Solar panel related question -- help please

in least sunny month.(kWh/sq m/day)

suspect you'd get a better answer in a better group. Anything with solar in the name or alt.energy.homepower. Or get an answer from the panel vendor.

I've gone looking for that definition in the past, and again today. I never found anything that approached a definition. I chose to ignore it and do the calculations.

Are you sure you're quoting the right number? Your 2.96 sounds more like an annually averaged daily insolation than the least sunny month. Multiply that by 20% collection efficiency (round number) gives more like 0.6 kWh/day/m^2 for energy generated averaged over the year...much less in winter.

Given the way things tend to be measured at optimum, I'd expect that you'd need MPPT and optimum sun tracking and no buildings or trees to get that. And remember that the panel is rated with no rain or snow or or dust on it. And no bird crap or smog or smoke or nothin'. And no losses in the conversion to what you actually need. Local micro-climates can have a siginificant effect. Those won't show up in average insolation numbers.

Solar sucks if you have any alternatives.

"mike"

.... Phil

diation in least sunny month.(kWh/sq m/day)

...my opinion: Collection Efficiency = (Electrical Energy generated in kW) / (Energy 'available' in light per sq m).

Although, don't know why it is called 'collection efficiency' instead of 'conversion efficiency'. Maybe there is one more factor involved which would include reflectors and/or lensing.

There is a maximum amount of energy in sunlight per sq meter. From memory, it was around 0.35 W per sq cm - measured for straight in sun at the equator.

As I said, it is my opinion that CE relates to how well you can convert that supplied energy into electricity. Also, from memory I remember solar cells at the time being in the range of having 1-3% conversion efficiency. But, absolutely not even near 10%. Had something to do with the limited spectral sensitivity range of the silicon. The sunlight had a lot of energy in frequencies the silicon couldn't 'see'.

After reading wiki, it appears that the 0.35W number more reflects what to expect from a solar cell.

radiation in least sunny month.(kWh/sq m/day)

The number I always heard was 1kW/m^2 = 1mW/mm^2 (1 mm^2 is nice for laser stuff, since its about the size of a laser spot, and also about the size of your eye pupil.)

George (millimeter) Herold

After reading wiki, it appears that the 0.35W number more reflects what to expect from a solar cell.

That would give 3500 watts per m^2. Fails the reality check.

I think the insolation runs about 1 kw / m^2 at the equator (I did not google it though).

That's about right, although I take issue with 1mW/mm^2. ;-)

Jeroen (SI) Belleman

No. I don't understand it myself. However, that's never stopped me from guessing.

I subscribe to Home Power Magazine. I vaguely recall mention of PGF in a recent issue. Other than that, I've only seen it on panel spec sheets.

in least sunny month.(kWh/sq m/day)

It's a conglomeration of various inefficiencies found in solar power generation. As near as I can guess(tm), the collection efficiency is a magically derived number intended to make it easy to calculate the number of solar sells, err... cells, needed to supply some minimally acceptable power level.

Reading between the lines and Googling, it seems to be the ratio of what would be considered normal output (including all the losses) to maximum output under ideal conditions. It my guess(tm) is correct, that's a rather dumb way to size a system since the losses and inefficiencies vary substantially with the system type, local conditions, inverter type, etc. Some of the factors are age variable, such as cell deterioration, battery charge efficiency, and growth of nearby trees.

Some hint as to what you're trying to accomplish would perhaps produce an answer that fits your context. The only real help I can offer is to suggest that you ignore PGF and use one of numerous online solar calculators to size your system. Also, do a sanity check. Drive around your area and note that addresses of home solar installations. Ask the owners for performance statistics. Many such systems are publicly available online, usually on the controller vendors web site.

Of course, anything you can actually buy is obsolete. Nano antenna panels are theoretically >80% efficient:

Correct, but that's only at noon, with the sun directly overhead.

in least sunny month.(kWh/sq m/day)

I'm not exactly familiar with the term "PGF", but from your definition, it seems pretty straigh forward...

If the PV module's efficiency is the same as collection efficiency and the module's efficiency is the typical 15% like normal mono-silicon modules are, and the average sun hours is say, 5 hours, (5 kW-Hours at for a typical 1kW/m^2), then I would think this PGF would be 0.15 X

You could just replace that 5 kW-hours with how many there are on the day you are calculating I suppose.

boB K7IQ

Looks correct to me:

By coincidence, todays lunch time discussion was over installing solar panels on a mountain top radio site to reduce the cost of utility power. My simplificated back of the envelope guesswork was:

Start with a single commodity 130 watt panel for about $150. Average solar insolation in Santa Cruz CA is about 4.5 kWh/sq-meter/day. (goes down to 2.0 in December): or the equivalent of about 4.5 hrs of useful illumination per day to produce 4.5 * 130 = 585 watt-hrs per day. Commercial electricity averages about $0.23/kw-hr, so one panel will save no more than: $0.23/kw-hr * 0.585 kw-hr/day = $0.14/day if I ignore all the system inefficiencies and losses. Therefore, each panel might save about $4/month in electricity. With the panel (alone) costing $150, it will take 150/4 = 37.5 months to pay for the panel in savings. Of course, this is the best case. Reality tends to be much worse.

Grin, no Jeroen was just poking fun at me for an earlier post about units. (I was hoping to 'twig' him with mW/mm^2.)

George H.

Oh. In college, we had the irritating habit of inventing our own units of measure. For example, coeds were rated in milliHelens. Helen of Troy was said to have launched 1000 ships, so one milliHelen would launch one ship. Negative milliHelens were also allowed for coeds that could sink a ship.

We also had units of measure for evaluating instructors. One not so memorable instructor, Dr Green, successfully failed the entire class of about 40 students for not meeting his ever changing minimum standard of excellence. Thereafter, the tendency for an instructor to fail one student was measured in Greens.

Somewhat later, I was working on an RF power amplifier design, that incorporated such a complicated system of feedback, that it would not neatly fit into the traditional Class A, AB, B, C, D, etc designations. We therefore named it after our fearless leader, the Class M amplifier. Unfortunately, that appears in the product literature, owners manual, and a magazine article which required far too many explanations to be considered a good idea.

I created a similar problem when I decided that everything above the water line should be measured in feet, while everything below should be measured in fathoms. I thought that specifying the cable length for a sonar transducer in fathoms was cute, but nobody else did. Both production and purchasing could not consistently multiply by six.

So, if you really want to get his attention, just name 1 mw/sq-mm as one Jeroen. He will either consider it an honor, or assault you violently at your next meeting.

Full disclosure and confession: I still use micro micro Farads. Old habits are difficult to break.

diation in least sunny month.(kWh/sq m/day)

A particular solar panel product will specify the output power , Pout, meas ured at standard test conditions (STC) which are defined as 1000 W/m^2, 25o C cell temperature, and 1.5 Atm air pressure. They also specify a derating factor for the variation of Pout with temperature, usually something on the order of -0.5%/oC, and a tolerance for the output power. The PGF is an ins tallation specific conversion factor specifying the daily panel output elec trical energy in Wh, or kWh, per watt, or kW, unit of solar panel output po wer. The specifics used to compute the PGF are average cell temperature, av erage irradiation of the cell in units of 1000 W/m^2, average meteorologica l conditions, average insolation. Once the PGF is known, all you have to do is divide the minimum daily energy required of the solar (adjusted for ob vious system losses like inverter efficiency, charging efficiency, panel po inting errors, panel dirt accumulation...) by the PGF to derive the total r equired panel STC power. Then divide by the per panel STC output to derive the number of panels.

It's not that deep, it's a number used by installers in fixed geographic ar eas. They can lump all the stuff that stays the same on all their jobs into the PGF. Things like the efficiencies of their inverters, presumptions of cable losses from just eyeballing the building dimensions, the low ball wor st case insolation, the charging efficiencies of the battery system if any, typical panel power output reductions to be expected from soot, pollen or whatever in their area, etc... all that junk is pretty much the same every time. The only custom work that needs to be done is to size the energy requ irements of the customer. Then PGF makes the energy->panel power conversion . Manufacturers don't spec energy, they spec power, and systems are designe d around energy and not power. So PGF gets them there.

radiation in least sunny month.(kWh/sq m/day)

Thanks for the links. As the 'panel generation factor' is used to determine the number of solar panels to be used at a given site, the following is an alternative scheme that some of us have put together to determine the number of panels at the same site. Let us suppose that the electrical parameters for a given panel are: Wp=75 Watts Isc=4.8 A Voc=21.4 Volt Im=4.4 A Vm=16.8 Volt Active solar panel area AA = 0.605 sq. meter

Define FF = (Vm*Im)/(Voc*Isc) = 0.718

Then, at STP Efficiency= (Voc*Isc*FF)/(It*AA)*100% = 12.36% (at STP It=1000 Watts per sq. m)

Now let us suppose that the average insolation at the site is: 500 Watts per sq. m Actual efficiency at 500 Watts per sq. meter = 12.36*0.5/100.0 = 0.0618 = 6.2 % Power produced by the single panel=0.062*500*AA(0.605) = 18.75 Watt per hr So for a 8 hour period(brightest possible sunlight), the total power generated by the panel = 8*18.75 = 150 Watts per day So if the total power requirement is ww Watts, the total number of panels is ww/150

How does the reasoning look like ? Are there any glaring mistakes ? Please provide your comments

gaccckkkk!!!

slipped a 'few' decimal points.

Don't forget the cost of the security system to thwart 'solar panel' thieves.

Snip

Also use the proper derating factor on performance statistics gleaned from any solar system owner, that spent his hard earned money.

Mikek :-)