Problem 107-8    Source:  Problem prepared by the website author.

A 440 V, 6 pole and 60 Hz three-phase induction motor, connected in the Y configuration , supplies a constant load of 120 N . m at a speed of 1,180 rpm. Ventilation losses are 600 W and friction losses are 400 W. Determine:

a) the mechanical power, P_mec, and the nominal power, P_n ;

b) the rotor copper losses, P_jr ;

c) the induction torque, τ_ind ;

d) the new speed of the machine if we double the value of the load.

Solution of the Problem 107-8   

Item a

Initially we will calculate the value of ω_r.

Using the eq. 107-33 we can calculate the rated power of the motor, or:

Carrying out the calculation, we get:

Using the eq. 107-31 we can calculate the mechanical power of the motor, or:

Carrying out the calculation, we get:

Item b

Let's calculate the synchronous speed of the machine. Using the eq. 107-02, we have:

With this value we can calculate the motor slip using eq. 107-04.

Carrying out the calculation, we get:

Knowing the slip, it is possible to calculate the air-gap power through eq. 107-30. Like this:

Carrying out the calculation, we get:

Now we can calculate the rotor copper losses using eq. 107-27, or:

Item c

To solve this item we will calculate the synchronous speed of the machine.

The torque or induced torque of the machine is given by eq. 107-32.

Carrying out the calculation, we get:

Item d

By carefully observing the torque x speed curve of an induction motor, we can see that in a region close to the slip in which the machine operates, the motor slip increases approximately linearly with the increase in load and the mechanical speed of the rotor also decreases approximately linearly with load. Then, it is possible to write the following relation:

Since, by the problem statement, τ_new = 2 x τ_old   we easily conclude that s_new = 2 x s_old = 0.0334 .

So, transforming the eq. 107-04 we get: