Calculating Power: How Force and Speed Affect Work

[Vibhor]

Homework Statement

Homework Equations

The Attempt at a Solution

Power is given by F.v , but I am not sure how to calculate F in this case .Again 'v' is the speed of the point of application of force F . I also do not understand how force applied on blades is causing work .

Any help is appreciated .

Thanks

 

[billy_joule]

What's the equation for kinetic energy?

 

[Vibhor]

What's the equation for kinetic energy?

I think you meant "expression" of KE . It is ½mv² . Could you respond to my doubts in the OP ?

 

[billy_joule]

I think you meant "expression" of KE . It is ½mv² .

Nope, I meant equation (ie contains an equality)
E_K = 1/2mv²

Could you respond to my doubts in the OP ?

Forces aren't really relevant here. Let's look at the definition of power:
P = E/t
We know that the winds energy is E_K = 1/2mv²
And that some fixed fraction of that is converted to power.
A bit of algebra and we can see what relationship Power has to wind speed..

 

[Vibhor]

@haruspex , @Doc Al Could you please share your ideas about doubts raised in OP . I can do this problem from energy concept . Interestingly the solution provided uses force approach .

 

[PeroK]

@haruspex , @Doc Al Could you please share your ideas about doubts raised in OP . I can do this problem from energy concept . Interestingly the solution provided uses force approach .

Power is also Energy/second. Personally, I would think of the air as a series of particles impacting the generator. I can't see any reason to think about forces when you're given the facts about energy.

 

[haruspex]

Interestingly the solution provided uses force approach .

That seems strange. Doesn't that require making assumptions about how the energy is transferred from wind to turbine? Can you describe the solution in more detail?

 

[Vibhor]

Can you describe the solution in more detail?

Please see the attached image .

 

[haruspex]

Please see the attached image .

Maybe I'm missing something, but I see no compelling logic in the F=vdm/dt step. What does dm/dt mean here? What mass is changing? I could understand F=d(mv)/dt, i.e. the rate at which the air loses momentum, and from that you can certainly get the rest of the steps.
As others have posted, this is a circuitous route, given the problem statement.

 

[Vibhor]

Maybe I'm missing something, but I see no compelling logic in the F=vdm/dt step. What does dm/dt mean here? What mass is changing? I could understand F=d(mv)/dt, i.e. the rate at which the air loses momentum, and from that you can certainly get the rest of the steps.
As others have posted, this is a circuitous route, given the problem statement.

Ok .

An alternative solution is also provided . Please see the attached image . Do you think this one makes sense ?

 

[haruspex]

Ok .

An alternative solution is also provided . Please see the attached image . Do you think this one makes sense ?

Yes, that's quite simple and persuasive.

 

[Doc Al]

Yes, that's quite simple and persuasive.

I agree.

 

[David Lewis]

Power is also Energy/second.

You're mixing a physical quantity with a unit of measure.
You may have energy divided by time, or joules per second.

 

[Vibhor]

I agree.

I am not sure about the first line .Can we assume particles in wind to be uniformly distributed ?

 

[Vibhor]

@Doc Al ,what do you think about the OP ?

 

[Nidum]

http://www.reuk.co.uk/Calculation-of-Wind-Power.htm

 

[Doc Al]

I am not sure about the first line .Can we assume particles in wind to be uniformly distributed ?

I think that's a reasonable assumption.

@Doc Al ,what do you think about the OP ?

I think the problem is fine. The way I'd do it is to track a parcel of air that hits the fan. The volume will be x*A; the rate at which the air hits the fan will be v*A. (A is cross-sectional area.) That should allow you to see that the energy in the parcel is proportional to v², and the power is proportional to v³.

 

[Doc Al]

http://www.reuk.co.uk/Calculation-of-Wind-Power.htm

@Vibhor: It looks like Nidum's link uses the same approach I stated above. Read it!

 

[Vibhor]

I think the problem is fine.

I meant what I had written in the "The attempt at a solution" .What do you have to say about issue raised in OP?

 

[Doc Al]

I meant what I had written in the "The attempt at a solution" .What do you have to say about issue raised in OP?

Ah, sorry...

Power is given by F.v , but I am not sure how to calculate F in this case .Again 'v' is the speed of the point of application of force F . I also do not understand how force applied on blades is causing work .

You can calculate the force by considering the rate of change of momentum of the air as it hits the fan. (That's the approach used in one of the solutions you posted.) The air pushes the blades, making them turn, thus doing work. (They are angled.) That approach should allow you to get the v³ proportionality.

 

[Vibhor]

(That's the approach used in one of the solutions you posted.)

So , do you agree completely with the official solution posted in #8 ?

 

[Doc Al]

So , do you agree completely with the official solution posted in #8 ?

It's OK. (Though I prefer the energy approach.)

My version of that force solution would be along these lines:
$F = \frac{d(mv)}{dt} = \frac{d(\rho x A v)}{dt} = \rho A v^2$

Again, I am thinking in terms of a voume of air moving at speed v.

 

[Vibhor]

You can calculate the force by considering the rate of change of momentum of the air as it hits the fan. (That's the approach used in one of the solutions you posted.) The air pushes the blades, making them turn, thus doing work. (They are angled.) That approach should allow you to get the v³ proportionality.

It's OK. (Though I prefer the energy approach.)

My version of that force solution would be along these lines:
$F = \frac{d(mv)}{dt} = \frac{d(\rho x A v)}{dt} = \rho A v^2$

Again, I am thinking in terms of a voume of air moving at speed v.

Thanks a lot Doc Al

 

[TSny]

In the power formula $P = \vec{F} \cdot \vec{u}$, the velocity $\vec{u}$ is the velocity of the point of application of the force. Wouldn't this be the velocity of a point of the blade? Of course, the speed of a point of the blade depends on the distance from the axis of rotation. So I guess you would need to use some average speed $\overline{u}$ for the points of the blade. Also, the force is probably different for different points of the blade. So, again you would need to use an average $\overline{F}$. Even if you can argue that $\overline{F}$ is proportional to the square of the wind speed, $v^2$, you would still need to argue that $\overline{u}$ is proportional to $v$ in order to get the power to be proportional to $v^3$.

The energy method avoids all this, but does rely on the assumption that the rate at which energy is transferred to the blades is proportional to the rate of flow of kinetic energy of the wind.

 

[Vibhor]

In the power formula $P = \vec{F} \cdot \vec{u}$, the velocity $\vec{u}$ is the velocity of the point of application of the force. Wouldn't this be the velocity of a point of the blade? Of course, the speed of a point of the blade depends on the distance from the axis of rotation. So I guess you would need to use some average speed $\overline{u}$ for the points of the blade. Also, the force is probably different for different points of the blade. So, again you would need to use an average $\overline{F}$. Even if you can argue that $\overline{F}$ is proportional to the square of the wind speed, $v^2$, you would still need to argue that $\overline{u}$ is proportional to $v$ in order to get the power to be proportional to $v^3$.

Thanks

The energy method avoids all this, but does rely on the assumption that the rate at which energy is transferred to the blades is proportional to the rate of flow of kinetic energy of the wind.

Why is this an assumption ? Isn't this what is happening Or is it some simplified model ?

 

[TSny]

Why is this an assumption ? Isn't this what is happening Or is it some simplified model ?

Seems to me it's an assumption. The problem statement refers to it as an assumption. Why should the wind transfer a fixed fraction of its energy to the blades (where the fraction is independent of the speed of the wind and the speed of the blades)? Turbulence, etc.

 

[Vibhor]

Ok . I misunderstood your point about assumption.