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 loads could technically be added to the network, it is not possible to add a power or current load to a short-circuited bus. This is because having I = (S/U)* with U=0 cannot be solved.

>>> import roseau.load_flow as rlf

>>> def create_network() -> rlf.ElectricalNetwork:
...     # Define the ground and make it the reference of potentials
...     ground = rlf.Ground(id="Gnd")
...     rlf.PotentialRef(id="PRef", element=ground)
...     # Create three LV buses
...     bus1 = rlf.Bus(id="Bus1", phases="abcn", nominal_voltage=400)
...     bus2 = rlf.Bus(id="Bus2", phases="abcn", nominal_voltage=400)
...     bus3 = rlf.Bus(id="Bus3", phases="abcn", nominal_voltage=400)
...     # Connect the neutral of the first bus to the ground
...     rlf.GroundConnection(ground=ground, element=bus1)
...     # Create a voltage source at the first bus
...     rlf.VoltageSource(id="Src", bus=bus1, voltages=400 / rlf.SQRT3)
...     # Add LV lines
...     lp1 = rlf.LineParameters.from_catalogue("U_AL_3x240+95")
...     lp2 = rlf.LineParameters.from_catalogue("U_AL_3x150+150")
...     rlf.Line(id="Line1", bus1=bus1, bus2=bus2, parameters=lp1, length=1.0, ground=ground)
...     rlf.Line(id="Line2", bus1=bus2, bus2=bus3, parameters=lp2, length=2.0, ground=ground)
...     # Create the network
...     en = rlf.ElectricalNetwork.from_element(bus1)
...     return en
...

>>> # Create the 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["Bus3"].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, 2.5934809855243657e-13)
>>> en.res_lines

line_id	phase	current1	current2	power1	power2	potential1	potential2	series_losses	series_current	violated	loading	max_loading	ampacity
Line1	a	338.25+35.15j	-338.25-35.1j	78115.2-8117j	-63659.5+17359.1j	230.94-0j	191.47-31.45j	14455.7+9251.67j	338.25+35.12j	False	0.88	1	388
Line1	b	-338.13-35.08j	338.17+35.06j	46059.1+63575.1j	-31612.4-54338.4j	-115.47-200j	-76.01-168.56j	14446.8+9245.92j	-338.15-35.07j	False	0.88	1	388
Line1	c	-0.12-0.07j	0.08+0.05j	0-32.34j	-0+21.11j	-115.47+200j	-115.46+200.02j	0+0j	-0.1-0.06j	False	0	1	388
Line1	n	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	False	0	1	388
Line2	a	338.25+35.1j	-338.22-35.05j	63659.5-17359.1j	-16017.7+35850.7j	191.47-31.45j	57.72-100.02j	47641.8+18501.7j	338.24+35.07j	True	1.13	1	300
Line2	b	-338.17-35.06j	338.22+35.05j	31612.4+54338.4j	16017.7-35850.7j	-76.01-168.56j	57.72-100.02j	47630.1+18497.1j	-338.2-35.04j	True	1.13	1	300
Line2	c	-0.08-0.05j	0+0j	0-21.11j	0+0j	-115.46+200.02j	-115.45+200.03j	0+0j	-0.04-0.02j	False	0	1	300
Line2	n	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	False	0	1	300

Looking at the line results of the second bus of the line Line2, which is Bus3 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 also possible to create short-circuits between more than two phases. Let’s create a short-circuit between phases “a”, “b”, and “c”.

>>> en = create_network()
>>> en.buses["Bus3"].add_short_circuit("a", "b", "c")
>>> en.solve_load_flow()
(1, 3.979039320256561e-13)
>>> en.res_lines

line_id	phase	current1	current2	power1	power2	potential1	potential2	series_losses	series_current	violated	loading	max_loading	ampacity
Line1	a	358.46-160.14j	-358.46+160.18j	82782.9+36982.8j	-63514.6-24659.9j	230.94-0j	173.32-8.66j	19268.3+12331.7j	358.46-160.16j	True	1.01	1	388
Line1	b	-317.92-230.37j	317.95+230.34j	82782.9+36982.8j	-63514.6-24659.9j	-115.47-200j	-94.16-145.77j	19268.3+12331.7j	-317.94-230.35j	True	1.01	1	388
Line1	c	-40.54+390.51j	40.51-390.53j	82782.9+36982.8j	-63514.6-24659.9j	-115.47+200j	-79.16+154.43j	19268.3+12331.7j	-40.52+390.52j	True	1.01	1	388
Line1	n	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	False	0	1	388
Line2	a	358.46-160.18j	-358.46+160.22j	63514.6+24659.9j	0-0j	173.32-8.66j	-0+0j	63514.6+24665.8j	358.46-160.22j	True	1.31	1	300
Line2	b	-317.95-230.34j	317.98+230.32j	63514.6+24659.9j	0+0j	-94.16-145.77j	-0+0j	63514.6+24665.8j	-317.98-230.32j	True	1.31	1	300
Line2	c	-40.51+390.53j	40.48-390.54j	63514.6+24659.9j	-0-0j	-79.16+154.43j	-0+0j	63514.6+24665.8j	-40.48+390.54j	True	1.31	1	300
Line2	n	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	False	0	1	300

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()
>>> # The ground MUST be passed as a keyword argument
... gnd = en.grounds["Gnd"]
... en.buses["Bus3"].add_short_circuit("a", ground=gnd)
>>> en.solve_load_flow()
(1, 4.985456492079265e-13)
>>> en.res_lines

line_id	phase	current1	current2	power1	power2	potential1	potential2	series_losses	series_current	violated	loading	max_loading	ampacity
Line1	a	358.46-160.14j	-358.46+160.18j	82782.9+36982.8j	-63514.6-24659.9j	230.94+0j	173.32-8.66j	19268.3+12331.7j	358.46-160.16j	True	1.01	1	388
Line1	b	0.12-0.07j	-0.08+0.05j	0-32.34j	-0+21.11j	-115.47-200j	-115.49-200j	0+0j	0.1-0.06j	False	0	1	388
Line1	c	-0.12-0.07j	0.08+0.05j	0-32.34j	-0+21.11j	-115.47+200j	-115.46+200.02j	0+0j	-0.1-0.06j	False	0	1	388
Line1	n	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	False	0	1	388
Line2	a	358.46-160.18j	-358.46+160.22j	63514.6+24659.9j	0-0j	173.32-8.66j	0+0j	63514.6+24665.8j	358.46-160.22j	True	1.31	1	300
Line2	b	0.08-0.05j	-0-0j	0-21.11j	0+0j	-115.49-200j	-115.51-200j	0+0j	0.04-0.02j	False	0	1	300
Line2	c	-0.08-0.05j	0+0j	0-21.11j	0+0j	-115.46+200.02j	-115.45+200.03j	0+0j	-0.04-0.02j	False	0	1	300
Line2	n	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	0+0j	False	0	1	300

>>> en.res_grounds

ground_id	potential
Gnd	0+0j

Here the potential at phase “a” of bus Bus3 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 constant-power or constant-current load on a short-circuited bus:

>>> try:
...     load = rlf.PowerLoad("Load", bus=en.buses["Bus3"], powers=[10, 10, 10])
... except RoseauLoadFlowException as e:
...     print(e)
...
The power load 'Load' is connected on bus 'Bus3' that already has a short-circuit.
It makes the short-circuit calculation impossible. [bad_short_circuit]

At least two phases or a phase and a ground must be given when creating a short-circuit:

>>> try:
...     en.buses["Bus3"].add_short_circuit("a")
... except RoseauLoadFlowException as e:
...     print(e)
...
For the short-circuit on bus 'Bus3', expected at least two phases or a phase and a ground.
Only phase 'a' is given. [bad_phase]