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1.3 Open and Short Circuit Test

The document describes procedures for conducting open and short circuit tests on transformers, detailing the measurements and calculations involved. It includes formulas for determining core and copper losses, as well as equivalent impedance referred to the low and high voltage sides. Several examples illustrate the application of these tests and calculations for different transformer specifications.

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Robert Rumaguera
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613 views8 pages

1.3 Open and Short Circuit Test

The document describes procedures for conducting open and short circuit tests on transformers, detailing the measurements and calculations involved. It includes formulas for determining core and copper losses, as well as equivalent impedance referred to the low and high voltage sides. Several examples illustrate the application of these tests and calculations for different transformer specifications.

Uploaded by

Robert Rumaguera
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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OPEN CIRCUIT TEST

Note: Low voltage side is supplied with rated voltage while high side is left open
circuited.

POC = wattmeter reading = core loss, PCO


IOC = ammeter reading = open circuit current
EOC = voltmeter reading = rated voltage
OPEN CIRCUIT TEST

Eoc2 Eoc2
Rm = Xm =
Poc Q oc
Soc = E oc I oc Poc = Pco

Q oc = S oc 2 − Poc2
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.

Psc = wattmeter reading = copper loss, Pcu


Isc = ammeter reading during the test
Esc = voltmeter reading during the test
SHORT CIRCUIT TEST

Esc
Psc = Pcu (rated) ZeH =
P Isc
R eH = sc
Isc 2 XeH = ZeH 2 − R eH 2

ReH, XeH = equivalent resistance, reactance and impedance respectively referred


to the high voltage side of the transformer.
OPEN/SHORT CIRCUIT TEST
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
ZeH = = = 16.07 ZeL = = = 0.1607 Ω
Isc 4.17 a2 (2400/240)2

Note: R m and X m which can be calculated from the low side are too large compared to R eL and XeL
OPEN/SHORT CIRCUIT TEST
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:
Psc 100 Esc 100
R eH = 2
= =4 ZeH = = = 20
Isc 52 Isc 5

X eH = Z eH 2 − R eH 2 = 202 − 42 = 8 6

R eH + jXeH 4 + j8 6
ZeL = = = 0.16 + j0.784 Ω
a2 (2000/400)2
OPEN/SHORT CIRCUIT TEST
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 Esc 100
R eH = 2 = = 4.36 ZeH = = = 10
Isc 102 Isc 10

X eH = Z eH 2 − R eH 2 = 102 − 4.362 = 9
OPEN/SHORT CIRCUIT TEST
SOLUTION:

S 100k
I1 = = = 15.15 𝐴
V1 6600
cos−10.8 = 36.870

I1 = 15.15∠ − 36.870

6600
V1= I1 Ze1 + aV2 = 15.15∠ − cos−10.8 4.36 + j9 + ∗ 330
330
VP = 6735 V

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