,

Prepared by : (1) Patel Sandip A. ( ) (2) Patel Tarun R. ( ) (3) Patel Yash P. ( ) (4) Prajapati Jigar V. ( ) (5) Prajapati Urvashi H. ( ) TOPIC : O. C., S. C. AND SUMPNER TEST

1.O PEN C IRCUIT T EST (O.C T EST )

 Figure shows the O.C test on transformer.  The primary voltage of the transformer is connected to a rated ac voltage by using a variac.  A voltmeter is connected across the primary winding to measures a primary winding.  The ammeter is used to measure a primary current.

 The secondary is open circuited because it is an open circuit test.  Note that the ac supply voltage is applied at the low voltage side which is primary side.  At the secondary side higher voltage is used.  The wattmeter is used to measure the input power.

 Procedure : 1. Connect the circuit as shown in figure. 2. Keep the variac at its minimum voltage position. 3. Switch on the power supply and adjust the variac to the rated voltage. 4. Now measured a primary current, voltage and power using voltmeter, ammeter and wattmeter. 5. The ammeter gets primary current I o and wattmeter gets power Wo.

6. The observation table for the O.C test is as follows. 7. The no load current Io is very small as compared to the full load primary current. 8. As I2 is zero, the secondary copper loss is zero. Voltmeter reading Ammeter reading Wattmeter reading V1 (rated voltage) Io (ampere) Wo (watt)

9. The primary copper loss will be negligible because of Io is small. 10. There for the total copper loss is very small and can be assumed to equal to zero. 11. Hence the wattmeter reading Wo represents the iron losses Wo = Pi = Iron losses

S HORT C IRCUIT T EST (S.C T EST )

 Variac is used to adjust the input voltage to the rated voltage.  The voltmeter is connected to measure the primary voltage.  The ammeter measures the short circuit rated primary current.  The wattmeter measures the short circuit input power.  The secondary is short circuited.

 Procedure : 1. Connect the circuit as shown in figure. 2. Short circuit the secondary which is low voltage high current, low resistance winding. 3. Keep the variac at its minimum voltage and switch on the ac supply.

4. By increase the primary voltage gradually note down the wattmeter, voltmeter and ammeter reading. 5. The observation table is shown Voltmeter Reading Ammeter Reading Wattmeter Reading Vsc Volts Isc Ampere Wsc Power

 Parameter calculation :  The primary and secondary currents are the rated currents.  Therefore the total loss is the full load copper loss.  The iron losses are a function of applied voltage. As the applied voltage in S.C the iron loss will be negligibly small.  Hence the wattmeter reading Wsc = Full load copper loss = Pcu

 We can calculate the parameters R1, X1 and Z1of the equivalent circuit from the short circuit.  We know that Wsc = Vsc Isc Osc  Hence the short circuit power factor is given by cos Osc = Wsc / Vsc Isc  But wattmeter reading Wsc indicated the full load copper loss Wsc = copper loss = I’sc X R1 R1 = Wsc / I’sc

 Efficiency calculation from O.C and S.C test :  We can obtained the value of efficiency at any power factor cos O, or at any load which is a fraction of full load. efficiency(n) = x V2 I2 cos O / (x V2 I2 cos O) + Wo + x’ Wsc

 Calculation of regulation :  The percentage regulation (%R) is given by, %R = I1 R1 cos O + I1 X1 sin O / V1 X100 %R = I2 R2 cos O + I2 X2 sin O / V2 X100  We can the obtain the parameters such as R1, X1, R2, X2 from the S.C test of the transformer.  Where as the voltage V1, V2, and current I1, I2 in the above equation known from the given transformer.

 Polarity of transformer :  The transformer works on the principle of mutual inductance.  To indicate the winding direction easily the dot convention is used.  If two coils are mutually coupled then the terminals bearing the dots will be having in phase induced voltage.  The dots are marked on the primary and secondary winding to indicate the similar polarity of the two winding.

S UMPNER OR BACK TO BACK TEST : The open circuit test and short circuit test carried out over a transformer are useful to obtain its equivalent circuit. But the o.c. and s.c. tests are not useful for the heat run test in which the temperature rise in the transformer is to be monitored by fully loading it continuously. This is because in o.c. test the transformer is subjected only to the core loss will in s.c. test it is subjected only to the copper loss but the transformer is not subjected simultaneously to both this losses.

The solution to this problem is to carry out the sumpner test. The sumpner test can be carried out simultaneously on to identical transformers. The set up for sumpner test is shown in fig. this test is also called as back to back test.

S ET UP : As shown in fig. the primary windings of the two transformers are connected in parallel with each other across the rated voltage supply V1. The secondary windings of the two transformers are connected in series opposition as shown. The indication of correct connection is that the voltage across the terminals T2 and T4 is zero. A low voltage supply (V2 ) is connected across the series connection of the two secondaries to inject a current Ifl at low voltage into the secondary voltage windings.

O PERATION : With V2 assumed to be zero : According to superposition theorem,V2 source is assumed to be short circuited, then the two transformers will appear in open circuit to source V1. This is due to the fact that the secondary windings are in phase opposition and so no current can flow through them. So the total current drown from source V1 is the sum of no load currents of the two transformers i.e. 2Io.

So the reading of wattmeter W1 is given by, W1 = 2Po = 2Pi = 2 × core loss of each transformers Table shows the reading of various meters on the primary side to V2 = 0. V1 I1 W1 Rated voltage 2Io 2Po

With V1 assumed to be zero : Now imagine that V1 is short circuited. So V1= 0. this is equivalent to two transformers in series connected across V2 and are short circuited on their primary side. So the impedance seen by the source V2 will be 2 ZΩ. the source voltage V2 is adjusted such that a full load current Ifl is circulated on the secondary side. So I2 = Ifl.

And the second wattmeter W2 reading will be 2 Pc i.e. twice the full load copper loss of each transformer. W2 = 2 Pc = 2 × full load copper loss of each transformer. table shows the reading of various meters on the secondary side for V1 = 0. V2 I2 W2 Adjusted to get I2 = Ifl Ifl 2 Pc

C ONCLUSION : In the sumpner test, even though the transformers are not supplying any load, full iron loss takes place in their cores (2 Po) and full copper loss takes place in that windings (2 Pc). So the net power input is given by, Pт = W1 + W2 = 2 (Po + Pc) Hence the heat run test can be conducted on the two transformers when no actual load is supplied but only the losses are being supplied. It is possible to conduct this test on the three phase transformers as well.

THANK YOU