2 methods of obtaining Ipf?

[Dannywhite]

got a question in my assignment but unsure how to do it as i havent had much experience testing. any help appreciated.

 

[SparkyChick]

What do you think the answers are and why?

 

[Dannywhite]

id say by doing the calculation Ze / voltage but also by doing it with the tester.. basic knowledge i know but any of what i said correct? lol

 

[Andy-1960]

Removed

 

[SparkyChick]

You're nearly right on one method of acquiring Ipf. You can calculate it using measurements as you've stated, but the maths is I=V/R so U0/Ze, but that will only give you the prospective earth fault current (L-E fault), which may be less than the prospective short circuit current (L-N fault). So to calculate it, you would need to measure the L-N loop impedance, do another calc and use whichever is higher.

This is for single phase, 3 phrase is different.

So, measurement/calculation is one method of obtaining it. There is another method for obtaining it. What might that be?

 

[Dannywhite]

what would the L-N fault calc be? and i would say by phoning up the supplier! but not exactly sure about that as they probably wouldnt know

 

[SparkyChick]

The L-N fault calc is the same (I=V/R). You measure the loop impedance across L and N on the incomer, and divide U0 by that value. That will give you the prospective short circuit current.

And you're bang on with the second method.

So, measurement/calculation and enquiry

 

[Dannywhite]

thank you very much! im just lacking common sense

 

[DavidSpark]

Measuring it is correct, what test and type of instrument would be used?

Calculating it is also another method as you said, V=IR therefore I = V/R in this instance V=Uo and R=Ze.

There is one final way of obtaining an Ipf however I think (I may be wrong someone please correct me if so) it may be a bit more of a generic value for the area. That's a huge hint for you.

 

[hightower]

There is one final way of obtaining an Ipf however I think (I may be wrong someone please correct me if so) it may be a bit more of a generic value for the area. That's a huge hint for you.

The OP has already mentioned enquiry if that's what you're meaning

 

[Richard Burns]

Perhaps he means the guidance that except in large city centres the fault current is not likely to exceed 16kA; and so a consumer unit to BS61439-3 which is type tested to 16kA could be used (subject to the origin protection type) (Annex ZB).
However this is a bit of a bypassing the issue approach, but it is probably what you would get if you managed to get a response to your enquiry.

 

[DavidSpark]

The OP has already mentioned enquiry if that's what you're meaning

In the space I loaded the thread and replied, someone else had already replied and the OP responded. Oops.. shouldn't have left it so long!

 

[gazdkw82]

When doing the calc is it always best to use nominal voltage or the voltage the system is actually operating on?

 

[SparkyChick]

You should use the stated nominal voltage of the supply. This is constant. The actual supply voltage is changing constantly as a result of the ever changing supply/demand characteristics of the grid.

As a bonus, the ever changing supply voltage is why you can't realistically measure voltage drop using a voltmeter (measure at the origin, then by the time you measure at the end of the circuit, the supply voltage is likely to have changed).