Question

A 50 kW, 120 V, long shunt compound generator is supplying a load at its maximum efficiency and the rated voltage. Armature resistance is 50 m$\Omega$, series field resistance is 20 m$\Omega$, shunt field resistance is 40 $\Omega$, rotational loss is 2 kW. The maximum efficiency of the generator is _____ %

• A. 82.35

Answer 1

The correct option is A 82.35
$\text{The shunt field current is,}{I}_f=\frac{120}{40}=3A$

$\text{For maximum efficiency}$

$I_{Lm}^2(R_a+R_s)=P_r+I_f^2(R_a+R_s+R_f)$

$I_{Lm}^2(0.05+0.02)=3^2(0.05+0.02+40)+2000$

or $I_{Lm}=183.64A$

$\text{Thus the power output at maximum efficiency is:}$

$P_o=120\times 183.64=22086.8W$

$\text{The total copper loss is}$

$P_m=I_o^2(R_a+R_s)+I_f^2{R}_f$

$=(183.64{)}^2\times 0.07+(3{)}^2\times 40$

$=2720.65W$

$\text{The power developed at maximum efficiency is}$

$P_d=P_o+P_m$

$=22036.8+272065=24757.45W$

$\text{Thus power input:}$

$P_{in}=24757.45+2000=26757.45W$

$\text{Hence, the maximum efficiency is:}$

$\eta =\frac{22-36.8}{26737.43}=0.8235or82.35\%$