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# Solved Problems: Electrical Engineering and Instrumentation - Transformer

Electrical Engineering and Instrumentation - Transformer - Solved Problems: Electrical Engineering and Instrumentation - Transformer

PROBLEMS

A 400 kVA transformer has a primary winding resistance of 0.5 ohm and a secondary winding resistance of 0.001 ohm. The iron loss is 2.5 Kw and the primary and secondary voltages are 5 kV and 320 V respectively. If the power factor of the load is 0.85, determine the efficiency of the transformer (i) on full load and (ii) on half load.

Solution

Rated output = 400 kVA = 400x103 kVA

Full load secondary current, I2 = Rated output/V2 = 1250 A

Total resistance referred to secondary, re2 = r2+r1(V2/V1)2 = 0.033 ohm Full load copper loss, Pc = I22 re2 = 51.5625 Kw

Iron loss, Pi = 2.5 x103 watts

(i) Transformer efficiency at full load and 0.85 pf = 86.2%

(ii) Transformer efficiency at half load and 0.85 pf = 91.69%

2. Find all day efficiency of a transformer having maximum efficiency of 98% at 15 kVA at unity power factor. Compare its all day efficiencies for the following load cycles:

a. Full load of 20 kVA, 12 hours per day and no load rest of the day. b. Full load, 4 hours per day and 0.4 full load rest of the day.

Assume the load to operate on upf all day

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