Excercises Electric Machinery Fundamentals

Chapter 5 Problem 5-7



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib

Description

A 208-V Y-connected synchronous motor is drawing 50 A at unity power factor from a 208-V power system. The field current flowing under these conditions is 2.7 A. Its synchronous reactance is $1.6\,\Omega$. Assume a linear open-circuit characteristic.



In [2]:

    
Vt = 208   # [V]
Ia =  50   # [A]
Xs =   1.6 # [Ohm]
Ra =   0   # [Ohm]
PF2 =  0.8
If_1 = 2.7 # [A]

(a)

Find $\vec{V}_\phi$ and $\vec{E}_A$ for these conditions.

(b)

Find the torque angle $\delta$ .

(c)

What is the static stability power limit under these conditions?

(d)

How much field current would be required to make the motor operate at 0.80 PF leading?

(e)

What is the new torque angle in part (d)?

SOLUTION

(a)

The phase voltage of this motor is $V_\phi = 120 V$, and the armature current is $\vec{I}_A = 50\,A \angle 0°$ . Therefore, the internal generated voltage is:

$$\vec{E}_A = \vec{V}_\phi - R_A \vec{I}_A - jX_S \vec{I}_A$$



In [3]:

    
Vphi = Vt / sqrt(3)
EA = Vphi - Ra*Ia - Xs*1j*Ia
EA_angle = arctan(EA.imag/EA.real)
print('''
EA = {:.0f} V ∠{:.1f}°
=================='''.format(abs(EA), EA_angle/pi*180))









    



EA = 144 V ∠-33.7°
==================

(b)

The tor que angle $\delta$ of this machine is



In [4]:

    
delta = EA_angle
print('''
δ = {:.1f}°
=========='''.format(delta/pi*180))









    



δ = -33.7°
==========

(c)

The static stability power limit is given by

$$P_\text{max} = \frac{3V_\phi E_A}{X_S}$$



In [5]:

    
Pmax = (3*Vphi*abs(EA)) / Xs
print('''
Pmax = {:.1f} kW
=============='''.format(Pmax/1000))









    



Pmax = 32.5 kW
==============

(d)

A phasor diagram of the motor operating at a power factor of 0.78 leading is shown below.

Since the power supplied by the motor is constant, the quantity $I_A cos \theta$ , which is directly proportional to power, must be constant. Therefore,



In [6]:

    
theta1 = 0 # [rad]
theta2 = arccos(PF2)
Ia2 = Ia*cos(theta1) / cos(theta2)
IA2 = Ia2 * (cos(theta2)+sin(theta2)*1j)
IA2_angle = theta2
print('IA2 = {:.1f} A ∠{:.2f}°'.format(abs(IA2), IA2_angle/pi*180))









    



IA2 = 62.5 A ∠36.87°

The internal generated voltage required to produce this current would be:

$$\vec{E}_{A2} = \vec{V}_\phi - R_A \vec{I}_{A2} - jX_S \vec{I}_{A2}$$



In [7]:

    
EA2 = Vphi - Ra*IA2 - Xs*1j*IA2
EA2_angle = arctan(EA2.imag/EA2.real)
print('EA2 = {:.0f} V ∠{:.1f}°'.format(abs(EA2), EA2_angle/pi*180))









    



EA2 = 197 V ∠-24.0°

The internal generated voltage $E_A$ is directly proportional to the field flux, and we have assumed in this problem that the flux is directly proportional to the fiel d current. Therefore, the required field current is:

$$I_{F2} = \frac{E_{A2}}{E_{A1}}I_{F1}$$



In [8]:

    
If_2 = abs(EA2)/abs(EA) * If_1
print('''
If_2 = {:.2f} A
============='''.format(If_2))









    



If_2 = 3.69 A
=============

(e)

The new torque angle $\delta$ of this machine is



In [9]:

    
delta2 = EA2_angle
print('''
δ_2 = {:.1f}°
============'''.format(delta2/pi*180))









    



δ_2 = -24.0°
============