Solar Array swa sizing

[bigspark17]

Any help on solar array swa sizing? 4kw array needed around 20M swa from house to array.

thanks

 

[pc1966]

I don't think it is any different from sizing a 4kW load really, just power in the opposite direction!

I recently sorted a few issues with a friend's 4kW solar system, can't say I was impressed with the installation in some ways but they had used 4mm 3C SWA for the feed. Not necessarily as it is needed from a CCC aspect, maybe to reduce losses (so more money long-term), or because they had a lot of it for covering all jobs they did, or based on bonding requirements. Now had they only earthed the armour properly...

They also has rotary isolators for both the AC circuit at both ends (by house CU, also by inverter in the garden) and for the pair of DC feeds (2 strings of pannels, not symmetric, but resulting in 4 poles to switch) from the panels to the inverter, along with a warning label to switch off AC first so inverter shuts down.

I don't have the latest amendment 2 regs yet, but the size needed might also be based on minimum CPC size from a bonding aspect, so that should be checked based on your customers supply type, etc.

 

[JD6400]

You should aim for 1% volt drop , as the inverter can see the drop and compensate for it , possibly putting the output of the inverter over voltage , so the smaller the volt drop the better.

The 4 pole DC isolators are to do with breaking the DC as it helps with arcing , you loop each cable through it so the live and neutral each break 2 contacts at a time .

 

[JD6400]

From what I have read , the main thing with amendment 2 is making sure any component used is fed from the right direction if it states it is directional in the manufacturer's instructions

 

[pc1966]

Meeting a 1% drop is quite hard going, looks like 10mm needed here?!

Edit: Playing with this on-line calculator has 4mm meeting about 1.5% drop and 6mm meets 1.02% which is near-as-damn-it:

Cable Calculator

Cleveland Cable Company - Cable Calculator

www.clevelandcable.com

Really should be checked against the regs' tables or another calculator though.

Edit #2: Trying T&E or 2C SWA gives slightly different results, but seems a sane choice is going to be around the 6mm size.

 

[JD6400]

I have never got my head around how the inverter can see the volt drop and then add to it , but apparently it does .
I have found that it is normally one of the indicator's if the system was installed to a decent standard if the installer has run a dedicated cable for the solar when it is installed on satellite buildings on the site and not just chanced it and used the existing submain's .
If I remember correctly , it at least , used to be a requirement to show your calculations for the feed cables to be 1% or under , for the MCS paperwork.

 

[pc1966]

I have never got my head around how the inverter can see the volt drop and then add to it , but apparently it does .

I guess if the inverter is periodically altering the drive current / phase angle and monitoring its terminal voltages then it could determine the supply impedance rather like a MFT does Zs, etc. True, that is not measuring just the feed cable but the whole network, but I guess in most cases the supply impedance (i.e. leading to PSCC) from the 60-100A grid is going to be far lower than the end of sub-main for a 10-20A feed.

Presumably it has to do something like that anyway so it knows to shut down if the grid supply is lost, so clearly the inner workings of the on-grid inverters are much more complicated than a simple "AC from caravan battery" style of inverter!

 

[marconi]

Any help on solar array swa sizing? 4kw array needed around 20M swa from house to array.

thanks

Is your question regarding the 230V ac or the direct current produced by the 4kW array; I read your question as the latter.

The way to size the swa is to find out the no load voltage produced by each panel of the array, draw out how the panels are connected together and then determine the maximum no load output voltage of the total array. If the panels are all in series to produce one string then maximum output voltage is panel voltage x number of panels.

But sometimes sub-sets of panels are wired in series to produce strings, and then these stings are wired in parallel. In this case the total voltage is the number of panels in a string x panel voltage.

One you know the maximum voltage you need to know that the array produces its maximum power of 4kW when the array no load voltage is halved - the solar power converter uses a technique called maximum power point tracking MPPT to maximise the product of the panel output voltage x panel current = Panel Power output.

So, if the peak array no current voltage was 400Volts dc, peak power is when the array voltage is 200V (not exactly true because panels are not linear but this is a complication you can ignore).

Thus 4000 = 200 x I, where I is the current flowing between the array and the solar power converter. I then at the MPPT is 20Amps

Now use this on-line calculator to select the size of swa cable which I have done for an underground 2 core SWA cable:

Voltage Drop Calculator | TLC Electrical - https://www.tlc-direct.co.uk/Technical/Charts/VoltageDrop.html?ad_position=&source=adwords&ad_id=355954183600&placement=&kw=swa%20cable%20size%20calculator&network=g&matchtype=b&ad_type=&product_id=&product_partition_id=&campaign=ROAS_Cable&version=finalurl_v3&gclid=Cj0KCQjwmPSSBhCNARIsAH3cYgY9IHGqAlf93qzwrdCe-Ki_kgLmTaQu1ORljSS9aVaCqCLL81fTcmYaAnqqEALw_wcB

It suggest 2.5mm2. The voltage drop at 20A is 7.6V - call it 8V. The power wasted as heat over its length at 4kW array output is 20 x 8 = 160Watts which is 160/4000 or 4% transmission loss. If you used a 4mm2 cable the loss would be 2.5/4 x 4% = 2% or about 80Watts since the resistance of a cable conductor is proportional to its cross sectional area and power transmission loss is proportional to R since Ohmic loss is IsquaredR.

Now cost up the required length of 2.5 or 4. Does the extra cost of 4mm2 justify itself by the value of the energy saved from lower power transmission loss. Compare the value of the energy saved over a year with the opportunity cost of investing the extra purchase cost at current interest rate. This is a Kelvin's Law for economic cable investment. See:

https://www.allumiax.com/blog/how-to-choose-the-most-economic-size-and-type-of-cable

 

[pc1966]

Trying a simple spread sheet for cable low costs for the economics is hard to do as the solar power per year is not constant, but assuming the "average" is 1/4 max for the purpose of I2R calculation, and using the resistance values alone from the OSG I get the following:

Electricity cost​	27.9​	p/kWh​	​	​	​
Solar power (max)​	4​	kW​	​	​	​
Solar max current​	17.4​	A​	​	​	​
Solar average/max for I2R​	0.25​	(ratio)​	​	​	​
Solar average power​	1​	kW​	​	​	​
Solar average current​	4.35​	A​	​	​	​
Expected lifetime​	20​	years​	​	​	​
Cable length​	20​	m​	​	​	​
Cable size​	2.5mm​	4mm​	6mm​	10mm​	​
Resistance​	14.8​	9.2​	6.2​	3.7​	Ohm/km​
Loss​	5.60​	3.48​	2.34​	1.40​	W​
Lost annual power​	47.8​	29.7​	20.0​	12.0​	kWh/year​
Annual cost​	13.3​	8.3​	5.6​	3.3​	£/year​
Over lifetime​	267.0​	165.9​	111.8​	66.7​	£​
Volt drop (max)​	5.15​	3.20​	2.16​	1.29​	V​
Volt drop (max %)​	2.24​	1.39​	0.94​	0.56​	%​

Without messing around on assumed future electric cost and interest rates, etc, I would add the estimated lifetime loss cost to the install cost and pick the cheapest (if customer thinks it through).

 

[pc1966]

Adding 16mm column and putting in Superlec prices for LSZH 3C SWA I get the following:

Cable size​	2.5mm​	4mm​	6mm​	10mm​	16mm​	​
Resistance (per unit length)​	14.8​	9.2​	6.2​	3.7​	2.3​	Ohm/km​
Resistance (installed)​	0.296​	0.184​	0.124​	0.074​	0.046​	Ohm​
Loss power​	5.60​	3.48​	2.34​	1.40​	0.87​	W​
Lost annual energy​	47.8​	29.7​	20.0​	12.0​	7.4​	kWh/year​
Annual cost​	13.3​	8.3​	5.6​	3.3​	2.1​	£/year​
Over lifetime​	£266.95​	£165.94​	£111.83​	£66.74​	£41.49​	​
Volt drop (max)​	5.15​	3.20​	2.16​	1.29​	0.80​	V​
Volt drop (max %)​	2.24​	1.39​	0.94​	0.56​	0.35​	%​
​	​	​	​	​	​	​
Cable cost (per unit)​	£1.74​	£2.36​	£3.08​	£4.75​	£7.09​	£/m (Superlec LSZH Apr 2022)​
Cable cost (installed)​	£34.80​	£47.20​	£61.60​	£95.00​	£141.80​	​
Total cost (cable + loss)​	£301.75​	£213.14​	£173.43​	£161.74​	£183.29​	​

So cheapest over lifetime (ignoring interest on borrowed money to install adding to install cost, ignoring likely future tariff increases adding to wasted power cost) it seems the optimum choice is 10mm.

At 10 year pay-back it is 6mm coming out best, if power less also thinner works out marginally better, etc, etc.

Maybe not need 3C so 2C SWA and using armour as CPC makes even more sense for 10mm case, etc.

 

[pc1966]

I may have really over-estimated the average solar power, but equally I have probably underestimated future electricity costs as well but a large amount!

 

[marconi]

I may have really over-estimated the average solar power, but equally I have probably underestimated future electricity costs as well but a large amount!

There are on line estimating tools which can produce annual and month by month figures for pv generation.

I use this one:

PVGIS (PV-GIS) - https://photovoltaic-software.com/pv-softwares-calculators/online-free-photovoltaic-software/pvgis

 

[pc1966]

There are on line estimating tools which can produce annual and month by month figures for pv generation.

I have seen some tools such as that and they offer you the expected annual power. But the problem here is the I2R losses depend on the current squared, so the shape of the illumination waveform has an impact on the relationship between average solar generation and average cable loss.

E.g. if we get 10% of the panel capacity on average (roughly what I just saw on one site for south England) that could be:

100% generation for 10% of the time, so I2 term is 0.1 * (17.4)^2 = 302.7
50% generation for 20% of the time, so I2 term is 0.2 * (17.4 / 2)^2 = 15.1
25% generation for 40% of the time, so I2 term is 0.4 * (17.4 / 4)^2 = 7.6

Due to Earth's rotation the average daylight per year is 50% so 10% panel capacity averaged per year really means 20% average illumination per day, so probably 30-40% sunny cloud-free time and allowing for sine-shaped illumination arc over the day, etc. (but of course that varies summer/winter, and so on).

Above I had assumed 100% time at 1/4 power = 18.9 for I2 term so loss cost maybe high by 20-50% or something.

 

[pc1966]

I think the TL;DR answer is a 1% VD goal is probably close to the most economic cable choice!

 

[marconi]

The trick of course is to connect the panels to achieve the highest off load pv array voltage - to keep current lower for a given power- while being mindful of the inverter input voltage limit and staying under the Uo/U voltage spec of swa cable noting that for example 600/1000V are ac rms figures so higher figures for dc by a factor of sqrt2 Thus 850/1400V dc.

Uo voltage between live conductor and earth.
U voltage between live conductors.

In practice, peak panel power output is normally at a voltage whigh is between 70 and 80 % of the no current open circuit voltage not at 50% I mentioned in my opening post. I tend to do pessimistic calculations so it performs better than expected rather than worse. Real panel voltage/current/irradiation curves tend to allow thinner rather than thicker cables cos the IsquaredR energy loss is lower Than the 50% assumption would calculate.