97.MET October 2020 Q.7(b)

In a 25 KVA, 3300/233 V, single phase transformer, the iron and full-load Cu. losses are respectively 350 and 400 watts. Calculate the efficiency at half-full load 0.8 power factor.

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Given

KVA = 25
V1 = 3300 
V2 = 233 
Iron loss = 350 W
F.L Cu loss = 400 w
Cos Φ = 0.8

To find

1.Half load efficiency
2.Full load efficiency

Solution

1. Full load output at 0.8 pf = 25 x 0.8
                                            = 20 KW

    At half load output = 20 / 2
                                  = 10 KW

    
    Cu loss at half load is 1/4th of that full load.
    At half load Cu loss = 400 / 4
                                    = 100 W
                                    = 0.1 KW
    
    Iron loss remains same = 0.35 KW

    Efficiency = Output / (Output + Cu loss + Iron loss)
    
    Half load efficiency = H.L output / (H.L output + H.L Cu loss + Iron loss)
                                   = 10 / (10 + 0.1 + 0.35 )
    Half load efficiency = 95.69 %

2. Full load efficiency = F.L output / (F.L output + F.L Cu loss + Iron loss)
                                   = 20 / (20 + 0.4 + 0.35)
    Full load efficiency  = 96.38 %