by www.reasonablepower.com
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Generally we start with a question, such as:
What is the wind load on a flat plate, which is stationary in the wind, which is round, which has a diameter of 14 ft., and which has a 50 mile per hour wind hitting it?
ANSWER:
Area = A = (pi)(7 ft)(7 ft) = 153.938040026 square feet
or in SI units (1 square foot = 0.09290304 square meter); Area = 14.301311890 square meter
Wind Speed = v = 50 mph
or in SI units (1 mile/hour = 0.44704 meter/second); Wind Speed = 22.352 meter/second
Using the formula:
where
Fd is the force of drag (or in this case Force Against the flat plate),
p is the density of the air,
v is the speed of the air against the object,
A is the area of the object which the air is blowing against,
Cd is the drag coefficient,
The density of air at sea level is about 1.2 kg/m^3 (1.2 kilo grams per cubic meter) .
p = 1.2 kilogram/cubic meter (for this example)
The drag coefficient for a flat plate in open air is about 1.28
Cd (which is a dimensionless constant)= 1.28 (for this example)
Thus:
Fd = (1/2)(1.2 kilograms/cubic meter)(22.352 meter/second)(22.352 meter/second)(14.301311890 square meter)(1.28)
Fd = 5,487.44114 kilogram meter / second^2 = 5,487.44114 Newton
or if we go back to previous units (1 newton = 7.233013851 pound foot/square second);
Fd = 39,690.7378 pound foot/second^2
or
F = 39,690 pounds
If (for example) we are working with a wind powered generator which has 4' long blades, and each blade starts at 3' from the center, we want to subtract the Fd of a 6' diameter flat plate from the Fd of the 14' flat plate.
First we substitute the areas:
Area = A = (pi)(3 ft)(3 ft) = 28.274333882 square feet
or in SI units (1 square foot = 0.09290304 square meter); Area = 2.626771572 square meters
Fd = (1/2)(1.2 kilograms/cubic meter)(22.352 meter/second)(22.352 meter/second)(2.6267 square meter)(1.28)
Fd = 1,007.8973 kilogram meter / second^2 , which is the same as 1,007.8973 Newton
or if we go back to previous units (1 newton = 7.233013851 pound foot/square second);
Fd = 7,290.1355 pound foot / second^2
or
F = 7,290 pounds
Then we subtract the two forces:
39,690 pounds - 7,290 pounds = 32,400 pounds [or pounds force] [or pound foot / square second]
This 32,400 pounds is the force on the blades if the blades are being held still. When the blades turn, at different speeds there is different force on the blades. For now, we are simply looking at the force on the blades while they are stationary.
We can use this in both calculating the wind load and in calculating the center shaft torque for horsepower.
If (for example) we are working with a wind powered generator which has it's blades (currently) set at a 45 degree angle, then (in an ideal theoretical situation) we could use the hypotenuse formula and a little bit of trigonometry for right triangles to calculate the lift and drag. The force of the wind as drag against the blades forcing them back would be (the wind load at 90 degrees, which is the hypotenuse) multiplied by (the sine of 45 degrees) = (32,400.561 pounds)x(0.70711) = 22,910.760 pounds. Since the blades are set at a 45 degree angle, the force turning the blades is the same amount as the force of drag against the blades. Notice that the (ideal theoretical) wind load against the blades, which are not yet being allowed to turn, is now 22,910.760 pounds instead of 32,400.561 pounds because the blades are on an angle.
If we want to verify this, we multiply the squares of the lift and drag at 45 degrees, take the square root of that, and then check that against the wind load at 90 degrees.
(22,910 pounds)(22,910 pounds) + (22,910 pounds)(22,910 pounds) = (524,868,100) + (524,868,100) = 1,049,736,200
Square root of 1,049,736,200 = 32,399.627 , or rounded up it is 32,400 which was the wind load at 90 degrees, so it is correct.
We rounded our numbers and estimated as we went along in this example, so close is fine.
Thus:
We have about 22,910 pounds force attempting to turn the blades. That is a lot!
And, we have about 22,910 pounds of wind load.
If (for example) we are working with a wind powered generator which has it's blades centered at 100' above the ground, to get the cantilever torque (of the blades, and only the blades) which is attempting to bend the support at it's base, we multiply the wind load against the height.
That, in this case, is the wind load of the blades (set at a 45 degree angle), and it does not include the support, nor any of the structure.
Cantilever Torque (by the blades alone while they are still stationary and while they are being hit by a 50 mile per hour wind) attempting to bend the support at it's base = Tc = Force x distance = (22,910.76 pounds) x (100')
Tc = 2,291,076. foot pounds.
If (for example) we want to know how much torque the wind load against the blades generates against the shaft, while the blades are not yet being allowed to turn, we spread out the force per area of the blades and per distance from the center of the shaft and calculate that.
A little example of this,
calculated at a lower wind speed (from 1 mph to 5 mph) is:
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Wind Power Facts