Determination of
Induction-Motor Parameters
• DC Test
– Determines R1
– Connect any two stator leads to a variablevoltage DC power supply
– Adjust the power supply to provide rated
stator current
– Determine the resistance from the voltmeter
and ammeter readings
ECE 441
1
RDC
VDC

I DC
ECE 441
2
For a Y-Connected Stator
RDC  2 R1, wye
R1, wye
RDC

2
ECE 441
3
For a Delta-Connected Stator
RDC
R1  2 R1 2

 R1
R1  2 R1 3
R1  1.5RDC
ECE 441
4
Determination of
Induction-Motor Parameters
• Blocked-Rotor Test
– Determine X1 and X2
– Determines R2 when combined with data from
the DC Test
– Block the rotor so that it will not turn
– Connect to a variable-voltage AC supply and
adjust until the blocked-rotor current is equal
to the rated current
ECE 441
5
ECE 441
6
Simplified Equivalent Circuit
Neglect the exciting current under blocked-rotor
conditions – remove the parallel branch
ECE 441
7
IEEE test code recommends that the blocked-rotor
test be made using 25% rated frequency with the test
voltage adjusted to obtain approximately rated current.
A 60-Hz motor would use a 15-Hz test voltage.
The calculated reactance is corrected to 60-Hz by
multiplying by 60/15.
Calculated resistance is correct.
ECE 441
8
R1  R2  RBR ,15
Z BR ,15 
VBR ,15
RBR ,15 
PBR ,15
I BR ,15
2
I BR
,15
R2  RBR ,15  R1
ECE 441
9
2
2
Z BR ,15  RBR

X
,15
BR ,15
2
2
X BR ,15  Z BR

R
,15
BR ,15
60
X BR ,60  X BR ,15
15
X BR ,60  X 1  X 2
ECE 441
10
How Is the Blocked-Rotor
Impedance Divided?
X BR ,60  X 1  X 2
If the NEMA-design letter of the motor is known, use
Table 5.10 to divide the impedances. Otherwise,
divide the impedances equally.
ECE 441
11
Determination of
Induction-Motor Parameters
• No-Load Test
– Determine the magnetizing reactance, XM and
combined core, friction, and windage losses.
– Connect as for blocked-rotor test (next slide).
– The rotor is unblocked and allowed to run
unloaded at rated voltage and rated
frequency.
ECE 441
12
Electrical connection for the No-Load Test is the same
as for the Blocked-Rotor Test
ECE 441
13
Determination of
Induction-Motor Parameters
• At no-load, the speed is very close to
synchronous speed – the slip is =0,
causing the current in R2/s to be very
small, and will be ignored i the
calculations.
• IM>>Ife, so I0 = IM.
ECE 441
14
The equivalent circuit for the no-load test is shown.
Ignore
ECE 441
15
S NL  VNL I NL
2
2
S NL  PNL
 QNL
2
2
QNL  S NL
 PNL
2
QNL  I NL
X NL
X NL
QNL
 2
I NL
X NL  X 1  X M
Substitute X1 from the blocked-rotor test to determine
the value of XM.
ECE 441
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Example 5.16
• The following data were obtained from noload, blocked-rotor, and DC tests of a
three-phase, wye-connected, 40-hp,
60-Hz, 460-V, design B induction motor
whose rated current is 57.8A. The
blocked-rotor test was made at 15 Hz.
ECE 441
17
Blocked-Rotor
No-Load
DC
Vline = 36.2V
Vline = 460.0V
VDC = 12.0V
Iline = 58.0A
Iline = 32.7A
IDC = 59.0A
P3phase = 2573.4W
P3phase = 4664.4W
a) Determine R1, X1, R2, X2, XM, and the combined
core, friction, and windage loss.
b) Express the no-load current as a percent of rated
current.
ECE 441
18
Convert the AC test data to corresponding phase values
for a wye-connected motor.
2573.4W
PBR ,15 
 857.80W
3
36.2V
VBR ,15 
 20.90V
3
I BR ,15  58.0 A
4664.4W
PNL 
 1554.80W
3
460V
VNL 
 265.581V
3
I NL  32.7 A
ECE 441
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Determine R1
RDC
VDC 12.0V


 0.2034
I DC 59.0 A
R1, wye
RDC

 0.102 / phase
2
Determine R2
Z BR ,15 
VBR ,15
RBR ,15 
PBR ,15
I BR ,15
2
I BR
,15
20.90V

 0.3603 / phase
58.0 A
857.8W

 0.2550 / phase
2
(58 A)
R2  RBR ,15  R1, wye  0.2550  0.102  0.153 / phase
ECE 441
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Determination of X1 and X2
2
2
2
2
X BR ,15  Z BR

R

(0.3603)

(0.255)
 0.2545
,15
BR ,15
X BR ,60
60
60

X BR ,15  (0.2545)  1.0182
15
15
From Table 5.10, for a design B machine,
X1 = 0.4XBR,60 = 0.4(1.0182) = 0.4073Ω/phase
X2 = 0.6XBR,60 = 0.6(1.0182) = 0.6109/phase
ECE 441
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Determination of XM
S NL  VNL I NL  (265.581V )(32.7 A)  8684.50VA
2
2
QNL  S NL
 PNL
 (8684.50) 2  (1554.8) 2  8544.19VARS
X NL
QNL 8544.19
 2 
 7.99
2
I NL
(32.7)
X NL  X 1  X M
X M  X NL  X 1  7.99  0.4073  7.58 / phase
ECE 441
22
Determination of combined friction, windage, and core
loss:
2
PNL  I NL
R1, wye  Pcore  Pf ,w
1554.8  (32.7) (0.102)  Pcore  Pf , w
2
Pcore  Pf , w  1446W / phase
b) Express the no-load current as a percent of rated
current.
% I NL
I NL
32.7

100% 
100%  56.6%
I rated
57.8
ECE 441
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