How much current and voltage does CS6P-240P produce on an overcast day?

Here, near Olympia, WA, it is mostly cloudy during winter. I understand that it is impossible to know how much light there is and how much overcast the sky is. I am just looking for ball park figures. Is it less than 24V? How low is the current?

I plan to get 12 panels. Would they be able to still put at least some juice in the 24V battery bank on a fully cloudy day? Would getting 4 extra panels offset the effect of the "missing" sun?

Thanks much.

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A solar module produces a 'curve' of power throughout the day. When exposed to any amount of light, it will produce voltage and amperage. If you do a search for IV curve, you'll find lots of useful science.

You are correct about adding panels to make up for 'missing' sun. To determine how many panels you need, first you figure out your loads and how much of that you want to offset. Then you design your sustem around the average Sun Hours for Olympia (of a similar site). 1 sun hour is equal to 1000W of energy hitting a 1 meter square area for 1 hour.

The entire world is mapped for sun hours, and it's fairly easy to find. Here's a sample for Olympia from NASA's data:

Output of a solar panel is measured at standard test conditions (STC) of 1000W/Square meter at 25 degrees Celsius. Using the available sun hours, you can then figure out how much power you are likely to get out of your array on a yearly basis.

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You can even skip ahead and use your Iphone or Android phone to calculate power output.

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Here is the IV curve for a Canadian Solar 255

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Thank you, SF, To understand better the curves, please clarify for me the first line in the table. You said that the design of a system should be based on sun hours and the table is using "day". That is confusing to me. Thanks again, Luby

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Sure. In the table, for January (I), it shows 1.03 kWhr per square meter of surface. So for that day, only 1030 watts was available. Does that make sense?

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Thank you again! I understand what you wrote. Is the 1030W available that day per hour or is it a combined wattage for the whole day? How do I use this figure with the curves which are laid out in W/h? Should I devide this amount by 24 hours (which I think would be dumb, since there is no sun during the night) or by 12 hours? I hope I am explaining my confusion (and unknowingness).

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OK. The (basic) math would look like this:

The 1030W available is for that daily period of daylight (average), for the month.The 240W module has an area of 1.6M. So 1030 x 1.6 = 1648W available for the daylight period, hitting the module.The module is 14.92% efficient at converting the sun's energy into DC power, so: 1648 x 14.92 = 245.9W per 240W module.
This is an explanation from someone who employs third party software every time he sizes a system - so use with caution! But I think it answers your question in that even in the longest, darkest month of the year, each 240W module can put out about 246W of DC power.Hope this is useful.
Edit: December is longer and darker at .88 kWhr. Makes Portland, OR seem downright tropical.