Excercises Electric Machinery Fundamentals

Chapter 6 Problem 6-3



In [1]:

    
%pylab inline
%precision 4









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.4f'

Description

A three-phase 60-Hz induction motor runs at 715 r/min at no-load and at 670 r/min at full load.

Note:

The no-load speed is near but not identical with the synchronous speed. You will always have some losses that the machine needs to overcome. Hence the speed of the rotor will never reach synchronous speed even with no-load.



In [2]:

    
fe       =  60.0 # [Hz]
n_noload = 715.0 # [r/min]
n_m      = 670.0 # [r/min]

(a)

How many poles does this motor have?

(b)

What is the slip at rated load?

(c)

What is the speed at one-quarter of the rated load?

(d)

What is the rotor’s electrical frequency at one-quarter of the rated load?

SOLUTION

(a)

This machine produces a synchronous speed of:

$$n_\text{sync} = \frac{120f_e}{p}$$



In [3]:

    
p = 120*fe / n_noload
p









    Out[3]:





10.0699

So nearest even number of poles is:



In [4]:

    
p=floor(p/2)*2 # nearest even number



In [5]:

    
p=floor(p/2)*2 # nearest even number
print('''
p = {:.0f} 
======'''.format(p))









    



p = 10 
======

which produces a syncronous speed of



In [6]:

    
n_sync = 120*fe / p
print('n_sync = {:.0f} rpm'.format(n_sync))









    



n_sync = 720 rpm

(b)

The slip at rated load is:

$$s = \frac{n_\text{sync} - n_m}{n_\text{sync}} \cdot 100\%$$



In [7]:

    
s = (n_sync - n_m) / n_sync
print('''
s = {:.2f} %
=========='''.format(s*100))









    



s = 6.94 %
==========

(c)

The motor is operating in the linear region of its torque-speed curve, so the slip at $\frac14$ load will approximatly be:



In [8]:

    
s_c = 1/4 * s
s_c









    Out[8]:





0.0174

The resulting speed is:



In [9]:

    
n_m_c = (1 - s_c) * n_sync
print('''
n_m_c = {:.0f} r/min
================='''.format(n_m_c))









    



n_m_c = 708 r/min
=================

(d)

The electrical frequency at $\frac14$ load is:

$$f_r = sf_e$$



In [10]:

    
fr_d = s_c * fe
print('''
fr_d = {:.2f} Hz
=============='''.format(fr_d))









    



fr_d = 1.04 Hz
==============