7 Examples of Per Unit Method Short Circuit Calculations

The per unit method short circuit calculations, allow engineers to solve power system problems by hand. But also, to verify software outputs.

I’m going to cover the most important calculations for different power system scenarios. You’ll learn how to apply the per unit method to the following sources of short circuit currents:

  1. Electric utility systems
  2. Generators
  3. Synchronous motors
  4. Induction motors

Before we start, let’s talk about short circuit currents and the per unit method. This will highlight why these calculations are so important.

Background on short circuit currents

incoming line into hydroelectric facility

To no one’s surprise, power engineers design power systems to be safe and reliable. This includes protection from short circuits. Because short circuits can cause the following problems if not accounted for in a design:

  • Power outages
  • Interruption of essential facilities and services
  • Equipment damage (e.g. equipment turning into a ball of flames)
  • Personnel injury or even fatalities
  • Facility fires

Clearly, you want to avoid short circuits. But, no matter how careful you are, they do happen from time to time. Some of the causes are the following:

  • Loose electrical connections
  • Rodents in equipment
  • Voltage surges
  • Insulation deterioration
  • Moisture or dust accumulation on equipment
  • The intrusion of objects in electrical equipment (e.g. tools and tape)

Also, the following happens when a short circuit occurs in a power system:

  • The short circuit current will flow from all the connected power sources to the fault location.
  • At the fault location, arcing can lead to fires and burning.
  • System voltages drop in proportion to the short circuit current magnitude. At the fault location, the voltage can drop to zero for a max fault.
  • All equipment carrying short circuit current will experience thermal and mechanical stress. The stress depends on the current magnitude and duration of the current flow.

Basic per unit representations

Let’s establish the typical per unit relationships for 3\phi systems. First, we start with the following base values:

  • Base volts = line-to-line volts
  • Base kVA = three-phase kVA

Next, the following are the basic per unit relationships:

Per unit volts = \frac{Actual \: volts }{Base \: volts}

Per unit amps = \frac{Actual \: amps}{Base \: amps}

Per unit ohms = \frac{Actual \: ohms}{Base \: ohms}

Just as important, the following are the derived base value equations:

Base amps = \frac{Base \: kVA (1000)}{\surd(3) (Base \: volts)} = \frac{Base \: kVA}{\surd(3)(Base \: kV)}

Base ohms = \frac{Base \: volts}{\surd(3) (Base \: amps)} \: or \: = \frac{Base \: kV^{2}(1000)}{Base \: kVA}

Finally, to convert from percent (per) on an old base to per unit on a new base do the following:

X_{pu} = \frac{per \: X}{100}(\frac{New \: Base \: kVA}{Old \: Base \: kVA})(\frac{New \: Base \: volts}{Old \: Base \: volts})^{2}

Important Note: the per unit system has four base values. These are base kVA, base volts, base ohms, and base amps. If you’re given any of the two base values, you can derive the other two.

Commonly, you assign KVA and voltage to a study’s base values. Then, base ohms and amps are derived for each of the voltage ratings in the power system.

#1 Electric utility systems

The electric utility delivers short circuit current from its system generator(s). These are the generators utilities use to generate power for their customers. Of course, these generators don’t directly feed customers.

Rather, a utility connects to its customers through transformers. A transformer simply changes the system voltage and current magnitude. But, they do not generate voltage or current.

So, the short circuit a customer receives through a transformer depends on the following:

  • The transformer’s secondary voltage rating, and impedance.
  • The impedance of the generator(s).
  • The impedance of the circuit from the generator(s) to the transformer’s primary terminals. Also, the impedance of the circuit from the transformer’s secondary terminals to the fault location.

In Section #2, we’ll discuss generators in greater detail.

Per unit method short circuit calculation

Here we convert the 5 below examples to per unit on a 100,000 kVA base.

Example #1: the available 3\phi short circuit is 750,000 kVA or 750 MVA.

X_{pu} = \frac{kVA_{b}}{kVA_{sc}} = \frac{100,000}{750,000} = 0.13

Example #2: the available 3\phi short circuit is 35,500 amps at 12,470 volts.

X_{pu} = \frac{kVA_{b}}{\surd(3)(I_{sc})(kV)} = \frac{100,000}{\surd(3)(35,500)(12.47)} = 0.13

Example #3: the equivalent utility reactance is 0.5 per unit on a 500,000 kVA base.

X_{pu} = (X_{pu \: old})\frac{kVA_{b}}{kVA_{old}} = (0.5) \frac{100,000}{500,000} = 0.1

Example #4: the equivalent utility reactance is 0.55 ohms per phase at 12,470 volts.

X_{pu} = (X) \frac{kVA_{b}}{1000 \times kVA^{2}} = (0.55) \frac{100,000}{1000(12.47)^{2}} = 0.35

Example #5: a 5,000 kVA transformer with an impedance of 7% on its kVA rating. Assume the impedance is all reactance.

X_{pu} = \frac{per \: X_{T}}{100}(\frac{kVA_{b}}{XFMR \: kVA}) = \frac{7}{100}(\frac{100,000}{5,000}) = 1.4

#2 Generators

generator stator

Prime movers drive generators. Some prime mover examples include the following:

  • Turbines
  • Diesel engines
  • Gas turbine
  • Water wheels

Now, what happens when a short circuit occurs on the circuit the generator feeds?

The generator will continue to produce voltage. This is because the generator’s DC field excitation is still maintained. Also, the prime mover will continue to drive the generator at a normal speed.

Think of a hydroelectric facility. Water from the reservoir will continue flowing and turning the turbine. The turbine then activates the generator. As a result, the generator voltage produces a short circuit current. The current flows from the generator to the fault location with the following limits:

  • Generator impedance
  • Impedance of the circuit between the generator and the fault location

Even more, short circuits can happen at a generator’s terminals. In this scenario, the generator’s impedance is the only limit to the outputted current.

For the generator per unit method short circuit calculation, see the next section. Because the generator and synchronous motor calculations are alike.

Important Note: a synchronous generator and synchronous motor are very similar. They’re both synchronous machines. The synchronous generator converts mechanical energy to electrical energy. While the synchronous motor converts electrical energy to mechanical energy. 

#3 Synchronous motors

Synchronous motors are very similar to generators as we just learned. In fact, their construction is near identical.

With a synchronous generator, there’s a stator winding where alternating current flows. This produces a rotating magnetic field. But, there’s also a field excited by direct current.

The machine uses the direct current to produce torque through this second magnetic field. Also, this field allows the rotor to synchronize with the speed of the rotating magnetic field of the stator. So, this other magnetic field needs to be more stationary relative to the main rotating magnetic field.

Now, a normal synchronous motor operates by drawing AC power from its connected line. It then converts electric energy to mechanical energy. When a short circuit happens though, the system voltage drops considerably. The motor then quits delivering energy to the connected mechanical load. In return, the motor slows down.

But, the inertia of the motor rotor and load will continue to drive the synchronous motor. It’s like how the prime mover continues to drive the generator in a hydroelectric facility. So at this point, the synchronous motor acts as a generator. It delivers a short circuit current for many cycles after the short circuit occurs.

Important Note: when a motor is run from an external prime mover, it becomes a synchronous generator. This can happen when the 3\phi power supply goes away and the DC field supply remains. In our case, as the inertia after power shutoff continues spinning the “motor.”

The amount of short circuit produced depends on the following:

  • Synchronous motor impedance
  • Impedance of the circuit between the motor and the fault location

Per unit method short circuit calculation

Here we convert the below example to per unit on a 100,000 kVA base.

Example: a 1,000 HP, 0.8 pf, synchronous motor has a subtransient reactance (X^{''}_{d}) of 20%.

Motor kVA = 0.746 x 1,000/0.80 = 932.5 kVA

X^{''}_{pu} = \frac{per \: X^{''}_{d}}{100}(\frac{kVA_{b}}{Motor \: kVA}) = \dfrac{20}{100}(\frac{100,000}{932.5}) = 21.4

#4 Induction motors

Like the synchronous motor, the induction motor behaves similarly. After a short circuit occurs, the inertia of the motor rotor and load will continue to drive the induction motor.

One big difference exists though. The induction motor doesn’t have a supporting DC field winding. BUT, a flux exists in the induction motor when it normally operates.

The generated field is by induction through the stator, versus a DC winding. This flux then acts in the same way as the DC field winding of a synchronous motor. Also, the motor rotor flux remains normal as long as the external power source applies voltage.

But, when a short circuit occurs, the external voltage source goes away. The flux of the motor rotor can’t instantly change though. The inertia of the rotating parts of the motor will continue to drive the motor. This then leads to a generated voltage in the stator winding.

As a result, a short circuit current flows from the motor to the fault location. This lasts for a few cycles until the rotor flux decays to zero. The amount of short circuit produced depends on the following:

  • Motor impedance
  • Impedance of the circuit between the motor and the fault location

Per unit method short circuit calculation

Here we convert to ohms per phase at 4,160 volts.

Example: a 4160 volt rated motor control center has 2,000 HP at 0.8 pf of connected motor load. The motors have a reactance of 20% on a kVA rating of 2,000.

Motor kVA = 0.746 x 2,000/0.80 = 1,865 kVA

X = \frac{per \: X_{m}}{100}(\frac{kV^{2}1000}{Motor \: kVA}) = \frac{20}{100}(\frac{4.16^{2} \times 1,000}{1,865}) = 1.86 ohms per phase at 4160 volts


Learning per unit method short circuit calculations are super important for power engineers. Because you can better understand short circuit calculations, and power systems in general. This includes knowing how to size a current limiting reactor for a substation.

Just be sure you learn the basics and all the sources of short circuit currents very well. Otherwise, you can make simple hard to catch mistakes.

What are your thoughts on per unit method short circuit calculations? What source of short circuits do you see the most?


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3 thoughts on “7 Examples of Per Unit Method Short Circuit Calculations”

  1. Hello! Thank you for this article! I found it helpful. I do have a couple questions, though:

    1. In examples #6 and #7, it appears that the HP rating was used in place of Motor kVA. Can you please clarify? Shouldn’t this have been converted to kVA first?

    2. In the second “Important Note” I think the energy types are switched. “The synchronous generator converts electric energy to mechanical energy. While the synchronous motor converts mechanical energy to electric energy.” This does not agree with what the article says later about synchronous motors.

    • Hi Krista,

      Great eye!

      1) Yes, it should have been – I made the revision. Generally, what I had originally written is more conservative; thus, why you’ll typically come across this calculation without the conversion.

      2) You’re absolutely right. Typo on my end. Thanks for the catch!

  2. Date: 12-JANUARY-2023 (Thursday)

    Here’s greetings of a MERRY CHRISTMAS and a happy New Year 2023 to all !, Indeed, the PER UNIT Method of calculating or solving short circuit current fault at any point in the electrical power system circuit is realy powerful and relatively simple to apply and one which I have learned to use in my old company and one which is common to Western countries. However, there is another method which is bein gused in the Middle East countries which is based in the IEC-60909 std. and originated and implemnted in Eurpope, which is basically employing Impedance Method referring from Thevenin’s Theorem. But this latter is more complicated to use since there are factors which must be considered before actually finding the various S.C. faults type and values and applying voltage factor c. But for me, I think the PER UNIT Method is more practical and easy and easy to check back the S.C. fault resulting value by changing the KVAbase reference with another value to check back the answers. T.Y. !, by FMSJR.


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