Problem 107-10    Source:   Problem 5.10 - page 272 - SEN, P.C. -    Book: Principles of Electric Machines and Electronics - 3rd Edition - Ed. Wiley - 2014.

A three-phase induction motor of 60 Hz and 6 poles, rotates at 1,140 rpm and consumes 90 kW from the power source. Ventilation and friction losses are 2,100 W and stator and core copper losses are 4,800 W. . Determine:

a) the power in the air gap, P_gap ;

b) the rotor copper losses, P_jr ;

c) the mechanical power delivered to the load, P_mec ;

d) the torque on the load, τ_load ;

e) the engine efficiency.

Solution of the Problem 107-10   

Item a

To determine the power in the air gap we can use the eq. 107-21. Soon:

Where we use the fact that P_iron + P_E = 4,800 W.

Item b

We can calculate the rotor copper losses using eq. 107-27. First we must calculate the slip of the motor. As the motor has 6 poles, then the synchronous speed is 1,200 rpm. And the slip is:

Using the eq. 107-27, we have:

Item c

Knowing the value of P_gap we can calculate the mechanical power in the rotor, or:

Item d

To calculate the torque on the load, we must first calculate the nominal (or output) power of the machine and ω_r. So, as we know, the output or nominal power is the difference between mechanical power and rotational losses. Soon:

And the angular speed of the rotor, ω_r, is:

And to find the value of the torque in the load we will use the eq. 107-33.

Doing the math, we find:

Item e

And the motor efficiency is given by eq. 107-23. Soon:

Doing the math, we find: