Open Circuit and Short Circuit Tests

Open Circuit and Short Circuit Tests

AU: Dec.-06,07,08,10,15,16, 19, May-08,10,11,12,13,16,17, Nov.-06, Oct.-02

• The efficiency and regulation of a transformer on any load condition and at any power factor condition can be predetermined by indirect loading method. In this method, the actual load is not used on transformer. But the equivalent circuit parameters of a transformer are determined by conducting two tests on a transformer which are,

1. Open Circuit Test (O.C. Test)

2. Short Circuit Test (S.C. Test)

Key Point: The parameters calculated from these results are effective in determining the regulation and efficiency of a transformer at any load and power factor condition, without actually loading the transformer.

• The advantage of this method is that without much power loss the tests can be performed and results can be obtained. Let us discuss in detail how to perform these tests and how to use the results to calculate equivalent circuit parameters.

1. Open Circuit Test (O.C. Test)

• The experimental circuit to conduct O.C. test is shown in the Fig. 6.18.1.

• The transformer primary is connected to a.c. supply through ammeter, wattmeter and variac. The secondary of transformer is kept open. Usually low voltage side is used as primary and high voltage side as secondary to conduct O.C. test.

• The primary is excited by rated voltage, which is adjusted precisely with the help of a variac. The wattmeter measures input power. The ammeter measures input current. The voltmeter gives the value of rated primary voltage applied at rated frequency.

• Sometimes a voltmeter may be connected across secondary to measure secondary voltage which is V₂ = E₂ when primary is supplied with rated voltage. As voltmeter resistance is very high, though voltmeter is connected, secondary is treated to be open circuit as voltmeter current is always negligibly small.

• When the primary voltage is adjusted to its rated value with the help of variac, readings of ammeter and wattmeter are to be recorded.

The observation table is as follows

V₀ = Rated voltage, W₀ = Input power, I₀ = Input current = No load current

• As transformer secondary is open, it is on no load. So current drawn by the primary is no load current I₀. The two components of this no load current are,

I_m = I₀ sin ϕ₀, I_c = I₀ cos ϕ₀

Where cos ϕ₀ = No load power factor

And hence power input can be written as,

W₀ = V₀ I₀ cos ϕ₀

The phasor diagram is shown in the Fig. 6.18.2.

• As secondary is open, I₂ = 0. Thus its reflected current on primary is also zero. So we have primary current I₁ = I₀. The transformer no load current is always very small, hardly 2 to 4 % of its full load value. As I₂ = 0, secondary copper losses are zero. And I₁ = I₀ is very low hence copper losses on primary are also very very low. Thus the total copper losses in O.C. test are negligibly small. As against this the input voltage is rated at rated frequency hence flux density in the core is at its maximum value. Hence iron losses are at rated voltage. As output power is zero and copper losses are very low, the total input power is used to supply iron losses. This power is measured by the wattmeter i.e. W₀. Hence the wattmeter in O.C. test gives iron losses which remain constant for all the loads.

W₀ = Pi = Iron losses

Calculations: We know that,

Key Point: The no load power factor cos 'ϕ₀ is very low hence wattmeter used must be low power factor type otherwise there might be error in the results. If the meters are connected on secondary and primary is kept open then from O.C. test we get R'₀ and X'₀ and with which we can obtain Ro and X₀ and knowing the transformation ratio K.

2. Short Circuit Test (S.C. Test)

• In this test, primary is connected to a.c. supply through variac, ammeter and voltmeter as shown in the Fig. 6.18.2.

• The secondary is short circuited with the help of thick copper wire or solid link. As high voltage side is always low current side, it is convenient to connect high voltage side to supply and shorting the low voltage side.

• As secondary is shorted, its resistance is very very small and on rated voltage it may draw very large current. Such large current can cause overheating and burning of the transformer. To limit this short circuit current, primary is supplied with low voltage which is just enough to cause rated current to flow through primary which can be observed on an ammeter. The low voltage can be adjusted with the help of variac. Hence this test is also called low voltage test or reduced voltage test. The wattmeter reading as well as voltmeter, ammeter readings are recorded. The observation table is as follows,

• Now the currents flowing through the windings are rated currents hence the total copper loss is full load copper loss. Now the voltage applied is low which is a small fraction of the rated voltage. The iron losses are function of applied voltage. So the iron losses in reduced voltage test are very small. Hence the wattmeter reading is the power loss which is equal to full load copper losses as iron losses are very low.

• Thus we get the equivalent circuit parameters R_1e X_1e and Z_1e. Knowing the transformation ratio K, the equivalent circuit parameters referred to secondary also can be obtained.

• Important note: If the transformer is step up transformer, its primary is L.V. while secondary is H.V. winding. In S.C. test, supply is given to H.V. winding and L.V. is shorted. In such case we connect meters on H.V. side which is transformer secondary though for S.C. test purpose H.V. side acts as primary. In such case the parameters calculated from S.C. test readings are referred to secondary which are R_ze' Z_2e and X_2e. So before doing calculations it is necessary to find out whether the readings are recorded on transformer primary or secondary and accordingly the parameters are to be determined. In step down transformer, primary is high voltage itself to which supply is given in S.C. test. So in such case test results give us parameters referred to primary i.e. R_1e' Z_1e and X_1e.

Key Point: In short, if meters are connected to primary of transformer in S.C. test, calculations give us R_1e and Z_1e. If meters are connected to secondary of transformer in S.C. test calculations give us R_2e and Z_2e.

3. Calculation of Efficiency from O.C. and S.C. Tests

We know that,

Thus for any p.f. cos ϕ₂ the efficiency can be predetermined. Similarly at any load which is fraction of full load then also efficiency can be predetermined as,

4. Calculation of Regulation

• From S.C. test we get the equivalent circuit parameters referred to primary or secondary.

• The rated voltages V₁, V₂ and rated currents (I₁) F.L. and (I₂) F.L. are known for the given transformer. Hence the regulation can be determined as,

where I₁, I₂ are rated currents for full load regulation.

For any other load the currents I₁, I₂ must be changed by fraction n.

I₁, 1₂ at any other load = n (I₁) F.L., n (I₂) F.L.

• Thus regulation at any load and any power factor can be predetermined, without actually loading the transformer.

Ex. 6.18.1 Draw the equivalent circuit for a single phase 1100/220 V transformer on which the following results were obtained:

i) 1100 V, 0.5 A, 55 W on primary, secondary being open circuited

ii) 10 V, 80 A, 400 W on L.V. side, H.V. being short circuited.

Calculate the voltage regulation and efficiency for the above transformer when supplying 100 A at 0.8 p.f. lagging. AU: Dec.-06, 07, 19, May-08, 17 Marks 8

Sol. :

Ex. 6.18.2 The following data were obtained on 20 kVA, 50 Hz, 2000/200 V distribution transformer:

Draw the approximate equivalent circuit of the transformer referred to the HV and LV sides respectively. AU: Dec.-16, May-11, Marks 16

Sol.:

Ex. 6.18.3 Obtain the equivalent circuit of a 200/400 V, 50 Hz, 1-phase transformer from the following test data :

O.C. test: 200 V, 0.7 A, 70 W on L.V. side

S.C. test: 15 V, 10 A, 85 W - on H.V. side

Calculate the secondary voltage when delivering 5 kW at 0.8 p.f. lagging, the primary voltage being 200 V. AU May-13, Marks 16

Sol. :

Ex. 6.18.4 The primary of the transformer is rated at 10 A and 1000 V. The open circuit reading are V₁ = 1000V, V₂ = 500 V, I = 0.42 A, P_oc =100Wcircuit readings are 11 Psc=400 W.The short circuit ewadings are I₁ = 10A, V₁ = 125V and P_sc =400 W Draw the equivalent circuit for the transformer. Predict the output voltage for the load impedance Z_L  = 19+ j12 ohms and draw the phasor diagram. AU Dec.-15, Marks 16

Sol. :

Ex. 6.18.5 The parameters of a 10 kVA, 500/250 V, 50 Hz, single phase transformer are as follows:  

Primary resistance = 0.2 Ω, primary reactance = 0.4 Ω Secondary resistance = 0.5Ω , secondary reactance = 0.1 Ω Exciting circuit resistance and reactance = 1500 Ω, 750 Ω, respectively. Find out the results of open circuit test and short circuit test.

Sol. :

Ex. 6.18.6 A 100 kVA, 6600 V/330 V, 50 Hz single phase transformer took 10 A and 436 W at 100 V in a short circuit test, the figures referring to high voltage side. Calculate the voltage to be applied to the high voltage side on full load at power factor 0.8 lagging when the secondary terminal voltage is 330 V. AU: Dec.-10, Marks 12

Sol. : Refer example 6.18.3 for the proceduce and verify the answer: V₁ = 6734.666 V.

Review Questions

1. What are the tests required to draw the equivalent circuit of a single phase transformer? How are they conducted? AU: Oct.-02, May-10, 16, Marks 16

2. Explain how the efficiency of a single phase transformer is estimated from the open circuit and short circuit test. AU Dec.-06, Marks 8

3. With neat circuit diagrams, explain the open-circuit and short-circuit tests conducted on single-phase transformer. How to obtain regulation and efficiency from these tests ? AU: Dec.-08, May-12, Marks 16

4. A 5 kVA, 500/250 V, 50 Hz, single phase transformer gave the following readings,

O.C. Test: 500 V, 1 A, 50 W (L.V. side open)

S.C. Test: 25 V, 10 A, 60 W (L.V. side shorted)

Determine:

i) The efficiency on full load, 0.8 lagging p.f.

ii) The voltage regulaion on full load, 0.8 leading p.f.

iii) The efficiency on 60 % of full load, 0.8 leading p.f.

iv) Draw the equivalent circuit referred to primary and insert all the values in it.

(Ans. % η = 97.32 %, % R = -1.95 %, % n = 97.103 %)

5. The open circuit and short circuit tests on a 10 kVA, 125 / 250 V, 50 Hz, single phase transformer gave the following results :

O.C. test: 125 V, 0.6 A, 50 W (on L.V. side)

S.C. test: 15 V, 30 A, 100 W (on H.V. side)

 Calculate:  i) Copper loss on full load, ii) Full load efficiency at 0.8 leading p.f.

iii) Half load efficiency at 0.8 leading p.f.,

iv) Regulation at full load, 0.9 leading p.f.

(Ans.: % η on full load = 97.23%, % non half load = 97.69 %, % R = 1.8015 %)

6. The open circuit and short circuit tests on a kVA, 250 / 125 V, 50 Hz, single phase transformer gave the following results :

O.C. Test: 250 V, 0.7 A 90 W (H.V. side) and

S.C. Test: 12 V, 30 A, 90 W (L.V. side)

Calculate: i) The full load efficiency ii) The voltage on L.V. side when supplying full load current both at 0.8 leading power factor.

(Ans.: % η= 94.11 %, Terminal V2 = 131.0952 V)

7. From the following test data, calculate % efficiency and % regulation at half-load and 0.8 Slagging p.f. for 1.5 kVA, 220 V/110 V, single phase, 50 Hz transformer.

O.C. test: 220 V, 0.7 A, 32 W and S.C. test: 9 V, 6.82 A, 44 W

(measuring instruments connected on H.V. side in both the tests)

(Ans. % R = 2.028 %, % η = 93.3125 %)

8. A 40 kVA, 400/200 V, 50 Hz, single phase o transformer gave the following results.

Open circuit test: 400 V, 5 A, 500 W and Short circuit test: 10 V, 50 A, 150 W

i) Draw the equivalent circuit of the transformer and show the values of resistance and reactance interms of primary side.

ii) Calculate the terminal voltage, when the transformer delivers the solid output at a p.f. of 0.8 lagging on the L.V. side.

(Ans. : i) R_1e = 0.06Ω, Z_1e = 0.2Ω, X_1e = 0.1907 Ω ii) 191.88 V)

9. A 1-phase; 50 kVA, 2400/120 V, 50 Hz transformer gave the following results.

O.C. test with instruments on the L.V. side 120 V, 9.65 A, 396 W

S.C. test with instruments on the H.V. side 92 V, 0 20.8 A, 810 W

Calculate:

i) The equivalent circuit constants and draw the equivalent circuit.

ii) The efficiency when rated kVA is delivered to a load having a power factor of 0.8 lagging.

iii) The voltage regulation.

(Ans. i) 1.8662 Ω 4.416 Ω, 4.00222, 97.07 %, 3.38%)

10. A 100 V / 200 V, 50 Hz, single phase transformer gave the following test results

(No load H.V. side) 1000 V, 0.24 A, 90 W

 (Short circuit L.V. side) 50 V, 5 A 110 W

The regulation of the transformer at full load at power factor 0.8 lag is 4.46 %.

Calculate i) Rating of the transformer

ii) Equivalent circuit of transformer with circuit constants

iii) kVA load for maximum efficiency

iv) Voltage to be applied on H.V. side on full at u.p.f. when the secondary terminal voltage is 200 V.

(Ans.: 4.4'Ω, 8.89 Ω, i) 5 kVA iii) 4.52 kVA iv) 1022 V)

11.A 20 kVA single phase transformer has voltage rating of 1100/110 V. During short circuit test it gives following readings,

60 V, 18 A, 560 W L.V. side shorted

Find the power factor at which the regulation is zero, on load. 

(Ans. : 0.8556 leading)

 12.A 100 kVA, 6600 / 240 V, 50 Hz, single phase transformer takes 5 A and 109 W when 50 V are applied in a short circuit test to the h.v. side and low voltage side shorted. Find the voltage to be applied to the H.V. side on full load at 0.8 p.f. lagging when the secondary terminal voltage is 240 V.AU: Oct.-02 

(Ans.: 6734.666 V)

13. Obtain the approximate equivalent circuit parameters of a 200/400 volts, 50 Hz single phase transformer from the following test results:

 O.C. test: 200 volts, 0.7 amps, 77 watts on L.V. side

S.C. test: 12 volts, 10 amps, 100 watts on H.V. side

Also calculate the secondary voltage when delivering 5 kW at 0.5 lagging power factor, the primary voltage being 200 volts. 

(Ans. : 371.0315 V)