Sample fault current calculations help contractors understand the source of power flow problems

A power system functions normally until after the occurrence of a fault in the system. The good news is fault events can be minimized or avoided through diligent electrical design, accurate record keeping information on equipment/devices/motors, proper installation, and use of agency-certified equipment.

There are three major sources of fault current: an electric utility power system, a generator, and a motor. Short circuit faults are called shunt faults. An open-circuit condition is known as a series fault. Any phase/circuit to ground condition is called a ground fault. Among all faults, a balanced 3-phase short circuit is the most critical and serious. However, it is one of the least likely of faults to occur. The elements in a power distribution system that limit or impede the fault current value include: cables, transformers, and reactors.

The NEC requires protection to personnel and electrical systems against damage during short circuit conditions. Generally, circuit breaker ratings are determined for the worst-case fault situation.

You typically perform short circuit calculations when working with complex and interactive power distribution systems; however, manual calculations can be used for more simplified systems. A short circuit calculation determines the amount of current that can flow at certain points in the distribution system. An electrical device or piece of equipment can then be selected for appropriate rating (withstand or interrupting rating) based on these calculations.

Let's work through a few simple examples to show how you can quickly and easily calculate fault currents.

We'll first use the concept of admittance to calculate the fault current in the system shown in **Fig. 1** (click here to see **Fig. 1**). Admittance is a measure of how easily a circuit or device will allow a current to flow. It is the inverse of impedance, which is defined as a measure of opposition to current.

Step 1: Because the electric utility can't provide us with fault information at this particular site, we assume it to be infinite. We calculate the maximum amount of power that the transformer (XFMR) will allow to flow to its secondary (i.e., load side) using this formula.

XFMR let through power = XFMR kVA rating ÷ [% Impedance ÷ 100]

= 5,000kVA ÷ [5 ÷ 100]

= 100,000kVA

Step 2: Now we can calculate the cable let-through power. This is defined as the amount of power that the cable would let through from an infinite source to the load side of the XFMR. The formula we use for this step is as follows:

Cable let through power = [1,000 x (kV phase-phase)^{2}] ÷ [cable impedance (ohms) per phase, per 1,000 ft x total distance (ft)]

= [1,000 x (12kV)^{2}] ÷ {0.15 ÷ 1,000 ft} x 100,000 ft

= 9,600kVA let through

Step 3: Next, we calculate the total let-through fault power by using the following formula:

Net fault power = 1 ÷ [(1 ÷ XFMR let through power) + (1 ÷ cable let through power)]

= 1 ÷ [(1 ÷ 100,000kVA) + (1 ÷ 9,600kVA)]

= 8,759kVA

Step 4: Now we can find the fault current using the following formula:

Fault current = net fault power ÷ (secondary XFMR voltage rating x √3)

= 8,759kVA ÷ (12kV x √3)

= 421A

Let's work through another sample calculation where the electric utility fault power level is known (click here to see **Fig. 2**). The given value is 50,000kVA. In this situation, we assume cable length to be minimal and therefore neglect its effect since the impedance is minimal.

Step 1: XFMR let through power = 2,000kVA ÷ [5 ÷ 100]

= 40,000kVA

Step 2: Net fault power = 1 ÷ [(1 ÷ 40,000kVA) + (1 ÷ 50,000kVA)]

= 22,222kVA

Step 3: Fault current = 22,222kVA ÷ (0.48kV x √3)

= 26,729A

For our third and final example calculation, we'll work a problem where generator data is available to us

(click here to see **Fig. 3**). When working with generators, we introduce a new value known as sub-transient reactance, which is referred to as X"d. This value is typically shown on the generator nameplate or can be obtained directly from the manufacturer.

Step 1: The short circuit kVA available at the generator is calculated using the following formula:

Generator fault power (MVA) = generator MVA rating ÷ X"d

= 800 MVA ÷ 0.17

= 4,706 MVA

Step 2: XFMR let through power = 1,000 MVA ÷ [10 ÷ 100]

= 10,000 MVA

Step 3: Net fault power = 1 ÷ [(1 ÷ 10,000 MVA) + (1 ÷ 4,706 MVA)]

= 3,200 MVA

As you can see, simplified short circuit fault calculations can be performed using a basic understanding of sources of fault power and current and impedance values that impede the short circuit power flow.

*Bodhankar, P.E., is an electrical engineer with CHA in Albany, N.Y. He can be reached at nbodhankar@chacompanies.com.*