Calculating R1 R2 and finding a suitable CPC

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

  1. tillie

    tillie New Member

    If you measure the cicuit then depending on your conduit route it will not be any longer than 15mtrs ( I measured it at 12 mtrs ).

    Personally I would use 0.1 secs.

    The equation t = k2 x s2 / I2 will give you the maximum time allowed before a cable becomes thermally damaged not the actual disconnection time.

    In your calculation of R1 + R2 you have only included one conductor.

    Regards
     
    Deleted member 176520 likes this.
    • Calculate Actual R1+R2 = 7.41+7.41x30x1.20 / 1000 = 0.275652Ω + 0.11Ω = 0.385652Ω
    • 230/0.38 = 605A
    • 605² x 0.1 = 36602.5
    • Square Root of 36602.5 =191.3177984/115 = 1.66mm²
    • Cable Size 1.66mm²
    • Chosen cable:2.5mm²
    I will leave it as 30m, then once the drawings have been looked at, I will then change it to the actual measurement. Thanks for the help.
     
  2. ajohn

    ajohn Well-Known Member

    7.41+7.41x30 is for 31m of cable. ;) Or that's how I read it. Not the same as 30x1.2/1000(7.41+7.41) which is for two 30m lengths of 2.5mm cable.

    :) I find the rest interesting. It more or less says that 30m of 2.5mm twin and earth isn't any good. :) Good grief. Sounds like it'll cost me a fortune when our electrics are upgraded and checked as the earth wire isn't big enough even for 12m.

    John
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    Deleted member 176520 likes this.
  3. unphased

    unphased Screwfix Select

    So what cable are you actually using? Singles in conduit? 2.5mm2 6242Y (T&E) is manufactured with a 1.5mm2 cpc. You can get it with a 2.5mm2 cpc but it is a special and you would pay a premium for it. Or are you using another cable type? Just curious.
     
  4. dobbie

    dobbie Well-Known Member

    Readily available on the internet from Ireland and the earth is also sheathed.
     
  5. Coloumb

    Coloumb Well-Known Member

    Yes but it would be right on the very limit of the Zs requirements, so in all probability it wouldn't work like that. Plus there are limiting factors for the cpc size if the ctt. is supplying fixed equipment or sockets so as to limit the touch voltage. In fact you would more likely have to fit an RCD. The point I was trying to make was that the time and fault current are related, so you wouldn't design a circuit by saying "what the maximum amount of time I'm allowed to clear the fault" and work forwards from that.
     
  6. Coloumb

    Coloumb Well-Known Member

    In your first example the cpc size is shown as too small but in reality you could use 2.5mm twin with a 20/16a breaker. The issue is that the let through energy limits of the breaker need to taken into account. This would reduce the disconnection time to ~.01 and the cpc size of 1.5mm would be fine. IMO it's not correct to treat the time as a constant. It leads to misleading results.
     
  7. ajohn

    ajohn Well-Known Member

    Some aspects of this thread bemuse me but pass on why other than precise calcs would be more complicated.

    But this is the info manufacturers provide. Eaton in this case as I thought that would presented neatly

    ZsMCBs.jpg

    TripTimeMCBs.jpg

    The 2nd one is particularly interesting as that states they must trip in 0.02sec at 5 times their rated current. I also shows that the magnetic aspect of the trip comes into action cleanly at the same point. They can't be persuaded to trip in 0.1sec. Then there is the other one, Zs figures. A 16amp one will trip at 230/3=76,7amps, 4.8x It's rated current. Funnily enough that gives a circa 5sec trip time. It's of interest as that one must include the resistance of the mcb itself and very very little more is needed for the trip time to plummet rather abruptly.

    John
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  8. ajohn

    ajohn Well-Known Member

    LOL I don't, it worries me. Take another aspect - the word adiabatic and what it actually means

    "An adiabatic process occurs without transfer of heat or mass of substances between a thermodynamic system and its surroundings. In an adiabatic process, energy is transferred to the surroundings only as work.[1][2] The adiabatic process provides a rigorous conceptual basis for the theory used to expound the first law of thermodynamics, and as such it is a key concept in thermodynamics.

    Some chemical and physical processes occur so rapidly that they may be conveniently described by the term "adiabatic approximation", meaning that there is not enough time for the transfer of energy as heat to take place to or from the system.[3]"

    The approximation used on cable R looks to assume that the copper may reach 120C. Heat transfer isn't fast especially with things like PVC so why use pure jargon rather than say this approximate equation can be used and why it can only be approximate and it's limitations.

    :) Next time I want some cable I must ask for twin & cpc :) :) again jargon, it's an earth wire ( or should be if it's actually connected to it )

    Why is jargon introduced ? Some probably know but it can also interfere with learning.

    John
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  9. ajohn

    ajohn Well-Known Member

    Why not start by saying that cpc and live use the same size of wire.

    so max R = 2.62/1.2 = 2R18 may as well apply the correction factor here.
    R/m = 2R18 /60 = 0R036/m even 1mm is better than that

    Then try what power cable size can be used with a 16A mcb which is 2.5/1.5mm^2

    Total R is 30x0R0071+30x0R0121 = 0R213 + 0R363 = 0R576 way way less than the mcb Zs giving a fault current of about 400 amps = around 25 x the rating of the mcb

    Or for min cpc ;) not that I think you need it but maybe to verify with the equation
    2R62-0R213=2R4
    Max R/m - 2R4/30 = 0R080/m

    :) I use a cranky calculator so do check. :) I mean it even has an X<->Y and brackets if I need them. I've used electronics notation R shows where the decimal point is.

    If totally wrong can some one point out why.

    John
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  10. peter palmer

    peter palmer Well-Known Member

    Why has the graph got 2 curves on it joined at the top and bottom?
     
  11. Coloumb

    Coloumb Well-Known Member

    I think it's to show the discrimination range against bigger breakers. Not actually a time/current chart per se.
     
  12. Thanks for the help. I’ve spoken to the tutor and he said that I need to work out my Zs which is 230/0.8 to get Max Zs.

    Still unsure how to work out the R1 R2. As he said what I had done was not quite right.
     
  13. ajohn

    ajohn Well-Known Member

    I'm sure what I have just posted is correct as you can't do better than the manufacturers data and you have the one that matters Zs of the breaker. It seems from the guide that you can only use 2.5mm on the power so that sets one of the R's. If you ;) use 2.5mm twin & cpc ;) the earth wire included sets the resistance of the cpc. That gives the total circuit resistance and the prospective min fault current, way way over what is needed for a 5 sec break.

    The last calc I did is the max R/metre for the cpc. If you need diameter you'll have to convert it. So one R is the power feed, 2.5mm and the other is simply what is needed to make that = the Zs of the breaker. That leads to a < 5 sec break time on a worst case mcb. If you need shorter than that things get difficult as in real terms there isn't sufficient info available to do it. The fact that a factor of 1.2 is used would do it but that seems to be for wire temperature increase. If someone chose to have a fault current a little over 8 times the rating it will break as fast as it is able to. That's what the little bump at the bottom of the curve shows.

    It is a time current curve. In this case rather than having a separate chart for each one they specify it as ratios of the rating. However if some one wanted to use a curve for a specific make of breaker OK no difference really. All that can be read from them is similar to what can be read on this one. Very short break times need currents over 5 times the rating. A current that is 5 times the rating will open in under 5 secs. That ties in with doing the same thing via the Zs of the breaker. It's aimed at breaking in under 5 secs.

    :) Really the way the initial question was asked it would have been better to post the lot as to me it's still not clear exactly what you need to use in the equation - what would be used for the job - 2.5 T&W or some fictitious cabe that meets the power aspects of the breaker and just meets the Zs requirements of the breaker. There are several reasons why some one might be asked to do that as daft as it is. There doesn't seem to a single example of using an mcb on the web. Plenty on using fuses, Mcb's have an entirely different method of operating. Part thermal and part magnetic. The magnetic one seems to be most important. Some could for instance go for a ratio of twice the rating of the mcb. That would give an opening time between about 8 and 40 secs.

    The reason for the 2 curves is the spec - the range of 5 sec opening times can be from 3 to 5 times the stated rating. That is the spec for an mcb. The dotted line represents magnetic and mechanical variations. The flat line at the bottom at 10mSec is due to the fact that it takes time for things to move. That is part mechanical so at some point once it's there more current can't speed it up any more.

    John
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  14. This is the Question as its presented :
     

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  15. ajohn

    ajohn Well-Known Member

    Zs gives the max total R for the fault loop. 16A mcb means 2.5mm line. That will take up some of the resistance limit as per your calcs. The question doesn't say anything about minimum csa for the cpc so try 2.5mm t&e. Calc the total resistance including the 1.5mm cpc in that. If less than Zs all ok and you have your r1 and r2 so stick them through the formulae.

    The other check that should be done if a length is available is volts drop on the line.. That could cause you to go up in line size. I'd assume that you would use 16A for that, the rating of the mcb.

    If in the unlikely event the sum of the resistance was greater than the Zs of the mcb you would simply pick a larger size for the cpc. That is unlikely to happen with standard cables used within their rating. ;) Why - well there would be no point in making them if it could happen. The limit on twin and earth looks to be voltage drop. You could calculate that based on a 2.5mm line.

    Or if you wanted you could assume separate wires. In that case 2.5mm line and it looks like a 1mm cpc would be more than sufficient but that isn't a solution any one would be likely to use in practice.

    Personally for my own curiosity I would also use the equation with a cpc that just meets the Zs of the mcb, No need for a wire size, it's just what's left assuming a 2,5mm line.

    John
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    Deleted member 176520 likes this.
  16. ajohn

    ajohn Well-Known Member

    The only other thing I can add concerns disconnect time. If the fault loop resistance accounting for derating = mcb Zs it will be less than 5 secs. Past that increasing current just produces the same disconnect time as per the curves with no oddities that matter. So for short disconnect times the fault loop resistance must give a fault current of > 5 x the mcb's rating. A curve might show that a factor of 6x it's rating will definitely do that. So anything higher will do. The limit will be the break capacity of the mcb. Seems that's 6KA so would need a fault loop resistance of 230/6000 = 0R038 to reach that.

    John
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  17. ajohn

    ajohn Well-Known Member

    :) Thanks for posting this. It's interesting. :oops: Say you also have an rcb somewhere. Looks like that may well trip first / same time etc. That in turns suggests rcbo's are a better idea as the mcb's main purpose in life seems to be if the rcb doesn't trip. In fact the guide suggests using an rcb if the cpc link has too much resistance to suite an mcb. It now also states that mcb's trip at between 3 and 5 times their rating.

    John
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    Deleted member 176520 likes this.
  18. I’ve been told to use this calculation for finding R1+R2

    Max Zs - Ze / Table I3 factor x length times overall by a 1000.

    Working that out

    Zs 16A is 2.73
    Ze 0.11
    Factor 1.2
    Length using a scale ruler is 12m

    It comes out at 187 which seems wrong.

    Also for the time we were told that 0.1 seconds is fine.
     
    Last edited by a moderator: Jun 12, 2019
  19. ajohn

    ajohn Well-Known Member

    2.73-0,11= 2.62 apply factor 2.62 divided by1.2=2.18 That includes the factor for the wire that will be used so don't apply it again

    12m of 2.5mm = 12 x 0.00741 = 0.0889

    12m of 1.5mm = 12 x 0.0121 = 0.1452

    Those are your R's. The mohms/m I have used are from the guide.

    So the prospective min fault current is 230 / (0.0889+0.1452) = 230 / 0.2341 = 982A

    All that you can say about trip time is that at currents indicated by the Zs of the breaker which in this case is 230 / 2.73 = 83,2A mcb's are guaranteed to trip in under 5 secs and that given typical mcb curves 982A is so much greater than that and 61 times it's rating it will trip in a very short time, probably less than 0.1 sec. The curve I posted is typical. However Appendix I of the guide gives a formulae for calculating it. Frankly I can't see why as they are too variable for that to be accurate as some one fitting one wont know where it is in range for 3 to 5 times it's ratings 5 sec break time. :) but sounds like you should use it. It must also be in the BS.

    Then use the R's in your formulae but if it includes derating factors use R values that don't already include it. I don't know what the formulae is or involves but would expect it to try and account for wire heating up.

    John
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