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JohnnyHS (Petroleum) (OP)
20 Oct 09 6:43
Hi guys,

I'm generating the company forms for the electrical installation certificate schedule of test results to the 17th Edition BS7671:2008 and have the following questions/statements that you fine sparkies may confirm/clarify for me...

The Prospective Fault Current

This was measured by a test engineer (who had the test equipment to measure this) from a contractor and the figure was 0.68kA. As this figure is measured at the main distribution board incoming Ipresume, for any future installations connected to this main distribution board for testing can I use this figure for future Prospective Fault Current values rather than having to measure this again?
And how do you measure this anyway!? smile

Johnny H-S BSc(Hons) MInstMC

Engineers like to solve problems. If there are no problems handily available, they will create their own problems.

rbulsara (Electrical)
20 Oct 09 8:17
We don't measure prospective fault current, we calculate it.

Someone is messing with you. I would be interested in how the 'test' was done and how the value of 0.68KA was arrived for a main distribution switchboard. Seems very low, by a factor of 100.

Taking a cue from you own signature line, I would say leave the electrical forms to sparkies and not create your own problems.

Rafiq Bulsara

JohnnyHS (Petroleum) (OP)
20 Oct 09 8:39
thanks you for your reply rbulsara! unfortunately i cannot leave this job for the sparkies!

Please have a look at this forum. It looks like they actually measure for the prospective fault current - any comments? In any case could you please tell me the formula for calculating it? In theory i can check whether the figure is correct then.

Also the "factor of 100" in your comment very low - is that correct? looking in the on site guide to BS7671 it states

"in general the fault current is unlikely to exceed 16.5kA"

have any of you guys filled the Electrical Installation Certificate (BS7671) in. We have these results

nominal voltage : 415V
Freq            : 50Hz
Prospective Fault Current : 0.68 kA
External Loop Impedance : 0.35 ohm


Johnny H-S BSc(Hons) MInstMC

Engineers like to solve problems. If there are no problems handily available, they will create their own problems.

rbulsara (Electrical)
20 Oct 09 8:47
I have nothing more to add.  

Rafiq Bulsara

JohnnyHS (Petroleum) (OP)
20 Oct 09 8:50
you get a star for that last post! lol! anyone else?!

Johnny H-S BSc(Hons) MInstMC

Engineers like to solve problems. If there are no problems handily available, they will create their own problems.

davidbeach (Electrical)
20 Oct 09 9:57
Looks like an attempt to measure the ground fault current of an impedance grounded system.  Kinda silly, but if it is truly impedance grounded you shouldn't kill too many people.  But that tells you nothing about the prospective fault current of any fault that involves more than one phase.

Actual prospective fault current must be calculated as any attempt to measure anything involving more than one phase, or a system that is solidly grounded, will result in injury, death, and destruction.

To use any hand-held/portable instrument to measure the current of a impedance grounded system as a verification that the impedance is in place seems to have far too high a risk level.

Obviously something is lost in translation across the pond.
jghrist (Electrical)
20 Oct 09 11:21
Apparently, this is a loop impedance test made to ensure that the earth fault current will be high enough to trip the ground fault interrupter (or residual current detector RCD in UK terms).  See
davidbeach (Electrical)
20 Oct 09 11:30
But to call it a test of prospective fault current, without other qualification, is highly misleading. The will be much higher fault currents.  
JohnnyHS (Petroleum) (OP)
20 Oct 09 12:47

I have got my paws on some guidance notes and it states for the prospective fault current(Ipf) section that

a) this figure is to be acquired by enquiry or by measurement


b) The maximum prospective fault current recorded should be the greater of either the short circuit current or the earth fault current.

So how to measure/calculate these two...

short circuit current?

Is this a case of connecting a special instrument (mentioned in my second post link) to the supply live and neutral and measuring the current that flows?!

earth fault current?

divide the applied voltage by the earth fault loop impedance = prospective earth fault current

415 / 0.35 = 1,185A = 1.185kA

or if that measurement (in my form by contractor - first post) 0.68kA is correct then the earth fault loop impedance would have to be around 0.610 ohm by calculation to get 0.68kA on the form (not 0.35 ohm as reported which would give 1.185 by caluclation). i presume it was measured which may account for the difference as stated by jghrist..


Any comments/views all?

oh PS this is whats causing all the fuss! smile

the forms on the right hand side. BS 7671:2008 forms (431 k) Updated March 2008 Page 6!)


Johnny H-S BSc(Hons) MInstMC

Engineers like to solve problems. If there are no problems handily available, they will create their own problems.

PHovnanian (Electrical)
20 Oct 09 14:49
In the notes on BS 7671:2008, there is the following:

"Prospective fault current (PFC). The value recorded is the greater of either the short-circuit current or the earth fault current. Preferably determined by enquiry of the distributor."

From what I gather, the measured/calculated values of either 0.68 kA or 1.185 kA are for earth faults. The short circuit current may be higher, but like the note says "Preferably determined by enquiry of the distributor." That's how its usually done on this side of the pond. Actually testing available short circuit current is dangerous, expensive, and not done with equipment typically carried in an electrician's tool bag.

If I were doing this, I'd contact the local power company and ask for this data, worded exactly the way this form requests it. But I don't work in that jurisdiction.

And, for future reference, take anything posted by people from far away with a big grain of salt when it deals with regulations, codes, etc. These are usually very specific, with peculiar definitions. And although we can give generic engineering advice, we can really get you into trouble with the local inspectors.
FreddyNurk (Electrical)
20 Oct 09 18:48
I think jghrist has it right, it reads to me (particularly with the level of fault indicated) that they're checking that the cable impedance isn't too high to allow the protective device to trip.

I admit I know nothing of the BS7671, though I do note that in Australia, the latest electrical standard, AS/NZS3000 now mandates that 'fault loop impedance' be checked to ensure that there will be enough current to operate the protection.

Which is rather different to calculation (measurement?) of the prospective fault level at the incomer of a site.  
ScottyUK (Electrical)
20 Oct 09 20:37
The loop impedance value at the service entrance is required so it can be included within the overall loop impedance calcs to verify that the protection will clear a fault and in particular that the 0.4s and 5.0s clearance times for sockets and fixed equipment final circuits respectively can be met. If those times can't be met using a standard MCB then an RCD or RCBO will be required.

There are a number of instruments which can measure the available fault current at a point on the circuit using a loop impedance test. Obviously that value is valid only at the time it is measured and changes on the system could modify it. I haven't dismantled one of the instruments or studied one in detail but I assume it uses some sort of pulsed load and measures volt-drop in the network caused by the momentary application of a relatively heavy load resistor and implies the impedance from that. I'm just guessing so that could be way wide of the mark. These instruments aren't available for most of the stuff I work on because the fault levels are too high in an industrial environment, the risk to the user is far too high even if they were available, and we have a good model of the local network to calculate the available fault current. Having said that, they're fairly popular with the domestic and light commercial guys where the incident energy is much lower.


We don't have many impedance-grounded LV installations in the UK, all our public system is solidly grounded.

If we learn from our mistakes I'm getting a great education!

davidbeach (Electrical)
20 Oct 09 21:08
So, obviously nobody is measuring the prospective fault current.  Measuring loop impedance is one thing, actually measuring fault current is something entirely different.  The instrument may even do the calculations to get from loop impedance to fault current, but that still isn't measuring the fault current.

If solidly grounded, I'm surprised you'd have concerns about 0.4 or 0.5 second clearing, must have some really high impedance circuits.
ijl (Electrical)
21 Oct 09 1:21

Quote (JohnnyHS):

divide the applied voltage by the earth fault loop impedance = prospective earth fault current 415 / 0.35 = 1,185A = 1.185kA
My guess is that 415V is the phase-phase voltage, which means that the above result must be divided by square root of three
 I = 415V / 0.35Ohm / sqrt(3) = 684.6A ≈ 0.68kA

ScottyUK (Electrical)
21 Oct 09 1:51
Hi David,

The 0.4s includes all the internal building wiring, so on (say) a 32A ring circuit fed from 240V single phase it's surprisingly easy to get up close to the 0.4s without outrageously long circuits. It's a lot worse if the property is near the end of the distribution feeder with a significant external loop impedance: don't forget we tend to have relatively large padmount transformers serving a number of streets rather than a polemount serving three or four properties as seems to be the norm in the US (or my perception of the norm anyway).

Somewhere I have the calcs from the bungalow I rewired a good few years ago - I'll post them and you'll see what I mean.

If we learn from our mistakes I'm getting a great education!

davidbeach (Electrical)
21 Oct 09 11:14
We'll also use pad mount transformers where distribution is underground, but typically 112.5kVA is as big as many utilities will go to keep the available fault current at the nearest service below 10kA.  To minimize voltage drop, no service transformer serves any great distance, probably >95% of all residential services are within 250 feet of the service transformer.  Even at the receptacle, there is an expectation of an instantaneous trip, but no requirement that it must be so.
ScottyUK (Electrical)
21 Oct 09 11:37
Ours are frequently in the 250kVA - 1MVA range with service cables which radiate perhaps ¼ mile out from the sub. Hopefully Marmite will notice this thread and put some more accurate numbers to this - I'm not a distribution engineer so those numbers are educated guesses rather than statements of fact.

If we learn from our mistakes I'm getting a great education!

davidbeach (Electrical)
21 Oct 09 17:31
Wow, so what kind of fault duty at the house nearest the transformer?
ScottyUK (Electrical)
21 Oct 09 21:37
Highest I've seen recently - on one of those meters funnily enough - was just over 8kA. I'd done some work at a friend's place which needed inspection under Part P of our building regs, which I'm not certified to sign off, so another friend did the testing for me. The service fuse would have been either 80A or 100A - it was under a utility seal, so I don't know - and the sub was practically on the other side of the garage wall.

If we learn from our mistakes I'm getting a great education!

davidbeach (Electrical)
21 Oct 09 22:27
Oh, but you're at nearly twice the voltage.  240V 1 phase here and 400V 3 phase there?
ScottyUK (Electrical)
22 Oct 09 2:24
You got it. Our public utilities operate solidly earthed 400/230V systems. Residential premises rarely require a 400V 3-phase service although there are cases where they do have such installations.

Yes, a 1MVA transformer at 208/120V would certainly deliver some punch if a fault occured.

If we learn from our mistakes I'm getting a great education!

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