Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[rcgldr]

variable propeller pitch

A propeller uses an AOA versus prop disk that decreases as radius increases, normally set so that the propeller pitch (the effective advanced distance per revolution) is the same at all radius (except for the hub).

It might be easier to figure the math out if you knew the pitch numbers for the BB propeller. There's a physical pitch based on geometry, and the effective pitch based on air flow. Some props have the pitch measured in a static flow (no headwind) situation, but there is an induced headwind in the vicinity of the prop. Some simplified static thrust calculators ignore the geometrical pitch of a prop. If there was a headwind speed that corresponded to the geometric pitch times revolutions per unit of time, then the propeller would not increase the air speed except for the twisting (torque) of the air flow.

- - -

In the meantime, I've run into a math problem trying to determine a limit. For an ideal (no losses) sailcraft that diverts apparent wind directly aft of the sailcraft (diverting it to become an apparent headwind), the limit of the flow aft of the sailcraft wrt ground approaches wind speed x cos(θ) as the sailcraft speed approaches infinity.

Here are the formulas:

θ = sailcraft heading wrt wind. θ = zero means directly downwind.
v = sailcraft speed
w = true wind speed
aw = apparent wind speed (wrt sailcraft)
da = diverted apparent wind (the wind flow off the aft end of the ideal sail) wrt sailcraft
dw = diverted apparent wind wrt ground

ac = apparent cross wind = w sin(θ)
ah = apparent head wind = v - w cos(θ)

aw = sqrt(ac² + ah²)
aw = sqrt((w sin(θ))² + (v - w cos(θ))²)
aw = sqrt( w² + v² - 2 w v cos(θ) )

after the idealized sail diverts the apparent wind, it is in the same direction as the sailcraft, so da has the same magnitude as aw, just a different direction.

da = sqrt( w² + v² - 2 w v cos(θ) )

The speed of the diverted wind relative to ground is

dw = v - da = v - sqrt( w² + v² - 2 w v cos(θ) )

using a spreadsheet to test the formula, it turns out that

limit v -> ∞ of dw = w cos(θ)

However I'm not able to directly solve this limit.

Continuing, the downwind component of dw = dw cos(θ), so

limit v -> ∞ for dw cos(θ) = w cos²(θ)

The true wind is slowed down by w - dw cos(θ), so

limit v -> ∞ for w - dw cos(θ) = w - w cos²(θ) = w sin²(θ)

This limit is approached from above, with slightly higher numbers at lower speeds. The point of this is that the true wind is slowed by at least wind speed x sin²(θ) regardless of the idealized sailcraft speed. This is getting back to the point that for an ideal sailcraft, there is some finite power input but zero power consumption, so there's no mathematical limit to the sailcrafts speed.

 

[RonL]

If it has not already been considered, the building of counter rotating props will eliminate some torque problems, double the math fun and maybe add just a little more efficiency. Make the lead prop about 10% smaller than the following prop, then consider a slightly adjustable mast that let's angle be adjusted (a little like timing a brushed electric motor). Just a couple of quick thoughts.

Ron

 

[OmCheeto]

If it has not already been considered, the building of counter rotating props will eliminate some torque problems, double the math fun and maybe add just a little more efficiency. Make the lead prop about 10% smaller than the following prop, then consider a slightly adjustable mast that let's angle be adjusted (a little like timing a brushed electric motor). Just a couple of quick thoughts.

Ron

Oh dear hey-Zeus...

Dear Ron,

I'm trying to simplify the math, not make it more complicated.​

sincerely,

Om

 

[RonL]

Oh dear hey-Zeus...

Dear Ron,

I'm trying to simplify the math, not make it more complicated.​

sincerely,

Om

Sorry Om, I was trying to not get carried away. Could have mentioned what has already been said about leverage of energy and suggested use of transfer to a center opening in the prop(s) where a high speed impeller produces a high velocity air discharge which en-trains air flow from the larger prop area, giving it a more low pressure zone to move into. The thrust through the system would not meet resistance from the surroundings as quickly.

Some people love math wish I could do better at it, but I am making some progress

I better quit

Ron

 

[RonL]

I designed and built the thing, and you've lost me. What heat is being converted to work?
Yes he did. He's also done some great analysis and some really nice animations.

Sorry to not answer sooner, I have a hard time converting my thoughts into words that make sense to most people. Here is what my mind is seeing, the weight of the cart and it's motion is converted into energy to drive the propeller, the air resistance in front of the propeller is reduced in an increasing lower value as RPM's go higher (in my mind a slight vacuum), the thrust generated by the propeller into the forward moving wind is in a slightly compressed state (warmer) and by some standard a more solid mass and as it moves away from the spinning propeller it should start to appear somewhat like a mushroom (if it could be seen) this increases the area of applied force by some amount.
In my thoughts (within reason) a heavier cart might actually produce a stronger energy transfer, provided wind friction area can be held to a minimum, resulting in a faster speed.

Hope this makes some sense to someone, would it fall into a (Non-Equalibrium Thermodynamic system) ?

Ron

 

[rcgldr]

Here is what my mind is seeing, the weight of the cart and it's motion is converted into energy to drive the propeller.

A summary of earlier posts.

From a ground frame of reference, the thrust from the propeller is slowing down the wind, and slowing down the wind is the source of power that propels the cart. When the cart is accelerating, the increase in it's kinetic energy is part of the power consumed from the power sourced by slowing down the wind.

A backwards force exerted from the ground to the wheels creates a torque that the wheels use to drive the propeller. There's an effective gearing between wheels and propeller that multiplies force and divides speed. This only works if there's a tailwind since the propeller thrust speed is less than the ground speed, relative to the cart.

 

[ThinAirDesign]

In my thoughts (within reason) a heavier cart might actually produce a stronger energy transfer, ...

Nope. All other things being the same, a heavier cart would be slower.

JB

 

[OmCheeto]

Ok, I simplified the FBD to the point where I hope those like me* can see where the offset wheelbase comes from:

the left up arrow indicates the force imparted onto propeller by the wind.

the right down arrow indicates the reaction force of the wheel-propeller interface which has to be resisted.

ergo, there is a counterclockwise torque on the vehicle.

this is why no one should take me seriously when I only have about 30 minutes to put together diagrams and thoughts.

*idiot!

 

[rcgldr]

Ok, I simplified the FBD to the point where I hope those like me can see where the offset wheelbase comes from.

The offset wheelbase was done to widen the wheelbase on the left as roll prevention from the counter clockwise torque from the air. The wheelbase could have been extended equally on both sides and would provide the same roll prevention (actually a tiny bit more due to the small amount of mass in the wheel moved to the right). One of the posts mentioned that the wheelbase was too close to the legal limit on width for a towed vehicle, and another post mentioned that the right axle wasn't extended simply because it wasn't needed.

Any torque from the wheelbase, whether offset or equal, is a reaction to a torque applied to the wheelbase from the vehicle. The reason for a widened wheelbase is to prevent the vehicle from rolling over onto it's side due to the vehicle torque.

 

[mender]

I thought that one of the main reasons was to equalize the dynamic load on the rear tires for better traction and more even wear.

 

[RonL]

I thought that one of the main reasons was to equalize the dynamic load on the rear tires for better traction and more even wear.

I think things are going on that may not have been considered, weight is key to traction, and energy transfer is taking place between traction and propeller resistance. This torque resistance will tend to lift one side and add pressure to the other and at speed, I suspect if some way to measure each wheel for turns, you will find they will all have a different reading.
Because of energy transfer the drive wheels will slip just a little during each turn, based on how much actual force is pushing down on them.
Omcheeto's drawing has the qualities of a true Professor.

Ron

 

[ThinAirDesign]

I thought that one of the main reasons was to equalize the dynamic load on the rear tires for better traction and more even wear.

While those were side bennies, the sole reason was with our higher than expected efficiencies (and thus speeds) we almost tipped it over in Ivanpah.

JB

 

[ThinAirDesign]

Omcheeto's drawing has the qualities of a true Professor.

Ron

Perhaps -- but it's still quite nebulous at best and most likely just plain wrong.

JB

 

[mender]

While those were side bennies, the sole reason was with our higher than expected efficiencies (and thus speeds) we almost tipped it over in Ivanpah.

JB

Must have missed that!

(edit)

 

[ThinAirDesign]

Must have missed that!

(edit)

http://www.fasterthanthewind.org/2010/04/so-heres-short-version-of-story-behind.html

 

[OmCheeto]

Perhaps -- but it's still quite nebulous at best and most likely just plain wrong.

JB

I agree. Kind of.

 

[RonL]

Perhaps -- but it's still quite nebulous at best and most likely just plain wrong.

JB

Before I say anything else, I must make it clear that to an extreme, I am IMPRESSED and RESPECTFUL of what has been done here and say well done to all involved.

If it sounded as if I was implying a lack of thought by all or anyone involved, please forgive, it was in no way intended. The attention to wheel hub design, shows clearly a very high level of thought processing.

My nebulous thoughts may be wrong and out of context, they are for the most part based on 40 years of contract jobs and sales of skid steer loaders, where spinning, sliding, skidding and hydraulic levered load increase to eliminate slippage, have all been put to use in making a machine utilize a maximum efficiency of energy transfer.

I'll ease out now and return to my own compressed air/electric build and research.
Thanks for the u-tube releases, they have confirmed many of the things I could never say in good descriptive wording.

Best wishes

Ron

 

[ThinAirDesign]

Best wishes

Ron

Ron, no offense was taken and I apologize if it appeared so. I was merely commenting on the fact that without clearer definitions, Om's drawing was rather ambiguous.

As to the wheel RPMs that you mentioned, there is no differential in the BB and thus under most normal circumstances the RPMs will match exactly. There are ratchets on the drive axle which allow us to maneuver the cart around without lockup, but again this in straight operation will not allow one driving wheel to rotate faster than the other.

I love skid steers (I own a tracked version - ASV RC-85).

JB

 

[RonL]

Ron, no offense was taken and I apologize if it appeared so. I was merely commenting on the fact that without clearer definitions, Om's drawing was rather ambiguous.

As to the wheel RPMs that you mentioned, there is no differential in the BB and thus under most normal circumstances the RPMs will match exactly. There are ratchets on the drive axle which allow us to maneuver the cart around without lockup, but again this in straight operation will not allow one driving wheel to rotate faster than the other.

I love skid steers (I own a tracked version - ASV RC-85).

JB

Thanks, I wasn't sure as I thought the drawing was more clear than my comments If everything is OK I'll leave well enough alone.

I left equipment sales about the time mini-excavators started showing up in the US with rubber tracks, the rubber tracks caught on quickly with the skidders.
I made a move into heavier equipment, but can never think of being without a small do almost anything machine, my motto, "being without a Bobcat, is like working with a broken arm", at least if one is outside doing more than simple yardwork. (guess I'm still a "black and white" guy). I did look up the machine you mentioned, it looks like a really productive unit.

I'll keep a watch for new and improved progress in the wind machines.

Ron

 

[Llyricist]

One more of the many esoteric things I learn from this discussion. I had never heard of a skid steer before, now I know what one is.

 

[spork]

I did look up the machine you mentioned, it looks like a really productive unit.

I can't speak to how productive it it - but it's a friggin' blast to drive it up and down steep slopes in the woods!

I'll keep a watch for new and improved progress in the wind machines.

I posted an update to our blog in the last day or two ( www.fasterthanthewind.org ). I've got 4 more days worth of glassing the turbine blades (final coat) before finishing and painting. Then I've got a bit of transmission work to do. As you may have guessed we're not doing anything like the pace we did last year. JB is on the road full-time, and I'm squeezing my efforts in between work and flying. But there's MUCH less to do. I still hope to do some runs in the next several weeks.

 

[RonL]

I can't speak to how productive it it - but it's a friggin' blast to drive it up and down steep slopes in the woods!



I posted an update to our blog in the last day or two ( www.fasterthanthewind.org ). I've got 4 more days worth of glassing the turbine blades (final coat) before finishing and painting. Then I've got a bit of transmission work to do. As you may have guessed we're not doing anything like the pace we did last year. JB is on the road full-time, and I'm squeezing my efforts in between work and flying. But there's MUCH less to do. I still hope to do some runs in the next several weeks.

Hi spork,

http://www.skidsteerhistory.com/How it began.htm

The joy of being paid to work with a machine like the skid loader, is hard to put in words. My first machine was the 1968 M-610 Bobcat, I have witnessed the actual life of these machines and looking at what has developed from those early years until now is hard to believe.
I won't live to see the results of what you guy's are doing, but I think you are the Keller's of wind powered (or assisted) transportation.

Ron

 

[A.T.]

I posted an update to our blog in the last day or two ( www.fasterthanthewind.org ). I've got 4 more days worth of glassing the turbine blades (final coat) before finishing and painting.

Will you post the blade geometry anytime soon?

 

[spork]

Will you post the blade geometry anytime soon?

I'd be happy to. I didn't realize anyone wanted to see it.

c Upwind Car SJM 2X 10 mph wind
vc=(2.0+1)*10*1.467 !vo (f/s)
omega=150 !RPM
rho=.00226 !Density (slug/ft^3)1500 ft MSL
vsnd=1116.5 ! Speed of sound (f/s)
rt=9.25 !Tip radius (ft)
rh=0.5 !Hub radius (ft)
nblade=2 !Blade number
ao=5.73 !Lift curve slope of 2-d section
apzl=-5.0 !zero lift angle of attack (deg)
ld=67 !Section max L/D (6412 @350000 Re)
cl=0.9 !CL at max L/D

! Thrust= 149.975052(lbs) Torque= 351.829559(ft-lb) Power= 10.0482292(HP)
! etat= 0.837301373

rad(ft) chord(ft) Beta(deg) thick(%)
0.937500 1.015620 73.071310 18.490000
1.112500 1.096130 69.482840 17.130000
1.287500 1.176640 66.095340 15.960000
1.462500 1.257150 62.914090 14.940000
1.637500 1.337660 59.938620 14.040000
1.812500 1.418170 57.164120 13.240000
1.987500 1.498680 54.582660 12.530000
2.162500 1.579190 52.184180 12.000000
2.337500 1.662220 49.957450 12.000000
2.512500 1.726720 47.890720 12.000000
2.687500 1.774310 45.972170 12.000000
2.862500 1.806830 44.190270 12.000000
3.037500 1.826130 42.534050 12.000000
3.212500 1.833980 40.993140 12.000000
3.387500 1.832030 39.557910 12.000000
3.562500 1.821790 38.219500 12.000000
3.737500 1.804570 36.969740 12.000000
3.912500 1.781530 35.801190 12.000000
4.087500 1.753660 34.707050 12.000000
4.262500 1.721830 33.681180 12.000000
4.437500 1.686740 32.717960 12.000000
4.612500 1.649010 31.812360 12.000000
4.787500 1.609130 30.959730 12.000000
4.962500 1.567530 30.155920 12.000000
5.137500 1.524530 29.397150 12.000000
5.312500 1.480420 28.679970 12.000000
5.487500 1.435410 28.001270 12.000000
5.662500 1.389660 27.358200 12.000000
5.837500 1.343310 26.748190 12.000000
6.012500 1.296450 26.168870 12.000000
6.187500 1.249120 25.618090 12.000000
6.362500 1.201350 25.093890 12.000000
6.537500 1.153150 24.594480 12.000000
6.712500 1.104470 24.118210 12.000000
6.887500 1.055260 23.663560 12.000000
7.062500 1.005430 23.229150 12.000000
7.237500 0.954870 22.813710 12.000000
7.412500 0.903410 22.416060 12.000000
7.587500 0.850840 22.035120 12.000000
7.762500 0.796880 21.669880 12.000000
7.937500 0.741180 21.319430 12.000000
8.112500 0.683230 20.982910 12.000000
8.287500 0.622320 20.659520 12.000000
8.462500 0.557410 20.348540 12.000000
8.637500 0.486800 20.049270 12.000000
8.812500 0.407420 19.761090 12.000000
8.987500 0.312520 19.483400 12.000000
9.162500 0.178680 19.215650 12.000000

 

[chingel]

I think it is important to understand that it's not just that the propeller is driven by the wheels, it's that if you engage the propeller the wind starts pushing harder on the propeller and the extra wind push drives it forward.

If you engage the propeller at windspeed, it will use as much energy as it puts drag on the wheels, no matter if you multiply the force or not, but the important thing is that the wind will start pushing harder, because it also has to push on the screwing propeller blade.

 

[spork]

I think it is important to understand that it's not just that the propeller is driven by the wheels, it's that if you engage the propeller the wind starts pushing harder on the propeller and the extra wind push drives it forward.

If I were in a motorglider and gliding directly downwind, would you say the tailwind pushes harder on the propeller when the prop is engaged? How about if I were gliding upwind when I engaged the prop?

The motorglider and its prop would see no difference at all (tailwind, headwind, or no wind).

Our cart operates exactly as the motorglider in a tailwind. Th spinning prop interacts with the air it's in in exactly the same way as an airplane prop in flight interacts with the air it's in.

 

[A.T.]

...it will use as much energy as it puts drag on the wheels,...

Comparing energy with a force makes no sense.

...no matter if you multiply the force or not,...

It does matter. Without multiplying the force, the propeller thrust cannot be greater than the wheel drag.

 

[Llyricist]

I'd be happy to. I didn't realize anyone wanted to see it.

c Upwind Car SJM 2X 10 mph wind
vc=(2.0+1)*10*1.467 !vo (f/s)
omega=150 !RPM
rho=.00226 !Density (slug/ft^3)1500 ft MSL
vsnd=1116.5 ! Speed of sound (f/s)
rt=9.25 !Tip radius (ft)
rh=0.5 !Hub radius (ft)
nblade=2 !Blade number
ao=5.73 !Lift curve slope of 2-d section
apzl=-5.0 !zero lift angle of attack (deg)
ld=67 !Section max L/D (6412 @350000 Re)
cl=0.9 !CL at max L/D

! Thrust= 149.975052(lbs) Torque= 351.829559(ft-lb) Power= 10.0482292(HP)
! etat= 0.837301373

rad(ft) chord(ft) Beta(deg) thick(%)
0.937500 1.015620 73.071310 18.490000
...
9.162500 0.178680 19.215650 12.000000

Is that still a NACA 64XX airfoil?

Nevermind, I see it referenced in there.

 

[A.T.]

I'd be happy to. I didn't realize anyone wanted to see it.

c Upwind Car SJM 2X 10 mph wind

Thanks. I just tested it quickly. And in theory it looks like you could reach 2x with:

transmission eff: 0.93
aero drag coefficient: 0.22
rolling drag coefficient: 0.01
frontal area[m^2]: 1.8
mass[kg]: 295

 

[spork]

Thanks. I just tested it quickly. And in theory it looks like you could reach 2x with:

transmission eff: 0.93
aero drag coefficient: 0.22
rolling drag coefficient: 0.01
frontal area[m^2]: 1.8
mass[kg]: 295

Thanks. That's what I'm hoping for (on the optimistic side). What prop efficiency did you use, and what speed ratio (i.e. gearing)?

 

[A.T.]

Thanks. That's what I'm hoping for (on the optimistic side). What prop efficiency did you use, and what speed ratio (i.e. gearing)?

I used a polar that Llyricist created for NACA6412 @450000 Re, so it not quite accurate. I took the gearing that gives 150RPM at 20mph. The inverted prop efficiency at 2xWS given by JavaProp was 81%.

Just a quick test, I will play around more when i have time.

 

[Llyricist]

I used a polar that Llyricist created for NACA6412 @450000 Re, so it not quite accurate. I took the gearing that gives 150RPM at 20mph. The inverted prop efficiency at 2xWS given by JavaProp was 81%.

Just a quick test, I will play around more when i have time.

If you go here:

https://skydrive.live.com/redir.aspx?cid=27579d7d745f6b79&resid=27579D7D745F6B79!103&authkey=qc!eaR1FyTA%24

You can get AF_6.xml, which is the 6412 at Re 350,000

Change the number after the underscore to match the next higher that you actually have.

 

[A.T.]

If you go here:

https://skydrive.live.com/redir.aspx?cid=27579d7d745f6b79&resid=27579D7D745F6B79!103&authkey=qc!eaR1FyTA%24

You can get AF_6.xml, which is the 6412 at Re 350,000

Thanks. It didn't change the results much though. But lowering the Cd to 0.2 and Crr to 0.006 brings you to 2.17 x windspeed.

 

[rcgldr]

The inverted prop efficiency at 2xWS given by JavaProp was 81%.

Isn't the new "prop" going to be used as a turbine to drive the BB upwind? If so, wouldn't the efficiency be limited by Betz law to about 59%? I'm thinking more like 50% overall.

 

[spork]

Isn't the new "prop" going to be used as a turbine to drive the BB upwind? If so, wouldn't the efficiency be limited by Betz law to about 59%? I'm thinking more like 50% overall.

The new "prop" is indeed going to be a turbine - but the Betz limit really doesn't pertain to efficiency per-se. Of course efficiency can be defined a million different ways, but for our purpose I think it makes sense to look at how much wind energy is converted to useful work (torque x rotary speed of turbine) divided by the total energy lost by the wind.

What Betz tells us is that a stream-tube of air of the same diameter as the turbine, can only give up 59% of it's energy to the turbine. This is because the capture area is smaller than the diameter of the turbine (i.e. some of the air ends up going around rather than through), and you can't bring the wind to a full-stop at the turbine (so the air in the wake of the turbine still has some energy).

 

[rcgldr]

The new "prop" is indeed going to be a turbine - but the Betz limit really doesn't pertain to efficiency per-se. Of course efficiency can be defined a million different ways, but for our purpose I think it makes sense to look at how much wind energy is converted to useful work (torque x rotary speed of turbine) divided by the total energy lost by the wind.

As mentioned in a previous post, I'm thinking of efficiency in terms of power output versus power input, and power as force times speed. Using the cart as a frame of reference, power input = relative air speed x force, power output = relative ground speed x force. If overall efficiency was 55%, then with a 2:1 air:ground effective gear ratio, then output force would be (.55 x 2 =) 1.1 x input force, so 0.1 x force factor left over to compensate for rolling resistance and aerodynamic drag with the BB cart moving upwind at 1 x wind speed.

I was thinking that Betz law applied even to open wind mill generators, which would be similar to the turbine usage on the BB cart, but perhaps I'm wrong on this.

 

[spork]

I was thinking that Betz law applied even to open wind mill generators, which would be similar to the turbine usage on the BB cart, but perhaps I'm wrong on this.

It does apply to open windmill generators, and does apply to the cart. But it's really not a measurement of turbine "efficiency" - or at least not a relevant measure of efficiency (since I suppose we can define it as a sort of efficiency if we choose to).

Betz law says that we can't get all the energy out of a specific reference volume of air. It doesn't limit how much of the extracted energy can be put to good use. The latter is the measure of efficiency that matters to us.

 

[rcgldr]

Betz law says that we can't get all the energy out of a specific reference volume of air. It doesn't limit how much of the extracted energy can be put to good use.

I'm wondering how relatively large the turbine will need to be in order to achieve 1x or greater upwind speed with the BB cart. The "efficiency" I'm wondering about is the power the turbine applies to the wheels to drive the cart and turbine against the wind, versus the turbine drag force times the relative (wrt cart) head wind speed.

 

[A.T.]

The inverted prop efficiency at 2xWS given by JavaProp was 81%.

Isn't the new "prop" going to be used as a turbine to drive the BB upwind?

Yes, it's the inverse propeller efficiency:

shaft_power / (turbine_drag * airspeed)

If so, wouldn't the efficiency be limited by Betz law to about 59%?

That's the limit for turbine efficiency defined in the in rest frame of the turbine:

shaft_power / total_kinetic_energy_flow_through_disc

I'm thinking more like 50% overall.

Actually the turbine efficiency of this rotor is much lower, around 12%. But this is not the relevant efficiency for an upwind cart. They are not trying to extract as much energy as possible from a certain volume of air, but rather to achieve the maximal speed. They want as much shaft_power as possible with as little turbine_drag as possible, while for a stationary turbine producing power the turbine_drag is irrelevant.

Imagine you could achieve 100% turbine efficiency and stop the air relative to the turbine. For an upwind car that means you would have not only slowed down the air relative to the ground, but accelerated it in the opposite direction, wasting energy on this. So maximal turbine efficiency is not your goal here. The inverse propeller efficiency is a better indicator, but the cart performance also depends on the other efficiency parameters.

 

[rcgldr]

It's the inverse propeller efficiency:
shaft_power / (turbine_drag * airspeed)

versus Betz law:
shaft_power / total_kinetic_energy_flow_through_disc

This is what I was wondering, if the power to the wheels versus the drag x airspeed could be greater than Betz's 59% factor. If the efficiency is 70% or higher, then greater than 1x wind speed upwind should be achievable with the BB cart.

 

[A.T.]

This is what I was wondering, if the power to the wheels versus the drag x airspeed could be greater than Betz's 59% factor.

Yes sure. For a very slow rotating turbine the inverse propeller eff. is mainly a function of the L/D ratio of the blades. But there are practical limits and minimum power requirements that have to be balanced against this.

 

[chingel]

If I were in a motorglider and gliding directly downwind, would you say the tailwind pushes harder on the propeller when the prop is engaged? How about if I were gliding upwind when I engaged the prop?

The motorglider and its prop would see no difference at all (tailwind, headwind, or no wind).

Our cart operates exactly as the motorglider in a tailwind. Th spinning prop interacts with the air it's in in exactly the same way as an airplane prop in flight interacts with the air it's in.

I'm not sure how a motorglider works, how does it make it's propeller spin?

With the cart that has a screw in a wooden block, when you push the wooden block and the screw isn't turning, you only push a certain amount, but if the screw is turning, you also have to push on the turning parts if the screw isn't slipping in the wood, basically you will be pushing harder when you want to keep the block at the same speed. I think it's the same way with the wind cart. The extra energy to go faster than the wind when you engage the propeller at windspeed doesn't come from multiplying the force, it just makes sure the propeller keeps turning the right way, but because the wind starts pushing harder, ie it also has to work on the rotating parts.

 

[A.T.]

I'm not sure how a motorglider works,

Neither does the air, that interacts with its propeller.

The extra energy to go faster than the wind when you engage the propeller at windspeed doesn't come from multiplying the force, it just makes sure the propeller keeps turning the right way, .

The gearing (multiplying the force) is not only making the propeller turn the right way, it makes the entire cart accelerate the right way. What is "extra energy" anyway? And why do you keep taking about engaging the propeller at windspeed? It is engaged all the time.

 

[Redbelly98]

Betz law says that we can't get all the energy out of a specific reference volume of air. It doesn't limit how much of the extracted energy can be put to good use. The latter is the measure of efficiency that matters to us.

I've been following this discussion as much as I can. Put another way:

• A certain volume of air contains an amount of energy x.
• That air loses an amount of energy y in the turbine.
• An amount of energy z gets converted to mechanical energy.
• z ≤ y ≤ x

Betz law says that y/x (or z/x for that matter) can't be more than 59%. However, z/y is the measure of efficiency used here.

 

[A.T.]

I've been following this discussion as much as I can. Put another way:

• A certain volume of air contains an amount of energy x.
• That air loses an amount of energy y in the turbine.
• An amount of energy z gets converted to mechanical energy.
• z ≤ y ≤ x

All those kinetic energies depend on the reference frame. They are some frames where the turbine puts energy into the air. Your z ≤ y ≤ x is valid only in some reference frames.

Betz law says that y/x (or z/x for that matter) can't be more than 59%.

Betz law applies only in the rest frame of the turbine, which is the rest frame of the upwind cart. In that accelerating frame the kinetic energy contained in a volume of air is not constant. So you can stay well below that 59% at a constant turbine efficiency (around 12%), and still get more and more shaft_power, because the air moves faster and faster. However, that increase in turbine power is countered by the increasing power demand at the wheels, so the theoretical net power is approximately constant and depends mainly on the true wind velocity.

 

[spork]

That's the limit for turbine efficiency defined in the in rest frame of the turbine:

shaft_power / total_kinetic_energy_flow_through_disc

I believe the Betz limit doesn't look at the actual energy flow through the disk, but rather the energy that would flow through the disk if that flow were not disturbed by the disk. I think it uses the disk reference area applied to the free-stream. In reality, some of the energy from the reference area ends up flowing around the disk due to the high pressure upstream of the disk.

I'm not sure how a motorglider works, how does it make it's propeller spin?

That's where the motor comes in.

 

[chingel]

The gearing (multiplying the force) is not only making the propeller turn the right way, it makes the entire cart accelerate the right way. What is "extra energy" anyway? And why do you keep taking about engaging the propeller at windspeed? It is engaged all the time.

Yes you are right the gearing also determines the speed at which the propeller and also the cart will go. It has to have an advantage so that the wind pushing on the propeller wouldn't make it go the other way.

With the extra energy I meant the energy to go faster than windspeed once you are already at windspeed. If you just multiply force by levers, there is no extra energy, but the extra energy comes from the wind which I think starts to push more.

I talked about engaging the propeller at windspeed, because I wanted to understand what happens in such a situation. The only difference I think compared to when it is engaged all the time, is that the propeller already has some momentum at windpseed. The cart should still work when you engage the propeller at windspeed, the propeller just needs some time to speed up, maybe making the cart slow down a little until the propeller is up to speed.

 

[Redbelly98]

All those kinetic energies depend on the reference frame.
.
.
.
Betz law applies only in the rest frame of the turbine...

D'oh, of course!

 

[A.T.]

With the extra energy I meant the energy to go faster than windspeed once you are already at windspeed.

That's not "extra energy". Thats the same energy that is used below wind speed.

but the extra energy comes from the wind which I think starts to push more.

There is no "extra energy". All the energy comes from the true wind being reduced.

 

[chingel]

I mean extra energy compared to when the propeller is not spinning. When you engage the propeller, the wind also works on the propeller blade that is turning, acting like a screw in a wooden block.