ENS5361 Power Systems 2
Design Project: Short-Circuit Calculation in PowerFactory
Due: Friday the 19th of May 2023 at 17:00
Learning Outcomes
By the completion of this project, you will be able to gain familiarity with the basic short-circuit calculation features of the DIgSILENT PowerFactory software.
Instructions
This Microsoft Word document is to be used as a template for the writing up the report of this design project. All requested answers should be typed, and high-resolution images placed if required, all in the indicated areas of this document. Reports with scanned hand-written content will not be accepted!
Show all working. Answers to questions should be expressed clearly.
Make sure the placeholder is replaced by your surname in the name of this Word document before submission.
Submit your report by 5pm, 19 May 2023, via Canvas, by using the plagiarism-prevention service Turnitin. Extensions will not be granted without reasonable ground. Please refer to the ECU Policy on Assessment for details on late submission penalties. Do NOT use an assessment coversheet. A student declaration is now part of the submission process and coversheets impact Turnitin similarity reports.
The total number of available marks is 40 (as indicated).
Introduction
DIgSILENT PowerFactory has been introduced in ENS3206/ENS6143 Power Systems 1 and is still available in the computer labs in Building 5. You can prepare for this project by revisiting the software fundamentals: you can redo the Getting Started tutorials accessible via the Help menu.
In this project, you will learn how to use it for Short-Circuit Calculation. Fault calculations will be executed with different methods and the results will be analysed.
In the first part of the project activity, the equipment of a 20 kV grid shall be verified and dimensioned. You will get to know the most important short-circuit quantities according to IEC 60909.
In the second part of the project activity, you will investigate the grounding of the network and the effect of the transformer neutral. Different treatments of the transformer neutral will be compared.
In the last part of the project activity, you will execute short-circuit calculations according to the complete method and compare the results to IEC 60909.
Verification of Busbars Sizing
Import and activate the project provided ENS5361_2023_START.
Have a look at the project overview. In this project you will investigate a medium-voltage network.
Run the short-circuit calculation cases listed below and enter the results in Table 1:

Place your answer here (2 marks) ?
Table 1
Base case Coupling closed
Max. initial short-circuit power Sk” [MVA]
Max. initial short-circuit power Ik” [kA]
Highest peak short-circuit current ip [kA]
Highest mechanical Loading, Ip of a busbar [%]
Highest thermal Loading, Ith of a busbar [%]
Overloaded network components (loading = 100%)
Configuration allowed?
Base Case: Maximum 3-Phase Short-Circuit Calculation at all busbars according to IEC 60909 for the given network topology.
Click Calculate Short-Circuit in the toolbar.
Set the calculation Method to IEC 60909.
Choose as a Fault Type a 3-Phase Short-Circuit.
Calculate should be set to Max. Short-Circuit Currents.
Select Fault Location At All Busbars.
Execute the calculation and enter the results in Table 1 (use Network Model Manager ? Busbar ? Flexible Data).
Note: In the short-circuit calculation, a distinction is made between two loadings: the thermal and mechanical load. The colouring of the graphic is done according to the higher of the two loadings and can be set up in the Diagram Colouring Scheme for the short-circuit calculation.
Coupling closed: Change the network topology by closing the couplings of the double busbar systems in the network.
Close the circuit breaker of substation SS1 (20kV) to connect its two busbars.
Also close the circuit breaker that couples the busbars in substation SS3 (20kV).
Execute a 3-Phase Short-Circuit according to IEC 60909 at All Busbars again and enter the results in Table 1.
Are there any overloaded elements now?
What is the reason for the higher loading in the second case?

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Because some busbars are overloaded in a certain switching state, their busbar types need to be adapted. Select suitable busbar types from the project library to sufficiently size the busbars.

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Verification of Cable Sizing
In the following, the effect of the short-circuit duration on the thermal loading of the cables will be investigated.
Open the Calculate Short-Circuit command and execute a Max. 3-Phase Short-Circuit according to IEC 60909 on all Busbars and Junction Nodes. Set the Fault Clearing Time (Ith) in the command to 0.3 s. Can you observe any thermal loadings over 100%?
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The short-circuit quantity for the thermal loading of the cables is the thermal equivalent short-circuit current Ith. Display Ith in the result boxes.
In the following analyse the influence of the short-circuit duration on the thermal load capacity of the cables. Are there any overloaded lines at a Fault Clearing Time of 0.6s?
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How does the thermal loading of the lines change in relation to the Fault Clearing Time? Use Comparing of Results (see 19.5 in the User Manual) for analysis.
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Display the following calculation parameters of the lines for further analysis:
c:Ithr (Rated Short-Time (1s) Current (Conductor))
c:Ithrtk (Rated Short-Time Current (Tk))
c:Ithmax (Max. Thermal equivalent Short-Circuit Current)
Note: The rated short-circuit current in data sheets is typically specified for a fault duration of 1s. To be able to determine the thermal load capacity of cables for other fault durations, the rated short-circuit current is adjusted to the fault duration Tk. Accordingly, cables have a higher rated short-circuit current for shorter fault durations.
How is the thermal loading of the lines calculated?
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The lines in the medium-voltage network shall be designed for a max. 3-phase short circuit for the duration of 1s. Execute a Short-Circuit Calculation for the Max. 3-Phase Short-Circuit Currents with a Fault Clearing Time of 1 s.
Analyse the thermal loading of all lines. Which lines exceed their thermal rating? What cross section do they have (Type ? Cable Analysis)?
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Select suitable line types that can withstand a Max. 3-Phase Short-Circuit over a period of 1s. What is the minimum cross-section that the lines should have?
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For the analysis of individual lines, a specific line can also be selected as fault location. Calculate a Max. 3-Phase Short-Circuit on line SS1-SS2-1 in the 20 kV grid, where the short-circuit location should be at 70% of the line length (from the network supply). To define the short-circuit, right-click on the line and select Calculate ? Short-Circuit....
Analyse the influence of the short-circuit location on the thermal loading of the line. Select busshc instead of bus1 or bus2 for Ith in the Variable Selection (in the Bus and Phase drop-down menu). Use Comparing of Results for analysis. Please note that the bus shc corresponds to the short circuit location. Buses bus 1 and bus 2 represent the beginning and the end of the line, respectively.
At which short-circuit location can the highest thermal loading of the line be observed? Why?
Place your answer here (2 marks) ?
Dimensioning of a Circuit Breaker
In the following a circuit breaker is to be dimensioned. The breaking current for which a circuit breaker must be designed is determined by the time curve of the short-circuit current and the
desired breaking time.
Analyse the influence of the Break Time (in Short-Circuit Calculation) on the breaking current Ib. To do this, display Ib in the results boxes. Use Comparing of Results for the analysis. How does the breaking current Ib change in relation to the break time?
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Observe the breaking current Ib at the substations SS2 and SS3 as a function of the break time. Are there any differences?
Why is Ib at SS3 independent of the break time?
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By assigning a type to a circuit-breaker, it is possible to detect an overloading of the circuit breaker in the event of a short-circuit. Add a new type to the circuit-breaker S0.1.1 of the substation SS1. The circuit breaker has the following data:
Rated Current = 0.4 kA
Rated Breaking Current (under Short-Circuit VDE/IEC) = 10 kA
Peak Short-Circuit Current (under Short-Circuit VDE/IEC) = 25 kA
The circuit breaker should be designed to clear a short-circuit after t = 100 ms. For verification, calculate a Max. 3-Phase Short-Circuit on All Busbars with the corresponding Break Time.
Right-click on substation SS1 and select Show Detailed Graphic of Substation. Display the peak loading (m:BrkIpload:bus1) and breaking loading (m:BrkIbload:bus1) in the result box of the circuit breaker.
Is the circuit breaker sufficiently dimensioned?
Place your answer here (1 mark) ?
Single-Phase Short-Circuit and Floating Neutral
Single-phase short-circuits occur particularly frequently in medium-voltage networks. The type
of network grounding significantly influences the single-phase fault. In this part of the project,
different connections of the transformer star neutral will be analysed. A floating neutral is most likely to be used in small medium-voltage networks.
Navigate to the page Basic Data ? Grounding/Neutral Conductor and make sure that the LV Side Star Point of the two pairs of 110/20 kV transformers at substation SS1 and SS3 are Not connected.
Execute a Single Phase to Ground short-circuit at SS1 first, then at SS3 (use User Selection). Use the IEC 60909 method.
Note the resulting initial short-circuit current Ikss at SS1 as well as the voltages of phase A, B and C in Table 2:

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Table 2
Floating Neutral Ground-Fault Compensation Solidly Grounded Neutral
Ikss:A [kA]
Ikss:B [kA]
Ikss:C [kA]
U:A [kV]
U:B [kV]
U:C [kV]
What is the value of the real and imaginary part of the zero-sequence impedance (R0, X0) at both fault locations (SS1 and SS3)? Enter the results in Table 3:

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Table 3
Substation SS1 Substation SS2
R0 [ohm]
X0 [ohm]
Single-Phase Short-Circuit and Ground-Fault Compensation
For the compensation of a single phase to ground fault, a compensation coil can be installed at the neutral of the transformer.
The zero-sequence reactances determined in Table 3 represent the capacities of the lines to earth. The current can flow via these capacities in the event of a single-phase fault to the ground.
By installing a coil at the neutral of the transformers, this capacitive ground fault current can be compensated.
First, install a coil at the star point of one of the transformers at SS1.
Open the transformer T-SS1.
Navigate to the page Basic Data ? Grounding/Neutral Conductor and connect the Star Point of the transformer on the LV Side.
Enter the internal grounding impedance of the transformer in a way that the network is compensated. Therefore, use the zero-sequence impedance (R0, X0) at the short-circuit busbar from Table 3. To compensate the current that flows via the capacities of the lines, the reactance of the coil must equal a third of the zero-sequence impedance.
Calculate a single phase to ground short-circuit at SS1 again and check if the fault current is smaller.
Enter the values observed at substation SS1 in Table 2.
What is the reason for the differences compared to the network with floating neutrals?
Place your answer here (2 marks) ?
With parallel transformers, as in substation SS1, both of the neutrals are grounded and thus loaded, not only one. Determine the grounding impedance of the two parallel neutrals in order to compensate the ground fault current.
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Resistance, Re:
Reactance, Xe:
Single-Phase Short-Circuit and Solidly Grounded System
The problem with previous grounding concepts is that the currents are very small in the event of a fault. This can lead to problems when detecting faults, as the protection devices cannot detect them. For this reason, a large number of networks are operated with low-resistance neutral grounding. In the event of a short-circuit, the current is set to be higher than the load current. This is also done via the neutral.
Connect the star point of all transformers to ground without any impedance.
Enter the values for SS1 in Table 2.
The Complete Method and The Full-Size Converter Model
While the short-circuit calculation according to IEC 60909 is used for maximum short-circuit currents and the dimensioning of equipment, the complete method can be used to determine the short-circuit currents more precisely.
In contrast to the short-circuit calculation according to IEC 60909, the fault location has a significant impact on the short-circuit current contribution of inverter-based generators (Full-Size Converter), such as wind turbines. By considering a sensitivity factor (K Factor), the contribution of Full-Size Converters can be altered depending on the fault location to simulate their realistic behaviour. The short-circuit current contributions of wind turbines to short-circuits at different fault locations will be analysed in this section.
Select the Short-Circuit Model Full size converter for the two wind turbines WT and WT 2 in the network – on the page Short-Circuit Complete - with the following data:
K Factor = 2
Max. Current = 1 p.u.
Calculate a Max. 3-Phase Short-Circuit according to the complete method at substation S-Wind.
Examine the short-circuit contributions of wind turbine WT for the fault locations S-Wind and SS2 (10 kV) (busbar at the generator Gen SS2). Note down the results in Table 4.

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Table 4
S-Wind SS2 (10 kV)
Initial short-circuit current (m:Ikss) in kA
Voltage at the terminal (n:u) in p.u.
Investigate the impact of the K Factor in the short-circuit model full-size converter. Use a K Factor of 1 and repeat the calculation according to Table 5.

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Table 5
S-Wind SS2 (10 kV)
Initial short-circuit current (m:Ikss) in kA
Voltage at the terminal (n:u) in p.u.
What influence does the fault location have on the short-circuit contribution of the wind turbines?
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How does the K Factor affect the injected short-circuit current?
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Comparison of the Complete Method and IEC 60909
In the next part of the exercise the complete method will be compared with the calculation
according to IEC 60909.
Calculate a Max. 3-Phase Short-Circuit at the busbar SS2 (10 kV) with both methods.
Observe the sub-transient short-circuit current m:Ikss of the wind turbine WT and enter the results in Table 6.

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Table 6
IEC 60909 Complete method
Initial short-circuit current (m:Ikss) in kA
Voltage at the terminal (n:u) in p.u.
How do the two calculation methods compare?
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Calculation of Multiple (Simultaneous) Faults
A typical application is the calculation of double ground faults:
Calculate a simultaneous Single Phase to Ground Fault at the 20 kV lines SS1-SS2-1 (fault on phase a) and SS1-SS2-2 (fault on phase b) by applying the option Multiple Faults for the short-circuit method complete.
For this mark both lines and select the right-mouse option Calculate ? Multiple Faults… After that double-click on each one of the listed Short-Circuit Event and select the fault type.
Afterwards press Close. The short-circuit dialog appears. Note that the Method complete, and the option Multiple Faults are selected automatically. Press Execute.
List the current distribution after the calculation.
Place your answer here (1 mark) ?
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