Short Circuits and Fault Levels
An overview, in electrician’s terms, with some examples
The basic electrical theorem (Ohm’s Law) says the amount of current that will flow through a short circuit depends on two variable values,
1. the system voltage and
2. the connected total impedance of the current flow path from the source to the point of the fault.
The typical system voltages are very familiar to all of us. The connected total impedance of the shortcircuit current flow path needs a little clarification, however.
This impedance normally includes the feeder conductors’ resistance and reactance, any transformers’ impedances (going from the point of fault back to the energy source), and any other equipment connected in the path of current flow.
Fig. 1 is a very simplified one-line, with the following: a power source, transformer, and an overcurrent protective device (OCPD) having a specific short-circuit current interrupting rating and a load with a fault.
Let’s talk about the power source first. In many short-circuit current calculation examples, you’ll see references like “Assume the power source has infinite capacity” or “The source has an infinite bus.” What does this mean, and why is it important to our sample calculation?
All that is being said in this case, is: “The source voltage has no or an extremely low internal impedance”. Since the source has been assumed to have infinitesimal impedance of its own, the corresponding shortcircuit current will be at its worst case.
Now let’s look at the transformer. The impedance determining the amount of short-circuit current on its secondary, is made up of two separate impedances: Its own impedance plus that of the secondary conductors which run to the point of the fault.
The transformers own impedance is the amount of its own opposition to the flow of short-circuit current through it. Now, all transformers have impedance, and it’s generally expressed as a voltage percentage.
This is the percentage of normal rated primary voltage, that must be applied to the transformer to cause full-load rated current to flow in the short-circuited secondary. (You remember that from TAFE) For instance, if a 11kV to 415V transformer has an impedance of 5%, this means that 5% of 11kV, or ≈550V, applied to its primary will cause rated load current flow in its secondary. If 5% of primary voltage will cause such current, then 100% of primary voltage will cause 20 times (100 divided by 5) full-load-rated secondary current to flow through a solid short circuit on its secondary terminals.
Obviously, then, the lower the impedance of a transformer of a given kVA rating, The higher the amount of short-circuit current it can deliver.
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