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Solar Energy Collection
Mike T., Swampeast MO
Member Posts: 6,928
Need assistance please.
Have local, government hourly solar energy data in watts per square meter at zero degrees of tilt.
Need to convert to various degrees of collector tilt. Is this possible?
Yes, realize it will take the latitude, sun angle (maybe even hourly sun angle). Have the equations for finding the sun angle at any time on any day, but don't know how or if it's possible to convert for degree of tilt of a solar radiation collector.

Call me crazy, but I want to "imagineer" a solar system integrated with a Vitodens. Have the data logs for the boiler and have very accurate estimates of boiler output in certain conditions. With some reasonably accurate estimates of solar <I>system</I> efficiency I should be able plot the two together and determine just what can be done.
Thanks for any help or references!
Have local, government hourly solar energy data in watts per square meter at zero degrees of tilt.
Need to convert to various degrees of collector tilt. Is this possible?
Yes, realize it will take the latitude, sun angle (maybe even hourly sun angle). Have the equations for finding the sun angle at any time on any day, but don't know how or if it's possible to convert for degree of tilt of a solar radiation collector.

Call me crazy, but I want to "imagineer" a solar system integrated with a Vitodens. Have the data logs for the boiler and have very accurate estimates of boiler output in certain conditions. With some reasonably accurate estimates of solar <I>system</I> efficiency I should be able plot the two together and determine just what can be done.
Thanks for any help or references!
0
Comments

You can do this
With a big of trigonometry. Oh boy!
Assume the "base radiation" refers to the radiation impinging on the solar collector when the collector is directly facing the sun. The adjusted radiation is the effective radiation at an xdegree tilt.
Adjusted Radiation = Base Radiation * cos(x)
cos(x) being the cosine of x.
Michael0 
solar data
Try this link for maps with differing tilts, tracking, etc. http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/atlas/
For software to do this:
http://www.fchart.com/
Here too:
http://www.retscreen.net/ang/g_solarw.php
Also ran across this site, but watch the math!:
http://www.courses.ait.ac.th/ED06.22/course1/lecs/module2/m23o98.html
THE books for solar thermal:
J. A. Duffie and W.A. Beckman, Solar Engineering of Thermal Processes, John Wiley and sons, 1980
P.J. Lunde, Solar Thermal Engineering Space Heating and Hot Water Systems, John Wiley, 1980.
good luck!0 
Trig is Fun
So, I first find the sun angle at a given hour on a given day. Then make correction for the difference between a flat collector (info I have) as compared to an ideal collector that directly faces the sun using:
Adjusted Radiation = x * cos(0) where "x" = the base radiation.
Then recompute for the new collector angle using the formula you provided.
Does that sound right?0 
Indeed
Yes, so long as the "cos(0)" there refers to the cosine of the actual angle in degrees.
Keep in mind (your data may already take this into account) that you can only use the simple formula when talking about the variability in the tilt of the solar panel with respect to the sun, not of the sun with respect to the solar panel. This sounds redundant, I know.
That is, as the sun moves during the day, you will have to account for two things:
1) Adjustment for the orientation of the collector versus the sun (as per the formula above), PLUS
2) Adjustment for the quantity of solar energy under ideal conditions. This is because as the angle of the sun changes, the amount of loss of energy will probably vary with respect to the amount of the atmosphere it has to penetrate to reach your home.
I hope that makes sense.
Michael0 
more variables
Plus the reflective coefficient of the cover glass, and the absorber plate, etc. And of course the weather, usually determined by historical data. It's a bit more complex than simple trig I'm afraid.
Brad0 
Thanks Michael!
Think I got it.
On February 15, the sun's declination is:
23.45° * sin(((46 + 284)/365) * 360°)
= 13.3°
So, the tilt of a flat collector with respect to the sun is:
latitute  declination
37.3°  (13.3°) = 50.6°

Say in some given hour on that day, measured solar radiation availability was 372 watts/sq.meter parallel to the ground.
Were the collector tilted at 50.6°:
Adjusted Radiation = Base Radiation * cos(x)
372 = x * cos(50.6°)
x = 372 /.634
Base Radition = 587

Now say the collector is actually tilted 37° (latitude):
50.6° (ideal)  37° (actual) = 13.6°
Adjusted Radiation = Base Radiation * cos(x)
x = 587 * cos(13.6°)
x = 570

Time of day doesn't matter because I'm not trying to follow the sun as it crosses the sky and the data available came from flatsurface measurement, right?
Now, the only thing left is to find the ideal collector placement where it has very little shading from OctMar, but somewhat more the rest of the year.0 
True
I think one has to hope that the government data takes everything into account except the details of the plate itself and how you tilt it.
Which means I'd have to retract my statement about caring of the sun with respect to the collector  the government data should take care of that.
Michael0 
looks pretty good
From a theoretical standpoint.
A few things:
Part I: You can't use a daysofyear approach to calculating the declination of the sun because it's too nonlinear. Your best bet it to consult charts of declination. Heck, you can look them up, though that makes it difficult to consult an excel spreadsheet. You might want to look up hardcoded altitudes with respect to your particular latitude, which I take it is 37.3.
For example, I believe that for your particular latitude on February 15, the angle of the sun with respect to the ground was closer to 50.1 degree ballpark.
Part II: The thing I'd worry about here (I don't know if it's a legit worry, I haven't thought about it enough) is that this will generate slightly varying results throughout the day/year depending on the azimuth of the sun. You may have thought about this already, but I don't really have the time to think too much since, well, I'm actually at work and am supposed to be on the phone right now...
Here's an example. If you tilt the collector directly at the sun (50.6 degrees in your example) at noon, your radiation increases by a factor of 1.57. But at 3pm, is this necessarily the case? I don't know, but I suspect the answer is no. The factor may be greater or smaller. Secondarily, do we care? It may be too small to matter. I don't have the answers to these right now.
Part III: Same as above.
===========
Time of day doesn't matter in one respect because you've got wholeday data (I'm assuming) from the government sources.
However, again, it might matter because tilting the panel upwards might result in different angles over time depending on the actual movement of the sun. The analysis would more likely be spot on if it were, say, noon for 12 hours followed by midnight for 12 hours so you didn't worry about the sun's path  this is because the adjustments are simplified noontime calculations.
Michael0 
two angles from flat
Mike,
With hourly data, you need to compute two angles in comparing the flat panel to your possible panel position. There is the elevation angle, which it looks like you have, and there is the orientation angle. As you tilt your panel up, you also have to look at the horizontal orientation angle between the plate and the sun. The solar track is computed in terms of elevation and declination.
So you get the W/m^2 by dividing the cos(azimuth) into the flat measurement, just like you said. Once you pick an angle of tilt and orientation, then it looks like this:
collected power = base power * cos(azimuth  tilt) * cos(declination  orientation) * efficiency
Efficeincy is a complex function of outside temperature, collector inlet temp and flow. This also assumes you have a flat plate collector. Evacuated tubes have a slightly more complex function for efficiency because the tubes bend light to neighbors when there is an angle between delcination and orientation.
lots of fun
jerry
0 
Here's a sample of the data:
Date, Time, Radiation (watts sq/meter)
2152005, 0600,0
2152005, 0700, 1
2152005, 0800, 33
2152005, 0900, 102
2152005, 1000, 197
2152005, 1100, 472
2152005, 1200, 555
2152005, 1300, 566
2152005, 1400, 525
2152005, 1500, 435
2152005, 1600, 305
2152005, 1700, 157
2152005, 1800, 16
2152005, 1900, 0
All days are roughly similar, growing towards noon then diminishing afterwards. Energy available on flat ground, right?
Assuming a flatplate collector facing due south, believe I'm compensating for the tilt of the collector to the South as compared to flat ground.
Do I still have to compensate for EW movement of the sun when the panel won't "track" from east to west? It [appears] that such is already reflected in the data.
p.s. Realize I haven't even BEGUN to calculate that actual energy I can collect and realize that efficiency will change with outdoor temp, collector temp, HX efficiency, etc., etc.0 
jerry's right
if you're using data on an hourly basis rather than on a daily average basis, you'll have to correct for the azimuth of the sun as well.
You don't need to correct for the azimuth of the sun for a flat collector because it's a twodimensional problem. It's also less important to correct for the azimuth if you have daily data because the pluses and minuses I believe average out. But if your panel isn't flat and it's not tracking the sun, you'll need the additional correction factor, which should amount to another cosine function.
Michael0 
Terribly Confused
You don't need to correct for the azimuth of the sun for a flat collector because it's a twodimensional problem.
Am concerned with a flat collector facing due south. Can vary the tilt of the panel as related to the declination of the sun. Isn't that a twodimensional problem?0 
Terribly Confused
You don't need to correct for the azimuth of the sun for a flat collector because it's a twodimensional problem.
Am concerned with a flat collector facing due south. Can vary the tilt of the panel as related to the declination of the sun. Isn't that a twodimensional problem?
It's also less important to correct for the azimuth if you have daily data because the pluses and minuses I believe average out.
Hasn't the reduced value of the data when it's not noon already compensated for the sun crossing the sky?0 
I thought I'd restart the thread on the left
Since the messages were getting sort of long in the vertical dimension.
I thought about it a bit yesterday and decided you can't even just do two cosines. Unfortunately, it's fairly complicated. I'm sure there's software out there that's already done to do it, but I don't know of it. I'll put something together later.
An interesting note about the solar data you collected  was this a cloudy morning? The solar energy in the afternoon (say, 3pm) was much higher than 9am, despite the sun being in the exact same place, just mirrorimaged.
Anyway, this is a good exercise in remembering my dot products and cross products and so forth, which I enjoy, so I'll try to make something that'll do the calculations soon.
Michael
0 
why is htis not a dual cosine problem
We are looking at the two axis projected area problem. The difference between the spherical projection and planar is fairly small as 93M mile solar radius.
So we use the planar projection of the horizontal surface measured normal to the solar radiation to compute the radiation density. Then we need to project the selected panel placement onto that same normal plane. I've done this before for graphics shading, and I'm fairly sure it's a separable 2 cosine problem. It really is just a trig problem, nothing more. There are times when running the points through a transform matrix is necessary, but this case is simple enough to avoid that.
One has to decide what to choose as the center of the time window for the measurement to do the projection computation. One also has to realize that this is a computation of what would have been collected rather than what will be. (You know that disclaimer they add to mutual fund advertisements.)
enjoy, jerry
0 
it's been a while
It's been a while since I've had to do any math since I don't do it for a living and it's been, well, longer than I can remember since I've had to do any for school.
Anyway, again, I just thought a little bit about it, but an offhand calculation seemed to indicate that the dual cosines didn't work.
Let's say you take a panel facing the suns rays and rotate it 45 degrees in two orthogonal directions (altitude and azimuth). cos(45)*cos(45) = 0.5 correction factor.
However, rotating 45 degrees in two orthogonal directions I believe only generates an angle of around 54.7 degrees between the sun's rays and the normal vector to the panel. cos(54.7) = 0.58 correction fator.
Haven't checked my math yet, so I could be wrong.
Michael
0 
Thanks Michael and Jerry
Have been busy with other work last couple of days so haven't had time to play with the geometry. Probably this weekend.
0 
Can't be 100% sure about the weather that day, but it appears to have been very sunny. Air temp measured on the South stayed above 80° for a few hours. Boiler went into shutdown late afternoon through mid evening.0 
\"What Would Have Been\"
is EXACTLY what I want to determine with some reasonable degree of accuracy. Solar won't be happening this year or likely even next. Have to save the $$ and want to get at least another year of system datalogs. Will have the Vitodens set up for outdoor temp diversity next year. Am currently switching to reduced mode (quite reduced by the way) whenever outdoor temp is above 40°. Have had a few days of cloudy weather with temps in the 40s and despite no real problem with space temp maintenance, that "Eurocave" feeling that ME mentioned, is coming back...0 
well, here's a prototype
No guarantees. I threw some stuff together to approximate what you would get with your panel in different locations.
The interpretation I've taken for the updown and left/right angles is important for proper usage of the spreadsheet. The updown tilts about a horizontal axis. However, the "east/west" left/right angle is not rotating towards the east and towards the west, but rather clockwise and counterclockwise, that is, as if spinning on a vertical stick.
The only inputtable values are in bold and green highlighted. I didn't bother making the spreadsheet userfriendly because, well, I dunno. I'm not friendly, I guess.
Michael0 
Thanks Michael. Will let you know how it works out. Seems pretty friendly to me...0
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