Excercises Electric Machinery Fundamentals

Chapter 2 Problem 2-7



In [1]:

    
%pylab inline
%precision 4









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.4f'

Description

A 30-kVA 8000/230-V distribution transformer has an impedance referred to the primary of

$Z_{EQ} = 20\,\Omega + j100\,\Omega$.

The components of the excitation branch referred to the primary side are

$R_C = 100\,k\Omega$ and
$X_M = 20\,k\Omega$

(a)

If the primary voltage is 7967 V and the load impedance is $Z_L = (2.0 + j0.7)\Omega$, what is the secondary voltage of the transformer? What is the voltage regulation of the transformer?

(b)

If the load is disconnected and a capacitor of $X_L = -j3.0\,\Omega$ is connected in its place, what is the secondary voltage of the transformer? What is its voltage regulation under these conditions?



In [2]:

    
Zeq =  20.0 + 100.0j   # [Ohm]
Rc  = 100.0e3          # [Ohm]
Xm  =  20.0e3          # [Ohm]
Zload = 2.0 + 0.7j     # [Ohm]
Xload = -3.0j          # [Ohm] 
Vp = 7967.0            # [V]

SOLUTION

(a)

The easiest way to solve this problem is to refer all components to the primary side of the transformer. The turns ratio is:



In [3]:

    
a = 8000/230.0
a









    Out[3]:





34.7826

Thus the load impedance referred to the primary side is:



In [4]:

    
Zloadp = a**2 * Zload  
Zloadp









    Out[4]:





(2419.6597353497164+846.8809073724007j)

The referred secondary current is $I_s' = \frac{V_P}{Z_{EQ} Z_L'}$:



In [5]:

    
Isp =  Vp/ (Zeq + Zloadp)
print('Isp = {:.3f} A ∠{:.2f}°'.format(
            abs(Isp), angle(Isp, deg=True)))









    



Isp = 3.044 A ∠-21.21°

and the referred secondary voltage is $V_s' = I_s' Z_L'$:



In [6]:

    
Vsp =  Isp * Zloadp
print('Vsp = {:.1f} V ∠{:.2f}°'.format(
            abs(Vsp), angle(Vsp, deg=True)))









    



Vsp = 7804.5 V ∠-1.92°

The actual secondary voltage is thus



In [7]:

    
Vs = Vsp /a
print('Vs = {:.1f} V {:.2f}°'.format(
            abs(Vs), angle(Vs, deg=True)))
print('===================')









    



Vs = 224.4 V -1.92°
===================

And the voltage regulation with $VR = \frac{V_P-V_S'}{V_S'}$ is therefore:



In [8]:

    
VR = (abs(Vp) - abs(Vsp)) / abs(Vsp) * 100
print('VR = {:.2f} %'.format(VR))
print('===========')









    



VR = 2.08 %
===========

(b)

The easiest way to solve this problem is to refer all components to the primary side of the transformer. The turns ratio is:



In [9]:

    
a  # already defined earlier...









    Out[9]:





34.7826

Thus the load impedance referred to the primary side is:



In [10]:

    
Xloadp = a**2 * Xload  
Xloadp









    Out[10]:





-3629.4896030245745j

The referred secondary current is $I_s' = \frac{V_P}{Z_{EQ} X_L'}$:



In [11]:

    
Isp =  Vp/ (Zeq + Xloadp)
print('Isp = {:.3f} A ∠{:.2f}°'.format(
            abs(Isp), angle(Isp, deg=True)))









    



Isp = 2.257 A ∠89.68°

and the referred secondary voltage is $V_s' = I_s' X_L'$:



In [12]:

    
Vsp =  Isp * Xloadp
print('Vsp = {:.2f} V ∠{:.2f}°'.format(
            abs(Vsp), angle(Vsp, deg=True)))









    



Vsp = 8192.60 V ∠-0.32°

The actual secondary voltage is thus



In [13]:

    
Vs = Vsp /a
print('Vs = {:.1f} V ∠{:.2f}°'.format(
            abs(Vs), angle(Vs, deg=True)))
print('====================')









    



Vs = 235.5 V ∠-0.32°
====================

And the voltage regulation with $VR = \frac{V_P-V_S'}{V_S'}$ is therefore:



In [14]:

    
VR = (abs(Vp) - abs(Vsp)) / abs(Vsp) * 100
print('VR = {:.2f} %'.format(VR))
print('============')









    



VR = -2.75 %
============