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Determination of Equivalent Circuit Parameters of Single Phase Induction Motor

The document describes methods to experimentally determine the parameters of the equivalent circuit model of a single-phase induction motor. It involves three tests: 1) A DC test to determine stator resistance. 2) A locked rotor test to determine equivalent resistance and reactance. 3) A no-load test to determine magnetizing reactance and rotor resistance. Parameters are calculated from measurements of voltage, current, power and slip using equations derived from the motor equivalent circuit models under different operating conditions.

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Reeta Dutta
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405 views6 pages

Determination of Equivalent Circuit Parameters of Single Phase Induction Motor

The document describes methods to experimentally determine the parameters of the equivalent circuit model of a single-phase induction motor. It involves three tests: 1) A DC test to determine stator resistance. 2) A locked rotor test to determine equivalent resistance and reactance. 3) A no-load test to determine magnetizing reactance and rotor resistance. Parameters are calculated from measurements of voltage, current, power and slip using equations derived from the motor equivalent circuit models under different operating conditions.

Uploaded by

Reeta Dutta
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Determination Of Equivalent Circuit Parameters Of Single Phase Induction

Motor
It is possible to find the parameters of the equivalent circuit of the single phase induction motor
experimentally as shown in Fig.1.

I1

R1
2

X1
2

Rw
2

X2
2

1R2 Forward
2 s

Xm
2

1
2
R Backward
2 2 s

V1

Rw
2

R1
2

Xm
2

X1
2

X2
j
2

Fig.1: Equivalent circuit of single phase induction


motor. For this purpose, three tests should be conducted:

1- The DC Test:
The DC resistance of the stator can be measured by applying DC current to the terminals of the
main winding and taking the reading of the voltage and the current (or using ohmmeter) and
determine the DC resistance as fallows:
R=
I

DC

(1)

DC

DC

Then, the AC resistance is given by:


R 1=1.15RDC

(2)

2- The Locked Rotor Test


When the rotor is locked (i.e. prevented from running), sb = s f = 1. The secondary impedances
become much less than the magnetizing branches and the corresponding equivalent circuit becomes
that of Fig.2.

IBL

R1
2

X1
2

X2
2

R
2

V BL

R
2

X1
j X2
2

R1
2

2
Fig.2: Approximate equivalent circuit of the single

phase induction motor at standstill The circuit in Fig.2 can be rearranged to the equivalent
circuit that is shown in Fig.3.
IBL R1

jX1

jX2

V BL
R
Fig.3: Rearranged approximate equivalent circuit of the single phase induction motor at
standstill.
2

The readings to be obtained from this test are:


a) Single phase power PBL
b) Phase voltage VBL
c) Phase current IBL
Then, Req , Zeq , and Xeq can be obtained using the following equations:
R=

e q

BL
I

P
2

BL

V
Z BL
I

(4)

eq

BL

X eq Z eq R eq

(5)

Separation of X1 , X2 , R1 , and 2R can be done as follows:


X
X
1

Xeq

(6)

R2 Req R1

(7)

3- The No Load Test


When the induction motor is allowed to run freely at no load, the forward slip sf approaches zero
and the backward slip sb approaches 2 (sf s,sb 2 s). The secondary forward impedance
becomes very large with respect to the magnetizing branch, while the secondary backward
impedance becomes very small if compared with the magnetizing branch. Accordingly, the
equivalent circuit corresponding to these operating conditions can be approximated by that of Fig.4.

INL

R1
2

X1
2

Rw
2

j Xm
2

Forward

E1 F

V NL

X1j
R1
2

R
4

Backward

X2
2

Fig.4: Approximate equivalent circuit of the single phase induction motor at no load.

The circuit in Fig.4 can be rearranged to the equivalent circuit that is shown in Fig.5.

R1

INL

jX1

Rw
2

Xm
2

Forward

E1 F

V NL
R 2

X
2

Backward

Fig.5: Rearranged approximate equivalent circuit of the single phase induction motor at no load
The readings to be obtained from this test are:
d) Single phase power PNL
e) Phase voltage VNL
f) Phase current I NL

Then, Rw , and Xm , can be obtained as fallows:


P coremechanical P NL I NL( R 1
4

V N L I N L K R 1

1i(X

2 ))

(9)

Note: (INLI NL , cos1

R
w

2
P

4
N L

VN L IN L

E
1 F

(10)

coremechanical

R 2
2w

Rw

(11)

Im = I NL I w
2

X=2
m

E1 F
Im

(12)
(13)

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