Transformers- Testing

 
 

No-load test:
• As the name suggests, the secondary is kept open circuited and nominal value of the input voltage is applied to the primary winding and the input current and power are measured.
• This test is performed to determine core or iron loss, P_i and no-load current I₀.
• This test is helpful in determination of magnetizing component I_m’ energy component I_e and so no-load resistance R₀ being given as V₁/I_e and no-load reactance given as V₁/I_m.
  
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• In this test secondary (usually high voltage) winding is left open, all metering instruments (ammeter, voltmeter and wattmeter) are connected on primary side and normal rated voltage is applied to the primary (low voltage) winding, as illustrated below








• Since no load current I0 is very small, therefore, pressure coils of watt meter and the volt meter should be connected such that the current taken by them should not flow through the current taken by them should not flow through the current coil of the watt meter




Short Circuit Test:

• This test is performed to determine the full-load copper loss and equivalent resistance and reactance referred to secondary side.




• In this test, the terminals of the secondary (usually the low voltage) winding are short – circuited, all meters (ammeter, voltmeter and wattmeter) are connected on primary side and a low voltage, usually 5 to 10 % of normal rated primary voltage at normal frequency is applied to the primary, as shown in fig below. 




• The applied voltage to the primary, say V_s’ is gradually increased till the ammeter A indicates the full load current of the side in which it is connected. The reading W_s of the wattmeter gives total copper loss (iron losses being negligible due to very low applied voltage resulting in very small flux linking with the core) at full load. Le the ammeter reading be I_s.







Sumpner’s Test or Back to Back Test

• This test is mainly performed to determine temperature rise.




• In this test two transformers are loaded fully, in similar way as two dc machines in a regenerative test, and the power required from the supply is just to meet the iron and full load copper losses of the two transformers.




• It needs two identical transformers









Practice Objectives:

1. In a single-phase transformer, the magnitude of leakage reactance is twice that of resistance of both primary and secondary. With secondary short-circuited, the input power factor is
a. 1/2
b. 1/5
c. 2/5
d. 1/3

2. When short-circuit test on a transformer is performed at 25 V, 50 Hz, the drawn current ‘I1’ is at a lagging p.f. angle of Φ1. If the test is performed at 25 V, 25 Hz and the drawn current “I2’ is at a lagging p.f. angle of Φ2, then
a. I2 > I1 and Φ2 < Φ1 b. I2 > I1 and Φ2< Φ1 c. I2 < I1 and Φ2 > Φ1
d. I2 < I1 and Φ2 > Φ1

3.A 20 kVA, 2000/200 V, 1-phase transformer has nameplate leakage impedance of 8% Voltage required to be applied on the high - voltage side to circulate full - load current with the low voltage winding short-circuited will be
a. 16 V
b. 56.56 V
c. 160 V
d. 568.68 V

4. In a 100 kVA, 1100/220 V, 50 Hz single -phase trans former with 2000 turns on the high -voltage side, the open- circuit test result gives 220 V, 91 A, 5kW on low voltage side. The core-loss component of current is, approximately
a. 9.1 A
b. 22.7 A
c. 45.0 A
d. 91 A

5. A 1 kVa, 200/100 V, 50 Hz, single-phase transformer gave the following test results on 50 Hz :
OC (LV side) : 100 V, 20 watts
SC (HV side) : 5 A, 25 watts
It is assumed that no-load loss components are equally divided. The above tests were then conducted on the same transformer at 40 Hz.
Tests results were:
OC (HV):160 V, W1 watts.
SC (LV): 10 A, W2 watts
neglecting skin effect, W1 and W2 will be

a. W1 = 16 watts, W2 = 25 watts
b. W1 = 25 watts, W2 = 31.25 watts
c. W1 = 20 watts, W2 = 20 watts
d. W1 = 14.4 watts, W2 = 25 watts

6. Match List I (Test) with List II (Quantities) and select the correct answer using the code given below the lists:
List I
A. O C Test
B. S C Test
C. Sumpner’s Test
D. Load Test
List II
1. Copper loss and iron loss
2. Total losses
3. Iron loss
4. Copper loss
Codes;
---A B C D
a. 3 4 1 2
b. 2 1 4 3
c. 3 1 4 2
d. 2 4 1 3


Solved Problems

1. A single-phase 10-kVA 480/120-V transformer is to be used as an auto-transformer tying a 600-V distribution line to a 480-V load. When it is tested as a conventional transformer, the following values are measured on the primary (480-V) side of the transformer:

Open-circuit test Short-circuit test
V_OC = 480 V V_SC = 10.0 V
I_OC = 0.41 A I_SC = 10.6 A
V_OC = 38 W P_SC = 26 W
Find the per-unit equivalent circuit of this transformer when it is connected in the conventional manner. What is the efficiency of the transformer at rated conditions and unity power factor? What is the voltage regulation at those conditions?

Solution:
The base impedance of this transformer referred to the primary side is


The open circuit test yields the values for the excitation branch (referred to the primary side):



The excitation branch elements can be expressed in per-unit as




The short circuit test yields the values for the series impedances (referred to the primary side):




The resulting per-unit impedances are



The per-unit equivalent circuit is



At rated conditions and unity power factor, the input power to this transformer would be P_IN = 1.0 pu.
The core losses (in resistor R_C) would be




The copper losses (in resistor R_EQ) would be



The output power of the transformer would be



and the transformer efficiency would be



The output voltage of this transformer is




The voltage regulation of the transformer is