Voltage drop question

Dad has a kiln in a workshop at the bottom of his garden, rated at

11.25kW, so maximum current 49A assuming 230V (perhaps notionally 47A given its age, the rating plate mentions 240V/415V rather than 230V/400V, it can be wired for 1ph or 3ph, he runs it on 1ph).

The main part of the run from the house is 35m of 16mm^2 SWA, so expected to give a voltage drop around 5V.

He's keen to reduce the voltage drop (thinks this will reduce the duty cycle, I suppose it will, but I have my doubts if it will be a significant effect).

I know it is normally frowned on to parallel-up conductors to give increased current carrying capacity ...

But given that the SWA that is installed actually is four core, would it still be bad practice to parallel the live and neutral cores *just* to achieve equivalent of 32mm^2 thereby reducing voltage drop? To clarify, the MCB(s) would not be uprated, just done to halve the current in each core, if somehow one core were to go open-ciruit, the current in the remaining core would end up the same as it presently is.

I can envisage that physically getting 32mm^2 of strands securely into the henley blocks at the house end and the terminals of the consumer unit in the workshop might be a challenge.

This would also involve switching the CPC from one of the cores to the steel armour. MCB trip time calculations certainly aren't in my day job, given this is not handheld equipment and is effectively a submain, does a 5s disconnect time apply, rather than 0.4s?

I can see that 35m of 16mm^2 gives R1 as 0.403 ohm for single core (or 0.202 ohm for double cores) R2 as 1.085 ohm PME supply assumed Ze 0.35 ohm

Assume worst case where one core has failed prospective fault current = 230V / (0.35 + 0.403 + 1.085) = 125A

For a 50A type C MCB, a quick glance at the curves, I think this gives a disconnect time of approximately 70 seconds, oh dear, I suppose that alone is enough to rule out the changes he has in mind !!!

Seeing that, I went on to check the existing situation which uses one

16mm^2 core each for live, neutral and CPC fault current = 230V / (0.35 + 0.403 + 0.403) = 198A from the 50A type C curve this still gives a disconnect time of 20s !

So even the existing CPC doesn't meet the disconnect time for a 50A type C MCB, even getting an actual measurement of Ze rather than assuming

0.35 ohms, couldn't possibly result in a low enough Zs to allow a type C MCB.

But reverting to type B would meet the disconnect time (I think he only recently changed to type C at the house end because of poor fault discrimination).

OK, so I think the disconnect time issue has trumped the original question about using parallel conductors, any thoughts (or mistakes in my calculations)?

Reply to
Andy Burns
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Ooops, I think I picked up the wrong curve on the type B MCB graph, 198A would still take 20s to trip a 50A type B (the same time as a 50A type C, is that correct?)

So is the only way to make this safe change to a 50A fuse instead of MCB?

Reply to
Andy Burns

Well its not going to be much even if you can halve the drop...

The reason for this is that it can be very difficult to ensure fault protection for the individual conductors. Since in this scenario you are not using the parallel conductors to achieve a higher design current, but only reduce the voltage drop, its not going to be an issue here (assuming the fault protection is adequate now!)

While a little non standard, I can't see any particular problem so long as everything is labelled clearly, so a future maintainer can see what is going on.

Ought not be a problem - 35mm^2 tails are not uncommon.

Nothing wrong with that. (I assume that the armour is connected to the CPC anyway, even if paralleled with a core?)

Yup, OSG 3.5.2 (411.3.2.3 BS7671) would apply as the circuit exceeds 32A

Erm, well its late/early and I might be not thinking right, but it seems to be you are a tad pessimistic on those figures.

16mm^2 should have a resistance of 1.15 mOhm/m, so R1 would be that times 35, or 0.0403 Ohms - i.e. an order of magnitude better.

R2 for the armour would be 50 mm^2 of steel with a resistance of 3.1 mOhms/m - or 0.1085 ohms.

With high current circuits, its probably better to actually measure rather than work from the worst case standardised figures, since in reality the value is often significantly lower, and this can make design simpler.

However redoing that with the above figures: 230 / (0.35 + 0.04 + 0.11) = 460 which is much closer.

Here is where you really want to use a lower measured Ze... ;-)

Well... if we go with the 460A, then that just about scrapes in at 5 secs on:

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did not specify the material the SWA was insulated with, but if we take it as a 90 degree C XLPE material, then we can check the adequacy of the CPC with the adiabatic equation and a k factor of 143:

s = sqrt( 460^2 x 5 ) / 143 = 7.19 mm^2

Since we have a copper equivalent of 50 / 2.255 = 22mm^2 then we are well in.

(Is this cable also exporting an equipotential zone to the room with the kiln? Because that will complicate things if it needs to be a main bonding conductor as well!)

The dual core solution would perform better - even with the loss of one core (since the loss will likely only affect half of the round trip).

It might not be as bad as that, since if you are using the armour in parallel with a core then you have a better result than above.

Fault discrimination with what?

Reply to
John Rumm

Could you simply measure the voltage at each end of the cable, under load...? That would tell you what the actual volt drop is. Then you'd know if it was worth worrying about...

I've heard that kiln elements go higher resistance with age - if it's really old then it might be worth replacing the element(s) - though that could result in an increased voltage drop.

What is it about the installation that is worrying your Dad ? Adrian

Reply to
Adrian Brentnall

that method is asking for wrong results. Run a wire temporarily so you can measure the Vdrop directly with a multimeter.

NT

Reply to
NT

Well there is 250W (ish) of loss in the cable out of 11250W or about

2% is that worth worrying about?

I'd measure the drop to start with to see what reality is. Just need a meter at the kiln measure the volts with it off and then with it on. The load might change as the kilns gets up to working temperature so might be best to monitor the volts through a fireing.

Reply to
Dave Liquorice

All this does sound like overkill to me considering the minimal gains. I'd not want to be the person paying the bill though!

Brian

Reply to
Brian Gaff

You can meaure it entirely at the kiln end, by switching the kiln on and off and looking at the difference.

You really want professional (fused) test leads for this though, or if you have a socket too, use a plug-in power meter on its voltage setting.

Reply to
Andrew Gabriel

I would start by measuring Ze. Unless it is really low or the kiln is running 24/7 then I would not bother trying to lower the voltage drop.

I'll have to check the calcs later, but it seems that they do meet the 5 seconds disconnection time.

Reply to
ARWadsworth

Yes, thats what I though

thanks

I think so, long time since I looked at the setup

Ah, the table I used was ohms/km rather than milliohms/m, and it was even later/earlier when I tried it, so I think zero hit the deck somewhere, I did say I wasn't used to the calcs!

Yes I appreciate that, but a) it's 50 miles away, b) I don't have the kit, c) with my wrong numbers it looked like even if it was zero it wouldn't help

OK.

sounds better

I had assumed it was XLPE, he's not replied to a few of my Q's yet.

I figured the 22m^2 equivalent, but wasn't aware of the adiabatic equation at all.

Yes, I don't think he has an earth rod for the workshop, can't remember if he has plumbed water down there either ...

Ah yes forgot to count that.

He has 50A type B MCB in the C/U down at the workshop end, and was getting trips, but the house end was tending to trip every time, so he replaced house end with a type B. In the end he found the reason for that (elements wired in parallel that should have been in series, were tripping after several hours, I think his controller uses slow start to build up to full heat overnight).

Thanks again John.

Reply to
Andy Burns

I don't know, I think he's under the impression that the elements are ending up running at full load for long periods, he started off thinking he had over 10V of drop, I think he's probably only got about 5V anyway, and could maybe get it down to 2-3V ... I'm dubious about what effect it'll have

Yes, he's asked me to take a meter with me when I next visit, I'm tempted to take a spare UPS, and monitor it from a gash laptop overnight

Reply to
Andy Burns

He assures mum that it doesn't take much once it gets up to temperature, but then he tells me a slightly different story, I've given him my old CurrentCost clamp meter, but he's never told me how much a firing costs ...

He does sell some of his work, and makes other work to order, but I suspect it's not fully costed!

Reply to
Andy Burns

He thinks it will be struggling to sustain top temperature, and shortening the life of elements more quickly in the process of trying.

Reply to
Andy Burns

Not sure if his controller insists in ramping up over several hours, I realise the same measurement still applies, just less convenient to take readings,

I've conveniently ignored the lighting circuit (lower tolerance on voltage drop, probably scrapes in even with existing configuration) and sockets (I think he avoids running the pugger on its 1ph->3ph convertor while the kiln is running and I doubt the wheel uses much power).

Anyway, yes there is a socket circuit, so I could leave a meter (or UPS) plugged in there rather than poking about with unfused leads.

Reply to
Andy Burns

You could turn that into underfloor heating:-)

Reply to
ARWadsworth

That's easy for you, with your shiny new toy. I only have a multimeter (not true RMS) and an ancient wind-up Robin insulation/continuity tester (nearly a decade out of calibration).

It's a bee he's got in his bonnet ... If I can persuade him it'll only make a 2-3V improvement he might drop the idea, but then again, since it seems allowable to double-up in these circumstances, he might want to suck it and see.

Thanks, It did seem counter-intuitive that a dead short on such a fat cable (even of that length) wouldn't pop the breaker in under a minute, serves me right for trying to think at 3am.

Reply to
Andy Burns

You can guestimate the Ze.

Measure the voltage at the house CU with no load. Then measure the voltage at the house CU and with a known load. Ze = (Vnoload-Vload)/current

It's how I calculated Tim Watts Ze and was within 0.02 ohms.

The problem is that if you have a Ze of say 0.15 ohms then there will always be a voltage drop of around 7V on your incoming supply with a 47A load even if you moved the kiln to next to the CU. I realise that you do not pay for this voltage drop as you would on your SWA cables.

Now if your Ze is 0.02 ohms it would be worth doing something about the SWA voltage drop.

Reply to
ARWadsworth

I could borrow one, I was thinking you'd need the proper kit to measure it, rather than estimating it that way.

Yes, I see ...

Reply to
Andy Burns

I'd say it was unlikely that a few volts drop would make any significant difference..

I'm guessing it's a pottery kiln ? Does it have a temperature controller, or does it rely on cones to control the firing ?

My glass kiln (computer-controlled - and smaller at only 14kw) uses a surprisingly-small amount of electricity per firing - as the heating elements are turned on and off to achieve the required ramp and hold times.

I would think that he's probably worrying about nothing...

Adrian

Reply to
Adrian Brentnall

forget that about the power on my kiln - it's 3kw or thereabouts - duty cycle comments still apply! Adrian

Reply to
Adrian Brentnall

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