12.MET September 2018 Q.7(B)

A 25 kVa signal phase transformer 2200:200V has a primary and secondary resistance of 1Ω and 0.01 Ω respectively. Find the equivalent secondary resistance and full load efficiency at 0.8pf lagging, if the iron losses of the transformer are 80% of the full load copper losses.

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Given

KVA = 25000
V1 = 2200
V2 = 200
R1 = 1 ohm
R2 = 0.01 ohm
PF = 0.8
Losses = 80%

To find

1. Secondary resistance
2. Full load efficiency

Solution

Voltage transformation ratio, K = V2 / V1
                                                  = 200 / 2200
                                                  = 0.09

1. Secondary resistance, R02  = R2 + (K2 x R1)
                                                 = 0.01 + (0.092 x 1 )
                                                 = 0.018 ohm

Secondary Full load current I2 = KVA / V2
                                                 = 25000 / 200
                                                 = 125 A

Full load Cu loss = I22  x R02
                            = 1252  x 0.018
                            = 281.25 W

Iron loss = 80 % of full load Cu loss
               = 0.8 x 281.25
               = 225 W

Total loss = 281.25 + 225
                = 506.25 W

Full load output = 25000 x 0.8 = 20000 W

2. Full load efficiency = 20000 / (20000 + 506.25)
                                = 97.53 %