ELE2117
Rotating Machines & Transformers
Assignment #7
LECTURER
DR. SHASHI PERSAUD
DETERMINATION OF INDUCTION MOTOR EQUIVALENT CIRCUIT
PARAMETERS FROM THE STATOR RESISTANCE, NO-LOAD AND
LOCKED ROTOR TESTS
STUDENT NAME:
STUDENT ID:
DATE:
Instructions
Instructions
This assignment is based on a number of lab tests performed on a 208V, 60Hz Squirrel
Cage Induction Motor, by the course lecturer on behalf of the ELE2117 class of 2022.
Carefully read the assignment and answer all questions.
You can print this document and enter your results in the spaces provided, or you can
NEATLY answer the questions by hand, on your own note paper.
Notes: You only need to submit pages 9 – 23.
This is a very important assignment and will be graded accordingly.
Submission




Create a cover title page for your Lab Assignment. Clearly state your name, institution,
course, student number and date on the title page.
Save/scan your completed assignment document as a pdf file. Proofread and tidy your
assignment script before submitting it as a pdf document.
Upload your pdf to the Dropbox link provided by Wednesday, 28th December 2022 at
11:59pm. Marks will be deducted for late assignments.
Save your completed Assignment 7 as a pdf file with this name format:
Lastname_Firstname ELE2117 Assignment 7
Upload your pdf file to the Dropbox link below before 11:59pm (Guyana time) on
Wednesday, December 28, 2022. This link will close shortly after 11:59 am.
Dropbox link: https://www.dropbox.com/request/NzITjZhTHNWMKLIg5Jkf
Goodluck!
2
Objectives
1. To determine the exact equivalent circuit impedance parameters for the
LabVolt squirrel cage induction motor from the:
a. Stator resistance test
b. no-load test, and
c. locked-rotor test.
Equipment
1. Labvolt Squirrel Cage Induction Motor module
2. Labvolt Prime Mover / Dynamometer module
3. Variable three-phase supply
4. Labvolt DACI, Fluke 43B Analyzer and Fluke 179 Multimeters
5. Connecting leads
6. Calculator
Photograph of the labVolt SCIM
Front face schematic of the labVolt SCIM
3
Background
The simplest and most widely used rotor for induction motors is the so-called squirrel-cage
rotor, from which the squirrel-cage induction motor gets its name. The squirrel-cage rotor
consists of a laminated iron core, which is slotted lengthwise around its periphery. Solid bars
of copper or aluminum are tightly pressed or embedded into the rotor slots. At both ends of
the rotor, short-circuiting rings are welded or brazed to the bars to make a solid structure. The
short-circuited bars, because their resistance is much less than the core, do not have to be
specially insulated from the core. In some rotors the bars and end rings are cast as a single
integral structure for placement on the core. The short-circuiting elements actually form
shorted turns that have high currents induced in them by the stator field flux. Compared to
the intricately wound and arranged wound rotor or the armature of the DC motor, the
squirrel-cage rotor is relatively simple. It is easy to manufacture and is essentially troublefree in actual service.
In an assembled squirrel-cage induction motor, the periphery of the rotor is separated from
the stator by a very small air gap. The width of this air gap, in fact, is as small as mechanical
clearance needs will permit. This insures that the strongest possible electromagnetic
induction action will take place. When power is applied to the stator of a practical induction
motor, a rotating magnetic field is created. As the field begins to revolve, its flux lines cut the
shorted turns embedded around the surface of the squirrel-cage rotor and generate voltages in
them by electromagnetic induction. Because these turns are short-circuited with very low
resistance, the induced voltages cause high currents to circulate in the rotor bars. The
circulating rotor currents then produce their own strong magnetic fields. These local rotor
flux fields produce their own magnetic poles, which are attracted to the rotating field. Thus,
the rotor revolves with the main field.
The starting torque of the basic squirrel-cage induction motor is low, because at rest the rotor
has a relatively large inductive reactance (XL) with respect to its resistance (R). Under these
conditions we would expect the rotor current to lag rotor voltage by almost 90. We thus say
that the power factor in the circuit is low. This means that the motor is inefficient as a load
and cannot derive really useful energy for its operation from the power source. Despite the
inefficiency, more torque is developed and the motor begins to turn. As it starts turning, the
difference in speed between rotor and rotating field, or slip, goes from a maxim of 100 % to
some intermediate value, say 50 %. As the slip decreases in this manner, the frequency of the
voltages induced in the rotor decreases, because the rotating field cuts conductors at a
decreased rate; this, in turn, causes the overall inductive reactance in the circuit to decrease.
As inductive reactance decreases, the power factor begins to increase. This improvement is
reflected as an increase in torque and a subsequent increase in speed.
When the slip drops to some value between 2 and 10 %, the motor speed stabilizes. This
stabilization occurs because every tendency for the motor speed to increase to where slip will
drop below 2 % is naturally offset by the fact that, as the rotor approaches within 2 % of the
synchronous speed, the effects of reduced induction overcome the previous tendency to
increase torque as the motor is speeded up from start. Thus, the motor exhibits an automatic
speed control characteristic similar to that of the DC shunt motor.
Source: LabVolt SCIM training module (also see Wildi chapter 13).
4
Testing
The stand test procedure for three-phase induction motors is specified in the IEEE standard
112-2004, titled IEEE Standard Test Procedure for Polyphase Induction Motors and
Generators (https://engineering.purdue.edu/~dionysis/EE452/Lab12/IEEEstd_112.pdf).
The standard specifies a power supply with maximum total harmonic distortion of 0.05%, a
maximum voltage unbalance of 0.5%, and supply frequency within +/- 0.5%.
The standard identifies the following tests and how they should be performed and the results
interpreted:
(i)
The winding resistance test which should be done at ambient winding temperature
and for all terminal combinations.
a. “With the machine at ambient temperature, measure the terminal-to-terminal
winding resistance with the machine connected in the configuration to be used in
the efficiency testing. Measure and record all combinations, i.e., T1-T2, T2-T3,
and T3-T1, to assure that the specific precise value needed in further analyses will
be available. Also measure and record the ambient temperature.”
(ii)
The idle (no-load) test for the determination of core loss and windage and friction
losses.
a. “This test is performed by running the machine as a motor at rated voltage and
frequency with no connected load. When separation of no-load losses is to be
accomplished, run this test and read temperature, voltage, current, and power
input at rated frequency and at voltages ranging from 125% of rated voltage down
to the point where further voltage reduction increases the current.”
b. “The measured input power is the total of the losses in the motor at no-load.
These losses consist of the stator I2R, friction (including brush-friction loss on
wound-rotor motors), windage, and core losses.”
(iii)
Load tests for the for the determination of efficiency, power factor, speed, current,
and temperature rise,
a. The machine is coupled to a load machine and is subjected to loads at four load
points approximately equally spaced between not less than 25% and up to and
including 100% load, and two load points suitably chosen above 100% load but
not exceeding 150% load. A spread in load test points is necessary to determine
the efficiency accurately over the entire load range of the machine and more than
six load points may be used if desired.
b. Readings of electrical power, current, voltage, frequency, speed or slip, torque,
stator winding temperature or stator winding resistance, and ambient temperature
shall be obtained at each load point. In loading the machine, start at the highest
load value and move in descending order to the lowest.
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(iv)
The locked rotor test (impedance test), which it emphasizes needs to be performed
safely and quickly, in recognition of the mechanical hazards present and the need to
limit winding temperature rise below the rated 40oC, and at reduced frequency to
minimize stray (unaccounted for) losses.
The standard identifies four methods for the calculation of the equivalent circuit
parameters, of which method 1 is best applicable here.
a. The rotor of a wound-rotor motor shall be blocked so that it cannot rotate freely,
but can be moved; and the impressed voltage shall be increased gradually until a
current of approximately rated value is obtained. Voltage and current on all
phases shall be read and recorded, and the voltage in the different phases shall be
balanced. Holding the same voltage, the rotor shall be turned slowly and the
minimum and maximum values of current during a complete cycle shall be
recorded. The rotor shall then be blocked for the impedance test on the position
that gives a current equal to the average of the minimum and maximum values
previously recorded.
For the locked-rotor test, take simultaneous readings of voltage and current in all
phases and of power input at several levels of voltage in order to establish the
value with special care in the neighborhood of full-load current. The stator
winding temperature or stator winding resistance shall also be recorded. Care
shall be taken not to overheat the windings. Taking the highest readings first and
the lower readings in succession will help to equalize the temperature.
b. Determine the rotor resistance, R2, and the total leakage reactance, X1 + X2, from
these data. The calculations start by assuming a relationship between X1 and X2.
When design details are available, use the calculated ratio X1/X2. Otherwise, use
X1 = X2 for Design A, Design D, and wound rotor motors, X1 = 0.67X2 for
Design B motors, and X1 = 0.43 X2 for Design C motors.
Note: The subscript 1 refers to the stator, and the subscript 2 refers to the rotor.
6
The exact phase equivalent circuit of the Induction Motor referred to the stator
Rs
X’R
Xs
Rm
Xm
𝑅𝑅′
𝑠
Figure 1: Exact phase equivalent circuit of an induction motor referred to the stator.
Figure 1 above shows the exact phase equivalent circuit of an induction motor. An
explanation of the significance of each of the impedance parameter (Rm, Xm, Rs, Xeq, and
R’r/s) is provided following:
Rm
The magnetizing branch resistance Rm is a primarily a resistive representation of hysteresis
and eddy current losses in the laminated stator core. Friction and windage losses are also
lumped together with core losses, unless advanced methods are used to isolate these from the
losses measured at no load. In some simplified models, Rm may be omitted from the
equivalent circuit as it is usually significantly higher than the stator core reactance Xm.
Xm
The magnetizing branch reactance Xm models the reactive power requirements of the stator
core needed to produce the rotating magnetic field at rated voltage and frequency. Both Rm
and Xm are defined as the no load losses (PNL and QNL) of the motor as they always present
when motor is energized, irrespective of the motor loading.
7
Rs
This is the actual resistance of the stator windings that can be measured by the stator DC
resistance test, or with an ohmmeter in smaller motors that have higher winding resistance.
Once the stator is energized current will flow through the stator coil resulting in heat losses.
The losses are referred to the stator winding loss or the stator as copper losses, because
windings are made out of copper wires, PCu. Stator copper losses increase quadratically
above the no-load value as more current is drawn from the source when the motor is loaded.
XS
This is the leakage inductive reactance of the stator windings. It results from the stator
leakage flux that does not link the rotor coils (conductors), i.e., is not mutually coupled with
the rotor. Xs is a source of reactive power loss due to the I2X effect. Like stator the copper
loss, it increases quadratically above the no-load value as more current is drawn from the
source when the motor is loaded.
R’r/s
This slip-dependent resistance represents both the actual rotor resistance referred to the stator
side and the resistive effect of the mechanical load on the rotor. It usually appears in the
equivalent in the R/s form but cam be decomposed into its two components as shown:
𝑅𝑅′ = 𝑅′ + (1 − 𝑠) 𝑅′
𝑅
𝑅
𝑠
𝑠
R’r
This is the resistance of the rotor winding referred to the stator side. Since there is no access
to measure the resistance of the rotor windings in a Squirrel Cage Induction Motor (SCIM)
this value is calculated from the Locked-Rotor Test. It is possible to measure this value in a
Wound Rotor Induction Motor (WRIM) from the brushes using a DC resistance test.
(1-s)R’r/s
This is resistive effect of the mechanical load in the equivalent circuit. As the load in the
motor changes the slip (s) also changes. This component is low at high mechanical loads
which results in the higher rotor referred currents required to match the higher load.
Reference: Class presentation PowerPoint – Induction Motor Equivalent Circuit and Tests
8
Part 1: Stator resistance measurement
1. Wye-connect the stator and perform a DC resistance test across each pair of motor
input terminals.
Apply rated current verify your results (several measurements) before recording the
results in the table below.
Note: This was done and the results are presented in columns 2 and 3 from left.
Complete Table 1 below to find the average stator resistance per phase.
Table 1: Stator resistance test results
Terminals
VDC
IDC
1-2
37.54 V
1.494 A
25.13 ohms
12.565 ohms
2-3
38.48 V
1.501 A
25.64 ohms
12.82 ohms
3-1
38.81 V
1.500 A
25.87 ohms
12.94 ohms
RS(avg) =
12.77 ohms
RLL
RS
Note: The subscripts 1, 2 and 3 refer to the stator terminals of LabVolt squirrel
cage induction motor module. Terminals 4, 5 and 6 are WYE-connected.
Part 2: Locked Rotor (LR) test
In this test the rotor of the induction motor must be locked (not turning). To accomplish this:
 The dynamometer (mechanical load) will be coupled to the induction motor and set to
apply maximum torque to the motor shaft.
 With the rotor locked (not turning), the stator supply voltage will be gradually raised
from zero until rated current flows in the induction motor stator winding.
 Readings will then be taken of stator line-to-line voltages, line currents and threephase power. Use the two-wattmeter function of the Data Acquisition and Control
Interface (DACI) to measure power.
9
Figure 2 below shows the experimental schematic for the locked-rotor test.
Note that, in addition to the main two-wattmeter connection, there is an additional current
and an additional voltage measuring element. This is to allow for the measurements of all
line currents and line-to-line voltages.
+
A
IA
Mechanical
brake
+
EAB
0-208V
3 phase
IB
ECA
B
C
IC
ECB
+
+
Figure 2: Schematic – Locked-rotor test
1.
Inspect the front panel of the induction motor module and record the stator
winding full-load current.
IFL = 1.2 A
2.
Connect the locked-rotor test circuit described and shown in the above
schematic.
3.
Connect the dynamometer to 120 VAC and adjust to the maximum load torque
(3 Nm) so that the rotor cannot turn.
1.
Set the variable three-phase supply voltage to 0 V.
Slowly increase the stator supply voltage until rated current (max 1.5A) flows
in the stator. Note that the rotor should not be turning during this test.
2.
Measure the 3ph power, the three (3) applied line-to-line voltages, the three (3)
line currents and the rotor speed.
 Record your results in Table 2 on the next page.

Calculate the average line-to-line voltage and line currents.

Calculate the 1ph values and enter in the right column.
10
Table 2: Locked Rotor test results
LR Quantity
Measured 3ph value
P
136.3 W
VAB
76.08 V
VCB
76.13 V
VBA
76.69 V
Average voltage
3.
IA
1.509 A
IB
1.508 A
IC
1.520 A
Average current
1.512 A
45.43 W
44.05 V
1.512 A
Calculate the single-phase apparent power input to the motor
𝑆𝐿𝑅(1𝑝ℎ) =
4.
76.30 V
1ph value
𝑉𝐿𝐿(𝑎𝑣𝑔)
√3
𝑥 𝐼𝐿(𝑎𝑣𝑔) =
Calculate the single-phase reactive power input to the motor.
𝑄𝐿𝑅(1𝑝ℎ) = √(𝑆𝐿𝑅(1𝑝ℎ))2 − (𝑃𝐿𝑅(1𝑝ℎ))2 =
In the space provided below, draw the reduced phase equivalent circuit of the
induction motor at locked-rotor. (Hint: Ignore the magnetizing branch.)
11
4.
What is the value of slip at locked-rotor?
sLR =
ns  nr
=
ns
5.
With reference to your locked-rotor equivalent circuit, calculate R’R. Recall the
value of RS from step 2.
𝑅𝑅′ =
6.
𝑃𝐿𝑅(1𝑝ℎ)
2
𝐼𝐿𝑅(𝐿𝑁)
− 𝑅𝑆
With reference to the locked-rotor equivalent circuit, calculate the value of XS.
Assume a NEMA class B rotor design (k = 0.4).
𝑋𝑆 + 𝑋′𝑅=
𝑄𝐿𝑅(1𝑝ℎ)
2
𝐼𝐿𝑅(𝐿𝑁)
=
𝑋𝑆 = 𝑘(𝑋𝑆 + 𝑋𝑅′ ) =
𝑋𝑅 = (1 − 𝑘)(𝑋𝑆 + 𝑋𝑅′ ) =
12
Part 3: No-load (NL) test
In this test the unloaded, wye-connected, induction motor will be energized at approximately
rated voltage and measurements of the line-to-line voltages, line currents and three-phase
power recorded. The motor should not be coupled to any other device!
The schematic diagram for the no-load test is shown below. Note that, in addition to the main
two-wattmeter connection, there is an additional current and an additional voltage measuring
element. This is to allow for the measurements of all line currents and line-to-line voltages.
+
A
IA
+
EAB
0-208V
3 phase
IB
ECA
Uncoupled
rotor
B
C
IC
ECB
+
+
Figure 3: Schematic diagram - No-load test
5.
Note: The wiring diagram is the same as for the Locked Rotor test but with the
brake and belt removed.
6.
Build the no-load test circuit.
Set the supply to approx. 208 V line-to-line (120 V line-to-neutral).
Measure the 3ph power, the three (3) applied line-to-line voltages, the three (3)
line currents and the rotor speed.
Record your results in the table on the next page.
Calculate the average line-to-line voltage and line currents.
Calculate the 1ph values and enter in the right column.
13
Table 3: No Load test results
NL Quantity
Measured 3ph value
P
41.74 W
VAB
206.0 V
VCB
205.3 V
VBA
206.4 V
1ph value
Average voltage
IA
0.649 A
IB
0.638 A
IC
0.641 A
Average current
7.
Calculate the single-phase apparent power input to the motor
𝑆𝑁𝐿(1𝑝ℎ) =
8.
𝑉𝐿𝐿(𝑎𝑣𝑔)
√3
𝑥 𝐼𝐿(𝑎𝑣𝑔) =
Calculate the single-phase reactive power input to the motor.
𝑄𝑁𝐿(1𝑝ℎ) = √(𝑆𝑁𝐿(1𝑝ℎ))2 − (𝑃𝑁𝐿(1𝑝ℎ))2 =
9.
In the space neatly provided below, draw the phase equivalent circuit of the
induction motor at no-load. (Hint. Omit the load circuit)
14
10.
Calculate the no-load power factor and current angle.
pfNL =
 = cos-1(pfNL) =
11.
Calculate the voltage across the magnetizing branch (VM).
VM = VNL(LN) – INL(RS + jXS)
=
12.
Calculate the stator copper loss per phase at no-load
P(Cu)NL = (INL)2Rs =
13.
Calculate the magnetizing power (iron) loss per phase.
PM = PNL(1) - P(Cu)NL
=
14.
Use your circuit above to calculate the values of RM and XM.
𝑅𝑀 =
15.
| 𝑉𝑀2 |
𝑃𝑀
=
Calculate the stator copper loss per phase at no-load
Q(Cu)NL = (INL)2Xs =
16.
Calculate the magnetizing reactive power per phase.
QM = QNL(1) - Q(Cu)NL
=
17.
𝑋𝑀 =
|𝑉2 |
𝑀
𝑄𝑀
=
15
Part 4: The exact and approximate equivalent circuits
Enter the calculated equivalent circuit parameters in the exact and approximate
equivalent circuits below






Figure 4: Exact phase equivalent circuit with calculated parameters





Figure 5: Approximate (simplified) phase equivalent circuit with calculated
parameters
16
Part 5: Equivalent circuit (approximate) calculations
In the previous sections, the equivalent circuit parameters of the induction motor were
determined from the stator resistance, the no-load and the locked-rotor tests. These were used
to create an exact equivalent circuit of each phase of the motor.
The exact equivalent circuit allows for a more accurate prediction of the SCIM performance
at different values of slip, but this accuracy comes with higher calculation complexity.
Calculation complexity can be reduced with a great loss of accuracy by simplifying the
equivalent circuit.
In simplifying the equivalent circuit, we move the magnetizing branch in parallel with the
input terminals. This effectively decouples the equivalent circuit into parallel magnetizing
and load branches, each of which can be analyzed independently.
We will now use this simplified equivalent circuit to predict the behaviour of the motor at
various values of slip.
Problem #1
(a) Calculate the torque developed when the motor is operating at rated voltage and rated
speed of 1670 rpm.
(b) Calculate the stator line current and power factor.
(c) Calculate the real and reactive power input to the motor
(d) Calculate the motor full-load efficiency.
Solution #1
(a) The following steps demonstrate how the equivalent circuit is used to solve this
problem.
Calculate the total impedance of the series combination of Rs, X and R’R/s.
𝑍
𝑒𝑞
′
= 𝑅 + 𝑅𝑅 + 𝑗𝑋 =
𝑆
𝑒𝑞
𝑠
Find the rotor referred current (IR’).
𝐼′ =
𝑅
𝑉𝑆(𝐿𝑁)
=
𝑍𝑒𝑞
17
Calculate the three-phase power delivered to the rotor (PR).
′
𝑃 = 3|𝐼′ |2 𝑅𝑅 =
𝑅
𝑅
𝑠
Calculate the three − phase torque developed by the rotor
𝑇=
9.55𝑃𝑅
𝑛𝑆
=
(b) Find the magnetizing branch current (IM)
𝐼𝑀 =
𝑉𝑆(𝐿𝑁)
𝑅𝑀
+
𝑉𝑆(𝐿𝑁)
𝑗𝑋𝑀
=
Calculate the stator line current (Is)
𝐼𝑆 = 𝐼𝑀 + 𝐼𝑅′ =
(c) Calculate the real and reactive power input to the motor
𝑆𝑆 = 3𝑉𝑆(𝐿𝑁)𝐼𝑆∗ = 𝑃𝑆 + 𝑗𝑄𝑆
=
(d)
% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑃𝑚𝑒𝑐ℎ
𝑃𝑆
𝑥 100% =
18
Problem #2
Calculate (a) the start-up torque, and (b) the breakdown torque of the induction motor when
operating at rated voltage.
Solution #2
The following steps demonstrate how the equivalent circuit is used to solve this problem.
(a) Start-up torque
To solve this problem, we must first recognize that the motor experiences slip at start-up is
1pu. The problem is then essentially a repeat of the previous problem.
′
𝑍
= 𝑅 + 𝑅𝑅 + 𝑗𝑋 =
𝑆
𝑒𝑞
1
𝐼′ =
𝑉𝑆(𝐿𝑁)
=
𝑍𝑒𝑞
𝑒𝑞
𝑅
′
𝑃 = 3|𝐼′ |2 𝑅𝑅 =
𝑅
𝑅
𝑠
𝑇=
9.55𝑃𝑅
𝑛𝑆
=
(b) Breakdown torque
To solve this problem, we must first calculate the breakdown slip.
𝑠𝐵𝐷 =
𝑅𝑅′
|𝑅𝑆 + 𝑗𝑋𝑆|
=
′
𝑍
= 𝑅 + 𝑅𝑅 + 𝑗𝑋 =
𝑆
𝑒𝑞
1
𝐼′ =
𝑉𝑆(𝐿𝑁)
=
𝑍𝑒𝑞
𝑒𝑞
𝑅
′
𝑃 = 3|𝐼′ |2 𝑅𝑅 =
𝑅
𝑅
𝑠
𝑇=
9.55𝑃𝑅
𝑛𝑆
=
19
Part 6: Determination of the Torque vs Speed characteristic
We will now use the equivalent circuit to determine the torque versus speed characteristic of
the induction motor at varying values of slip.
1. Using the methodology of problem #1 and #2 perform the calculations for the slip
conditions in Table 4. Show your work and results in Table 4 below.
Table 4 Equivalent circuit calculations
nr
(rpm)
Slip
(s)
1800
0
1440
0.2
1080
0.4
720
0.6
360
0.8
0
1
Z
IR’
T
20
2. NEATLY and ACCURATELY plot a graph of Torque versus speed, corresponding to
the calculated values in Table 4 and the value of breakdown torque calculated
previously.
Label the axes and use an appropriate scale.
Clearly indicate values of Torque, speed and slip on the axes.
Show LR, BD, NL and FL points on your graph.
Torque (Nm)
180
360
540
720
900
1080
1260
1440
1620
1800
Speed (rpm)
Figure 6: Torque vs. speed characteristics of the squirrel cage induction motor.
21
3. NEATLY and ACCURATELY plot a graph of the magnitude of rotor current versus
speed, corresponding to the calculated values in Table 4.
Label the axes and use an appropriate scale.
Clearly indicate values of Current and speed.
Show LR, BD, NL and FL points on your graph.
Rotor referred current (I’R)
180
360
540
720
900
1080
1260
1440
1620
1800
Speed (rpm)
Figure 7: Stator Current vs. Speed characteristics of the squirrel cage induction motor.
End of assignment!
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