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 ο 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 × πππ‘ππ ππππππ πππ 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: π 01 I1 = I0 Iw V1 Iµ π 0 π01 Phasor Diagram: πΌ2′ = 0 V1 I0 π0 Open circuit Iw ο¦0 I0 ο¦ Iµ Input power π0 = π1 πΌ0 πππ ∅0 ο The no-load power factor, πππ ∅0 = π0 π1 πΌ0 The core loss component of no-load current πΌπ€ = πΌ0 πππ ∅0 The magnetizing component of no-load current πΌπ = πΌ0 π ππ∅0 ο Core loss resistance, π 0 = π1 πΌπ€ ο Magnetizing reactance, π0 = ο π1 πΌπ ο 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 ο 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 × πππ‘ππ ππππ πππ 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 π 01 π01 Short circuit 2 Input power πππΆ = πΌππΆ π 01 ο π 01 = π01 πππΆ 2 πΌππΆ ο πππΆ = ο πΌππΆ But, π01 = ο π01 = 2 2 π 01 + π01 2 2 π01 − π 01 ο 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₯πππ΄πππ ∅ % ππππππππππ¦ ππ‘ π₯ % ππππ, ο¨π₯ = × 100 2 π₯πππ΄πππ ∅ + π0 + π₯ πππΆ Where, kVA = Rating of transformer πππ ∅ = Power factor of load π0 = Wattmeter reading in OC test = Iron or Core loss in transformer in kW πππΆ = Wattmeter reading in SC test = Copper loss in transformer at full load in kW DETERMINATION OF REGULATION OF TRANSFORMER: Regulation can be calculated as – πΈ2 − π2 πΌ2 π 02 πππ ∅ ± πΌ2 π02 π ππ∅ πππππππ‘πππ π£πππ‘πππ ππππ’πππ‘πππ = × 100 = × 100 πΈ2 πΈ2 π1 − πΈ1 πΌ1 π 01 πππ ∅ ± πΌ1 π01 π ππ∅ πππππππ‘πππ π£πππ‘πππ ππππ’πππ‘πππ = × 100 = × 100 π1 π1 + 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.
