Solution Electrical Numerical 2

Numerical 2: 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. 

Solution: Given

Power rating = 25 KVA

Voltage Ratio = 3300/233 V

Iron Loss = 350 W

copper loss = 400 W

P.F = 0.8 

Iron losses are constant =350 W or 0.35 kW. 

But Copper losses are variable and are proportional to sq. of current. Thus at half load copper losses will be 1/4th of full load copper losses.i.e 100 W or 0.1 kW.

Efficiency

$\displaystyle \displaystyle \mathrm{\eta = \frac{Out\,put}{In\,put}\ = \frac{Out\,put}{Out\,put + Losses}\ }$ 

Efficiency at half load

$\displaystyle \displaystyle \mathrm{\eta_{1/2} = \frac{\frac{1}{2}Out\,put}{\frac{1}{2}Out\,put + Losses\, at\, half\, load}\ }$ 

$\displaystyle \displaystyle \mathrm{\eta_{1/2} = \frac{\frac{1}{2}25}{\frac{1}{2}25 + 0.35+0.1}\ =\ 0.9569}$ 

or 95.69%.