**Short Circuit Current Calculation:**

**What is the importance of Short circuit current calculation?**

Short circuit current calculation studies are important for every Electrical engineer to estimate the value of fault currents and hence to find the following details.

- To determine the switchgear rating for
**protective relaying** - To determine the voltage drop during the starting of large motors.
- To determine the rating of the protective equipment, MCCs, and
**Breaker**panels.

**Why MVA method is preferable compared to other methods?**

We can also find the short circuit parameters using Ohmic and Per Unit Methods. The conversion formulas used for both these methods are complex and not easy to memorize.

In the MVA method, it is not necessary to convert impedances from one voltage to another as in the Ohmic method. And it is not required any base MVA value as in the Per Unit method. So the calculations using MVA Method are simple hand calculations and also quick.

## Short circuit current calculation using MVA method:

The following is the procedure

- Convert the typical
**single line diagram**to an equivalent MVA diagram. - Simplification of an equivalent MVA diagram into a single short-circuits MVA value at the point of fault.

This can be easily achieved with the following three steps.

**Step-1:** **Convert all single line components to short circuit MVA’s.**

In practice, the MVA method is used by separating the circuit into components and calculating each component with its own infinite bus. Equipment such as generators, motors, transformers, etc., is normally given their own MVA and impedance or reactance ratings.

The short circuit MVA of each component in the given SLD is equal to its MVA rating divided by its own per unit impedance or reactance.

**Step-2:** **Combine individual MVA values.**

1) Series MVA’s are combined as resistances in parallel.

2) Parallel MVA’s are added arithmetically.

**Step-3: Reduce MVA diagram into a single short-circuits MVA value at the point of fault.**

Reduce MVA diagram by simplifying the equivalent MVA diagram using the MVA quantities obtained in the previous step.

**Short Circuit Current Calculation-Example: **

Consider an example Power system network as shown in the below SLD.

**SLD Components Data:**

**1.Generator-A:**

10 MVA, 10% reactance

**2.Generator-B:**

5 MVA, 7.5% reactance

**3.Transformer:**

15 MVA, 5% reactance, 11/33KV

**4.Transmission Line:**

Impedance Z = 5+j20 ohms

**For this Network find the short circuit MVA and fault current values fed to the symmetrical fault between phases if it occurs at points F1 and F2 that is**

**At the****high voltage terminals of the transformer**F1**At the****load end of the transmission line**F2.

## Solving the given Network using MVA Method:

Let’s see how to calculate the fault current using MVA Method at Points F1 and F2.

The solution is given below, solved using the above mentioned procedure in three steps.

**Step-1:**

**Convert all single line components in the given SLD to short circuit MVA’s.**

**1. Generator-A:**

10 MVA, 10% reactance

**Short Circuit of MVA of Generator-A MVA1 = MVA/Sub-transient reactance of generator in per unit**

MVA1 =10/0.1=100

**2.** **Generator-B:**

5 MVA, 7.5% reactance

**Short Circuit of MVA of Generator-B MVA2 = MVA/Sub-transient reactance of generator in per unit**

MVA2 =5/0.075=66.67

**3. Transformer:**

15 MVA, 5% reactance, 11/33KV

**Short Circuit of MVA of Transformer MVA3 = MVA/Impedance in per unit**

MVA3=15/0.05=300

**4. Transmission Line:**

Impedance Z = 5+j20 ohms

Z= sqrt(5*5+20*20)

Z= sqrt(25+400)

Z=sqrt(425)

Z=20.615 ohms

Voltage rating of Transmission line= 33KV

**Short Circuit of MVA of Transmission Line MVA4 = KV ^{2}/Impedance in ohms**

MVA4= 33*33/20.615 = 52.83

**SLD Equivalent MVA Diagram:**

Using the above Short circuit MVA values of each component in the SLD, draw the MVA diagram as shown below.

**Step-2:**

**Combine individual MVA values.**

Two Generators are connected in parallel.

**Combined MVA1-2**= MVA1 + MVA2 = 100 + 66.67 =166.67

**Step-3:**

**Reduce MVA diagram into a single short-circuits MVA value at the point of fault to find SC MVA and SC Current values.**

**1. Short Circuit MVA and Short Circuit Current Calculation for Fault F1:**

MVA1-2 is in series with MVA-3

**Total Short circuit MVA up to the fault F1**= Combined MVA1-2-3= (MVA1-2 * MVA3)/ (MVA1-2 + MVA3)

MVA1-2-3= (166.67 *300)/ (166.67 +300) =107.144

**Total Short circuit MVA up to the fault F1**=107.144**Short Circuit Current at F1 = Total Short circuit MVA up to the fault*1000/ (1.732 * KV) =**107.144*1000/ (1.732*33) =1874.58A

**2. Short Circuit MVA and Short Circuit Current Calculation for Fault F2:**

MVA1-2-3 and MVA-4 are in series.

**Total Short circuit MVA up to the fault F2**= Combined MVA1-2-3-4= (MVA1-2-3* MVA4)/ (MVA1-2-3 + MVA4)

MVA1-2-3-4= (107.144*52.83)/ (107.144 +52.83) =35.38

**Total Short circuit MVA up to the fault F2**=35.38**Short Circuit Current at F2 = Total Short circuit MVA up to the fault*1000/ (1.732 * KV) =**35.38*1000/ (1.732*33) =619A

In this way, we can find the short circuit MVA and current values for any type of network and any type of fault using the simple MVA method quickly and easily.

PhalanxguardJust a note, you’ve multiplied 33kV by sqrt 3, but 33kV is already the line voltage, so I don’t think this is necessary. Please correct me if I’m wrong!

phanibabuPost authorThe power for a three-phase is P = √3× V line × I line × cos(theta). Hence above explanation is correct.

Faisalits true that for three phase power system S(MVA) = √3 x Vline x Iline

its derived from the following:

S1 = Vphase1 x Iline

S2 = Vphase2 x Iline

S3 = Vphase3 x Iline

Total S = S1 + S2 + S3 = 3 (Vphase x Iline)

we know Vphase = Vline / √3

Total S = 3 (Vline/√3 x Iline)

but 3/√3 = √3

therefore total S (MVA) = √3 x Vline x Iline