The
resistance of a conductor is directly proportional
to its length (L) as R ∝ L. Thus doubling its length will double its
resistance, while halving its length would halve its
resistance. Also the
resistance of a conductor is inversely proportional
to its cross-sectional area (A) as R ∝ 1/A
This article on Professional Electrician.com may answer your other query re: RCD breaking current relationship with PFC
Code:
/technical/prospective-fault-current/
The experts at NICEIC provide us with more critical best practice advice.
Either by enquiry, measurement or calculation, prospective fault current (Ipf) should be determined at every relevant point within an installation (Regulation 434.1). The values obtained should be compared against the relevant installation design criteria to confirm that the fault rating of each device is not less than the maximum prospective fault current that may occur at that particular point in the installation (Regulation 434.5.1).
For many smaller installations, especially domestic installations, verification is normally a matter of confirming that all installed protective devices have a greater rated braking capacity than the maximum value of prospective fault current determined for the installation.
The maximum value is the higher of the maximum prospective short-circuit fault current (line-to-neutral) and the maximum prospective earth fault current (live to earth), and must be recorded in the relevant section of an NICIEC electrical certificate or report.
Where the maximum prospective fault current is determined by testing, the ‘live’ test should be performed at the origin of the installation where the fault current is highest, such as at the distribution board or consumer unit or other switchgear supplied directly from the distributor’s equipment.
With the main switch ‘OFF’ the short circuit test is performed as shown in Fig 1. An earth fault current test should also be made and the higher of the readings taken should be recorded as the maximum value of prospective fault current.
The earthing conductor, main protective bonding conductors and circuit protective conductors should all be connected during these tests, because the presence of these and any other parallel paths to Earth may reduce the impedance of the fault loop and so increase the prospective fault current.
Conditional rating
Where a consumer unit to BS EN 61439-3 is installed in domestic or similar premises, the arrangement of the devices and the outer enclosure (of the consumer unit) is designed to safely withstand a prospective fault current of up to 16 kA.
In such circumstances, the maximum prospective fault current should be recorded as 16 kA in the relevant part of the electrical certificate or report, and undertaking prospective fault current testing at any point in the installation would be considered unnecessary.
In some cases, prospective fault current testing may need to be carried out at other relevant points in the installation, particularly at the furthest points, to confirm that the fault current is sufficient to cause operation of the protective device before the permitted limiting temperature of any conductor or cable is exceeded (Regulation 434.5.2).
The relationship between the protective device and the cable
For any relevant disconnection time the maximum energy withstand of the cable must be greater than the let- through energy of the protective device. If this is not the case, there is the risk that under fault conditions the conductor or cable may suffer thermal damage and as a result it may have to be replaced. This thermal relationship can be stated as:
S2 k2 ≥ I2t
To illustrate this relationship, Fig 2 shows a 16 mm2 cable incorporating an uninsulated 6mm2 protective conductor supplying a load. The circuit has been installed using a 70 °C thermoplastic insulated and sheathed flat cable (copper) and is protected by a BS 88-3 type (C) fuse, rated at 63 A. (Assuming a TN earthing system for the installation, a disconnection time of 5 s is permitted by Regulation 411.3.2.3 of BS 7671).
The maximum thermal withstand of the cable depends on its cross-sectional area S and also K which is a factor that takes account of the material properties of conductors and insulation.
From Table 54.3 of BS 7671 the value of k for the cable shown in Fig 2 is 115, and so, the maximum thermal withstand of the cable can be calculated as follows:
Code:
Maximum thermal cable withstand = S2 k2 = (6² x 115²) = 476 100 J
As shown by the calculations in Fig 2, the cable is exposed to a higher fault current at position (A) than at position (B). However, a fault occurring at position (B) represents the most onerous condition because the lower fault current leads to a longer disconnection time, which results in a higher let-through energy I2t.
Table 1 is based on Appendix 3 (Fig 3A1) of BS 7671, and shows the time/current disconnection values for a BS 88-3 fuse rated at 63 A, from which the respective let-through energy I²t has been calculated.
As the required disconnection time increases from 0.1 s to 5 s, the minimum current required to disconnect the 63 A fuse decreases (inverse characteristic of the fuse). For a fault of 320 A (position B) the energy let-through of the protective device exceeds 476 100 J, the maximum thermal withstand of the cable.
It should be noted that only the earth fault current is determined in the example because the protective conductor is of a reduced cross-section so it will have a higher resistance and therefore a lower prospective fault than that resulting from a short-circuit. However, where necessary the short-circuit fault current should also be determined.
![[ElectriciansForums.net] I HAVE TWO QUESTIONS](https://electricians.s3.eu-west-1.amazonaws.com/data/attachments/45/45879-a14dd88f61c15b325521b139ee2dc833.jpg "[ElectriciansForums.net] I HAVE TWO QUESTIONS")
![[ElectriciansForums.net] I HAVE TWO QUESTIONS](https://electricians.s3.eu-west-1.amazonaws.com/data/attachments/45/45880-29d7025dba00051c9824521dd6d257f7.jpg "[ElectriciansForums.net] I HAVE TWO QUESTIONS")
