ELECTRICAL MACHINES – IUNIT – IV

TRANSFORMERS

PART – A (2 MARKS)

1. Write the emf equation of a transformer. The emf equation of a transformer is given by E1 4.44 f N1 m for primary winding and E2 4.44 f N2 m for secondary winding where f = frequency of AC supply, Hz N1 = No. of turns in primary winding N2 = No. of turns in secondary winding m = Maximum or peak value of flux in the core, weber 2. Why is the rating of a transformer specified in KVA?An important factor in the design and operation of electrical machines is the relation between the life of the insulation and operating temperature of the machine. Therefore, temperature rise resulting from the losses is a determining factor in the rating of a machine. We know that copper loss in a transformer depends on current and iron loss depends on voltage. Therefore, the total loss in a transformer depends on the volt-ampere product only and not on the phase angle between voltage and current i.e., it is independent of load power factor. For this reason, the rating of a transformer is in kVA and not kW.

3. What is voltage regulation of a transformer?The voltage regulation of a transformer is the arithmetic difference (not phasor difference) between the no-load secondary voltage (0V2) and the secondary voltage V2 on load expressed as percentage of no-load voltage i.e.

%age voltage regulation =

where 0V2 = No-load secondary voltage = K V1V2 = Secondary voltage on load

4. Draw the approximate equivalent circuit of transformer?

Where R2’, X2’ are the resistance and reactance of the secondary winding referred to the primary. R2’ = R2/K2 and X2’ = X2/K2

Neglecting the no load branch, since the no load current is very small, we get the approximate equivalent circuit referred to the primary side as

5. Power factor of transformer is low under no load, why?

The power factor of a transformer is low under no load because the secondary side is open as there is no load. The no load current of the transformer is mainly used for setting up the flux in the core. This no load current (Io) lags the primary input voltage by a large angle due to the inductance of the primary winding as the resistance of the winding is very less. Hence the power factor of the transformer under no load is low.

The no load current, Io can be resolved into two components, Im (magnetizing component) and Iw ( loss component). The power factor on no load cosΦo = Iw/Im. Since Iw is very small compared to Im (mainly magnetizing current), the power factor is low on no load.

6. What is meant by leakage reactance in transformers?

Leakage reactances. Both primary and secondary currents produce flux. The flux which links both the windings is the useful flux and is called mutual flux. However, primary current would produce some flux which would not link the secondary winding. Similarly, secondary current would produce some flux that would not link the primary winding. The flux such as 1 or 2 which links only one winding is called leakage flux. The leakage flux paths are mainly through the air. The effect of primary leakage flux 1 is to introduce an inductive reactance X1 in series with the primary winding. Similarly, the secondary leakage flux 2 introduces an inductive reactance X2 in series with the secondary winding. There will be no power loss due to leakage reactance. However, the presence of leakage reactance in the windings changes the power factor as well as there is voltage loss due to IX drop.

ELECTRICAL MACHINES – I

UNIT – IVTRANSFORMERS

PART – B (12 MARKS)

1. (a) Derive the EMF equation of transformer. (6) (b) From the phasor diagram of the transformer on load, derive the expressions for voltage regulation at lagging and leading power factors. (6)

2. Draw and explain the phasor diagrams of a transformer on no load and load conditions.

Phasor diagram on no load

Consider a practical transformer on no load i.e., secondary on open-circuit as shown in Fig. (i). The primary will draw a small current I0 to supply (i) the iron losses and (ii) a very small amount of copper loss in the primary. Hence the primary no load current I0 is not 90° behind the applied voltage V1 but lags it by an angle 0 < 90° as shown in the phasor diagram in Fig. (ii).No load input power, Wo = V1 Io cos o

As seen from the phasor diagram in Fig. (ii), the no-load primary current I0 can be resolved into two rectangular components viz. (i) The component IW in phase with the applied voltage V1. This is known as active or working or iron loss component and supplies the iron lossand a very small primary copper loss. Iw = Io cos o(b) The component Im lagging behind V1 by 90° and is known asmagnetizing component. It is this component which produces themutual flux in the core.Im = Io sin o

Transformer on load (Practical Transformer)

Phasor diagram. Fig. shows the phasor diagram of a practical transformer for the usual case of inductive load. Both E1 and E2 lag the mutual flux by 90°. The current I'2 represents the primary current to neutralize the demagnetizing effect of secondary current I2. Now I'2 = K I2 and is opposite to I2. Also I0 is the no-load current of the transformer. The phasor sum of I'2 and I0 gives the total primary current I1. Note that counter e.m.f. that opposes the appliedvoltage V1 is E1. Therefore, if we add I1R1 (in phase with I1) and I1 X1 (90° ahead of I1) to E1, we get the applied primary voltage V1. The phasor E2 represents the induced e.m.f. in

the secondary by the mutual flux . The secondary terminal voltage V2 will be what is left over after subtracting I2R2 and I2X2 from E2. Load power factor = cos 2 Primary power factor = cos 1 Input power to transformer, P1 = V1I1 cos 1 Output power of transformer, P2 = V2I2 cos 2

Phasor diagram of transformer on load (lagging pf)

3. Obtain the parameters of the approximate equivalent circuit of a single phase, 4 KVA,200/400V,50Hz transformer for which the following are the test results:

O.C test: 200V, 0.75A, 70W on LV side S.C test 15V, 10A, 80W on H.V side

4. Explain the effects of resistance and leakage reactance of the winding of a transformer.

Winding resistances. Since the windings consist of copper conductors, it immediately follows that both primary and secondary will have winding resistance. The primary resistance R1 and secondary resistance R2 act in series with the respective windings as shown in Fig.1 below. When current flows through the windings, there will be power loss as well as a loss in voltage due to IR drop. This will affect the power factor and E1 will be less than V1 while V2 will be less than E2.

Fig. 1 – Winding Resistances

Leakage reactances. Both primary and secondary currents produce flux. The flux which links both the windings is the useful flux and is called mutual flux. However, primary current would produce some flux which would not link the secondary winding (See Fig.2). Similarly, secondary current would produce some flux that would not link the primary winding. The flux such as 1 or 2 which links only one winding is called leakage flux. The leakage flux paths are mainly through the air. The effect of these leakage fluxes would be the same as though inductive reactance were connected in series with each winding of transformer that had no leakage flux as shown in Fig 1. In other words, the effect of primary leakage flux 1 is to introduce an inductive reactance X1 in series with the primary winding as shown in Fig.1 . Similarly, the secondary leakage flux 2 introduces an inductive reactance X2 in series with the secondary winding. There will be no power loss due to leakage reactance. However, the presence of leakage reactance in the windings changes the power factor as well as there is voltage loss due to IX drop.

Fig. 2 . Leakage FluxAlthough leakage flux in a transformer is quite small (about 5% of ) compared to the mutual flux , yet it cannot be ignored. It is because leakage flux paths are through air of high reluctance and hence require considerable e.m.f. It may be noted that energy is conveyed

from the primary winding to the secondary winding by mutual flux which links both the windings.

5. Develop the equivalent circuit of single phase transformer from the first principle and hence deduce the approximate equivalent circuit.Exact Equivalent Circuit of a Loaded TransformerFig. 1 shows the exact equivalent circuit of a transformer on load. Here R1 is the primary winding resistance and R2 is the secondary winding resistance. Similarly, X1 is the leakage reactance of primary winding and X2 is the leakage reactance of the secondary winding. The parallel circuit R0 X0 is the no-load equivalent circuit of the transformer. The resistance R0 represents the core losses (hysteresis and eddy current losses) so that current IW which supplies the core losses is shown passing through R0. The inductive reactance X0 represents a loss-free coil which passes the magnetizing current Im. The phasor sum of IW and Im is the no-load current I0 of the transformer.

Fig. 1 Exact Equivalent Circuit of a Loaded Transformer

Simplified Equivalent Circuit of a Loaded TransformerThe no-load current I0 of a transformer is small as compared to the rated primary current. Therefore, voltage drops in R1 and X1 due to I0 are negligible. The equivalent circuit shown in Fig. 1 above can, therefore, be simplified by transferring the shunt circuit R0 X0 to the input terminals as shown in Fig. 2 below. This modification leads to only slight loss of accuracy.

Fig. 2 Simplified Equivalent Circuit of a Loaded TransformerEquivalent circuit referred to primaryIf all the secondary quantities are referred to the primary, we get the equivalent circuit of the transformer referred to the primary as shown in Fig. 3 (i). This further reduces to

Fig. 3 (ii). Note that when secondary quantities are referred to primary, resistances/reactances/impedances are divided by K2, voltages are divided by K and currents are multiplied by K.

Fig. 3 Equivalent circuit referred to primary

When the primary quantities are referred to primary, resistances/reactances/impedances are multiplied by K2, voltages are multiplied by K and currents are divided by K.

Approximate Equivalent Circuit of a Loaded TransformerThe no-load current I0 in a transformer is only 1-3% of the rated primary current and may be neglected without any serious error. The transformer can then be shown as in Fig. 4. (i) . This is an approximate representation because no-load current has been neglected. Note that all the circuit elements have been shown external so that the transformer is an ideal one.

Fig. 4 Approximate Equivalent Circuit of a Loaded Transformer

Approximate Equivalent circuit of transformer referred to primaryIf all the secondary quantities are referred to the primary, we get the equivalent circuit of the transformer referred to primary as shown in Fig. 5. Note that when secondary quantities are referred to primary, resistances/reactances are divided by K2, voltages are divided by K and currents are multiplied by K. The equivalent circuit shown in Fig. (7.26) is an electrical circuit and can be solved for various currents and voltages. Thus if we find V'2 and I'2, then actual secondary values can be determined as under:Actual secondary voltage, V2 = K V'2Actual secondary current, I2 = I'2/K

Hence the approximate equivalent circuit is deduced.