In a 50 KVA, 1100/220 V transformer, the iron and copper losses at full load are 350 W and 425 W respectively. Calculate - 1. Full load with unity power factor - 2. Half load with unity power factor - 3. Full load with 0.8 pf lagging -

Given:

KVA = 50

V1= 1100V

V2= 220 V

P1= 350 W

Pw (FL) = 425W

Efficiency of full load and cos ф= 1

ƪ Ƞ FL= [KVA * 103 * cos ф/ (KVA * 103 * cosф) + P1 +Pw (FL)] * 100

= [50 * 1031 / (50 103*1) + 3.50 + 4.25)] * 100

= 98.47

Efficiency at half load and cosф = 1

Pw HL = 1/4 * Pw(FL)

= 1/4 * 425

= 106.425W

ƪ Ƞ HL = [(0.5 * KVA * 103 * cos ф)/ ((0.5 * KVA * 103 * cosф) + P1 + Pw (HL)] * 100

= [0.5 * 50 * 103* 1]/ [(0.05 * 50* 103 * 1) + 350 + 106.25] * 100

= 98.219

Efficiency at half load and cos ф = 0.8

ƪ Ƞ HL = (0.5 * 50 * 103 * 0.8) / (0.5 * 50 * 103 * 0.8) + 350 + 106.25

= 97.77 ƪ