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Topic Title: Cable operating temperatures
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Created On: 24 June 2015 10:26 PM
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 24 June 2015 10:26 PM
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hello, long intro, small question

I wonder if you might be able to help with something. this is not urgent at all and is a curiosity.

But first, briefly because some of you ask me, It hasn't half been a strange eight weeks or so. Not me you understand, the others. Our troubled friend is now up North with his Mum and is OK as far as I know but not great. We had a steak together last Tue, me with my arm in a bucket of ice all evening, and two texts since. Fingers crossed. I did not break my arm in the fall down the stairs and am making a miracle recovery from being black and blue. Thanks for asking. AFAIK all good.

So, the behaviour pattern is in evidence and I am thick into text books. An odd thing but after troubled times I do the self-improving thing. As it happens one of those drove me to signing up for a year at electrician college with a view to packing it in at the end of the first year. 16 years later.....

I've done a thing about whether a circuit not protected by an RCD can feed anything back into an RCD protected circuit and have concluded that yes it can so voltages can be transferred. That had been on my mind for a while because of bathrooms often having the only RCBO in a board otherwise full of MCBs and a shared earth bar or even a shared neutral bar. But that's not the question. (It is a good dabble though and I think it is important).

Now though, I'm digging into cable operating temperatures and there's something I can't find.

I'd like to understand how a cable reaches temperature and how long it takes.

Is there a formula involving 'delta t' ?

Why? I use Amtech as a diagnostic tool and oft model a problem circuit into it in order to establish the operating temperature of a cable. If I shove an overload down a ring final circuit with a 70 degree cable on it. How long before that cable reaches 70 degrees? If I do the same at full capacity but not an overload? and so on.

I'm in all sorts of interesting stuff here but the truth is that none of them are brave enough to venture a formula so they all come across as a bit flimsy.

Like I say, not urgent but if you have any references mad person in self improving mode would appreciate them.

Not going out again...ever...


Edited a bit.

Edited: 24 June 2015 at 10:40 PM by Zs
 25 June 2015 12:09 AM
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Ah, well the reason it is not in the book is ' it's a bit complicated'.
' it's actually !"£!"£y complicated in full depth' would be nearer the mark.
Such lucid insights don't sell text books.

To a first approximation, cable warming up is a bit like filling a bath with a leaky plug.

I guess you wanted a bit more than that really - I suggest you pour yourself a drink and relax - I've got one in for myself here already.

The copper of the wire is heated by the energy arriving - this is the current flowing, times the volts dropped, and has the units of joules, named after the famous beer maker. A joule is a watt for a second, and there are 3600,000 of them in a kilowatt hour, and in round numbers about 700,000 in a small Mars bar if burnt 100% efficiently.
This heat arriving may be seen as like the water entering the bath.
Now, as energy comes in, the temperature rises (like the water level in the bath) the heat starts to flow out into the surroundings, flowing faster as the temp rises (like water pressure rises, forcing more out faster through the leak )

In the Adiabatic case (- equivalent to a well fitting bath plug ) the heat arrives and can't leave, so the temperature ramps up steadily.
That is one very simple and utterly unrealistic extreme - unless the rate of heating is near infinite.
The other noddy case when the heat has been on ( that bath has been filling and leaking) for ages, and rate of arrival and rate of exit are now exactly equal and the system has reached equilibrium - that is annex 4 of BS7671, steady temp of 70 degrees on the center of the copper core.

Now to solve for any other condition needs knowledge of heat capacity (how much energy in joules is needed to warm up a block of stuff by how many degrees) and the thermal conductivity (how many watts flow through a known area for a given temperature difference pushing it along)
The ratio of these quantities , in the right units, gives a time constant for a particular material. Us long haired (well not so much now but once, long ago) physics types like to talk about the a "thermal relaxation time" - sounds like the time to chillax perhaps, but also the time for a system of heat sources and thermal conductors to reach equilibrium.
Take a sip of that drink now if I were you.
This is only the first layer of the onion. Thermal conductivity, printed and available in books of constants like Kaye and Laby
( Online here ) are not truly constant. In fluids (including air around a cable ) once you have a bit of heat, the density of the hot region falls due to expansion, and the now lighter region of warm air or water or whatever rises, and cold stuff replaces it. This is convection, and as the air speed is a highly non-linear function of temperature, then the rate of heat removal is much faster at higher temperatures than at low ones, and the curve is higher order (rate of rise changes getting ever steeper) than linear.
Then in many materials there are phase changes, where heat goes in and something happens to the material, but the temperature does not actually increase much at all. Any one who cooks in a Bain Marie knows this - no matter how much fire wood you put under the scrambled egg in the water bath, the temperature remains pretty much at 100 degrees, until the water has all boiled away. Now some waxes and plastics exhibit a molecular re-organisation some temperatures, that may give a different heat capacity above and below the critical temperature, as well as a latent heat of crossing that temperature.
PTFE for a laugh does this at 25 degrees C, and above and below transition is a white slippy solid plastic, but the electric properties have changed on either side. Oh and both electrical and thermal conductivity of metals gets worse with increasing temperature, more or less in lockstep, as it is the free electrons that do all the electric stuff and abou 95% of the thermal, and they start falling over each other more when the molecular lattice is agitated.

Take another sip or two of that drink if I were you.
Actually, as a related aside, some high power electronics in missiles for example, has a very short design life and the normal approach of heatsink (and fan if you don't like convection) is not needed, just a little one-time block of wax - it holds the active device at a low temp until it has all melted, and by then it has "arrived at the final destination" as the ol' sat-nav would say. Umm.

So much waffle, now time to thrash some example numbers.

Take rather more than a sip of that drink if I were you - and hang on tight, maths is NOT my greatest talent and this may be off by pi, root 2 or my age in seconds. (not a lot of people know that..)

Take a length of 1mm2 T and E.
Lets have giggle and pull 1 core out and hit it with an utterly incredible PJ tronics overload of 100A and let's play dangerous, and put that plastic insulated core in a bucket of water
This is deliberately an extreme example, and even then the answers may surprise.

Assume 18 milliohms per meter. (lets neglect change of resistance with temp for a minute.)

1.8 volts, so 180 watts will be dissipated in a 1m length. (lets just consider that 1m for now, terminated into something heavier at the ends.)
The copper in that 1 core weighs in at 1000 cubic millimetres, or 1.0 cc - at a weight of 9 grams or so.
heat capacity 400millijoules per degree per gram, so if it was wearing a woolly jumper and feeling all adiabatic it would kick off getting hot at a rate of about 45degrees a second, - or solid to melted metal in under 30 seconds. But of course its not adiabatic, some heat gets out in that sort of time, so in reality the rate of rise slows down after a bit, and it takes a touch longer to melt.....

But with a 1mm surround of PVC, with water at 100 deg around that, its quite different. We can imagine modelling the insulating effect of that jacket by unzipping a trapezoidal skin of plastic of 1mm thick, 1m long and on average about 3mm wide (if we could lay it flat and make it into a rectangle ) we have 3000 mm square and 1mm thick and we wish to pump out 180 watts thorough that area..
PVC thermal conductivity of 0.1 watts per meter kelvin, means instead of a 1meter cube, we have 1mm = 1/1000 metres thick, and 3000mmsq of area - about 1/300 of a square meter, so our 0.1 becomes more like 3 watts per degree of difference between the 'inside' of the PVC skin and the outside over the full length.
So in steady state, our 180 watts is a 60 degree gradient between the copper and the outer of the first layer of PVC. Now in reality PVC softens by 100 derees and melts promptly by 150-160 So our piece of 1mm will hold, until the inner reaches say 150 degrees, and we need to hold the outer at 80-90.

But how long for the PVC alone to reach that equilibrium - well its about 1 degree per joule per gram for most plastics when well away from phase changes, and we have 3000 cubic mm (thats 3.0 cc, or about 5 grams, so at two and a half a joules to raise the volume of plastic by 1 degree, so at 180 watts we see the temperature rise at 36 degrees a second - comparable to the copper alone, so the two together its about 18 degrees per second ... or 3 to 4 seconds for our 60 degree between copper and water to be well on the way to established. (strictly one time constant is ~60% of the way to the final value being 1-(1/e) and all that.)

In water, some time before the water actually boils, the PVC has had it, unless you are up a mountain. Iced water would be fine indefinitely at 100A until the ice melted, and the temp began to climb.

but lets assume instead its in my pint of beer, now at about 15 degrees. (personal sacrifices I make for science.. ) heat capacity is 4 watts per cc, per degree rise.
but I've drunk enough that now it more like quarter of a liter, and therefore the temp rise is a handy one degree C per 1000 joules dissipated. Well with our 180 watt input it will warm up at a degree per 5.5 seconds or so. So we have over 4 minutes before my drink reaches the temperature of 80 degrees, where it will plateau until all the alcohol is gone, before rising again towards 100, again at about 5.5 secs per degree..

There is nothing for it at this point, but to conclude either, 1mm pvc covered is unsuited to 100A, for more than a few minutes or we need a continuous flow beer chilling system. I know where my money is.
Equally we can get through about half of Queens Bohemian Rhapsody, with 100A of lighting on a 1mm core cable cooled only by half a glass of beer. Should have chosen something by Debuie Harry, and we'd have made it until the lights went off.

Substitute a 100kg concrete slab for that beer, and maybe a slightly less drastic overload than that 100A on 1mm, and you could be several hours to reach an equilibrium, and it will depend on shape, orientation relative to vertical, moisture content... And a 1mm wire will probably be fine at 100A when in contact with briskly convecting transformer oil, pretty much all day long so long as the oil does not boil, and for that we and fit a Buchholz trip that knocks the power off if the pressure rises too high.
(So the network transformers always have wire that looks too thin for the current rating when the oil is out. Solid insulated ones look quite a bit different.)

so it's not a one liner in the book ;-)

edited to correct a really daft mathematical error.

regards Mike

Edited: 25 June 2015 at 04:04 PM by mapj1
 25 June 2015 12:53 AM
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You answered the how it gets to the temperature question brilliantly but sadly you raised another squillion questions.

I'll ask you them soon, if I can work out how to, but I'm settling on musing the leaky bath plug or the working bath plug because I completely get what you described there. I've given it a plastic bath in my head because a posh one would absorb some of the heat and I can't cope with that just yet You see, I don't think this is adiabatic in terms of without heat loss. Rather it's kind of anti-adiabatic except for the box it's in such as the insulation or the trunking or both.

Too many variables? Do we just go for the simple?

It begs a question about the BS7671 annexe though and indeed about Amtech. I guess I had subconscious suspicions about those being based on something constant and you just touched on it. Our cables are not constant so diversity comes in to play and then even gets a bit messed up if a circuit is on half tilt for a few seconds a minute or full tilt for only a minute, twice a day. If you get my drift?


I've not been sipping lately in case I needed to drive at short notice so have a glug of that super science enhancing brew for me.

Fantastic. I thank you for a brilliant post. I want to know a bit more. but you've given me something to go and investigate.

So, no Delta t? I was convinced there would be one of those.

 25 June 2015 01:05 AM
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Well, please come back when the questions have formulated themselves. I'll come back after the beer has worn off. There are a few typos in there but I'll nuke them later.

regards Mike
 25 June 2015 05:08 AM
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The simple way I used to approach this with C & G 238 students was to use Mike's bath tub but leave it there.

My question to the students would be - how long does it take a water supply with a 10 litre per minute flow rate to fill a 100 litre bath.

The answer of 10 minutes is quickly returned.

Second question - how long would it take if I pulled the plug out and then filled the bath with some water escaping via the open plug.

Various responses about diameter of plug hole - height of water above the plug at any given time and then the group finally realises that the adiabatic equation is there to make their lives simple .

If you really must know then start here
Then here.
That should put you off.

As an aside
One of our number is studying an electrical engineering degree - I have provided some help with the earlier stages of his study, but he recently ask for help with a probelm that involves Laplace - I used to teach this in a basic way for HND students studying electrical network theory - but that was nearly 25 years ago. I an afraid I had to decline to help as I have long since lost the ability to work at that level.


Geoff Blackwell

Edited: 25 June 2015 at 05:34 AM by GeoffBlackwell
 25 June 2015 08:07 AM
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For your degree student I note that J.J. D'Azzo and C.H. Houpis " Feedback Control System Analysis and Synthesis"
was the recommended "light read" ha!! for La Place transforms and so on when I worked at UOY.
I don't recommend it for those without at least A level maths as a start.
A firm grip of complex planes, Argand diagrams and all that goes with it is a pre-requisite. Nowadays though there is a
lot of very good stuff on the web. It is worth noting that many American universities tend to produce student hand outs
and stuff that assume a mathematical level somewhere near begin to mid A-level as their eduction system is quite a bit different
in how specialist it expects people to be at different ages.

This is sometimes helpful to UK students as a bit of a refresher, or if some of it was not so well stuck in the first place.
Is that time stamp real? This is the bulletin board that never sleeps !

regards Mike
 25 June 2015 08:13 AM
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I started asking about this on here, around a slightly different scenario (thermal equivalent current, max demand etc, impact of showers and so forth), some months ago! Some regular and respected people on here were giving me bits to use and thoughts to consider.

e.g. the effect of a shower load (40A) for only 10 or 15 mins on a piece of distribution 10mm2 cable that may have a steady current at 30A in an ambient of 30degC to see if the shower can be negated from the max demand comsideration (and expecially in a multishower scenario). In effect, if the temperature (and time to heat/cool) is not going to rise to a point of damaging the insulation etc, then it could be negated....the 'time step' as those helpful minds on here pointed out at one point is a factor.

I'm still not sure about it all or that I'm competent to explain it, but much better informed than I was.

Its nice someone else is asking related as I see it, but I have to say the answers above do get across the possible complexity of it all from all the work/reading I've tried to do on the subject.

It isn't so simple, unless its made simple in some fashion as said above. Joule heating, heat capacity (of a material), thermal conductivity, environment (cooling etc), density of the material can all be factors I messed around with depending on what variables one has. I made a stab at a simple exercise, but it could never ever be considered anywhere near a rigid treatment and plus, I dont have it in me I don't think, not yet anyway :-)

Edited: 25 June 2015 at 08:23 AM by psychicwarrior
 25 June 2015 09:17 AM
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Zs - being a bear of little brain, I'm a lot happier with pictures than complex maths - so I find graphs of conductor temperature over time easier to picture - e.g. "Fig 2" in The curve gets stretched around as variables change - with less amps per sq mm (i.e. larger cables) it takes more time, with higher currents it gets taller quicker and so on, but the basic shape - starts increasing fast, gradually slowing down until it reaches equilibrium is the same. Of course if the current is switch off part way through it then starts to cool down (by the reverse curve - fairly fast to start with the more gradually until it reaches equilibrium with ambient temperature)
- Andy.
 25 June 2015 11:04 AM
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Originally posted by: mapj1
Take a length of 1mm2 T and E.

The copper in that 1 core weighs in at 100 cubic millimetres, or 0.1 cc - at a weight of 9 grams /cc that is just under 1 gram.


But how long for the PVC alone to reach that equilibrium - well its about 1 degree per joule per gram for most plastics when well away from phase changes, and we have 3000 cubic mm (thats 0.3 cc, or about 0.5 grams, so half a joule to raise the volume of plastic by 1gram, so at 180 watts we rise at 360 degrees a second - comparable to the copper alone, so the two together its about 180 degrees per second ... or 1/3 second for our 60 degree between copper and water to be well on the way to established.

Haven't you got a factor of 10 error in both of those?

1mm^2 times a metre long is 1000 cubic mm.

while with the plastic you correctly get 3000 cubic mm, but that's 3 cubic cm not 0.3.

(I wouldn't normally nit-pick, but the result of 1mm^2 melting in 3 seconds at 100A was a rather nice number to know, except I think it should be 30secs).

And to add to the fun, you haven't considered the change in resistance of the wire as it heats up.

If the current remains constant then the power just goes up linearly I^2 * R. But the current probably won't stay constant. If the load is resistive (do we model a short-circuit fault as resistive?) the current will go down; however if it's a regulated load that draws constant energy regardless of the voltage (like an electronic power supply or simply a thermostatically controlled heater) the current will go up.
 25 June 2015 01:28 PM
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ahhh. late night and too much bear garden.

Quite so 1000mm long and 1mmsq is indeed 1cc = 9 grams.

I'm sure a bit of 1mmsq is some way in between BS3036 30A and 45A fuse wire, making for a rather optimistic ring main repair, and rather a thin cooker one. (It's been a while mind) so I think you are right 30 secs, not 3 sec, which would be 9 or 10 times the dissipation, lets say 3 times as many amps - I really should have spotted that was rubbish.

When I have more time I will go back and double check the rest - please feel free to keep on looking - the nit picking is very much appreciated - let me know if you spot some more - tad embarrassing, yes, but I'd far rather be corrected.

regards Mike
 25 June 2015 01:29 PM
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Nice curves Andy, if you pardon the expression. I wish they'd turned the loads on and off a bit though.

Here's an example: You'll have to forgive the addition of the coiled cable in this.

Overnight, the plumbers on a job left the client with her immersion heater plugged in to an extension cable reel. One of those on a metal triangle frame

So 50 M of 1.5mm flex or the like with a 13A plugged heating appliance with a thermostat. About 5M uncoiled.( that document of yours points to the insulated section Andy, so I imagine a similar principal going on here)

In the morning, I picked up the reel to move it out of the way and the reel was too hot to touch, the cable itself touchable but hot.

During the night the immersion will have been on a steady tick-over, just keeping the water up to temperature so apart from after their evening showers and a shave I doubt that full current will have been a factor for more than a few minutes.

Notwithstanding the coiled cable and a less than perfect example because of the metal work etc. however, I'd have expected that to have cooled down considerably. But the intermittent use clearly kept it hot.

I don't have a non-coiled example apart from a fan heater flex warming up to the touch within 20 minutes or so of full use or the the kettle flex which never feels warm.

''And to add to the fun, you haven't considered the change in resistance of the wire as it heats up. ''

I agree arg. Slowly losing sight of the one simple equation here.

A large intermittent load is beginning to look as though it has its own stage on which to play and might not fall within the parameters used by the likes of Amtech and simple tables.

We looked at a circuit at the office job recently which was for one of the radiotherapy treatment places. the Linac (the bit they put you underneath) Now that is a very big load but isn't in actual use much of the time at all. I suppose it could run in all the cooling they have in that type of environment just fine. But then, for arguments sake, in the same void is a mamma cable for a heater battery in constant use during the winter...The stress on the cables, because of their temperature, is going to be different in the different seasons but with more factors than the expected summer/winter Amperage loadings.

There has to be a point at which knowing when for example, two cables are going to peak in temperature is relevant to design and safety. Not just knowing how hot they are expected to get.

Hence the question you see.

Not a great post but I think you will manage to decipher it.

 25 June 2015 07:05 PM
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Not sure if this may help.
From the IET design calc book

"D.2Temperatures (core and sheath) (Effect of load current on conductor operating temperature)
The conductor operating temperature at other than the full load current can be determined from the equation:"

Ib/It= sqrt tb-ta/tp-to

hence tb=Ib2/It2(tp-to)+ta

where:Ib = load current resulting in a conductor temperature tb at an ambient ta
It= tabulated current rating in Appendix 4 of BS 7671, resulting in a conductor temperature tp at ambient to.

This relationship assumes that temperature rise of a cable is proportional to the square of the current

Best regards
 25 June 2015 08:57 PM
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document of yours points to the insulated section Andy, so I imagine a similar principal going on here

Indeed - or grouping if you prefer.
the cable itself touchable but hot.

Saved by the intermittent nature of the load! - I've seen ones completely melted in the middle from smaller loads.

If it's of any help to anyone, this is what my head thinks what might happen when the load is cyclic - when the current is flowing I've coloured the curve red, when it's off I've coloured it blue.

firstly when it's on for a "long" time - i.e. the off comes after the cable reaches full temperature

and when it's on for a "shorter" time -

notice that the peak temperature is much lower.

- Andy.
 25 June 2015 09:51 PM
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I wonder if one or two of my posts are beginning to come home to roost with one or two here.
This is the analysis of time. thermal loss, thermal resistance and some thermodynamics, all of which are normally considered "difficult" subjects.
The unfortunate bit is that there is a formula, but it is very complex and needs immense skill to employ it! You will not find it in any textbook, and if you can apply it reasonably, you can derive it from first principles. But there is another problem, and that is that there are so many variables, which many would like to use as worst case values, which will make the result more like a lottery result! So Zs, it can either be like your shed, or you can use the values in the BYB! Remember though, it would take a hugely experienced person to begin to argue with your result, and I can only think of one or two who might be prepared to do so.

The whole thing is that the two limiting conditions (adiabatic and isothermal) never apply, there is always somewhere in between which is where real life exists. Judgement needs to be that of Solomon to get the exact answer. Mostly intermediate results are fairly safe, within limits!

You are not mad, not more than me anyway, but there is no exact answer unless you can exactly define the problem, which you can't. Always remember that no one else can either.


 25 June 2015 11:35 PM
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Dave, great post and lots of that resonates with me.

I am challenged all the time though Dave, even by the new kids. It might be the female thing I suppose and it gets to be a bit tiring but keeps me on my toes whilst looking things up and asking you, just to be sure before I speak it out loud. Passing observation that and not an issue at all. However, if I say 70 degrees you can bet that I'll be asked to back it up.

Back to the temperature thing though; I guess it gets to a point where the Science becomes the Art. Give me another 16 years and I might have turned a full circle but for now I'm rather interested in the in between that you mention because I reckon that is the important bit.

Great stuff. Am inspired by the responses.

Not mad...Dave Z says so. I'm happy with that.

 26 June 2015 12:34 AM
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It was once rumoured that Spike Milligan had a "not mad" certificate too.
I second Andy's slices of exponential truncating to lower peak amplitude triangle waves with a 'DC' offset at higher switching speeds as being the right idea, at least for a simple case, where only one time constant dominates.
Having measured temperature rise in RF cables professionally, it is fair to say it is right, but you can also get that fast triangle, not on a simple DCoffset set by the on to off ratio, but riding on the back of a much slower rising time constant, like the scales climbing up the tail and onto the back of a triceratops in certain situations. This occurs when the thing with the fast time constant, like the cable to its jacket, is mounted onto something much more massive and slower heating, like a building or a mast. Then you may see a time constant of seconds for the copper to the jacket, and minutes for the jacket to the tray, and hours for the tray to the outside. So if you switch in seconds, the wall and building only respond to the long term average, set by the on-time to off-time ratio, while the copper temperature rises and falls to follow every on and off. If you switch in half hours, the cable tray and the wall see the ups and downs, but the building bulk still only responds to the whole day average..

just to muddy the waters.

regards Mike

Edited: 26 June 2015 at 12:43 AM by mapj1

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