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AJJewsbury: 
 

In this context surely an earth fault will clear very quickly if an R.C.D. is installed at the origin of the S.W.A. cable, and disconnect the fault before serious damage due to cable  heating can occur.

Fuses and MCBs can be significantly faster than RCDs at high fault currents - an S type won't open in less than 40ms - at say 6kA that's 1,440,000 A2s so by the adiabatic would need over a 10mm2 copper conductor (if k=115).

By comparison a B-type MCB at 6kA should have an energy let-though of less than 54,000 A2s - so should be adequate with just a 2.5mm2 conductor from a fault protection perspective.

   - Andy.

But I am not considering high fault currents. The TT earthed stable will probably not allow a high L to E fault current. So the R.C.D. will be king. Fuses won't blow and M.C.B.s won't trip.

Z.

As I said in my reply, you will have to determine the maximum earth fault current, and determine the necessary overcurrent protection of the cpc in line with that.

First, don't use the measured Ze to the earth electrode alone to determine the maximum prospective fault current. When extraneous-conductive-parts are connected (if any), the effective combined earth electrode resistance may well be far lower than you measure doing the Ze test.

Example - suppose the Earth electrode resistance is 70 ohms, and effective resistance of extraneous-conductive-parts 10 Ohms. Combined resistance =70*10/(70+10) = 8.75 ohms. 
You thought you only had a fault current of 3.3 A, but in fact you have a fault current of 26.2 A.
Not a problem if your cable has a current-carrying capacity of at least 26 A (4 sq mm copper). But if the RCD didn't trip, would you be OK with 1.5 or 2.5 sq mm cpc, say for a fault near the origin?
I think in this case, we'd all agree the RCD would operate before the cpc had an issue.

However, what if the maximum earth fault current is much greater … for example:

(a) the effective combined earth electrode resistance of the extraneous-conductive-parts is 1 ohm? We've now got (at the origin) over 200 A of prospective fault current. We can't rely on that for protection against electric shock (we can only rely on the loop impedance through the means of earthing - the consumer earth electrode - for that) ,,, but we need to take it into account for overcurrent protection, and an RCD cannot provide overcurrent protection even for earth faults.

(b) the TT system is derived from a TN system, and unfortunately shares extraneous-conductive-parts with that system. Again, can't rely on that for protection against electric shock … but this time we definitely need to size our cpc's in a similar manner to TN systems for protection against overcurrent (earth fault current). Again, the RCD is not an overcurrent protective device according to BS 7671, it requires back-up of a suitable overcurrent protective device, and the cpc sized accordingly.


Not so clear-cut, is it? 

Zoomup: 
 

AJJewsbury: 
 

In this context surely an earth fault will clear very quickly if an R.C.D. is installed at the origin of the S.W.A. cable, and disconnect the fault before serious damage due to cable  heating can occur.

Fuses and MCBs can be significantly faster than RCDs at high fault currents - an S type won't open in less than 40ms - at say 6kA that's 1,440,000 A2s so by the adiabatic would need over a 10mm2 copper conductor (if k=115).

By comparison a B-type MCB at 6kA should have an energy let-though of less than 54,000 A2s - so should be adequate with just a 2.5mm2 conductor from a fault protection perspective.

   - Andy.

But I am not considering high fault currents. The TT earthed stable will probably not allow a high L to E fault current. So the R.C.D. will be king. Fuses won't blow and M.C.B.s won't trip.

Z.

As I said in my reply, you will have to determine the maximum earth fault current, and determine the necessary overcurrent protection of the cpc in line with that.

First, don't use the measured Ze to the earth electrode alone to determine the maximum prospective fault current. When extraneous-conductive-parts are connected (if any), the effective combined earth electrode resistance may well be far lower than you measure doing the Ze test.

Example - suppose the Earth electrode resistance is 70 ohms, and effective resistance of extraneous-conductive-parts 10 Ohms. Combined resistance =70*10/(70+10) = 8.75 ohms. 
You thought you only had a fault current of 3.3 A, but in fact you have a fault current of 26.2 A.
Not a problem if your cable has a current-carrying capacity of at least 26 A (4 sq mm copper). But if the RCD didn't trip, would you be OK with 1.5 or 2.5 sq mm cpc, say for a fault near the origin?
I think in this case, we'd all agree the RCD would operate before the cpc had an issue.

However, what if the maximum earth fault current is much greater … for example:

(a) the effective combined earth electrode resistance of the extraneous-conductive-parts is 1 ohm? We've now got (at the origin) over 200 A of prospective fault current. We can't rely on that for protection against electric shock (we can only rely on the loop impedance through the means of earthing - the consumer earth electrode - for that) ,,, but we need to take it into account for overcurrent protection, and an RCD cannot provide overcurrent protection even for earth faults.

(b) the TT system is derived from a TN system, and unfortunately shares extraneous-conductive-parts with that system. Again, can't rely on that for protection against electric shock … but this time we definitely need to size our cpc's in a similar manner to TN systems for protection against overcurrent (earth fault current). Again, the RCD is not an overcurrent protective device according to BS 7671, it requires back-up of a suitable overcurrent protective device, and the cpc sized accordingly.


Not so clear-cut, is it?