Question

A short shunt, cumulative compound generator is rated at 100 A, voltage across load is 240 V and the shunt field current is 3 A. It has an armature resistance of 50 m$\Omega$, a series field resistance of 10 m$\Omega$, a field diverter resistance of 40 m$\Omega$, and a rotational loss of 2 kW. The generator is connected to the load via a feeder $R_{fe}$ of 30 m$\Omega$ resistance. The efficiency when the generator is supplying the full load at rated voltage, is

• A. 86.82 %
• B. 78.12 %
• C. 81.87 %
• D. 89.86 %

Answer 1

The correct option is A 86.82 %

$\text{Output power,}{P}_o=240\times 100$

$=24000W$

$\text{Shunt field current,}{I}_f=3A$

$\text{Armature current,}{I}_a=100+3=103A$

$\text{Series field current,}{I}_{ag}=\frac{0.04}{0.04+0.01}\times 100=80A$

$I_d=100-80=20A$

$E_a=V_f+L_L{R}_{fe}+I_s{R}_s+I_a{R}_a$

$=240+100\times 0.03+80\times 0.01+103\times 0.05$

$=248.95V$

$V_f=E_a-I_a{R}_a$

$=248.95-103\times 0.05=243.8V$

$\text{Hence}{R}_f=\frac{243.8}{3}=81.267\Omega$

$\text{Copper losses:}$

$\text{Armature:}$	$I_a^2{R}_a={103}^2\times 0.05=530.45W$
$\text{Series field:}$	$I_s^2{R}_s={80}^2\times 0.01=64W$
$\text{Shunt field:}$	$I_f^2{R}_f=3^2\times 81.267=731.4,W$
$\text{Diverter resistance:}$	$I_d^2{R}_d={20}^2\times 0.04=16W$
$\text{Feeder resistance:}$	$I_L^2{R}_{fe}={100}^2\times 0.03=300W$
$\text{Total copper loss:}$	$P_{cu}=530.45+64+731.4+16+300$ $=1641.85W$

$\text{Thus, the power developed is,}{P}_d=P_o+P_{ck}=24000+1641.85$

$=25641.85W$

$\text{The power input is,}{P}_{in}=P_d+P_r$

$=25641.85+2000$

$=27641.85W$

$\text{Hence, the efficiency is,}$
$\eta =\frac{P_o}{P_m}=\frac{24000}{27641.85}=0.8682\left(or\right)86.82\%$