Solar Inter Row Spacing
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- čas přidán 17. 06. 2024
- 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.
- Jak na to + styl
![Intro to Solar Orientation [Solar Schoolhouse]](http://i.ytimg.com/vi/OR8EQ0DWpPw/mqdefault.jpg)
Best description of this I've seen yet. Thank you for the well put-together video!
Awesome, extremely helpful and very clear with how to do each calculation. Thanks!
What a great video Sr. thank you so much for sharing your knowleadge.
Subject matter well covered and thoroughly explained. Thank you !
I really like the way of you to describe that Inter Row Spacing of that array.
it's superb 👍👍
This was so exciting to watch, wonderfully explained, you have my deepest respect for making it so clear
very informative, looking forward to more of your videos.
Thank you King!! Excellent explanation, graphics, pacing, everything. Thanks so much.
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!
This is simply the best way to calculate Inter-Row Shading i've seen so efficient thanks a lot
Such a nice explanation, Thank you for the video Sir.
Amazing video, this one is a gem, so well explained!
You are excellent, this is a very understandable video, thank you for your efforts in teaching us! I will pray for you :)))
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.
Wow. This was an eye-opener. I always used the rule of 12 o'clock and the latitude of the location. Now I will be more precise.
that was quite useful . helped me understand the inter row spacing formula
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
It was wonderful to watch this video. As a renewable energy enthusiast, I really learned a lot. Thank you!
if you want renewable energy sources then use gravity as it's working anywhere in universe, so whole universe become full of energy forever
as we know that weight is only stuff that gives us opposite direction against gravity at universe level then we use this property in gravity defying uphill roller setup series which gives us electricity forever for everyone for millions of trillions of uncountable years for forever
keep talking about it because i am very poor too start it but every step i know about it
Amazing explanation. Thank you!
Wow! That was very useful! Thx
That's exactly what I have always thougth, long shadows before and after midday. Thats why I watch your video.
Very well explained. Thank you
Thanks for the great info. I am taking class in renewable energy ( solar) and this will help me
Thanks. Very clear, well presented.
excellent explanation--very helpful
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.
fantastic video thankyou. Very well explained.
Thank you very much! Making science simple!
Such a good process to calculation.
Awesome lesson, thank you
Yeah this is excellent and I have read a lot of descriptions of this before.
Thankyou! It was a great explanation
Thanks for sharing this useful video. And for the very correct first sentence about Engineers. :)
Just waoo... Amazed and thanks for this effort for us... Love from Pakistan
Excellent video sir. Thankyou.
Thank you so much. So very much helpful.
Nice and clear explanation.
Awesome ! Thanks for sharing
extremely helpful! thanks!
excellent way of explaining Thanks sir
You really know your shit.. Also you sell it very well!
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).
Thanks! Great stuff
This was super useful
This video is oddly satisfying
Appreciate it. Thank you
beautifully explained
thank you for sharing!
Excellent!!!
Great Job , thx
very useful!
do you have a south azimuth angle, I can only get east west azimuth, this would work like your azimuth
Thanks for sharing
Why do I multiply for the second calc instead of dividing?
thank you
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.
Put them in a row, instead of a column if possible.
Less space used and no problem of shadows.
Thank you very much sir
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?
Top Man.
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.
Here is a site where you can enter your own location data and it will create a chart for you: solardat.uoregon.edu/SunChartProgram.html
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.
V NICE SIR JI
Thank you sir ... very nice and informative video.
Please suggest what should be final distance 7.42 ft or 5.25 ft ?
7.42 of course.
Never realized how simple this solar thing was 🤪
How can get sun path chart for my place
that was cool...
well done
It just so happens that I'm at 30.71 degrees so this was perfect. lol
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?
Did you get the answer to this question? If so, please share.
i didnt type that.............or watch this video............
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
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.
its useful for my project (15 MW PV-PP)
so, you are no making the calculation for Kankakee,il, from that latitude should be somewhere around Houston,Tx
Jose Garcia ...No, not making a calculation FOR Kankakee. I'm FROM Kankakee. I was using the 30 degree latitude from the old NABCEP Study Guide.
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?
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
For method of only 1 triangle calculation; Azimuth angle must be 0 degree?
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.
I knew about this, cause the sun is changing angle between seasons,and in a day,and not much people think about it.
Grandpa, will you also discover hot water? 🤣🤣🤣🤣
Yes, Grasshopper! What will be discovered, and revealed, is how to heat water with a co-rotating acyclic dynamo!
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.
Could anybody pls explain to me why he substracts 90-45? What is a point and purpose?
Than q sir
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.
I need your help sir
Hi sir I am Asad iqbal from Islamabad Pakistan
Tiny panels?
Do you have, any thumb rule?
Could you please also share for how to select tilt angle
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.
tjwiltube the distance between solar arrays still more than 7.25 ft in your video please explain
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?
Why not just angle the next one 8 feet away?
Too many amperes for my cells to handle,
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
We found the lost arc.
im lost as to how the 9am mark was established
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 requires to be done for a worst case scenario- Winter Solstice! And if you miss shadow on that day, wait for another year ;)
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.
As a German, I find the hand gestures disturbing.
Thanks for the info!
What does being German have to do with the hand gestures?
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!
There's no 3 pm calculation?
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 Thank you, Sr.
@@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.
@@tjwiltube That explains :) But it's quite an interesting problem touching a real life scenario.
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.
clouds gather in my area all winter and spring. panels are pretty much useless.