Undergrad Power available to a wind powered vehicle traveling directly downwind

[hutchphd]

My question is super simple
What is the equation describing wind power available to vehicle assuming the known data above.

Unfortunately the answer is nt "super simple". If you wish to investigate I recommend proceeding from Betz's Law. First try it with no drag and the add your hypothetical drag. Have fun.

 

[electrodacus]

That limit irrelevant to DDWFTTW vehicles, for the reason stated here:

Not going to jump to this discussion until you agree with the equation I posted or provide the one you think is correct.
The equation is for theoretical max wind power available and more than that is not possible no matter the vehicle design.

So just need the equation if you disagree with the one I provided.
If you can not provide an equation for available wind power then you can not claim you understand how a wind powered vehicle works no matter the design.

 

[electrodacus]

Unfortunately the answer is nt "super simple". If you wish to investigate I recommend proceeding from Betz's Law. First try it with no drag and the add your hypothetical drag. Have fun.

This is your way of saying you do not know?
Betz's Law has nothing to do with the question or this example.

 

[hutchphd]

No it is my way of saying you don't know. In this case I don't care.

 

[electrodacus]

No it is my way of saying you don't know. In this case I don't care.

:)
In order to say I do not know you will need to know.
I do not think I can make my question simpler than it is.
I even provided a numerical value with the simple example. If you knew you will be able to provide the correct numerical value for power and the equation used to get there.

 

[erobz]

No its obvious you aren't comprehending the basic application of Newtons laws here, and you clearly don't deserve this with that horrible attitude.

Do you agree with this?

The force of drag acting on the cart in wind speed $w$ :

$$ F_D = \frac{1}{2} C_D \rho A ( w - v )^2$$

Yes or no?

 

[electrodacus]

No its obvious you aren't comprehending the basic application of Newtons laws here.

Do you agree with this?

The force of drag acting on the cart in wind speed $w$ :

$$ F_D = \frac{1}{2} C_D \rho A ( w - v )^2$$

Yes or no?

Do you understand that air is made out of particles that collide with the vehicle so the kinetic energy of the particle is transferred to the vehicle ?
It seems you posted an equation that I can not see. Is this a problem with my browser ?

 

[erobz]

Do you understand that air is made out of particles that collide with the vehicle so the kinetic energy of the particle is transferred to the vehicle ?
It seems you posted an equation that I can not see. Is this a problem with my browser ?

try refreshing

 

[electrodacus]

try refreshing

I did it is still blank.

 

[erobz]

I did it is still blank.

Maybe the gods have intervened and are stopping me from giving you the answer!

 

[electrodacus]

Maybe the gods have intervened and are stopping me from giving you the answer!

Seems like you got very close to writing the same equation I did.
Just multiply that by wind speed relative to vehicle w-v and you get the same equation I posted for power.

 

[erobz]

Seems like you got very close to writing the same equation I did.
Just multiply that by wind speed relative to vehicle w-v and you get the same equation I posted for power.

But that is incorrect. You multiply that by $v$, the velocity of the cart.

$$ P = F_D v = \frac{1}{2} C_D \rho A v ( w - v )^2$$

 

[Halc]

0.5 * air density * area * (wind speed - vehicle speed)³

This makes quite a few assumption. It for instance goes to zero when the two speeds are equal, but there are wind vehicles that can accelerate while moving exactly at speed going down wind, so there is still very much power available to it.

 

[electrodacus]

But that is incorrect. You multiply that by $v$, the velocity of the cart.

$$ P = F_D v = \frac{1}{2} C_D \rho A v ( w - v )^2$$

So you say (sorry I can still not see your equations maybe a problem with my browser).
P = 0.5 * air density * Area * coefficient of drag * (w-v)² * v
Where I say
P = 0.5 * air density * Area * coefficient of drag * (w-v)² * (w-v)

Seems we are super close we just need a few examples to see which one matches reality.

 

[electrodacus]

This makes quite a few assumption. It for instance goes to zero when the two speeds are equal, but there are wind vehicles that can accelerate while moving exactly at speed going down wind, so there is still very much power available to it.

That is exactly what I'm about to prove after we establish that the equation I provided is correct.
There is zero wind power available to a vehicle traveling direct down wind when vehicle speed equals wind speed.
The reason the vehicle you are likely referring to can exceed wind speed is because it uses energy storage to accelerate above wind speed for a limited amount of time proportional with the amount of stored energy.
But before getting there we need to agree to the equation describing the wind power available to vehicle.

 

[A.T.]

This limit is not relevant for the DDWFTTW cart, because the goal is not to extract as much energy per air volume as possible. The goal is to extract just enough energy to offset the losses and continue to accelerate. The goal is to be efficient, and for propellers that means: Accelerate a lot of air by a little bit. So the optimum is not to stop the air relative to the ground, but the opposite: disturb it as little as possible.

There is no hard theoretical limit on how efficient the cart can be made, and how many wind-speed multiples it can achieve.

Not going to jump to this discussion ...

And go on irrelevant tangents instead?

If you can not provide an equation for available wind power then you can not claim you understand how a wind powered vehicle works no matter the design.

If you think that maximizing extracted wind power is the goal, then you don't understand how high performance wind powered vehicles work.

 

[A.T.]

There is zero wind power available to a vehicle traveling direct down wind when vehicle speed equals wind speed.

Wrong, because by accelerating air backwards the vehicle still reduces the air-ground velocity difference (extracts wind energy), even when at wind speed going downwind.

... it uses energy storage to accelerate ...

Nope, see above.

 

[electrodacus]

If you think that maximizing extracted wind power is the goal, then you don't understand how high performance wind powered vehicles work.

This is a particular case of a direct downwind vehicle powered only by wind.
The only way for this vehicle to exceed wind speed is to store energy while below wind speed and then use that stored energy to accelerate above wind speed.
That is exactly what happens with those vehicles that means they will be above wind speed for a limited amount of time proportional with the amount of stored energy.
There is only one equation for available wind power to a wind powered vehicle traveling direct down wind.
The correct one is the one I provided and if you disagree you can provide the one you think is correct.

 

[A.T.]

The only way for this vehicle to exceed wind speed is to store energy...

Wrong, as already explained:

Wrong, because by accelerating air backwards the vehicle still reduces the air-ground velocity difference (extracts wind energy), even when at wind speed going downwind.

 

[electrodacus]

Wrong, as already explained:

Where is your equation for available wind power?
This is what I request here before going forward.
It seems to many people struggle with understanding energy conservation.

The equation for available wind power will not be different no matter the design of the wind powered vehicle.
If you look at the equation only area and coefficient of drag have anything to do with the vehicle design.

 

[A.T.]

It seems to many people struggle with understanding energy conservation.

There is nothing in Energy Conservation that prevents reducing the air-ground velocity difference (extracting wind energy) when moving downwind at wind speed, if you have contact to the ground too.

 

[russ_watters]

Where is your equation for available wind power?

[from post 5]
0.5 * air density * area * (wind speed - vehicle speed)3

Your answer doesn't match your question. The amount of power available in the wind is not a function of the vehicle speed it is a function of the wind speed relative to the ground only. Adding the term for vehicle speed inserts additional unstated constraints/assumptions about the nature of the vehicle and how/how effectively it extracts that power. It's a very different question.

To be specific, subtracting the wind vehicle speed assumes you are using a sail or turbine on the vehicle to extract power.

[typo strike through]

 

[electrodacus]

Your answer doesn't match your question. The amount of power available in the wind is not a function of the vehicle speed it is a function of the wind speed relative to the ground only. Adding the term for vehicle speed inserts additional unstated constraints/assumptions about the nature of the vehicle. It's a very different question.

To be specific, subtracting the wind vehicle speed assumes you are using a sail or turbine on the vehicle.

If vehicle is stationary then wind speed relative to vehicle will be the same as wind speed relative to ground.
Once the vehicle starts moving in the exact same direction as wind (downwind) the wind speed relative to vehicle drops and thus so is the wind power available to vehicle.
The vehicle design is irrelevant as the only important aspect is the vehicle area facing the wind and coefficient of drag.
So equation will be the same no matter the vehicle design.

 

[russ_watters]

Once the vehicle starts moving in the exact same direction as wind (downwind) the wind speed relative to vehicle drops and thus so is the wind power available to vehicle.

That isn't correct in general. It is true for some types of designs/vehicles but not true for others.

 

[electrodacus]

There is nothing in Energy Conservation that prevents reducing the air-ground velocity difference (extracting wind energy) when moving downwind at wind speed, if you have contact to the ground too.

Again provide the equation showing that there is any wind power available to a vehicle that travels direct downwind at the same speed as wind speed.
The equation I provided shows there is zero wind power available to vehicle in those conditions.

Vehicle has stored kinetic energy relative to the ground while it moves at same speed as wind speed so when you want to take energy at the wheel that energy will come from stored vehicle kinetic energy thus will result in vehicle slowing down if there is no other energy storage available as it is the case with Blackbird that stores energy in the pressure differential created by the large propeller.

 

[russ_watters]

Again provide the equation showing that there is any wind power available to a vehicle that travels direct downwind at the same speed as wind speed.
The equation I provided shows there is zero wind power available to vehicle in those conditions.

It just takes correcting your equation: 0.5 * air density * area * (wind speed)3

Since the wheels are coupled to the propeller/turbine the vehicle is harnessing the differential between ground speed and wind speed. That differential never changes regardless of cart speed.

Vehicle has stored kinetic energy relative to the ground while it moves at same speed as wind speed so when you want to take energy at the wheel that energy will come from stored vehicle kinetic energy thus will result in vehicle slowing down if there is no other energy storage available as it is the case with Blackbird that stores energy in the pressure differential created by the large propeller.

That's a run-on sentence that makes no sense. Obviously if you couple the turbine/propeller to the wheels, then one is propelling the cart and the other is powering it.

 

[electrodacus]

That isn't correct in general. It is true for some types of designs/vehicles but not true for others.

It is true for any vehicle design.
The equation is not dependent on vehicle design.

Imagine a simple sail vehicle (maybe just a cube on wheels).
The equation will perfectly describe what will happen to that vehicle.
Then imagine the exact same vehicle but you add an electrical generator at the wheel and a battery.
Now to know what happens to that vehicle you will also need to know how much power you extract at the wheel. And if you then use that stored energy in the battery you are able to exceed wind speed direct down wind but vehicle will eventually slow down below wind speed as that stored energy is being used up.
So you can not say the vehicle is powered by wind while it is accelerating using the energy stored in the battery.
That is exactly how blackbird vehicle works is just that instead of an electrochemical battery it uses pressure differential for energy storage.

 

[russ_watters]

It is true for any vehicle design.
The equation is not dependent on vehicle design.

Imagine a simple sail vehicle (maybe just a cube on wheels).
The equation will perfectly describe what will happen to that vehicle.

Um...yes, the equation applies to that design. But not to DWFTW designs.

That's the point. It's up to you where this goes from here - we can end the thread or you can try learning how the DWFTW vehicles work, but continuing to make the false claim is not one of your options.

You're trying to prove that they can't work based on an inaccurate understanding of how they work instead of learning how they actually work.

That is exactly how blackbird vehicle works is just that instead of an electrochemical battery it uses pressure differential for energy storage.

That is nonsense. "uses pressure differential for energy storage" isn't a thing.

 

[electrodacus]

It just takes correcting your equation: 0.5 * air density * area * (wind speed)3

Since the wheels are coupled to the propeller/turbine the vehicle is harnessing the differential between ground speed and wind speed. That differential never changes regardless of cart speed.

That's a run-on sentence that makes no sense. Obviously if you couple the turbine/propeller to the wheels, then one is propelling the cart and the other is powering it.

That equation will be true for a stationary vehicle or for a vehicle traveling perpendicular to wind direction.
The discussion is about a vehicle traveling direct downwind.

What you describe is an overunity device like those electric generators powered by an electric motor and electric motor is supplied by the electric generator and on top of that you can also power a light bulb.
As I already mentioned not understanding energy conservation.

 

[erobz]

So you say (sorry I can still not see your equations maybe a problem with my browser).
P = 0.5 * air density * Area * coefficient of drag * (w-v)² * v
Where I say
P = 0.5 * air density * Area * coefficient of drag * (w-v)² * (w-v)

Seems we are super close we just need a few examples to see which one matches reality.

Mine follows logically from fundamental physics, yours does not. You shouldn't need any examples to understand that.

There is a force $F_D = \propto ( w-v)^2$ ( from drag ) on the vehicle that is traveling along with the vehicle at velocity $v$. $P = F_D \, v$ is the power of that force. This is a basic application of Newtons Laws. The vehicle (which the force of drag is doing work on) is NOT traveling at speed $( w- v)$ w.r.t. a stationary frame (ground).