannot be used as gearbox ratio, due to the dynamic limiting
characterist ic and t he lim it imposed by the maximum permissible m ot or speed. The gear box
moment of inertia was neglected as a r esult of t he high load moment of inertia. The smallest m ot or
is the 1PA6103-4HG with nn=2300 RPM. This m otor also draws the lowest current, as it is the
most favorable regarding nlimit and iopt. Motors with nn=400 RPM and nn=2900 RPM cannot be used
as the smallest motor with nn=400 RPM is the 1PA6163-4HB, and t he sm allest motor with
nn=2900 RPM, 1PA6184-4HL.
4 Information for specific applications 09.99
518 Siem ens AG
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In this case, the optimum gearbox ratio is calculated using t he m otor torque when accelerating
i
MJiJM
mech
loadloadloadloadmotmot ⋅
⋅+⋅+⋅⋅=
η
αα
1
)( maxmaxmax
to
mechmot
load
load
load
opt J
M
J
i
η
α
⋅
+
=max
The following is true
Nm
D
wgmM Fload 45.183
2
34.0
02.081.95500
2=⋅⋅⋅=⋅⋅⋅=
2
maxmax 59.2
34.02
44.0
2−
=⋅=⋅= s
D
a
load
α
222 95.158)
2
34.0
(5500)
2
(kgm
D
mJload =⋅=⋅=
09.99 4 Information for specific applications
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SIMOV E RT MASTERDRIVES - Application Manual
0
20
40
60
80
100
120
0 500 1000 1500 2000 2500 3000
Motor speed in RPM
M mot, M perm. in Nm
M perm. (2300 RPM)
M perm. (1150 RPM)
M max (23 00 RPM)
M max (11 50 RPM)
Dynamic limiting char acteristics and motor torque characteristics (Mmot max when accelerating) for
motors with nn=1150 RPM and nn=2300 RPM
4 Information for specific applications 09.99
520 Siem ens AG
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4.13 Optimum traversing characteristics regarding the maximum motor
torque and RMS torque
4.13.1 Relationships for pure flywheel drives (high-inertia drives)
For pure flywheel drives (high-iner tia drives), optimum t r aversing characterist ics ar e t o be
determined regarding t he maximum motor t or que and RMS torque. As an example, a r otary table
drive will be investigat ed with the following traversing characteristics.
ω
load
ω
load max
ϕ
total
tbtv
ttotal
tp
t
α
load
α
load max
α
load max
-
t
Angle ϕtot., posit ioning time ttot. as well as the no-load interval tp are specified. The traversing
characterist ic should be symmetrical with tb=tv.
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SIMOV E RT MASTERDRIVES - Application Manual
Then the following is true
btot
tot
load tt −
=.
.
max
ω
for .
2totb tt ≤⋅
b
load
load tmax
max
ω
α
=
π
ω
⋅⋅⋅
=2
60
max
max i
nload
mot in RPM
The maximum m ot or torque when accelerating or deceler at ing is then given by
maxmax ))(( load
load
Gmotmot i
J
iJJM
α
⋅+⋅+=
2
.
.
))((
btotb
totload
Gmot ttti
J
iJJ −⋅
⋅+⋅+=
with
JGmoment of inertia, gearbox
load
Jmoment of inertia, load
igearbox ratio
The RMS motor tor que and the average m ot or speed are then given by
ptot
b
mot
ptot
bmot
rms tt t
M
tt
tM
M+
⋅
⋅=
+⋅⋅
=.
max
.
2max 2
2
)(2 60
)2(2
2
.
.
.
.max
max
ptot
tot
ptot
btotmotb
mot
average tt i
tt
ttnt
n
n+⋅⋅ ⋅⋅
=
+
⋅−⋅+⋅⋅
=
π
ϕ
The maximum m ot or torque and the RMS torque can be represented as a function of tb, with ttot., tp
and ϕtot. specified. The average speed is independent of tb and is given by:
2
.
2
.
.max 11
))((
btotbbtotb
tot
load
Gmotmot ttt
k
ttti
J
iJJM −⋅
⋅=
−⋅
⋅⋅+⋅+=
ϕ
b
ptot
motrms t
tt
MM ⋅
+
⋅= .
max 2
4 Information for specific applications 09.99
522 Siem ens AG
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It will now be investigated for which value of t b, the maximum motor torque is at a minimum. The
minimum for Mmot max is given by
)2(
)( 1
0.
22
.
max btot
btotbb
mot tt
ttt
k
dt
dM ⋅−⋅
−⋅
⋅−==
for
2.tot
bt
t=
The optim um t raversing character ist ic for the lowest mot or torque when accelerating or
decelerating is t her efore a triangular characteristic. However, f or a t r iangular characteristic, the
highest m ot or speed is also reached. If the maximum mot or torque and the maximum motor speed
are referred respect ively to the values for the triangular traversing char act eristic (tb=ttot./2) then the
following is obt ained:
2
..
max
max
)(
1
4
1
tot
b
tot
b
triangularmot
mot
tt
tt
M
M
−
⋅=
.
max
max
1
1
2
1
tot
b
triangularmot
mot
tt
n
n
−
⋅=
0
1
2
3
4
5
6
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
tb / t tot.
M max / M max triangular
Maximum mot or t or que as a funct ion of tb/ttot
09.99 4 Information for specific applications
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SIMOV E RT MASTERDRIVES - Application Manual
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
t b / t tot.
n mot max / n mot max triangular
Max. motor speed as a function of tb/ttot.
It will now be investigated, for which value of tb, the RMS motor t o r que has a minimum. The
minimum for Mrms is then given by
)
2
1
(
2
0max
max
.b
motb
b
mot
ptotb
rms t
Mt
dt
dM
ttdt
dM ⋅
⋅+⋅⋅
+
==
)
211
)( 2
(
22
.
22
.
.
.b
btotb
b
btotb
btot
ptot t
ttt
kt
ttt tt
k
tt ⋅
⋅
−⋅
⋅+⋅
−⋅ ⋅−
⋅−⋅
+
=
for
3.tot
bt
t=
The optim um t raversing character ist ic for the lowest RMS motor t or que is therefore a trapezoidal
traversing char acteristic with the same accelerat ing time, decelerating time and const ant velocity
time component s ( i. e. sam e r atios between them). I f the RMS motor torque is referred to the
values for the triangular traversing charact er ist ic ( tb=ttot./2), t hen the following is obtained:
2
..
.)(
1
4
2
tot
b
tot
b
tot
b
triangularrms
rms
tt
tt
tt
MM
−
⋅⋅=
4 Information for specific applications 09.99
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SIMOV E RT MASTERDRIVES - Application Manual
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
t b / t tot.
M rms / M rms triangluar
RMS motor torq ue as a function of tb/ttot.
4.13.2 Relationships for flywheel drives (high-inertia drives) with a constant load torque
If t he sam e example is used as basis as under Point 4.13.1, then, taking int o account a constant
load torq ue, the following changes ar e obtained. The m aximum m ot or torque is obt ained when
accelerating, as follows
i
M
ttti
J
iJJM load
btotb
totload
Gmotmot +
−⋅
⋅+⋅+= 2
.
.
max ))((
i
M
Mload
b+=
The RMS motor torque is t hen given by
ptot
b
load
vbtot
load
b
load
b
rms tt
t
i
M
Mtt
i
M
t
i
M
M
M+
⋅−+⋅−⋅+⋅+
=.
2
.
22 )()2()()(
09.99 4 Information for specific applications
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SIMOV E RT MASTERDRIVES - Application Manual
or with Mb=Mv,
ptot
tot
load
bb
rms tt
t
i
M
tM
M+
⋅+⋅⋅
=.
.
22 )(2
The expression f or the average speed does not chang e. As a r esult of t he const ant factor M load/i at
Mmot max, the minimum for the mot or torque is ag ain obtained at tb=ttot./2, i. e. for a triangular
traversing characteristic. The minimum for the RMS motor torque is then given by
2
.
1
btotb
bttt
kM −⋅
⋅=
.
2
2
.
2
.
)(
)(21 tot
load
btotb
ptot
rms t
i
M
ttt k
tt
M⋅+
−⋅ ⋅
⋅
+
=
and
22
.
.
2
.
2
2
.
2
2
.))(( )(2)(
)(
)(2
2
21
0
btotb
btotbbtot
load
btotb
ptot
b
rms ttt ttttt
i
M
ttt k
k
tt
dt
dM −⋅ −⋅⋅−−
⋅
+
−⋅ ⋅
⋅
⋅−
⋅
+
==
for
3.tot
bt
t=
Also here, nothing changes.
4 Information for specific applications 09.99
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4.13.3 Summary
Pure f lywheel drives (high- iner tia drives) without additional load torques
The optim um traversing charact eristic for the lowest motor tor que when accelerating or when
decelerating is a triangular traversing charact er ist ic. However, the highest motor speed is also
obtained. I f the ma ximum m otor tor que is available (e. g. as a result of the dynamic limit ing
characterist ic) , then with the triangular traversing char acteristic, t he shor test positioning t im e is
achieved.
The optim um traversing charact eristic regar ding the lowest RMS motor torque is then obtained for
a trapezoidal traversing characteristic with the same sub- division of accelerating time, decelerating
time and constant velocity time. Using this tr aversing characteristic, the lowest motor t em perature
rise is achieved and a motor speed which is 25% lower with respect t o the triang ular traversing
characterist ic; however, the m aximum motor tor que is 12.5% higher .
Flywheel drives (high- iner tia drives) with a constant load torque (e. g. friction torque)
It can be proven that the triangular traversing char acteristic results in the lowest motor tor que and
the trapezoidal charact er ist ic with tb=ttot./3 results in the minimum RMS value.
09.99 4 Information for specific applications
Siem ens AG 527
SIMOV E RT MASTERDRIVES - Application Manual
4.13.4 Example with 1PA6 motor
The optim um t raversing character ist ic for a rot ar y drive is to be det er m ined. This drive turns tr ee
trunk s through 90 deg r ees within 1 s. This is then f ollowed by a no-load interval of 0. 5 s. During
this no-load inter val, t he m agnetizing curr ent still continues to flow through the mot or.
Drive data
Load moment of inertia Jload = 1000 kg m2
Moment of inert ia, gearbox JG= 0.00831 kgm2
Gearbox ratio i = 80
Gearbox efficiency ηG= 0.95
Mech. efficiency ηmech = 0.9
Angle ϕtot. = 90 degrees
Positioning t im e ttot. = 1 s
No-load interval tp= 0.5 s
As this involves an induction motor, t he m inim um RMS torque is not investigated, but instead, t he
minimum RMS current . The condition tb=ttot./3 for the minim um RMS value is also approximately
true for the RMS current .
The mot or torque is given by
load
VZ
Gmech
load
Gmotmot iJ
iJJM
α
ηη
⋅
⋅⋅
+⋅+= )
)(
)((
with
)( load
signVZ
α
=
maxloadload
α
α
=when accelerating
maxloadload
α
α
−= when decelerating
0=
load
α
constant velocity motion, no- load interval
2
.
.
max btotb
tot
load ttt −⋅
=
α
4 Information for specific applications 09.99
528 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
The maximum m ot or torque when accelerating is given by
maxmax ))(( load
Gmech
load
Gmotmot iJ
iJJM
α
ηη
⋅
⋅⋅
+⋅+=
The following is valid for the mot or current ( without t aking into account saturation):
2
2222 1
)())(1()(
nnmot
mag
n
nmot
mag
nmot
mot
nmotmot kI
I
k
I
I
M
M
II ⋅+⋅−⋅⋅≈
kn=1for nmotmot nn ≤constant flux range
nmot
mot
nn
n
k=for nmotmot nn >field-weakening range
The following is valid for the RMS motor cur r ent and the average motor speed
ptot
iii
imotimot
rms tt
tt
II
I+
−⋅
+
≈∑−
−
.
1
2
1)()
2
(
RPM
tt i
n
ptot
tot
average 800
)5.01(2
8060
2
)(2 60
.
.=
+⋅⋅
⋅⋅
=
+⋅⋅ ⋅⋅
=
π
π
π
ϕ
The two limit ing values with tb=ttot./2 (triangular) and tb=ttot./3 (trapezoidal characteristic with tb=tv=tk)
is calculated using t he PATH configuring program. In this case, both cases ar e possible for the
same motor . The following results are obtained:
Triangular trapezoidal (with tb=tv=tk)
Mmot max =150.83 Nm Mmot max =169.7 Nm
nmot max =2400 RPM nmot max =1800 RPM
Irms =53.44 A I rms =49.76 A
Imot max =71 A Imot max =76.35 A
Pbr max =28. 05 kW Pbr max =23.67 kW
Pbr average =4.67 kW Pbr average =2.63 kW
Braking unit: braking unit:
P20=50 kW+ext. brake resistor P20=20 k W+ext. brak e r esist or
09.99 4 Information for specific applications
Siem ens AG 529
SIMOV E RT MASTERDRIVES - Application Manual
The same dr ive converter is also obt ained for both of the traversing character ist ics.
Motor:
1PA6 137-4HG.
Pn=29 kW, Mn=120 Nm, Mperm.=240 Nm, n n=2300 RPM, In=56 A, Imag=21 A, Jmot=0. 109 kgm2
Drive converter:
6SE7026-0ED61
PU n=30 kW, IU n=59 A, IU max=80.5 A
The mor e favorable case is the trapezoidal charact er ist ic with t b=ttot./ 3, as the RMS current is lower
here (i. e. a lower mot or temperat ur e rise), and, in addition, a lower rating brak ing unit is adequat e.
For the tr iangular traversing characterist ic, the drive operates in the field-weakening r ange due to
the higher speed. The resulting higher peak current is compensated by the lower maximum motor
torq ue. For the tr iangular traversing characteristic, a higher braking power is obtained, as the
speed increases more t han the maximum motor t orque decreases. The higher peak current for t he
trapezoidal traversing characteristic can be explained due to the higher maximum mot or torque,
and the therefore associated m otor saturat ion effects.
0
50
100
150
200
250
300
350
400
450
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
tb / t tot.
Torque in Nm
M mot max
M perm.
Maximum mot or t or que and dynamic limiting char act eristic as a funct ion of tb/ttot.
4 Information for specific applications 09.99
530 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
0
10
20
30
40
50
60
70
80
90
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
tb / t tot.
I in A
I rms
I n
RMS motor current and rated motor cur r ent as a function of tb/ttot.
0
0.5
1
1.5
2
2.5
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
Time in s
Angular velocity in 1 / s
Angular velocity f or t he trapezoidal traversing characteristic with tb=ttot./3
09.99 4 Information for specific applications
Siem ens AG 531
SIMOV E RT MASTERDRIVES - Application Manual
-8
-6
-4
-2
0
2
4
6
8
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
Time in s
Angular acceleration in 1/s^2
Angular acceler at ion for the t r apezoidal tr aversing characteristic with tb=ttot./3
-150
-100
-50
0
50
100
150
200
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
Time in s
Motor torque i n Nm
Motor torque for the trapezoidal traversing characteristic with tb=ttot./3
The mot or t orque when decelerating is lower as a result of the efficiency.
4 Information for specific applications 09.99
532 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
-30000
-20000
-10000
0
10000
20000
30000
40000
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
Time in s
Motor power in W
Motor output for the tr apezoidal tr aversing char acteristic with tb=ttot./3
0.00
10.00
20.00
30.00
40.00
50.00
60.00
70.00
80.00
90.00
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
Time in s
Motor current in A
Motor current for the t r apezoidal tr aversing char acteristic with tb=ttot./3
09.99 5 Index
Siem ens AG 533
SIMOV E RT MASTERDRIVES - Application Manual
5 Index
A
Accelerating force...............................................11
Accelerating power .............................................14
Accelerating time ..........................................14, 20
Accelerating time from the line supply..............395
Accelerating time, shortest .................................17
Accelerating torque.............................................14
Accelerating work................................................14
Acceleration........................................................11
Acceleration due to the force of gravity.................9
Acceleration torque.............................................26
Angle...................................................................11
Angle of rotation..................................................13
Angular acceleration .....................................11, 26
Angular deceleration...........................................26
Angular velocity...................................................11
Autotransformer................................................425
B
Brake application time.........................................38
Brake control.......................................................35
Brake motor ........................................................38
Brake resistor......................................................36
Braking energy............................................36, 491
Braking time........................................................20
Braking time, minimum.....................................406
Breakaway torque.............................................465
C
Cable drum .........................................................29
Capacitor battery...............................................284
Centrifugal drive................................................261
Connecting bar force.................................186, 213
Conservation of energy law...............................191
Constant power.................................................192
Contact force.....................................................313
Continuous braking power..................................36
Crank angle.......................................................185
Crank drive........................................................208
Crawl velocity............................................. 38, 140
Cross-cutter drive............................................. 254
Cut length................................................. 254, 272
Cut lenth, above synchronous.......................... 277
Cut lenth, sub-synchronous ............................. 275
Cut lenth, synchronous .................................... 276
Cycle time .......................................................... 36
D
DC current braking........................................... 463
Dead center...................................... 208, 294, 307
Deadtimes, changing ....................................... 139
Deceleration torque............................................ 26
Deforming work................................................ 184
Distance....................................................... 11, 13
Distance traveled ............................................. 144
Drag force .......................................................... 25
Drive wheel ........................................................ 25
Droop ............................................................... 199
E
Eccentric press ................................................ 184
Elevator drive..................................................... 83
Elevator height................................................... 83
Emergency off.................................................. 430
Energy.................................................................. 9
Energy equation............................................... 492
F
Feed force.......................................................... 18
Feed velocity...................................................... 18
Field-weakening range............................. 192, 392
Flywheel........................................................... 184
Force.................................................................... 9
Force-ventilated ................................................. 33
Four-jointed system.......................................... 306
Frame size, minimum....................................... 499
Friction angle, spindle........................................ 18
Friction coefficient.............................................. 27
Full-load steady-state output.............................. 29
5 Index 09.99
534 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
G
Gantry crane....................................................... 77
Gearbox ............................................................. 16
Gearbox efficiency ............................................. 16
Gearbox ratio ..................................................... 16
Gearbox ratio, optimum ................................... 503
H
Harmonics and disturbances fed back
into the line supply........................................ 483
High-bay racking vehicle.................................... 60
Hoisting drive................................................ 25, 34
hoisting force...................................................... 13
Hoisting force ..................................................... 29
Hoisting unit........................................................ 77
Hoisting work...................................................... 13
Holding brake............................................. 35, 491
I
Inclination angle ................................................. 92
Incline plane....................................................... 92
Intermittent duty.................................................. 32
Isolating transformer ........................................ 427
K
Kinetic buffering ....................................... 398, 474
Kinetic energy..................................................... 11
Knife roll ........................................................... 254
L
Leakage inductance.........................................418
Line failure........................................................ 398
Linear motion ..................................................... 11
Load distribution............................................... 197
Load duty cycle ............................................ 32, 36
Load equalization control ................................. 200
Load mass.......................................................... 29
Load sinking....................................................... 35
Load torque................................................ 16, 186
Load torque characteristic................................ 361
Load torque oscillation ..................................... 153
Load torque, square-law .................................. 387
M
Mass.....................................................................9
Master/slave drive ............................................199
Mechanical brake.............................................491
Mechanical efficiency.........................................26
Mechanically-coupled drives ............................197
Mesh welding machine.....................................323
Moment of inertia..........................................9, 161
Moment of inertia, roll.......................................106
Moment of inertia, spindle..................................18
Motor utilization ................................................499
Multi-motor drives.............................................474
N
No-load interval time ..........................................33
O
Oscillation frequency, mechanical....................174
Overhead wire height .......................................357
Overlap angle...................................................274
P
Pantograph.......................................................357
Partial load range .............................................418
Permissible acceleration ....................................28
Pitch angle, spindle ............................................18
Pivot drive.........................................................227
Pivot motion......................................................227
Position change, periodic .................................151
Position controller.............................................141
Position difference....................................141, 143
Positioning drives .............................................139
Positioning error ...............................................140
Positioning time................................................145
Positioning, closed-loop control........................139
Positioning, closed-loop control........................149
Positioning, fastest...........................................144
Positioning, open-loop control..................139, 147
Power ...................................................................9
Power adaption.................................................502
Power-on duration..............................................32
Press drive .......................................................184
Press force.......................................................184
09.99 5 Index
Siem ens AG 535
SIMOV E RT MASTERDRIVES - Application Manual
R
Ram distance....................................................186
Ram time (surge time)......................................187
Re-accelerating time.........................................398
Reciprocating compressor, single-cylinder.......161
Reduction factor kf..............................................10
Reel...................................................................103
Regenerative operation...............................36, 483
Resistance force.................................................25
Risk evaluation..................................................430
RMS motor current ...........................................195
RMS torque...................................................10, 32
Roll radius...........................................................13
Rotary table drive..............................................221
Rotational............................................................11
Rotational energy................................................11
Rounding-off .....................................................442
Rraction cycle .....................................................28
S
Saw drive..........................................................293
Self locking..........................................................19
Self-ventilated .....................................................33
Shaft output...........................................................9
Sinusoidal filter..................................................425
Slip........................................................................9
Slip frequency .......................................................9
Specific traction resistance.................................25
Speed (RPM)......................................................13
Spindle diameter.................................................18
Spindle drive ...............................................18, 238
Spindle efficiency................................................18
Spindle length .....................................................18
Spindle pitch .......................................................18
Spindle, horizontal.............................................246
Spindle, vertical.................................................238
Stall torque....................................................9, 192
Static friction....................................................... 27
Supplementary moment of inertia.................... 152
Surge loading................................................... 151
Switch-on duration ............................................. 32
Synchronous speed ............................................. 9
System losses.................................................. 453
T
Tangential force ....................................... 186, 213
Three-cylinder pump........................................ 167
Time constant, mechanical.............................. 152
Torque.................................................................. 9
Torque compensation ...................................... 174
Torque limit ...................................................... 174
Traction drive ............................................... 25, 34
Transformer at the drive converter output........ 425
Traversing curves, optimum............................. 520
Trolley .............................................................. 171
U
Unwind stand with closed-loop tension
control........................................................... 110
Unwinding ........................................................ 105
V
Velocity............................................................... 11
W
Web velocity..................................... 103, 254, 272
Wind force.......................................................... 26
Wind pressure.................................................... 26
Wind surface...................................................... 26
Winder control.................................................. 117
Winder drive..................................................... 103
Winder with closed-loop tension control .. 109, 117
Winding............................................................ 105
Winding power ................................................. 104
Winding torque................................................. 104
Work..................................................................... 9
6 Literature 09.99
536 Siem ens AG
SIMOV E RT MASTERDRIVES - Application Manual
6 Literature
/1/ Moder ne St r om richtertechnik
Peter F. Br osch
Vogel Fachbuch Kampr at h- Reihe
ISBN 3-8023-0241- 9
/2/ Grundlagen der elek trischen Antriebst echnik mit Berechnungsbeispielen
J. Vogel
Hüthig Verlag, 1991
ISBN 3-7785-2103- 9
/3/ AC-Servo-Antriebstechnik
Lehmann, R.
Franzis-Verlag, 1990
ISBN 3-7723-6212- 5
/4/ Elektrische Vorschubantr iebe für Werkzeugmaschinen
H. Groß
Siemens AG
ISBN 3-8009-1338- 0
/5/ Das 1x1 der Antriebsauslegung
W. Garbrecht, J. Schäfer
VDE Verlag, 1994
ISBN 3-8007-2005- 1
/6/ Lenze Formelsamm lung
Carsten Fräger
Lenze GmbH&Co KG Aerzen
/7/ Technische Formeln für die Praxis
Walter (u.a.)
Buch-und Zeit-Ver lagsgesellschaf t mbH Köln, 1992
ISBN 3-8166-0192- 8