Three-Phase Induction Motor

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Solution for Posting 30567

A three phase Y-connected 480 V (line to line) 30 hp 60 Hz four pole induction motor has the following equivalent circuit constants in Ohms-per-phase referred to the stator:

R1=0.21 R2= 0.20 X1=1.2 X2=1.1 Xp=39

The total friction windage and core losses may be assumed constant at 1340 Watts. The motor is connected directly to a 480 V source. Compute speed, shaft output torque in Nm, power in horsepower, efficiency, and terminal power factor for slips of 1, 2, and 3 %.

Solution:

The equivalent circuit of the induction motor:

Where R1 = 0.21 Ohm, X1 = 1.2 Ohm Z1 = R1+jX1 = 0.21 +j 1.2 Ohm
R2 = 0.2 Ohm, X2 = 1.1 Ohm Z2 = R2/S+jX2 = 0.2/S +j 1.1 Ohm
Xm = 39 Ohms Zm = j 39 Ohm

Slip S = (n1-n)/n1
Where Synchronous speed n1 = (120f1)/pole numbers
f1 is the system frequency f1 = 60Hz, total pole numbers is 4 poles.
Thus, n1 = (120*60)/4 = 1800 rpm

The motor speed n = (1-S)*n1

The motor voltage per phase
It is given that the motor line-to-line voltage is 480V, and the motor is Y-connected, so per phase voltage is 480/1.732 = 277V. Please note the induction motor equivalent circuit is for per phase.

The total impedance Z = Z1 +Z2//Zm = (R1+jX1) + [(R2/S+jX2)*jXm]/(R2/S+jX2 ...