Short Circuit Current Calculation-MVA Method

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.

  1. To determine the switchgear rating for protective relaying
  2. To determine the voltage drop during the starting of large motors.
  3. 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 on Three phase Transmission Line

Short circuit current calculation using MVA method:

The following is the procedure

  1. Convert the typical single line diagram to an equivalent MVA diagram.
  2. 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.

One Line Diagram of given Power system Network
One Line Diagram

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

  1. At the high voltage terminals of the transformer F1
  2. 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 = KV2/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.

SLD Equivalent MVA Diagram to find Short circuit current calculation
SLD Equivalent MVA Diagram

Step-2:

Combine individual MVA values.

Two Generators are connected in parallel.

Combined MVA Diagram to find Short circuit current calculation
Combined MVA Diagram

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

Reduced MVA Diagram for Fault-F1
Reduced MVA Diagram for Fault-F1

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.

Reduced MVA Diagram for Fault-F2
Reduced MVA Diagram for Fault-F2

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.

7 thoughts on “Short Circuit Current Calculation-MVA Method

  1. Phalanxguard

    Just 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!

    Reply
    1. phanibabu Post author

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

      Reply
  2. Faisal

    its 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

    Reply
  3. BAKANGANA MABUNA EDDY

    This study case is very simple and good for the purpose of explaining the method. Question, can you connect an induction motor at the end and explain how the fault at F1 will be calculated taking into account the contribution of that motor?

    Reply
    1. phani babu Post author

      Hi,,,
      Thanks for asking. The End load can be anything..by using the MVA method you can calculate the short circuit current in the same method as explained above.

      Reply
  4. Manideep

    Hii sir , ur explanation is good but I have a doubt that what is the short circuit current calculation by kva method

    Reply
    1. phani babu Post author

      Hi Manideep,
      Both KVA and MVA short circuit current calculation methods are same. If you take KVA as the base value instead of MVA it is KVA short circuit current calculation method.

      Reply

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