Excercises Electric Machinery Fundamentals

Chapter 6

Problem 6-11


In [1]:
%pylab inline


Populating the interactive namespace from numpy and matplotlib

Description

The input power to the rotor circuit of a six-pole, 60 Hz, induction motor running at 1100 r/min is 5 kW.


In [2]:
fse  =   60 # [Hz]
n_nl = 1100 # [r/min]
p    =    6
  • What is the rotor copper loss in this motor?

SOLUTION

This synchronous speed of this motor is:

$$n_\text{sync} = \frac{120f_{se}}{p}$$

In [3]:
n_sync = 120*fse / p
print('n_sync = {:.0f} r/min'.format(n_sync))


n_sync = 1200 r/min

The slip of the rotor is: $$s_{nl} = \frac{n_\text{sync}-n_{nl}}{n_\text{sync}} \cdot 100\%$$


In [4]:
s_nl = (n_sync - n_nl) / n_sync
print('s_nl = {:.2f} %'.format(s_nl*100))


s_nl = 8.33 %

The air gap power is the input power to the rotor, so:


In [5]:
Pag = 5000 # [W]

The power converted from electrical to mechanical form is: $$P_\text{conv} = (1-s)P_{AG}$$


In [6]:
Pconv = (1-s_nl) * Pag
print('Pconv = {:.0f} W'.format(Pconv))


Pconv = 4583 W

The rotor copper losses are the difference between the air gap power and the power converted to mechanical form, so:

$$P_\text{RCL} = P_{AG}-P_\text{conv}$$

In [7]:
Prcl = Pag - Pconv
print('''
Prcl = {:.0f} W
============'''.format(Prcl))


Prcl = 417 W
============