In [1]:
%pylab inline
A 50-kW, 460-V, 50-Hz, two-pole induction motor has a slip of 5 percent when operating a full-load conditions. At full-load conditions, the friction and windage losses are 700 W, and the core losses are 600 W. Find the following values for full-load conditions:
In [2]:
fse = 50.0 # [Hz]
p = 2
Pout = 50e3 # [W]
s = 5e-2
Pfw = 700.0 # [W]
Pcore = 600.0 # [W]
Pmisc = 0.0 # [W]
In [3]:
n_sync = 120*fse / p
print('n_sync = {:.0f} r/min'.format(n_sync))
Therefore, the shaft speed is:
$$n_m = (1-s) n_\text{sync}$$
In [4]:
n_m = (1 - s) * n_sync
print('''
n_m = {:.0f} r/min
================'''.format(n_m))
In [5]:
w_m = n_m *(2.0*pi/1.0) * (1/60)
tau_load = Pout / w_m
print('''
tau_load = {:.1f} Nm
==================='''.format(tau_load))
The induced torque can be found as follows:
$$P_\text{conv} = P_\text{OUT} + P_\text{F\&W} + P_\text{core} + P_\text{misc}$$Remember that the $P_\text{core}$ losses can not really be assigned to the stator or the rotor side only, but to both. This basically means that if the $P_\text{core}$ losses are given they should be considered anyway.
In [6]:
Pconv = Pout + Pfw + Pcore + Pmisc
Pconv
Out[6]:
In [7]:
tau_ind = Pconv / w_m
print('''
tau_ind = {:.1f} Nm
=================='''.format(tau_ind))
In [8]:
fr = s*fse
print('''
fr = {:.1f} Hz
==========='''.format(fr))