Question

A 3.3 kV, 20-pole, 50 Hz, 3-phase, star connected induction motor has a slip-ring rotor of resistance $0.025\Omega$ and stand still reactance of $0.28\Omega$ per phase. The motor has a speed of $294rpm$ when full load torque is applied. The ratio of maximum to full load torque is ________

• A. 1.57
• B. 9.18
• C. 4.16
• D. 2.34

Answer 1

The correct option is D 2.34

$s_{Tmax}=\frac{0.025}{0.28}=0.09$

Maximum torque,$T_{max}=\frac{3}{{\omega }_s}\times \frac{0.5V^2}{x_2^{\prime }}$ $...\left(i\right)$

Full load torque, $T_{f1}=\frac{\frac{3}{{\omega }_s}\times V^2\left(\frac{r_2^{\prime }}{S_{f1}}\right)}{{\left(\frac{r_2^{\prime }}{S_{f1}}\right)}^2+x_2^{\prime 2}}$ $...\left(ii\right)$

Synchronous speed, $N_s=\frac{120\times 50}{20}=300rpm;$

$N=294rpm$

Slip,
$S_{f1}=\frac{300-294}{300}=0.02$

Dividing equation (i) by (ii),
$\frac{T_{max}}{T_{f1}}=\frac{0.5[{\left(\frac{r_2^{\prime }}{S_{f1}}\right)}^2+x_2^{\prime 2}]}{x_2^{\prime }\left(\frac{r_2^{\prime }}{S_{f1}}\right)}$

$=\frac{0.5[{\left(\frac{0.025}{0.02}\right)}^2+{0.28}^2]}{0.28\times \left(\frac{0.025}{0.02}\right)}$

$=2.34$