Transformer Testing

The circuit constants, efficiency and Voltage regulation of a transformer can be determined by open circuit and short circuit tests.

These tests are very convenient as they provide the required information without actually loading the transformer.

The power required to carryout these tests is very small as compared with full load output of the transformer.

Open Circuit or No load test

To determine the iron losses (or core losses (or) fixed losses) and parameters R₀ and X₀ of transformer.

The rated voltage is applied to the primary (usually LV winding) while the secondary is left open circuited.

The primary voltage (V₁), No load current (I₀), No load input power (W₀) are measured.

Normal rated Voltage is applied to the primary, normal iron losses will occur in the transformer core. Hence wattmeter will record the iron losses and small copper loss in the primary.

The No load current I₀ is very small (usually 2-10% of rated current), copper losses in the primary under no load condition are negligible as compared with iron losses.

Hence, wattmeter reading practically gives the iron losses in the transformer. Iron losses are same at all loads.

Iron losses, P₁ = Wattmeter reading (W₀)

No load current = Ammeter reading (I₀)

Applied Voltage = Voltmeter reading (V₁)

Input Power (W₀) = V₁ I₀ cos θ₀.

In OC test enables to determine iron losses and parameters R₀ and X₀ of the transformer.

Short Circuit or Impedance test

This test is used to determine R₀₁ (or R₀₂), X₀₁ (or X₀₂) and full load copper losses of the transformer.

In this test, (usually LV Winding) is short circuited by a thick conductor and variable LV is applied to the primary.

The LV input is gradually raised till at voltage V_SC, full load current I₁, flows in the primary. Then I₂ in the secondary also has FL Valve since I₁ / I₂ = N₂ / N₁

Under these conditions, the copper loss in the windings is the same as that on FL.

There is no output from the transformer under short circuit conditions. Therefore, input power is almost copper loss.

Iron loss is negligibly small since the voltage V_SC is small. Hence the wattmeter will practically register the full load copper losses in the transformer winding.

Full load Copper loss P_C = Wattmeter Reading (W_S)

Applied Voltage = Voltmeter reading (V_SC)

F.L. Primary current = Ammeter reading (I₁)

Where R₀₁ - Total resistance of transformer referred to Primary.

Total impedance referred to Primary, Z₀₁ = V_SC / I₁

Advantages of OC & SC Tests:

(i) Power required to carryout these tests is very small as compared to the Full load output of the transformer. In OC test, the power required is equal to iron loss and In SC test, the power required is equal to Full load copper loss.

(ii) To determine efficiency at any load and PF without actually loading the transformer.

(iii) To determine parameters R₀₁ and X₀₁(or R₀₂ & X₀₂) to calculate voltage drop and hence voltage regulation of transformer.

Example 1.13:

A single phase, 50 Hz transformer has a FL secondary current of 500A, the primary current being one fifth of this valve. The transformer Parameters are, R₁ = 0.6 Ω; R­₂ = 0.03 Ω; X₁ = 2 Ω and X₂ = 0.06 Ω. If the secondary is short circuited, find the Primary voltage required to circulate FL current. Neglect the No load current. What is the PF on short circuit?

Solution :

Neglecting No load current, the equivalent circuit of transformer on short circuit as referred to primary is as shown in Fig. 1.36

Primary Voltage required to circulate FL current.

Example 1.14:

A 200 kVA, 2000/440V, 50 Hz, single phase transformer gave the following test results:

O.C. test: 2000V, 1.8A, 1.75 kW on HV side

S.C. test: 13V, 300V, 1 kW on LV side

Obtain the equivalent circuit as referred to HV side.

Solution :

Component of No load current corresponding to core loss

Example 1.15:

The LV winding of a 300 kW, 11000/2200V, 50 Hz transformer has 190 turns and a resistance of 0.06 Ω. The HV side has 910 turns and resistance of 1.6 Ω. When LV winding is short circuited, the FL current is obtained with 550V applied to HV side. Calculate the equivalent resistance and leakage reactance referred to HV side.

Assume F.L. efficiency 98.5%

Solution :

Polarity test:

Polarity test is used to confirm the correct connection of the line and neutral connections. Polarity test is must for transformers when parallel operation is done.

In determining the relative polarity of the two windings of a transformer, the two windings are connected in series across a voltmeter. One of the windings is excited from a voltage source. If the polarities of the windings are as marked on the diagram, the voltmeter reading is V=V₁~V₂. If the voltmeter reads (V₁+V₂), the polarity markings of one of the windings must be interchanged.

Sumpner (or) Back to Back Test:

This test is conducted simultaneously on two identical transformers and provides data for finding the efficiency, regulation and temperature rise.

Power required to conduct this test is equal to the losses of the two transformers. The transformers are tested under FL condition with less power.

The primaries of the two transformers are connected in parallel across the rated supply (V₁), while two secondaries are connected in phase opposition. The voltage across T₂T₁ must be zero due to phase opposition, otherwise it will be double the rated secondary voltage. No circulating current in the loop formed by the secondary because their induced emf's are equal and opposite. An auxillary LV transformer (T) can be adjusted to give a variable voltage and hence current in the secondary loop circuit.

Operation:

(i) The secondaries of the transformers are in phase opposition. With switch S₁ closed and S₂ open, there will be no circulating current (I₂=0) in the secondary loop circuit. Because, the induced emf's in the secondaries are equal and in opposition. The current drawn from the supply is 2I₀ where I₀ is the No load current of each transformer. The reading of W₁ equal to the corelosses of the two transformers.

W₁ = Core losses of the two transformers.

(ii) Now switch (S₂) also closed and output voltage of the regulating transformer is adjusted till Full load current I₂ flows in the secondary loop circuit. The primary current I₁ circulates in the primary winding only and will not pass through W₁ (wattmeter 1). The full load currents are flowing through the primary and secondary winding. Now, the reading of wattmeter will be equal to the full load copper losses of the two transformers.

W₂ - Full load copper losses of two transformers.

W₁ + W₂ = Total losses of two transformers at FL.

Important Points:

(i) The Wattmeter (W₁) gives the core losses of the two transformers while wattmeter (W₂) gives the FL copper losses (or at any other load current I₂) of the two transformers. Therefore, the power required to conduct this test is equal to the losses of the two transformers.

(ii) Although transformers are not supplying any load, yet FL iron-loss and FL Cu losses are occuring in them.

(iii) There are two Voltage Sources (Supply Voltage and regulating transformers) and there is no interference between them. The supply voltage gives only 2I₀, while regulating transformer supplies I₂ and hence I₁ (-kI₂).

Advantages:

1. The power required to carry out the test is small.

2. Transformers are tested under FL condition.

3. The iron losses and FL copper losses are measured simultaneously.

4. I₂ can be adjusted to any current value. The cu loss can be found at FL (or) any load.

5. The temperature rise can be noted.

Example 1.16:

In a Back to Back, the wattmeter W₁ reads 4 kW and W₂ (at full current) reads 6 kW. Find the FL efficiency of each transformer. The transformers are rated at 200 kVA. Assume PF to be unity.

Example 1.17:

Two-similar 250 kVA, 1φ transformers gave the following results when tested by back to back method.

Mains wattmeter = 5 kW(W₁); Py series circuit wattmeter W₂ = 7.5 kW (at FL current)

(ii) Find the efficiency of each transformer at 75% FL and 0.8 PF lead.

Solution :

Wattmeter W₁ gives the iron losses of the two transformers and W₂ gives the FL copper losses of two transformer

Total losses of each transformer at 75% full load:

=P_i + (0.75)² P_c

= 2.5+ (0.75)² × 3.75 = 4.61 kW

Input to each transformer at 75% Full load and 0.8 P.F. lead

= 150 +4.61 = 154.61 kW

Efficiency n of each transformer is given by:

Separation of No load losses:

The core losses (or iron losses) consist of hysteresis loss and eddy current loss

Where,

B_m= Maximum fluxdensity; f-frequency; K_e, K_h - constants.

For a given machine maximum flux density (B_m);

P_h & f and P_e & ƒ²

(or) P_h= af and P_e = bf² where 'a' and 'b' are constants.

Total corelosses, P_i = af + bf².

Hence if the total core loss for given B_m is known at two frequencies, the constants a and b. Can be calculated. Knowing the values of a and b, the hysteresis loss component and eddy current loss component of the core loss can be determined.

P/f and ƒ curve = P_i­ = af + bf²

P/f = a + bf

The total core losses are meausred at various frequencies while the other factors upon which core losses depend are maintained constant. If the graph is plotted between P_i/f and f, it will be a straight line with slope tan φ = b (Fig.1.40). The constants a and b can be evaluated. Hence the hysteresis and eddy current losses at a given can be found out.

Example 1.18:

The Iron loss in a certain transformer is 80 kW at 25 Hz and 204 W at 60 Hz. The Maximum flux density being the same. Calcualte the total iron losses at 100 Hz at the same maximum flux density.

Solution :

Total iron losses = (P_i) = af + bf²

Total iron loss at 100 Hz can be obtained by putting the values of a, b and f

Example 1.19:

The Iron loss in a transformer core at normal flux density was measured at 30 Hz and 50 Hz. The results being 30W and 54W. Calculate the hysteresis and eddy current loss at 50 Hz.

Solution :

Total Iron loss (P_i) = af + bf²

For the first case; 30 = 30a + b x (30)²

For the second case; 54 = 50a + b x (50)²