The corresponding poly-phase induction motor circuit is shown in Fig. 24, where
V1 = applied voltage per phase
R1 = stator resistance/phase
R2 = rotor resistance/phase
X1 = stator leakage reactance/phase
X2 = rotor standstill leakage reactance/phase
K = turn-ratio of secondary to primary
R0 = no-load resistance/phase
X0 = no-load reactance/phase
Fig. 24. Equivalent circuit of an induction motor
It is illustrated in the figure that when the circuit is transferred to the left, the circuit and calculations are much simpler because of the negligible inaccuracy. Such a process is called induction motor equivalent circuit as the approximate value.
Maximum Power Output (shown in figure no. 25)
In order to obtain the maximum power output, let’s differentiate the mentioned equation by equating the first derivative to 0.
By the mentioned equation we can say that gross mechanical power output will maximum when the standstill leakage impedance is equal to the equivalent load resistance RL of the motor Z01.
In a 3-phase induction motor maximum torque occurs at a slip of 15 per cent. The equivalent secondary resistance of the motor is 0.07 Ω/phase. If the gross power output is 9.5 II W. Calculate.-
(i) Equivalent load resistance,
(ii) Equivalent load voltage, and
(iii)Current at this slip.
Slip at maximum torque, s = 15 per cent
The equivalent secondary resistance of the motor,
R2 = 0.07 Ω/phase
Gross power output, Pg = 9.5 kW or 9500 W
- Equivalent load resistance, RL:
We know that
= (please calculate the value)
- Equivalent load voltage, V:
V is a fictitious voltage drop which is equal to the consumption in the secondary load connected, i.e., rotor, V = I2RL
But, gross power, Pg =
= please calculate the value
- Equivalent load current: