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OC and SC test for PDF

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Testing of Transformer by
Open-circuit and Short-circuit
Test
By : Dr. Atul R. Phadke
Associate Professor in Electrical Engineering
Government College of Engineering, Karad (Maharashtra)
OPEN CIRCUIT TEST:
Circuit Diagram: Auto
transformer
W0
M
L
I0
A
a1
1 Phase
AC Supply
V
C
A1
V
Open
V1
a2
LV
V
V2
A2
HV
In this test, one of the windings of transformer is kept open.
Other winding is connected to the supply of rated voltage and frequency through an
auto-transformer.
The rated voltage is applied and the readings of wattmeter W0, ammeter I0 and
voltmeter V1 are recorded.
As the secondary is open circuited, a very small current called the no-load current
flows in the primary.
This no-load current I0 is 2 to 6% of the rated (full load) current.
Voltmeter can be connected on secondary side to measure secondary voltage V2.
Generally, the high voltage side is kept open and the rated voltage is applied to the
low voltage side.
OPEN CIRCUIT TEST:
Circuit Diagram:
Auto
transformer
W0
M
L
I0
A
a1
1 Phase
AC Supply
V
C
A1
V
Open
V1
a2
LV
V
V2
A2
HV
In this test, secondary is kept open, therefore, output power is equal to zero.
Input power W0 = Output power + losses
 W0 = Losses (Core loss + Copper loss)
Secondary current I2 = 0 and primary current (no-load current I0 ) is negligible.
For example, if the no-load current is 5% of rated current, copper loss = 0.052 ×
π‘Ÿπ‘Žπ‘‘π‘’π‘‘ π‘π‘œπ‘π‘π‘’π‘Ÿ π‘™π‘œπ‘ π‘ , i.e., 0.25% of rated copper loss.
Hence, the copper loss in open-circuit test is negligible.
Therefore, the wattmeter reading W0 = Core or Iron loss.
DETERMINATION OF PARAMETERS OF NO LOAD CURRENT BRANCH:
Equivalent Circuit under no-load test:
𝑅01
I1 = I0
Iw
V1
Iµ
𝑅0
𝑋01
Phasor Diagram:
𝐼2′ = 0
V1
I0
𝑋0
Open
circuit
Iw
0
I0

Iµ
Input power π‘Š0 = 𝑉1 𝐼0 π‘π‘œπ‘ ∅0
 The no-load power factor, π‘π‘œπ‘ ∅0 =
π‘Š0
𝑉1 𝐼0
The core loss component of no-load current 𝐼𝑀 = 𝐼0 π‘π‘œπ‘ ∅0
The magnetizing component of no-load current πΌπœ‡ = 𝐼0 𝑠𝑖𝑛∅0
 Core loss resistance, 𝑅0 =
𝑉1
𝐼𝑀
 Magnetizing reactance, 𝑋0 =

𝑉1
πΌπœ‡

Note: If LV winding is connected to supply, R0 and X0 are referred to the lowvoltage side.
SHORT CIRCUIT TEST:
Circuit Diagram:
Auto
transformer
ISC
A
WSC
L
M
A1
1 Phase
AC Supply
C
V
a1
Short circuit
V VSC
A2
HV
a2
LV
In this test, one of the windings of transformer is short circuited.
Other winding is connected to the supply through an auto-transformer.
The applied voltage is adjusted such that the full load current flows through both
the windings.
The readings of wattmeter WSC, ammeter ISC and voltmeter VSC are recorded.
As the secondary is short circuited, a very small voltage is sufficient to circulate
rated currents.
VSC is 2 to 6% of the rated voltage.
Ammter can be inserted on secondary side to measure secondary current I2.
Generally, the low voltage side is short circuited and voltage is applied to the high
voltage side.
SHORT CIRCUIT TEST:
Circuit Diagram:
Auto
transformer
ISC
A
WSC
L
M
A1
1 Phase
AC Supply
C
V
a1
Short circuit
V VSC
A2
HV
a2
LV
In this test, secondary is short circuited, therefore, output power is equal to zero.
Input power WSC = Output power + losses
 WSC = Losses (Core loss + Copper loss)
As Vsc is about 1 to 6% of rated voltage, core flux is also 1 to 6% of rated value.
Core loss is approximately proportional to the square of core flux. For example, if
the core flux is 5% of rated flux, core loss = 0.052 × π‘Ÿπ‘Žπ‘‘π‘’π‘‘ π‘π‘œπ‘Ÿπ‘’ π‘™π‘œπ‘ π‘ , i.e., 0.25% of
rated core loss.
Hence, the core or iron loss in short-circuit test is negligible.
Therefore, the wattmeter reading WSC = Copper loss at full load if ISC = full load
current.
DETERMINATION OF PARAMETERS OF EQUIVALENT CIRCUIT:
Equivalent Circuit under short circuit test:
ISC
VSC
𝑅01
𝑋01
Short
circuit
2
Input power π‘Šπ‘†πΆ = 𝐼𝑆𝐢
𝑅01
 𝑅01 =
𝑍01
π‘Šπ‘†πΆ
2
𝐼𝑆𝐢

𝑉𝑆𝐢
=

𝐼𝑆𝐢
But, 𝑍01 =
 𝑋01 =
2
2
𝑅01
+ 𝑋01
2
2
𝑍01
− 𝑅01

Note: If HV winding is connected to supply, R and X are referred to the high-voltage
side.
DETERMINATION OF EFFICIENCY OF TRANSFORMER:
Efficiency can be calculated as –
π‘₯π‘˜π‘‰π΄π‘π‘œπ‘ ∅
% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 π‘Žπ‘‘ π‘₯ % π‘™π‘œπ‘Žπ‘‘, π‘₯ =
× 100
2
π‘₯π‘˜π‘‰π΄π‘π‘œπ‘ ∅ + π‘Š0 + π‘₯ π‘Šπ‘†πΆ
Where,
kVA = Rating of transformer
π‘π‘œπ‘ ∅ = Power factor of load
π‘Š0 = Wattmeter reading in OC test = Iron or Core loss in transformer in kW
π‘Šπ‘†πΆ = Wattmeter reading in SC test = Copper loss in transformer at full load in kW
DETERMINATION OF REGULATION OF TRANSFORMER:
Regulation can be calculated as –
𝐸2 − 𝑉2
𝐼2 𝑅02 π‘π‘œπ‘ ∅ ± 𝐼2 𝑋02 𝑠𝑖𝑛∅
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘Ÿπ‘’π‘”π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› =
× 100 =
× 100
𝐸2
𝐸2
𝑉1 − 𝐸1
𝐼1 𝑅01 π‘π‘œπ‘ ∅ ± 𝐼1 𝑋01 𝑠𝑖𝑛∅
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘Ÿπ‘’π‘”π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› =
× 100 =
× 100
𝑉1
𝑉1
+ in case of inductive and resistive loads and − in case of capacitive loads
ADVANTAGES AND DISADVANTAGES:
Disadvantages –
1) Performance of transformer at actual loading conditions, i.e., temperature
rise and noise at full load cannot be checked.
2) This method gives less accurate results.
Advantages –
1) Power consumption in these tests is quite small as compared to the direct
loading test.
2) The method is suitable for testing of large transformers also.
3) Complete prameters of equivalent circuit can be determined.
4) The regulation of transformer at any load can be found from the
parameters of equivalent circuit.
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