Question

A 3-phase induction motor has 5% slip at full load torque and at normal voltage. The rotor resistance and rotor standstill leakage reactance per phase are 0.1 $\Omega$ and 0.6 $\Omega$ respectively. What should be the percentage reduction in supply voltage if the motor has to develop full-load torque at $\frac{3}{4}th$ of normal full-load speed. (Neglect stator impedance)

• A. 20.35
• B. 36.57
• C. 63.43
• D. 79.64

Answer 1

The correct option is A 20.35

$T_{ef1}\propto \frac{V_1^2}{{\left(\frac{r_2}{s}\right)}^2+x_2^2}\times \frac{r_2}{s}\propto \frac{V_1^2}{{\left(\frac{0.10}{0.05}\right)}^2+(0.6{)}^2}\times \frac{0.10}{0.05}\propto 0.4587V_1^2$

$\text{When speed is}\frac{3}{4}th\text{of normal full speed}$

$\text{Full load speed = (1 - s)}{N}_s$

$\text{Full load speed = 0.95}{N}_s$

$\frac{3}{4}th\text{of full load speed = 0.7125}{N}_s$

$\text{New slip}=\frac{N_s-N_r}{N_s}=0.2875$

So, $T_{ef2}\propto \frac{(x{V}_1{)}^2}{{\left(\frac{0.10}{0.2875}\right)}^2+(0.6{)}^2}\times \frac{0.01}{0.2875}\propto 0.7231\left(x{V}_1^2\right)$

$\text{Now as both torques are equal,}$
$0.4587V_1^2=0.7231x^2{V}_1^2$
$x^2=0.6343$

$x=0.7964$

$\text{\% reduction in voltage}=\frac{(1-0.7964)}{1}\times 100=20.35\text{\%}$