Question

A 400 V, 50 Hz, 4-pole, 1440 rpm star connected squirrel cage induction motor is controlled by $\frac{V}{f}$ control as frequency is reduced to 80% of rated value, the percentage slip at reduced frequency will be

• A. 4.5%
• B. 3.4%
• C. 4%
• D. 5%

Answer 1

The correct option is D 5%

Given motor has synchronous speed,
$N_s=\frac{120\times 50}{4}=1500rpm$

$\therefore Slip,s=\frac{1500-1440}{1500}=\frac{60}{1500}=4\%$

Slip speed, $s{N}_s=0.04\times 1500=60rpm$

New frequency, $f^{\prime }=\frac{80}{100\times 50}=40Hz$

The synchronous speed at new frequency ( 40 Hz)
$=\frac{120\times 40}{4}=1200rpm$

$\text{The slip speed}\left(s{N}_s\right)\text{remains constant in}\frac{V}{f}\text{control method}$

$N_s=1200-60=1140rpm$

$s=\frac{N_s-N_r}{N_s}=\frac{1200-1140}{1200}\times 100=5\%$