Special Applications

Application Manual for NEMA Motors
Section 10
Special Applications
Page
Part 1
Power Factor Correction
1
Part 2
Methods of Starting 3AC Induction Motors
5
Part 3
Duty Cycles and Inertia
23
Part 4
Horsepower Determination
30
Part 5
Formulas and General Data
34
Application Manual for NEMA Motors
Section
10
Part Section 1 10
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Date
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Application Manual for NEMA Motors
Power Factor Correction
Power factor is the ratio of actual power used in a circuit expressed in watts or kilowatts to the power
which is apparently being drawn from the line (apparent power) expressed in volt amperes or kilovolt
amperes. Power factor is generally expressed in percent.
Low power factor increases the power company’s cost of supplying actual power because more
current must be transmitted than is actually used to perform useful work. This additional transmitted
current increases the cost incurred by the power company and is directly billed to the industrial
consumer by means of power factor clauses in the rate schedules. Low power factor reduces the load
handling capability of the industrial plants electrical system as well as the load handling capabilities
of the power company’s generators, transmission lines and transformers.
When the volt ampere product (KVA) exceeds the actual power (KW), a component known as reactive
power (KVAR) is present. The operating current consists of two parts: one which results in useful
work and another known as reactive current which merely bounces energy back and forth. Both
generate heat in the wires or conductors. The reactive current is always present in inductive load
devices. Although reactive current is actually part of the total current indicated by amp meter reading,
reactive power does not register on a kilowatt hour meter.
The inductive reactance of the A.C. induction motor causes the motor current to lag behind the motor
voltage, and thereby causes the power factor to drop below unity. This can be offset by the addition of
capacitors connected across the motor terminals on the load side of the motor starter. Power factor
correction capacitors are generally connected to the motor in such a manner that they are
automatically removed from the system as the power source is removed from the motor. The capacitor
causes the current to lead the voltage which tends to offset the lagging current caused by the motor
reactance thereby improving the system power factor. The capacitive current in the capacitors
opposes the inductive current in the induction motor.
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Power Factor Correction
Over correction of power factor by the addition of excessive capacitance is dangerous to the motor
and driven equipment; therefore, it is undesirable. Over correction of power factor by the addition of
excessive capacitance must be avoided because:
1. Over correction of power factor may cause damage to the A.C. induction motor as well as the
driven equipment. The power factor correction capacitors are electrical energy storage devices.
When the motor is de-energized, the capacitor which remains connected in parallel to the motor
can maintain the motor voltage. If the motor is re-energized after a short time, the motor voltage
and the line voltage may be additive, and dangerously high currents and torques may result.
Although this condition is present only for a short time, it can result in dangerously high
transient currents and torques which in turn can cause physical damage to the motor’s power
transmission devices and the driven load. The important consideration is time. The motor voltage
will normally decay in five seconds or less, therefore, the motor should not be re-energized
before five seconds has elapsed.
2. Regenerative effect on motors. The motor power factor should not be corrected on motors
connected to loads which are capable of causing the motor to rotate at speeds above
synchronous motor speed when the motor is de-energized. The addition of power factor
correction capacitors to motors connected to overhauling loads must be done with extreme care.
Although the motor and its connected capacitors are de-energized, continued rotation of the
rotor combined with the stored energy in the capacitor causes the motor to act as an alternator
(A.C. generator). The voltage thus generated in the A.C. motor severely strains the capacitors
and may be sufficient to cause destruction of the capacitors. When power factor correction is
required on A.C. motors connected to overhauling loads, contact the factory for technical
assistance.
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Power Factor Correction
3. Excessive capacitance applied to an A.C. induction motor will correct the power factor of other
inductive loads connected to the same power supply source. Since the same power supply
system is used to supply power to many inductive loads, over correction at one load point will
correct the power factor at another load point. The user that over corrects the power factor of
this inductive load does not receive additional financial benefits from the power company even
though he is aiding the power company by allowing it to supply power having a leading current
(current leads the voltage) to other customers on the same supply system.
For the above reasons, the power factor of an A.C. induction motor should not be corrected above
95%. The amount of capacitance expressed as KVAR required for power factor correction can be
determined by use of the following formula:
.746 x HP
EFF
X
Factor from adjoining table = KVAR to correct power factor
EFF = Motor efficiency as a decimal
Example: KVAR required to correct power factor of a 60 HP,
1800 RPM, 460 volt motor from 87 to 95 =
.746 x 60
.917
X
.238 = 11.6
This value will probably
fall between standard
sizes available. The next
lower standard size is
usually selected.
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Application Manual for NEMA Motors
Desired Power Factor in Percent
A = Uncorrected Motor Power Factor in Percent
A
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
85
.714
.679
.645
.613
.580
.549
.518
.488
.459
.429
.400
.372
.343
.316
.289
.262
.235
.209
.183
.156
.130
.104
.078
.052
.026
-
86
.741
.706
.672
.640
.607
.576
.545
.515
.486
.456
.427
.399
.370
.343
.316
.289
.262
.236
.210
.183
.157
.131
.105
.079
.053
.027
-
87
.769
.732
.698
.666
.633
.602
.571
.541
.512
.482
.453
.425
.396
.369
.342
.315
.288
.262
.236
.209
.183
.157
.131
.105
.079
.053
.026
-
88
.794
.759
.725
.693
.660
.629
.598
.568
.539
.509
.480
.452
.423
.396
.369
.342
.315
.289
.263
.236
.210
.184
.158
.132
.106
.080
.053
.027
-
89
.822
.787
.753
.721
.688
.657
.626
.596
.567
.537
.508
.480
.451
.424
.397
.370
.343
.317
.291
.264
.238
.212
.186
.160
.134
.108
.081
.055
.028
-
90
.850
.815
.781
.749
.716
.685
.654
.624
.595
.565
.536
.508
.479
.452
.425
.398
.371
.345
.319
.292
.266
.240
.214
.188
.162
.136
.109
.082
.056
.028
-
91
.878
.843
.809
.777
.744
.713
.682
.652
.623
.593
.564
.536
.507
.480
.453
.426
.399
.373
.347
.320
.294
.268
.242
.216
.190
.164
.137
.111
.084
.056
.028
-
92
.905
.870
.836
.804
.771
.740
.709
.679
.650
.620
.591
.563
.534
.507
.480
.453
.426
.400
.374
.349
.321
.295
.269
.243
.217
.191
.167
.141
.114
.086
.058
.030
-
93
.939
.904
.870
.838
.805
.774
.743
.713
.684
.654
.625
.597
.568
.541
.514
.487
.360
.434
.408
.381
.355
.329
.303
.277
.251
.225
.198
.172
.145
.117
.089
.061
.031
-
94
.971
.936
.902
.870
.837
.806
.775
.745
.716
.686
.657
.629
.600
.573
.546
.519
.492
.466
.440
.413
.387
.361
.335
.309
.283
.257
.230
.204
.177
.149
.121
.093
.063
.032
-
95
1.005
.970
.936
.904
.871
.840
.809
.779
.750
.720
.691
.663
.634
.607
.580
.553
.526
.500
.474
.447
.421
.395
.369
.343
.317
.291
.265
.238
.211
.183
.155
.127
.097
.066
.034
-
Application Manual for NEMA Motors
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Application Manual for NEMA Motors
Methods of Starting Three Phase A.C. Induction Motors
The purpose of all motor starters is to provide a means of connecting the motor to the power supply
thereby accelerating the motor and connected load from standstill to normal operating speed.
The following items should be considered when selecting a motor starter for a specific three phase
A.C. induction motor and motor load.
1. The power source (phase, voltage, frequency).
2. The starting torque requirements of the load.
3. Power source restrictions concerning amperage draw.
Three phase A.C. Induction motor starters can be classified into three basic categories.
A. Full Voltage Type
The full voltage type of across-the-line starter simply connects
the motor directly to the power source.
B. Reduced Voltage Type
Reduced voltage starters cause a voltage, lower than that of
the power source, to be impressed on the motor terminals in
order to reduce motor inrush current and starting torque.
Example:
a. Autotransformer
b. Series Resistor
c. Series Reactor
C. Increment Type
Increment starters use various motor reconnecting techniques to reduce motor inrush current
and starting torque. Normal line voltage is maintained at the motor terminals.
Example:
a. Part Winding
b. Star Delta
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Full Voltage Start
General
Across-the-line starting is the most basic and widely used method of starting squirrel cage induction
motors and is, therefore, used as a basis for comparing other starting methods.
Operation
A pilot device (such as a start push button) closes the line contactor to connect the motor directly to
the line.
Advantages
1. The across-the-line starter is the most simple A.C. motor starting device and, therefore, the least
expensive. It provides reliable, trouble-free operation with low maintenance costs.
2. The across-the-line starter allows A.C. motor to develop its maximum starting torque.
Caution:
1. The high inrush current (approximately 6 to 7 times the name-plate full load current) may be in
excess of that allowed by the power source.
2. The high motor starting torque may cause excessive shock loading to the driven equipment.
3. The high starting current may cause a temporary reduction in motor terminal voltage. This
voltage drop will reduce the motor starting torque by the square of the voltage ratio. An
excessive voltage drop may cause dimming of lights or cause magnetic relays to “trip out”.
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Application Manual for NEMA Motors
Reduced Voltage Start Autotransformer
General
The autotransformer starter is classified as a reduced voltage starter. It is a device with which the
applied motor voltage can be reduced below that of the line voltage. Both motor starting current and
torque, therefore, will be reduced below those values obtained with across-the-line starting.
Any standard three phase induction motor may be used with an autotransformer starter. The starter
portion of the autotransformer start connects the motor leads to the reduced voltage output winding of
the autotransformer. After a preset time delay (normally 10 to 20 seconds) the starter connects the
motor leads to the full line voltage.
Operation
Two autotransformer starter designs are used, the open-circuit transition and the closed-circuit
(Korndorfer) transition types. Both manual and magnetic open or closed circuit transition
autotransformers are available. During switching from reduced voltage starting to full applied line
voltage operation, however, a voltage is continuously applied to the motor terminals from the moment
of reduced voltage starting and during the switching to full line voltage operation:
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Reduced Voltage Start Autotransformer
Advantages:
1. Starting torque per starting amp ratio equal to that of the across-the-line starter.
2. The most desirable starting current and starting torque can be selected by means of
reconnecting the motor leads to the 50%, 65% or 80% output taps of the autotransformer. The
characteristics of the motor load and allowable accelerating times establish the best tap
connection.
% Tap
50
65
80
% LT
25
42
64
% LRA
27
45
66
Where:
% LT =
Starting torque expressed as a percentage of the value encountered during
across-the-line starting.
% LRA = Starting current drawn from the power lines expressed as a percentage of the value
encountered during across-the-line starting. This value includes the approximate required
auto-transformer magnetization current.
NOTE:
Both % locked torque and % locked rotor current vary approximately as the square of the
voltage applied to the motor.
3. Limited motor noise and vibration during starting.
4. On the closed circuit transition type starter, voltage transients during the transition period are
minimized which reduces the possibility of unacceptable performance of other electrical
components within the same plant.
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Reduced Voltage Start Autotransformer
Caution:
1. The autotransformer output tap may have to be changed to a higher percentage voltage value if
the load torque and/or inertia exceeds that which the motor can accelerate within the required
starting period.
2. The transfer from reduced voltage to full voltage operation should be delayed until the motor
speed is high enough to insure that the current change during switching will not exceed power
company requirements.
3. The starter as well as the motor should be evaluated for applications requiring starting. For
autotransformer starters, NEMA states that one 15 second starting period every four minutes for
a total of four per hour is acceptable. The majority of standard induction motors are capable of
four 15 second starting periods per hour.
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Primary Resistor or Reactor Starting
Operation
Resistor type starters introduce a resistor bank, in series with the motor windings.
The initial current surge through the motor is limited by the resistors. Simultaneously, a voltage drop
develops across the resistors, reducing the voltage applied to the motor. At reduced voltage, the
motor torque capability is reduced.
As the motor begins to accelerate, it produces a counter emf, opposing the applied voltage and
further reducing the initial current surge. As the current surge is reduced, the voltage drop across the
resistor bank diminishes while that of the motor is increased. This increases motor torque while
current inrush is diminishing. The net result is a smooth and gradual accelerating cycle without open
transients in the motor windings.
An adjustable timing device on the starter is preset to initiate a run contactor at the proper time
during the accelerating period. This function closes a run contactor, shorting across the resistors. The
start contactor then opens, removing the resistors from the circuit.
Reactor type starters follow the same sequence for starting, with the exception that the reactors
remain shorted after the final stage of acceleration. As implied, the reactors are core wound devices,
having adjustable voltage taps. These devices limit the initial motor current surge by an inherent
tendency to oppose a sudden changing condition of current and voltage.
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Primary Resistor or Reactor Starting
Advantages
The resistor or reactor controllers are of the closed transition type since the lines to the motor are not
opened during transfer to the run condition.
Resistors supplied provide 65% voltage but have taps at 80% to allow for adjustment. Reactors have
50%, 65% and 80% taps. All controllers are connected for 65% voltage as standard.
Current drawn from the line upon starting is reduced to approximately the value of the tap used.
Starting torque is reduced to approximately the value of the tap squared. For example, with
connections to the 65% taps, current inrush will be approximately 65% of full-voltage. Starting torque
will be about 42% of that developed under full-voltage starting.
Any standard motor may be used. No special windings or connections are needed.
Caution:
The motor will not start if the breakaway torque required of the load exceeds that which the motor
can develop on the starting connection.
Full impact of inrush current and torque are then experienced at transfer from the start to run
connection.
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Increment Starting (Part Winding Start – 1/2 Winding Method)
General
Part winding starting, 1/2 winding type, is the most commonly used method of increment starting of
A.C. induction motors. The 1/2 winding method of part winding starting requires the use of a specific
motor starter with an A.C. induction motor having two parallel stator windings suitably connected
internally for part winding starting.
The starter must be capable of energizing, with full line voltage, 1/2 of the motor winding, then after a
slight time delay (not to exceed 3 seconds) energizing the complete winding in parallel with the first
half of the motor winding.
Advantages:
1. Starting current is reduced to 60 to 65% of the value encountered if the motor were started
across-the-line.
2. Starting torque is approximately 45 to 50% of the value encountered if the motor were started
across-the-line.
3. Continuous connection of the motor to line during the transition period minimizes voltage
fluctuation during the transition period.
4. Can be applied to most 4, 6 and 8 pole motor designs.
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Increment Starting (Part Winding Start – 1/2 Winding Method)
Caution:
1. Motor to be part winding started must be designed and built properly to insure that two parallel
stator windings are provided.
2. The motor will not start if the torque demanded by the load exceeds that developed by the motor
on the first step or when 1/2 of the motor winding is energized. When the second half of the
winding is energized, the normal starting torque (same as across-the-line starting) is available to
accelerate the load.
3. By use of two-step start winding starting, the inrush current is divided into two steps thus
providing the power source’s line voltage regulators sufficient time to compensate for voltage
drop caused by motor starting.
4. Motor heating on first-step operation is greater than that normally encountered on across-theline start. Therefore, elapsed time on the first step of the part winding start should not exceed
three seconds.
5. When the first half of the winding is energized, a slight increase in electrical noise and vibration
may be encountered.
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Increment Starting (Star-Delta)
General
Star-Delta starting is a method of increment starting used with a three phase A.C. induction motor to
reduce the initial values of motor starting current and torque compared to those values obtained with
across-the-line starting.
Both ends of each phase winding is brought into the motor conduit box. The starter is designed to
connect these windings in star on the first step. After a preset time delay, the starter will disconnect
from the star configuration and reconnect in delta for continuous operation.
The voltage impressed across each phase of the motor winding during the first step (Star connection)
of a Star-start Delta-run motor is lower than the voltage which would be impressed across each
phase of the motor if across-the-line starting were used. This lower voltage results in lower starting
current and torque.
Three phase A.C. induction motors having Star-Delta winding connections are popular in Europe
because supply voltages of 220 and 380 are used. Motors wound for 220/380 volts may be started
across-the-line on either 220 or 380 volts or may be operated at Star-start Delta-run on 220 volts.
Operation
Two types of Star-Delta starters are used, the open-circuit transition and the closed-circuit transition
types. Both types connect the motor windings in Star on the first step. After a predetermined time
interval, the timer causes the starter contactors to reconnect the motor windings in the Delta or “run”
connection.
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Increment Starting (Star-Delta)
Advantages:
1. Starting torque per starting amp ratio equal to that of the across-the-line starter.
2. Starting current is reduced to approximately 33% of the value encountered if the motor were
started across-the-line.
3. Soft start - Starting torque is approximately 33% of the value encountered if the motor were
started across-the-line. This low starting torque is often desirable to softly accelerate loads
having high inertia and low retarding torque (Example - centrifuges and unloaded compressors).
The low starting current during the starting period allows the motor to withstand a longer
acceleration time than an equivalent sized across-the-line start motor. The Star-Delta motor,
however, will accelerate only slightly more inertia than the across-the-line motor since the
thermal capacity of both motors is the same. To accelerate a specific inertia, the Star-Delta
motor will produce a lower temperature rise within the motor than an equivalent sized acrossthe-line start motor because the longer acceleration time allows the heat in the motor to be more
efficiently dissipated to the housings and surrounding atmosphere.
4. Can be adapted to most 2, 4, 6; and 8 pole motor designs.
5. May be started as frequently as an across-the-line start motor if the retarding torque of the load
is negligible.
6. Limited motor noise and vibration during starting.
7. On the open-circuit transition type, no resistors are required.
8. On the closed-circuit transition type, voltage fluctuation during the transition period is minimized.
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Increment Starting (Star-Delta)
Caution:
1. Motor started on Star connection and operated on the Delta connection must be specifically
designed for Star-Delta.
2. The motor will not start if the torque demanded by the load exceeds that developed by the motor
on the Star connection. When the motor is connected in Delta, normal starting torque is
available to start load.
3. The transfer from Star to Delta should be delayed until the motor speed is high enough to insure
that the current change during switching will not exceed power source requirements. Generally,
the starter timer should be set so that switching from Star to Delta occurs at 80 to 90% of fullload speed.
4. On the open-circuit transition type, line voltage fluctuation can result during the transition period
due to sudden current changes.
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Application Manual for NEMA Motors
Summary
Figure (1) summarizes in chart form the motor starting performance with these various starting
methods.
To this point, we have reviewed these various starting methods and their effect on inrush current, line
current and starting torque. Our major concern is the motor; to get it started and up to speed as
rapidly as the load permits. Figure (2) shows why this is necessary. This is a typical speed current
curve, and illustrates that inrush current remains high throughout most of the accelerating period.
If the power system on which this motor is to operate cannot stand this inrush at full voltage, and a
means of reduced voltage or increment start is elected, then we must be aware of the effect on
current and torque as displayed in Figure (1).
Bear in mind also, the motor does not recognize the type of starter being used. It interprets what is
received at the motor terminals as applied voltage, and that which appears at the shaft as a torque to
be overcome. It is recognizing two factors; voltage and load torque.
An increasing number of specifications are being written which state that the motor must be capable
of starting with 90% or 80%, or 70% voltage, and must be capable of momentary operation with a
voltage dip to 90%, or 80% or 70% voltage.
This introduces three points for consideration.
1. Will the motor develop enough torque at start to initiate rotation of the load?
2. Will the motor be capable of maintaining rotation during accelerating period?
3. Will the motor have sufficient torque to sustain rotation during periodic voltage dips?
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Application Manual for NEMA Motors
Summary
Figure (3) displays a family of motor speed torques for terminal voltages of 100%, 90%, 80% and
70% rated voltage. The speed torque requirement of a centrifugal pump is also shown as a typical
load curve.
1. The motor will break away and begin rotating as long as more torque is generated at locked
rotor than the load requires. In this example the motor will start under any one of the four
voltage conditions.
2. As long as the motor is generating more torque than required by the load the motor will
continue to accelerate. This will continue until the torque developed by the motor is equal to the
torque required by the load. Acceleration will stop and the motor will attempt to operate the
driven device at this speed. If the speed is too low for the driven device the torque condition of
either the motor or the driven device must be changed to increase the speed to acceptable
value.
3. When the voltage dip occurs the motor performance will be in accordance with the speed torque
curve for this reduced value. The motor will continue to operate as long as the intersection of
the load curve and the motor speed torque curve occurs above the breakdown torque point of
the motor. The closer this operating point approaches the breakdown point the quicker the
motor will overheat and therefore the shorter time the motor can successfully withstand the
voltage dip.
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Application Manual for NEMA Motors
General Comparison of Characteristics of Various Methods of Motor Starting
Reduced-Voltage Starting
Autotransformer
Primary
Resistance
Reactor
Part Winding
Wye (Star) – Delta
50%
Tap
65%
Tap
80%
Tap
65%
Tap
80%
Tap
50%
65%
80%
2-Step
3-Step
Starting current drawn
from line as % of that
which would be drawn
upon full-voltage starting.
25%
42%
64%
65%
80%
50%
65%
80%
50%
25%
33.3%
Starting torque developed
as % of that which would
be developed on fullvoltage starting.
25%
42% 64%
Increase slightly
with speed.
42% 64%
25%
42% 64%
Increase greatly with speed.
50%
12.5%
33.3%
Smoothness of
acceleration
Second in order of
smoothness
Smoothest of reduced-voltage type.
As motor gains speed, currennt
decreases. Voltage drop across
resistor or reactor decreases and
motor terminal voltage increase.
Allowable accelerating
times (typical)
30 seconds
Starting current
and torque
5 seconds
15 seconds
Adjustable within limits of various taps.
Fourth in order Third in order of
of smoothness. smoothness.
2-3 Seconds
Limited by
motor design.
45-60 Seconds
Limited by motor
design.
Fixed
Application Manual for NEMA Motors
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Typical Current vs Speed
Squirrel Cage Induction Motor
700%
600%
% FULL LOAD CURRENT
500%
400%
300%
200%
100%
0%
0%
10%
20%
30%
40%
50%
60%
70%
% SYNCHRONOUS SPEED
80%
90%
100%
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Typical Motor Speed Torque Curves at Various Voltages and Load Curve
250%
% FULL LOAD TORQUE
200%
150%
100% VOLTAGE
90%
100%
80%
70%
50%
CENTRIFUGAL PUMP
0%
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
Application Manual for NEMA Motors
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Application Manual for NEMA Motors
Duty Cycles and Inertia
Horsepower Variations
Many machines work on a definite duty cycle that repeats at regular intervals. If the values of power
required during the cycle, and the length of their durations are known, then the rating of the motor
required can be calculated by the root-mean square (RMS) method:
Multiply the square of the horsepower required for each part of the cycle by the duration in
seconds. Divide the sum of these results by the effective time in seconds to complete the whole
cycle. Extract the square root of this last result. This gives the rms horsepower. If the motor is
stopped for part of the cycle, only 1/3 of the rest period should be used in determining the
effective time for open motors (enclosed motors use 1/2 of the rest period). This is due to the
reduction in cooling effect when motor is at rest.
Example:
Assume a machine operation, where an open motor operates at a 8 HP load for 4 minutes, 6 HP
load for 50 seconds, 10 HP load for 3 minutes, and the motor is at rest for 6 minutes.
RMS HP =
(82 x 240) + (62 x 50) + (102 x 180) =
240 + 50 + 180 + 360
3
59.6 = 7.7 HP
Use 7.5 HP Motor
Application Manual for NEMA Motors
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Application Manual for NEMA Motors
Accelerating Time
For fast repeating cycles involving reversals and deceleration by plugging, the additional heating
due to reversing and external WK2 loads must be taken into consideration. Consequently a more
elaborate duty cycle analysis than devised is required. It is necessary to know the torque, time
and motor speed for each portion of the cycle, such as acceleration, running with load, running
without load, deceleration, and at rest.
For such detailed duty cycles it is sometimes more convenient to calculate on the basis of torque
required rather than on the horsepower basis.
Accelerating Time
time, sec. =
WK 2 x (change in rpm)
308 x torque (lb-ft) average from motor
The above formula can be used when the accelerating torque is substantially constant. If the
accelerating torque varies considerably, the accelerating time should be calculated in increments: the
average accelerating torque during the increment should be used and the size of the increment used
depends on the accuracy required. The following equation should be used for each increment:
time, sec. =
WK2 x rpm
308 x (T-TL)
WK2
=
moment of inertia, lb-ft2 of system (motor + load)
RPM
=
motor speed
TL
=
motor torque, lb-ft at a given speed
T
=
load torque, lb-ft at the same speed
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Accelerating Time
Moment of Inertia (WK2)
The radius of gyration K, depends on the shape of the object, and the axis of rotation. An
unsymmetrical object will have a different K depending on the orientation of the axis of rotation. In the
formula for calculating “accelerating time” the product of weight W and the square of the radius of
gyration K2 appears. This will hold true in any formula whenever the moment of inertia is of concern.
Formulas for WK2, based on specific weights of metals, can be found in Table 1.
In Table 2, WK2 of solid cylinders per inch of length is given.
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Accelerating Time
(a) Circluar Cylinder
Rft.
(b) Hollow Circluar Cylinder
Rft.
Lft.
Dft.
Lft.
D1ft.
D2ft.
WK2 = 170.4 w LD4
w = weight of material
WK2 =170.4 w L X
(D24 – D14)
WEIGHT OF MATERIAL
lbs/cu. in.
1. Magnesium 0.0628
2. Aluminum 0.0924
3. Cast Iron 0.260
4. Steel 0.282
5. Copper 0.318
6. Bronze 0.320
7. Lead 0.411
TABLE 1
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Application Manual for NEMA Motors
WK2 of Steel Shafting and Disc
To determine the WK2 of a given shaft or disc multiply the WK2 given below, by the length of the shaft,
or thickness of disc, in inches. To determine inertias of solids of greater diameter than shown below,
multiply the nearest tenth of the diameter by 104 or move decimal point 4 places to the right and
multiply by length as above. For hollow shafts, subtract WK2 of inside diameter from WK2 of outside
diameter and again multiply by length.
Diameter
(inches)
WK2
(LB.Ft.2)
Diameter
(inches)
3/4
1
1-1/4
1-1/2
1-3/4
2
2-1/4
2-1/2
2-3/4
3
3-1/2
3-3/4
4
4-1/4
4-1/2
5
5-1/2
6
6-1/4
6-1/2
6-3/4
7
7-1/4
7-1/2
7-3/4
8
8-1/4
8-1/2
8-3/4
9
9-1/4
9-1/2
9-3/4
10
10-1/4
0.00006
0.0002
0.0005
0.001
0.002
0.003
0.005
0.008
0.011
0.016
0.029
0.038
0.049
0.063
0.079
0.12
0.177
0.25
0.296
0.345
0.402
0.464
0.535
0.611
0.699
0.791
0.895
1.000
1.13
1.27
1.41
1.55
1.75
1.93
2.13
10-1/2
10-3/4
11
11-1/4
11-1/2
11-3/4
12
12-1/4
12-1/2
12-3/4
13
13-1/4
13-1/2
13-3/4
14
14-1/4
14-1/2
14-3/4
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
WK2
(LB.Ft.2)
2.35
2.58
2.83
3.09
3.38
3.68
4.00
4.35
4.72
5.11
5.58
5.96
6.42
6.91
7.42
7.97
8.54
9.15
9.75
12.61
16.07
20.21
25.08
30.79
37.43
45.09
53.87
63.86
75.19
87.96
102.3
118.31
136.14
155.92
177.77
Diameter
(inches)
WK2
(LB.Ft.2)
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
66
72
78
84
90
96
201.8
228.2
257.2
288.8
323.2
360.7
401.3
445.3
492.8
543.9
598.8
658.1
721.4
789.3
861.8
939.3
1021.8
1109.6
1203.1
1302.2
1407.4
1518.8
1636.7
1761.4
1898.1
2031.9
2178.3
2332.5
2494.7
3652.5
5172.0
7125.0
9584.0
12629.0
16349.0
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Equivalent WK2
Equivalent WK 2
=
WK 2 x
N 2
Nb
W = weight of the rotating part in pounds
K = radius of gyration of the rotating part in feet
N = speed of the rotation part in RPM
Nb = speed of the motor shaft in RPM
It should be noted that if there is gearing or belting between the rotating part and the motor, this must
be taken into account, and the figure used for the WK2 of the rotating part must be adjusted to a WK2
equivalent to direct connection to the motor shaft. By the use of the above equation, it is possible to
calculate the WK2 of a system including several rotating parts, which are rotating at different speeds.
The WK2 of each part is adjusted to its equivalent WK2, and the equivalent WK2 figures are added
together to obtain the WK2 of the whole system.
High WK2 Acceleration
When starting a high WK2 load, step-starting may be required to maintain the torque necessary to
accelerate the mass. Care must be taken in choosing resistor capacity to dissipate the heat resulting
from the starting currents.
Data required to determine the correct motor for a high WK2 load.
1. WK2 of load
2. Torque required
3. Duty cycle
Application Manual for NEMA Motors
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Equivalent WK2
Example:
200 lbs-ft for 5 seconds accelerates load to 1150 rpm; 50 lbs-ft for 10 seconds does work required;
200 lbs-ft for 5 seconds decelerates load to standstill; 12 seconds at rest (1/3 off-time is used for
open motors).
Total cycles time = 5 + 10 + 5 + 12 = 24 seconds
3
rms torque =
(2002 x 5 ) + (502 x 10) + (2002 x 5)
24
rms torque =
133.1 lb. ft.
HP
=
rms torque x 1150
5250
HP
=
133.1 x 1150 = 29.15
5250
Therefore, use motor rated 30 HP, 1150 rpm
Note: The above example is calculated on the basis of rms torque rather than rms HP. The additional
heating due to the nature of the duty cycle has been taken into consideration.
Information Required for Proper Selection Involving High Inertia or Duty Cycle
It is recommended that the Little Rock Electrical Application Department be supplied with the
following information:
1. Load WK2 at motor shaft.
2. Number of starts, stops or reversal per unit of time.
3. HP load and length of each operating period.
4. Length of standing idle periods.
5. Method of stopping.
6. Special torque requirements of motor, such as need to break away heavy friction load from rest;
need to bring heavy inertia load up to speed (or down to stop) in specified period of time; need to
have high pull-out torque to carry momentary overloads.
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Horsepower Determination
Determination of HP Requirements
The horsepower required can often be determined from the factual information or power requirements
for specific operations. Where the force o torque required is known, one of the equations below may
be used to calculate the horsepower required by constant load characteristics.
Power for Transition
HP
=
Force (lbs) x ft. per min.
33,000
Power for Rotation
HP
=
Torque (lbs.-ft.) x RPM
5,250
Power to Drive Pumps
HP
=
Gal. per min. x total head (inc. friction) x specific gravity
3,960 x eff. of pump
2
Where friction head (ft) = pipe length (ft) x [velocity of flow (fps)] x .02
5,367 x eff. of pump
Power to Hoist a Load
HP
=
Weight (lbs) x feet per min. x sin 0
33,000
0 = Angle of hoist with horizontal
Power to Drive Fans
HP
=
Cu. ft. gas per min. x water gage pressure (in.)
6,350 x efficiency
Efficiencies for fans range between 50 and 80 percent.
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Formulas and General Data
To Find
Amperes when
HP is known
Amperes when
KW is known
Amperes when
KVA is known
Kilowatts
Input (KW)
Kilovolt
Amperes (KVA)
Horsepower
Output (MP)
=
=
=
=
=
=
Three Phase Formulas
746 x HP
1.73 x E x Eff x PF
1000 x KW
1.73 x E x PF
1000 x KVA
1.73 x E
1.73 x E x I x PF
1000
1.73 x E x I
1000
1.73 x E x I x Eff x PF
746
PF
EFF
T
f
I
E
KW
KVA
=
=
=
=
=
=
=
=
HP
N
Ns
P
=
=
=
=
Power Factor as a Decimal
Efficiency as a Decimal
Torque in LB-FT
Frequency in Hz
Current in Amperes
Voltage in Volts
Power in Kilowatts
Apparent Power in Kilovolt
Amperes
Output in Horsepower
Motor Sped in RPM
Synchronous Speed in RPM
Number of Poles
NEMA KVA Code Letters
Synchronous Speed - Frequency Number of Poles of AC Motors
Code KVA/HP
Ns = 120 x f f = P x Ns P = 120 x f
P
120
Ns
Horsepower - Torque - Speed
HP = T x N
5250
T = 5250 x HP
N
N = 5250 x HP
T
Locked Rotor Current ILR From
Nameplate Data - Three Phase
ILR = 577 x HP x KVA/HP
E
A
B
C
D
E
F
G
H
J
K
0 - 3.14
3.15 - 3.54
3.55 - 3.99
4.0 - 4.49
4.5 - 4.99
5.0 - 5.59
5.6 - 6.29
6.3 - 7.09
7.1 - 7.99
8.0 - 8.99
Code
KVA/HP
L
M
N
P
R
S
T
U
V
9.0 - 9.99
10.0 - 11.19
11.2 - 12.49
12.5 - 13.99
14.0 - 15.99
16.0 - 17.99
18.0 - 19.99
20.0 - 22.39
22.4 & UP
Example: For a 100 HP motor, 3 phase,
460 volts, KVA Code G
1LR = 577 x 100 x (5.6 to 6.29)
460
1LR - 702 to 789 Amperes
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Application Manual for NEMA Motors
Formulas and General Data
Effect of Line Voltage on Locked Rotor Current (Approx.)
Locked Rotor Amperes (LRA) is directly proportional to applied voltage.
LRA = LRA at nameplate voltage X
Applied Voltage
Nameplate Voltage
Example:15 HP - 1800 RPM - 254T Frame - RGZ - Standard Efficiency Motor is rated 116 AMPS
Locked Rotor at 460 Volts. What are the Locked Rotor AMPS at 495 Volts?
LRA = 116 x 495 = 125 AMPS
460
General Approximations (Rules of Thumb)
At
At
At
At
3600 RPM a motor develops 1.5 LB-FT of Torque per HP at Rated HP Output
1800 RPM a motor develops 3.0 LB-FT of Torque per HP at Rated HP Output
1200 RPM a motor develops 4.5 LB-FT of Torque per HP at Rated HP Output
900 RPM a motor develops 6.0 LB-FT of Torque per HP at Rated HP Output
At 575 Volts a 3 Phase motor draws 1.00 AMP per HP at Rated HP Output
At 460 Volts a 3 Phase motor draws 1.25 AMP per HP at Rated HP Output
At 230 Volts a 3 Phase motor draws 2.50 AMP per HP at Rated HP Output
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Temperature Conversion Table
Find known temperature in ˚C/˚F column. Read converted temperature in ˚C or ˚F column.
˚C
-40.0
-37.2
-34.4
-32.2
-29.4
-26.6
-23.8
-20.5
-17.8
-15.0
-12.2
-9.4
-6.7
-3.9
-1.1
1.7
4.4
7.2
10.0
12.8
15.5
18.3
21.1
23.9
26.6
29.4
˚C/˚F
-40.0
-35.0
-30.0
-25.0
-20.0
-15.0
-10.0
-5.0
0.0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.0
50.0
55.0
60.0
65.0
70.0
75.0
80.0
85.0
Fahrenheit to Centigrade
(˚F - 32) 5 = ˚C
9
˚F
-40.0
-31.0
-22.0
-13.0
-4.0
5.0
14.0
23.0
32.0
41.0
50.0
59.0
68.0
77.0
86.0
95.0
104.0
113.0
122.0
131.0
140.0
149.0
158.0
167.0
176.0
185.0
˚C
32.2
36.0
37.8
40.5
43.4
46.1
48.9
51.6
54.4
57.1
60.0
62.7
66.5
68.3
71.0
73.8
76.5
79.3
82.1
85.0
87.6
90.4
93.2
96.0
98.8
101.6
˚C/˚F
90.0
95.0
100.0
105.0
110.0
115.0
120.0
125.0
130.0
135.0
140.0
145.0
150.0
155.0
160.0
165.0
170.0
175.0
180.0
185.0
190.0
195.0
200.0
205.0
210.0
215.0
Centigrade to Fahrenheit
(˚C x 9) + 32 = ˚F
|
5
˚F
194.0
203.0
212.0
221.0
230.0
239.0
248.0
257.0
266.0
275.0
284.0
293.0
302.0
311.0
320.0
329.0
338.0
347.0
356.0
365.0
374.0
383.0
392.0
401.0
410.0
419.0
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Formulas and General Data
To Find
Amperes when
HP is known
Amperes when
KW is known
Amperes when
KVA is known
Kilowatts
Input (KW)
Kilovolt
Amperes (KVA)
Horsepower
Output (MP)
=
=
=
=
=
=
Three Phase Formulas
746 x HP
1.73 x E x Eff x PF
1000 x KW
1.73 x E x PF
1000 x KVA
1.73 x E
1.73 x E x I x PF
1000
1.73 x E x I
1000
1.73 x E x I x Eff x PF
746
PF
EFF
T
f
I
E
KW
KVA
=
=
=
=
=
=
=
=
HP
N
Ns
P
=
=
=
=
Power Factor as a Decimal
Efficiency as a Decimal
Torque in LB-FT
Frequency in Hz
Current in Amperes
Voltage in Volts
Power in Kilowatts
Apparent Power in Kilovolt
Amperes
Output in Horsepower
Motor Sped in RPM
Synchronous Speed in RPM
Number of Poles
NEMA KVA Code Letters
Synchronous Speed - Frequency Number of Poles of AC Motors
Code KVA/HP
Ns = 120 x f f = P x Ns P = 120 x f
P
120
Ns
Horsepower - Torque - Speed
HP = T x N
5250
T = 5250 x HP
N
N = 5250 x HP
T
Locked Rotor Current ILR From
Nameplate Data - Three Phase
ILR = 577 x HP x KVA/HP
E
A
B
C
D
E
F
G
H
J
K
0 - 3.14
3.15 - 3.54
3.55 - 3.99
4.0 - 4.49
4.5 - 4.99
5.0 - 5.59
5.6 - 6.29
6.3 - 7.09
7.1 - 7.99
8.0 - 8.99
Code
KVA/HP
L
M
N
P
R
S
T
U
V
9.0 - 9.99
10.0 - 11.19
11.2 - 12.49
12.5 - 13.99
14.0 - 15.99
16.0 - 17.99
18.0 - 19.99
20.0 - 22.39
22.4 & UP
Example: For a 100 HP motor, 3 phase,
460 volts, KVA Code G
1LR = 577 x 100 x (5.6 to 6.29)
460
1LR - 702 to 789 Amperes
6/97
Application Manual for NEMA Motors
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Page
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6/97
Application Manual for NEMA Motors
Formulas and General Data
Effect of Line Voltage on Locked Rotor Current (Approx.)
Locked Rotor Amperes (LRA) is directly proportional to applied voltage.
LRA = LRA at nameplate voltage X
Applied Voltage
Nameplate Voltage
Example:15 HP - 1800 RPM - 254T Frame - RGZ - Standard Efficiency Motor is rated 116 AMPS
Locked Rotor at 460 Volts. What are the Locked Rotor AMPS at 495 Volts?
LRA = 116 x 495 = 125 AMPS
460
General Approximations (Rules of Thumb)
At
At
At
At
3600 RPM a motor develops 1.5 LB-FT of Torque per HP at Rated HP Output
1800 RPM a motor develops 3.0 LB-FT of Torque per HP at Rated HP Output
1200 RPM a motor develops 4.5 LB-FT of Torque per HP at Rated HP Output
900 RPM a motor develops 6.0 LB-FT of Torque per HP at Rated HP Output
At 575 Volts a 3 Phase motor draws 1.00 AMP per HP at Rated HP Output
At 460 Volts a 3 Phase motor draws 1.25 AMP per HP at Rated HP Output
At 230 Volts a 3 Phase motor draws 2.50 AMP per HP at Rated HP Output
Section
10
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Date
Application Manual for NEMA Motors
Temperature Conversion Table
Find known temperature in ˚C/˚F column. Read converted temperature in ˚C or ˚F column.
˚C
-40.0
-37.2
-34.4
-32.2
-29.4
-26.6
-23.8
-20.5
-17.8
-15.0
-12.2
-9.4
-6.7
-3.9
-1.1
1.7
4.4
7.2
10.0
12.8
15.5
18.3
21.1
23.9
26.6
29.4
˚C/˚F
-40.0
-35.0
-30.0
-25.0
-20.0
-15.0
-10.0
-5.0
0.0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.0
50.0
55.0
60.0
65.0
70.0
75.0
80.0
85.0
Fahrenheit to Centigrade
(˚F - 32) 5 = ˚C
9
˚F
-40.0
-31.0
-22.0
-13.0
-4.0
5.0
14.0
23.0
32.0
41.0
50.0
59.0
68.0
77.0
86.0
95.0
104.0
113.0
122.0
131.0
140.0
149.0
158.0
167.0
176.0
185.0
˚C
32.2
36.0
37.8
40.5
43.4
46.1
48.9
51.6
54.4
57.1
60.0
62.7
66.5
68.3
71.0
73.8
76.5
79.3
82.1
85.0
87.6
90.4
93.2
96.0
98.8
101.6
˚C/˚F
90.0
95.0
100.0
105.0
110.0
115.0
120.0
125.0
130.0
135.0
140.0
145.0
150.0
155.0
160.0
165.0
170.0
175.0
180.0
185.0
190.0
195.0
200.0
205.0
210.0
215.0
Centigrade to Fahrenheit
(˚C x 9) + 32 = ˚F
|
5
˚F
194.0
203.0
212.0
221.0
230.0
239.0
248.0
257.0
266.0
275.0
284.0
293.0
302.0
311.0
320.0
329.0
338.0
347.0
356.0
365.0
374.0
383.0
392.0
401.0
410.0
419.0
6/97