# Excercises Electric Machinery Fundamentals

## Problem 6-4



In [1]:

%pylab inline




Populating the interactive namespace from numpy and matplotlib



### Description

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]



#### (a)

• The shaft speed $n_m$

#### (b)

• The output power in Watts

#### (c)

• The load torque $\tau_\text{load}$ in Newton-metres

#### (d)

• The induced torque $\tau_\text{ind}$ in Newton-metres

#### (e)

• The rotor frequency in Hertz

### SOLUTION

#### (a)

The synchronous speed of this machine 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 = 3000 r/min



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))




n_m = 2850 r/min
================



#### (b)

The output power in Watts is 50 kW (stated in the problem).

#### (c)

$$\tau_\text{load} = \frac{P_\text{OUT}}{\omega_m}$$


In [5]:

w_m = n_m *(2.0*pi/1.0) * (1/60)
print('''




===================



#### (d)

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]:

51300.0


$$\tau_\text{ind} = \frac{P_\text{conv}}{\omega_m}$$


In [7]:

tau_ind = Pconv / w_m
print('''
tau_ind = {:.1f} Nm
=================='''.format(tau_ind))




tau_ind = 171.9 Nm
==================



#### (e)

The rotor frequency is:

$$f_r = sf_{se}$$


In [8]:

fr = s*fse
print('''
fr = {:.1f} Hz
==========='''.format(fr))




fr = 2.5 Hz
===========