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
===================   =============