In [1]:
%pylab inline
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]
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))
In [4]:
n_b = (10.5 + 1) / 1 # n_b = Sio/Sw
print('Sio/Sw = {:.1f}'.format(n_b))
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))
In [6]:
print('''
Sio/Sw = {:.1f}
=============
'''.format(n_b))
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))
The voltage across the common winding ( $N_C$ ) is:
In [8]:
Vnc = Vl_p
print('Vnc = {:.0f} V'.format(Vnc))
and the voltage across the series winding ( $N_{SE}$ ) is:
In [9]:
Vnse = Vh_p - Vl_p
print('Vnse = {:.0f} V'.format(Vnse))
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))