Short Circuit Current Calculation in Substation Design

Quiz

This chapter is meant for explaining the concept of calculating Short-Circuit Current in Electrical Substation Design. It is vital for an electrical substation design engineer to calculate the short-circuit current in the system, as it is a very important factor in making the substation safe and economical.

Read this chapter to learn the basics of short circuit and short circuit current along with its formula and step-by-step calculation in substation design.

What is Short-Circuit Current Calculation in Substation Design?

The process of determining the value of electric current that flows within the electrical system of a substation when two or more electrical conductors are accidentally joined together or connected to the ground providing very low impedance path for current flow, is referred to as Short-Circuit Current Calculation.

The calculation of short-circuit current in substation design is very important as it allows design engineers to select correct size of circuit breakers, CTs, busbars, earthing, and protection settings.

In electrical substation design, we generally calculate the symmetrical three-phase fault to determine worst-case mechanical and thermal stresses, and we also calculate asymmetrical faults using sequence networks.

What is a Short-Circuit?

In electrical systems, a short-circuit is nothing but a fault condition which creates an unwanted low-impedance path for the flow of electric current between two points in the substation system. It can be a connection between two phase conductors or between a phase conductor and ground.

In substation systems, short-circuit can be either bolted, having negligible impedance, or arcing, having a finite impedance.

Short-circuit causes following key consequences in an electrical substation system −

Huge amount of electric current flow
Voltage collapse near the short-circuit location
Mechanical and thermal stress on system equipment

What is Short-Circuit Current?

In electrical substation, the short-circuit current is the amount of electrical current that flows when a short-circuit fault occurs at a particular location in the system.

For a 3-phase bolted short-circuit fault, the short-circuit current is given by supply voltage divided by the Thevenin's equivalent impedance of the system seen from the fault point, i.e.,

$$\mathrm{I_{sc(pu)} \:=\: \frac{1}{Z_{th(pu)}}}$$

This is the short-circuit current in per-unit form, with a pre-fault voltage equal to 1 pu.

In actual RMS value, it can be given as

$$\mathrm{I_{sc} \:=\: I_{sc(pu)} \:\cdot\: I_{base}}$$

Where,

$$\mathrm{I_{base} \:=\: \frac{S_{base}}{\sqrt{3} \:\times\: V_{base}}}$$

Why Short-Circuit Current Calculation is Important?

In electrical substation design, the following are major reasons highlighting the importance of short-circuit calculation −

It helps selecting correct short of circuit breaker, switchgear, and CTs.
It also allows to select correct busbar and conductor sizes as mechanical stresses are directly proportional to square of system current.
It allows for designing effective earthing protection system.
It enables engineers to configure protection relay settings.
It helps to assess thermal and dynamic withstand capacities of transformer, cable, and equipment ratings.
It helps ensuring compliance with standards and safe operation of the substation.

Short-Circuit Current Formula

For three-phase symmetrical short-circuit fault, the short-circuit current can be calculated by using Thevenin's theorem as

$$\mathrm{I_{sc} \:=\: \frac{V_{th}}{Z_{th}}}$$

Here,

$\mathrm{V_{th}\:}$ is the Thevenin's equivalent phase-voltage at the fault point in the system
$\mathrm{Z_{th}\:}$ is the Thevenin's equivalent impedance seen from the fault point in the system.

For asymmetrical short-circuit faults, the short-circuit current can be calculated using sequence network formulae as,

For balanced three-phase fault −

$$\mathrm{I_{3ph} \:=\: \frac{V}{Z_{1}}}$$

Here, V is the phase-to-neutral voltage.

For single line to ground fault −

$$\mathrm{I_{LG} \:=\: \frac{3V}{Z_{0} \:+\: Z_{1} \:+\: Z_{2}}}$$

For line-to-line fault −

$$\mathrm{I_{LL} \:=\: \frac{V}{Z_{1} \:+\: Z_{2}}}$$

Here, $\mathrm{\:Z_{0}Z_{1},\:}$ and $\mathrm{\:Z_{2}\:}$ are the zero, positive, and negative sequence impedances respectively.

Calculation of Short-Circuit Current in Substation Design

The step-by-step process for calculating the short-circuit current in substation design is explained below −

Step 1 - Firstly, collect following system data

Nominal bus voltage
Rating and %Z of transformer
Rating and sub-transient reactance of generator
Length and impedance of feeders and lines
Short-circuit MVA
Ratings of circuit breakers and CTs
Per unit data from equipment nameplates

Step 2 - Select a common base-MVA and base-voltage −

Now, select a common base-MVA $\mathrm{\:(S_{base})\:}$ which is typically equal to rating of largest equipment and select a common base-voltage $\mathrm{\:(V_{base})\:}$ which is equal to nominal bus voltage.

Step 3 - Convert all impedances to the selected common base MVA and base voltage as follows −

For changing base MVA only −

$$\mathrm{Z_{pu(new)} \:=\: Z_{pu(old)} \:\times\: \frac{S_{base(new)}}{S_{base(old)}}}$$

For changing base voltage only −

$$\mathrm{Z_{pu(new)} \:=\: Z_{pu(old)} \:\times\: \left(\frac{V_{base(new)}}{V_{base(old)}}\right)^{2}}$$

For changing both base MVA and base voltage −

$$\mathrm{Z_{pu(new)} \:=\: Z_{pu(old)} \:\times\: \frac{S_{base(new)}}{S_{base(old)}} \:\times\: \left(\frac{V_{base(new)}}{V_{base(old)}}\right)^{2}}$$

Step 4 - Now, construct the equivalent sequence network −

For 3-phase faults, we need positive sequence network only, hence we just find series sum of positive sequence impedances up to the fault point in the system.
For asymmetrical faults i.e., line to ground fault or double line fault, or double line to ground fault, we need to construct zero, positive, and negative sequence networks and have to interconnect them according to symmetrical component procedure.

Step 5 - Determine Zth (pu) at the fault location by adding series impedances

For three-phase faults,

$$\mathrm{Z_{th(pu)} \:=\: Z_{src(pu)} \:+\: Z_{trans(pu)} \:+\: Z_{line(pu)} \:+\: \cdots}$$

Step 6 - Calculate the per-unit short-circuit fault current and convert into amperes

Per unit fault current, with pre-fault voltage equals to 1 pu −

$$\mathrm{I_{sc(pu)} \:=\: \frac{1}{Z_{th(pu)}}}$$

Converting to amperes,

$$\mathrm{I_{sc} \:=\: I_{sc(pu)} \:\times\: I_{base} \:=\: I_{sc(pu)} \:\times\: \frac{S_{base}}{\sqrt{3}V_{base}}}$$

Step 7 - Calculate the required short-circuit MVA

$$\mathrm{S_{sc}(MVA) \:=\: \sqrt{3}V_{LL}(kV) \:\times\: I_{sc}(kA)}$$

After getting theoretical understanding of short-circuit current calculation in substation design. Let us now discuss a solved numerical example to get practical understanding of the same.

Numerical Example

An electrical substation has an incoming supply at 33 kV with a short-circuit level of 1000 MVA at the 33 kV busbar. The substation has a 33/11 kV transformer of rating 10 MVA and with a percent impedance of 10% on its 10 MVA base.

Calculate the three-phase bolted short-circuit fault current at the 11 kV (low-voltage) busbar. Use a common base $\mathrm{\:S_{base} \:=\: 10\:MVA}$.

Solution â Given data,

Short-circuit level at 33 kV busbar $\mathrm{S_{sc(33kv)} \:=\: 1000\: MVA}$

Transformer Ratings = 33/11 kV, 10 MVA, $\mathrm{\:\%Z_{tr}\:}$ = 10% = 0.10 on 10 MVA base

$$\mathrm{S_{base} \:=\: 10\:MVA}$$

Incomer impedance on the common 10 MVA and 33 kV base in pu:

$$\mathrm{Z_{in(pu)} \:=\: \frac{S_{base}}{S_{sc(33kv)}} \:=\: \frac{10}{1000} \:=\: 0.01\:pu}$$

Referring incomer impedance to 11 kV bus, i.e. same $\mathrm{\:S_{base}\:}$ and different voltage base:

$$\mathrm{Z_{in(11kv)} \:=\: Z_{in(33kV)} \:\times\: \left(\frac{V_{33}}{V_{11}} \right)^{2}}$$

$$\mathrm{\Rightarrow\: Z_{in(11kV)} \:=\: 0.01 \:\times\: \left(\frac{33}{11} \right)^{2} \:=\: 0.01 \:\times\: 9 \:=\: 0.09\: pu}$$

Now, the impedance of transformer on 11 kV base:

As a transformer appears the same in pu on both sides when $\mathrm{\:S_{base}\:}$ is common and voltage bases follow the voltage ratio. Hence,

$$\mathrm{Z_{tr(pu)} \:=\: 0.10}$$

Thus, the equivalent Thevenin's impedance at the 11 kV busbar will be,

$$\mathrm{Z_{th(11kV,\:pu)} \:=\: Z_{tr(pu)} \:+\: Z_{in(11kV)} \:=\: 0.10 \:+\: 0.09 \:=\: 0.19\: pu}$$

Then, short-circuit fault current in per unit and amperes will be,

In per unit form,

$$\mathrm{I_{sc(pu)} \:=\: \frac{1}{Z_{th(11kV,\:pu)}} \:=\: \frac{1}{0.19} \:=\: 5.263\: pu}$$

The base current at 11 kV,

$$\mathrm{I_{base} \:=\: \frac{S_{base}}{\sqrt{3}V_{base}} \:=\: \frac{10\: MVA}{\sqrt{3} \:\times\: 11\: kV} \:=\: 524.87\: A}$$

Hence, the short-circuit fault current at 11 kV will be,

$$\mathrm{I_{sc} \:=\: I_{sc(pu)} \:\times\: I_{base} \:=\: 5.263 \:\times\: 524.87 \:=\: 2762.39\: A}$$

Therefore, the three-phase bolted short-circuit fault current at the 11 kV (low-voltage) busbar in the given substation is 2.762 kA.

Also, the short-circuit MVA at 11 kV side will be,

$$\mathrm{S_{sc}(MVA) \:=\: \sqrt{3}V_{LL}(kV) \:\times\: I_{sc}(kA)}$$

$$\mathrm{\Rightarrow\: S_{sc}(MVA) \:=\: \sqrt{3} \:\times\: 11 \:\times\: 2.762 \:=\: 52.62\: MVA}$$

From this calculation, we can interpret that the circuit breakers and CTs on the 11 kV side of the substation must have interrupting/thermal capability above 2.8 kA.

Conclusion

In conclusion, the short-circuit current is one of the very important considerations in substation design which helps electrical design engineers to select correct sizes of circuit breakers, CTs, busbars, conductors, earthing and protection system. In short, it plays an important role in designing a safe and economical substation.

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