Abb technical guide no.07 revc

Technical guide No. 7
Dimensioning of a drive system
ABB drives

2 Dimensioning of a drive system | Technical guide No. 7

Technical guide No. 7 | Dimensioning of a drive system 3
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Specifications subject to change without notice.
3AFE64362569 REV C EN 28.4.2011
Technical guide No. 7
Dimensioning of a drive system

Contents
Chapter 1 - Introduction ............................................................................7
General..............................................................................................7
Chapter 2 - Drive system ...........................................................................8
Chapter 3 - General description of a dimensioning procedure..................9
Chapter 4 - Induction (AC) motor.............................................................11
4.1 Fundamentals.............................................................................11
4.2 Motor current .............................................................................13
4.2.1 Constant flux range .............................................................14
4.2.2 Field weakening range.........................................................15
4.3 Motor power ..............................................................................16
Chapter 5 - Basic mechanical laws .........................................................17
5.1 Rotational motion .......................................................................17
5.2 Gears and moment of inertia .......................................................20
Chapter 6 - Load types............................................................................22
Chapter 7 - Motor loadability...................................................................25
Chapter 8 - Selecting the frequency converter and motor ......................26
8.1 Pump and fan application (Example) ............................................27
8.2 Constant torque application (Example) ........................................29
8.3 Constant power application (Example) ........................................31
Chapter 9 - Input transformer and rectifier..............................................35
9.1 Rectifiers....................................................................................35
9.2 Transformer................................................................................36
Chapter 10 - Index...................................................................................38

Chapter 1 - Introduction
General
Dimensioning of a drive system is a task where all factors have
to be considered carefully. Dimensioning requires knowledge
of the whole system including electric supply, driven machine,
environmental conditions, motors and drives, etc. Time spent at
the dimensioning phase can mean considerable cost savings.

Chapter 2 - Drive system
A single AC drive system consists typically of an input transformer
or an electric supply, frequency converter, an AC motor and load.
Inside the single frequency converter there is a rectifier, DC-link
and inverter unit.
Figure 2.1 A single frequency converter consists of 1) rectifier, 2) DC-link,
3) inverter unit and 4) electric supply.
In multi-drive systems a separate rectifier unit is commonly used.
Inverter units are connected directly to a common DC-link.
Figure 2.2 A drive system which has 1) a separate supply section,
2) common DC-link, 3) drive sections and 4) electric supply.

This chapter gives the general steps for dimensioning the motor
and the frequency converter.
1) First check the initial conditions.
In order to select the correct frequency converter and motor,
check the mains supply voltage level (380 V to 690 V) and
frequency (50 Hz to 60 Hz). The mains supply network’s fre-
quency doesn’t limit the speed range of the application.
2) Check the process requirements.
Is there a need for starting torque? What is the speed range
used? What type of load will there be? Some of the typical
load types are described later.
3) Select the motor.
An electrical motor should be seen as a source of torque.
The motor must withstand process overloads and be able to
produce a specified amount of torque. The motor’s thermal
overloadability should not be exceeded. It is also necessary
to leave a margin of around 30% for the motor’s maximum
torque when considering the maximum available torque in the
dimensioning phase.
4) Select the frequency converter.
The frequency converter is selected according to the initial
conditions and the selected motor. The frequency converter’s
capability of producing the required current and power should
be checked. Advantage should be taken of the frequency
converter’s potential overloadability in case of a short term
cyclical load.
Chapter 3 - General description of a
dimensioning procedure

Dimensioning phase Network Converter Motor Load
1) Check the initial
conditions of the
network and load
2) Choose a motor
according to:
- Thermal loadability
- Speed range
- Maximum needed
torque
3) Choose a frequency
converter according to:
- Load type
- Continous and
maximum current
- Network conditions
fN
= 50 Hz, 60 Hz
UN
= 380 to 690 V
Tload
T
n min n max
Tload
T
TS
n min n max
Imax
IN
n min n max
TS
General description of a dimensioning procedure
Figure 3.1 General description of the dimensioning procedure.

Chapter 4 - Induction (AC) motor
Induction motors are widely used in industry. In this chapter
some of the basic features are described.
4.1 Fundamentals
An induction motor converts electrical energy into mechanical
energy. Converting the energy is based on electromagnetic
induction. Because of the induction phenomenon the induction
motor has a slip.
The slip is often defined at the motor’s nominal point (frequency
( fn ), speed ( nn ), torque ( Tn ), voltage ( Un ), current ( In ) and
power ( Pn )). At the nominal point the slip is nominal:
where ns
is the synchronous speed:
When a motor is connected to a supply with constant voltage
and frequency it has a torque curve as follows:
Figure 4.1 Typical torque/speed curve of an induction motor when
connected to the network supply (D.O.L., Direct-On-Line). In the picture
a) is the locked rotor torque, b) is the pull-up torque, c) is the maximum
motor torque, Tmax and d) is the nominal point of the motor.
(4.1)
(4.2)

Torque
Speed
Induction (AC) motor
A standard induction motor’s maximum torque ( Tmax, also
called pull-out torque and breakdown torque) is typically
2-3 times the nominal torque. The maximum torque is available
with slip smax which is greater than the nominal slip. In order
to use an induction motor efficiently the motor slip should be
in the range - smax ... smax. This can be achieved by controlling
voltage and frequency. Controlling can be done with a frequency
converter.
Figure 4.2 Torque/speed curves of an induction motor fed by a frequency
converter. Tmax is available for short term overloads below the field
weakening point. Frequency converters, however, typically limit the
maximum available torque to 70% of Tmax.
The frequency range below the nominal frequency is called a
constant flux range. Above the nominal frequency/speed the
motor operates in the field weakening range. In the field weak-
ening range the motor can operate on constant power which
is why the field weakening range is sometimes also called the
constant power range.
The maximum torque of an induction motor is proportional to
the square of the magnetic flux ( Tmax ~ ψ 2
). This means that the
maximum torque is approximately a constant at the constant
flux range. Above the field weakening point the maximum torque
decrease is inversely proportional to the square of the frequency
( Tmax ~ ).

Constant ﬂux range
Speed
Field weekening range
Flux
Tmax
Voltage
Figure 4.3 Maximum torque, voltage and flux as a function of the relative
speed.
4.2 Motor current
An induction motor current has two components: reactive current
( isd ) and active current ( isq ). The reactive current component
includes the magnetizing current ( imagn ) whereas the active cur-
rent is the torque producing current component. The reactive
and active current components are perpendicular to each other.
The magnetizing current ( imagn ) remains approximately constant
in the constant flux range (below the field weakening point). In
the field weakening range the magnetizing current decrease is
proportional to speed.
A quite good estimate for the magnetizing current in the constant
flux range is the reactive ( isd ) current at the motor nominal point.
Figure 4.4 Stator current ( is ) consists of reactive current ( isd ) and active
current ( isq ) components which are perpendicular to each other. Stator
flux is denoted as ψs
.

, when 0.8 * Tn ≤ Tload
≤ 0.7 * Tmax
4.2.1 Constant flux range
Below the field weakening point the current components can be
approximated as follows:
The total motor current is:
It can be seen that with zero motor torque the active current
component is zero. With higher torque values motor current
becomes quite proportional to the torque. A good approximation
for total motor current is:
Example 4.1:
A 15 kW motor’s nominal current is 32 A and power factor is
0.83. What is the motor’s approximate magnetizing current at
the nominal point? What is the total approximate current with
120% torque below the field weakening point.
Solution 4.1:
At the nominal point the estimate for the magnetizing current is:
The approximate formula for total motor current with 120% torque
gives:
The approximate formula was used because torque fulfilled the
condition 0.8 * Tn ≤ Tload ≤ 0.7 * Tmax
(4.5)
(4.3)
(4.4)
(4.6)

4.2.2 Field weakening range
Above the field weakening point the current components also
depend on speed.
Total motor current is:
The motor current can be approximated quite accurately within
a certain operating region. The motor current becomes proportional
to relative power. An approximation formula for current is:
Approximation can be used when:
and
(4.8)
(4.7)
(4.10)
(4.9)
(4.11)
(4.12)
In the field weakening range the additional current needed in
order to maintain a certain torque level is proportional to rela-
tive speed.
Example 4.2:
The motor’s nominal current is 71 A. How much current is needed
to maintain the 100% torque level at 1.2 times nominal speed
(Tmax = 3 * Tn).
Solution 4.2:
The current can be calculated by using the approximation for-
mula:

4.3 Motor power
The motor’s mechanical (output) power can be calculated from
speed and torque using the formula:
Because motor power is most often given in kilowatts
(1 kW = 1000 W) and speed in rpm revolutions per minute,
1 rpm = rad/s), the following formula can be used:
The motor’s input power can be calculated from the voltage,
current and power factor:
The motor’s efficiency is the output power divided by the input
power:
Example 4.3:
The motor nominal power is 15 kW and the nominal speed is
1480 rpm. What is the nominal torque of the motor?
Solution 4.3:
The motor’s nominal torque is calculated as follows:
Example 4.4:
What is the nominal efficiency of a 37 kW (Pn = 37 kW,
Un =380 V, In =71 A and cos(ϕn) = 0.85) motor?
Solution 4.4:
The nominal efficiency is:
(4.13)
(4.14)
(4.15)
(4.16)

Chapter 5 - Basic mechanical laws
5.1 Rotational motion
One of the basic equations of an induction motor describes the
relation between moment of inertia ( J [kgm2
]), angular velocity (
ω [rad/s]) and torque ( T [Nm]). The equation is as follows:
In the above equation it is assumed that both the frequency and
the moment of inertia change. The formula is however often
given so that the moment of inertia is assumed to be constant:
Torque Tload represents the load of the motor. The load consists
of friction, inertia and the load itself. When the motor speed
changes, motor torque is different from Tload . Motor torque can
be considered as consisting of a dynamic and a load component:
If the speed and moment of inertia are constants the dynamic
component ( Tdyn ) is zero.
The dynamic torque component caused by acceleration/decel-
eration of a constant moment of inertia (motor’s speed is changed
by Δn [rpm] in time Δt [s], J is constant) is:
The dynamic torque component caused by a variable moment
of inertia at constant speed n[rpm] is:
If the moment of inertia varies and at the same time the motor is
accelerating the dynamic torque component can be calculated
using a certain discrete sampling interval. From the thermal
dimensioning point of view it is however often enough to take
into account the average moment of inertia during acceleration.
(5.1)
(5.2)
(5.3)
(5.4)
(5.5)

Basic mechanical laws
Example 5.1:
The total moment of inertia, 3 kgm2
, is accelerated from a speed
of 500 rpm to 1000 rpm in 10 seconds. What is the total torque
needed when the constant load torque is 50 Nm?
How fast will the motor decelerate to 0 rpm speed if the motor’s
electric supply is switched off?
Solution 5.1:
The total moment of inertia is constant. The dynamic torque
component needed for acceleration is:
Total torque during acceleration is:
If the motor’s electric supply is switched off at 1000 rpm the
motor decelerates because of the constant load torque (50 Nm).
Following equation holds:
Time to decelerate from 1000 rpm to 0 rpm:
Example 5.2:
Accelerating of a fan to nominal speed is done with nominal
torque. At nominal speed torque is 87%. The fan’s moment of
inertia is 1200 kgm2
and the motor’s moment of inertia is 11 kgm2
.
The load characteristics of the fan Tload is shown in figure 5.1.
Motor nominal power is 200 kW and nominal speed is 991 rpm.

Speed
Torque
Figure 5.1 Torque characteristics of a fan. Speed and torque are shown
using relative values.
Calculate approximate starting time from zero speed to nominal
speed.
Solution 5.2:
Motor nominal torque is:
The starting time is calculated by dividing the speed range into
five sectors. In each sector (198.2 rpm) torque is assumed to be
constant. Torque for each sector is taken from the middle point
of the sector. This is quite acceptable because the quadratic
behaviour is approximated to be linear in the sector.
The time to accelerate the motor (fan) with nominal torque can
be calculated with formula:

Direction of energy
Acceleration times for different speed sections are:
0-198.2 rpm
198.2-396.4 rpm
396.4-594.6 rpm
594.6-792.8 rpm
792.8-991 rpm
The total starting time 0-991 rpm is approximately 112 seconds.
5.2 Gears and moment of inertia
Gears are typical in drive systems. When calculating the motor
torque and speed range gears have to be taken into account.
Gears are reduced from load side to motor side with following
equations (see also figure 5.2 ):
Figure 5.2 A gear with efficiency η. Gear ratio is n1
:n2
.
(5.6)
(5.7)
(5.8)

Also all the moments of inertia ( J [kgm2
]) within the system have
to be known. If they are not known they can be calculated which
is rather difficult to do accurately. Typically machine builders can
give the necessary data.
Example 5.3:
A cylinder is quite a common shape for a load (rollers, drums,
couplings, etc.). What is the inertia of a rotating cylinder
(mass=1600 kg, radius=0.7 m)?
Solution 5.3:
The inertia of a rotating cylinder (with mass m [kg] and radius r
[m]) is calculated as follows:
In the case of a gear, the moment of inertia to the motor shaft
has to be reduced. The following example shows how to reduce
gears and hoists. In basic engineering books other formulas are
also given.
Example 5.4:
Reduce the moment of inertia to the motor shaft of the following
hoist drive system.
Figure 5.3 A Hoist drive system used in example 5.4.
Solution 5.4:
The total moment of inertia consists of J1=10 kgm2
,
J2=30 kgm2
, r=0.2 m and m=100 kg.
The moment of inertia J2 and mass m are behind a gearbox with
gear ratio n1:n2=2:1.
The moment of inertia J2
is reduced by multiplying with the square
of the inverse of the gear ratio. The mass m of the hoist is re-
duced by multiplying it with square of the radius r and because
it is behind the gearbox it has to be multiplied with the square
of the inverse of the gear ratio, too.
Thus the total moment of inertia of the system is:

Chapter 6 - Load types
Certain load types are characteristic in the industrial world.
Knowing the load profile (speed range, torque and power) is es-
sential when selecting a suitable motor and frequency converter
for the application.
Some common load types are shown. There may also be com-
binations of these types.
1. Constant torque
A constant torque load type is typical when fixed volumes are
being handled. For example screw compressors, feeders and
conveyors are typical constant torque applications. Torque is
constant and the power is linearly proportional to the speed.
Figure 6.1 Typical torque and power curves in a constant torque
application.
2. Quadratic torque
Quadratic torque is the most common load type. Typical
applications are centrifugal pumps and fans. The torque is
quadratically, and the power is cubically proportional to the
speed.
Figure 6.2 Typical torque and power curves in a quadratic torque
application.

3. Constant power
A constant power load is normal when material is being rolled
and the diameter changes during rolling. The power is constant
and the torque is inversely proportional to the speed.
Figure 6.3 Typical torque and power curves in a constant power
application.
4. Constant power/torque
This load type is common in the paper industry. It is a com-
bination of constant power and constant torque load types.
This load type is often a consequence of dimen-sioning the
system according to the need for certain power at high speed.
Figure 6.4 Typical torque and power curves in a constant power/torque
application.
5. Starting/breakaway torque demand
In some applications high torque at low frequencies is needed.
This has to be considered in dimensioning. Typical applica-
tions for this load type are for example extruders and screw
pumps.
Load types

Figure 6.5 Typical torque curve in an application where starting torque is
needed.
There are also several other load types. They are however hard to
describe in a general presentation. Just to mention a few, there
are different symmetrical (rollers, cranes, etc.) and unsymmetrical
loads. Symmetry/non-symmetry in torque can be for example
as a function of angle or time. These kinds of load types must
be dimensioned carefully taking into account the overloadability
margins of the motor and the frequency converter, as well as the
average torque of the motor.
Load types

T / Tn
Relative speed
Chapter 7 - Motor loadability
Motor thermal loadability has to be considered when dimension-
ing a drive system. The thermal loadability defines the maximum
long term loadability of the motor.
A standard induction motor is self ventilated. Because of the self
ventilation the motor thermal loadability decreases as the motor
speed decreases. This kind of behaviour limits the continuous
available torque at low speeds.
A motor with a separate cooling can also be loaded at low
speeds. Cooling is often dimensioned so that the cooling effect
is the same as at the nominal point.
With both self and separate cooling methods torque is thermally
limited in the field weakening range.
Figure 7.1 A standard cage induction motor’s typical loadability in
a frequency controlled drive 1) without separate cooling and 2) with
separate cooling.
An AC motor can be overloaded for short periods of time without
overheating it. Short term overloads are mainly limited by Tmax
(check the safety margin).
Generally speaking, a frequency converter’s short term load-
ability is often more critical than the motor’s. The motor thermal
rise times are typically from 15 minutes (small motors) to several
hours (big motors) depending on the motor size. The frequency
converter’s thermal rise times (typically few minutes) are given
in the product manuals.

Chapter 8 - Selecting the frequency
converter and motor
The motor is selected according to the basic information about
the process. Speed range, torque curves, ventilation method and
motor loadability give guidelines for motor selection. Often it is
worth comparing different motors because the selected motor
affects the size of the frequency converter.
When selecting a suitable frequency converter there are several
things to be considered. Frequency converter manufacturers
normally have certain selection tables where typical motor pow-
ers for each converter size are given.
The dimensioning current can also be calculated when the torque
characteristics is known. The corresponding current values can
be calculated from the torque profile and compared to converter
current limits. The motor’s nominal current gives some kind of
indication. It isn’t however always the best possible dimensioning
criteria because motors might for example be derated (ambient
temperature, hazardous area, etc.).
The available supply voltage must be checked before selecting
the frequency converter. Supply voltage variations affect the
available motor shaft power. If the supply voltage is lower than
nominal the field weakening point shifts to a lower frequency
and the available maximum torque of the motor is reduced in
the field weakening range.
The maximum available torque is often limited by the frequency
converter. This has to be considered already in the motor selec-
tion phase. The frequency converter may limit the motor torque
earlier than stated in the motor manufacturer’s data sheet.
The maximum available torque is also affected by transform-
ers, reactors, cables, etc. in the system because they cause a
voltage drop and thus the maximum available torque may drop.
The system’s power losses need to be compensated also by the
frequency converter rating.

8.1 Pump and fan application (Example)
Some stages in pump and fan application dimensioning:
– Check the speed range and calculate power with highest
speed.
– Check the starting torque need.
– Choose the pole number of the motor. The most economic
operating frequency is often in the field weakening range.
– Choose motor power so that power is available at maximum
speed. Remember the thermal loadability.
– Choose the frequency converter. Use pump and fan rating. If
the pump and fan rating is not available choose the frequency
converter according to the motor current profile.
Example 8.1:
A pump has a 150 kW load at a speed of 2000 rpm. There is no
need for starting torque.
Solution 8.1:
The necessary torque at 2000 rpm is:
It seems that 2-pole or 4-pole motors are alternative choices
for this application.
Figure 8.1 Motor loadability curves in a pump and fan application.
Comparison of 1) 2-pole and 2) 4-pole motors.
1) motor p=2
For a 2-pole motor the loadability at 2000 rpm according to the
loadability curve is about 95%. The motor nominal torque must
be at least:
Selecting the frequency converter and motor

The corresponding nominal power must then be at least:
A 250 kW (400 V, 431 A, 50 Hz, 2975 rpm and 0.87) motor is
selected. The nominal torque of the motor is:
The motor current at 2000 rpm speed (constant flux range) is
approximately:
The minimum continuous current for the frequency converter is
then 384 A.
2) motor p=4
For a 4-pole motor the loadability at 2000 rpm is 75%.
The minimum nominal torque of the motor is:
The minimum power for a 4-pole motor is:
A 160 kW motor (400 V, 305 A, 50 Hz, 1480 rpm and 0.81)
fulfills the conditions. The approximated current at a speed of
2000 rpm (66.7 Hz) is:
The exact current should be calculated if the selected frequency
converter’s nominal current is close to the approximated motor
current.
A 4-pole motor requires less current at the pump operation
point. Thus it is probably a more economical choice than a
2-pole motor.

8.2 Constant torque application (Example)
Some stages in dimensioning of a constant torque application:
– Check the speed range.
– Check the constant torque needed.
– Check the possible accelerations. If accelerations are needed
check the moments of inertia.
– Check the possible starting torque required.
– Choose the motor so that torque is below the thermal load-
ability curve (separate/self ventilation?). Typically the nominal
speed of the motor is in the middle of the speed range used.
– Choose a suitable frequency converter according to the di-
mensioning current.
Example 8.2:
An extruder has a speed range of 300-1200 rpm. The load at
1200 rpm is 48 KW. The starting torque requirement is 200 Nm.
Acceleration time from zero speed to 1200 rpm is 10 seconds.
The motor is self-ventilated and the nominal voltage is 400 V.
Solution 8.2:
The constant torque requirement is:
A suitable motor is a 4-pole or a 6-pole motor.
Figure 8.2 Motor loadability curves in a constant torque application.
comparison of 1) 4-pole and 2) 6-pole motors.

1) Motor p=4
At 300 rpm speed the thermal loadability is 80%.
The estimated minimum nominal torque is:
The minimum motor nominal power is:
A suitable motor is for example a 75 kW (400 V, 146 A, 50 Hz,
1473 rpm and 0.82) motor. The motor nominal torque is:
Motor current is approximately (T/Tn ≈ 0.8):
According to the calculated motor current a suitable frequency
converter can be selected for constant torque use.
The starting torque requirement (200 Nm) is not a problem for
this motor.
If the motor’s moment of inertia is 0.72 kgm2
the dynamic torque
in acceleration is:
Thus the total torque during acceleration is 391 Nm which is
less than the nominal torque of the motor.
2) Motor p=6
At speeds of 300 rpm and 1200 rpm the motor loadability is 84%.
Thus the minimum nominal torque of the 6-pole motor is:
The minimum value of the motor nominal power is:

A suitable motor could be for example a 55 kW (400 V, 110 A,
50 Hz, 984 rpm and 0.82) motor. The motor nominal torque is:
The dimensioning current can be approximated at a speed of
1200 rpm:
The nominal (continuous) current of the frequency converter
must be over 96 A.
The starting torque requirement is less than motor’s nominal
torque.
If the inertia of the motor is 1.2 kgm2
the dynamic torque in ac-
celeration is:
The total torque needed during acceleration is 397 Nm which is
less than the nominal torque of the motor.
A 6-pole motor current is 19 A smaller than with a 4-pole motor.
The final frequency converter/motor selection depends on the
motor and frequency converter frame sizes and prices.
8.3 Constant power application (Example)
Some stages in dimensioning of a constant power application:
– Check the speed range.
– Calculate the power needed. Winders are typical constant
power applications.
– Dimension the motor so that the field weakening range is
utilized.
Example 8.3:
A wire drawing machine is controlled by a frequency converter.
The surface speed of the reel is 12 m/s and the tension is 5700
N. The diameters of the reel are 630 mm (empty reel) and 1250
(full reel). There is a gear with gear ratio n2
:n1
=1:7.12 and the
efficiency of the gear is 0.98.
Select a suitable motor and converter for this application.

Solution 8.3:
The basic idea of a winder is to keep the surface speed and the
tension constant as the diameter changes.
Figure 8.3 Basic diagram of a winder.
In rectilinear motion the power is: P = Fv
In rotational motion the power is: P = Tω
The relation between surface speed and angular velocity is:
Torque is a product of force and radius: T = Fr
By using the above formulas the motor can be selected:

The gear must be taken into account before choosing the motor.
Speeds, torques and power have to be reduced:
1) Motor p=2
If a 2-pole motor is selected loadability at a speed of 1305 rpm
is about 88% and 97% at 2590 rpm. The minimum nominal
power of the motor is:
A 200 kW (400 V, 353 A, 50 Hz, 2975 rpm and 0.86) motor is
selected. The motor nominal torque is:
The dimensioning current is calculated according to a torque
of 511 Nm:
2) Motor p=4
If a 4-pole motor is selected it can be seen from the loadability
curve that loadability at a speed of 1305 rpm is about 98% and
about 60% at 2590 rpm. The minimum nominal power of the
motor is:

A 90 kW (400 V, 172 A, 50 Hz, 1473 rpm and 0.83) is selected.
The motor nominal torque is:
Dimensioning in this case is done according to the motor current
at 1305 rpm. The motor current is:
With a 2-pole motor the field weakening (constant power) range
was not utilized which led to unnecessary overdimensioning.
zA 4-pole motor is a better choice for this application.

Torque
Line current
Chapter 9 - Input transformer and rectifier
There are several types of input rectifiers. The rectifier type might
limit the operation.
A conventional rectifier is a 6 or 12 pulse diode rectifier. Diode
rectifiers only support motoring loads where the power flow is
one way only.
In certain processes where the load can also be generating, the
energy needs to be absorbed. For short generating loads the
traditional solution has been a braking resistor where the power
generated has been transformed into heat losses. If however
the load is generating all the time, a true 4-quadrant rectifier is
needed.
Both the input transformer and the rectifier are dimensioned ac-
cording to the motor shaft power and system losses. For example
if high torque at low speed is delivered the mechanical power is
nevertheless quite low. Thus high overloads do not necessarily
mean high power from the rectifier point of view.
Figure 9.1 Line current in a constant torque application. Line current is
small at low speed.
9.1 Rectifiers
Rectifiers are dimensioned according to motor shaft power. A
single drive’s input rectifier can be selected using the approxi-
mation formula:
In drive systems where there is a common DC-link, there can
be motoring and generating power at the same time. Rectifier
power is then calculated approximately as follows:
(9.1)
(9.2)

9.2 Transformer
An input transformer’s power can be calculated as follows:
In the above formulas:
Ptotal is the total motor shaft power
k is the transformer loadability (k-factor)
1.05 stands for transformer voltage drop (impedance)
ηr is the rectifier efficiency
cos(α) is the rectifier control angle (=1.0 for diode
rectifier)
ηc is the AC choke (if there is one) efficiency
ηi is the inverter efficiency
ηm is the motor efficiency
Typically total shaft power is multiplied by a coefficient
1.2 - 1.35.
Example 9.1:
In a constant torque application the maximum shaft power
needed is 48 kW at a speed of 1200 rpm. A 55 kW motor and
70 kVA inverter unit was selected.
Specify the rectifier and input transformer. A 6-pulse diode sup-
ply is used (efficiency 0.985), there is a DC-choke in the DC-link,
inverter efficiency is 0.97 and motor efficiency is 0.95.
Solution 9.1:
For the rectifier the estimated power is:
Input transformer and rectifier
(9.3)
The choke efficiency is included in the inverter efficiency. Be-
cause of diode supply unit cos(α) =1. The power of the input
transformer (k=0.95) is:

Symbols
4-quadrant 35
A
acceleration 17, 18, 29, 30, 31
AC motor 8
active current 13, 14
angular velocity 17, 32
C
centrifugal pumps 22
constant flux range 12, 13, 28
constant power 12, 23, 31, 34
constant torque 22, 23, 29, 30, 35, 36
coupling 21
cubically 22
cyclical load 9
D
DC-link 8, 35, 36
decelerate 18
drum 21
E
efficiency 16, 20, 31, 36
electric supply 7, 8, 18
F
fan 18, 19, 22, 27
field weakening range 12, 13, 15,
25, 26, 27, 31
flux range 12, 13, 14, 28
frequency 8, 9, 10, 11, 12, 17, 22,
24, 25, 26, 27, 28, 29, 30, 31, 32,
33, 34
friction 17
G
gear 20, 21, 31, 32
generating 35
I
induction 11, 12, 13, 14, 15, 16,
17, 25
induction motor 11, 12, 13, 17, 25
input transformer 8, 35, 36
inverter 8, 36
K
kilowatt 16
L
load 8, 9, 10, 12, 14, 17, 18, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29,
30, 33, 35, 36
load profile 22
load type 9, 22, 23, 24
locked rotor torque 11
M
maximum torque 9, 12, 26
mechanical 11, 16, 17, 18, 19, 20,
21, 35
motor 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 24,
25, 26, 27, 28, 29, 30, 31, 32, 33,
34, 35, 36
motoring 35
N
nominal point 11, 13, 14, 25
O
overloadability 9, 24
P
power 9, 11, 12, 14, 15, 16, 18,
22, 23, 26, 27, 28, 30, 31, 32, 33,
34, 35, 36
power factor 14, 16
pull-out torque 12
pull-up torque 11
Q
quadratically 22
quadratic torque 22
R
reactive current 13
rectifier 8, 35, 36
rectifier unit 8
roller 21, 24
S
separate cooling 25
shaft power 26, 35, 36
slip 11, 12
speed 9, 11, 12, 13, 15, 16, 17,
18, 19, 20, 22, 23, 25, 27, 28, 29,
30, 31, 32, 33, 35, 36
speed range 19, 20, 22, 27, 29,
31
starting torque 9, 24, 27, 29, 30,
31
supply 7, 8, 9, 11, 18, 26, 36
supply voltage 9, 26
T
thermal loadability 25, 27, 29, 30
torque 9, 10, 11, 12, 13, 14, 15,
16, 17, 18, 19, 20, 22, 23, 24, 25,
26, 27, 28, 29, 30, 31, 33, 34, 35,
36
transformer 8, 26, 35, 36
V
voltage 9, 11, 12, 13, 16, 26, 29,
36
Chapter 10 - Index

Abb technical guide no.07 revc

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3AFE64362569REVCEN28.4.2011#15649

Abb technical guide no.07 revc

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Abb technical guide no.07 revc