Short-circuit calculations

Let’s see how we can make a short-circuit calculation.

We will start by creating a simple network composed of two LV lines. As usual with short-circuit calculations, we won’t add any loads.

Note

While impedance and current loads could technically be added to the network, it is not possible to add a power load on the same bus as the one we want to compute the short-circuit on. This is because having I = (S/U)* with U=0 is impossible.

>>> import numpy as np
... import roseau.load_flow as rlf

>>> def create_network():
...     # Create three buses
...     source_bus = rlf.Bus(id="sb", phases="abcn")
...     bus1 = rlf.Bus(id="b1", phases="abcn")
...     bus2 = rlf.Bus(id="b2", phases="abcn")
...     # Define the reference of potentials
...     ground = rlf.Ground(id="gnd")
...     pref = rlf.PotentialRef(id="pref", element=ground)
...     ground.connect(bus=source_bus)
...     # Create a LV source at the first bus
...     un = 400 / np.sqrt(3)
...     source_voltages = [un, un * np.exp(-2j * np.pi / 3), un * np.exp(2j * np.pi / 3)]
...     vs = rlf.VoltageSource(
...         id="vs", bus=source_bus, phases="abcn", voltages=source_voltages
...     )
...     # Add LV lines
...     lp1 = rlf.LineParameters.from_geometry(
...         "U_AL_240",
...         line_type=rlf.LineType.UNDERGROUND,
...         conductor_type=rlf.ConductorType.AL,
...         insulator_type=rlf.InsulatorType.PVC,
...         section=240,
...         section_neutral=120,
...         height=rlf.Q_(-1.5, "m"),
...         external_diameter=rlf.Q_(50, "mm"),
...     )
...     line1 = rlf.Line(
...         id="line1", bus1=source_bus, bus2=bus1, parameters=lp1, length=1.0, ground=ground
...     )
...     lp2 = rlf.LineParameters.from_geometry(
...         "U_AL_150",
...         line_type=rlf.LineType.UNDERGROUND,
...         conductor_type=rlf.ConductorType.AL,
...         insulator_type=rlf.InsulatorType.PVC,
...         section=150,
...         section_neutral=150,
...         height=rlf.Q_(-1.5, "m"),
...         external_diameter=rlf.Q_(40, "mm"),
...     )
...     line2 = rlf.Line(
...         id="line2", bus1=bus1, bus2=bus2, parameters=lp2, length=2.0, ground=ground
...     )
...     # Create network
...     en = rlf.ElectricalNetwork.from_element(source_bus)
...     return en
...

>>> # Create network
... en = create_network()

Phase-to-phase

We can now add a short-circuit. Let’s first create a phase-to-phase short-circuit:

>>> en.buses["b2"].add_short_circuit("a", "b")

Let’s run the load flow, and get the current results.

Note

If you get an error saying roseau.load_flow.RoseauLoadFlowException: The license is not valid. Please use the activate_license(key="..."), make sure you follow the instructions in Solving a load flow.

Note

All the following tables are rounded to 2 decimals to be properly displayed.

>>> en.solve_load_flow()
(1, 3.339550858072471e-13)
>>> en.res_branches

branch_id	phase	type	current1	current2	power1	power2	potential1	potential2
line1	a	line	374.19+65.47j	-374.2-65.22j	86414.44-15119.6j	-69427.92+23726.69j	230.94-0j	190.79-30.15j
line1	b	line	-373.43-65.15j	373.71+64.99j	56149.99+67164.05j	-39212.61-58608.72j	-115.47-200j	-75.38-169.94j
line1	c	line	-0.88-0.32j	0.61+0.24j	37.17-214.38j	-22.32+155.56j	-115.47+200j	-116.82+208.22j
line1	n	line	0.16-0.01j	-0.13-0j	0j	-0.17+1.03j	0j	1.38-8.15j
line2	a	line	374.2+65.22j	-374.11-64.94j	69427.92-23726.69j	-15076.23+41188.79j	190.79-30.15j	57.67-100.09j
line2	b	line	-373.71-64.99j	374.11+64.94j	39212.61+58608.72j	15076.23-41188.79j	-75.38-169.94j	57.67-100.09j
line2	c	line	-0.61-0.24j	-0j	22.32-155.56j	-0-0j	-116.82+208.22j	-119.55+224.61j
line2	n	line	0.13+0j	-0j	0.17-1.03j	-0j	1.38-8.15j	4.18-24.45j

Looking at the line results of the second bus of the line line2, which is bus2 where we added the short-circuit, one can notice that:

• the potentials of phases “a” and “b” are equal;

• the currents and powers in phases “a” and “b” are equal with opposite signs, i.e. the sum of the currents is zero;

• the currents and powers in these two phases are very high;

which is expected from a short-circuit.

Multi-phase

It is possible to create short-circuits between several phases, not only two. Let’s first remove the existing short-circuit then create a new one between phases “a”, “b”, and “c”.

>>> en = create_network()
>>> en.buses["b2"].add_short_circuit("a", "b", "c")
>>> en.solve_load_flow()
(1, 6.572520305780927e-13)
>>> en.res_branches

branch_id	phase	type	current1	current2	power1	power2	potential1	potential2
line1	a	line	364.42-152.4j	-364.45+152.64j	84159.75+35195.32j	-62323.26-24107.78j	230.94-0j	169.06-4.66j
line1	b	line	-329.25-298.27j	329.5+298.09j	97671.94+31407.98j	-74421.29-19633.88j	-115.47-200j	-94.56-145.13j
line1	c	line	-35.27+450.66j	35.03-450.73j	94203.88+44984.19j	-73584.22-31005.25j	-115.47+200j	-80.99+156.96j
line1	n	line	0.11-0.01j	-0.08-0.01j	0j	-0.5+0.64j	0j	6.47-7.18j
line2	a	line	364.45-152.64j	-364.48+152.85j	62323.26+24107.78j	3461.67-1626.3j	169.06-4.66j	-6.49+7.18j
line2	b	line	-329.5-298.09j	329.7+297.94j	74421.29+19633.88j	1.41+4300.23j	-94.56-145.13j	-6.49+7.18j
line2	c	line	-35.03+450.73j	34.78-450.79j	73584.22+31005.25j	-3463.08-2673.93j	-80.99+156.96j	-6.49+7.18j
line2	n	line	0.08+0.01j	-0j	0.5-0.64j	-0j	6.47-7.18j	19.44-21.56j

Now the potentials of the three phases are equal and the currents and powers add up to zero at the bus where the short-circuit is applied.

Phase-to-ground

Phase-to-ground short-circuits are also possible. Let’s remove the existing short-circuit and create a new one between phase “a” and ground.

>>> en = create_network()
>>> # ground MUST be passed as a keyword argument
... en.buses["b2"].add_short_circuit("a", ground=en.grounds["gnd"])
>>> en.solve_load_flow()
(1, 2.464140003155535e-13)
>>> en.res_branches

branch_id	phase	type	current1	current2	power1	power2	potential1	potential2
line1	a	line	95.83-188.13j	-95.86+188.37j	22130.38+43446.19j	-16871.5-29433.8j	230.94+0j	160.32-7.98j
line1	b	line	0.96-0.74j	-0.65+0.52j	36.74-277.43j	-10.48+232.63j	-115.47-200j	-163.66-224.36j
line1	c	line	-0.81-0.43j	0.55+0.33j	8.47-212.03j	-29.32+150.27j	-115.47+200j	-159.37+177.78j
line1	n	line	0.24-0.25j	-0.21+0.22j	0j	4.52+15.58j	0j	-48.11-24.34j
line2	a	line	95.86-188.37j	-95.99+188.69j	16871.5+29433.8j	0j	160.32-7.98j	0j
line2	b	line	0.65-0.52j	0j	10.48-232.63j	0j	-163.66-224.36j	-265.1-275.72j
line2	c	line	-0.55-0.33j	0j	29.32-150.27j	0j	-159.37+177.78j	-252.37+130.63j
line2	n	line	0.21-0.22j	0j	-4.52-15.58j	0j	-48.11-24.34j	-149.45-75.72j

>>> en.res_grounds

ground_id	potential
gnd	0+0j

Here the potential at phase “a” of bus b2 is zero, equal to the ground potential. The currents in the other phases are also zero indicating that the current of phase “a” went through the ground.

Additional notes

The library will prevent the user from making mistakes, for example when trying to add a power load with the short-circuit, or when forgetting parameters.

>>> try:
...     load = rlf.PowerLoad("load", bus=en.buses["b2"], powers=[10, 10, 10])
... except RoseauLoadFlowException as e:
...     print(e)
...
The power load 'load' is connected on bus 'b2' that already has a short-circuit.
It makes the short-circuit calculation impossible. [bad_short_circuit]

>>> try:
...     en.buses["b2"].add_short_circuit("a")
... except RoseauLoadFlowException as e:
...     print(e)
...
For the short-circuit on bus 'b2', expected at least two phases or a phase and a ground.
Only phase 'a' is given. [bad_phase]