A short circuit, also known as a fault, occurs when energized conductors inadvertently come into contact with one another or another unintentional conductive path. Short circuits often allow dangerous amounts of current to flow through a circuit, far greater than what they are designed for. Overcurrent protective devices like breakers and fuses have to be used to clear the faults and ensure safety. The 2019 edition of NFPA 70B chapter 9 covers the requirements of studies like short circuit analysis.
In order for an electrical system to be capable of reliably and efficiently transferring power from point A to point B, the conductors and transformers that make up the pathway from the source (say a gas power plant) to the load (say the dishwasher in your house) have to have a very low impedance. The load by comparison, will have a much higher impedance. By Ohm's Law, this means that the load is usually what determines how much current flows. Usually, this is great, but in a short circuit event the low impedance of the system turns the problem upside-down.
Let's say the insulation on the wires leading up to your dishwasher breaks down after years of use. The wires then come into contact with one another and create a short circuit. Now, the load is no longer what is limiting the amount of current, but the transmission system itself. By Ohm's law, since the impedance is low and the voltage of the source has remained the same, the current will increase (often drastically)!
The diagrams below show an example of a fault as viewed from a circuit perspective. The model below is based on the infinite bus assumption. What this means is that there has been no additional impedance included between the source and the first transformer. This assumption may seem unreasonable, but it's often very useful and will always yield conservative results. Moreover, the diagram below is for a three-phase fault. Instead of assuming that a single wire has broken down its insulation and come into contact with some other conductive pathway, we are assuming that all phase conductors have had their insulation broken down and are now in contact with one another. This condition will generally yield the highest fault currents in a system.
Example Circuit for Fault Current Calculations
The phase of the fault current can be important for more sophisticated applications, but we'll worry only about the magnitude of the current here. Applying Ohm's Law to the circuit above, the magnitude of the fault current |I | can be calculated:
|I |= |Vn | / | Zt + Zc |
Where:
Vn is the secondary source voltage from line to neutral in Volts. Even if a source were delta-connected, the line-line voltage divided by √(3) is still applicable.
Zt is the transformer positive sequence impedance in Ohms
Zc is the conductor impedance in Ohms
The transformer positive sequence impedance is often given as a percentage (and is often referred to as simply "the transformer's impedance"). We can convert the percentage impedance of the transformer to a value in Ohms by understanding the basis for that percentage. The conversion is:
Zt = Z% ( VL ^ 2 / S )
Where:
Zt is the transformer impedance in Ohms, usually assumed to be purely reactive (purely inductive)
Z% is the transformer nameplate impedance percentage
VL is the nominal line-line voltage rating of the transformer being referenced to (the secondary voltage for infinite bus short circuit current calculations)
S is the three-phase apparent power rating of the transformer
Infinite bus symmetrical fault calculations only make sense when applied to the secondary of a transformer. If we tried to calculate the infinite bus fault current before the transformer, the result would be...well, infinite. There would be no impedance between the voltage source and the fault location and Ohm's law would say to expect an infinitely large current flow. Of course, we know this is not real. In reality, there will be source impedances from the transmission system, upstream transformers, generators themselves, and more. The example below shows how these calculations can be applied to find the fault current with limited information on a project.
Asymmetrical Fault Current - Discussions this far have been surrounding the symmetrical fault current, the current that flows in a three-phase fault as a steady-state condition. For a short period of time after the fault, though, the fault current can actually be much higher than the symmetrical value. The asymmetrical fault current is the sum of the symmetrical fault current and a transient DC offset. This offset decays exponential just like in an RL circuit. The rms asymmetrical fault current is given by:
Ia = Is √( 1 + 2 e^( -2 R t / L ) )
Where:
Ia is the asymmetrical fault current
Is is the symmetrical fault current
e is Euler's constant
R is the system resistance in Ohms
L is the system reactance in Henries
t is the time in seconds after the fault event
This equation can be re-written to utilize a time basis of cycles and an X/R ratio instead of the resistance and inductance as well. The results are the same: The maximum asymmetrical rms fault current is equivalent to 173% of the symmetrical fault current. Depending on the system's particular impedance and the time at which the fault occurs, the actual maximum effective asymmetrical fault current could be much closer to the symmetrical fault.
A Plot of the Asymmetrical Fault Current and its Components
Interpretation - The symmetrical and asymmetrical fault current each have their place in a good design. It is advisable that conductors be sized based to withstand the symmetrical fault current, since that is the standard method used in ICEA and NEC documents. Meanwhile, the system overcurrent protection scheme should be designed accounting for both the symmetrical and asymmetrical fault current. We need the system to clear a fault as quickly as possible, regardless of whether it is symmetrical or has an offset. Overcurrent devices are often rated in terms of their symmetrical fault current and are tested at a particular X/R ratio. If the X/R ratio of the system exceeds the tested value, then additional derating is required to ensure the design will operate safely. Equipment is often rated for duty based on symmetrical fault currents with special caveats for maximum X/R ratios.
Example: Compute the symmetrical fault current seen on the secondary (low voltage) of a 6.9kV : 480V transformer with a nameplate apparent power rating of S=100kVA and a nameplate impedance of Z=4%. What is the peak rms asymmetrical fault current possible?
Solution: For this example, we are neglecting any secondary conductor impedance and just focusing on the transformer. This means we first have to compute the transformer impedance.
Referenced to 480V, the transformer impedance is:
Zt = .04 * ( 480^2 / 100000 ) = .09216 Ohms
Then, the fault current magnitude is given by the effective line-neutral voltage and Ohm's law:
I = ( 480 / √(3) ) / (.09216) = 3005.6A
The peak asymmetrical fault current is equivalent to √3 times the symmetrical fault current:
√3 * 3005.6 = 5205A
Comments