%pylab notebook

"""
Explanation: Excercises Electric Machinery Fundamentals
Chapter 2
Problem 2-15
End of explanation
"""


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

"""
Explanation: 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.
End of explanation
"""


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

"""
Explanation: (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:
$$\frac{V_H}{V_L} = \frac{N_C + N_{SE}}{N_C}$$
End of explanation
"""


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

"""
Explanation: (b)
The power advantage of this autotransformer is:
$$\frac{S_{IO}}{S_W} = \frac{N_C + N_{SE}}{N_{SE}}$$
End of explanation
"""


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

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


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

"""
Explanation: (c)
As determined in (b), the power advantage of this autotransformer system is:
End of explanation
"""


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

"""
Explanation: (d)
The voltages across each phase of the autotransformer are:
End of explanation
"""


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

"""
Explanation: The voltage across the common winding ( $N_C$ ) is:
End of explanation
"""


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

"""
Explanation: and the voltage across the series winding ( $N_{SE}$ ) is:
End of explanation
"""


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

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


