As mentioned in Part 1, the value of the fault loop impedance shall be able to generate enough fault current to activate any protective device.
What does this mean? The earth fault current must be high enough to trigger the automatic disconnection of a protective device within an specified time. If the value of the earth fault current is not within the tripping current of the protective device, then there will be no means of protective device will provide protection against touch voltage. At the location of the fault, there is a possibility that the full voltage is present in the equipment enclosure posing hazard to people with the increased touch voltage higher than 50 Vac.
If in cases where there is no way to increase the fault current, say the distance of the fault from the supply is too far - cable impedance is limiting the earth fault current, the only solution will be using a residual current (earth leakage) protection. It is not always advisable to have this solution though. There are applications that when a residual current protection is used, annoying erratic device operation is possible particularly when the protection device setting is not correct.
In the event that a reduction of the earth fault-loop impedance required to ensure correct disconnection times of the protective device is not possible due to distance factor, additional bonding to earthing system may be used.
To calculate the fault loop impedance
Zs x Ia ≤ Uowhere:
Zs = the impedance of the earth fault-loop comprising the source, the active conductor up to the point of the fault and the return conductor between the fault and the source
Ia = the current require to cause automatic operation of the disconnecting protective device within the required disconnecting time
Uo = the nominal AC rms voltage to earth (230V for a 400V line-to-line
Note:Return path will comprise both protective earthign and neutral conductors
Example, a 400V, 3ph load is supplied by a 25 meters long, 2.5mm2 cable and protected by a 6A MCB. Calculate the fault-loop current, if there is a bolted fault at the load. Check if the fault current is sufficient to provide protection in specified time.
Substituing values into the formula
Zs x Ia ≤ Uo
Zs ≤ 230 / 6
Zs ≤ 38.3 Ohms
For a 2.5mm2 cable, the typical impedance is
Z = 9.44 ohms / km
The total cable fault-loop impedance for the example will be
Z = 2 x 25 x 9.44 / 1000 (The 2 in means that it will be the active conductor and neutal conductor)
Z = 0.472 ohms
If we divide Zs by Z then we get 81 meters which is the maximum length of 2.5 mm2 cable to used in the application.
In part III, we shall be discussing how to coordinate circuit breakers with earth fault currents.