The Betz Limit - Again

[finnsawyer]

I found I was scammed.  The pictures shown in the link below were supposed to show the spreading of the airflow past a wind turbine.  In reality it shows no such thing.  The rotor is actually a type of propeller, and is being spun by a motor.

     http://allentech.net/about/images/windtunnel-flow-280w.gif

This spreading of the airflow is important as it is integral to the derivation of the Betz limit.  According to Betz' derivation the air flow far downstream from the wind turbine has a velocity 1/3 that of the velocity far upstream of the turbine.  The comments of Mr. Jacobs do lead some credence to this as he mentions sufficient slow down of the air such that a match lit behind the turbine "will hardly go out".  He also states that the blades fling the air outward.  You can find a derivation of Betz' Law here:

     http://www.windpower.org/en/stat/betzpro.htm

There is also available on the site a simple graph that purports to show the pressure distribution before and after the turbine's rotor.  The pressure increases above ambient until the flow reaches the rotor and then starts from below ambient just behind the rotor and rises back to ambient, where I use the term ambient for the air pressure in the airflow a large distance before the turbine.  The velocity of the flow behind the rotor decreases with distance behind the rotor until it reaches the 1/3 value.

There are some issues with this picture.  For one, the decrease to 1/3 initial velocity behind the rotor means the radius of air flow has increased by 73% to move the same mass of air per second.  Considering that a tube going from the radius of the rotor "a" to 1.73xa contains twice the volume of air than the original cylinder of radius "a" and twice the mass, what physical process makes that air get out of the way?  Secondly, what physical process causes the drop in pressure behind the turbine?  And thirdly, why should nature require a slow down in the air flow once it has passed beyond the turbine, anyway?  Nature acts to keep the air moving, not slow it down.  I would be interested in any thoughts about the issue.  

[scottsAI]

finsawyer,

F=ma

Work is force applied over a distance.

Power is the rate of work. P = F * V

To extract power a pressure difference across the blades MUST exist, this creates the (de)acceleration needed to get Force. Considering hydro, nobody seems to have a problem understanding how it works.

Wind is much the same, the compressibility of the air causes confusion.

The pressure must be the lowest at the back side of the blades. Pressure increases as you move away from the back of the blades back up to ambient, only way it could work. Since the air has been decelerated it must be moving slower! QED.

Solidity of the blades restricts air flowing through thus forming on the Front side of the blades a dome shaped high pressure zone, Just as on the back side is a cone (dome) shaped low pressure zone. I see many diagrams like your second link. It defies logic. Low pressure area will be filled by a higher pressure.

The first picture looks right, if a motor driven propeller the upside would be low pressure and would be sucking in the smoke trail. Here it looks like it's bumped up, just like I would expect from a wind generators. The extra air not going through the blades must go around them. This causes a turbulence.

Have fun,

Scott.

[thefinis]

It is all magic just like electricity.

Like Scott touched on the compressibility of air throws a big wrench in making this easy to figure. Water goes through or goes around or speeds up but air as the pressure increases decreases in volume for the same mass. It is kind of like a water hose with 10 lbs of pressure and one with 30 lbs of pressure. The 30 lb will flow more water but the water moves faster. Now with air the air stacks up (increase in pressure and mass per volume) and flows through the blades to a lower pressure area behind the blades but as it does it speeds up (pressure differential) and then expands in the lower pressure area. In effect it is moving more mass at same speed through the same area.

How the heck the blades suck power out of the flow is for someone else to explain cause it still seems like magic to me. Okay I'll try the blades get the power from being shoved out of the way so the air can pass through and the speed and the density of the air are part of the formula for how hard it shoves the blades.

Somehow the Betz limit comes in when figuring in how much air you can shove through an area while slowing it, with only the surrounding air moving at a set speed and density to contain it.

Heck finsawyer are you just stirring the pot. You are one of the old timers here. I tried to keep my reply simple so some of the newer ones might understand. Did I miss your point for posting?

Finis

[finnsawyer]

I don't think it's magic and Scott's answer just begs the question.  Why is the air at lower pressure on the back side?  Here's my take on it.  We need to take a walk on the windy side.

There was some talk lately about putting wind turbines in wind tunnels.  It was generally considered a bad idea to make the size of the WT such that it filled up the entire wind tunnel.  But what if we did?  That is, we construct a closed wind tunnel with a circular cross section.  In such a tunnel, neglecting friction losses, the velocity must everywhere be the same and the pressure does not change either.  We now place a WT in the middle of the tunnel such that the blades extend right to the wall.  The fact that the WT is in the tunnel does not change the requirement that the mass flow of air through the tunnel must still be the same at all planes at right angles to the tunnel including as it passes through the blades of the turbine, even though we could have changes locally in the velocity profile.  We may, for instance lock the blades so they can't rotate.  There will then be a pressure differential between the front of the blades and the back due to the resistance to flow presented by the blades.  The action of the wind tunnel fans will lower the pressure behind the blades and raise it in front of the blades.  There will also be some rotational torque or force on the blades due to their pitch and angle of attack.  If we let blades rotate and connect the turbine to a generator can we get power out?  You bet.  This may not be obvious due to the fact that the air velocities are the same some distance in front of the blades as some distance behind the blades.  The velocity cubed will be the same at both ends of the tunnel.  However we must recognize that the internal energy of the air has changed.  The air in front of the WT satisfies Bernoulli's Equation with one constant and the air behind the WT with a different lower constant.  There is a change in energy due to the pressure differential.  We may write the power change due to the air flow through the turbine as:

     (P1 - P2)xAxV.  

P1 is the pressure before the WT, P2 the pressure after the WT, A the area of the WT or the wind tunnel, and of course V the velocity at either end of the tunnel where the air flow is smooth.

We now consider the case of a WT in a pipe as before but now in a uniform air flow such as a wind field.  The resistance of the turbine to air flow will cause a pressure build up in front of the turbine that will extend to the upstream end of the pipe.  There the pressure will cause some air to be deflected around the pipe and will cause a speed up in the air flow along the outside of the pipe. This faster flow will have a reduced pressure, which when the flow proceeds past the back end of the pipe will result in a reduced pressure behind the WT.  This is the "Venturi" effect.  This is the cause of the reduced pressure behind the blade plane.  We may then make the pipe shorter without significantly changing things too much until the pipe length equals the width of the WT's rotor.  After that we enter a new regime, the one where Mr. Jacobs flinging of the air off the end of the blades becomes possible.

So, the up shot of all this is that the air deflected around the WT causes a speed up of the air flow past the turbine.  This faster low pressure flow creates a lower pressure region behind the turbine.    

[scorman]

you do realize that Betz came up with his formula in the early 1900's and that his mathematical analysis was based on a single simple solid disc blocking the wind?

it may hold for a single rotor, but two or three rotors in different planes ..all bets (pun intended) may be off

[finnsawyer]

I don't think the analysis of Betz, at least what I've seen presented, mirrors reality very well.  In point of fact, the speed of the air leaving the rotor could have any value as long as the faster moving air going past the rotor satisfies the rest of the mass flow equation.  That is, the speed up of the air deflected around the rotor just compensates for the slower flow behind the rotor.  Such a condition could extend out to infinity, but won't since there will be a shear in the air flow behind the rotor tips.  While such a shear can exist for some distance, eventually the two flows will mix and go to some average value or may even result in turbulence.  The inertia of the air flowing past the rotor will act somewhat like a pipe wall and keep the air behind the rotor from expanding outward.  In fact the pressure distribution in that flow will be such as to cause the flow to expand inward.

I don't like the flinging part of Mr. Jacobs' statements.  To me this implies an air flow outward off of the tips.  Some air escapes before it passes over the entire length of the blade.  So, some power is lost.  On the other hand, such a flow will reduce the pressure behind the rotor since the flow rate or speed past the rotor must increase to accommodate it.  It might turn out to be a wash.  In any case, I can see an argument for a ring or hoop around the rotor to stop such a flow.  In that case the speed of the air through the rotor can be assumed constant.  By assuming different ratios for the air diverted around the rotor the pressure in front of the rotor can be determined from Bernoulli's Equation.  The question then becomes, "What is the resulting speed of the air flowing past the rotor and what would be the pressure behind the rotor?"  Or to put it another way, "What would be a reasonable assumption for that flow and pressure?"

This pressure change power equation has been a paradigm shift more me.  I had a problem seeing how one could get the power out from that pressure differential.  Now it's clear.  The ultimate object of this is to determine how to design the blades.  A pressure difference across the rotor will act to speed up the air as it moves across the blades.  So, the blades must remove energy from the air flow at just the right rate to keep the speed constant.  At least that's the ideal case.  Making the blades wider will increase the pressure in front of the rotor and spill more air.  It will also increase drag.  Where is the "sweet spot"?

Well, that's as far as my thinking has got me now.  We'll see what the future brings.  

[richhagen]

Just thinking here, nothing firm -

Well, you are removing kinetic energy from the air, slowing it down along the axis of the velocity of the wind.  It stands to reason that for far down stream, if the density ends up the same as that of the surrounding air on the backside of the turbine, it would have to occupy more area in a plane perpendicular to the velocity of the wind that is larger in an amount that is a function of the resultant reduction in velocity along the axis of the velocity of the wind.  

Of course this may not look like a simple picture of a cone as there is probably a bit of turbulence and mixing going on there as the air that passed through and around the swept area moves away from the back of the turbine.  

Anyway, just some thoughts, and there are probably multiple related accurate ways to look at this that should reach the same conclusion.  Rich

[electrondady1]

fin,

thanks for sharing your thoughts on this subject

i find it surprising that there has been no examination and no hard data available on it.

all it would require is a functional windmill and a large number of sensors

arrayed around it both upstream and downstream .

comparing real world variations in air pressure and velocity.

beyond the scope of farmers from Michigan or landlords from Ontario.

but i thought, "raison d'etre" for sandia labs.

(it's recently come to my attention that they are heavily involved in research on rail guns)

oh well,

i have always been under the impression that when the  velocity of a column of air is reduced the static pressure increases.

and that when velocity  is increased air pressure decreases.

i follow your reasoning on how the air in front of the rotor must speed up in order to

bypass the restriction .

but shouldn't the air down stream of the rotor with reduced velocity have an increase in pressure ?

[finnsawyer]

"i have always been under the impression that when the  velocity of a column of air is reduced the static pressure increases.

and that when velocity  is increased air pressure decreases.

i follow your reasoning on how the air in front of the rotor must speed up in order to

bypass the restriction .

but shouldn't the air down stream of the rotor with reduced velocity have an increase in pressure ?"

As I mentioned, the lower pressure behind the rotor is due to the Venturi effect.  When a tube of air is moving, the pressure inside the tube will equal the pressure of the moving air stream unless something acts to disturb it.  Well, air flowing through the rotor will act in that fashion, but since the flow is fairly low it may not have much effect.  In any case a balance will be reached.  A rise in pressure means that the surrounding flow must slow down.  The fast stream must give up kinetic energy.  In other words the fast flowing tube of air performs the same function as the fans do behind the rotor in the case of the closed wind tunnel system.  It pumps down the pressure behind the rotor to some degree.  The energy in the air being deflected is not completely lost but acts to sustain that lower pressure, which in turn helps determine the power output.

The air entering the rotor already has a reduced velocity, but it is at a higher pressure.  In fact, ideally its velocity doesn't change as it passes through the rotor.  Now, to keep the velocity the same but to go from a state of high pressure to a state of low pressure the internal energy of the air must become less.  That's where the blades come in.  As the air passes over the blades, energy is imparted to the blades at some rate, which we recognize as output power.  We immediately recognize that this will not be a perfect process.  For one thing air foils do not exhibit a uniform lift force over their entire width.  Left to their own devices the air foils will turn over.  I hope this explanation is of some help.      

[finnsawyer]

Well, it's time to add some more.  By making some (reasonable?) assumptions it is possible to calculate the maximum power ratio.  We start by recognizing that in the undisturbed air in front of the WT we may write according to Bernoulli's Equation that:

       Pi + .5xpxv1^2 = C1,

where Pi is the pressure of the undisturbed flowing air, p is its density, and V1 is the its velocity.  We assume an ideal rotor such that the velocity of the air passing through the rotor is everywhere kxV1.  Since this reduced flow as it enters the rotor has done no work it must satisfy the same form of Bernoulli's Equation.  That is we can write:

       P1 + .5xpx(kV1)^2 = C1 = Pi + .5xpxV1^2.

Solving for P1 we get:

       P1 = Pi + .5xpxV1^2x(1 - k^2).  

For the deflected part of the air flow we recognize that if all the flow were deflected, the speed of the air passing immediately outside the rotor would be increased by 50%.  We assume the relationship is linear.  Since a fraction of

(1 - k)xV1 is deflected, then we assume the speed up is 1/2 that fraction.  So, we get:

       V2 = V1x(1 + (1 - k)/2) = V1x(3 - k)/2.

Again no work has been done.  So, we may write from Bernoulli's Equation:

       P2 + .5xpxV1^2((3 - k)/2)^2 = Pi + .5xpxV1^2.

       P2 = Pi + .5xpxV1^2x[1 - ((3 - k)/2)^2.

We may then write for P1 - P2:

       P1 - P2 = .5xpxV1^2x[1 - k^2 - 1 + ((3 - k)/2)^2].

The output power from the ideal rotor is given by Pout = (P1 - P2)xkxAxV1.  So, we may write pout as:

      Pout = .5xpxAxV1^3xkx[((3 - k)/2)^2 - k^2].

Expanding the quadratic:  

      Pout = .5xpxAxV1^3xkx[9/4 - 3k/2 + k^2/4 - k^2].

Simplifying:

      Pout = .5xpxAxV1^3x(9k/4 - 3k^2/2 - 3k^3/4).

The incident power  is given by: Pin = .5xpxAxV1^3.

The power ratio PR = Pout/Pin = 9k/4 - 3k^2/2 - 3k^3/4.

The value of k for which this is a maximum is found from differentiating PR with respect to k and setting the result to zero.

      d(Pr)/dk = 9/4 - 3k - 9k^2/4 = 0.

One solution for this equation is k = .535.  So, 53.5% of the incident air flows through the rotor.  The maximum value of PR, that is, the best possible efficiency, then is equal to:

      PRmax = 0.6596.

This is about 10% greater than the Betz Limit.

[finnsawyer]

If we use a value of k = .85 in the expression for PR = 9k/4 - 3k^2/2 - 3k^3/4, we get PR = .368.  What this may be telling us is that if we design our blades with the assumption that all of the incident air passes through the blades, we may in effect be forcing the wind turbine to operate at the high air flow low efficiency part of the power transfer curve.

[TAH]

I spent a lot of time looking at the jib effect on a sail. From all that I could tell the main gain seemed to be from increasing the effeceincy of the sails shape. A sailboat with a solid wing sale looked like it had about the same lift as a sail/jib combo

[finnsawyer]

I can see that with the main sail the mast helps set up the proper air flow to cause the sail to billow out.  That is, the leading edge of the sail will tend to move toward the low pressure at the side of the mast giving an overall air foil effect.  A jib doesn't or needn't have that large circular structure at the leading edge.  Of course sail boats are quite limited in how close they can sail to the wind.  Which brings me to ice boats.  How do you think an ice boat would do with just a jib?

[TAH]

I looked at ice boats too. If I remember right they can do up to something like 10x wind speed. I think that is mostly gain from having very little load compared to a displacement hull that most sail boats have. I think I decided in the end that the jib mostly shaped the wind to make the main work better. The mast and the tightness of the main make a big difference. The fastest sail boats have solid sails that are shaped like a wing. I don't think I ever saw a jib with one of these. I don't know if it would make difference there. I started putting together some double blade setups to see if there was a difference and I didn't find any, but I was just working with two stacked props so the tips were further apart and closer at the base. To make it work like a jib I am pretty sure it should be the other way. The other thing is the the alignment of the props. The jib is downwind of the main but has a leading edge ahead of it. This would mean the front prop would be bugger and slightly behind the jib prop.

[finnsawyer]

I mentioned the ice boats, because I thought their behavior would be closer to what actually takes place with a wind turbine.  My thought with the jib on the ice boat would be only with the jib.  What would the performance be?

Your idea of the two sails together leads into some thoughts I've had.  The wind tunnel tests with airfoils are essentially static in that the air foil doesn't move and does no work.  In the case of the sail boat the speeds are slow enough that the wind tunnel tests are close enough.  For the ice boat perhaps not.  For the wind turbine, definitely not.  The blades are moving quite fast and doing work.  The motion of the blades is such that they will always be tending to compress and slow down the fast moving air stream over the convex surface.  There will also be a tendency to reduce the pressure on the flat or concave side.  Well, something has to change, since power is being produced.  That conservation of energy thing.  One would expect the air flow to slow down.  That's where the pressure difference or gradient comes for the wind turbine.  That pressure gradient acts to keep the air flow moving.  This changes the whole paradigm for the blade design.  Maybe what happens with the output turbine of a jet engine would be closer to what we want.  Maybe the wake considerations don't apply at all.  A multibladed turbine with curved blades and perhaps even a knife edge leading edge might actually give better results.  I doubt I would ever be in a position to test any of this.  But the object must be to pull energy out with the blade and tend to slow down the air flow at just the right rate so that the tendency to speed up the flow due to the pressure differential just balances.

[scorman]

GeoM,

I have spent a considerable amount of time researching this topic.

Yes the iceboat has the advantage of virtually no friction vs a sailboat sliding in water ..a catamaran is a planing hull and moves a lot faster than a standard displacement hull ...friction again

" My thought with the jib on the ice boat would be only with the jib."

a smaller single sail ie the jib will have less lift ...period!

To visualize what may be happening for wind turbines, I have used JavaFoil which is a GUI modeling program

Here is the flow diagram of a single NACA4425 airfoil at an AOA of 6 degrees:

Here is the same airfoil doubled up as a "jib/mainsail"

note the small window showing numbers for Cl and Cd ..also note the flow pattern in between the two blades and the point at which the lower blade would start to show "flow separation" on the topside ( hint ..it is laminar in between the blades)

I have a lot more of these images (and higher res) at diff AOA and diff spacings/lead of multiple blades:

http://s145.photobucket.com/albums/r203/scorman1/blade%20design/flow%20diagrams/

That JavaFoil program generates Cl,Cd and L/D in chart form for any AOA and RE#:

http://www.otherpower.com/images/scimages/7526/NACA4425polar.txt

In the article I posted by Arvel Gentry on this subject, he discusses similar effects.

I had him on the phone for 3/4 hour and he claimed they did experiments on lift on up to 8 stacked wings.

But what does he know? He was only the teamleader on the project that designed the wings of the space shuttle and an avid racing sailor!

The only attempt I have seen on implementing this type of multiple rotor was Bill Allison back in late 70's:

http://s145.photobucket.com/albums/r203/scorman1/Wind/Allison%20article/

I am NOT in total agreement with Allison's implementation or conclusions, but generating real data will be the bottom line. As I mentioned before, I am going with a three blade x 2 rotor design.

If you have specific qs on any of the above ..fire away

Stew Corman from sunny Endicott

BTW, Betz's argument is a very simple one ...he started with a simple equation for flow of air through a disc area ..took the first derivative, and then solved for the peak  at slope = zero, which is the maximum efficiency.

If you don't like the equation, then you won't like the answer.

[TAH]

One issue with your simulation is that the jib is normally not the same shape or size as the main. The slip between is also adjust depnding on the wind angle and conditions. A turbine with a jib blade would always be tacking at the same angle to the wind so the slip and size of the jib blade could be fixed.

[scorman]

TAH,

you just hit on the complexity of the design of a multi-rotor

Instead of just pitch angle for a particular planform and profile, you have the gap spacing between rotors, and the index angle which determines the lead.

If you look at my link to the other flow diagrams, you'll see that pinching the gap gives more drag than lift!

My conundrum is compounded by the fact that the modeling may NOT be accurate at all.

Those flow diagrams are for the 75% station only, since the gemometric spacing varies with radial distance.

If you did a search on Allison's patents, his first one was for a conventional airfoil,

but the one shown at a later time in the article is a triangular blade. The polar data for that design is terrible ..yet it flew!!

BTW, there are high performance jet wings that use a jib style leading edge for enhanced lift ..I can provide references.

Stew

[finnsawyer]

You seem to have missed my comment about the fast moving air foil affecting the air flow when it is doing work.  Can JavaFoil deal with that?  In the extreme, all the work done on the air foil due to the pressure differential would end up going into countering drag.  We need a solution somewhere in between, where we allow some external work to done.  It would be interesting to see that kind of simulation.  I suspect the flow lines would break down very quickly for say a TSR of 7.  That's where the pressure differential across the wind turbine would come into play.

As far my analysis is concerned versus Betz', I can only go on what I read.  My Model doesn't need a spreading of the air flow down stream.  Half the flow is through the turbine.  The rest goes past.  There is no problem with the mass equation.  Both flows could continue down stream indefinitely.  Of course, they won't.  But that's no concern.  What matters is what happens at the turbine.  Mathematically, to find the maximum or minimum of a curve you differentiate with respect to the variable of interest and set the result to zero.  That's what I did.  My result was 10% higher than Betz'.  Well, nice if it holds, but it may not.  The assumption that the speed of the air flow just beyond the radius of the rotor, which determines the pressure behind the rotor, satisfies a linear equation may be faulty.  Rather than being linear the curve may belly up or sag down.

As far as measurements are concerned, there is no reason that one couldn't measure the pressure distribution in front of, along side of, and behind the rotor.  A multi stage manometer would be useful for that.  Those are expensive, but one could build their own.  We did stuff like that on a summer job I had one time.

[finnsawyer]

"comparing real world variations in air pressure and velocity.

beyond the scope of farmers from Michigan or landlords from Ontario."

I don't think it is, anymore.  To measure pressure all you would need would be what I call a differential manometer, which can be made from some clear tubing and water.  At 15 pounds air the height of a water column that it can support is about 34.5 feet.  But we only want to measure a pressure differential, the greatest value of which would be when we go from say a wind speed of 15 mph (22 feet per second) to either a dead stop or a speed up of about 50%.  My analysis using Bernoulli's Equation indicates that a reasonable length U shaped tube and a few yardsticks should do.  I invite any and all to run the numbers.  It should work as long as the wind speeds are fairly low.  But the theory is independent of wind speed.  I suppose one would want to use rigid tubing at the pressure sensing ends of the device.