Question

A 300 kVA transformer has a primary winding resistance of 0.4 Ω and a secondary winding resistance of 0.0015 Ω. The iron loss is 2 kW and the primary and secondary voltages are 4 kV and 200 V respectively. If the power factor of the load is 0.78, determine the efficiency of the transformer on full load.

Answer 1

) Rating = 300 kVA = V1 I1 = V2 I2

Hence primary current, I1 = = = 75 A
and secondary current, I2 = = = 1500 A

Total copper loss = I12 R1 + I22 R2, (where R1 = 0.4  and R2 = 0.0015 )

= (75)2(0.4) + (1500)2(0.0015)

= 2250 + 3375 = 5625 watts

On full load, total loss = copper loss + iron loss

= 5625 + 2000

= 7625 W = 7.625 kW

Total output power on full load = V2 I2 cos 2

= (300  103)(0.78) = 234 kW

Input power = output power + losses = 234 kW + 7.625 kW = 241.625 kW

Efficiency,  =  100%

=  100% = 96.84%

(b) Since the copper loss varies as the square of the current, then total

copper loss on half load = 5625  = 1406.25 W

Hence total loss on half load = 1406.25 + 2000

= 3406.25 W or 3.40625 kW

Output power on half full load = (234) = 117 kW

Input power on half full load = output power + losses

= 117 kW + 3.40625 kW = 120.40625 kW

Hence efficiency at half full load,

 =  100%

=  100% = 97.17%