Question

A 120 V, 60Hz, 1/3-hp, 4-pole, single phase induction motor has following circuit parameters:
$R_1=2.5\Omega ,$ $X_1=1.25\Omega ,$ $R_2=3.75\Omega ,X_2=1.25\Omega ,$ and $X_m=65\Omega$. The motor runs at a speed of 1710 rpm and has a core loss of 25 W. The friction and windage loss is 2 W. The efficiency of the motor is_________.

• A. 81.48%
• B. 72.36%
• C. 61.41%
• D. 65.86%

Answer 1

The correct option is D 65.86%

$\text{Synchronous speed,}$ $N_s=\frac{120\times 60}{4}=1800rpm$

$\text{Slip,}$ $s=\frac{1800-1710}{1800}=0.05$

$\overrightarrow{Z_f}=\frac{j65[\frac{3.75}{0.05}+j1.25+j1.25]0.5}{\left(\frac{3.75}{0.05}\right)+j(1.25+65)}$

$=(15.822+j18.524)\Omega$

$\overrightarrow{Z_b}=\frac{j65[\frac{3.75}{1.95}+j1.25+j1.25]0.5}{\left(\frac{3.75}{1.95}\right)+j(1.25+65)}$

$=(0.925+j0.64)\Omega$

$\overrightarrow{Z_{in}}=2.5+j1.25+15.822+j18.524+0.925+j0.64$

$=(19.247+j20.414)\Omega$

$\overrightarrow{I_1}=\frac{120}{19.247+j20.414}=4.277\angle -{46.69}^{\circ }A$

$\text{Input power,}$ $P_{in}=Re[120\times 4.277\angle {46.69}^{\circ }]=352.1W$

$\text{Forward current,}$ $\overrightarrow{I_{2f}}=\frac{j65}{\left(\frac{3.75}{0.05}\right)+j(1.25+65)}\times 4.277\angle -{46.69}^{\circ }$

$=2.778\angle {1.86}^{\circ }A$

$\text{Backward current,}$ $\overrightarrow{I_{2b}}=\frac{65}{\left(\frac{3.75}{1.95}\right)+j(1.25+65)}\times 4.277\angle -{46.69}^{\circ }$

$=4.195\angle -{45.02}^{\circ }A$

$\text{Forward air gap power,}$
$P_{agf}={2.778}^2\times 0.5\times \frac{3.75}{0.05}=289.4W$

$\text{Backward air gap power,}$
$P_{agb}={4.195}^2\times 0.5\times \frac{3.75}{1.95}=16.9W$

$\text{The net air gap power is,}$
$P_{ig}=289.4-16.9=272.5W$

$\text{The gross power developed is,}$
$P_d=(1-0.05)\times 272.5=258.9W$

$\text{Net power output is,}$
$P_0=258.9-25-2=231.9W$

$\text{Efficiency,}$ $\eta =\frac{231.9}{352.1}=0.6586\left(or\right)65.86\%$