Resolved Problems about Induction Motors

Problem 107-4 Source: Adapted from example 6-2 - page 322 - CHAPMAN, Stephen J. - Book: Fundamentals of Electric Machines - 5th edition - Ed. McGrawHill - 2013.

A three-phase induction motor of 480 V, 60 Hz and 50 HP, consumes 60 A from the mains with FP = 0.85 delayed. The stator copper losses are 2 kW and the rotor copper losses are 700 W. Friction and windage losses are 600 W, core losses are 1800 W. Find the following quantities:

a) The power in the air gap, P_gap ;

b) The converted power P_mec ;

c) The output power P_n ;

d) The efficiency, η, of the engine.

Solution of the Problem 107-4

Item a

From the problem data it is possible to calculate the input power of the motor using eq. 107-22. In this way, we have:

P_in = √3 x 480 x 60 x 0.85 = 42,400 W

Usando a eq. 107-21 é possível calcular a potência no entreferro.

P_gap = 42,400 - 2,000 - 1,800 = 38,600 W

Item b

Usando a eq. 107-28 calculamos a potência mecânica do rotor, ou:

P_mec = 38,600 - 700 = 37,900 W

Item c

A potência de saída do motor podemos calcular usando a eq. 107-31.

P_n = 37,900 - 600 = 37,300 W

Como é comum expressar a potência de saída de um motor de indução em HP, vamos fazer essa transformação.

P_n = 37,300 / 746 = 50 HP

Item d

E para encontrarmos a eficiência (ou rendimento) do motor vamos usar a eq. 107-23.

η = 37,300 / 42,400 = 0.88 = 88%

Back to Top

Previous Problem

Next Problem