Power loads (P)¶
They represent loads for which the power is considered constant, i.e. it is independent of the voltage.
ZIP equation: \(S = s \times V^0 + 0 \times V^1 + 0 \times V^2 \implies S = \mathrm{constant}\)
Equations¶
The equations are the following (star loads):
\[\begin{split}\left\{
\begin{aligned}
\underline{I_{\mathrm{abc}}} &= \left(\frac{\underline{S_{\mathrm{abc}}}}{\underline{V_{\mathrm{abc}}}
-\underline{V_{\mathrm{n}}}}\right)^{\star} \\
\underline{I_{\mathrm{n}}} &= -\sum_{p\in\{\mathrm{a},\mathrm{b},\mathrm{c}\}}\underline{I_{p}}
\end{aligned}
\right.\end{split}\]
And the following (delta loads):
\[\begin{split}\left\{
\begin{aligned}
\underline{I_{\mathrm{ab}}} &= \left(\frac{\underline{S_{\mathrm{ab}}}}{\underline{V_{\mathrm{a}}}-\underline
{V_{\mathrm{b}}}}\right)^{\star} \\
\underline{I_{\mathrm{bc}}} &= \left(\frac{\underline{S_{\mathrm{bc}}}}{\underline{V_{\mathrm{b}}}-\underline
{V_{\mathrm{c}}}}\right)^{\star} \\
\underline{I_{\mathrm{ca}}} &= \left(\frac{\underline{S_{\mathrm{ca}}}}{\underline{V_{\mathrm{c}}}-\underline
{V_{\mathrm{a}}}}\right)^{\star}
\end{aligned}
\right.\end{split}\]
Example¶
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_((1000 - 300j) * np.ones(3), "VA"))
# Create a network and solve a load flow
en = rlf.ElectricalNetwork.from_element(bus1)
en.solve_load_flow()
# Get the powers of the loads in the network
en.res_loads["power"]
# | | power |
# |:--------------|---------------------------:|
# | ('load', 'a') | 1000-300j |
# | ('load', 'b') | 1000-300j |
# | ('load', 'c') | 1000-300j |
# | ('load', 'n') | -5.57569e-31+ 2.25385e-32j |
# Get the voltages of the network
en.res_buses_voltages.transform([np.abs, ft.partial(np.angle, deg=True)])
# | | ('voltage', 'absolute') | ('voltage', 'angle') |
# |:---------------|--------------------------:|-----------------------:|
# | ('bus1', 'an') | 230.94 | 9.69325e-22 |
# | ('bus1', 'bn') | 230.94 | -120 |
# | ('bus1', 'cn') | 230.94 | 120 |
# | ('bus2', 'an') | 229.414 | -0.113552 |
# | ('bus2', 'bn') | 229.414 | -120.114 |
# | ('bus2', 'cn') | 229.414 | 119.886 |
# Modify the load value to create an unbalanced load
load.powers = rlf.Q_(np.array([5.0, 2.5, 0]) * (1 - 0.3j), "kVA")
en.solve_load_flow()
# Get the powers of the loads in the network
en.res_loads["power"]
# | | power |
# |:--------------|-----------------------:|
# | ('load', 'a') | 5154.28 - 1581.93j |
# | ('load', 'b') | 2494.65 - 668.07j |
# | ('load', 'c') | 0 |
# | ('load', 'n') | -148.93 + 3.78664e-14j |
# Get the voltages of the network
en.res_buses_voltages.transform([np.abs, ft.partial(np.angle, deg=True)])
# | | ('voltage', 'absolute') | ('voltage', 'angle') |
# |:---------------|--------------------------:|-----------------------:|
# | ('bus1', 'an') | 230.94 | 4.05321e-24 |
# | ('bus1', 'bn') | 230.94 | -120 |
# | ('bus1', 'cn') | 230.94 | 120 |
# | ('bus2', 'an') | 215.746 | -0.253646 |
# | ('bus2', 'bn') | 229.513 | -121.963 |
# | ('bus2', 'cn') | 235.788 | 121.314 |