Open circuit and short circuit test on Single-Phase  transformer.

Experiment No.: – 08

Aim of the Experiment: –

To perform the open circuit and short circuit test of a single phase transformer and to
draw the equivalent circuit after determining its constants.

Apparatus Required: –

Sl. No.NameSpecificationQuantity
1Single Phase Transformer5 KVA, 115V/230V, 50Hz1 Nos.
2Single Phase Auto-transformer10 KVA, (0-300)V, 30A1 Nos.
3Wattmeter10A/20A LPF1 Nos.
4Wattmeter15A/30A UPF1 Nos.
5Voltmeter(0-250)V, MI1 Nos.
6Voltmeter(0-75)V, MI1 Nos.
7Ammeter(0-5)A, MI1 Nos.
8Ammeter(0-30)A, MI1 Nos.
9Connecting WiresPVC Insulated CopperAs Per Required
Table No.: – 8.1

Circuit Diagram: –

Fig. 8.1

Theory: –

The performance of a transformer can be calculated on the basis of its equivalent circuit
which contains four main parameters, the equivalent resistance R01 as referred to primary( or secondary R02), the equivalent leakage reactance X01 as referred to primary, the core-loss conductance G0 and the magnetizing susceptance B0. These constants or parameters can be easily determined by two test i.e. Open circuit test and short circuit test. These are very economical and convenient, because they furnish the required information without actually loading the transformer.

The purpose of O.C. test is to determine no load loss or core loss and no load current I0 which is helpful in finding X0 and R0. One winding of the transformer whichever is convenient but usually high voltage winding is left open and the other is connected to its supply of normal voltage and frequency. A wattmeter (W), Voltmeter (V) and ammeter (A) are connected in the low voltage winding i.e. primary winding in the present case. With normal voltage applied to the primary, normal flux will be setup in the core, hence normal iron loss will occur which are recorded by the wattmeter. As the primary no load current I0 is small, Cu loss is negligibly small in primary and nil in secondary. Hence, the wattmeter reading represents practically the core loss under no load condition.

For short circuit test, one winding usually the low voltage winding, is solidly short circuited by a thick conductor (or through an ammeter which may serve the additional purpose of
indicating rated load current). A low voltage (usually 5 to 10% of normal primary voltage) at correct frequency (though for Cu losses it is not essential) is applied to the primary and is continuously increased till full- load current is flowing both in primary and secondary (as indicated by the respective ammeters). Since, in this test, the applied voltage is a small percentage of the normal voltage, the mutual flux ɸ produced is also a small percentage of its normal value. Hence, core loss is very small with the result that the wattmeter reading represents the full load Cu loss or I2R loss for the whole transformer i.e. both primary Cu loss and secondary Cu loss. If Vsc is the voltage required to circulate rated load current, then Z01= Vsc/I1. A two winding transformer can be represented by means of an equivalent circuit as shown below

Fig. 8.2

Procedures: –

Open Circuit Test

  1. Connect circuit as shown in the circuit diagram. Open circuit the secondary and apply full load voltage to the primary through a Variac. The cupper loss is negligible since there is only no load current is flowing. Hence power consumed is the core losses of the core.
  2. Note voltmeter, ammeter and wattmeter readings.

Short Circuit Test

  1. Connect as shown in the circuit diagram. Short circuit the secondary and apply a low
    voltage to the primary through an auto transformer. The iron losses are negligible
    since the flux will be very low on account of the primary and secondary.
  2. Increase the voltage gradually till full load current flows in the primary.
  3. Note voltmeter and ammeter and wattmeter reading.

Observation Table for Open Circuit Test: –

Sl. No.V (in Volts.)I0 (in Amp.)W (in Watts)Iw (in Amp.)Im (in Amp.)Cos ɸ0
1
Table No.: – 8.2

Calculations:

\[\color\red{W=V\times{I_0}\times{cos\phi_0}}\] \[\color\red{I_W=\dfrac{W}{V}}\] \[\color\red{I_M=\sqrt{I_0^2-I_W^2}}\] \[\color\red{R_0=\dfrac{V}{I_W}}\] \[\color\red{X_0=\dfrac{V}{I_M}}\]

See the no load phasor diagram below

Observation Table for Open Circuit Test: –

Sl. No.V (in Volts.)I (in Amp.)W (in Watts)
1
Table No.: – 8.3

Calculation:

Let the total equivalent resistance of primary and secondary referred to primary side be R1
ohms and the total equivalent leakage reactance referred to primary side be X1 ohms.

ohms and the total equivalent leakage reactance referred to primary side be X1 ohms.

\[\color\red{W_c \;=\; I_2R_1}\] \[\color\green{hence\;}\color\red{R_{01}\;or\;R_{sc}\;=\dfrac{W_c}{I^2}}\] \[\color\green{Also\;}\color\red{\dfrac{V}{I}\;=\;Z_1}\] \[\color\green{and}\] \[\color\red{X_{01}\;or\;X_{sc}\;=\;\sqrt{Z_1^2-R_1^2}\;}\color\green{ohms.}\]

Precautions: –

  1. Don’t switch on power supply without concerning teachers.
  2. Single Phase Auto transformer must be kept at minimum potential point. Before switch on the experiment.

Conclusion: –

To be written by Student.

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