**What is a Short Circuit Study?**

A short circuit study calculates the short circuit capacity at designated locations within a power system or power delivery infrastructure.

**Why Do We Make Short Circuit Study?**

The resulting data from the Short Circuit Study may also be utilized to select electrical equipment ratings, may serve as the basis for protective device coordination study and Arc-Flash hazard analysis.

**Assumptions for MVA Method Short Circuit Calculations**

**For Short Circuit calculations**

- The Power Utility, generators and all motors are sources of electrical energy or short circuit currents.
- Transformers, reactors and cables limit short circuit currents.
- Capacitors and static loads such as heaters and lighting do not contribute to short circuit current.
- If actual motor impedance are not known
- All motors 37kW or less are lumped and assigned an impedance Z = 25%.
- All motors abover 37kW are lumped and assigned an impedance Z = 17%.

- For motors,
**1 HP = 0.75 kW = 1 KVA**

## Short Circuit KVA of Circuit Elements

**Utility**: KVA_{SC} = Utility FAULT DUTY (KVA)

Example:

Fault Duty = 0.04 pu @ 100MVA

KVA_{SC} = (100/0.04) x 1000 = 2,500,000 kVA

**Generator**: KVA_{SC} = (100 x KVA_{G}) / %Z = KVA_{G} / X"_{d}

Example:

Generator 50 MVA, 11 000 V, X"_{d} = 0.113

KVA_{SC} = (50 x 1000) / 0.113 = 442,478 kVA

**Motor**: KVA_{SC} = (100 x KVA_{M}) / %Z = KVA_{M} / X"_{d}

Example:

Motor 1500 HP, 4000V, FLA = 193, X"_{d} = 0.167.

KVA_{SC} = 1500 / 0.167 = 9000 kVA

**Transformer**: KVA_{SC} = (100 x KVA_{T}) / %Z = KVA_{T} / Z_{pu}

Example:

Transformer 132kV / 11kV, 3PH, 50/56MVA @ 55^{O}C, 66.5 / 74.5 MVA @ 65^{O}C, OA/FA, Z = 9% @ 50MVA

KVA_{SC} = 50 x 1000 / 0.09 = 555,555 kVA

**Reactor**: KVA_{SC} = (1000 x KV^{2}) / Z (ohms)

Example:

Reactor 11 kV, 0.125 ohms

KVA_{SC} = 11^{2} x 1000 / 0.125 = 1,523,520 kVA

**Cable**: KVA_{SC} = (1000 x KV^{2}) / Z (ohms)

Example:

Cable : 3/C - 185 mm^{2}, 400V, 150 m, R = 0.0258 / km, X = 0.027 / km

Z = (0.0258^{2} + 0.027^{2})^{0.5} x 150 / 1000 = 0.0056 ohms

KVA_{SC} = 0.40^{2} x 1000 / 0.0056 = 28,571 kVA

In the next part of this tutorial, we shall be having examples.

Sir Good day,

Regarding the impedance motors that you have mentioned which if the value is not given, we have to assume those impedance rating. What partcular standard it is included on?

You can check IEC60909.

If I have 5 x 50kVA generators, 3 ph, 440V, 60Hz, what should be the busbar s.c rating. I do not have the %Z or X’d

The complex short-circuit MVA method is an alternative to the per unit method if X/R ratios need to be calculated. It is much simpler and less susceptible to errors.

Caution should be applied in applying the kVA method as the X/R ratio at the fault point is not calculated. X/R ratio is important in determining the asymmetrical magnitude of the symmetrical current. The calculated X/R ratio should not be greater than the circuit breaker test X/R ratio. There are cases where a breaker’s symmetrical current rating (at an X/R ratio) are fine but the asymmetrical rating (momentary or closing & latching duties) is not. For estimation or conceptual design, MVA may work. For detailed design, recommend to use the per-unit method with separate X and R values.

hello i’m selamat from indonesia

how to calculate MVAsc of transformer 160 KVA, 12/0.38 kV, Z% = 4, Exciting current (Io%) 2.3, Conection transformer DYn11 and how to get R1, L1 and R2, L2 , Rm and Xm information of them?

Thanks a lot before

hello im ryan.

im having some difficulty in assuming what to use as %Z if most of the loads are lightings. in some hand books that iv read. if most of the loads are lightings, we shall assume 50% of the total load shall contribute to the fault current but they didnt say what %Z should be assumed?

in your example of cable short circuit calculation, KVAsc=0.40 2 x 1000/0.0056=28,571 kva. Where is that 0.40 coming from?

in your example of cable short circuit calculation, KVAsc = 0.40 2 x 1000 / %Z. Where is that 0.40 comes from.

Hi Kurt,

It’s my bad.

I did not realize that when I changed my theme, the superscripts & subscripts are not working. I have fixed it now and you might be able to get a better idea on the calculation.

Anyway the previously shown as 0.40 2 should have actually been 0.40^2. The 0.40 came for the 400 volts or 0.40 kV.

MVA method provides the 3-phase symmetrical fault current. Check your standards (IEEE, IEC, AS, etc) to determine the methodology in the selection of circuit breakers based on symmetrical fault currents.

sure its the basis, but im not sure if the result in MVA method is Ipeak and how to get the Isym or vice versa. making and breaking of breakers are base on Ipeak and Isym.

Wondering if the result in MVA method can be used in selecting ACB’s/breakers. The making and breaking rating of ACB’s/breakers are expressed in KA.

Surely the fault current results from the calculation should be the basis in selecting breaker breaking capacity.

sir, can you please help me in my thesis preparation?just badly needed of your expertise. i need to prepare for my PEE. Thanks a lot, i’ll appreciate if you can reply.

This is a good reference. How can use this to calculate the available short circuit fault current if our source is only a prime power generator of 650 KVA, 415/240V,3PH?

Ask the manufacturer for the X’d of the generator. Typically X’d ranges from 17% to 20%. Assuming it is 17%, the fault MVA will be 650/0.17 = 3824KVA or 3.824MVA