Question

A 25 kVA transformer has an iron loss of 200 W and a full-load copper loss of 350 W. Calculate the transformer efficiency for the following load conditions: a. Full load at 0.8 power factor lagging b. 70% of full load at unity power factor c. 40% of full load at 0.6 power factor lagging

Answer 1

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%