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flex_m

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Hi all, just working through an assignment at the moment, got everything done other than two questions worded "select an appropriate minimum sized CPC and confirm your selection by calculation". I've already calculated my tabulated current, volt drop and live conductor size for each question.

Our teacher said we should be using the adiabatic equation, but we haven't been given a Ze, in my mind I won't be able to calculate the PSC because I can't just assume the Ze. It's TN C S, but if I assume the maximum Ze of 0.35 and it is actually lower, the PFC will be higher and my calculation will be incorrect.

Am I missing something obvious here or should I be using a different equation?
 
Have you been given any details about the incoming supply?

In the UK the standard supply characteristics that represent the normal maximum, and are what will be quoted if you ask the DNO, for small supplies are 16kA for PFC (and single phase PSCC) and 25kA for three phase PSCC
These values can be higher in some situations, but generally they are the designs maximum limits for a uk supply, in the absence of any other information I would use those values for your calculations and maybe add a not to explain why you used that value. The tutor should be more interested in knowing that you understand and can carry out the calculation correctly


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I would say that if you do not know the actual fault current that will flow then, for design purposes, you can determine or look up the fault current that will cause disconnection in the maximum permissible time for that circuit and protective device.
E.g. for a distribution circuit protected by a BS88-3 63A fuse the fuse will disconnect in 5 seconds with a fault current of 320A.
These figures are in the tables in appendix 3 of BS7671.

A higher fault current will disconnect faster and so will not heat up the cable for so long and so allow a smaller cross sectional area cable to be used.
This appears contradictory as surely a higher current requires a thicker cable; but it is the combination of time and the current that are relevant.

So for the case above of 320A over 5s the adiabatic equation shows that a 5mm² csa conductor would be required.
If you look at graph 3A1 then the same fuse with a fault current of ~810A will disconnect in 0.1s and so require a 1.79mm² csa conductor.

If you have the manufacturers I²t values for your protective device then this will give a more accurate minimum size conductor and does not require a value for current or time as this is combined in the I²t value given and this value can be used directly in the adiabatic equation.
 
Hi all, just working through an assignment at the moment, got everything done other than two questions worded "select an appropriate minimum sized CPC and confirm your selection by calculation". I've already calculated my tabulated current, volt drop and live conductor size for each question.

Our teacher said we should be using the adiabatic equation, but we haven't been given a Ze, in my mind I won't be able to calculate the PSC because I can't just assume the Ze. It's TN C S, but if I assume the maximum Ze of 0.35 and it is actually lower, the PFC will be higher and my calculation will be incorrect.

Am I missing something obvious here or should I be using a different equation?

Any drawings with the assignment?
 
Hi all, just working through an assignment at the moment, got everything done other than two questions worded "select an appropriate minimum sized CPC and confirm your selection by calculation". I've already calculated my tabulated current, volt drop and live conductor size for each question.

Our teacher said we should be using the adiabatic equation, but we haven't been given a Ze, in my mind I won't be able to calculate the PSC because I can't just assume the Ze. It's TN C S, but if I assume the maximum Ze of 0.35 and it is actually lower, the PFC will be higher and my calculation will be incorrect.

Am I missing something obvious here or should I be using a different equation?

I have always been told if data is missing make an assumption and articulate this within your answer, as long as the working out is correct then there will be no issues, they just want to see you can work through the calculation by the sounds of it.
 
Thank you all for taking the time to answer, it's gonna be a couple of days till i have the time to open the regs book again and have another wack at it. I'm going to have to make an assumption somewhere as no information on protective devices has been given other than the circuits are protected by bsen 60898 mcb's. I think the best assumption to make is of the Ze, im thinking 0.1 would be a reasonable choice.

Simonslimline - no drawings of the circuit, only a layout. I've calculated the line conductors as 1 and 1.5 and the circuits are 20 and 8m respectivly.
 
For BSEN60898 type B MCBs then you can state that using the adiabatic equation the maximum size of cpc bunched in a cable would be at the minimum current (at x5 the current rating) for "instantaneous" disconnection in a maximum time of 0.1s
this gives the following minimum cpc sizes
32A
0.44mm²
20A
0.27mm²
16A
0.22mm²
10A
0.14mm²
6A
0.08mm²

Therefore in all cases a 1.0mm² cpc would be adequate for your circuit at any current rating up to 32A, however minimum sizes are stipulated in table 52.3 and so for power a 1.5mm² cpc would be used.
 
Cheers mate, that does make sense, but surely once in operation a pcf in the region of 0.5kA could flow, does that not make the calculation and hence conductor size incorrect?
 
OK I can see your point here for MCBs.
If the fault current rises and the time to trip stays the same then the csa of the cpc will increase.

The problem with the graph in Appendix 3 is that they effectively stop at 0.1s however MCBs will trip slightly faster as the fault current gets higher, but this is not shown on those graphs.
However the minimum csa will not increase above the 1.5mm² cable as this is why it is at that size.

View attachment Eaton MCB trip curves.pdf
E.g, this graph from Eaton shows that a B32 trips at 5 X rating in 0.02s giving a 0.02mm² csa and at 500A (15.6 on the graph) a trip time of 0.014s and so a csa of 0.51mm².

Using Hager let through energy values at 6kA for a B32 the I²t is 24000 which gives a csa from the adiabatic of 1.37mm².
Still less than 1.5mm².

(lost the post three times while typing so I hope it makes sense)
 
Thanks so much for explaining this, it makes a lot more sense now! The problem i'm stuck with now is how to prove by calculation the csa I've selected, although all the information you've provided makes sense, I'm still going to need a PFC to put in my equation, I'm thinking assuming a Ze would be the easiest route.

If I use 6KA, I will need 1.5mm cpc for my 1mm line :sick:
 
Go for it, I think the fact that you are demonstrating the use of the adiabatic equation correctly will be more critical in this case than the actual value of cpc obtained.
If you have made an assumption on the value of Ze and that assumption is stated clearly in your working then you cannot go wrong.

(for a 6A MCB (say for lighting) Hager gives an I²t of 10,500 that leads to a 0.89 mm² minimum csa for a cpc)
 

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