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Adiabatic short-circuit compliance on very short short-circuits
Hi, This is an abstract question - but I'm just interested to see what I've misunderstood here... Imagine a domestic installation, with a PSC of 6kA at the incoming supply, fused at 100A. Right next to the incoming supply, there's a consumer unit, which contains one 20A MCB. The CU is connected to the incoming supplies with very short 25mmsq tails (I've ignored their contribution to lowering the fault level.) This 20A MCB supplies a short (let's say 0.5m) piece of 2.5mmsq T&E, which supplies a single socket next to the consumer unit. My calculations suggest that the L-N fault current at the socket is going to be about 5kA, and for adiabatic I^2t or (kS)^2 compliance, we would need to break a short circuit in around 3ms max to avoid damaging the 2.5mm cable. This seems to be well off the bottom of graphs for MCB response times. All examples I can find around the place of using the adiabatic temperature rise equations give answers which are conveniently above 0.1ms, which allows slightly glib commentary about MCBs always tripping in under 100ms, etc. Although this was a completely contrived example, it actually applies to any S/C fault on any final circuit which occurs sufficiently close to the CU. Something like this: http://groups-beta.google.com/group/...e=source&hl=en contains an example adiabatic compliance calculation, but it appears to me that it only considers a short occurring at the end of the final circuit, not close to the beginning. (It's also considering L-PE rather than L-N, but I don't think that's an important distinction here) I'm clear why *load* is considered only at the end of a circuit, but not why fault conditions are being calculated for the end of the circuit. Given that fuse and MCB time vs. current graphs don't tend to go down to single-millisecond levels, should one actually be looking at I2t let-through graphs and comparing that with k2S2? TIA, Will (hoping that Andy and Andrew aren't both on holiday!) |
In article ,
"Will Dean" writes: Hi, This is an abstract question - but I'm just interested to see what I've misunderstood here... Imagine a domestic installation, with a PSC of 6kA at the incoming supply, fused at 100A. Right next to the incoming supply, there's a consumer unit, which contains one 20A MCB. The CU is connected to the incoming supplies with very short 25mmsq tails (I've ignored their contribution to lowering the fault level.) This 20A MCB supplies a short (let's say 0.5m) piece of 2.5mmsq T&E, which supplies a single socket next to the consumer unit. My calculations suggest that the L-N fault current at the socket is going to be about 5kA, and for adiabatic I^2t or (kS)^2 compliance, we would need to break a short circuit in around 3ms max to avoid damaging the 2.5mm cable. Really you want Andy for this, but the 5kA sounds wrong to me. Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms. Add to that the 0.5m of 2.5mmsq T&E, which will have a resistance of 0.09 ohms, and we have 0.13 ohms. 240V across 0.13 ohms is only 1.846kA. Being a physicist rather than an electronic engineer, I'll work out the temperature rise from first principles rather than using the method in BS7671. Let's assume fastest MCB trip time possible is half a cycle, 0.01 sec. 500cm of 2.5mm^2 cable during a fault current of 1.846kA for 0.01 sec. Mass of copper is 8.96g/cm^3 * 0.025cm^2 * 500cm * 2 = 224g, Specific heat capacity of copper = 0.385J/g/K... Power dissipated per 500cm: P = I^2 * R = 1846^2 * 0.09 = 306kW Energy dissipated per 500cm during 0.1 sec: 306kW * 0.01 = 3060J. Temp rise = 3060 / 0.385 / 224 = 35.5 Centigrade Thus a fault current of 1.846kA for 0.1 sec will result in a temperature rise of about 35.5 Centigrade in the copper conductors. If cable was previously within its correct operating temperature range (i.e. = 70 C), it will survive this without damage (i.e. not exceed 160 C for PVC). -- Andrew Gabriel |
[cancaled earlier posting due to mistake in calculation]
In article , "Will Dean" writes: Hi, This is an abstract question - but I'm just interested to see what I've misunderstood here... Imagine a domestic installation, with a PSC of 6kA at the incoming supply, fused at 100A. Right next to the incoming supply, there's a consumer unit, which contains one 20A MCB. The CU is connected to the incoming supplies with very short 25mmsq tails (I've ignored their contribution to lowering the fault level.) This 20A MCB supplies a short (let's say 0.5m) piece of 2.5mmsq T&E, which supplies a single socket next to the consumer unit. My calculations suggest that the L-N fault current at the socket is going to be about 5kA, and for adiabatic I^2t or (kS)^2 compliance, we would need to break a short circuit in around 3ms max to avoid damaging the 2.5mm cable. Really you want Andy for this, but the 5kA sounds wrong to me. Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms. Add to that the 0.5m of 2.5mmsq T&E, which will have a resistance of 0.09 ohms, and we have 0.13 ohms. 240V across 0.13 ohms is only 1.846kA. Being a physicist rather than an electronic engineer, I'll work out the temperature rise from first principles rather than using the method in BS7671. Let's assume fastest MCB trip time possible is half a cycle, 0.01 sec. Specific heat capacity of copper = 0.385J/g/K Density of copper = 8.96g/cm^3 So, mass of copper in 500cm of 2.5mm^2 cable is 8.96g/cm^3 * 0.025cm^2 * 500cm * 2 = 224g Power dissipated per 500cm: P = I^2 * R = 1846^2 * 0.09 = 306kW Energy dissipated per 500cm during 0.01 sec: 306kW * 0.01 = 3060J. Temp rise = 3060 / 0.385 / 224 = 35.5 Centigrade Thus a fault current of 1.846kA for 0.01 sec will result in a temperature rise of about 35.5 Centigrade in the copper conductors. If cable was previously within its correct operating temperature range (i.e. = 70 C), it will survive this without damage (i.e. not exceed 160 C for PVC). -- Andrew Gabriel |
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"Andrew Gabriel" wrote in message
.. . Really you want Andy for this, but the 5kA sounds wrong to me. Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms. Yup, agree that. Add to that the 0.5m of 2.5mmsq T&E, which will have a resistance of 0.09 ohms, I don't get that - 90 mOhm. I think it's more like 7 mOhm. Cheers, Will |
In article ,
"Will Dean" writes: "Andrew Gabriel" wrote in message .. . Really you want Andy for this, but the 5kA sounds wrong to me. Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms. Yup, agree that. Add to that the 0.5m of 2.5mmsq T&E, which will have a resistance of 0.09 ohms, I don't get that - 90 mOhm. I think it's more like 7 mOhm. Yes, I got it wrong. It should be 0.009 ohms -- Table 4D2B claims 18 mOhms/metre. I'm not sure where your 7 mOhm comes from, although it's a lot nearer than my 90 mOhms ;-) There's another mistake too -- I calculated the temperature rise for the mass of copper in 5m of cable rather than 0.5m. These two combined make the result way off. I'll do the calculation and post again. -- Andrew Gabriel |
"Andrew Gabriel" wrote in message
.. . Yes, I got it wrong. It should be 0.009 ohms -- Table 4D2B claims 18 mOhms/metre. I'm not sure where your 7 mOhm comes from, although it's a lot nearer than my 90 mOhms ;-) I suspect the 9 vs 7 is temperature related. Table 9A in the OSG gives 14.82/2 = 7.41 Using 16.8nOhm-metres (the first figure I stumbled across on the internet - I don't know what temperature it's for) gives something around 7-ish, too. There's another mistake too -- I calculated the temperature rise for the mass of copper in 5m of cable rather than 0.5m. A physicist, you said? Hmm. Cheers, Will |
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Will Dean wrote:
"Andrew Gabriel" wrote There's another mistake too -- I calculated the temperature rise for the mass of copper in 5m of cable rather than 0.5m. A physicist, you said? Hmm. Physicist mathematician :-) patronising:-) I think he did very well /patronising:-) Owain |
"Andrew Gabriel" wrote in message
.. . So you are always going to get some length of 2.5mm^2 T&E which is inadiquately protected if the PSC at supply end is greater than 2936A. In the case of your 6kA PSC supply and 2.5mm^2 T&E, that's going to be the first 2.3 metres of cable (if I did the calculation right;-) OK, we're sort of agreeing now. Given that all circuits of = 2.3 metres are susceptible to faults WITHIN the first 2.3 metres, this situation must pertain for almost every final circuit in real life. I can only assume that one must check I2t let-through figures for the protection device, rather than disconnection time graphs. What I don't understand is why some of the examples I've seen seem to include the final circuit's own loop impedance in this calculation - that seems to be a mistake. Cheers, Will |
On Fri, 05 Aug 2005 18:06:10 +0100, Will Dean wrote:
"Andrew Gabriel" wrote in message .. . So you are always going to get some length of 2.5mm^2 T&E which is inadiquately protected if the PSC at supply end is greater than 2936A. In the case of your 6kA PSC supply and 2.5mm^2 T&E, that's going to be the first 2.3 metres of cable (if I did the calculation right;-) OK, we're sort of agreeing now. Given that all circuits of = 2.3 metres are susceptible to faults WITHIN the first 2.3 metres, this situation must pertain for almost every final circuit in real life. In real life PSC of 6kA are uncommon although theoretically possible. A plausible scenario would be a cleaners' socket in a meter cupboard on the ground floor of a block of flats containing the sub-station in the basement. I can only assume that one must check I2t let-through figures for the protection device, rather than disconnection time graphs. What I don't understand is why some of the examples I've seen seem to include the final circuit's own loop impedance in this calculation - that seems to be a mistake. I thought the resistance of the MCB might come to our rescue but I've just measured one and it's DC resistance is less than 0.01 ohms as measured by good test gear. Likewise the meter might help out a bit? However the AC impedance might be quite a bit more as MCBs do get a little warm in operation. On such short circuits what fault could you have that both gave rise to a full short circuit and which did not implicate the cable. -- Ed Sirett - Property maintainer and registered gas fitter. The FAQ for uk.diy is at http://www.diyfaq.org.uk Gas fitting FAQ http://www.makewrite.demon.co.uk/GasFitting.html Sealed CH FAQ http://www.makewrite.demon.co.uk/SealedCH.html |
"Ed Sirett" wrote in message
n.co.uk... In real life PSC of 6kA are uncommon although theoretically possible. A plausible scenario would be a cleaners' socket in a meter cupboard on the ground floor of a block of flats containing the sub-station in the basement. I quite agree, but the socket is an irrelevance really - I'm talking about any S/C fault which occurs close to the start of a final circuit, and therefore doesn't benefit from the reduction in fault level which the final circuit gives. I thought the resistance of the MCB might come to our rescue but I've just measured one and it's DC resistance is less than 0.01 ohms as measured by good test gear. Likewise the meter might help out a bit? However the AC impedance might be quite a bit more as MCBs do get a little warm in operation. Pure reactance ("AC impedance") wouldn't create any heat. On such short circuits what fault could you have that both gave rise to a full short circuit and which did not implicate the cable. Well, I'm really talking about cable faults in this case (e.g. someone bangs something large and metallic right through the cable). What I'm fishing for is the relevance/application of the adiabatic calculations in such situations. Will |
In message ,
wrote: I have been wary of MCBs in high PFC places ever since. Sub station was through the wall to the CU. I suspect the PSC was in the region of 10 kA. And as you will notice, most MCBs are rated to break 6kA, some (notably the Wylex wire-replacement types) are lower. To butt in a bit late, just to back up another poster, I've never measured a PSC higher than about 1.6kA at the end of the meter tails, and the majority are below 1kA. Possibly has something to do with not living in the centre of London, but although it's an interesting question, I'm not sure how relevant it is in "real life". Taking the calculation that the first 2.3m or so of 2.5mm2 cable is "at risk" and the fact that in order to create a short this close to a CU the most likely method is a nail through the cable, is it really a problem if the heat dissipation damages the cable? You're going to be replacing the nail-damaged length anyway... Hwyl! M. -- Martin Angove: http://www.tridwr.demon.co.uk/ Two free issues: http://www.livtech.co.uk/ Living With Technology .... Get 'em by the balls and their hearts and minds will follow. |
Will Dean wrote:
This is an abstract question - but I'm just interested to see what I've misunderstood here... Hi Will, sorry to come in a bit late to this. Hope you're still around... Imagine a domestic installation, with a PSC of 6kA at the incoming supply, fused at 100A. Right next to the incoming supply, there's a consumer unit, which contains one 20A MCB. The CU is connected to the incoming supplies with very short 25mmsq tails (I've ignored their contribution to lowering the fault level.) This 20A MCB supplies a short (let's say 0.5m) piece of 2.5mmsq T&E, which supplies a single socket next to the consumer unit. My calculations suggest that the L-N fault current at the socket is going to be about 5kA, and for adiabatic I^2t or (kS)^2 compliance, we would need to break a short circuit in around 3ms max to avoid damaging the 2.5mm cable. All figures agreed. This seems to be well off the bottom of graphs for MCB response times. [...] I'm clear why *load* is considered only at the end of a circuit, but not why fault conditions are being calculated for the end of the circuit. Given that fuse and MCB time vs. current graphs don't tend to go down to single-millisecond levels, should one actually be looking at I2t let-through graphs and comparing that with k2S2? Yes, you raise some interesting points that a designer does need to be aware of when the prospective fault level is high. Firstly, the reason that the fault calculation is usually done at the furthest point of the circuit is that that is /usually/ the worst case. When we're not at the bottom of the graph, as you put it, a lower fault current leads to higher I^2*t let-through and hence more risk of cable damage. There's a useful graphical approach he for any given cable CSA and 'k' value (k=115 for T&&E) you can superimpose an adiabatic withstand line for the conductor on the operating characteristic plot for the fuse or MCB. It's a log-log graph, so the cable's line is straight, with a gradient of -2 (t proportional to 1/I^2). You then see at a glance the minimum fault current at which the fuse or MCB will definitely protect the cable, i.e. the point of intersection of the two lines. With some (non-current-limiting) MCBs there may be a second intersection at high fault level. At higher currents the device's curve is now above the cable's line - meaning that the cable will not be protected. So, yes, you do need to refer to manufacturer's I^2*t data, which will almost certainly show that your bit of 2.5mm^2 wire is protected after all. Most MCBs (and HRC fuses) now are 'current-limiting' - meaning that at high current they will trip and quench the arc so quickly that the instantaneous current never has time to build up to its full prospective value. (Even if the fault occurs at the peak of the voltage waveform the inevitable inductance in the circuit means that the current has to rise from zero at a finite rate.) Secondly, you might find that at 5kA the supplier's main fuse beats the MCB - look at the characteristics for a 100A BS 1361 Type 2 fuse here http://www.bussmann.co.uk/images/Dat...S1361/LR85.pdf and extrapolate the line a little. Or use the I^2*t value given (57,300 A^2s) and conclude that this 100A fuse would protect a 2.5 mm^2 copper conductor. This becomes an important point once the PSSC exceeds the rated breaking capacity of the MCB - which is usually 6 kA or 9 kA (M6 or M9) - as the supplier's fuse is now acting as the backup protective device. Thirdly BS 7671 allows short lengths of conductor between a phase busbar and the input side of a fuse/MCB to be excused from fault protection provided that it's (a) 3 m long, (b) "erected in such a manner as to reduce to a minimum the risk of fault current" and (c) "erected in such a manner as to reduce to a minimum the risk of fire or danger to persons". [Reg. 473-02-02] OOI the highest PSSC I've encountered on a house installation is about 3.3 kA (Ze = 0.07 ohm). This is a property quite near the footway (short service cable) with the substation a stone's throw down the road. HTH -- Andy |
"Andy Wade" wrote in message ... Hi Will, sorry to come in a bit late to this. Hope you're still around... Well, I've just come back... Thanks for the intersting info, Andy. At the time I made the original post, I only had a 15th edition to hand. I have since bought a 16th (I have waited long enough in between for both my 15th and 16th to be brown-cover!), and it actually has explicit notes on the MCB graphs about checking with mfg data for very short times. As you rightly point out, the calculations for very small fractions of a cycle *are* different in nature anyway. Firstly, the reason that the fault calculation is usually done at the furthest point of the circuit is that that is /usually/ the worst case. When we're not at the bottom of the graph, as you put it, a lower fault current leads to higher I^2*t let-through and hence more risk of cable damage. Hmm, I'm not sure I'm convinced about this by the kinds of calculations I've done. Current is falling linearly with cable length, which brings I^2 down very quickly. Because one almost always seems to be on the magnetic part of the breaker curve, it's not clear to me that 't' is rising so quickly. Again I suppose it's down to looking at real-world breaker characteristics. OOI the highest PSSC I've encountered on a house installation is about 3.3 kA (Ze = 0.07 ohm). This is a property quite near the footway (short service cable) with the substation a stone's throw down the road. Yes, I'm aware that they tend in real life to be very low, but I wasn't really concerned about the realities of domestic installations here - I was merely trying to understand the proper way to do the calculations. Cheers, Will |
Will Dean wrote:
At the time I made the original post, I only had a 15th edition to hand. I have since bought a 16th (I have waited long enough in between for both my 15th and 16th to be brown-cover!), and it actually has explicit notes on the MCB graphs about checking with mfg data for very short times. Indeed so. As you rightly point out, the calculations for very small fractions of a cycle *are* different in nature anyway. Firstly, the reason that the fault calculation is usually done at the furthest point of the circuit is that that is /usually/ the worst case. When we're not at the bottom of the graph, as you put it, a lower fault current leads to higher I^2*t let-through and hence more risk of cable damage. Hmm, I'm not sure I'm convinced about this by the kinds of calculations I've done. Current is falling linearly with cable length, which brings I^2 down very quickly. Because one almost always seems to be on the magnetic part of the breaker curve, it's not clear to me that 't' is rising so quickly. You're right. For MCBs I^2*t increases with /falling/ current below the 'instantaneous' trip point (i.e. on the thermal-trip part of the characteristic) but increases with /rising/ current on the magnetic-trip part of the characteristic. There's a graph of I^2*t against I on page 130 of the /commentary/. [...] I was merely trying to understand the proper way to do the calculations. Me too... You would find the commentary interesting (well I did): http://www.iee.org/Publish/Books/Wir...?book=NS%20031 -- Andy |
"Andy Wade" wrote in message
... You're right. For MCBs I^2*t increases with /falling/ current below the 'instantaneous' trip point (i.e. on the thermal-trip part of the characteristic) but increases with /rising/ current on the magnetic-trip part of the characteristic. There's a graph of I^2*t against I on page 130 of the /commentary/. Interesting. That suggests that one should be very careful about where one does the adiabatic calculations - it's certainly a valuable footnote when demonstrating compliance by applying that calculation only at the far end of the circuit, if you've assumed that the magnetic + limited-let-through saves you at the near end. Me too... You would find the commentary interesting (well I did): http://www.iee.org/Publish/Books/Wir...?book=NS%20031 Grrrr. As I've just updated both my wiring regs and OSG, and bought BS7909 ( GBP4.30 per page!) in the last couple of weeks, my standards budget is rather blown at the moment. I'd like GN3, as well. But it's useful to have a recommendation for the book - some of the 'helper' books that go with the regs are poor, IMO. Paul Cook seems pretty good, in general, though. And I do get an IEE discount. I'll just about get the set up to date, and they'll issue a '17th edition', no doubt. Cheers, Will |
Will Dean wrote:
But it's useful to have a recommendation for the book - some of the 'helper' books that go with the regs are poor, IMO. Err, yes. They could combine all the GNs into one book for a start. Also we now have the OSG (essential but poorly constructed) /and/ the Electrician's Guide to the building Regs. There's a huge amount of overlap between these two and they cry out to be combined. Paul Cook seems pretty good, in general, though. Yes, although Brian Jenkins, who did the commentary on the 15th Edition, is a better writer, IMHO. I'll just about get the set up to date, and they'll issue a '17th edition', no doubt. There may not ever be a 17th Ed. - just further editions of BS 7671, I guess. Then BS EN 6xxxx and maybe BS EN ISO xxxxxx. Who knows? -- Andy |
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