Excercises Electric Machinery Fundamentals

Chapter 5 Problem 5-10



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib

Description

A synchronous machine has a synchronous reactance of $1.0\,\Omega$ per phase and an armature resistance of $0.1\,\Omega$ per phase.

If $\vec{E}_A = 460\,V\angle-10°$ and $\vec{V}_\phi = 480\,V\angle0°$, is this machine a motor or a generator?
How much power P is this machine consuming from or supplying to the electrical system?
How much reactive power Q is this machine consuming from or supplying to the electrical system?



In [2]:

    
Ea         = 460         # [V]
EA_angle   = -10/180*pi  # [rad]
EA         = Ea * (cos(EA_angle) + 1j*sin(EA_angle))
Vphi       = 480         # [V]
VPhi_angle =   0/180*pi  # [rad]
VPhi       = Vphi*exp(1j*VPhi_angle)
Ra         =   0.1       # [Ohm]
Xs         =   1.0       # [Ohm]

SOLUTION

This machine is a motor, consuming power from the power system, because $\vec{E}_A$ is lagging $\vec{V}_\phi$

It is also consuming reactive power, because $E_A \cos{\delta} < V_\phi$ . The current flowing in this machine is:

$$\vec{I}_A = \frac{\vec{V}_\phi - \vec{E}_A}{R_A + jX_s}$$



In [3]:

    
IA = (VPhi - EA) / (Ra + Xs*1j)
IA_angle = arctan(IA.imag/IA.real)
print('IA = {:.1f} A ∠ {:.2f}°'.format(abs(IA), IA_angle/pi*180))









    



IA = 83.9 A ∠ -12.96°

Therefore the real power consumed by this motor is:

$$P =3V_\phi I_A \cos{\theta}$$



In [4]:

    
theta = abs(IA_angle)
P = 3* abs(VPhi)* abs(IA)* cos(theta)
print('''
P = {:.1f} kW
============'''.format(P/1e3))









    



P = 117.7 kW
============

and the reactive power consumed by this motor is:

$$Q = 3V_\phi I_A \sin{\theta}$$



In [5]:

    
Q = 3* abs(VPhi)* abs(IA)* sin(theta)
print('''
Q = {:.1f} kvar
============='''.format(Q/1e3))









    



Q = 27.1 kvar
=============