three-phase.pythree-phase.py
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#!#
#!# =================================================
#!#  Three-phased Current: Y and Delta configurations
#!# =================================================
#!#
#!# This examples shows the computation of the voltage for the Y and Delta configurations.
#!#

####################################################################################################

import math

import numpy as np
import matplotlib.pyplot as plt

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from PySpice.Unit import *

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#!# Let use an European 230 V / 50 Hz electric network.

frequency = 50@u_Hz
w = frequency.pulsation
period = frequency.period

rms_mono = 230
amplitude_mono = rms_mono * math.sqrt(2)

#!# The phase voltages in Y configuration are dephased of :math:`\frac{2\pi}{3}`:
#!#
#!# .. math::
#!#  V_{L1 - N} = V_{pp} \cos \left( \omega t \right) \\
#!#  V_{L2 - N} = V_{pp} \cos \left( \omega t - \frac{2\pi}{3} \right) \\
#!#  V_{L3 - N} = V_{pp} \cos \left( \omega t - \frac{4\pi}{3} \right)
#!#
#!# We rewrite them in complex notation:
#!#
#!# .. math::
#!#  V_{L1 - N} = V_{pp} e^{j\omega t} \\
#!#  V_{L2 - N} = V_{pp} e^{j \left(\omega t - \frac{2\pi}{3} \right) } \\
#!#  V_{L3 - N} = V_{pp} e^{j \left(\omega t - \frac{4\pi}{3} \right) }

t = np.linspace(0, 3*float(period), 1000)
L1 = amplitude_mono * np.cos(t*w)
L2 = amplitude_mono * np.cos(t*w - 2*math.pi/3)
L3 = amplitude_mono * np.cos(t*w - 4*math.pi/3)

#!# From these expressions, we compute the voltage in delta configuration using trigonometric identities :
#!#
#!# .. math::
#!#   V_{L1 - L2} = V_{L1} \sqrt{3} e^{j \frac{\pi}{6} } \\
#!#   V_{L2 - L3} = V_{L2} \sqrt{3} e^{j \frac{\pi}{6} } \\
#!#   V_{L3 - L1} = V_{L3} \sqrt{3} e^{j \frac{\pi}{6} }
#!#
#!# In comparison to the Y configuration, the voltages in delta configuration are magnified by
#!# a factor :math:`\sqrt{3}` and dephased of :math:`\frac{\pi}{6}`.
#!#
#!# Finally we rewrite them in temporal notation:
#!#
#!# .. math::
#!#  V_{L1 - L2} = V_{pp} \sqrt{3} \cos \left( \omega t + \frac{\pi}{6} \right) \\
#!#  V_{L2 - L3} = V_{pp} \sqrt{3} \cos \left( \omega t - \frac{\pi}{2} \right) \\
#!#  V_{L3 - L1} = V_{pp} \sqrt{3} \cos \left( \omega t - \frac{7\pi}{6} \right)

rms_tri = math.sqrt(3) * rms_mono
amplitude_tri = rms_tri * math.sqrt(2)

L12 = amplitude_tri * np.cos(t*w + math.pi/6)
L23 = amplitude_tri * np.cos(t*w - math.pi/2)
L31 = amplitude_tri * np.cos(t*w - 7*math.pi/6)

#!# Now we plot the waveforms:
figure = plt.figure(1, (20, 10))
plt.plot(t, L1, t, L2, t, L3,
         t, L12, t, L23, t, L31,
         # t, L1-L2, t, L2-L3, t, L3-L1,
)
plt.grid()
plt.title('Three-phase electric power: Y and Delta configurations (230V Mono/400V Tri 50Hz Europe)')
plt.legend(('L1-N', 'L2-N', 'L3-N',
            'L1-L2', 'L2-L3', 'L3-L1'),
           loc=(.7,.5))
plt.xlabel('t [s]')
plt.ylabel('[V]')
plt.axhline(y=rms_mono, color='blue')
plt.axhline(y=-rms_mono, color='blue')
plt.axhline(y=rms_tri, color='blue')
plt.axhline(y=-rms_tri, color='blue')
plt.show()

#fig# save_figure(figure, 'three-phase.png')

8.4.1. Three-phased Current: Y and Delta configurationsΒΆ

This examples shows the computation of the voltage for the Y and Delta configurations.

import math

import numpy as np
import matplotlib.pyplot as plt

from PySpice.Unit import *

Let use an European 230 V / 50 Hz electric network.

frequency = 50@u_Hz
w = frequency.pulsation
period = frequency.period

rms_mono = 230
amplitude_mono = rms_mono * math.sqrt(2)

The phase voltages in Y configuration are dephased of \(\frac{2\pi}{3}\):

\[\begin{split}V_{L1 - N} = V_{pp} \cos \left( \omega t \right) \\ V_{L2 - N} = V_{pp} \cos \left( \omega t - \frac{2\pi}{3} \right) \\ V_{L3 - N} = V_{pp} \cos \left( \omega t - \frac{4\pi}{3} \right)\end{split}\]

We rewrite them in complex notation:

\[\begin{split}V_{L1 - N} = V_{pp} e^{j\omega t} \\ V_{L2 - N} = V_{pp} e^{j \left(\omega t - \frac{2\pi}{3} \right) } \\ V_{L3 - N} = V_{pp} e^{j \left(\omega t - \frac{4\pi}{3} \right) }\end{split}\]
t = np.linspace(0, 3*float(period), 1000)
L1 = amplitude_mono * np.cos(t*w)
L2 = amplitude_mono * np.cos(t*w - 2*math.pi/3)
L3 = amplitude_mono * np.cos(t*w - 4*math.pi/3)

From these expressions, we compute the voltage in delta configuration using trigonometric identities :

\[\begin{split}V_{L1 - L2} = V_{L1} \sqrt{3} e^{j \frac{\pi}{6} } \\ V_{L2 - L3} = V_{L2} \sqrt{3} e^{j \frac{\pi}{6} } \\ V_{L3 - L1} = V_{L3} \sqrt{3} e^{j \frac{\pi}{6} }\end{split}\]

In comparison to the Y configuration, the voltages in delta configuration are magnified by a factor \(\sqrt{3}\) and dephased of \(\frac{\pi}{6}\).

Finally we rewrite them in temporal notation:

\[\begin{split}V_{L1 - L2} = V_{pp} \sqrt{3} \cos \left( \omega t + \frac{\pi}{6} \right) \\ V_{L2 - L3} = V_{pp} \sqrt{3} \cos \left( \omega t - \frac{\pi}{2} \right) \\ V_{L3 - L1} = V_{pp} \sqrt{3} \cos \left( \omega t - \frac{7\pi}{6} \right)\end{split}\]
rms_tri = math.sqrt(3) * rms_mono
amplitude_tri = rms_tri * math.sqrt(2)

L12 = amplitude_tri * np.cos(t*w + math.pi/6)
L23 = amplitude_tri * np.cos(t*w - math.pi/2)
L31 = amplitude_tri * np.cos(t*w - 7*math.pi/6)

Now we plot the waveforms:

figure = plt.figure(1, (20, 10))
plt.plot(t, L1, t, L2, t, L3,
         t, L12, t, L23, t, L31,
         # t, L1-L2, t, L2-L3, t, L3-L1,
)
plt.grid()
plt.title('Three-phase electric power: Y and Delta configurations (230V Mono/400V Tri 50Hz Europe)')
plt.legend(('L1-N', 'L2-N', 'L3-N',
            'L1-L2', 'L2-L3', 'L3-L1'),
           loc=(.7,.5))
plt.xlabel('t [s]')
plt.ylabel('[V]')
plt.axhline(y=rms_mono, color='blue')
plt.axhline(y=-rms_mono, color='blue')
plt.axhline(y=rms_tri, color='blue')
plt.axhline(y=-rms_tri, color='blue')
plt.show()
../../_images/three-phase.png