Excercises Electric Machinery Fundamentals

Chapter 2 Problem 2-15



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib

Description

An autotransformer is used to connect a 12.6-kV distribution line to a 13.8-kV distribution line. It must be capable of handling 2000 kVA. There are three phases, connected Y-Y with their neutrals solidly grounded.



In [2]:

    
Vl  = 12.6e3 # [V]
Vh  = 13.8e3 # [V]
Sio = 2000e3 # [VA]

(a)

What must the $N_C / N _{SE}$ turns ratio be to accomplish this connection?

(b)

How much apparent power must the windings of each autotransformer handle?

(c)

What is the power advantage of this autotransformer system?

(d)

If one of the autotransformers were reconnected as an ordinary transformer, what would its ratings be?

SOLUTION

(a)

The transformer is connected Y-Y, so the primary and secondary phase voltages are the line voltages divided by 3 . The turns ratio of each autotransformer is given by:

$$\frac{V_H}{V_L} = \frac{N_C + N_{SE}}{N_C}$$



In [3]:

    
a = (Vh/sqrt(3)) / (Vl/sqrt(3))
n_a = 1 / (a-1)  # n_a = Nc/Nse
print('''
Nc/Nse = {:.1f}
=============
'''.format(n_a))









    



Nc/Nse = 10.5
=============

(b)

The power advantage of this autotransformer is:

$$\frac{S_{IO}}{S_W} = \frac{N_C + N_{SE}}{N_{SE}}$$



In [4]:

    
n_b = (10.5 + 1) / 1  # n_b = Sio/Sw
print('Sio/Sw = {:.1f}'.format(n_b))









    



Sio/Sw = 11.5

Since 1/3 of the total power is associated with each phase, the windings in each autotransformer must handle:



In [5]:

    
Sw = Sio / (3*n_b)
print('''
Sw = {:.1f} kVA
==============
'''.format(Sw/1000))









    



Sw = 58.0 kVA
==============

(c)

As determined in (b), the power advantage of this autotransformer system is:



In [6]:

    
print('''
Sio/Sw = {:.1f}
=============
'''.format(n_b))









    



Sio/Sw = 11.5
=============

(d)

The voltages across each phase of the autotransformer are:



In [7]:

    
Vh_p = Vh / sqrt(3)
Vl_p = Vl / sqrt(3)
print('''
Vh_p = {:.0f} V
Vl_p = {:.0f} V
'''.format(Vh_p, Vl_p))









    



Vh_p = 7967 V
Vl_p = 7275 V

The voltage across the common winding ( $N_C$ ) is:



In [8]:

    
Vnc = Vl_p
print('Vnc = {:.0f} V'.format(Vnc))









    



Vnc = 7275 V

and the voltage across the series winding ( $N_{SE}$ ) is:



In [9]:

    
Vnse = Vh_p - Vl_p
print('Vnse = {:.0f} V'.format(Vnse))









    



Vnse = 693 V

Therefore, a single phase of the autotransformer connected as an ordinary transformer would be rated at:



In [10]:

    
print('''
Vnc/Vnse = {:.0f}/{:.0f}   Sw = {:.1f} kVA
===================   =============
'''.format(Vnc, Vnse, Sw/1000))









    



Vnc/Vnse = 7275/693   Sw = 58.0 kVA
===================   =============