Question

The rotor current at start of a three-phase, 460 volt, 1710 rpm, 60 Hz, four-pole, squirrel cage induction motor is six times the rotor current at full load. The speed at which the motor develops maximum torque in rpm Is __________.

• A. 1242

Answer 1

The correct option is A 1242
$n_s=\frac{120\times 60}{4}=1800rpm$

$s_{fL}=\frac{1800-1710}{1800}=0.05$

$T\propto \frac{I^2{R}_2}{s}$

Thus, $\frac{T_{st}}{T_{fL}}={\left|\begin{array}{c}\frac{I_{2st}}{I_{2FL}}\end{array}\right|}^2.s_{FL}$

$\frac{T_{st}}{T_{fL}}=6^2\times 0.05$

$\Rightarrow \frac{T_{st}}{T_{fL}}=1.8$

Also, $\frac{T_{st}}{T_{max}}=\frac{2s_{Tmax}}{1+S_{Tmax}^2}$

$\frac{T_{FL}}{T_{max}}=\frac{2s_{Tmax}\cdot s_{FL}}{s_{Tmax}^2+s_{FL}^2}$

From equation (i) and (ii)

$\frac{T_{st}}{T_{fL}}=\frac{s_{Tmax}^2+s_{FL}^2}{s_{FL}+s_{FL}\times s_{Tmax}^2}$

$1.8=\frac{s_{Tmax}^2+0.0025}{s_{FL}+s_{FL}\times s_{Tmax}^2}$

$s_{Tmax}^2+0.0025=0.09+0.09s_{Tmax}^2$

$s_{Tmax}=0.31$

$\text{Speed of maximum torque}$
$=(1-0.31)\times 1800$
$=1242rpm$