Correct panel spacing - in rows of tilted panels??

[(unknown)]

Hello,

   I have a question that probably has been ask before but I did a search and turned up nothing. So here goes I'm going to install solar panels in 3 rows and each row is mounted on a flat roof so I'll need to tilt them up in the back. the tilt angle is 18 degree's so the back of the panel is roughly 20" up from the roof deck. So how far back  should the next row of panels be to not be shaded from the first row?? The site is located 40 degree's north latitude.The panels are orientated due south so thats a plus. My goal is to maximize the roof area so I could fit as many as i can up there. I was thinking that the height x 1.8 would be the correct formula to start the next row. Is this the answer for my location? or is this to short? I know it will work for the summer months but the winter is also a priority. My geometry skills from high school only go so far....(25 yrs ago)     Thanks in advance for your comments!!!

[clflyguy]

Paul,

  I don't know if I'm posting twice or not but it's been over an hour and I don't

see my response up yet so I'll repeat.

  At 40 degrees Lat. for 30" high panels you need a min. of 66" & for 60" panels

you need 11' (132") measured from hinge point to hinge point. If your panels are

of some other size you can extrapolate from these measurements.

  I'm glad I had to re-post, as I caught an error in my initial measurement!

[Ungrounded Lightning Rod]

The earth's axial tilt is about 23 1/2 degrees.  (23.44).

Add your latitude and that's how much your panels will be tipped up when facing the sun in the dead of winter, and the angle of the sunlight from vertical at the winter solstice as well.

Subtract off the angle of your roof.  Then a little trig will tell you what the ideal spacing would be assuming the sun is a point source.

Add a tad (say a half-inch) for the sun being a disk.  (Make a pinhole, hold it out under the sun, and measure the size of the disk and the distance to the pinhole.  That ratio will be constant and you want half the disk's diameter as extra spacing.  Which is probably trivial for reasonably sized panels.)

[Ungrounded Lightning Rod]

The earth's axial tilt is about 23 1/2 degrees...  Add your latitude...  Subtract off the angle of your roof.  Then a little trig will tell you what the ideal spacing would be assuming the sun is a point source.

Spacing of panels is length of panels times 1/cos(that angle) plus the sun-is-a-disk tweak.

[Paul h]

Hi,

   Thanks for your response. I don't quiet understand the "30" high panel and the 66" from hinge to hinge" part? I'm going to assume that the 30" is the height from the floor to the back of the raised panel and not the panel length? Is the 66" hinge to hinge measurement would also mean from the front of each panel in each row? this is how understand it.  Thanks Paul

[Paul h]

Hi,

  I'm just not that familiar with trig to figure out what that equation all means.I so get the pinhole example and the circle factor added to the measurement. Beyond that i'll have to do the figures to learn the formula that you stated. Thanks though...Paul

[Ungrounded Lightning Rod]

The three main trig functions have a mnemonic to make them easy to remember:

soh cah toa.

Sin(angle) = Opposite / Hypotenuse

Cosine(angle) = Adjacent / Hypotenuse

Tangent(angle) = Opposite / Adjacent

Where:

 - Hypotenuse is the long side of the right triangle.

 - Adjacent is the short side that's next to the angle you're doing trig functions on.

 - Opposite is the short side that's not attached to the angle you're doing trig functions on.

You don't use the others (secant, cosecant, cotangent) very often.  When you do they can be computed in terms of sine, cosine, and tangent.  (secant = 1/cosine)

[GaryGary]

Hi,

This gets a little more complicated because the sun drops in elevation before and after noon -- so using the noon sun elevation of 26.5 degrees on shortest day of the year at 40 lat will result in shading the 2nd row before and after noon.

The spacing equation out of "solar Energy Handbook", Kreider is:

D = sin (SunElev + ColTilt)*(Lc)/sin(SunElev)

Where:

D is distance between the front edge of the first row of collectors and the front edge of the next row.

ColTilt is the collector tilt angle from horizontal (18 deg in your case)

SunElev is the elevation angle of the when it is the lowest you want to design for (see below)

Lc is the length of your collector -- I'll assume 96 inches -- you can redo the calc if its more or less.

To figure out the sun elevation, get a SunChart free here:

http://solardat.uoregon.edu/cgi-bin/SunChart.cgi

for your location.  This is a great thing to have on hand -- it gives you the solar bearing and elevation for every minute of the year.

To calc the minimum SunElev I'll assume you want the collector to work without shading form 9am to 3pm on Dec 21 (shortest day of year and lowest sun).  For 40 deg lat, the sunchart gives a SunElev of just under 15 degs at 3pm on Dec 21.  

The solar elevation at noon on Dec 21 at 40 lat is 26.5 degs, so the sun drops from 26.5 elev at noon to 15 degs at 3pm and 9am.  

So, D is then:

D = sin(15 deg +18 deg )(96 inches)/sin(15 deg) = 200 inches

Remember this measured from where the front edge of the first row of collectors hits the roof to where the front edge of where the 2nd row of collectors hits the roof.

This scales directly with the lenght of your collector, so if your collector is really 48 inches long, D would be (48/96)(200) = 100 inches.

If this seems like too much spacing, you could just shorten the spacing up a bit and tolerate some shading on the 2nd and 3rd rows.

For example, if you were willing to tolerate some shading in mid winter -- say from about Nov 1 to the end of Feb, the 200 inches goes down to 155 inches.  

This may not be a big sacrifice, as the 18 deg tilt is not very optimal for winter anyway.

Hope that makes sense?

Gary

[clflyguy]

Yes, you understand correctly.

  I happened to have done the research before

for my 30 degree latitude and still had the info handy. I did all the stuff

(graphically)that ULR explained to you, took into account your 10 degree higher

latitude and gave you the answer based on two very common panel dimensions, since

you asked for a distance and not info on how to figure it out for yourself.

  Personally, I am not very good at math but I visualize well and have a terrific

CAD program that many times has allowed me to overcome being mathematically

challenged.