Available Fault Current: What is it? (And How To Calculate It)

Contents

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Key learnings:

Available Fault Current Definition: Available Fault Current (AFC) is defined as the maximum current available during a fault condition, also known as available short-circuit current.
Importance of AFC Marking: AFC must be marked with a calculation date as per the 2011 NFPA 70: NEC section 110.24.
Fault Current Calculation: To calculate fault current, use the system voltage, conductor constant, and length of the service entrance conductor.
Example Calculation of AFC: In a 480V system, the AFC can be calculated using a given formula and specific parameters, resulting in 18,340A.
Reducing Fault Current: Methods like increasing cable length, using current limiting reactors, and current limiting devices can effectively reduce fault current.

What is Available Fault Current?

Available fault current (AFC) is defined as the highest current available during a fault. It is the maximum current that can flow to electrical equipment when a fault occurs. AFC is also known as available short-circuit current.

The term ‘Available Fault Current’ was introduced in the 2011 NFPA 70: National Electric Code (NEC) in section 110.24 (latest version of the code).

This section requires marking the maximum available fault current along with the date the fault current calculation was done.

The rating marked as available fault current is not an equipment rating. But it is the maximum amount of unwanted current that will flow on equipment in case of fault occurs.

The term short-circuit current rating (SCCR) is different from available fault current. For all equipment or circuits must not have SCCR less than the AFC.

Marking AFC helps electricians choose the correct equipment rating to comply with NEC sections 110.9 and 110.10.

Available Fault Current Formula

NEC 110.24 mandates labeling available fault current. Before calculating AFC for equipment in dwellings, you need the AFC rating at the secondary terminals of the utility transformer feeding the dwelling.

In most cases, the rating of available fault current is provided by the utility and it is labeled at the secondary terminal of the utility transformer.

According to this rating, the available fault current is calculated for all equipment. The calculation for all equipment is different as it depends on the circuit impedance.

Follow the below steps to calculate the available fault current;

Find the system voltage ( $E_{L-L}$ )
Find the conductor constant (C) from the table
Find the length of the service entrance conductor (L)
Now, using the above values, calculate the value of the multiplier (M) using the below equations.

$F = \frac{1.73 \times L \times I}{C \times E_{L-L}}$

$Multiplier\ (M) = \frac{1}{1+F}$

To find the available fault current at premises, this multiplier (M) is multiplied by the available fault current labeled at the secondary terminal of the utility transformer.

How to Calculate Available Fault Current

Let’s take an example to understand how to calculate available fault current.

For that we consider a three-phase system having 480V line-line voltage. And the conductor constant C for this system is 13900.

The available fault current at a secondary winding of the utility transformer is 35000A, and the length of the service entrance conductor is 100ft.

E_L-L = 480V

C = 13,900

I = 35,000A

L = 100ft

Now, put these values in the above equation.

$F = \frac{1.73 \times L \times I}{C \times E_{L-L}}$

$F = \frac{1.73 \times 100 \times 35,000}{13,900 \times 480}$

$F = 0.9075$

$Multiplier (M) = \frac{1}{1+F}$

$M = \frac{1}{1+0.9075}$

$M = 0.524$

$I_{sc} = I \times M$

$I_{sc} = 35,000 \times 0.524$

$Available \, Fault \, Current \, I_{sc} = 18,340A$

How to Reduce Available Fault Current

In a low-resistance circuit, the current can increase drastically, potentially damaging the circuit. Therefore, it’s necessary to use methods or equipment to reduce the short circuit current or available fault current.

Mainly there are three methods to reduce available fault current:

Increasing cable length
Using current limiting reactors
Using current limiting devices

Increase Cable Length

We can reduce the fault current by increasing the length of the cable.

In the above example, we have taken the length of the cable as 100ft. Instead of that, let’s calculate the fault current with a 200ft length cable and other parameters remaining as it is.

By calculating the above steps, you can find the available fault current is 12460A. Hence, it is reduced by increasing the length of the cable.

Observing the relationship between F and L, we can see that they are directly proportional to one another.

Using Current Limiting Reactor

The reactors can be used at any place in the distribution system to reduce the fault current. However, it will degrade the voltage profile during normal operation. But it is the cheapest method to reduce the fault current.

In most cases, a current limiting reactor is connected in series with the circuit or system.

Using Current Limiting Devices

This is a non-linear device that is used in two states; conducting (zero resistance) and limiting (high resistance).

When the circuit is operating in normal conditions, this device simply conducts and there is no change in the voltage profile.

But as the current increases, this device increases the resistance in the circuit to limit the magnitude of the fault current.

current limiting devices — Current Limiting Devices