Answer:
a. 97.32 percent
b. 97.92 percent
c. 95.91 percent
Explanation:
given parameters
apparent power, S = 25 kVA
Iron Loss, Piron = 200 W
copper Loss at full load, P cufl = 350 W
let the transformer efficiency at full load be E
let the real power be P out
a.
at full load and 0.8 power factor
P out = Scos∅
= 25 x 10³ x 0.8
=20000 W or 20 kW
efficiency at full load is
E = (P out)/(P out + P iron + P cufl) x 100%
= (20000)/(20000 + 200 + 350) x 100%
= 97.32%
b.
at 70% of full load at unity power factor of 1
Pout = 70% x Scos∅
= 0.7 x 25 x 10³ x 1
= 17,500W or 17.5kW
copper loss at 70% full load
P cu0.7fl = (70%)² x 350
= (0.7)² x 350
= 171.5 W
Iron loss remain the same, P iron = 200 W
Efficiency of transformer at 70% of full load
E = (P out0.7)/(P out0.7 + P iron + P cufl0.7) x 100%
= (17500)/(17500 + 200 + 171.5) x 100%
= 97.92%
c.
at 40% of full load at a power factor of 0.6
P out = 40% x Scos∅
= 0.4 x 25 x 10³ x 0.6
= 6000 W or 6 kW
copper loss at 40% full load
P cu0.7fl = (40%)² x 350
= (0.4)² x 350
= 56 W
Iron loss remain the same, P iron = 200 W
Efficiency of transformer at 40% of full load
E = (P out0.4)/(P out0.4 + P iron + P cu0.4) x 100%
= (6000)/(6000 + 200 + 350) x 100%
= 95.91%
