Question

A 10 kVA, 2500/250 V single phase transformer has full load ohmic losses of 300 W. The
maximum possible voltage drop in the transformer secondary voltage is 20 V. Find voltage
regulation of this transformer for rated kVA output at 0.8 pf lagging.

Answer 1

Answer:

18.5 kW

In a distribution transformer (constant voltage), iron losses are essentially constant, and copper losses losses vary with load squared (I^2R). Losses related to efficiency are:

efficiency = output/input = load / (load + losses)

reordering:

losses = load * (1 - efficiency)/efficiency

If the efficiency is 90% at 500kVA and also 90% at half-load (250kVA), then the total looses are:

full load: 55.6kW

half load: 27.8kW

If we use the variable, Cl for core loss, and I2R for copper losses then the following two equations can be developed:

Cl + I2R_full = 55.6 kW [1]

Cl + I2R_half = 27.8 kW [2]

Then noting that I2R_half = I2R_full/4, because the losses vary by the square of current and one-half the current becomes 1/4 when squared:

Cl + I2R_full/4 = 27.8 kW

Rearranging:

I2R_full = 4 * (27.8 kW - Cl)

Substituting back in [1]:

Cl + 4 * (27.8 kW - Cl) = 55.6 kW

111 kW - 3*Cl = 55.6 kW

Cl = -55.6 kW / -3 = 18.5 kW