Question

15.A 3 phase induction motor has a 4 pole, star connected stator winding and runs on a 220 V, 50 Hz supply. The rotor resistance per phase is 0.1 Ω and rotor reactance of 0.9 Ω. The ratio of stator to rotor turns is 1.75. The full load slip is 5%. Calculate for this load :
(i)The load torque in kg-m. (ii) speed at maximum torque (iii) rotor emf at maximum torque.

Answer 1

Answer:

(a) Neglecting stator impedance, the power developed using equation 8.17 is given as

P=\frac{{V_{1}}^{2}R_{2}^{‘}(1-s)s}{{R}'{_{2}}^{2}+s^{2}{X}'{_{2}}^{2}}P=

R

′

2

2

+s

2

X

′

2

2

V

1

2

R

2

‘

(1−s)s

Referring the quantities to rotor side the equation becomes

\frac{V_{2}^{2}R_{2}(1-s)s}{R_{2}^{2}s^{2}X_{2}^{2}}=\frac{0.67^{2}\times 200^{2}\times 0.1\times 0.96\times 0.04}{0.1^{2}+0.04^{2}\times 0.9^{2}}

R

2

2

s

2

X

2

2

V

2

2

R

2

(1−s)s

=

0.1

2

+0.04

2

×0.9

2

0.67

2

×200

2

×0.1×0.96×0.04

\simeq \frac{68.95}{0.01}=6895≃

0.01

68.95

=6895 watts

or \frac{6895}{735}=9.38