Author Topic: Blade theory questions  (Read 12623 times)

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SmoggyTurnip

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Blade theory questions
« on: May 27, 2006, 04:22:36 PM »
On page 6 of Hugh's blade design worksheet he says"


"To satisfy Betz, the wind in each part of the swept

area of the rotor the wind must be slowed down to 1/3

of its upstream velocity, and this slowing is done by

the THRUST force, which is closely related to the LIFT

force.

For Betz, Thrust = (4/9)rhoAV^2"


On page 5 he says


"Wind through the rotor = (2/3)V (Following Betz'z therom)"


My questions:


1) If you slow the wind down to 1/3 of its upstream velocity

   shouldn't the wind through the rotot = (1/3)V.


2)Where does this come from?


     "For Betz, Thrust = (4/9)rhoAV^2"


Thanks in advance,

SmoggyTurnip


.

« Last Edit: May 27, 2006, 04:22:36 PM by (unknown) »

hvirtane

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Re: Blade theory questions
« Reply #1 on: May 27, 2006, 11:07:04 AM »
1) If you slow the wind down to 1/3 of its upstream velocity

shouldn't the wind

through the rotor = (1/3)V.


Hugh is correct.

If you want to satisfy Betz,

at the rotor the wind velocity is V-(1/3)V = (2/3)V


--


2)Where does this come from?

"For Betz, Thrust = (4/9)rhoAV^2"


That comes from the equation to calculate

the thrust at the rotor:


At the rotor

F = m(Vr) = [(Vr)x(rho)xA]x(Vr)] =

[(2/3)V x (rho) x A x (2/3)V] =

(4/9) x (rho) x A x V x V


F = the thrust

Vr = the velocity of the wind at the rotor

V = the velocity of the wind before the rotor

m = the mass of the air moving through the rotor

rho = the density of the air

A = the area, which the wind rotor 'covers'.


Again the calculation is done to satisfy

Betz theorem locally at the rotor.


- Hannu

« Last Edit: May 27, 2006, 11:07:04 AM by hvirtane »

finnsawyer

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Re: Blade theory questions
« Reply #2 on: May 28, 2006, 09:32:24 AM »
The Betz Theorem states that the most energy you can capture to allow sufficient air flow behind a wind mill is 59.6 percent of the energy in the incident air flow.  That is, you must leave 40.4 percent of the energy in the air stream.  Since the energy or power depends on the cube of the speed of the air flow, it is a simple calculation to determine that the air leaving the mill must have a speed .739 times that of the incident air at the Betz limit.  If you slow it down more you will cause a high pressure bubble that will eventually extend in front of the blades.  This will cause the air to flow around the entire mill and to bypass the blades.


The "Thrust Force" looks like it is the axial force that pushes the mill backwards toward the tower.  The problem is that that force is static.  That is, for a given wind velocity it doesn't change.  It also doesn't cause anything to move.  Basic physics states that a force must move (something) through a distance to do work.  The axial thrust on the mill or blade assembly can not account for the change in wind speed as the air flows through the blades.  So, just what is this "Thrust Force"?

« Last Edit: May 28, 2006, 09:32:24 AM by finnsawyer »

hvirtane

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Re: Blade theory questions
« Reply #3 on: May 28, 2006, 02:39:00 PM »
I'm sorry, my explanation above was cryptic and

a bit wrong indeed.


We should maybe go a bit

deeper to understand the equation.


--


Another explanation:


First the basic equation for the force.


(1) F = m x a


, where a is the mass and a the acceleration,

the change of the (wind, air) speed.  


(I should not write m x v

as I did above!)


From the equation (1) we get:


(2) F = m x (v/t)


v/t is the change of

the wind speed (air speed) at a given time.


We need as well the equation for the mass:


(3) m = (rho) x A x s


(, where s is the distance the air moves

at a given time)


We can give the value of s using

the values of v and t. So we get:


(4) m = (rho) x A x v x t


Now at the rotor,


(5) F = ma = [(rho) x A x Vr] x t x [(Vr)]/t =

[(2/3)V x (rho) x A x (2/3)V] =

(4/9) x (rho) x A x V x V


F = the thrust

Vr = the velocity of the wind at the rotor

V = the velocity of the wind before the rotor

m = the mass of the air moving through the rotor

rho = the density of the air

A = the area, which the wind rotor 'covers'.


Is this explanation any clearer?


- Hannu

« Last Edit: May 28, 2006, 02:39:00 PM by hvirtane »

SparWeb

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Re: Blade theory questions
« Reply #4 on: May 28, 2006, 11:33:40 PM »
I can follow your explanation, Hannu, but I learned aircraft propellor theory in school.  I only have to turn a few things around in my mind's eye to understand the Betz limit.  It would help others greatly to have a picture of the effect a wind turbine has on the air flowing through it.


ASCII drawing:

                 __________

                /

               /

             /|        ->

---->---/ |

   ->        |        ->       Flow expands coming out of the prop disk

   ->      <|=

___->___ |

            \ |        ->

             \|

               \       ->

                \__________


The air going through a wind turbine is slowed down as it passes through the blades.  This happens because the blades remove some of the energy, and to satisfy conservation of energy, that energy is lost from the momentum of the air.


Now we go to Bernoulli's principle: a streamline of air flow will get larger as it slows down, smaller as it speeds up.  Airplane props and wings speed air up by forcing the stream lines to be smaller.  Wind turbines do the opposite: they slow the air down, therefore the streamlines expand.


The wind turbine blades will extract more and more energy by speeding up.  Under the torque load of an alternator (or their own bearings) they cannot speed up beyond a certain point, because to do so would require more energy than is available in the wind.  Extracting more energy would slow down so much air behind the prop that it could't get out of the way of the incoming air to keep the cycle going.


That's Betz's limit.  Hannu did the math better than I can.  (I'll stick to elegant pictures).

« Last Edit: May 28, 2006, 11:33:40 PM by SparWeb »
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hvirtane

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Re: Blade theory questions
« Reply #5 on: May 29, 2006, 03:02:35 AM »
my mind's eye to understand the Betz limit


Actually I think that Betz limit might not be valid at all.

I will try writing about that some points down later.

Now I have to go to do something else.


One thing, which might be especially important for wind mill 'propellers'

is the matter that different size wind turbines are behaving themselves quite differently,

if we don't change the number of the blades.




- Hannu

« Last Edit: May 29, 2006, 03:02:35 AM by hvirtane »

SmoggyTurnip

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Re: Blade theory questions
« Reply #6 on: May 29, 2006, 06:13:35 AM »
Very excellent explanations! Thanks for taking the time.


So I guess it must be a typo in the pdf file the quote


"To satisfy Betz, the wind in each part of the swept

area of the rotor the wind must be slowed down TO 1/3"


Should read


"To satisfy Betz, the wind in each part of the swept

area of the rotor the wind must be slowed down BY 1/3"


.

« Last Edit: May 29, 2006, 06:13:35 AM by SmoggyTurnip »

hvirtane

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Re: Blade theory questions
« Reply #7 on: May 30, 2006, 05:31:56 AM »
"To satisfy Betz,

the wind in each part

of the swept area of the rotor

the wind must be slowed down TO 1/3"


In my opinion this is exactly correct.

The wind rotor is extracting (in principle)

2/3 of the wind speed (power) available.

And because of that the wind speed goes

down to 1/3 of the original wind speed.


At the rotor the wind speed is 2/3 of

'the original' wind speed, because

of the original wind speed

only 1/3 is left after the rotor.


- Hannu

« Last Edit: May 30, 2006, 05:31:56 AM by hvirtane »

acme12

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Re: Blade theory questions
« Reply #8 on: August 14, 2014, 06:00:01 AM »
I'm sorry, my explanation above was cryptic and

a bit wrong indeed. <p>
We should maybe go a bit

deeper to understand the equation.<p>
--<p>
Another explanation:<p>
First the basic equation for the force. <p>
(1) F = m x a<p>
, where a is the mass and a the acceleration,

the change of the (wind, air) speed.  <p>
(I should not write m x v

as I did above!) <p>
From the equation (1) we get: <p>
(2) F = m x (v/t) <p>
v/t is the change of

the wind speed (air speed) at a given time. <p>
We need as well the equation for the mass: <p>
(3) m = (rho) x A x s<p>
(, where s is the distance the air moves

at a given time) <p>
We can give the value of s using

the values of v and t. So we get:<p>
(4) m = (rho) x A x v x t<p>
Now at the rotor, <p>
(5) F = ma = [(rho) x A x Vr] x t x [(Vr)]/t =

[(2/3)V x (rho) x A x (2/3)V] =

(4/9) x (rho) x A x V x V<p>
F = the thrust

Vr = the velocity of the wind at the rotor

V = the velocity of the wind before the rotor

m = the mass of the air moving through the rotor

rho = the density of the air

A = the area, which the wind rotor 'covers'.<p>
Is this explanation any clearer? <p>
- Hannu

Is the throust force the force that is transferred from propeller to the tower in wind direction? And is the thrust force the force that mainly bends the propeller blades, and is much higher on the blade tip then at the blade root where is "0"?

Is there any simple formula to calculate the moment (torque), that the pitch control bearing must handle?

Thanks

Marko





Flux

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Re: Blade theory questions
« Reply #9 on: August 14, 2014, 11:33:49 AM »
Thrust mainly depends on the energy extracted from the prop. When stationary the area you are concerned with is the total blade area facing the wind and is small.

The worst case would be if you take the blade swept area as if it was a disc. As you can't extract all the energy ( Betz and all that) the real thrust loading will be below the disc loading by a fair bit but for working calculations it is best to used the swept area as a disc to have a safety factor.

In real working conditions most of the energy extraction does come from the blade tips so there is a lot of truth in what you say but the factor is not zero at the blade root. You are probably justified in using the blade area rather than swept area right at the centre.

Flux

acme12

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Re: Blade theory questions
« Reply #10 on: August 14, 2014, 04:16:51 PM »
Thank you Flux

Lets take the worst case, where blade extract the 2/3 of avaliable energy from rotor center to blade tip.
8339-0

The center of graviti (in this case energy) for a  small section on disk with radius 100 % is at near 66% of radius. So I figured out this formula:

Moment at pitch bearing is near: Mpb= F/Nx0,66 = (4/9) x (rho) x A x V x V / N x 0,66


F = the thrust

Vr = the velocity of the wind at the rotor

V = the velocity of the wind before the rotor

m = the mass of the air moving through the rotor

rho = the density of the air

A = the area, which the wind rotor 'covers'.

N = number of blades

0,66 = center of graviti / energy

Mpb = Moment at pitch bearing



Fast calculation for turbine 5,2 m (homebrew 17 ft) in diameter, at 15 m/s (33,5 mph)

A= 21,23 m2
rho = 1,2 kg/ m3
V= 15 m/s
N = 3 (three blades)
 
F= 2547,6 N
Mpb is then = 560 Nm

That is like one small man with 56 kg is standing on blade 1m from the rotor center. I dont know if I am right, but it looks it is OK. Is it?

joestue

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Re: Blade theory questions
« Reply #11 on: August 14, 2014, 04:58:46 PM »
check that moment by calculating it from power delivered to the alternator by the blade.


35mph is 51 feet per second, times tip speed ratio of 7 is 359, divided by 17, divided by pi is 6.72 rotations per second, or 403 rpm.
 
560 Nm times 403 rpm is 23.6 kilowatts.

35mph is 15.6m/s, so 17 foot diameter works out to 22Kw at a theoretical efficiency of .46.
so you are correct, but in practice you might get half of that.
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acme12

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Re: Blade theory questions
« Reply #12 on: August 15, 2014, 03:05:45 AM »
Hi joestue
You should multyply 560 Nm by 3 (number of blades). P-> over 70 kW (error, too high). What I want is to calculate the moment on a hub pitch bearing and not the torque on the shaft.

P= M x w
TSR =7
w(386 RPM) = 40rad/s
Cp=0,4 (please correct)
V=15 m/s

P=Cp*rho/2*pi/4*D^2* V^2= 0,4*2865 = 1146 wats
M= 28,6 Nm
Something is wrong here, but I dont know where (see results on Dan Lenox site, page 8): http://www.briery.com/wind_turbine/build_log8.html



 

joestue

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Re: Blade theory questions
« Reply #13 on: August 15, 2014, 04:28:41 AM »
ok, i don't understand what you mean by hub pitch bearing torque. (torque is the same as moment, afaik)
i'm pretty sure if you reduce your units correctly, you will arrive at the answer you got, which checks out..

now if you're looking for a different unit then of course it won't match up. the number of blades matters not, when it comes to energy extracted.
« Last Edit: August 15, 2014, 04:38:50 AM by joestue »
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acme12

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Re: Blade theory questions
« Reply #14 on: August 15, 2014, 06:00:02 AM »
The "Thrust Force" looks like it is the axial force that pushes the mill backwards toward the tower.  The problem is that that force is static.  That is, for a given wind velocity it doesn't change.  It also doesn't cause anything to move.  Basic physics states that a force must move (something) through a distance to do work.  The axial thrust on the mill or blade assembly can not account for the change in wind speed as the air flows through the blades.

Sorry joestue, my english is realy bad. Please read the text above from finnsawyer for better explanation of thrust force. If I am thinking right, there are three forces that blade and connection between blade and hub must handle. All these forces acting in different directions. First one is centrifugal force. Second is the thrust force (read the text above), and third is the force / resistance of generator through shaft. Without last one there would be no generated power while rotation. The thrust and centrifugal force are not the forces that makes power since there is no movement in any way or in way of shaft rotation, but I think are the most problematic for blade stability.
I do not exclude that I am completely wrong.
Regards Marko

acme12

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acme12

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Re: Blade theory questions
« Reply #16 on: August 15, 2014, 07:27:13 AM »
I am slow learning ;D. I watched the picture again. Eureka: It is the lift force that I am looking for. But this means it is out of my reach. 

But faster the blade rotates, the biger thrust force is and smaller the force that rotates the propeller. if I am right here, and maybe Hannu is wrong. Or not if lift force at higher speed blade is lower.



« Last Edit: August 15, 2014, 07:35:11 AM by acme12 »

kitestrings

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Re: Blade theory questions
« Reply #17 on: August 15, 2014, 01:50:33 PM »
amce12,

Quote
Fast calculation for turbine 5,2 m (homebrew 17 ft) in diameter, at 15 m/s (33,5 mph)

A= 21,23 m2
rho = 1,2 kg/ m3
V= 15 m/s
N = 3 (three blades)
 
F= 2547,6 N
Mpb is then = 560 Nm

I was landing the above formulas and conversions into a spreadsheet to better follow the units (some of which I rarely use).  I follow all except how you solved for F above.  I must of missed something.

Joestue,

[harvesting] "Half" seems about right.  Where did the 46% figure come from?


If we compare to one of the popular blade calculators at 5.2m, 35 mph, (w/ CoL of .8 and prop eff .3)  I get a thrust of 2907N at 412 rpm, and just under 16 kW.  Seems to be in district.

Now this all assumes that the blades are fully loaded right up to 35 mph.  Most likely a variable pitch system is going to start below that point.  On a furling machine the area will start to wane, and thereby reduce thrust, nearer to 20-25 mph, correct?

~kitestrings

kitestrings

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Re: Blade theory questions
« Reply #18 on: August 16, 2014, 08:25:43 AM »
Quote
Lets take the worst case, where blade extract the 2/3 of avaliable energy from rotor center to blade tip.


This is actually the best case.  Worst/highest loading, but best/most efficient harvest of energy.

Quote
Actually I think that Betz limit might not be valid at all.

I will try writing about that some points down later.

Hannu, I hope you get back to this thought.  You've peaked my interest.

~kitestrings

joestue

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Re: Blade theory questions
« Reply #19 on: August 16, 2014, 02:58:24 PM »
i might be halucinating but i thought .46 was the betz limt.

in any case acme12 was looking for solving the moment exerted on the blades, which cancel out at the hub, so that's cleared up.
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kitestrings

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Re: Blade theory questions
« Reply #20 on: August 18, 2014, 09:11:11 AM »
I thought the Betz limit was 59.6%.  I must of missed something here.  ~ks

Flux

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Re: Blade theory questions
« Reply #21 on: August 18, 2014, 09:46:22 AM »
I think Betz original figure is 16/27, usually now expressed as 59.3%

This is from momentum theory. there are those that suspect the figure can be a bit higher as the momentum theory is very basic, but until someone measures a turbine producing more than 59.3% there is little point in worrying, it's near enough for now.

I don't think this helps the original poster, I am not quite sure what he is asking but seems to be something to do with the loading on the pitch mechanism. How loading is distributed along the blade I wouldn't to hazard a guess. The figures for thrust loading seem in order. I once measured thrust loading very crudely and got about 50% disc loading. This probably means something like CP x disc loading then adding in some blade inefficiency

Flux

kitestrings

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Re: Blade theory questions
« Reply #22 on: August 19, 2014, 04:28:43 PM »
Thanks.  I probably spend too much time analyzing these things, but I find its more enjoyable than cross-words.

Having tossed this around a bit more I think the assumptions are reasonable , & if anything closer, to the theoretical limits anyway.  In acme's "fast calculation" he is assuming 15 mps, or 33.5 mph.  Joestue is using 35 mph.  I'd initially used the same.  If you work the numbers at 33.5 mph it works out to about 22.7 kW, of which we might get half, say 11 kW.  Similarly, Alton's blade calculation worksheet shows 13.2 kW for this diameter rotor at 15 mps and 5.2 m.

What the safety factor should be on these things, well that's another matter.

~ks

joestue

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Re: Blade theory questions
« Reply #23 on: August 19, 2014, 08:34:59 PM »
i think he was trying to find the thrust exerted on a single blade.
this is why in his aug 15 post he suggests i multiply the torque by three.. (an error)

anyhow the torque at the root of the blade is probably insignificant, but the moment exerted on the hub certainly isn't. in which case, divide the torque produced by the turbine by three, and you get bending moment for each blade at the root in the driven direction.
however you have to add the thrust (divide thrust by three) as a vector to the torque, to find the direction and measure of the moment exerted on the hub by each blade.
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acme12

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Re: Blade theory questions
« Reply #24 on: August 21, 2014, 01:27:50 PM »
damn, is there any way to delite previous message?  (Your wish is my command: DamonHD)

Kitestrings, sorry for cross-words, messing and metric units.
Joestue hit's the pin head in last message.
The picture below shows the static calculation for cantilever beam. I think, we can easily transfer this for calculating the momentum obtained in single blade/hub connection, or as folllow Ml. Lets just think that q (load in metric units) is W(in picture) = is load on the whole turbine devide by N (number of blades). And just think that the load is increasing from support (hub) to blade tip, which is correct, if we had a perfect blade airfoil turning. 




qx (lift thrust load on one blade in wind direction)

qy (lift rotating load on one blade)

ql (lift load on single blade)

L (lenght of the blade)

w (angular speed)

N (number of blade)

F (thrust force on the on the whole turbine, like Hannu said = (4/9) x (rho) x A x V x V)

P (power delivered to generator)

load in x way:

qx= F / N = (4/9) x (rho) x A x V x V / N


Moments for one blade:

Ml = 2/3 x ql x L

Mx= 2/3 x qx x L

My= 2/3 x qy x L

The power delivered to the rotor P is:
P=M x w
or
P= Myx w x N

If we know the power delivered to generator and rotational speed RPM then My[/sub and qy is easy to get:

My = P / (w x N)

2/3 x qy x L = P / (w x N)

qy = P / (w x N x L x 2/3)

Then simple Pythagorean theorem comes in:

ql2= qx2 + qy2

and

Ml2= Mx2 + My2

Now I need a power calculator :)
For serious size turbine, given formulas are probably bad. But for my use, if I will find a windy location, will probably be just fine. Certainly there are shortcuts and other much better formulas.

Regards
« Last Edit: August 22, 2014, 05:30:24 AM by DamonHD »

SparWeb

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Re: Blade theory questions
« Reply #25 on: August 23, 2014, 08:56:38 PM »
Acme12,
Kitestrings,

I regret not seeing this sooner, but I was away on vacation.

This is a subject I really enjoyed learning as I was learning it.  I hope you guys are having as much fun.

A few resources, though rather technical:

http://wind.nrel.gov/designcodes/papers/
http://www.nrel.gov/wind/model_analysis.html

Depends on how deep you want to go.
When I was in school I did an analysis for a final exam much like this, though that case was for an aircraft propeller thrust. 
Later when my interest was piqued for wind turbines I discovered that the work is basically the same.
If you are analyzing it in pieces, or segments along the span of the blade, then I found that about 10 pieces is all you need to get reasonably accurate results.  Later when I got my hands on more powerful software I was able to analyse much more refined sections, then do nearly pure integrations, but none of that really helps the fact that the input data is approximate, making the ouput approximate on any case.

Developing the distribution of load along the span as you have shown it, then using trig to turn the lift vector into the normal vector, will give you the result you're looking for.  If doing it in piece-wise analysis, then actually you should just use a sum of moments produced by the series of individual loads.  But it won't be hard to do it the other way if you prefer.

Watch the relative wind vector, it changes along the span.  There is also the problem that it does NOT obey the TSR that you want the turbine to use, and that in the real world there is an "inflow" factor that messes with everything by several degrees.  You will find the inflow described in the Lissaman/Walker paper in the link above.

In fact, I recommend you scan through the Lissaman/Walker paper as soon as possible, you seem ready to start taking it in already.  Then drill down into the details as you please and as suits the level of detail you really are looking for.

There are other NREL papers that are of value but I don't have the links handy.  I want to hit "Post" before going off into internet-land to find them.
No one believes the theory except the one who developed it. Everyone believes the experiment except the one who ran it.
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SparWeb

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Re: Blade theory questions
« Reply #26 on: August 23, 2014, 09:05:14 PM »
www.nrel.gov/docs/fy06osti/38550.pdf

NREL TP-500-38550, Small Wind Research Turbine Final Report, D. Corbus and M. Meadors

Good reality check on thrust calculations.  They used a heavily instrumented Bergey Excel for the tests.
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Re: Blade theory questions
« Reply #27 on: August 23, 2014, 09:32:34 PM »
Dragging your calculations from the generator to the rotor is tricky...

Quote
...If we know the power delivered to generator and rotational speed RPM then...

Try to set up the conditions that you "input" to the turbine rotor so that you can solve all the way through.  Once you get power (expressed as torque@RPM) at the rotor shaft you can use that power to drive the theoretical generator.  And that's a whole different barrel of monkeys.

There's a quick way to determine thrust, which Hannu was discussing in the original thread you picked up, but it's become confused along the way.  Betz is as good as it gets as far as I can tell.

The wind speed ratio into rotor /out of rotor is 1/3 as they said, when the turbine is working at PEAK efficiency (ie. the Betz limit).  But don't confuse that for 2/3 power, or 2/3 thrust, or anything else.  It's just a speed ratio that results from the extraction of kinetic energy, and a theoretical one at best.  Remember that kinetic energy is calculated with the square of the velocity and you'll see that it can't be the same.  Then also consider the slight change in air pressure as it passes the rotor, and the very significant change in outward flow direction (like a tube that gets wider).  So thrust, like the power, has to take into account a lot of things before getting an accurate answer.

Fortunately, when making rough estimates, I find that the result is conservative, meaning the thrust is overestimated and one can design a support structure that will be adequate for the real loads because the design loads are over-sized.
No one believes the theory except the one who developed it. Everyone believes the experiment except the one who ran it.
System spec: 135w BP multicrystalline panels, Xantrex C40, DIY 10ft (3m) diameter wind turbine, Tri-Star TS60, 800AH x 24V AGM Battery, Xantrex SW4024
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acme12

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Re: Blade theory questions
« Reply #28 on: August 29, 2014, 04:07:30 PM »
SparWeb, very thank you for your input.
When I quickly flew Lissaman / Walker article I quickly realized that my knowledge of mathematics is no longer as it was when I was a student 13 years ago. I newer was a wizard in mathematics, but I was fairly well in integration i differentials.

If I would try to solve the problem in detail from begining, to me it would take too much time. But still, part of me would always be doubted if calculations are correct. For now, design loads from my rough estimates are much over-sized, what could be taken as a safety factor.

A few days ago I found article from Mr. Soren Gundtoft which he is generously sharing with the public: http://staff.iha.dk/sgt/Downloads/Turbines%20May4_2009_1.pdf
To this article belongs Excel spreadsheet: http://www.iha.dk/files/filer/pdf_engelsk/design_11.xls
Mr. Soren was solving the problem,  as you suggested me. He parted the blade into 7 elements. It can be divided into many parts but it is probably not necessary for DIY project.
When I will find the time I will flew over the Soren's work and try to connect the spreadsheet with the written part of the theory. I think this way is the quickest for me to learn how wind turbine works. I also found some literature in my home language, which we will be very helpful.

SparWeb

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Re: Blade theory questions
« Reply #29 on: August 31, 2014, 12:52:54 PM »
That is a good discovery.  It seems to cover the same subject and details, but of course just delivers the math and the program, not the explanation and details that expand the Wilson-Lissaman papers to more than 100 pages.

I was very interested to read the Schmitz model of wake rotation - so simple!
Seeing the serious difference in the prediction of the optimized blade angle and chord (Figures 3.10 and 3.11) is really interesting, too.

Thanks for linking that paper, good reading.
No one believes the theory except the one who developed it. Everyone believes the experiment except the one who ran it.
System spec: 135w BP multicrystalline panels, Xantrex C40, DIY 10ft (3m) diameter wind turbine, Tri-Star TS60, 800AH x 24V AGM Battery, Xantrex SW4024
www.sparweb.ca

guitardodds

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Re: Blade theory questions
« Reply #30 on: February 16, 2015, 04:29:23 PM »
Sorry for the thread resurrection guys, but I'm trying to track down the excel file that acme12 links to up there ^^^^^
The link is now dead, and I wondered if anyone downloaded it while it was live and would be so kind as to share with me.
I would massively appreciate it.
Barry

PS: I'm designing the blades for a 10 MW wind turbine as part of my final year MEng mechanical engineering degree and feeling a bit out of my depth.

SparWeb

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Re: Blade theory questions
« Reply #31 on: February 17, 2015, 02:02:51 PM »
You may not need to...

In the PDF, the source of the excel spreadsheet is given in the appendix B.
You could try to "cut-and-paste" into Excel.

After reading the excerpt, I bought this book listed in the references: Gasch, Twele,  Wind Power Plants
http://download.springer.com/static/pdf/125/bfm%253A978-3-642-22938-1%252F1.pdf?auth66=1424199550_ddb56ac4035847ef9d32dbe22ab479aa&ext=.pdf

Very good and straightforward analysis.  Wake rotation theory is much clearer to me now, and a tool I can actually use if I need to in the future.  Few translation mistakes; not enough to get in the way.
No one believes the theory except the one who developed it. Everyone believes the experiment except the one who ran it.
System spec: 135w BP multicrystalline panels, Xantrex C40, DIY 10ft (3m) diameter wind turbine, Tri-Star TS60, 800AH x 24V AGM Battery, Xantrex SW4024
www.sparweb.ca

guitardodds

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Re: Blade theory questions
« Reply #32 on: February 21, 2015, 06:50:10 PM »
Cheers, I'll give that a try. That book looks good, but £50 is a bit spendy for a poor student like me.