Sizing DC cables, PV Array to Inverter. Confused. Help please!

[infinity]

Recommended is 1% on ac and dc.

3% on Dc in guide? 1% AC

 

[yellowvanman]

3% on Dc in guide? 1% AC

Yes I see that now. The Logic PV Course recommends 1% on each, thats were I got my info from!

 

[screwdriver]

OK. Let’s try to get a consensus here.

From the DTI guide:

2.1.2: Use (total) Voc and Isc to specify ‘components’, ie switches, connectors, cable type, etc, using safety factors of 15% for voltage, and 25% for current.

2.1.4.1 Cable sizing: Apply the safety factors as above (plus any BS 7671 derating). It’s not specified which parameters to use for the calculations though, which I think it should.

As Screwdriver says in his excellent post #5, Imp and Vmp are used here.

I hold the DTI guide is wrong not to state this clearly.

2.1.4.1 also states, clearly, that the Voltage Drop between array and inverter should be less than 3% (one tick therefore a recommendation.) So working on a 2.99 % drop for your dc calculations is fine.

2.1.6 is titled ‘Main d.c. cable’ but is actually (I think) talking about panel interconnects and other local array wiring, which is confusing. Using, again Voc and Isc, it applies only to cable type; it’s nothing to do with cable sizing per se. This is where I went wrong in the beginning.

Does anyone have a referenced formula, in metres and mm[SUP]2[/SUP] (not, like most on the web, AWG & feet) for DC volt drop calcs in underground SWA please?

Imp and Vmp are only to be used to confirm voltage drop, the minimum cable size must be worked from isc x 1.25 x correction factors for bunching/heat/insulation etc as that is the current that will flow under fault conditions.

Basic formula for volt drop?

the chart is in the BS7671 for DC volt drop

metres x Imp x volt drop(in mV/A/M which can be found in BS7671 )

so a 150Metre long 2 core 10mm pvc swa cable carrying 8.43 amps is going to give:

150 x 8.43 x 4.4 = 5563.8mV or 5.56v of volt drop which you can work minimum system voltage by dividing by 3 and multiplying by 100, it comes to 185v which assuming your panels were 30vdc each would mean you would need more than 7 to have a high enough voltage to comply with voltdrop limit of 3%





The table for 2 core swa is

1.5mm 29mV/A/M
2.5mm 18
4 11
6 7.3
10 4.4
16 2.8
25 1.75
35 1.25

Now I haven't used a corrected Imp in my calculations for volt drop because I believe that the inverter will keep the panel at it's optimum power voltage/current and therefore will compensate for temperature differences my moving the mpp around to suit although I would have thought that for the majority of the time the vmp/imp is where we should be looking for normal running conditions, I really don't mind being corrected on this one, although the increase in current is so low in comparison to the voltages I doubt it would begin to make much difference unless you encountered a massive cable run!

 

[Gavin A]

4mm double insulated single core cable is the maximum you are likely to ever need to use on the DC side of a sub 4kWp installation unless it's in a field or something silly like that.

in open air it can take 32amps, and the volt drop is negligible particularly on this system because the voltage is so much higher than standard AC runs.

Most situations will be fine with 2.5mm, but as the price difference is minimal we tend to stick to 4mm cable across the board to avoid any potential for a cable ending up undersized by accident.

http://portal.segen.co.uk/reseller/docs/Datasheet 2491X Cable Spec.pdf

do the calcs anyway, but you'll swiftly come to the same conclusion.

 

[SolarCity]

Yep, we use 4mm as standard. We use 6mm every now and then.

 

[sharpener]

the volt drop is negligible particularly on this system because the voltage is so much higher than standard AC runs.

I've often seen this repeated in this forum but it depends totally on the array design.

Many 230 - 250W panels have 60 cells and so a Vmpp in the region of 30V. A 4kw array divided into 2 strings of 8 will therefore have string voltages of 240V and (neglecting skin effect losses on AC) for the same cross-section the AC and DC losses per m of cable will be much the same.

 

[Gavin A]

I've often seen this repeated in this forum but it depends totally on the array design.

Many 230 - 250W panels have 60 cells and so a Vmpp in the region of 30V. A 4kw array divided into 2 strings of 8 will therefore have string voltages of 240V and (neglecting skin effect losses on AC) for the same cross-section the AC and DC losses per m of cable will be much the same.

but on this system the voltage in operation will be somewhere between 350-400V, and most of the systems we design are in this range as this is the peak operating range of the inverters we use (and most inverters on the market).

 

[sharpener]

but on this system the voltage in operation will be somewhere between 350-400V.

Sorry, I was thrown off the scent by screwdriver's assumption of 30V per panel.

But I don't think you would get away with 4mm^2 in this instance because as the OP says it is a 150m run (so it is "in a field" to quote your earlier posting).

Maybe someone should do the definitive calculation using all the right figures to put us out of our misery.

 

[Mark42]

... Maybe someone should do the definitive calculation using all the right figures to put us out of our misery.

Here you are, now I understand which figure is which (NO thanks to the DTI guide, but many thanks to others on this forum) it’s easy:

Total current at max power, where panel Imp = 4.94A, 4.94 x 2 x 1.25 = 12.35A
Total voltage at max power, where panel Vmp = 50.6V, 50.6V x 8 x 1.15 = 466V

Total Isc where panel Isc = 5.35A, 5.35 x 2 x 1.25 = 13.4A (As Screwdriver said, we can all agree 10mm can handle this without getting out our tables!)

Total Voc where panel Voc = 60.5V, 60.5 x 8 x 1.15 = 560V (SWA 600/1000, no problem)

Cable volt drop calculation for the 165m DC run (now measured accurately) on 10mm 4-core (4.7 mA/V/m) is:

Vd = 4.7 x 165 x 12.35 / 1000 v = 9.58V

9.58/466 x 100% = 2.06%

So, as my gut was telling me all along, 10mm is fine.

Compared with running AC for 165m, engineering an AC volt drop of 1% max to prevent over-voltage tripping, it’s a huge saving in copper (and hence cash!).

I rest my case as to which option is better.

I just ordered 4-core 10mm at £4.07/m and some 16mm 4-core (for something else) at £5.68/m. Was I ripped off?

ps. I initially thought one uses a different Vd formula for DC. I was confused, and using Ohms/km rather than mA/V/m, and reckoning you had to double the cable length. The latter already takes account of the return path, and apparently reactance and capacitance in big-CSA cables, which is negligible for the smaller stuff, so the single phase figure can be used for DC. Please tell me if this is wrong!

Also, the calc above does not take account of any cable temp factor. Again, it’s negligible and not worth bothering with in this case since 10mm is clearly within spec.

I’m bored with all this thinking now. I’m going out to dig holes instead

Thanks, gentlemen, for all you helpful comments.

 

[sharpener]

Here you are, now I understand which figure is which (NO thanks to the DTI guide, but many thanks to others on this forum) it’s easy:

Total current at max power, where panel Imp = 4.94A, 4.94 x 2 x 1.25 = 12.35A
Total voltage at max power, where panel Vmp = 50.6V, 50.6V x 8 x 1.15 = 466V

Total Isc where panel Isc = 5.35A, 5.35 x 2 x 1.25 = 13.4A
Total Voc where panel Voc = 60.5V, 60.5 x 8 x 1.15 = 560V

Cable volt drop calculation for the 165m DC run (now measured accurately) on 10mm 4-core (4.7 mA/V/m) is:

Vd = 4.7 x 165 x 12.35 / 1000 v = 9.58V

I don't think this is quite right, 12.35A is for both strings (are you connecting them in parallel at source and if so why the four-core cable?), and also includes the 1.25 safety margin which you don't need when calculating voltage drop at Impp. This will put the volt drop down by at least 20% and possibly a further factor of two.

9.58/466 x 100% = 2.06%

Again, the 466V includes a 1.15 safety margin, removing this will put the actual % volt drop up by 15%.

 

[Mark42]

I don't think this is quite right, 12.35A is for both strings (are you connecting them in parallel at source and if so why the four-core cable?), and also includes the 1.25 safety margin which you don't need when calculating voltage drop at Impp. This will put the volt drop down by at least 20% and possibly a further factor of two.

Two strings of 8, paralleled at source, hence 2 x the current. Then the whole damn thing again for feeding into the second phase. Hence the 4-core.

Good point about ‘ignoring’ the safety factors. Thanks!

Again, the 466V includes a 1.15 safety margin, removing this will put the actual % volt drop up by 15%.

OK, understood. So is this (finally!) right?:

Vd = 4.7 x 165 x (4.94 x 2) / 1000 v = 7.66V

7.66/(50.6V x 8)x 100% = 1.89%

 

[Mark42]

And, for 6mm (7.9mA/V/m):

Vd = 7.9 x 165 x (4.94 x 2) / 1000 v = 12.88V

12.88/(50.6V x 8)x 100% = 3.2%

With a bit of jiggery-pokery with thermal coefficients, and the argument that this amount of energy would only pass at the very best times, it could be argued that 6mm is fine too. I wouldn’t bother, since the cable cost differential is probably not worth the losses over 25 years, but there it is. It could make a difference to profit on a commercial job.

 

[sharpener]

OK, understood. So is this (finally!) right?:

Vd = 4.7 x 165 x (4.94 x 2) / 1000 v = 7.66V

7.66/(50.6V x 8)x 100% = 1.89%

I think so!

 

[Gavin A]

Sorry, I was thrown off the scent by screwdriver's assumption of 30V per panel.

But I don't think you would get away with 4mm^2 in this instance because as the OP says it is a 150m run (so it is "in a field" to quote your earlier posting).

Maybe someone should do the definitive calculation using all the right figures to put us out of our misery.

ah right, I did look, but missed that bit.

 

[sharpener]

one all <g>