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Fault in a network with isolated neutral point

Model description

This example considers a fault in a network with isolated neutral. The model represents a 110/10 kV step-down substation and a 10 kV distribution network with three feeders. At time 0.2 s, a single phase earth fault (SPE) occurs in phase A at the end of cable line (CL) 4 of the third feeder, followed at time 0.5 s by an earth fault in phase B at the end of CL 2 of the first feeder, which creates a loop for fault currents to flow between phases A and B. At time 1 s, 100 ms apart, the circuit breakers on the first and third feeders trip, simulating the operation of the relay protection and eliminating the double earth fault. The example shows the process of running the model from the script development environment using command control and the visualisation of the simulation results. The model measures currents and voltages at the beginning of each feeder. Model appearance:

isolated_neutral_network_fault_1730467623955.png

The power supply is modelled by the Voltage Source (Three-Phase) block. Cable lines are modelled by the Three-Phase PI Section Line block. Transformers are modelled by the Three-Phase Transformer (Two Windings).html) block. Faults are modelled by the Fault (Three-Phase) block, in the settings of this block the type of fault can be selected using the Failure mode drop-down menu. The fault is tripped by the Circuit Breaker (Three-Phase) block, which simulates tripping from the relay protection. The load is modelled by the Wye-Connected Load block. System parameters [1,2]:

Element Parameter
System Balancing node
$U_{ном} = 121 кВ$
TR 1 TDN-16000/110/10
TR 2 TMG-2000/10/0.4
TR 3 TMG-2000/10/0.4
TR 4 TMG-1250/10/0.4
TR 5 TMG-400/10/0.4
TR 6 TMG-630/10/0.4
CL 1 3xAPVV 1x240; $L = 1 км$
CL 2 3xAPVV 1x120; $L = 1 км$
CL 3 3xAPVV 1x95; $L = 2 км$
CL 4 3xAPVV 1x70; $L = 1 км$
CL 5 3xAPVV 1x70; $L = 1.2 км$
Load 1 $S_{нагр} = 1.5+j1.125 МВА$
Load 2 $S_{нагр} = 1.5+j1.125 МВА$
Load 3 $S_{нагр} = 1+j0.8 МВА$
Load 4 $S_{нагр} = 0.35+j0.17 МВА$
Load 5 $S_{нагр} = 0.55+j0.34 МВА$

Running the model

Import the necessary modules for working with graphs and LaTex strings:

In [ ]: 
using Plots
using LaTeXStrings
gr();

Load model:

In [ ]: 
model_name = "isolated_neutral_network_fault"
model_name in [m.name for m in engee.get_all_models()] ? engee.open(model_name) : engee.load( "$(@__DIR__)/$(model_name).engee");

Running a loaded model:

In [ ]: 
results = engee.run(model_name);

Modelling results

To import the simulation results, the recording of the necessary signals has been enabled in advance and their names have been set. Convert the instantaneous values of currents and voltages of the variable results into separate vectors:

In [ ]: 
# вектор времени симуляции
sim_time = results["i_a_1"].time;
# вектора токов и напряжений в точке измерения 1
i_1 = hcat(results["i_a_1"].value,results["i_b_1"].value,results["i_c_1"].value,results["3I0_1"].value);
v_1 = hcat(results["v_a_1"].value,results["v_b_1"].value,results["v_c_1"].value);
# вектора токов и напряжений в точке измерения 3
i_3 = hcat(results["i_a_3"].value,results["i_b_3"].value,results["i_c_3"].value,results["3I0_3"].value);
v_3 = hcat(results["v_a_3"].value,results["v_b_3"].value,results["v_c_3"].value,results["3V0_3"].value);

Graphs of feeder 3 currents and voltages on LV busbars of the substation:

In [ ]: 
p1 = plot(sim_time, v_3[:,1:3]./1e3, label = [L"U_a" L"U_b" L"U_c"],
title = "Напряжения", linecolor = [:orange :green :red], ylabel = "U, кВ", xlabel="Время, c");
p2 = plot(sim_time, i_3[:,1:3], label = [L"I_a" L"I_b" L"I_c"], 
title = "Токи", linecolor = [:orange :green :red], ylabel = "I, А", xlabel="Время, c");
plot(p1, p2, layout=(2,1),legend = true, size = (750,440))
Out[0]:

When phase A is short-circuited to earth, the voltage in this phase relative to earth becomes zero, so that the voltage of the undamaged phases becomes equal to the line voltage. This is one of the disadvantages of insulated neutral networks. At a time of 0.5 s, in the event of an earth fault in phase B at the end of CL 2 of the first feeder, a single-phase fault transitions to a double-phase fault. The voltage of the faulted phases is significantly reduced due to the multiple increase of the fault currents.

Graph of the NP voltage on the substation LV busbars:

In [ ]: 
plot(sim_time, v_3[:,4]./1e3, label = L"3U_0", title = L"3U_0", linecolor = :orange, ylabel = "U, кВ", xlabel="Время, c")
Out[0]:

The oscillogram of the NC voltage corresponds to a short-circuit fault. The NC voltage on the busbars of the substation is used to determine the presence of a short-circuit fault in the network.

Zero sequence current (NSC) graph of the third feeder in the time interval from 0 to 0.4 s:

In [ ]: 
plot(sim_time[1:Int(indexin(0.4, sim_time)[1])],
i_3[1:Int(indexin(0.4, sim_time)[1]),4], title = L"3I_0",
linecolor = :orange, ylabel = "I, А", xlabel="Время, c", legend = false)
Out[0]:

The NC current is caused by capacitive currents between the undamaged phases and earth. If its value is small, as in this example, it is not necessary to trip the fault immediately. Due to the small value of the short-circuit current, the line voltages remain unchanged in magnitude and shifted in phase by an angle of 120°. These circumstances allow the power supply to be maintained until the fault is detected. Due to the delta connection of the windings of all transformers on the 10 kV side of the network, the windings are isolated from the ground and no fault current flows through the transformers.

Graph of NP currents of feeders 1 and 3 in the time interval from 0.4 to 1.5 s:

In [ ]: 
p1 = plot(sim_time[Int(indexin(0.4, sim_time)[1]):end], i_1[Int(indexin(0.4, sim_time)[1]):end,4],
title = L"3I_0(фидер\,1)", ylabel = "I, А", xlabel="Время, c")
p2 = plot(sim_time[Int(indexin(0.4, sim_time)[1]):end], i_3[Int(indexin(0.4, sim_time)[1]):end,4],
title = L"3I_0(фидер\,3)", ylabel = "I, А", xlabel="Время, c")
plot(p1, p2, layout = (2,1), legend=false, size = (750,440))
Out[0]:

When a second earth fault occurs in phase B, a fault current loop appears in phase B for the fault current to flow through earth between phases A and B. The large value of the fault current is due to the small resistance of the current flow loop consisting of the resistances of CL1, CL2, CL4 and TR1.

Addendum

Try to change the following model parameters yourself and investigate how this affects the currents and voltages during short-circuit faults:

  1. double the length of the cable lines;
  2. move the short-circuit block in phase B to busbar TR1.

Conclusion

This example showed the process of a single-phase ground fault to a double-phase ground fault. Engee model command tools were used to load and run the model from the script development environment. The measured currents and voltages were imported into the Workspace from the result variable and plotted on time plots.

References

  1. Reference book on designing of electric networks / edited by D.L. Faibisovich. - 4th edition, revision and addendum - M. : ENAS, 2012. - 376 с. : ill.
  2. Designer's handbook. Cables with cross-linked polyethylene insulation for voltage 6-35 kV. Kamsky Cable Ltd. URL: https://www.kamkabel.ru/netcat_files/userfiles/6-35-www.pdf?ysclid=m2j0odmzen889432059 (accessed 21.10.2024).

Blocks used in example