4. The Fault Power Method of Short-Circuit Calculation

The so-called MVA methods, applied by many experienced protection specialists who have mastered per unit and ohmic methods, employs fault powers to compute short-circuit currents. We have already used the method to some extent in the earlier examples involving utility short-circuit contributions, and we develop it at length here. The beauty of it lies in the fact that you use circuit element information in almost the same form that it appears on nameplates with almost no ohmic or per unit conversions. It is applied by calculating the admittance of each component of a circuit with its own infinite bus in terms of MVA. We can then combine pairs of circuit element MVAs in series and parallel according to the product-over-sum rule from basic network theory for combining the admittance and/or impedance of two circuit elements. Series MVA combinations are computed like impedances in parallel. Parallel MVA combinations are computed like impedances in series. In the following development, superscripts will indicate a specific circuit element and/or iterative fault power, and subscripts will indicate a positive, negative, or sequence fault power.

Series: MVAl and MVA2 = (MVAl xMVA2)(MVAl + MVA2)

Parallel: MVAl and MVA2 = MVAl + MVA2

The MVA method is best illustrated by example. Given the circuit of Figure 7, we want to determine the three-phase fault current at F:

Step 1: Convert all circuit elements to short-circuit MVAs. The short- circuit MVA of each circuit element is equal to its MVA rating divided by its own per unit impedance or reactance.

For the utility: 1000/1 = 1000 MVA

For the utility feeder: (34.5)2/5 = 238 MVA

For the utility transformer: 15/.07 = 214

For the customer motor: 5/0.2 = 25

Step 2: Combine MVAs. Since we have more than two circuit elements, we perform the calculations iteratively.

MVAl and MVA2 MVAx

1922381000

2381000

Let 192 MVA be the new MVAl. Then

MVAl and MVA2 MVAx

101214192

214192

Going this far with the calculation will allow you to determine the three-phase fault current at the 13.8KV bus without motor contribution.

Figure 7: MVA method example, MVAs with subscripts indicate sequence fault powers, MVAs with superscripts indicate iteration number.

Step 3: Convert MVA to symmetrical fault current

Ax

xI f 4430

2.133

10001011 (LLL without motor contribution)

We assume that fault current on the load side of the feeder breaker is the same as the fault current on the bus. But we want the fault on a feeder with a motor back-feeding fault current into it so we must combine the MVAs in parallel by adding them thus

MVAl + MVA4 = 101 + 25 = 126

Ax

xIsc 5526

2.133

1000126 (LLL with motor contribution)

At this point, it should be obvious how quickly the method may be applied, and, because of its iterative nature, how it lends itself to a computer solution. The method does not require a common MVA base as required in per unit methods. It is not necessary to convert impedances from one voltage to another as required by the ohmic method. Best of all, you do not need to deal with anything but large whole numbers. You may apply the method to compute single-line-to-ground, double- line-to-ground, and other shunt faults as well. Referring again to the circuit of Figure 7, we know that the fault at the 13.2-kV bus is 126 MVA. Assuming that the positive and negative impedances are equal, we can say that the positive sequence fault power is equal to the negative sequence fault power, so that

MVAl = MVA2 = 126

Now a single-line-to-ground fault on the 13.2-kV bus would have only the transformer and the motor contributing to zero sequence MVAs. The delta connection on the secondary of the transformer blocks any zero sequence power contribution from the utility. Therefore, our MVA block diagram may be redrawn to indicate flow of zero sequence fault power only. See Figure 8. MVAo

trans = MVAl = MVA 2 = 214 Assuming that the transformer zero sequence reactance is equal to its positive and negative sequence reactance is another common assumption in industrial practice. The zero sequence reactance of a motor is about one half its positive zero sequence reactance. Therefore

Figure 8 : Zero sequence fault power flow.

MVAomot = 5/0.1= 50 MVA

The total zero sequence fault power then is equal to the sum of the motor and transformer fault powers because of the parallel connection.

MVAotrans+MVAo

mot= 214 + 50 = 264

The single-line-to-ground fault power is obtained by the upper connection diagram shown in Figure 4c. This connection diagram follows from symmetrical component theory. Since these are three branches in parallel, the simplest approach is to take one branch out of the circuit and solve for its MVA value and then multiply the value by 3.

MVA1 and MVA2 = 126126

126126x= 63 MVA

MVA1,2 newand MVAo =

26463

26463x= 51 MVA

MVA= 3 x 51 = 153 MVA

Ax

xIs 6710

2.133

1000153lg (SLG with motor contribution and no neutral

impedance)

you wanted to limit the flow of ground-fault current with impedance of, for instance, 1.0 ohm, you could reformulate the network of Figure 9 with a reactor MVA of

MVA1740.1

)2.13( 2

Then MVAx

xMVA 11817451

174513

Ax

xIs 5189

2.133

1000181lg (SLG with motor contribution and neutral impedance)

Figure 9 MVA method-limiting fault-current example