Transformer Test, 1. Open circuit or no-load test, 2. Short circuit Test or Impedance Test Or On Load Test

Transformer Test

Transformer Test are classified in to two types –

1. Open circuit Test or No load Test

2. Short circuit Test or Impedance Test Or On Load Test

1. Open circuit Test or No load Test

In this test,. The secondary winding (typically high voltage) is left open, All metering devices (ammeter, voltmeter, and wattmeter) are connected on the primary side, and the primary (low voltage) winding is subjected to the normal rated voltage, as shown below.

Iron loss = Input power on no-load 

W0 watts (wattmeter reading) No load current = 0 amperes (ammeter reading) Angle of lag, = /Io Ie = and Im = √o – Caution: The pressure coils of the watt metre and the volt metre should be connected in a way that prevents the current they draw from passing through the watt meter’s current coil because the no load current I0 is so small.

V0 = Rated Voltage

Wo = Input power

I0 = Input current = no load current

Im = I0sinϕ0

Ic = I0cos ϕ0

cos ϕ0 = No load power factor

Hence power input can written as,

Wo = V0 I0 cos ϕ0

W0 = Pi = Iron losses

W0 = V0I0 cos ϕ0

cos ϕ0 = W0 / V0I0 = no load power factor

Ro = V0/Ic Ω

X0 = V0 /Im

2. Short circuit Test or Impedance Test Or On Load Test

This test is conducted to determine the equivalent resistance, reactance, and full-load copper loss on the secondary side. This test involves connecting the primary side of all metres (ammeter, voltmeter, and wattmeter), short-circuiting the secondary (typically the low voltage) winding terminals, and applying a low voltage to the primary that is typically 5 to 10% of the normal rated primary voltage at the normal frequency, as shown in the figure below.

The applied voltage to the primary,. Say Vs’ is gradually increased till the ammeter A indicates the full load current of the side in which it is connected. The reading Ws of the wattmeter gives total copper loss (iron losses being negligible due to very low applied voltage resulting in very small flux linking with the core) at full load. Le the ammeter reading be Is.

Equivalent impedance referred to primary= Commercial Efficiency and All day Efficiency

(a) Commercial Efficiency

Commercial efficiency is defined as the ratio of power output to power input in kilowatts.

(b) All-day Efficiency

The all day efficiency is define as. The ratio of output in kwh to the input in kwh during the whole day. Transformers used for distribution are connect for the whole day to the line but loaded intermittently. Thus the core losses occur for the whole day but copper losses occur only when the transformer is delivering the load current.

Therefore, the efficiency will be lower than commercial efficiency. If the transformer is not use to supply the load current throughout the day. The efficiency (commercial efficiency) will be maximum when variable losses (copper losses) are equal to constant losses (iron or core losses).sign is for inductive load and sign is for capacitive load Transformer efficiency, Where x is the ratio of secondary current I2 and rated full load secondary current.

Wsc = (Pcu) F.L. = full Load copper loss

Wsc = VscIsc cosϕsc

cosϕsc = VscIsc / Wsc

Wsc = I2sc R1e = Copper loss

Z1e = Vsc /Isc = Root(R21e+X21e)

X1e = Root(Z21e – R21e)