Question

(a) Draw the complete phasor diagram of the transformer under no-load conditions. (6)

(b) The following results were obtained on a 50 kVA transformer: open-circuit test - primary voltage, 3300 V; secondary voltage, 415 V; primary power, 430 W. Short circuit test - primary voltage, 124 V; primary current, 15.3; primary power, 525 W; secondary current, full-load value.

Calculate: (10)

(a) the efficiencies at full load and at half load for 0.7 power factor.

(b) the voltage regulations for power factor 0.7, (i) lagging,

(ii) leading;

(c) the secondary terminal voltages corresponding to (i) and (ii)

Answer 1

Part (a)

Phasor diagram of a transformer under no-load conditions:

The applied primary voltage V1 is taken as the reference phasor.
The no-load current I0 has two components: the magnetising component Im (in quadrature with V1, lagging by 90 degrees) and the core-loss (active) component Iw (in phase with V1).
I0 = Iw + Im, and I0 lags V1 by an angle phi0 (the no-load power factor angle), where cos phi0 = Iw/I0.
The flux phi is in phase with the magnetising current Im (neglecting hysteresis), and the induced e.m.f. E1 (and E2) lag the flux by 90 degrees. E1 is approximately equal and opposite to V1.
The diagram shows V1, I0, Im, Iw, phi, E1 and E2 with their phase relationships.

Part (b)

50 kVA transformer. OC test: V1 = 3300 V, V2 = 415 V, P = 430 W. SC test: V1 = 124 V, I1 = 15.3 A, P = 525 W, secondary current = full-load value.

Iron loss (from OC test) = 430 W.
Full-load copper loss (from SC test) = 525 W.
Full-load primary current = 50000/3300 = 15.15 A.

Part (a)

Efficiencies at 0.7 p.f.:

Full load: output = 50 x 0.7 = 35 kW. Losses = 430 + 525 = 955 W. Input = 35,955 W. Efficiency = 35000/35955 = 0.9734 = 97.34%.
Half load: copper loss = 525 x (0.5)^2 = 131.25 W. Total losses = 430 + 131.25 = 561.25 W. Output = 17.5 kW. Input = 18,061.25 W. Efficiency = 17500/18061.25 = 0.9689 = 96.89%.

Part (b)

Voltage regulation at 0.7 p.f.:

From SC test: equivalent impedance Z = 124/15.3 = 8.105 ohm. Equivalent resistance Req = P/I^2 = 525/15.3^2 = 525/234.1 = 2.243 ohm. Equivalent reactance Xeq = sqrt(8.105^2 - 2.243^2) = sqrt(65.7 - 5.03) = sqrt(60.67) = 7.79 ohm.
cos phi = 0.7, sin phi = 0.714.
(i) Lagging: %VR = I (Req cos phi + Xeq sin phi)/V1 x 100 = 15.15 (2.243 x 0.7 + 7.79 x 0.714)/3300 x 100 = 15.15 (1.570 + 5.562)/3300 x 100 = 15.15 x 7.132/3300 x 100 = 108.1/3300 x 100 = 3.28%.
(ii) Leading: %VR = 15.15 (1.570 - 5.562)/3300 x 100 = 15.15 x (-3.992)/3300 x 100 = -60.5/3300 x 100 = -1.83%.

Part (c)

Secondary terminal voltages:

Nominal secondary voltage = 415 V.
(i) Lagging: V2 = 415 (1 - 0.0328) = 415 x 0.9672 = 401.4 V.
(ii) Leading: V2 = 415 (1 + 0.0183) = 415 x 1.0183 = 422.6 V.

So efficiency = 97.34% (full load) and 96.89% (half load); regulation = 3.28% lagging and -1.83% leading; secondary voltages = 401.4 V (lagging) and 422.6 V (leading).

✓ Verified Model Answer (Text Solution)

Phasor diagram of a transformer under no-load conditions:

50 kVA transformer. OC test: V1 = 3300 V, V2 = 415 V, P = 430 W. SC test: V1 = 124 V, I1 = 15.3 A, P = 525 W, secondary current = full-load value.

Efficiencies at 0.7 p.f.:

Voltage regulation at 0.7 p.f.:

Secondary terminal voltages: