A few general questions

[R32]

New to this site so if this is in the wrong place please let me know where the appropriate place to post general questions are

If it is not possible to perform as actual I[SUB]pf[/SUB] measurement, what other method could be used to obtain the value of prospective fault current? I always assumed it could be obtained by calculation or even inquiry but can anybody think of another way?

Here is a question from an old exam paper I found through google

A two core SWA cable has a length of 10m and must support a load of not less than 40A without diversity being applied. Select a suitable cable and determine likely Zs at the MCB distribution board assuming the inverter output impedance is 0.65 at 35[SUP]o[/SUP]C

I have no idea where to start with this question could anybody help me try and understand a decent method for tackling a question like this by using this one as an example.
Many thanks in advance.

 

[telectrix]

from the load ( 40A ) you would expect to use 6mm cable. you next step would be to calculate R1+R2 of the 6mm. from that Zs can be calculated, as you know the output impedance. remember that you will be using the armour as cpc.( R2)

 

[R32]

Thank you telectrix. This is all very new to me so forgive me asking simple questions. To calculate R1+R2 is it the resistance of the cable in use (per meter) multiplied by the length of the loop (all times by the correction factor for the temp)? I have no idea how to calculate R1+Rn or whats involved in Zs

 

[Guest123]

Might help a bit........

http://www.electriciansforums.net/e...al-regulations/47059-cable-calc-template.html

 

[telectrix]

OK. Zs = (R1+R2) + Ze, where Ze is the exrternal impedance, R1 is the resistance of the L conductor, R2 the cpc.

from tables in BS7671, the impedance is given in mOhms/m so you multiply the mOhms/m by the length (in meters) to get R1+R2. ( using temperature correction factors as and when needed. this is easy when the cpc is a copper conductor same size as L and N, ( or sometimes smaller as with T/E). when using the armour od SWA as cpc, the calc. is a bit more complicated, as you have to allow for the different values for steel, and the csa.

 

[R32]

Thank you for your help! Just to clarify it would be 6mm by 6mm cable? so Zs = 0.65 + (R1+R2)

 

[telectrix]

your R1 is res. of L., R2 is res. of steel armour.

 

[R32]

Sorry this will probably drive you mad but I don't fully get it. I got R1+R2 as 6.16mΩ x 10m = 0.0616Ω as I assumed it would be 6mm by 6mm cable? So the outcome would be Zs = 0.65Ω + 0.0616Ω = 0.7116Ω

Are there any correction factors or things I am completely missing? The question states a temp. so I assumed it would have some impact on the calculation.
Thank you for the fast response really appreciated!

 

[telectrix]

take a look at a thread posted some time ago " can we use the armour as cpc"

 

[Inteificio]

Just being picky, but Zs does not EQUAL Ze + R1R2, it only approximates.

R1R2 is a measurement of resistance, whereas Zs and Ze would be impedance.

 

[Deleted member 26818]

Just being picky, but Zs does not EQUAL Ze + R1R2, it only approximates.

R1R2 is a measurement of resistance, whereas Zs and Ze would be impedance.

Yes quite correct, but the quetion is asking for the likely Zs, as calculated before impedance can be measured.
This is why we are required to make live tests to verify that the measured impedances are within design specs and that CPDs will operate.

 

[R32]

Sorry but can anybody help with post #8 on the previous page. I haven't got any sort of answer yet. Regards

 

[telectrix]

if it was a 3 core cable, then 1 core would be cpc and thus R1 would = R2, so you would use 6 and 6. however, as it's a 2 core, the steel armour is used as cpc, so you have to find the csa of the armour , apply correction to get the eqiuvalent in copper, then calculate R2.

 

[R32]

ahhh that makes perfect sense. Thanks. Would that be in the on site guide? the small red one is all I have right now.