General > General Technical Chat
Mess with your minds: A wind powered craft going faster than a tail wind speed.
electrodacus:
--- Quote from: IanB on December 21, 2021, 07:11:46 pm ---
--- Quote from: electrodacus on December 21, 2021, 06:07:52 pm ---I say that available wind power is defined by this equation 0.5 * air density * area * (wind speed - vehicle speed)^3
--- End quote ---
The equation 0.5 * (air density) * (area) * (wind speed)^3 gives the total flow of kinetic energy from wind moving through a particular cross section of area, and is applicable to wind turbine design as the air is moving through the turbine. A turbine could never extract more power than this from the wind (it will always be less than this). Note that this equation only works if the wind moves through the area of interest, like the disk area of a turbine.
Since a sail is a solid barrier, the wind does not, and cannot, move through the sail. Therefore, this equation cannot be applied to sails.
Another way of looking at this is to observe that the equation is (kinetic energy of wind per unit mass) x (mass flow of wind moving through control area). Since the mass flow of wind through the sail is zero, the energy flow must necessarily be zero.
When working with sails, you have to use force vectors and force balances.
--- End quote ---
This equation perfectly works for sails.
So for a sail cart that drives directly down wind this is the equation 0.5 * (air density) * (area) * (wind speed - vehicle speed)^3
And you can have a much more efficient wind turbine this way just have a sail on a cart and connect the electric generator to the wheels of the cart.
This can get you 2x the efficiency of typical propeller type wind turbine.
The problem is that you need a flexible cable connected to the cart and then every minute or so you need to bring the vehicle back to the start of the road.
It will not make much sense to do that even if for that part of the time efficiency is much higher.
Sail is the most efficient device to convert wind power in to kinetic energy.
IanB:
--- Quote from: electrodacus on December 21, 2021, 07:07:13 pm ---Sorry but you are just ignoring the fact that air is a compressible fluid.
Are you saying this graph below is completely wrong ?
https://en.wikipedia.org/wiki/Axial_fan_design
--- End quote ---
Yes. The given equations are no doubt fine, but the picture is wrong or misleading. That can happen with Wikipedia, since the articles do not get a rigorous peer review.
What are you going to trust more: an actual experiment, on video, where you can see the real world outcome, or a picture that someone posted on the internet?
IanB:
--- Quote from: electrodacus on December 21, 2021, 07:18:59 pm ---This equation perfectly works for sails.
--- End quote ---
So the equation is (kinetic energy of wind per unit mass) x (mass flow of wind moving through the sail).
What is the mass flow of wind moving through the sail?
Why do you say it works perfectly in this case?
fourfathom:
--- Quote from: electrodacus on December 21, 2021, 07:03:38 pm ---If you install the wind turbine on a vehicle and then drive direct down wind then equation of the power output for that wind turbine will again contain (wind speed - vehicle speed) since the wind speed relative to that moving wind turbine will be lower the faster the vehicle moves.
--- End quote ---
This is exactly correct, at least up to (vehicle speed = wind speed), for a turbine/generator bolted onto a moving vehicle. However, that is not what the rest of us are discussing; the Blackbird and related DDWFTTW vehicles.
electrodacus:
--- Quote from: Kleinstein on December 21, 2021, 07:17:31 pm ---
P_sail = 0.5 * air density * area * (wind speed - vehicle speed)^2* vehicle speed
With the active driven prop there can be even more power available, as the air may slow down even more. This is especially true for a speed close to or higher than the speed of the wind. This is because this way the wind can be slowed down to a speed below the vehicle speed.
--- End quote ---
Say wind speed is 10m/s and vehicle speed is half that 5m/s directly down wind
What sail will see is a 10 - 5 = 5m/s wind relative to it and vehicle thus correct equation for a 1m^2 sail will be
0.5 * 1.2 * 1 * (10-5)^3 = 75W
Using your equation 0.5 * 1.2 * 1 * ((10-5)^2 * 5) = 75W
Seems to provide the same value for half wind speed but
Vehicle speed of 2m/s
0.5 * 1.2 * 1 * (10-2)^3 = 307.2W
0.5 * 1.2 * 1 * ((10-2)^2 * 2) = 76.8W no longer correct
Vehicle speed or 8m/s
0.5 * 1.2 * 1 * (10-8)^3 = 4.8W
0.5 * 1.2 * 1 * ((10-8)^2 * 8.0) = 19.2W again incorrect.
Still even your incorrect formula will predict no wind power available when vehicle speed equal wind speed and above.
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