martinwinlow
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- Feb 6, 2021
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Title says it all, really... My calcs are saying my ~2m diameter turbine at 13m/s will impose a 750kg lateral load on the mast! I conclude my calcs are adrift...
Thoughts?
Thoughts?
Is There A 'Rule-Of-Thumb' Relationship Between A Wind Turbine's Output And The Force It Imposes On Its Mast?
- Thread starter martinwinlow
- Start date
I've found it tricky to estimate the thrust load on a wind turbine, too. I did try working it out, but found a lot of variables that need to be understood, and they aren't easy to quantify. If you go too conservative on each of them, you end up with big numbers like the OP has.
No, a "flat plate" estimate doesn't work, even though it's a simple way to estimate, the result is way too high. Except - surprise - a wind turbine in a runaway condition applies huge thrust loads on the tower that are much HIGHER than the flat-plate estimate. Several counter-intuitive things like that.
To get myself out of this indecision feedback-loop, I found some load-test data where a WT was instrumented by the NREL and run for a long time in real-world conditions. The report included thrust load measurements with some correlation to wind speed. You may be able to find reports like these on the NREL website (it's been reorganized so I think you need to use Sandia's servers now).
[EDIT: Found it: https://research-hub.nrel.gov/en/publications/small-wind-research-turbine-final-report]
Below is a chart from report TP-500-38550 (October 2005) the test of a Bergey Excel-S. It's rated at 10kW at 13 m/s, having a 5.6m (18.4 feet) diameter rotor. You're looking at data with a lot of scatter, so I wouldn't just rely on the average through the middle, but for design of the tower account for the maximums. If your WT has a diameter of 2 meters, then I'd scale it down by swept area. Square of 2/5.6 is a factor of 0.127. If the Excel's peak is 5500 Newtons, then your thrust goes up to 700 Newtons. Maybe use 1400 N to have a safety factor of 2. I have nothing to tell me if that's enough to address a potential runaway, but it's better than nothing.
Don't use "kg" for forces - misuse of the unit of mass not force can easily confuse.
No, a "flat plate" estimate doesn't work, even though it's a simple way to estimate, the result is way too high. Except - surprise - a wind turbine in a runaway condition applies huge thrust loads on the tower that are much HIGHER than the flat-plate estimate. Several counter-intuitive things like that.
To get myself out of this indecision feedback-loop, I found some load-test data where a WT was instrumented by the NREL and run for a long time in real-world conditions. The report included thrust load measurements with some correlation to wind speed. You may be able to find reports like these on the NREL website (it's been reorganized so I think you need to use Sandia's servers now).
[EDIT: Found it: https://research-hub.nrel.gov/en/publications/small-wind-research-turbine-final-report]
Below is a chart from report TP-500-38550 (October 2005) the test of a Bergey Excel-S. It's rated at 10kW at 13 m/s, having a 5.6m (18.4 feet) diameter rotor. You're looking at data with a lot of scatter, so I wouldn't just rely on the average through the middle, but for design of the tower account for the maximums. If your WT has a diameter of 2 meters, then I'd scale it down by swept area. Square of 2/5.6 is a factor of 0.127. If the Excel's peak is 5500 Newtons, then your thrust goes up to 700 Newtons. Maybe use 1400 N to have a safety factor of 2. I have nothing to tell me if that's enough to address a potential runaway, but it's better than nothing.
Don't use "kg" for forces - misuse of the unit of mass not force can easily confuse.
Solar Guppy
Red Cobra Guppy
@Solar Guppy
Do you go fishing yourself or do you buy fish at the supermarket?
Do you have disrespect for people who go fishing for themselves?
People who go fishing run the risk of coming home with nothing. Therefore nobody should go fishing. Right?
Do you go fishing yourself or do you buy fish at the supermarket?
Do you have disrespect for people who go fishing for themselves?
People who go fishing run the risk of coming home with nothing. Therefore nobody should go fishing. Right?
Exactly what my neighbours said about PV 20 years ago.
OzSolar
Whatever you did, that's what you planned.
Keep in mind that the force of wind is a function of the square of the difference in wind speed. I think I've got this right: 2X's the wind speed = 4X's the force.
In other words things go from almost, nothing to big. What's particularly tragic is that even though that most sites spend less than 1% of the year at wind speeds over 20 mph you've still to engineer around the design wind speed for your area which is usually 85 mph or higher.
Some wind turbines have furling systems, some don't. Those that do aren't reliable enough to safely count on 100% of the time so the only safe answer to engineer the tower on a broken furling system.
So yes, there is a relationship, the higher the output the higher the forces.
In other words things go from almost, nothing to big. What's particularly tragic is that even though that most sites spend less than 1% of the year at wind speeds over 20 mph you've still to engineer around the design wind speed for your area which is usually 85 mph or higher.
Some wind turbines have furling systems, some don't. Those that do aren't reliable enough to safely count on 100% of the time so the only safe answer to engineer the tower on a broken furling system.
So yes, there is a relationship, the higher the output the higher the forces.
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Use the maximum frontal area of the stem, hub, and blades, which is probably found looking directly at the blades. Assume the blades are stalled. Add whatever margin you want for safety (double is a good starting point).
I think the force varies due to gusts. If the blades are spinning slowly and a gust hits, the lateral force increases until the blades spin faster. The gust stops, so now the blades pull themselves into the wind (causing lower lateral force) until they slow down. I don't see how the force could ever exceed the value created by stalled blades pointing into the wind.
Warpspeed
Solar Wizard
For wind (or hydro) you need to be lucky enough to have have a suitable location.
If you don't have that, then its not going to work very well.
Simple as that.
brandnewb
Going for serious. starting as newb
we just need to give space for different scenarios. Different use cases. This blanket statement is not well thought off in my opinion.
Now if you were referring to all the micro wind turbine hype that got many of us into a pickle then yes I agree. But then at least make that distinction.,