Solution Electrical Numerical 17

Numerical 17: 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.

Solution: Equivalent secondary resistance:

$\displaystyle \mathrm{R = R_2 + \left ( \frac{N_2}{N_1} \right )^2 R_1\ }$ 

$\displaystyle \mathrm{R = \ 0.01 + \left ( \frac{200}{2200} \right )^2 \times 1\ =\ 0.018\ ohms }$

Full load efficiency:

full load current,

$\displaystyle \mathrm{I_2 = \frac{Power\ rating}{secondary\ voltage}\ =\ \frac{25000}{200}\ =\ 125\ A }$

full load cu-loss,

$\displaystyle \mathrm{I_2^2R =125^2 \times 0.02 =\ 312.5W }$ 

iron Loss = 80% of full load cu-loss = 0.8 x 312.5 = 250 W

total loss = 312.5 + 250 = 562.5 W

full load output = 25 x 0.8 =20 kW.

full load efficiency;

$\displaystyle \mathrm{\eta = \frac{20 \times 10^3}{20000 + 562.5}\times 100\ }$  = 97.26 %