OPEN CIRCUIT TEST
Note: Low voltage side is supplied with rated voltage while high side is left open circuited.
Ioc
A
Vrated
Poc
V Eoc
ɸ
N1
W
E1
N2
OPEN
E2
HV
ɸ
LV
POC = wattmeter reading = core loss, PCO
IOC = ammeter reading = open circuit current
EOC = voltmeter reading = rated voltage
jXeL
ReL
Im
V1
Rm
N1 : N2
I2=0
I1=0
jXm
E1
E2
Im = Ioc
V1 = Eoc
Rm
Rm
jXm
Voc 2
Voc 2
=
↔ Xm =
Poc
Qoc
Qoc = √Soc 2 − Poc 2
Soc = Voc Ioc ↔ Poc = Pco
SHORT CIRCUIT TEST
Note: Low voltage side is short circuited while the high voltage side is supplied with voltage adjusted so that
the high voltage side will draw rated current.
Isc
A
Psc
V Esc
Variable
supply
W
ɸ
N1
E1
HV
ɸ
E2
N2
SHORTED
LV
ReH
jXeH
N1 : N2
Im I = rated
1
V1
Rm
jXm
I2 = rated
E1
E2
HV
LV
Psc = wattmeter reading = copper loss, Pcu
Isc = ammeter reading during the test
Esc = voltmeter reading during the test
jXeH
ReH
I1 = Isc
V1=Esc
Pcu (rated) = Psc
Psc
Esc
R eH = 2 ↔ ZeH =
Isc
Isc
XeH = √ZeH 2 − R eH 2
ReH, XeH = equivalent resistance, reactance and impedance respectively referred to the high voltage side of
the transformer.
EXAMPLE: A short circuit test was performed on a 10 kVA, 2000/400 V single-phase transformer. The
instruments indicated 100 V, 5 A and 100 W. determine the equivalent complex impedance of the
transformer referred to the low voltage side.
SOLUTION:
R eH =
Psc
Isc
2
100
= 2 =4
5
ZeH
Esc 100
=
=
= 20
Isc
5
XeH = √ZeH 2 − R eH 2 = √202 − 42 = 8√6
ZeL =
R eH + jXeH
4 + j8√6
=
= 0.16 + j0.784 Ω
2
2
a
(2000/400)
EXAMPLE: In an open circuit test of a 10 kVA, 2400/240 volt, 60 cycle transformer, the low side volts
amperes and watts are found to be 240 volts, 0.75 A and 72 watts. The low or secondary side is short
circuited when 67 volts is applied to the high side, rated current of 4.17 A flows and the power is 146 watts.
Determine the equivalent impedance referred to the low side.
SOLUTION:
*Since only magnitude of impedance is asked,
Esc
67
ZeH =
=
= 16.07
Isc 4.17
ZeL
ZeH
16.07
= 2 =
= 0.1607 Ω
a
(2400/240)2
Note: R m and Xm which can be calculated from the low side are too large compared to R eL and XeL
EXAMPLE: A 100 kVA, 6600/330 V, 60 Hz, single phase transformer took 10 A and 436 watts at 100 V in a
short circuit test on the high side. Calculate the voltage to be applied on the high voltage side on full load
at 0.8 pf lagging when the secondary terminal voltage is 330 V.
SOLUTION:
Psc
436
R eH = 2 = 2 = 4.36
10
Isc
ZeH
Esc 100
=
=
= 10
Isc
10
XeH = √ZeH 2 − R eH 2 = √102 − 4.362 = 9
4.36Ω
j9Ω
I1
Vp
S=100kVA
E1
6600 : 330
S
100k
I1 =
=
= 15.15 𝐴
E1 6600
VP =E1 +I1 (Ze1 ) = 6600+(15.15∠ − cos −1 0.8)(4.36 + j9)
VP = 6735 V
E2 = VL
TRANSFORMER LOSSES
Core Loss – due to hysteresis and eddy current losses in the magnetic circuit (core) of the transformer
Pe = k e V1
2
V11.6
Ph = k h 0.6
f
Note: Core loss of a transformer is constant at any size of load.
EXAMPLE: The hysteresis and eddy-current losses in a 6.9 kV, 50 Hz transformer are 400 W and 200 W,
respectively. What will be the iron loss of this transformer if it is to be connected across a 6.6 kV, 60 Hz
source?
SOLUTION:
V11.6
Ph = k h 0.6
f
6.9k1.6
400 = k h ∗
500.6
k h = 3.015x10−3
Pe = k e V1 2
200 = k e ∗ 6.9k 2
k e = 4.2x10−6
@ V2 = 6.6kV and f = 60 Hz,
6.6k1.6
−3
Ph = 3.015x10 ∗
= 333.936 W
0.6
60
−6
Pe = 4.2x10 ∗ 6.6k 2 = 201.285 W
Piron = Ph + Pe = 333.936 + 201.285 = 535.2 𝑊
EXAMPLE: In a 400 V, 50 Hz transformer, the core loss is 2500 W. when supplied at 220 V, 25 Hz, the
corresponding loss is 850 W. Calculate the eddy current loss at normal frequency and p.d.
SOLUTION:
Pcore
V1.6
= Ph + Pe = k h 0.6 + k e V 2
f
@ V1 = 400V and f = 50 Hz,
4001.6
Pcore = k h
+ k e 4002
0.6
50
2500 = 1392.88k h + 160000k e
→ eqn. 1
@ V2 = 220V and f = 25 Hz,
2201.6
Pcore = k h
+ k e 2202
0.6
25
850 = 811.17k h + 48400k e
→ eqn. 2
k h = 0.2405 ; k e = 0.01353
@ VNormal = 400V and f = 50 Hz,
Peddy = k e V 2 = 0.0135 ∗ 4002
Peddy = 2165 𝑊