SHORT CIRCUIT TEST ON A SINGLE PHASE TRANSFORMER
APPARATUS: To find out copper loss of a transformer by performing short circuit test & to observe the dependence of copper loss on the current.
APPARATUS:
Sr. No. NAME RANGE NO. REQUIRED REMARKS
1 Single phase transformer
2 Variac.
3 Watt meter.
4 Volt meter
5 Ammeter.
THEORY: A transformer is a static apparatus used to transform a.c, electrical power from one voltage. It works on the principle of mutual inductance.
In a transformer, there are two types of losses:
(1) Copper loss
(2) Iron & core loss.
(1) COPPER LOSS: The copper loss is due to the resistance the transformer windings & the current passing through them.
Cu loss (Wc) = (I1)2R1 + (I2)2R2
= I2R -------------------(i)
Where R is the equivalent resistant of the transformer:
From Equation (i)
Cu loss (Wc) ? I2
(2) IRON LOSS: Iron loss is power wasted in the iron cod. The iron loss can be divided into two parts.
(a) EDDY CURRENT LOSS:
W = K (BMAX)2 f2 t2 V2 Watts.
Bmax = Maximum flux density in Wb/m2
F = Frequency of magnetic reversal.
T = Thickness of each lamination.
V = Volume of armature core in m3.
(b) Hysteresis loss:
Wh = ?Bmax f V watts
N = Steimetz hysteresis coefficient
Bmax = maximum flux density in Wb/m
F = frequency of magnetic reversal
V = volume of armature core in m
The iron loss depends upon the primary voltage.
The power equation of a transformer can be given as:
Input = Output + Cu loss + Iron loss
= V2 I2 cos + Wc + Wi
During short circuit test, the secondary winding is short circuited,
Therefore
V2 = 0
Hence during short circuit test,
Input = Wc + Wi -------(ii)
In a short circuit test a very small voltage is applied to the primary winding, hence the corresponding iron loss is also very small & can be neglected as compared to the copper loss.
Input =Wc = copper loss -----(iii)
From equation (iii) it is seen that input to the transformer during short circuit test is equal to the copper loss.
Hence wattmeter, which measures the input power during short circuit test, indicates the directly value of copper loss.
Precautions should be taken to see that during this test, currents in the windings should not exceed their rated values. The rated values of the current can be calculated as follows.
I1 = kva � 1000
Rated voltage v1
I2 = kva � 1000
Rated voltage v 2
PROCEDURE:
1. Set of the connections as per circuit diagram.
2. check that output voltage of the variac should be zero before switching on the supply to ensure the safety of the transformer from the sudden heavy rush of the short circuit current.
3. Gradually increase the voltage till full load current circulate in the windings.
4. Note down current of the wattmeter (wc) ammeter & voltmeter.
5. copper loss (Wc) v/s short circuit current(Ip)
OBSERVATION TABLE :
SR. NO.
PRIMARY VOLTAGE IN VOLTS.
PRIMARY CURRENT IN AMP. IP
COPPER LOSS IN WATTS WC
1
2
3
4
5
6
7
8
9
10
CALCULATION:
CONCLUSION: