camelot-forum.co.uk

Just wondering, hypothetically speaking.. Let's say I had a Jim Cracknell run of cable carrying 240V or so @ 1kW?

What would the V drop (resistance) be going from 1.5mm sq copper to the also off the shelf.2.5mm sq copper conductor? Say over a 20 mtr run?

Is it literally a 40% drop in resistance? Please consider this a teaching moment. I can find it out tomorrow but I thought it would be a nice opportunity for someone to teach the forum?

Cheers

Download the Megger app, has a nice handy cable calculator all free

TLC have a good calculator too.....
https://www.tlc-direct.co.uk/Technical/ ... 759dccbd35

Am i missing something with this?

Have Used both calculators and obtained different answers for similar situation.
One says 1.0mm the other says 1.5mm.

By coincidence the same length/loads as my a2a installation the other day.
I used 2.5mm which i had lying around, knowing it more than capable. at least my cable wont be getting warm then.

resybaby wrote: ↑Mon Sep 23, 2024 12:17 am Am i missing something with this?

Have Used both calculators and obtained different answers for similar situation.
One says 1.0mm the other says 1.5mm.

By coincidence the same length/loads as my a2a installation the other day.
I used 2.5mm which i had lying around, knowing it more than capable. at least my cable wont be getting warm then.

That looks good! Once back from Germany I'll try the Bongo power cable cobbled temporarily together to get power from workshop up to hot tub where the outdoor CU is. Just need to catch one or two pieces of late Autumn sunshine and I'll know from the Ketotek meter if its worth trenching in a cable. I hope so as it would be great to not hit the 253V trip and reset limit on the garage Hoymiles unit from the shove of the Enphase to get the power through..

If I do a suicide lead and remain fully aware I can switch that run on & off while seeing the change on the meter.

Immediate disassembled and back to commando sockets after though.

resybaby wrote: ↑Mon Sep 23, 2024 12:17 am Am i missing something with this?

Have Used both calculators and obtained different answers for similar situation.
One says 1.0mm the other says 1.5mm.

By coincidence the same length/loads as my a2a installation the other day.
I used 2.5mm which i had lying around, knowing it more than capable. at least my cable wont be getting warm then.

I notice one uses SWA and the other uses T&E, if you change the latter to SWA it also coes out with 1.5

The online calculators are rough and ready, don’t give all the correction factors for proper design but are handy.
Differ me in cable might be that I picked SWA, clipped direct, the other might default to something else.

Personally I wouldn’t use 1mm2 for external supplies in Swa.

Agreed, it's nothing but a pain in the arse if future demands require a bigger cable.
Oversized is good, as long as you don't go crazy.

Joeboy,

If I want to check the reality of a result, I compare doubling or halving one or both.

To halve the resistance, one would need to double the cross sectional area, which would double the current carrying ability (and power) at constant voltage.

The analogy would be of having two conductors in parallel, each carrying the same current and delivering the same power. However this may not always quite hold, for calculators/tables, simply because conductors come in standard incremental increases, and the conductor size has to always be rounded up! (ie minus zero, plus to the next higher size).

Also some calculators may not include every possible scenario. Perfect insulation of the conductor would mean the only heat loss could be at the ends, so the temperature at the centre of the run would continue to rise - likely past the maximum permissible (where the resistance will rise and the insulation sheaths will degrade).

So yes, the resistance would fall by 40%. A graph of resistance v cross sectional area is a straight line, so a proportional relationship (double one, halve the other in this case). We have not doubled the cross sectional area so the resistance has been reduced, but not by half.

Imagine three 0.5mm^2 conductors (equivalent to one 1.5mm^2 conductor), each carrying 333W. 5 of those 0.5mm^2 conductors will be able to carry 5*1/3kW = 1 2/3kW = 1.67kW

The relationship between parallel resistors and actual overall resistance is the sum of the reciprocals (whereas for the total of resistors in series is the sum of those resistors).

Without setting out all the Ohm’s Law related maths, increasing the cross sectional area from 1.5mm^2 to 2.5mm^2 conductors is a 67% increase of the former to the latter. The reciprocal of the 1.667 is 0.60 which is 60% of the original - a 40% decrease in resistance.

Someone else could likely explain it better than I can….