) 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%
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.
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