MVA Method Short Circuit Calculation
A Short Circuit Study is an important tool in determining the ratings of electrical equipment to be installed in a project. It is also used as a basis in setting protection devices. Computer software simplifies this process however, in cases where it is not available, alternative methods should be used. The per-unit and ohmic method are very tedious manual calculation. These hand calculations are very prone to errors due to so many conversion required. In per unit, base conversion is a normal part of the calculation method while in ohmic method, complex entities conversion.The easy way to do hand calculation is the MVA method.
In this example, we shall be presenting a short circuit study of a power system. Motors are are already lumped with ratings 37kW and below assigned an impedance value of 25% while larger motors are 17%. A 4MVA generator is also included into the system to augment the utility.
Figure 1
Utility: 33KV, 250 MVAsc （the 250MVA in this example is just assumed. Ask your utility for the actual fault level at your point of connection）
Transformer 1: 10 MVA, 33/11KV, 9% Z
11KV Bus
Generator: 4MVA, X\"d = 0.113
Transformer 2: 5 MVA, 11/6.6KV, 7% Z
Motor 1: 5MVA (Lumped), 17% Z
6.6KV Bus
Transformer 3: 2 MVA, 6.6KV/400V, 6% Z
Motor 3: 6.8 MVA (Lumped), 17% Z
400V Bus
Motor 4: 300 KVA (Lumped), 17% Z
Motor 5: 596 KVA (Lumped), 25% Z
In the event of a short circuit, the sources of short circuit current are
1. Utility
2. Generators
3. Motors
Static loads such as heaters and lighting do not contribute to short circuit.
The \"Equivalent MVA\" are:
Transformers and Motors
Generators
Cables and Reactors
In Figure 1, I have calculated the Equivalent MVAs of each equipment, writing it below the ratings.
Figure 2
Utility: MVAsc = 250MVA
Transformer 1: MVAsc = 10 / 0.09 = 111.11 MVA
11KV Bus
Generator: MVAsc = 4 / 0.113 = 35.4 MVA
Transnformer 2: MVAsc = 5 / 0.07 = 71.43 MVA
Motor 1: MVAsc = 5 / 0.17 = 29.41 MVA
6.6KV Bus
Transformer 3: MVAsc = 2 / 0.06 = 33.33 MVA
Motor 3: MVAsc = 6.8 / 0.17 = 40 MVA
400V Bus
Motor 4: MVAsc = 0.3 / 0.17 = 1.76 MVA
Motor 5: MVAsc = 0.596 / 0.25 = 2.38 MVA
Figure 3
Upstream Contribution
Starting from the utility, combine MVAs writing each one above the arrows.
At Transformer 1:
MVAsc @ 33KV = 250 MVA
MVAsc @ 11KV = 1/ (1 / 250 + 1 /111.11) = 76.87 MVA
At Transformer 2:
MVAsc @ 11KV = 76.87 + 35.4 + 29.41 = 141.68 MVA
MVAsc @ 6.6KV = 1/ (1 / 141.68 + 1 / 71.43) = 47.49 MVA
At Transformer 3:
MVAsc @ 6.6KV = 47.49 + 40 = 87.49 MVA
MVAsc @ 400V = 1/ (1 / 87.49 + 1 / 33.33) = 24.14 MVA
At 400V Motors
Motor 3: MVAsc = 24.14 x 1.76 / ( 1.76 + 2.38 ) = 10.26 MVA
Motor 4: MVAsc = 24.14 x 2.38 / ( 1.76 + 2.38 ) = 13.88 MVA
Downstream Contribution
Starting from the bottom (400V Bus), I combined MVAs writing each one below the arrows. In this bus, the motor contribution to short circuit is the sum of the MVAs of the lumped motors Motor 3 and Motor 4.
At Transformer 3:
MVAsc @ 400V = 1.76 + 2.38 = 4.14 MVA
MVAsc @ 6.6KV = 1/ (1 / 4.14 + 1 / 33.33) = 3.68 MVA
At Transformer 2:
MVAsc @ 6.6KV = 3.68 + 40 = 43.68 MVA
MVAsc @ 11KV = 1/ (1 / 43.68 + 1 / 71.43) = 27.11 MVA
At Transformer 1:
MVAsc @ 11KV = 27.11 + 29.41 + 35.4 = 91.92 MVA
Note: Two downstream plus the generator contribution.
MVAsc @ 33KV = 1/ (1 / 91.92 + 1 /111.11) = 50.3 MVA
To determine the Faults Current at any bus on the power system, add the MVA values above and below the arrows. The sum should be the same on any branch.
Example:
11 KV Bus:
From Transformer 1: MVAsc = 76.87 + 91.92 = 168.79 MVA
From Generator : MVAsc = 35.4 + 133.39 = 168.79 MVA
From Transformer 2: MVAsc = 141.68 + 27.11 = 168.79MVA
From Motor 1: MVAsc = 139.38 + 29.41 = 168.79 MVA
This is a check that we have done the correct calculation.
Ifault @ 11KV = 168.79 / (1.732 x 11) = 8.86 kA
All we have done above are three phase faults, you may ask, how about single phase faults?
For single phase faults, positive sequence, negative sequence and zero sequence impedances need to be calculated.
If = 3 (I1 + I2 + I0)
Examining the circuit in Figure 1, at the 400V Bus, on Transformer 3 contributes to the zero sequence current.
For transformers, the negative sequence and zero sequence impedance are equal to the positive sequence impedance.
Z1 = Z2 = Z0 or
MVA1 = MVA2 = MVA0
At the 400V Bus
1 / MVAsc =1/3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0)
1 / MVAsc = 1/3 (1 / 28.28 + 1 / 28.28 + 1 / 33.33 )
MVAsc =3 x 9.93 = 29.79 MVA
If = 29.79 / (1.732 x 0.4) = 43 kA
At 6.6KV Bus
1 / MVAsc = 1/3 (1 / MVAsc1 + 1 / MVAsc2 + 1 / MVAsc0)
1 / MVAsc = 1/3 (1 / 91.17 + 1 / 91.17 + 1 / 71.43)
MVAsc = 3 x 27.83 = 83.49 MVA
If = 83.49 / (1.732 x 6.6) = 7.26 kA
Conclusion:
This example illustrates that using the MVA Method of Short Circuit
Calculation, it will be very easy to calculate the fault current at any node within a power system.
1. can this method account for NGR grounded system say at 6.6kv level?
Yes you can. The NGR is a resistor so it’s MVA will be .
But remember that it does not contribute to the fault current but instead limits it.
