AERSP 583
Wind Turbine Aerodynamics

2a.7 The Rotor Disk Model

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We've just shown that the 2nd relationship between a and a' may be an extreme, but we don't know yet if this relation will result in maximized or minimized power. In order to better understand this, we have to look at the second derivative of  d2f / da2.

Check the second derivative...

d2fda2=d2ada2(1a)2dadaThis equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.33)
 

...from the first derivative...

dada=a1aThis equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.34)
 

...and the first derivative of eqn. (2a.30) ...

dada=1λr212a1+2a'This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.35)
 

Remember the first equation (2a.30) equated the pressure differences. We now also have its first derivative, and we will take its second derivative as well and obtain:

 d2ada2=1λr22(1+2a)(12a)2dada(1+2a)2This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
 =2λr2(1+2a)2[1+2a+(12a)dada]This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.36)
 

Now we use all of the above relations in the second deriviative:

 d2fda2=2(1a)λr2(1+2a)2[1+2a+(12a)dada]2dadaThis equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.37)
 

Assuming that the 1st derivative is equal to 0, we know that:

 dada=a1a>0This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.38)
 

Thus, we can indeed find that …

 d2fda2<0MaximumThis equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.39)
 
Click here for a video transcript.

Transcript: Maximum

If this is larger than zero, this last term here is less than zero. Agreed? 2(1-a) has a minus here. 2 is larger than zero. 1-a larger than zero. Lambda r square larger than zero. That's 1+squared larger than zero. Larger, larger, (1-2a), a is small enough, larger than zero. da'/da, I just looked at it. Larger than zero. So that's all larger than zero. There's the minus sign that makes it smaller than zero. This is all smaller than zero. So we can say at this point, indeed, the second derivative df by da square is less than zero. With the first derivative being equal to zero, second derivative less than zero, it means that we were really looking for a maximum in the power coefficient. That's all we wanted. That's probably one of the, most painful things that we're going to in the whole class, but it's in no text book. That's why I like to do that with you. So you know exactly where it comes from. Won't be everything like this. I promise.


So far we have eqns. (2a.30) and (2a.32), for a and a'. Equation (2a.30) was a result of equating the pressure drop across the disk:

 Δpa=Δpaa(1a)=λr2a(1+a)This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.40)
 

And equation (2a.32) from maximizing the integral for the pressure coefficient:

 CP,Maxa=(3a1)(14a)This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.41)
 

Now we have 2 equations and 2 unknowns. We can take a' from eqn. (2a.32), which is a sole function of a and plug it into eqn. (2a.30) to find an explicit relation for a.

 a(1a)=λr2(3a1)(14a)(1+(3a1)(14a))This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

 a(1a)=λr2(3a1)(14a)(14a+3a1)(14a)This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

 1a=λr2(13a)(14a)2This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

 λr2=(1a)(4a1)2(13a)This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.42)
 

That gives us now the axial inductor induction factor a as a function of λr , which is scaled tip speed ratio, along the blade for maximum power.

Now, what is the actual CP,Max?

 CP=8λ20λa(1a)λr3dλrThis equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.43)
 

We take the derivitive d/da of boxed equation (2a.42) above and obtain:

 2λrdλr=[6(4a1)(12a)2(13a)2]This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.44)
 

Now take eqns. (2a.32), (2a.42), and (2a.44) in the relation for CP,max, and we get the following:

 CP,Max=24λ2a1a2[(1a)(12a)(14a)(13a)]2daThis equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.45)
 

At the root where  λr = 0, this will define the lower bound of the integral of CP,max with respect to a.

 λr=0a1=0.25This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.46)
 

For the upper bound at the tip where λr becomes equal to the actual tip speed ratio λ, it defines the upper bound for a.

 λr=λa2=?This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.
(2a.47)
 
Click here for a video transcript.

Transcript: Upper and Lower Bounds

If lambda r is zero, how does the right hand side become zero? There are two options. One of the factors has to zero, right? a=1 doesn't work for us. a=1/4 yes that's the one that's going to make it zero. That defines here now the lower bound of the integral. So a1=1/4 or 0.25 For the upper bound at the tip, where lambda r becomes equal to the actual tip speed ratio, lambda, it defines the upper bound for a and that is what? So this becomes lambda square. I can look again at equation 3. And that's not an easy equation to solve for a. So we'll do that numerically.

In the table that you see here, given different lambdas, it solves equation 3 for the appropriate upper integration bound, which is a2, and then gives you the power coefficient that is associated with it. So we can look at this table and say, depending on the tip speed ration -the tip speed ratio was tip speed divided by incoming wind speed- what's the maximum power coefficient that you get? Out of all this messy analysis. Betz limit is 0.593 something, 16/27 is the exact answer. So you can see here clearly as you go up in tip speed ratio, you're getting fairly close to the Betz limit. In other words, the quicker the rotor spins, the closer you get to the Betz limit.

Think about me being a 2-bladed rotor. And I rotate very slowly, no tip speed ratio. There's a lot of mass flow passing through the rotor disk area that the blades do not work on. It's not surprising that if the tip speed ratio is small, you do not get much power. In fact, in the limit, I'm standing still, there's no torque being produced. There's no wake change in angle of momentum associated with that. But as I move quicker and quicker in rotor speed, in a given time, the blades they work on more mass flow that passes through the rotor disk area. As a consequence the power coefficient is going to go up. What is important here is to note that you get fairly close to the Betz limit with a tip speed ratio of 10 and not 1000. That's important.


The table below provides different values for the tip speed ratio, λ, and solves eqn. (2a.42) for the appropriate upper integration bound, a2. It then provides the associated maximum power coefficient. Note that the quicker the rotor spins (λ increases), the closer it gets to the Betz limit.

Values for the tip speed ratio λ, upper integration bound a2, and associated maximum power coefficient.
λ a2 CP,max
0.5 0.2983 0.289
1.0 0.3170 0.416
1.5 0.3245 0.477
2.0 0.3279 0.511
2.5 0.3297 0.533
5.0 0.3324 0.570
7.5 0.3329 0.581
10.0 0.3330 0.585