Mark42

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My brain is overheating trying to make sense of PV panel parameters and calculate the CSA of a DC feed cable to the inverter.

For my 16 HL Solar 250W panels, the spec is:

Imp 4.94A
Vmp 50.6V
Isc 5.35A
Voc 60.5V

Two strings of eight panels in parallel.

Therefore:

Total current at max power = 4.94 x 2 x 1.25 = 12.35A

Total voltage at max power = 50.6 x 8 x 1.15 = 466V

[Note 4.94A x 50.6V = 250W per panel, as spec, but allowing for max tolerance, 4.94 x 50.6 x 1.25 x 1.15 = 359W (ie unlikely!)]

However the DTI guide (Sec 2.1.6) uses Voc and Isc for their main dc cable size calculations:

Total Isc = 5.35 x 2 x 1.25 = 13.4A

Total Voc =60.5 x 8 x 1.15 = 560V

[Note 5.35A x 60.5V = 234W per panel, but allowing for max tolerances, 5.35 x 60.5 x 1.25 x 1.15 = 465W (ie ridiculous!)]

I believe the de-rating factors of 25% and 15% are to allow for possibly over-spec panels.

But surely the panels cannot be SIMULTANEOUSLY over-spec on both parameters of voltage AND current? And ALL of them, ALL of the time?

But the DTI guidance appears to indicate that they can be. Is this right, or am I misunderstanding something?

Also, surely Voc cannot simply be added up to give a 'real-world' figure for series strings can it? Are there not resistive losses?

So which voltage/current figures does one use to calculate DC volt drop? And why is only 1% recommended?

[*** EDIT: this is wrong. It's less than 3%, Ref: DTI Guide, Para 2.1.4.1 ***]

If calcs are made using the above absolute best performance figures, surely we don’t have to engineer the cabling to give less that 1% drop on the very few UK days when the panels are producing at the ABSOLUTE MAXIMUM possible? Why bother? If there is a few percent loss on those fantastic days, then so what?

I don’t get this at all.

The upshot is, what is the minimum CSA for a 150m DC feed cable, XLPE SWA, laid to rest in the cold Norfolk earth?
 
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You are applying numbers to the wrong situation.
- the max Voc is important because the cable has to be able to take that from an insulation perspective (current will be zero).
- the max Isc is important because the cables must be able to take this current for a sustained period without melting the insulation and sheath (voltage will be lower than normal).
- the max Vmpp/Impp are important as they are the max operating point of the panels and the conditions at which you need to design the voltage drop to be an appropriate percentage, normally 1%.
So they are three different situations not one.

Sunny design or similar will give you an answer to your sizing question

Regards
Bruce
 
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allowed volt drop per string is worked out under mpp which is normal operating conditions, it 3% of 466 which is just under 14v you would use 12.35a for the amps to calculate volt drop if you were using 2 core, it would be 10mm which gives volt drop of 8.7v (12.35a*150m*4.7mv)

you would now double check using the voc and isc (corrected) that the cable can take 460vdc and 13.4amps , I should hope you don't need to refer to the regs to see if 10mm takes 13.4amps!!!!

The panels can't be open circuit and short circuit at the same time!
 
To be totally clear, just found this!

2.1.4.1 Cable sizing

Cables must be rated, as a minimum, to the voltage and current ratings derived using the multiplication factors in 2.1.2.
Standard de-rating factors must also be applied (BS 7671).
Cables should be sized such that overall voltage drop at stc between the array and the inverter is <3%.
 
You are applying numbers to the wrong situation.
- the max Voc is important because the cable has to be able to take that from an insulation perspective (current will be zero).
- the max Isc is important because the cables must be able to take this current for a sustained period without melting the insulation and sheath (voltage will be lower than normal).
- the max Vmpp/Impp are important as they are the max operating point of the panels and the conditions at which you need to design the voltage drop to be an appropriate percentage, normally 1%.
So they are three different situation not one.

Sunny design or similar will give you an answer to your sizing question

Regards
Bruce

Thanks, Bruce, that’s great stuff!

But I want to understand the “Why?” of all this, not just plug numbers into someone else’s computer program and bow down, unquestioningly, to its sacred pronouncement. :)

Maybe the program is based on false premises, or ridiculous worst-case scenarios not applicable in every case.

That’s what engineering is all about, knowing when you don’t have to follow general rules like a parrot, and be able to state, robustly when necessarily, why that is.

Note that the DTI guide, in section 2.1.6, specifically states use Voc and Isc for cable calcs; it’s double–ticked too, indicating it’s a REGULATION, that mere mortals must follow.

Do you therefore agree that these are the wrong figures to use, in that Voc and Isc can never occur simultaneously in a real circuit?

I knew nothing about PV a week ago so hesitate to argue with ‘experts’ who write books so soon, but there it is …
 
I quite agree with you - understand what tools are doing and their limitations.

On the double tick, I suggest you are over-interpreting:
- you do use Voc, but only to check insulation levels;
- you do use Isc, but only to check it will not melt;
- and you use mpp voltage and curent for the voltage drop calc.

Edited to add look at an output curve for a panel and you will see Voc and Isc can never occur together. Where would the power come from?
 
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Thanks screwdriver, that's great stuff too. So it’s 3% 'allowed' for DC and I was wrong. But my comment about why engineer for only the very best days still applies.

My gut feeling always was that 10mm would be absolutely fine, considering the total ENERGY being transferred (regardless of voltage, current, day of week or whatever).

But I do try not to think with my gut.
 
You use voc and isc to calculate the maximum current and voltage the cable needs to take as a bare minimum.

the vmpp and Impp are there for you to work out everyday running conditions for volt drop which would generally be what I start with on a cable calc as it is quickest to get the right size for longer runs.

Same lines as BS7671 calcs as you work out volt drop on the load and not the rating of the mcb.
 
You may find that engineering for the best days will mean the inverter is always in limits with voltage and current so the tracker can work efficiently, I have seen cases where the AC has been under engineered ( 2.5mm 40meters through long loft) and the inverter has been tripping on the brightest days with output overvoltage so there has to be a limit set to how close you get to the edge!
 

On the double tick, I suggest you are over-interpreting:
- you do use Voc, but only to check insulation levels;
- you do use Isc, but only to check it will not melt;
- and you use mpp voltage and curent for the voltage drop calc.

Then it should bloody-well say so, because ‘over-interpreting’ is exactly what non-practical inspectors with their clipboards do!

… look at an output curve for a panel and you will see Voc and Isc can never occur together. Where would the power come from? (my emphasis) …

Exactly! Well said, Sir!
 
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You may find that engineering for the best days will mean the inverter is always in limits with voltage and current so the tracker can work efficiently, I have seen cases where the AC has been under engineered ( 2.5mm 40meters through long loft) and the inverter has been tripping on the brightest days with output overvoltage so there has to be a limit set to how close you get to the edge!

I agree, over-voltage tripping is a well-known phenomenon, and the reason I have (during my one-week steep learning curve!) come to definitely support long DC over long AC runs.

The problem of keeping voltage drop down on a long AC run on a good day, especially in an installation where the mains voltage tends to be on the high side, is a considerable engineering challenge, and one that can only be solved in one way: copper, and lots of it.

But I was talking about my long DC run. And so what if a little power is wasted on the best days? The chances are the inverter wouldn't be able to use it anyway.
 
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I always refer to sunny design version 2 and on there you can enter your cable sizes for both the ac and dc side and also length of run and the rest is worked out for you.Less than 1% on dc and ac side a thumbs up.
 
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?
 
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!
 
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.

https://www.electriciansforums.net/attachments/datasheet-202491x-20cable-20spec-pdf.120253/

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

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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.
 
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).
 
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.
 
... 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?
smile.png


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.
 
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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%.
 
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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%
 
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.
 
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.
 

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Mark42

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