This is a short, instructional video aimed at NABCEP PV Entry Level students and techs. It covers the NABCEP competency for calculating the inter-row spacing of solar-PV sub-arrays configured in a saw-tooth arrangement on a flat roof.
Einstein's observation 'If you can't explain it simply, you don't understand it well enough.' plainly applies here. Your explanation was so clear and easy to follow you obviously know this inside out. Many thanks for posting.
Aha! Now I have a more reasoned understanding of an installer’s reluctance to fit more than one row of panels on my flat garage roof with its 12’ x 24’ dimensions, with its longer dimension running east-west. Thank you for the wonderful presentation!
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I love this and I confidently say some of the best generating solar panels DIY installations I have ever seen have just been thrown together, LOL with zero calculations
You are a great educator. One thing I have never seen included in all these theoretical calculations is the effect of local weather. If the array is in an area where the morning in winter is frequently cloudy, there is a reduction in the possible output of the array. By turning the array more westerly an improvement in net daily output can be obtained. This is the situation in Adelaide South Australia lat. 35 south where a winters day frequently turns sunny in the afternoon. Having panels orientated north is a laboratory generated concept which does not include a real world situation. Thank you for your video.
Very appreciate the teacher trying to explain, still at his calculation, i would switch to Summer solistice sun set altitude angle instead of Winter solistice because summer is the time day is longest then we have more kW of peak-sun-hours. Starting point of calculation : top-right-hand-corner of solar array faces South (if Northern Hemisphere).
D=3/tan 22° (.136) D= tan/3 22° (7.33) If I have an object like a shed that I wanted to find the appropriate place for my array that is 14' and my solar elevation is 15° @ 9 am would my distance be 1.07'?
Great solar orientation vid without an aide of an app.. I NEED a clarification though. The 'fudging' solution is - randomly pick any time of day, at AZIMUTH 0 and using the location's relative solar elevation on the chosen time, and using the location's latitude. Given these parameters in the Trig calculation I will get the same answer? Any help?
With the angel of the sun at 45 degrees to the array at 9am and at 3pm the shadow cast by the panels would not hit the panels behind assuming that all sets of panels are the same width and if the panels are set back far enough for the 12 noon shadow.
I learned the other day that on cloudy days the optimal angle for a solar panel is pointing straight up at the sky. In areas where there are a significant number of cloudy days this complicates computing the optimal angle and spacing for solar panels.
Well-ll-ll...sorta, kinda, maybe. Here in the Midwest, in the winter, on a cloudy day, it's not uncommon for the zenith (straight up) to be the brightest spot in the sky. That said, the irradiance will be pathetic! Unless you're at the equator, you should never orient your modules straight up. Not only will this give you horrible performance over an extended period of time, your array will collect snow and dirt and stop producing altogether. Dual-axis trackers will occasionally tilt flat toward a bright spot in the clouds; but, not long after, they will tilt toward the actual position of the sun and dump any snow they've collected.
At latitudes much closer to the equator (more irradiance) , would you still base the row spacings on 9am to 3pm or would you go for say 7am to 5pm? Is there a rule of thumb for the useful length of day that takes latitude into consideration?
The two calculations need to be repeated anyways for a different latitude. I couldn't quite get the perspective with which the video is concluded i.e., doing one calculation instead of two - simplify to one calculation for an array placement that's repetitive?
Seems like you could just physically site the solar elevation angle (22 degrees in this case) from the top of the first solar panel to the ground behind. Position the second set of panels behind where the angle hits the ground. This method would eliminate the need for calculations and also correct for any ground slope. If you have adjustable panels, you would set the tilt to what it should be on Dec 21 when you do this. Any reason this won't work?
Yes, one can always space the racks far enough back, just based on local experience; BUT, the purpose of this video was to help NABCEP- Certification candidates to prepare for their exams.
No. The correct method is to use the two, individual trig calcs, as explained. However, trial-and-error, it is possible to find an azimuth, at your latitude, that will approximate a correct answer with only a single calc.
Excellent video that gives clarity on a lot of things in a simplistic way. I am looking for installing panels in similar pattern and looking for a sun path chart for 12.75 degrees. Trying different places on google, haven't found a decent one yet. Please share @tjwiltube a reference.
Very nicely explained, but confusion is, as per your assumptions, I feel, what you said is, if tree is 10 feet high or 20ft high, the shade at 45degree azimuth will remain at 5.25feet. Someone, please clarify this as you have nowhere taken at exact height of solar panels from ground in consideration.
In Natstat study guide problem, they take perpendicular on first triangle as 3 ft, if that perpendicular is what we have to measure in real scenario then that solves the confusion.
A simple heck is measure height of panel from back between ground and top and multiply it with factor 1.8. and then install your new string behind it. Let suppose your panel height is 3ft from back then (3*1.8=5.4). Now keep space between two strings approximately 5.4. thanks
What wrong with not having each individual unit where it is supposed to be? Angular aesthetic s? Variability would be easier on the land lost to development.
Hmm-m-m...this is a bit more involved than it may seem on the surface. First, the foundation principle is that maximum "density" of solar radiation strikes the surface of the PV module when the surface of the module is fully perpendicular to the direction of the sun's rays. Also, at noon, the solar radiation is traveling through less of the atmosphere than at any other time of the day, ignoring any arbitrary clouds. Thus, for a fixed PV array, setting the tilt angle equal of the location's latitude, and setting the array's azimuth to true south, will yield the highest, annual energy harvest. That said, this may not be the most ideal tilt angle. Example 1: Envision a flat-roofed commercial building with a large and active human population (each human continuously radiating about 100W of heat energy). The primary electrical load for this building is likely to be its air-conditioning load. Thus, the array is likely to be tilted way back, toward a flatter angle, to optimize the summer energy harvest, when the air-conditioning load will be at its maximum. Example 2: Envision an off-grid home with its only electric power source being a stand-alone, battery-based, PV system. Because there are more daily hours of sunlight in the summer than the winter, you would likely bias the array tilt toward a steeper angle in order to optimize the winter harvest, when daylight hours are at a minimum. This would lead to a less efficient summer harvest, but this would be compensated by the greater number of summer daylight hours. I hope that answers your question.
Hi, Jasmine. The physical system in the video is just being used as an example of a sawtooth array on a flat roof, and to provide a positional reference for the triangles. The numbers used in the calculation come from a sample problem and solution in an older version of the NABCEP study guide. The example calculation in the study guide was incomplete and created some confusion. The intention of the video is to show the correct calculations for the NABCEP sample data: 3-foot vertical height of the sub-arrays and sitting at 30-degrees N latitude. Does that make sense?
great there is a guy with some brain ..... I had that problem in my head and could not sleep ..... tonight I will sleep good ..... I have saved this video link, next time this issue comes up I will run that video again
if their is no sun, air, water, heat, then what is the power source, I tell you that it's gravity, gravity, gravity, gravity, gravity, gravity, gravity, gravity, gravity, gravity, gravity, gravity, gravity gravity gives great help so we used gravity defying uphill roller series setup from which we got the power
But you're calculating the sun's movement across the panel and the corresponding shadowing based on the 9 am elevation of the sun on 12/21. To be accurate, as the sun moves towards solar south, you need to recalculate the angle of the sun's striking the panel and the corresponding shadow cast at each progressive hour because the sun is getting higher in the sky. If the 9 am shadow misses the panel, the next step is to measure the 10 am shadow to see if it comes to bear on the next panel and so on. You can mark the length of each hour's shadow with markers or chalk. The final set of marks will look like an inverted arc on the ground, then simply arrange the next panels so they fall just outside of the shadow arc. An even faster way is to take a 9 am measurement, a noon measurement, and draw a line between them. This is less accurate but will give the path the shadow will travel across the ground.
This does not apply if you add infinite width to the next array. If you plan a solar park you'll have let's say 10x 100meter rows. Close enough to infinity from your perspective. Your calculation would only be necessary if for hat ever reason you wanted a sawtooth arrangement of only one* module for what ever length. *one meaning what ever width you choose, but negligent in the real world.
You don't need a 3:00 p.m. calculation. 9:00 a.m. is 3 hours east of noon. 3:00 p.m. is 3 hours west of noon. The sun's position at 9:00 a.m. and 3:00 p.m. will have nearly identical elevations and nearly identical azimuth angles away from the noon/south. Thus, the calculations will yield nearly identical results.
@@tjwiltube You're right... at 9:00 AM in winter you have the longest projected shadow of the year, so there's no need of 3 PM calculation, since the latter is shorter than the former, right?
Interesting, but not real world from what I have seen. Seems that it matters more when you add in price per sq ft of a lot or building roof vs what you can pull in per panel at a particular angle and what gives you more energy, more panels at less angle vs fewer at a greater angle. I have seen giant arrays that were almost flat. They must be able to get more from their roof or field with more panels vs having the correct angle and now add in and mono vs poly. It seems a website calculator similar to something like the Ikea kitchen cabinet 3d program with options to change parameters for available sq ft, size of the panel, # of panels, angle of the sun, mono vs poly, would be better to give you effeciency #s of different angles with more or less panels and which is better where space is a premium and thus you make the most power/$ per sq ft. Also use pvwatts.nrel.gov/pvwatts.php
It is real world within the context of the NABCEP exam. This video was made as an aid to passing the NABCEP exam relevant at the time the video was made.
BUT ... given that the sun is so much weaker in the winter months due to atmospheric absorption and often the sky is cloudy .... it maybe worthwhile putting in more panels, allowing some shading in winter in order to maximise production in summer. Do the math for that!
9am to 3pm sun times. divide in half is noon. look at the chart for noon june 21 and dec 21 average the difference and put up your panels. on the other hand; if you point one set of panels at 10:30am and another set of panels at 1:30pm; then you will have the sun almost all the day. i'm not an engineer obviously.
