At 20:00 on a Saturday with NCIS on the box? You must be kidding 
I'll see what I can do.
I'll see what I can do.
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At 20:00 on a Saturday with NCIS on the box? You must be kidding
I'll see what I can do.
Lol - I love that show!
I'm starting from the beginning (season 1) after wetting my appetite with a few episodes from season 3.
It's better than the new incarnation which is NCIS: Los Angeles though that's still good.
yes i know that but for any given fuse it gives a list of currents for time, what i'm asking is what is the time?
5 seconds for fixed equipment and 0.4 for sockets?
and the breakers chart isn't helpfull at all.. it says 0.1 - 5 seconds.. so what do we use there?
anyway.. a 50A breaker needs 250A to blow in 0.1-5 seconds..
so that would be ?
(sqrt (250 x 250 x t )) / 115
if "t" is 0.1 then S = 0.7mm but that doesn't sound right..
so if "t" is 5 then S = 4.9mm which sounds more reasonable for a 16mm feed..
For distribution circuits and final circuits exceeding 32A, disconnection times will be a maximum of 5s for TN (411.3.2.3) and 1s for TT systems (411.3.2.2).
When sizing a CPC for a circuit I normally use the PEFC at the origin of the circuit and select the time from the relevant time/current characteristic graph, carry out the adiabatic then check for thermal withstand (434.5.2).
eg
PEFC = 2kA
Disconnection time will be <0.1s
SQRT(2000x2000x0.1) = 632
632/115 = 5.5mm2 (6mm2 CPC)
K2S2 = 476100
I2t = 400000
K2S2 is greater than I2t
Doing the maths again for a PEFC of 1kA comes out with a CPC of 4mm2
When sizing a CPC for a circuit I normally use the PEFC at the origin of the circuit and select the time from the relevant time/current characteristic graph, carry out the adiabatic then check for thermal withstand (434.5.2).
eg
PEFC = 2kA
Disconnection time will be <0.1s
SQRT(2000x2000x0.1) = 632
632/115 = 5.5mm2 (6mm2 CPC)
K2S2 = 476100
I2t = 400000
K2S2 is greater than I2t
Doing the maths again for a PEFC of 1kA comes out with a CPC of 4mm2
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I usually just put in what the gaffer tells me to..
how do you know the PEFC at the origin when you're in the office?
how do you know the PEFC at the origin when you're in the office?
When I say I normally do it that way, it's only been once.
It was checking to see if a 6mm2 CPC was compliant as a flat was supplied via 16mm2 t&e from a 50A MCB - funnily enough, not that different from this thread!
On that occasion I has access to Ze and PEFC values.
The only thing I can think of (apart from using table 54.7) would be to use maximum values of Ze but that can't be right as it gives you very low CPC sizes that may not comply if the Ze is actually lower
For instance:
Ze = 0.2ohms
Fault current = 1150A
S = 3.16mm2 (4mm2 CPC)
Ze = 0.8ohms
Fault Current = 287A
S = 0.79mm2 (1mm2 CPC)
Both calculations end up with K2S2 being less than I2t
It was checking to see if a 6mm2 CPC was compliant as a flat was supplied via 16mm2 t&e from a 50A MCB - funnily enough, not that different from this thread!
On that occasion I has access to Ze and PEFC values.
The only thing I can think of (apart from using table 54.7) would be to use maximum values of Ze but that can't be right as it gives you very low CPC sizes that may not comply if the Ze is actually lower
For instance:
Ze = 0.2ohms
Fault current = 1150A
S = 3.16mm2 (4mm2 CPC)
Ze = 0.8ohms
Fault Current = 287A
S = 0.79mm2 (1mm2 CPC)
Both calculations end up with K2S2 being less than I2t
G
george765
thanks All
G
george765
Doh !!!!
I was saying 'thanks guys' cos I thought page 1 was the end of the thread, didn't realise that the thread had gone on to another two pages.
Very interesting thread too. I havn't got a copy of the regs so it was difficult to follow some of the postings, as I would like to have done.
After reading through the thread I have decided to run a 16mm earth in addition to the cpc. I do have a couple of overflow boiler pipes running (but not ending) near the CCU which I am Planning to bond too.
Thanks again all.
This is an excellent forum !
I was saying 'thanks guys' cos I thought page 1 was the end of the thread, didn't realise that the thread had gone on to another two pages.
Very interesting thread too. I havn't got a copy of the regs so it was difficult to follow some of the postings, as I would like to have done.
After reading through the thread I have decided to run a 16mm earth in addition to the cpc. I do have a couple of overflow boiler pipes running (but not ending) near the CCU which I am Planning to bond too.
Thanks again all.
This is an excellent forum !
Blimey - a 16mm earth in addition to the CPC? Oh well you've read the thread so we can't really help you further. By the way the boiler pipes will not require any bonding unless that's the point where the gas comes into your house.
Now back to that example someone asked me for.
If I was to connect a 1mm² piece of cable directly between the terminals of the supply, how hot could the cable potentially get before the cut-out fuse fails?
Well, we have two ways of measuring energy being put into a conductor:
Electrically: Q = I².t.R Joules [1]
Thermodynamically: Q = cp.m.δT Joules [2]
where
cp = specific heat at constant pressure (J/kg/°C)
m = mass of conductor (kg)
δT = temperature rise due to heat input (°C)
Rearranging [1] and [2] we can get δT = I².t.R/(cp.m) °C
If we put in values for the above:
I²t = 57,300 J/Ω
R = 18E-3 Ω/m
cp = 386 J/kg/°C
m = 0.0089 kg/m
δT = 300 °C
That is of course assuming the resistance of the copper isn't affect as it's heated up, no heat is lost to the environment, and we must remember to add on the ambient temperature, so it's more like 320 °C in total to touch.
Now back to that example someone asked me for.
If I was to connect a 1mm² piece of cable directly between the terminals of the supply, how hot could the cable potentially get before the cut-out fuse fails?
Well, we have two ways of measuring energy being put into a conductor:
Electrically: Q = I².t.R Joules [1]
Thermodynamically: Q = cp.m.δT Joules [2]
where
cp = specific heat at constant pressure (J/kg/°C)
m = mass of conductor (kg)
δT = temperature rise due to heat input (°C)
Rearranging [1] and [2] we can get δT = I².t.R/(cp.m) °C
If we put in values for the above:
I²t = 57,300 J/Ω
R = 18E-3 Ω/m
cp = 386 J/kg/°C
m = 0.0089 kg/m
δT = 300 °C
That is of course assuming the resistance of the copper isn't affect as it's heated up, no heat is lost to the environment, and we must remember to add on the ambient temperature, so it's more like 320 °C in total to touch.
It raises an interesting point though, and that is if you strike a nail through a cable at or near the consumer unit, assuming we were protected by MCBs then the cable would be very toasty
Your example differs to the potential problem you highlight somewhat!
G
george765
mfarrow";p="1469766 said:
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What fun we are having with this .
Just a few points to note:
1) The adiabatic equation is only valid over the time range 0.1 to 5 seconds – why – because below 0.1 seconds the current wave may be asymmetric and this cannot be represent by I2t as the (I) in that formula is the rms value of a sinewave. An asymmetric fault current is not that animal.
Beyond 5 seconds the equation produces unduly pessimistic results.
2) 434.2 (paragraph 2) requires that you confirm that the let through energy on sub 0.1 second faults does not exceed the withstand energy of the cable. This cannot be done by using the adiabatic equation, you have to obtain data for the CPD manufacturer or use the fairly high values quoted in British Standards.
3) Long duration faults caused by low fault currents usually allow more let through energy than high current faults. So it can be the fault on the end of a circuit that causes most damage. You should really check at both ends.
.Just a few points to note:
1) The adiabatic equation is only valid over the time range 0.1 to 5 seconds – why – because below 0.1 seconds the current wave may be asymmetric and this cannot be represent by I2t as the (I) in that formula is the rms value of a sinewave. An asymmetric fault current is not that animal.
Beyond 5 seconds the equation produces unduly pessimistic results.
2) 434.2 (paragraph 2) requires that you confirm that the let through energy on sub 0.1 second faults does not exceed the withstand energy of the cable. This cannot be done by using the adiabatic equation, you have to obtain data for the CPD manufacturer or use the fairly high values quoted in British Standards.
3) Long duration faults caused by low fault currents usually allow more let through energy than high current faults. So it can be the fault on the end of a circuit that causes most damage. You should really check at both ends
.Surely putting a nail through a cable between the line and neutral is the same as putting a cable between the L & N terminals?
If you mean do I have any examples of where the fault level was high enough for the MCB not to trip in time, then unfortunately I do not. It might not ever happen in real life, I'm just highlighting a potential problem. Don't forget the cable's resistance will be increased as well as it heats up (much like the impedance of a light bulb) so the fault level will drop significantly.
But lest we forget that the 0.1s trip time of an MCB equates to 5 full cycles at 50Hz, and a lot of damage can be done to the cable in that time. The DNO only needs to quote a fault level of 757A for that to happen with the above example (assuming they quote this in RMS, anyone know?).
The solution? HRC fuses of course, or ... [drumroll please] ... BS3036 re-wireables
.Thankfully for us, and the OP, we only usually consider fault levels on the load side. So yes, 6mm CPC should be ample. The boiler pipes, if they require it, should already be bonded through the rest of the pipework in the house to the main earthing terminal at the CU.
mfarrow";p="1470305 said:
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