Excercises Electric Machinery Fundamentals

Chapter 5 Problem 5-2



In [1]:

    
%pylab inline
%precision %.4g
import cmath









    



Populating the interactive namespace from numpy and matplotlib

Description

Assume that the motor of Problem 5-1 is operating at rated conditions.



In [2]:

    
Vt   = 480       # [V]
PF   =   0.8
fse  =  60       # [Hz]
p    =   8.0
Pout = 400 * 746 # [W]
Xs   =   0.6     # [Ohm]

(a)

What are the magnitudes and angles of $\vec{E}_A$ and $\vec{I}_A$ , and $I_F$?

(b)

Suppose the load is removed from the motor.

What are the magnitudes and angles of $\vec{E}_A$ and $\vec{I}_A$ now?

SOLUTION

(a)

The line current flow at rated conditions is: $$I_L = \frac{P}{\sqrt{3}V_TPF}$$



In [3]:

    
Pin = Pout
il = Pin / (sqrt(3) * Vt * PF)
il # [A]









    Out[3]:





448.6

Because the motor is $\Delta$-connected, the corresponding phase current is:



In [4]:

    
ia = il / sqrt(3)
ia # [A]









    Out[4]:





259

The angle of the current is:



In [5]:

    
Ia_angle = arccos(PF)
Ia_angle /pi *180 # [degrees]









    Out[5]:





36.87



In [6]:

    
Ia = ia * (cos(Ia_angle) + sin(Ia_angle)*1j)
print('''
Ia = {:.0f} A ∠{:.2f}°
=================='''.format(abs(Ia), Ia_angle / pi *180))









    



Ia = 259 A ∠36.87°
==================

The internal generated voltage $\vec{E}_A$ is:

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



In [7]:

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









    



Ea = 587 V ∠-12.2°
==================

The field current is directly proportional to $|\vec{E}_A|$, which = 480V when $I_F = 4 A$. The required field current is:

$$\frac{|\vec{E}_{A2}|}{|\vec{E}_{A1}|} = \frac{I_{F2}}{I_{F1}}$$



In [8]:

    
If1 = 4 # [A]
Ea1 = 480 # [V]
Ea2 = abs(EA)
If2 = (Ea2/Ea1) * If1
print('''
If2 = {:.2f} A
============'''.format(If2))









    



If2 = 4.89 A
============

(b)

When the load is removed from the motor the magnitude of $|\vec{E}_A|$ remains unchanged but the torque angle goes to $\delta = 0°$ . The resulting armature current is:

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



In [9]:

    
delta_b = 0*pi/180 # [rad]
EA_b = abs(EA) *(cos(delta_b) + sin(delta_b)*1j)
EA_b_angle = arctan(EA_b.imag/EA_b.real)

Ia_b = (Vt - EA_b) / (Xs*1j)
Ia_b_angle = arctan(Ia_b.imag/Ia_b.real)  # poosible warning might occur because of division by zero

print('''
EA_b = {:.1f} ∠{:>2.0f}°
Ia_b = {:.1f} ∠{:>2.0f}°
=================='''.format(abs(EA_b), EA_b_angle/pi*180, 
                             abs(Ia_b), Ia_b_angle/pi*180))









    



EA_b = 586.6 ∠ 0°
Ia_b = 177.6 ∠90°
==================






    



/usr/local/lib/python3.5/dist-packages/ipykernel/__main__.py:6: RuntimeWarning: divide by zero encountered in double_scalars