Question

A 2000 kVA, single-phase transformer is in circuit continuously. For 8 hours in a day, the load is 160 kW at 0.8 pf. For 6 hours, the load is 80 kW at unity pf and for the period remaining out of 24 hours, it runs on no-load. If the full-load copper losses are 3.02 kW, the total copper losses in 24 hours are:
1. 35.02 W
2. 24.16 W
3. 11.46 W
4. 38.40 W

Answer 1

Correct Answer - Option 1 : 35.02 W

Concept:

If the copper losses at full load are W_cu, then the copper losses at a fraction of full load (x) are given by

= x²W_cu

Calculation:

Given that, full load copper losses, W_cu = 3.02 kW

Full load kVA = 2000 kVA

At 160 kW, 0.8 pf, it operates for 8 hours.

Load kVA = 160/0.8 = 200 kVA

Fraction of full load = 200/2000 = 0.1

Copper losses in this case = 0.1² × 3.02 × 10³ = 30.2 W

At 80 kW, unity pf, it operates for 6 hours.

Load kVA = 80/1 = 80 kVA

Fraction of full load = 80/2000 = 0.04

Copper losses in this case = 0.04² × 3.02 × 10³ = 4.832 W

On no load, there are no copper losses.

Total copper losses = 30.2 + 4.832 = 35.032 W