Wind Load Calculation on Photovoltaic Panel

Great answer ba/ael. Many thanks from a poor Mechanical Engineer.

Do you know how does the factor "0.5" in your equation for q' vary with incidence angle?

According to NBC, it varies as shown in the attached sketch, taken from the User's Guide to NBC 2005. These are based on wind tunnel tests.

This figure is not applicable to the solar panels. It is meant for external forces on a small enclosed building with sloping roof.

The solar panels are plates supported on an open frame with wind flowing around all edges. NBC figures B23, B25 and B27 are more appropriate.

The Australian Building code treats open mono-slope roofs the best. I don't have a copy on hand at the moment. If someone else does they could post the appropriate excerpt.

Kdelta, does that mean the psf figures need to be higher or lower?

Spacecannon

The pressures would be higher than those calculated using Figure B7.

The force normal to the surface would be about 2.0 x q instead of the 1.3 factor suggested by others here.

Thank you very much, your a Scholar and a Gentleman.

'nuff said.. ga.

passingtongreen,

I think you are correct. There should be a suction on the leeward side. I'm not too sure what the value of that suction should be. Looking at the diagram, it could be as high as 0.7*q.

When in doubt, be conservative.

There are several rows of these, one behind the other at a distance of about 5m. Does suction still apply on all rows?

It might make a difference, a slightly diagonal wind might be intensified by funneling. I would just be conservative in the design. I think your worst case is with the wind in the opposite direction to the one you show. I would put a 0.8 pressure on the back and 0.5 suction on the front, making a total of 1.3. I would then, conservatively assume the shear and the overturning moment can reverse with wind reversal. Normally, you are permitted use either a lower load factor or a higher allowable stress for wind loads, depending on the design method, you could ignore this if it would make you more comfortable.

I must add that you have to evaluate the situation for yourself, my offering here is generic and could change if I was to become aware of all of the specifics of the problem.

The page is too fuzzy. Can you attach as pdf?

I noted that it was fuzzy too. I cannot attach as pdf because the CR4 editor will not accept that format. Sorry. I'll try a JPG.

That doesn't seem to be any better...sorry. My PDF is quite clear, but when I convert it to GIF or JPG, it gets fuzzy. Don't know how to solve the problem.

Dedalus,

If you contact me through CR4 and give me your email address, I will send you a pdf. Please do not post your email address on the forum.

Brilliant!

How can we find wind load blowing across (parallel to the road) a street light pole say 9 M long. Though the same can be calculated using the given formula, but I am not sure about the factor 0.5. What will be the factor in this case? Please help.

BB Raina

Poles are a special case, depending on their Reynolds Number, Natural Frequency and wind velocity, they may be subject to "Vortex Shedding". This is a sophisticated calculation and should be performed by a qualified Engineer

Wiki on vortex shedding

I tried again, but still no good.

Ba/el,

I am building a solar farm too and I was wondering if you could help me with wind loads also, Im a medical professional so the math is beyond me. I need to know what the wind loads are for a panel with a 20 degree tilt, front edge is 8 inches off the ground, back edge is about 43 inches off the ground. The panel size is 43.5" wide and 102.36" high. I tried to get this question answered in a previous post but you are the only one I have found that knows the answer. If you want a picture I can email it to you. The max wind around here for calc purposes is 95mph.

Please

Spacecannon

Just WOW, dudes, GAs, let me interpret this, its close to the answer I paid a local engineer a few thousand to calculate, but like in medicine I like to get a second opinion.

Thx didnt mean to co-opt the discussion,

Spacecannon

So let me get this right.

The overall drag coefficient I should use when the wind comes from the front is 0.7 (note #6) and when it comes from the rear is 1.3 (notes #4 & #13)?

Very comprehensive. I picked that up like a very hungry person. The OP should be very happy with that as well. GA, Ky.

As a Structural Engineer, my answers have been based on my knowledge of code requirements, however, if you think of the airflow over a moving vehicle, and the "vacuum" behind them that allows "drafting", you get an idea of what actually takes place.

The grand cycling tour of Spain, the Vuelta a España is taking place now, and that reminds me that the riders behind the leader, do only seven tenths as much work as the leader at their average flat speed, illustrating the leeward reduction of pressure.

Thanks passingtongreen

The second link did not open for me. Any suggestions? Not having much luck with links lately or Murphy is doing overtime, Ky.

I emailed it to you.

I found a way. Put it in bookmarks first and copied from there. Thanks for the offer Chris, I appreciate it.

The last Neanderthal, Ky.

I guarantee you that you are not the last neandertal... I know people who run and the sound of the word 'computer'...

You are actually high on the bell curve my friend.

Chris

Ky!, I THOUGHT I WAS THE LAST NEANDERTHAL.

Spacecannon aka Techosaur

Excellent references. The second one provides a huge amount of specific detail for a variety of roof types.

Thx Passingtongreen, good links, GA, Now to figure out what it means, you engineers use such big words and numbers too.

Spacecannon

Dear Mr. Dedalus

To day only I saw your posting on “Wind Load Calculation on Photovoltaic Panel” in the link https://cr4.globalspec.com/thread/59831/Wind-Load-Calculation-on-Photovoltaic-Panel

You have not given the area of the panel which is facing the force due to wind speed.

A formula is available to calculate the force due to high wind velocity and it is

(in psi) = (0.00034) x V^2 where V is the Velocity of air in Ft./Sec. ( I studied Engg. In FPS system)

You have given wind speed as 190 KM/Hr., which is equal to 118.75 Miles per hour and again equal to 174.16 Ft/Sec. ( simple conversion to Milesperhour and then Feet/Sec.)

Hence the pressure due to the wind speed of 190 KM/Hr., which is equal to 174.16 Ft/Sec. works out to (.00034) x ((174.16)^2) = 10.314 psi (or) 0.7262 Kg/cm^2 or 0.6875 bar. Etc,

You know the surface area of the panel ( which you have not posted in your quiery) and hence the load will be Area X Pressure and can be written as (Area in Sq.Inches) x (10.314 psi) = (Load in Pounds) (or) (Area in cm^2) x (0.762) = (Load in Kgs) if the pane is 90 Deg. to the wind direction and let the be W in Lbs. or Kgs.

You have shown the angle of the solar panel is 30 Deg. to the Horizontal and hence the force or Load to be resolved horizontally and vertically which is as simple as W X Sin30 and W Sin(90-30)

I think the above calclculation is clear to you.

DHAYANANDHAN.S

Mr. Dhayanandhan can you help out here?

I need to design a mechanical support for a solar panel and the results I'm having are ridiculously conservatives so I'm thinking I've done it wrong.

A have a solar panel which is 0,75x0,51m and it should stand against a 108km/h wind. It will be at an angle of 21º, but for a worst case scenario, I could design it for 90º.

The mechanical support is design with a set of 6 bolts, nuts and washers. The diameter for the washers is 15mm for external diameter and 5,3 for internal diameter.

So, if I've done it right, the Wind Load will be 1,2754 (air density) x 30 m/s x 1,5 = 573,93 N/m2. So, the force on the panel for a worst case scenario will be 573,93 N/m2 x 0,3825 m2 = 219,5 N

The area of a large M5 washers is 154 mm2, so the total area for 6 of them will be 928 mm2.

This results on a stress on the frame of 219 N/927 mm2 = 0,236 MPa.

I think this is stress is too low to be true, can anyone help me with it?

With best regards

Here is a study on the wind loads on solar panels...

https://www.osti.gov/servlets/purl/5350425/

https://www.wikihow.com/Calculate-Wind-Load

Thank you, I did those calculations.
But I'm really not sure if the stress on the washers and bolts are so low.

I would expect it to be at least 20Mpa, but I've calculated only 0,2 Mpa for the worst case scenario.

I think if it was so low, we would never have problems with wind on solar panels, so I don't trust my calculations.

Can anyone give me any advice if I did anything wrong?

.2 MPa is about 30 lb/force per square inch...I guess it depends on how many bolts you have and the size of the washers...

https://www.solar-frontier.com/eng/solutions/products/pdf/module_drawing.pdf

I'll use six 15mm diameter washers. But if the force is only 219N, a 2mm aluminum washer would do the job.
That's why I question the force calculation I did here. I've checked it quite a lot and I still think it is too low.

It seems to me the bolt load should equal the max load for the panel itself...so somewhere between 5000 and 6000 Pa ...

https://www.solarpowerworldonline.com/2016/07/mechanical-loading-tests-solar-panels/

..."ML tests have long been hailed as the de-facto tests for evaluating the mechanical strength of solar modules, especially with IEC 61215 having included the 5,400 Pa requirement for passing the standard. An ML test mounts a solar module flat on a standard mounting system, with 5,400 Pa of weight force placed on top to put stress on the solar module, shown in Figure 1. For a standard 60-cell module, this is equivalent to 916 kg of load on top of a single module. Pictures and power measurements are performed on the module before and after the ML test with the passing criteria of power degradation <5%.

The ML test is easy to set up, and it gives a good indication of general module strength; however, it is frequently mistaken for a test that estimates the reliability against settled snow on a pitched rooftop installation."...

6000Pa would be about .87 lbs of force per square inch....standard module about 60" x 40" = about 2400 sq inches = about 2000 lbs total load limit....4 bolts 500 lbs each...

http://www.derose.net/steve/resources/engtables/bolts.html