Calculating R1 R2 and finding a suitable CPC

Discussion in 'Electricians' Talk' started by Deleted member 176520, Jun 5, 2019.

  1. Cable Size: 2.5mm2 Ref Method B 24A VD 18
    Fault Current: 230/0.11 = 2090.90A

    Then to work out R1 + R2 = 19.51x30x1.20/1000 = 70.236Ω

    But then how do you do: thermal constraint (S) as you cant get the time as the Fault Current is too high, and we are not given any manufacturers details on MCBs.
    Last edited by a moderator: Jun 7, 2019
  2. Coloumb

    Coloumb Screwfix Select

    NO. For the fault current you need work out the Zs first. Your calculation would be for the fault current at the origin, not the end of the ctt.
  3. How would you do that? As to calculate Fault Current the OSG says:


    Would you then use the Zs given in BS7671 to do this?

    Would this also help:
    Regulation 434.5.2 of BS 7671:2018 (page 92) provides an equation for
    calculating the maximum operating time of the protective device
  4. Coloumb

    Coloumb Screwfix Select

    How can you calculate the fault current first if you don't know the Zs? have already done it previously...

    R1 + R2 + Ze = Zs (at the end of the ctt)

    so it follows...

    I = Uo/Zs
  5. peter palmer

    peter palmer Super Member

    The "t" part of the equation is how long the device takes to operate at any given amount of current. If you worked out that the fault current was say 200A for a given Zs you would then look at the curve graphs in 7671 for that particular device and that would give the time in which it would operate, that is your t value.

    Thats the way I understand it anyway but willing to be told otherwise, not that I've ever needed to do it that way, I just use whatever is on the van.
  6. 19.51x30x1.20/1000 = 0.70236 + 0.11 = 0.81236Ω
  7. Coloumb

    Coloumb Screwfix Select

    The Max Zs values for any given breaker are derived from the max amount of current that will clear a fault .4s so to limit the touch voltage to 50v. This will ALWAYS be less than the amount of current that occurs if the fault is a short to earth. This is why we would use a adiabatic equation to find the smallest size the cpc needs to be to clear the largest fault current.
    Deleted member 176520 likes this.
  8. Is what I have so far okay? So the next part would be the fault current:

    I = Uo/Zs 230/0.81 = 284A
  9. peter palmer

    peter palmer Super Member

    Yes but if the fault current is higher than the max Zs value gives then it will trip faster than 0.4 seconds, however I've just looked in the book of nonsense and the graph actually gives the same fault current value (80A) for a 16A MCB all the way from 0.1 seconds to about 12 seconds and the quick reference table says 80A from 1 to 10 seconds.

    Its fuses where the value changes all the way down to 0.1 seconds, for eg a 125A BS88 fuse will blow in 0.1 seconds with a fault current of 1750A but take 5 seconds if the current is only 680A. So if you did the adiabatic equation and you have a fault current of 680A you would use 5 for the "t" value in it.
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  10. peter palmer

    peter palmer Super Member





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  11. In order to work out t I've done

    t= 115²x2.5+1.5²/284² = 0.409947803

    284²x0.409947803= 32832.30959
    Square Root: 32832.30959/115 = 1.57562505
    Last edited by a moderator: Jun 7, 2019
  12. ajohn

    ajohn Screwfix Select

    Might it be that Zs for the breaker has to be used otherwise why include it. Something is needed as far as the phase is concerned anyway. Could be that they also have a specified max or min resistance. In that case the phase wire resistance could be calculated using the one that results in lowest fault current.

    So total known resistance is 2.73+0.11 = 2.84 ohms

    So assume the earth wire has zero resistance the fault current would be 230/2.84=80.98amps

    I can only assume that the breaker must achieve some current to break time requirement so the inclusion of a real earth wire would reduce the fault current and at some point the result would be unacceptable.

    Past that this might help 3 07 P 10-11 16th Cable Sizing 2 of 2.pdf

    ;) Couldn't help being curious but most info on the web relates to fuses. Those have a current to break time relationship. All I know about breakers is that at some current higher than their rating the break time doesn't change much as the current increased, it's short. Above it the time to break increases as the current is reduced so BS should give some usable figures some where that they must meet and some limitation on how long the break time can be.

    That can then be used to size a conductor. So another approach would be to find out what is an acceptable current for the breaker to open at and work back from that. Ohms law I=E/R or maybe there is an electricians version of that which gives csa directly.

    :) The use of Z gets me. Normally it's used for impedance which isn't the same as resistance however maybe your testers measure using 200ma at 50hz rather than dc.

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  13. ajohn

    ajohn Screwfix Select

    Actually I don't think that the 2.73ohms makes much sense. If 30m of wire it's way to high. Around 91 mohms per metre. 2.5mm is 7.4mohms per metre, even 0.5mm is only 36mohms / metre.

    Makes me wonder if a decimal point is in the wrong place. 2.73ohms at 16amps would be dissipating nearly 700w at 16amps - make a good low power immersion heater, green house heater etc.

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  14. peter palmer

    peter palmer Super Member

    I don't really get it either, as for the graphs, why is the time curve static for MCBs (80A for a 16A one) all the way up to 12 seconds or so where as for a fuse its still curved, surely the higher the fault current the quicker it will trip.
  15. I’m confused tbh. As I don’t know if what I had first was right, or what I last done today is right. Would anyone else mind having a go at the calculations i steps to compare with mine?
  16. ajohn

    ajohn Screwfix Select

    They are using log scales on one axis that I noticed so didn't look any more. I have seen a linear one some where or the other and the result is a curve eventually changing into more of a step down to a trip time which then varies less with increasing current. A common way of specking fuses is giving them an IT rating ;) or maybe it's I^2 T. That suggests that they fail at some point in time what ever the current is so is of more use in the region they are known to blow at.

  17. tillie

    tillie New Member

    I actually helped my apprentice with this question.

    You are doing the swimming pool/cafe project.

    I used a scale ruler and measured the circuit which was about 12 mtrs long .

    I first tried 2.5mm line conductor with a 1.5mm cpc which when calculted becomes

    ( 12.1 + 7.41 ) x 1.2 x 12 / 1000 = 0.28 + 0.11 ( ze ) = 0.39 230/0.39 = 589a fault current. Diconnection time = 0.1 secs

    589 x 589 x 0.1 squared /115 = 1.6mm

    So cpc too small.

    Try 2.5mm line conductor and a 2.5mm cpc when calculated becomes

    ( 7.41 + 7.41 ) x 1.2 x 12 / 1000 = 0.21 +0.11 ( ze ) = 0.32 230/0.32 = 718a fault current . Disconnection time = 0.1 secs

    718 x 718 x 0.1 squared / 115 = 1.9mm

    2.5mm cpc adequate.

    Deleted member 176520 likes this.
  18. That's the project. I have used 30m as my circuit length, the guidance was more than 15m. I got 2.5mm in the end. Mine was a little different, calculations.

    • Calculate Actual R1+R2 = 7.41x30x1.20 / 1000 = 0.26676Ω + 0.11Ω = 0.37676Ω
    • 230/0.37 = 621A
    • I then modified to get an exact time: t = k²S²/I² = 115²*2.5²/621² = 0.21s
    • 621² x 0.21 = 80984.61
    • Square Root of 82993.9 = 284.5779507/115 = 2.475mm²
    • Cable Size 2.5mm²
  19. Coloumb

    Coloumb Screwfix Select

    sorry ignore me
    Last edited: Jun 7, 2019
  20. chippie244

    chippie244 Super Member

    I love the sparkies arguments :):)

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