# Calculating R1 R2 and finding a suitable CPC

Discussion in 'Electricians' Talk' started by RDB85, Jun 5, 2019.

1. ### RDB85Member

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: Jun 7, 2019
2. ### ColoumbWell-Known Member

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. ### RDB85Member

How would you do that? As to calculate Fault Current the OSG says:

I=Uo/Zs.

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. ### ColoumbWell-Known Member

How can you calculate the fault current first if you don't know the Zs?

Look...you have already done it previously...

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

so it follows...

I = Uo/Zs

5. ### peter palmerWell-Known 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. ### RDB85Member

19.51x30x1.20/1000 = 0.70236 + 0.11 = 0.81236Ω

7. ### ColoumbWell-Known Member

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.

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8. ### RDB85Member

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 palmerWell-Known 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 palmerWell-Known Member

MCBs

Fuses

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11. ### RDB85Member

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: Jun 7, 2019
12. ### ajohnActive Member

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

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.

John
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13. ### ajohnActive Member

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.

John
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14. ### peter palmerWell-Known 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. ### RDB85Member

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. ### ajohnActive Member

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.

John
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17. ### tillieNew 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

Regards

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18. ### RDB85Member

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. ### ColoumbWell-Known Member

sorry ignore me

Last edited: Jun 7, 2019
20. ### chippie244Well-Known Member

I love the sparkies arguments