Simplified line¶
If the line is modeled with only series impedance, the model is simplified as there is no coupling with the ground. This is a common model of short lines, typically in distribution networks.
The corresponding diagram is:
Equations¶
With \(\underline{Y} = 0\), the equations become:
\[\begin{split}\left\{
\begin{aligned}
\underline{V_1} - \underline{V_2} &= \underline{Z} \cdot \underline{I_1} \\
\underline{I_2} &= -\underline{I_1}
\end{aligned}
\right.\end{split}\]
Usage¶
To create a simplified line, create an instance of LineParameter without providing y_shunt.
Here is a simplified line connecting a constant power load to a voltage source.
import functools as ft
import numpy as np
import roseau.load_flow as rlf
# Two buses
bus1 = rlf.Bus(id="bus1", phases="abcn")
bus2 = rlf.Bus(id="bus2", phases="abcn")
# A line
lp = rlf.LineParameters(id="lp", z_line=rlf.Q_(0.35 * np.eye(4), "ohm/km"))
line = rlf.Line(id="line", bus1=bus1, bus2=bus2, parameters=lp, length=rlf.Q_(1, "km"))
# A voltage source on the first bus
un = 400 / np.sqrt(3)
voltages = rlf.Q_(un * np.exp([0, -2j * np.pi / 3, 2j * np.pi / 3]), "V")
vs = rlf.VoltageSource(id="source", bus=bus1, voltages=voltages)
# The neutral of the voltage source is fixed at potential 0
pref = rlf.PotentialRef(id="pref", element=bus1, phase="n")
# A power load on the second bus
load = rlf.PowerLoad(
id="load", bus=bus2, powers=rlf.Q_(np.array([5.0, 2.5, 0]) * (1 - 0.3j), "kVA")
)
# The impedance matrix (in Ohm) can be accessed from the line instance
line.z_line
# array(
# [[0.35+0.j, 0. +0.j, 0. +0.j, 0. +0.j],
# [0. +0.j, 0.35+0.j, 0. +0.j, 0. +0.j],
# [0. +0.j, 0. +0.j, 0.35+0.j, 0. +0.j],
# [0. +0.j, 0. +0.j, 0. +0.j, 0.35+0.j]]
# ) <Unit('ohm')>
# For a simplified line, the property `with_shunt` is False and the `y_shunt` matrix is zero
line.with_shunt
# False
line.y_shunt
# array(
# [[0.+0.j, 0.+0.j, 0.+0.j, 0.+0.j],
# [0.+0.j, 0.+0.j, 0.+0.j, 0.+0.j],
# [0.+0.j, 0.+0.j, 0.+0.j, 0.+0.j],
# [0.+0.j, 0.+0.j, 0.+0.j, 0.+0.j]]
# ) <Unit('siemens')>
# Create a network and solve a load flow
en = rlf.ElectricalNetwork.from_element(bus1)
en.solve_load_flow()
# The current flowing into the line from bus1
en.res_branches[["current1"]].transform([np.abs, ft.partial(np.angle, deg=True)])
# | | ('current1', 'absolute') | ('current1', 'angle') |
# |:--------------|---------------------------:|------------------------:|
# | ('line', 'a') | 24.1958 | 16.4456 |
# | ('line', 'b') | 11.3722 | -105.263 |
# | ('line', 'c') | 0 | 0 |
# | ('line', 'n') | 20.628 | 168.476 |
# The current flowing into the line from bus2
en.res_branches[["current2"]].transform([np.abs, ft.partial(np.angle, deg=True)])
# | | ('current2', 'absolute') | ('current2', 'angle') |
# |:--------------|---------------------------:|------------------------:|
# | ('line', 'a') | 24.1958 | -163.554 |
# | ('line', 'b') | 11.3722 | 74.7366 |
# | ('line', 'c') | 0 | 0 |
# | ('line', 'n') | 20.628 | -11.5242 |
# The currents in the series components of the line
en.res_lines[["series_current"]].transform([np.abs, ft.partial(np.angle, deg=True)])
# | | ('series_current', 'absolute') | ('series_current', 'angle') |
# |:--------------|---------------------------------:|------------------------------:|
# | ('line', 'a') | 24.1958 | 16.4456 |
# | ('line', 'b') | 11.3722 | -105.263 |
# | ('line', 'c') | 0 | 0 |
# | ('line', 'n') | 20.628 | 168.476 |
# All currents are equal in magnitude for a simplified lines as no current escapes to the ground.
# The current from bus 2 has an opposite direction to the current from bus 1 as expected.
# The losses of the line can also be accessed. One can remark that there are no shunt losses
en.res_lines[["series_losses"]].transform([np.real, np.imag])
# | | ('series_losses', 'real') | ('series_losses', 'imag') |
# |:--------------|----------------------------:|----------------------------:|
# | ('line', 'a') | 204.904 | -2.66329e-15 |
# | ('line', 'b') | 45.2646 | -8.96306e-16 |
# | ('line', 'c') | 0 | 0 |
# | ('line', 'n') | 148.93 | 7.62657e-16 |
# With a simplified model, all the losses are caused by the series impedance of the line
res_lines = en.res_lines
total_losses = res_lines["power1"] + res_lines["power2"] # total = series + shunt
np.allclose(total_losses, res_lines["series_losses"])
# True