# Wind turbine energy capture

1. Apr 27, 2010

### groston

Consider a wind turbine placed alongside a road in a location with no wind. If a vehicle drives past (without loss of generality, at a constant speed) the turbine will spin and electrical power can be produced. The power produced is obviously supplied by the vehicle. As compared to the case where these is no turbine, the vehicle will necessarily consume additional fuel to maintain constant speed as some of the vehicle's power is captured by the turbine.

My question is this: What is the mechanism which causes the vehicle to consume more fuel?

2. Apr 27, 2010

### Staff: Mentor

Aerodynamic drag. The car is trying to push through the air and the wind turbine is pushing back.

3. Apr 28, 2010

### groston

Russ,

Your answer is too simplistic. How does the energy actually get transferred? Does the car 'push' a wall of air and the turbine 'push' back on this column? At what speed does the transfer happen - is it the speed of sound? How would you model this interaction?

4. Apr 28, 2010

### xxChrisxx

I think this is a real world case of "it makes no difference".

The car wouldn't acutually consume any more fuel (of a measurable level). The car is pushing the air anyway causing the loss. What happens to the wave of displaced air after makes no difference. The fact a turbine is sat by the road would have no measurable effect on the drag of the car.

The energy gets transferred becuase the displaced wave of air, flows over a turbine blade (just like a wing) causes a pressure difference and causes movement. This occurs at whatever speed the air was moving at when it 'came off' the car (nowhere near the speed of sound). It really is that simple.

It depends what you mean by model the interaction. You can use simple equations to estimate power transfer, or you can go to a full blown 3d CFD model showing the airflow and interaction.

EDIT: Just a personal comment on this, a wind turbine by the side of the road wouldn't really produce any usable power. Turbines are highly sensitive to flow, and generally like steady conditions hitting them at 90 degrees, anything other than that and the power produced begins to drop off quite quickly. This is impossible to achieve by powering it with the airflow coming off a car.

Last edited: Apr 28, 2010
5. Apr 28, 2010

### Staff: Mentor

Whether or not the effect is measurable (or the collected energy is measurable.....) isn't a useful thing to debate.

A road lined with trees or turbines has a higher drag on cars than a road lined with walls (or nothing). The road is basically a continuous conveyor belt full of cars and it carries a long column of air with it. Drag caused by objects on the road will slow this column of air.

6. Apr 28, 2010

### xxChrisxx

If you can't measure it, it may as well not exist.

If we must insist that the effect can't just be ignored that it's critical to determine how far away from the road/car the turbines are. At what point do we say the car is in free stream?

7. Apr 28, 2010

### groston

Agreed. And, I think that we can agree that any power generated (assuming no wind), comes from the passing vehicles, which means that the cars are producing additional power which is greater than the power being produced by the turbines (2nd Law), i.e., this is a bad idea.

My interest is understanding the mechanism. Your suggestion to 'estimate power transfer' misses the point - you are simply modeling the result, not the underlying physics which lead to the result. Your comment about CFD is more to the point, but my interest is not the actual model, but understanding the underlying physics.

I think that Russ' comment about drag and yours about the free stream might be the starting point. Does this seem like a reasonable way to think about the phenomenon: In a free stream condition, the air being dragged along by the car can be thought of as a bunch of 'springs' emanating from the vehicle whose distal ends are 'free'. When some object, tree, turbine, etc., is encountered, the distal end becomes 'fixed' and the 'spring' gets 'compressed', which pushes on the car.

(And this is where my comment about the speed of sound comes in. Once the object is encountered, at what rate does the 'spring' compress? Somehow, the interaction of the distal end needs to be 'communicated' back to the vehicle. Think about a cylindrical slug sliding in frictionless cylindrical tube of the exact same diameter. It is pushing a column of air in front of it. If the tube is suddenly capped, the column of air will become a spring. A simplistic, but probably reasonable model, would simply have the air pressure increase as a function of the trapped volume. A more interesting model would include 'waves' of higher density as shock waves move through the trapped air. My guess is that in a 'perfect' model, there would be a time lag between the time the cylinder is capped and the time the slug experiences the additional drag and that this time is related to the velocity of the shock wave which is somehow tied to the speed of sound.)

This raises another question: If the object sitting by the side of the road is a turbine with some diameter or simply a 'billboard' of the same size/shape, would the effect on the car be the same?

8. Apr 28, 2010

### xxChrisxx

I mean regardless of the cars power output and thermodynamics, from a purely practical point view a wind tubine just doesn't operate very well in changing conditions, ie a big puff of wind whenever a car goes past it. It requires a steady wind from 1 direction to work well.

Are you assuming that the wind tubines are infront of, alongside, or behind the car? Also do you know what happens to the airflow as a car travels through the air, (ie as it flows over and around the car)? Knowing this will help me gauge how to answer the bit below.

Putting the above into a real engineering context is meaningless. You mentioned some models, but those have assumptions that make them completely unrealistic in real world situations and are therefore pointless to answering the thread. For example, the above completely neglects that fact that air flows, it doesn't get compressed like in a cylinder. There are compressions and pressure waves but they are highly localised.

As the main drag from a car comes not from the higher pressure infront of it compressing the air, but the lower pressure behind it left in the wake. (in reality they act together to push the car back, but if you had to split them up the wake is farmore important than the front end)

May I suggest that if you want to learn more about the physics of the airflow, rather than a practical application, this thread gets moved (or you start a new one) to classical/general physics. May I also suggest that we start simple and build up to a rather complex scenario such as a car driving past a wind turbine.

Last edited: Apr 28, 2010
9. Apr 29, 2010

### dr dodge

if the drag on the vehicle is always present, and given standard conditions, fairly constant, based upon road design, why couldn't you use the air spilling off the highway to spin a turbine. Maybe long low ones that are paralell with the guard rails. As the air is forced off the highway in a "moderately lateral direction" it should spin, not a lot, but if it just ran LED marker lites its still something from wasted energy.

The cars do the work of moving the air mass, inertia carries the air mass, then the air mass does work to disipate its energy

I have done a few "ground turbulence" windmill experiments, and have a windmill that currently runs well, even when the wind is hardly there, because I am using the channeling effect of ground objects. I am pretty sure,( but have not actually measures yet) my windmill 6' off the ground runs 2-3 times the natural wind speed

a similar effect could be put into effect on the highways

dr

10. Apr 29, 2010

### xxChrisxx

Build cost, maintainance cost, reliability? Fact is it would be far cheaper and efficient simply to power the light.

It's all very well and good getting something that works in theory and controlled conditions but moving to something practical is far harder.

It's a fantastic idea, it's just not terribly practical. edit (currently)

11. Apr 29, 2010

### dr dodge

I agree its not practical at this time, but just like when oil prices go up, wells that are not profitable then go to the profitable column, if power costs continue to rise, as does grid demand, devices like this then may be more useful.

like the solar driveway/walkway markers, something for nothing is always good, just not always worth the cost to deploy. but as the cost to make it go down now its worth it

just an observation

dr

12. Apr 29, 2010

### minger

OK, I think I'll chime in on this. What I can seem to break this problem down to is almost the definition of a boundary layer. Consider an round object moving through some fluid. The object will experience both form drag (drag due to the planform shape) and viscous drag (due to shearing of the fluid).

What we're concerned here is of the viscous drag. One can then imagine a tube of large radius outside of the moving object. At what radius, will the tube have an effect on the moving object?

Certainly, if the tube is only marginally bigger than the object, then of course there will be viscous interaction between the object and the tube, however there will be some distance away where the effect is negligible. This is of course in a perfect world.

In real-life, the air billowing off of a moving vehicle is turbulent and highly vortical. For every "push" in the right direction, you'd get a pull in the wrong direction. Not only would this lead to very little energy being harnessed, but it would necessitate the turbine to having some sort of ratcheting mechanism. They are already expensive enough.

I guess in short, Russ is correct, the energy is coming from the car.

13. Apr 30, 2010

### Staff: Mentor

Chris, what I'm trying to avoid here is an accidental violation of conservation of energy and to make sure we address the meat of the OP's question. If a turbine on the side of the road generates 100W and of that, 50W is stolen from the car and 50W would have been dissipated anyway, that's 50W added to the car's engine load. If the engine is running at 30,000W of output, then an immeasurable 0.17% has been added to the load of the engine. You can't say that because it is too small to measure, it can be ignored because it isn't the fraction of the car's load that matters, it's the useful power and efficiency of the turbine that matters.

From the point of view of the turbine, this essentially works out to being a 67% efficient (33% efficient, but half of the output would have been wasted anyway) gas engine driving a fan, which drives a wind turbine. Now if you're the turnpike commission and you're just looking for a sneaky way to increase your rates by a tenth of a penny a mile, that's fine, but this isn't completely free (recovered) energy.
But that's the point - by definition, if the car is in free stream, the turbines are getting no wind from it.

The only way there could be a net savings here is if building and running the turbine is cheaper than running wires out to whatever road sign the turbine powers. Then it could end up cheaper for the drivers.

All that said, I'm pretty sure the OP is more interested in the aerodynamics - I just haven't had a chance to go into a lot of detail....

Consider a car drafting another car. The trailing car is being pulled forward by the lead car because there is an area of low pressure between the cars. But the lead car can be said to be "pushed" because the area of low pressure is smaller than if the trail car weren't there. If the trail car were off to the side, the drafting effect is reduced, but still present. If you instead put an obstruction off to the side of the road, the air behind the car is no longer able to follow the car and the area of low pressure behind the car is increased: that's increased pressure drag.

Now, for my thumb-wag calculation above, I assumed that half of the energy being absorbed by the turbine would have been dissipated anyway. I'm not really sure if that's true. I would think that some of it would have been dissipated anyway because the energy dissipates by rubbing together air molecules if nothing else is there, so you're dissipating the energy with a turbine before it can dissipate by rubbing air molecules together. But there has to be an increase in drag, because you're pulling the wake of the car away from the car, increasing the size of the low pressure area and decreasing its pressure.

14. Apr 30, 2010

### Staff: Mentor

You are correct. Pressure waves in air are sound waves, so the car pushes a column of air in front of it and the pressure tries to move away at the speed of sound. This creates a pressure gradient in front of the car.
A wind turbine doesn't stop the wind, a bilboard does: so a bilboard would be worse.

15. Apr 30, 2010

### Staff: Mentor

Without knowing exactly how much power would be produced this way, it is tough to tell if that's true (there is also an issue of safety). Solar panels are used to power a lot of roadside devices these days even though solar panels are rediculously expensive, particularly for small devices. So why are they used? Well if you have to run an electrical cable for several miles to get to your sign/light/traffic camera/etc., the cost of the cable can far outweigh the cost of a solar panel.

16. Apr 30, 2010

### minger

"Pressure waves" is sort of a misnomer here. When one decomposes the governing equations into eigenvalues and eigenvectors, you get the nature of the resulting propagating waves. In each direction, you get a left and right running acoustic wave, an entropy wave, and two vortical waves.

The two acoustic waves travel at $$u\pm c$$, while the rest simply travel at u, the mean flow. These are the eigenvalues. The eigenvectors describe how the waves "look". They are:
$$\zeta = \left\lbrace \begin{array}{c} \frac{\partial p}{\partial t} - \rho c\frac{\partial u}{\partial x} \\ c^2 \frac{\partial \rho}{\partial t} - \frac{\partial p}{\partial t} \\ \frac{\partial v}{\partial x} \\ \frac{\partial w}{\partial x} \\ \frac{\partial p}{\partial t} + \rho c\frac{\partial u}{\partial x} \end{array} \right\rbrace$$

Where the first and last are the acoustic waves, the second is the entropy wave, and the 3rd and 4th entries are the vortical waves.

I guess the important thing to note here is that entropy waves and acoustic waves look very similar. They are both a temporal gradient of pressure. The difference, aside from the speed of propagation is that the entropy wave has a density gradient while the acoustic wave has a velocity gradient.

I guess I'm just nitpicking here, but I don't think it would be accurate to say that under those conditions a "pressure wave" would propagate at the speed of sound away from the object.

17. Mar 30, 2012

### pfdovr

perhaps the answer to harnessing the 'wind' energy created by moving vehicles -- if there is an answer -- is to find where the energy is focused in one direction, such as a tunnel.