Science Forums

# Ideal Wind Turbine Power Analysis from the Reference Frame of the Air

## Recommended Posts

Here is my question at hand:

"From the reference frame of the AIR, (i.e. the XYZ coordinate system that is moving with respect to the ground at the same speed and direction as the wind), where does the energy come from to power an operational wind turbine that is bolted to the ground?"

To definitively answer this question in a quantifiable manner, here it is reformulated in the typical, "Given:/Find:/Solution:" format used in engineering textbooks.

Given: An ideal wind turbine is operating at an elevation of 2064 m (6772 ft). The density of the air is: ρ = 1 kg/m^3. The swept area of the turbine rotor disk is: S = 18 m^2. The velocity of the air relative to the ground, (i.e. "the wind"), is: Vwind = 3 m/s. The turbine is operating at its optimal pitch angle so: ΔV = 2 m/s, and the far downstream velocity is 1/3 of the influent velocity: V2 = 1/3 * Vwind = 1 m/s.

Find: The net power generated by the ideal turbine from: A.) the frame of reference of the ground, and B.) the frame of reference of the air. Assume that the fluid is inviscid and incompressible and that the flow is axial, steady, adiabatic, and laminar. The rotor is considered to be an ideal actuator disk which has zero aerodynamic drag, i.e. the blades have an infinite L/D ratio. There is no swirl in the effluent flow. In other words, this ideal turbine and the assumptions used for the analysis are identical to those utilized in the proof section of the following wikipedia article:  Betz's Law

Solution:

Part: A.) Here is my solution to the first part of the problem:

Note that the net power, (144 W), is 16/27 of the Power of the Wind, (Pwind = 1/2 ρ S Vwind^3 = 243 W). The part B.) solution, (solved from the AIR frame), should arrive at the same number for net power.

Part B.)?

Edited by ridgerunner
Reduced image contrast to improve readability of green velocities
##### Share on other sites

I have removed the posts placed by our resident troll, Superpolymath, and issued a warning, since I believe your effort deserves better replies that what he has provided.

Please give me some time to respond as I am very busy with some other matters, but I will respond as soon as possible.

Try to hold off on your Part B, at least until you get a response to Part A. Thanks

##### Share on other sites

Thank you. Although he (she?, it?) did present one reasonable critique; the graphic containing my solution to part A.) does fail to clearly identify the frame of reference and the associated XYZ coordinate axis. 🙂

##### Share on other sites

Sorry it has taken so long for my reply, but I have very little time available to devote to this forum just now.

OK, I have had a look at your math and there is nothing wrong with it, as long as you are considering an ideal turbine that is operating at the Betz limit of 59.3% coefficient of performance and has zero aerodynamic drag.

I can’t help but wonder what your purpose is in analyzing such a turbine, which does not exist and can never be built?

Practical wind turbines operate at about 50% of the Betz Limit, while even the best gas turbine generators achieve up to 85% of Betz. I would be quite satisfied if the dual fuel turbines on our ships achieved 85% of Betz, but they actually run closer to 75%.

I won’t post any math at this time; there is no need to and I don’t have the time to spare, plus there is no working LaTex here. But I will post math if and when the need arises.

Again, sorry it has taken so long for my reply, please go ahead and post your Part B and let’s see where you intend to go with this thread.

##### Share on other sites

17 minutes ago, OceanBreeze said:

... I can’t help but wonder what your purpose is in analyzing such a turbine, which does not exist and can never be built?...

I am not asking about some non-existent, impossible to build turbine. I am asking about how to compute the net power of the ideal turbine described in the problem statement, but doing so from the frame of reference of the air. (The turbine  is bolted to the ground.) In the air reference frame, the air mass has zero velocity and has no kinetic energy to give up; and yet the ideal turbine continues to generate 144 W of electricity. In other words, where is this energy coming from?

##### Share on other sites

You could have asked your question using an existing practical wind turbine with a coefficient of performance of 50% of Betz, for example. Although even 50% of Betz would be a very efficient wind turbine.

That is why I wonder why you went to all the trouble of specifying a turbine operating at the Betz limit, with no aerodynamic losses. Can you fault me for wondering why you did that?

The fact that you now claim that you “are not asking about some non-existent, impossible to build turbine” is strange, since that is exactly what you specified!

I will play along for a bit longer, but I will point out certain bits of information that you have decided to leave out of your question, such as the fact we are discussing a wind turbine that only exists as an impossible ideal.

As long as the only thing you have changed is to place your observer in the wind, and everything else remains the same; same turbine still bolted to the ground, still spinning and slowing down the incoming wind from 3m’s (wrt to the ground) to 1 m/s (wrt to the ground) and still producing 144 Watts of power.

All we need to do to balance the book on energy, as our windy observer will report that there is no energy available in the wind frame, as far as he is concerned; is say that the energy is coming from the ground moving under the still air at 3 m/s and dragging the generator along with it.

This isn’t some deep conundrum!

Before you think you have won a Kewpie doll for your post, I want you to consider what you wrote in that other (locked thread) on this same subject:

You wrote:

One brain-tweakingly unintuitive aspect of this vehicle is that when the vehicle operates at precisely wind speed, from the frame of reference of the cart, the air is stopped and has zero kinetic energy and the propeller is actually adding energy to the air speeding it up. But at the exact same time, an observer on the ground witnessing the same cart roll by will observe that the air behind the propeller is moving slower than the surrounding wind and the propeller is thus slowing down the air in this reference frame. Strange but true!

See if you can spot the difference between what you wrote then and now.

In case you can’t, or won’t see the difference, I will point it out to you:

In the latter case, you are admitting there is zero wind kinetic energy for the cart to draw upon to keep on working. Switching frames in this case, for example to the ground frame, gains you nothing!

Zero energy in the wind also means zero energy in the ground.

Frame switching is nothing more than an exercise in book keeping. If there is 243 Watts of wind energy with respect to the ground; in wind frame, there will be 243 Watts of “ground” energy with respect to the wind frame. It is just a shell game.

However, the reason why you and the rest of your gaggle of merry men will never be taken seriously here is because you think there can be Zero available wind energy with respect to your cart, but by switching the frame of reference to the ground, suddenly ground energy appears!

No! Frame switching does not create energy when there is none. All frame switching can do is attribute whatever energy is available in one frame, to another frame. If the energy is zero in one frame, it is zero in whatever frame you switch to.

If there is any argument to the contrary presented, it better have the backing of a very reliable reference source.

By the way, did anyone ever get up the courage to run the cart against a NB balloon blowing in the wind?

No? I didn’t think so!

I predict this thread will be closed before the next windy day!

##### Share on other sites

Firstly, I chose an idealized cart for the purpose of this thread discussion. I have no idea why this presents any problem; it greatly simplifies the equations and all the computed parameter values end up as smallish, tidy integers. I also chose an unrealistic value for the wind speed, 3 m/s (6.71 mph), which is likely too small for any real world turbine. As a fellow engineer, I'm sure you will agree that simplifying a problem is commonplace in engineering; besides, this is my thread and I can "frame" 🙂 the question in any way I see fit, yes?

You say:
"... As long as the only thing you have changed is to place your observer in the wind, and everything else remains the same; same turbine still bolted to the ground, still spinning and slowing down the incoming wind from 3m’s (wrt to the ground) to 1 m/s (wrt to the ground) and still producing 144 Watts of power. ..."

When you say "wrt to the ground," you are reverting back to the ground frame, which was part A.) of my question. In the rest frame of the air, the turbine is NOT slowing down the air from 3 m/s to 2 m/s; it is speeding it up from 0 m/s to 2 m/s, and is adding kinetic energy to the air at a rate of 72 W. Each second, the turbine, which is moving to the left along with the ground at 3 m/s, takes in 36 kg of air having zero velocity (KE1 = 0 J),  and puts out  36 kg of air having a velocity of 2 m/s (KE2 = 72 J). Meanwhile, the ground is losing 216 W of power to the base of the turbine, ((+72 N)(-3 m/s)), and the air is gaining a net of 72 W from the rotor ((-72 N)(-1 m/s)). The turbine nets 216 - 72 = 144 W. Here is  a summary of my answer to part B.):

Note that I am trying to be very careful with my semantics. If and when I use the term "Wind," I am referring to the "velocity of the air relative to the ground," i.e. Vwind = Vair - Vground, (which has the same direction and magnitude in all inertial reference frames). When I use the term "Air," I am referring to the velocity of the air specific to the current working reference frame. I have never said that energy can be extracted from the ground in the reference frame of the ground, because that is impossible; in the ground frame, the ground has zero velocity and has zero kinetic energy to give up. Likewise, in the air frame, the air has zero velocity and has zero kinetic energy to give up. What I *am* saying is that whenever the wind is blowing, (Vwind > 0), the air has kinetic energy in the ground frame, and the ground has energy in the air frame.

Note also that I am NOT discussing "the cart," which is a topic you have clearly stated is off limits on this forum. I am trying my best to adhere to your forum guidelines. In this thread, I am trying to point out, (in a hopefully interesting and possibly educational manner), that when conducting an engineering Work/Energy/Power analysis, the chosen frame of reference matters, (and the numbers that result may appear to be very strange and counterintuitive when looking at a physical event from a non-intuitive POV/FOR.) Looking at a ground based turbine from the reference frame of the air is just such an example. i.e. In the air frame, kinetic energy, (both rotational and linear), is being added to the air at the same time the air is adding (rotational) energy to the rotor/turbine. Feel free to check the math posted above which says this is true.

You say:
"... All we need to do to balance the book on energy, as our windy observer will report that there is no energy available in the wind frame, as far as he is concerned; is say that the energy is coming from the ground moving under the still air at 3 m/s and dragging the generator along with it. ..."

Yes. You then go on to discuss frame switching, and yes, you are absolutely correct that energy computations are merely a matter of bookkeeping. I agree! However, you then bring up the topic from the previous (now taboo) discussion and claim that I am making a completely different claim now than I did then. I disagree. Since you have now twice quoted a paragraph I wrote in the closed thread, allow me to simply offer the following food for thought:

"One brain-tweakingly unintuitive aspect of a stationary wind turbine is that when looked at from the frame of reference of the air, the air is stopped and has zero kinetic energy and the turbine rotor is actually adding energy to the air speeding it up. But at the exact same time, an observer on the ground witnessing the same turbine churning away will observe that the air behind the turbine is moving slower than the surrounding wind and the turbine is thus slowing down the air in this reference frame. Strange but true!"

Do you think that the above paragraph is "ridiculous," or inaccurate in any way?

Lastly, I firmly stand by what I said in the closed thread and what I am saying now. You can use theatrics with threats of shutting down the thread, and that is certainly within your power, but I would much prefer that, instead, you use sound, reasoned arguments, (i.e. applying Newtonian Mechanics using the common language of mathematics), to defend your stance. That is what I am trying to do. Note that at all times, I have refrained from name calling or raising any emotion at all - (IMO, emotion (other than humor) has no place in physics discussions!)

##### Share on other sites

Sorry RR, I still don't have time for this but will try to give you an answer tomorrow.

I should tell you up front you will not like my answer.

##### Share on other sites

Not to make a Federal issue out of this, but the reason why you’re choosing to use an idealized turbine that operates at the Betz limit of efficiency AND has no aerodynamic losses, presents a problem is simply that no such turbine exists; it is an illusion.

The research I have done on this subject, and on the people who have been pushing this on every forum that will allow it, indicates to me that your modus operandi relies on first creating an illusion, similar to every magician’s act, and then building on that illusion to fool your audience into believing in something that is impossible.

This is not the way a Physics discussion should be conducted and I want my objections made known right at the outset.

No, this is not your thread; you started it, but it does not belong to you. Other members, including myself, have every right to make objections, and even point and laugh if we see anything that seems ridiculous. Those are the ground rules here, plus I reserve the right to merge this thread with the other one that is already closed, if I decide it is just wolf in sheep’s clothing, to try and get around the ban on the “dwifty” topic.

Yes, I did revert back to the ground frame to stress my point that this is just the same wind turbine you described in Part A, with the turbine blades still turning at the exact same rpm and producing the exact same power. Again, I am going to interfere with any and all attempts by you to create any illusions that there is anything different happening just by placing your observer in a different frame of reference.

The change of frame allows you to analyze the turbine you introduced in Part A, from a frame where the observer sees the air at rest and the ground in relative motion. That is perfectly valid.

But other than that, absolutely nothing else has changed. The turbine is still producing 144 Watts of power, as before. I don’t know why you bothered to go through your calculations again; there is a much simpler way to show what all the relative air velocities are: In Part A the wind was blowing at 3 m/s wrt the ground frame, and in Part B the wind is at rest with respect to the air frame. The offset then is -3 m/s. This offset can be added to all of the other air velocities you calculated in part A: For example V2, which was + 1 m/s in Part A becomes -2 m/s in Part B. The minus sign indicates a change of direction.

You say you are not here to discuss “the cart” but I think we all know what your agenda is here, so I will take it upon myself to jump start that part of the conversation; once again leaving you no opportunity to create any illusions.

For now, I will keep the cart operating at the Betz limit, just to make a point. However, I am going to include aerodynamic drag, to introduce a bit of reality into this Physics discussion.

My approach will be to follow standard procedure (except for my allowance for Betz efficiency):

Power output of the turbine is given by this expression:

P = p A/4 [(V1 + V2) (V1^2-V2^2)]

Aerodynamic drag on the turbine is given by this expression:

D =p A/2 (V1^2 – V2^2)

As you know, p is air density (1 kg/m^3 that you specified) A is the area of the turbine blades (18 m^2) V1 is upstream wind velocity (free stream) and V2 is the down stream wind after slowed by the turbine blades.

To arrive at max Power, I differentiate the expression for Power with respect to V2 and set it equal to 0.

Since pA/4 is a constant, the differentiation is only done on the expression in brackets:

dP/dV2 = [(V1^2 -V2^2) + (-2V2) (V1 +V2)] = 0

(V1 + V2) (V1 - V2) = (2V2) (V1 + V2)

(V1 – V2) = (2V2)

Therefore, at Max Power V2 = V1 /3

Now, to that taboo subject of “le chariot notoire”

I add some wheels and neglect transmission losses (how generous I am!)

Force wheels = Power Max / Velocity cart

Net force on ze cart = Force wheels – Drag Force = 0 at max velocity (equilibrium)

Max Velocity Cart = Power Max / Drag Force

This greatly simplifies:

V cart Max = pA/4 (V1 + V2) (V1^2 – V2^2) / pA/2 (V1^2 – V2^2) = (V1 + V2) / 2

I am allowing for the Betz result V2 = V1 / 3

Result is V cart Max = 2/3 V1, in this case max V cart going upwind is 2 m/s

This is an impressive result and I understand the well-designed Aeolus upwind carts have done substantially better than Vmax = 2/3 Vwind, even though they do not operate at the Betz limit.

Part of the explanation is the speed trials are held on the Den Helder dike in Holland, where it is quite possible that very little, if any of the runs are directly into the wind as the dike is a long curving surface.

Mind you, I give full credit to these honest and very accomplished builders and racers and I accept that some of them, on that curved course, probably averaged close to or even briefly exceeded the wind speed.

But I have no respect at all for those who claim to have achieved impossible MULTIPLES of wind speed, both upwind and downwind, with a poorly designed cart and the only “evidence” presented is some data that was manipulated by the lone contestant.

When challenged to provide much better evidence by racing against a NB balloon blowing in the wind, all I hear is crickets and insults.

I will give you a chance to reply RR before I merge this thread with the other thread and the composite thread will then be closed.

##### Share on other sites

Firstly, I agree with almost everything you wrote up until the line: "Max Velocity Cart = Power Max / Drag Force."

Although what you are calling the: "Drag Force = D," is what I'm referring to as the: "Thrust Force = Ft," which is the force of the air acting on the rotor caused by the acceleration of the air through the control volume, (i.e. ΔV = V1 - V2.) In my example problem solution to part B.), Ft has a value of 72 N and is clearly shown on the free body diagrams.

Secondly, "le chariot notoire?" - Good one! 🙂

Lastly, thank you for not immediately shutting this thread down and allowing me to respond. I'm glad you seem willing to expand the discussion to include the more interesting case of the self-propelled directly upwind turbine cart. But note that it will take me a while to prepare a suitable response to your latest post.  I assume that since you appear to be busy, you'll have no problem with an asynchronous discussion having (possibly longish) time delays between responses. Note that I have no intentions of presenting any illusions, but rather, "Just the facts, Ma'am," as Joe Friday used to say.

As a teaser to what my response will include, note that when dealing with a self-propelled moving turbine cart, the force and power equations are more easily handled, and produce more tangible results, when written in terms of Vwind, Vcart and ΔV, rather than in terms of V1 and V2. i.e. (V1 = Vcart + Vwind), and (V2 = Vcart + Vwind - ΔV). Note also, that the optimal V2 = 1/3 V1, is only valid when the cart is stopped. My analysis indicates that as the cart speed increases, the optimal ΔV value also increases and approaches Vwind.

Cheers from Central Utah! 🙂

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.