a ltttle help please current carrying capacity DC cables commercial 6-800V DC

[Gavin A]

righto chaps and chappesses, I'm currently devising a spreadsheet to calculate the cable size needed for various sizes of cable for use on the DC circuits of commercial jobs where the DC voltage cable be 6-800V.

The reason being that a different spark has reported that an SWA cable we'd installed on a job a while back was running quite warm, and tbh I think we did the cable calcs on the job and I don't have a record for them, but as far as I could work out it ought to have been well inside the ratings... though having checked I see the ratings are for up to 90 deg C cable temp, so it could well be running warm even if inside the spec.


Anyway, one point that's confusing me is that the latest version of the solar installers guide says we need to work out the current carry capacity based on the Isc according to the requirements of BS7671 etc.

But in my head, I just can't get away from thinking that we ought really to be converting all this to watts, as surely we need to be taking account of the fact that while a single string circuit will only be carrying a current of around 8 amps, it's carrying that current at say 600V, meaning that it's carrying 4.8kW vs 1.8kW for an 8 amp 230V AC circuit.

So the way I've got the spreadsheet set up is to convert the ratings of the cables in BS7671 into an equivalent power rating in kW, then applied the derating factors etc, then compare the peak kW rating of the DC circuit.

If none of this matters and it really is just the current that matters, then fair enough, that makes things simpe, apart from the fact that it wouldn't even come close to explaining why I'm getting reports of a warm cable when the cable is 6mm2 carrying at most 16 amps (ok clipped direct SWA 4 core, possibly in direct sunlight, but even so it's still only 50% of the rated current for the cable in amps)


So can anyone clarify this for me, as to whether we should just be using the Isc figures in amps and ignoring the fact that it's running at 600V and therefore 2.5 x the actual power, as would be implied by the guidance, or am I right in thinking we should be accounting for the increased power flowing through the cable and doing the calcs in watts?

Sorry if this seems a bit basic, I'm just driving myself a wee bit made here trying to get my head around it, so a little assistance would be appreciated.

ps once I've sorted the spreadsheet, I don't mind posting it up for comment / others to use, as long as I've got it right.

 

[Gavin A]

hmm Napit technical reckon I should just be calculating it based on the current rating alone, ignoring voltage and power.


Which begs the question, why's my cable getting warm enough for another spark to question it if it's 50% inside even the derated current carrying capacity of the cable?

 

[telectrix]

all you need to consider is the current, not the wattage. obviously the current is dependent on voltage ( for the same wattage ), but it's current that warms the cable.

 

[SolarCity]

I would have said current rating alone also but I'm glad to see Napit confirming it.

As to why your cable is running hot - that really is a mystery. Presumably it is for solar and you wouldn't expect full power at this point in time anyway, nor expect the direct sunlight to have much of an impact.

If it is a spark that has flagged this up, presumably he has ruled out a lose connection?

 

[JulianC]

Yup, it's the current that affects the temperature. Said current being a function of the voltage and resistance generally (ignoring any black magic from them solar panel thingies).

Would be interesting to note the temperature of the cable. Even if it's sitting at 20 degrees, it will feel warm if the ambient is in single figures. Is the cable outside? Has it had a bit of sun on it?

 

[Gavin A]

ah... this'd be because the voltage essentially cancels out the resistance would it, ie resistance reduces proportionally to voltage increasing and vice versa for any given ampage.

right, I think that makes sense, which I think was how we'd been calculating it before, though this has always been a niggling doubt..... I'd kinda talked myself into assuming I must have had it wrong as that was about the only explanation I could come up with for why t'other spark was mentioning the warm cables.

tbh though it was first mentioned a while back, but I didn't have it on my urgent pile given that it was winter and the cables were outside, but we're intending to visit to do an update on the inverter anyway, and wanted to change out the cable at the same time if there actually was a problem.

I suspect it must just have been that the cable was on an outside wall in full sunlight, and therefore was hot mainly because it was a black cable in the sun, or something like that.

for reference I think the cable they meant is a 4 core 6mm2 SWA cable carrying to DC circuits of around 11amps, clipped direct to the outside south facing wall, with 3-4 SWA cables in total clipped next to each other on the wall, length of the run is around 25m to an isolator. AFAIK the warmth was on the cable in the middle of the run, not around any joints.

the cable's rated to 90 deg anyway, so shouldn't be an issue, I was more concerned because we've done the same with another system that's running at 16 amps, so if the 11amp one was getting warm then I'd have some serious worries about the other cable.

Looks like they're well inside the limits then anyway.

 

[telectrix]

resistance reduces proportionally to voltage increasing and vice versa for any given ampage.


no, the resistance only alters with temperature. it's the current that is inversely proportional to the voltage to produce the same wattage. e.g. 1000w @ 230V is 4.35A, but at 12V it's 83.33A.

​

 

[Gavin A]

I see my problem now, I've been attempting to apply the wrong law to the situation.

What I should have been looking at was Joules Law, which relates to the heat dissipation from a circuit of a given resistance.

My problem I think was that I knew this was what I was trying to find, but was looking in the wrong formula then trying to make Ohms law turn into Joules law through my own calculations rather than remembering what search term to use to find the formula someone much cleverer came up with a long time ago to make it easy for the rest of us.


So by my calculations the heat dissipated into the cable from this situation is minimal,

P = (16A x 16A) x 0.0718375 ohms (calculated for 25m of 6mm2 copper)
P= 18.4 Watts

so 18.4 watts per core of cable x 4 cores = 73.5 watts along a 25 m cable, which is absolutely nothing in heat terms.

I think this other spark is on a wind up, unless he was on about the AC cables and it's all got a bit lost in translation, alternatively it was just hot due to the sun.


ps I notice nobody else remembered the actual name of the law that was relevant here either, which makes me feel marginally less stupid for posting this thread.

 

[JulianC]

.

ps I notice nobody else remembered the actual name of the law that was relevant here either, which makes me feel marginally less stupid for posting this thread.

Tell me about it. I have forgotten so much basic stuff over the years it's frightening. Also known as 'I squared R' heating. Thank God for Thickipedia

 

[Gavin A]

Indeed.

anyway, that's a relief as I really didn't fancy having to go back to several jobs and replace several hundred meters of SWA cable because I'd ballsed up the calcs first time around and the cables were frying.


I think it might actually be a case of chinese whispers, and the issue is with the AC cable, which is relatively close to the limits, and as those limits are at 90 degrees for those cables, it's probably not surprising if it get's a little bit warm in summer. Might swap that out to save anyone worrying.

 

[telectrix]

i'd forgotten the name as well. then again, it was nearly 50 years ago. LOL. i generally refer to it as the PIIR equation, or substituting V for IR ( ohms law ) , P =IV, which we all know, of course.

 

[BruceB]

Coming late to this, it is indeed the current that is important.

To pick up one other point made in passing that the cable is rated to 90 degrees, that is not always as useful as you think because mostly the things the cable will be connected to are only rated to 70 degrees and therefore you cannot design to run the cable hotter than 70 degrees.

 

[telectrix]

good comment.

 

[Gavin A]

Coming late to this, it is indeed the current that is important.

To pick up one other point made in passing that the cable is rated to 90 degrees, that is not always as useful as you think because mostly the things the cable will be connected to are only rated to 70 degrees and therefore you cannot design to run the cable hotter than 70 degrees.

true.

We ought to still be reasonably well inside the limit even for 70 deg cable, though it starts getting a bit dicey if the cables were grouped together... which they weren't when we left them, but I can well imagine that they might be now as the spark for the building has been in the process of swapping a new board in, so I can imagine the tray is probably now piled high with cables all bunched nicely around ours to keep it nice and warm.

I guess maybe that's a lesson to cable size on the basis that someone at some point is going to add more cables bunched around our cable. Also, not to rely on CEF having the cable in stock next to the job that you ideally want just because they say they do when you ring them up to check in advance.

 

[Worcester]

@GavinA out of interest why do you use SWA?

We always run new, segregated (and labelled) traywork so as to keep the DC away from other cable runs. Our DC runs are regularly 100m long, and often we'll have 12 or more strings, so we take up a whole 100mm tray.