7 Examples of Per Unit Method Short Circuit Calculations

The per unit method short circuit calculations help simplify power systems. They identify if a piece of electrical equipment is safe from short circuits.

I’m going to go over some of these important calculations.

They’ll show you how you can apply the per unit method to various electrical equipment. As well as to different scenarios that you may come across in a power system analysis.

The calculations will be for the following sources of short circuit currents:

  • Electric utility systems
  • Generators
  • Synchronous motors
  • Induction motors

I’m going to also discuss how each of these sources comes into play with short circuits.

Before we get started though, let’s talk a little about short circuit currents. This will help highlight why these calculations are so important.

Background on short circuit currents

incoming line into hydroelectric facility

For starters, engineers aim to design all power systems to be safe and reliable. This includes protection from short circuits.

Because short circuits can wreak havoc if not properly accounted for in the design. The following can happen:

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

Clearly, short circuits are something you want to avoid at all costs. But, no matter how careful you are, they do happen for one reason or another. Some of the reasons include the following:

  • Loose electrical connections
  • Rodents in equipment
  • Voltage surges
  • Insulation deterioration
  • Moisture or dust accumulation on equipment
  • The intrusion of objects in 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 that carries short circuit current will experience thermal and mechanical stress. The stress depends on the current magnitude and duration of the current flow.

Before we dive into the short circuit sources and calculations, let’s touch on the per unit method as well.

Basic per unit representations

It’s important we first establish the basic per unit relationships.

For 3\phi systems, the nominal line-to-line system voltages are typically used as the base voltages.

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

Also, 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}

Now, the following are the derived 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}

Also, to convert from percent (per) on an old base to per unit on a new base use 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. There’s base kVA, base volts, base ohms, and base amps.

If you’re given any of the two values, you can derive the other two.

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

As I touched on earlier, we have four basic sources of short circuit current. Let’s now go over each one with per unit method short circuit calculations included.

Electric utility systems

The electric utility delivers short circuit current from the utility’s system generator(s). These are the generators utilities use to generate power for customers.

Keep in mind, a utility connects to its customers through transformers. Where a transformer simply changes the system voltage and current magnitude. But they don’t generate voltage or current.

Now, the short circuit a customer receives from a transformer depends on the following:

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

In the next section, we’ll discuss generators in greater detail.

Per unit method short circuit calculation

We want to 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.

Thus, 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. For simplicity, 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


generator stator

Prime movers drive generators. Some prime mover examples include:

  • Turbines
  • Diesel engines
  • Gas turbine
  • Water wheels

Now, let’s go over what happens when a short circuit occurs on the circuit the generator feeds.

The generator will continue to produce voltage. This is because the 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. This current will flow from the generator to the fault location.

What’s more, the flow of the short circuit has only two limits:

  • The impedance of the generator
  • The impedance of the circuit between the generator and the fault location

Further, how about a short circuit at the generator’s terminals? The generator’s own impedance will be the only limit for the current from the generator.

For the generator per unit method short circuit calculation, see the next section. As the generator and synchronous motor calculation will be alike.

Important Note: a synchronous generator and synchronous motor are very similar. They’re both synchronous machines. 

The synchronous generator converts electric energy to mechanical energy. While the synchronous motor converts mechanical energy to electric energy. 

Synchronous motors

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

They have a field excited by direct current. Then there’s a stator winding where alternating current flows. This produces a rotating magnetic field.

So, what’s the point of the DC field? We need another magnetic field to produce torque. Also, to allow the rotor to synchronize with the speed of the rotating magnetic field of the stator. So, this 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.

Then when a short circuit happens, the system voltage drops considerably. The motor then quits delivering energy to the connected mechanical load. Thus, the motor begins to slow 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.

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:

  • The impedance of the synchronous motor
  • The impedance of the circuit between the motor and the fault location

So you can see how a synchronous motor behaves like a generator in certain scenarios.

Per unit method short circuit calculation

We want to 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%.

X^{''}_{pu} = \frac{per \: X^{''}_{d}}{100}(\frac{kVA_{b}}{Motor \: kVA}) = \dfrac{20}{100}(\frac{100,000}{1,000}) = 20

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 continues 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:

  • The impedance of the motor
  • The impedance of the circuit between the motor and the fault location

Per unit method short circuit calculation

We want to convert to ohms per phase at 4,160 volts.

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

X = \frac{per \: X_{m}}{100}(\frac{kV^{2}1000}{Motor \: kVA}) = \frac{20}{100}(\frac{4.16^{2} \times 1000}{2000}) = 1.73 ohms per phase at 4160 volts.


Learning per unit method short circuit calculations is important. Whether you do short circuit studies, or you just want to better understand power systems. It comes in handy!

Just be sure you learn the basics very well. This includes learning about all the sources of short circuit current.

Otherwise, you can make simple mistakes. Hidden mistakes that’ll make you bang your head on the wall in frustration.

In the end, you can make the simplest calculations all the way to the most complex. For example, learning how to size a current limiting reactor for a substation.

What are your thoughts on per unit method short circuit calculations?


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