Question

A 3-phase, 4-pole, 400 V, 50 Hz squirrel cage induction motor when working at rated voltage and rated frequency develops a full load torque at 1440 rpm. Now motor is fed from 30 Hz for same air gap flux.The speed of motor for same torque will be____rpm

• A. 840

Answer 1

The correct option is A 840
$\text{For same air gap flux,}\frac{V}{f}\text{ratio should be constant}$

$\text{Flux,}\varphi =\frac{V}{f}=\text{constant}$

$\frac{V_2}{f_2}=\frac{V_1}{f_1}$

$V_2=\frac{400}{50}\times 30=240V$

Synchronous speed of motor for 50 Hz source,
$N_s=\frac{120\times 50}{4}=1500rpm$

$Slip,s_1=\frac{1500-1440}{1500}=0.04$

At small value of slip, the electromagnetic torque is given by
$T=\frac{3}{{\omega }_s}\times \frac{V^{2}s}{r_s}$

$T\propto \frac{V^2}{f}.s$

$\frac{V_2^2}{f_2}.s_2=\frac{V_1^2}{f_1}.s_1$

$\frac{(240{)}^2}{30}{s}_2=\frac{{400}^2}{50}\times 0.04$

$\text{Slip,}{s}_2=0.067$

Synchronous speed of motor at 30 Hz source,
$N_{s2}=\frac{120\times 30}{4}=900$ rpm

Rotor speed, N = (1 - 0.067) $\times$ 900
N = 840 rpm

$\text{Alternate Method:}$
$\text{For same air gap flux,}\frac{V}{f}\text{ratio should be constant}$

$\text{Flux,}\varphi =\frac{V}{f}=\text{constant}$

$\frac{V_2}{f_2}=\frac{V_1}{f_1}$

Slip speed will be same i.e., (For same torque)

$N_{s1}-N_{r1}=N_{s2}-N_{r2}$

$N_{s1}=\frac{120\times 50}{4}=1500rpm$

$N_{r1}=1440rpm$

$N_{s2}=\frac{120\times 30}{4}=900rpm$

$\Rightarrow$ $1500-1440=900-N_{r2}$

$\Rightarrow$ $N_{r2}=840rpm$