### Rotor Torque and Power

Our next task is to find an expression for the incremental power coefficient d *C _{P}*. Before proceeding, though, let us define some additional dimensionless parameters that will he helpful in the analysis and useful to a thorough understanding of the effects of wake rotation.

Angular Induction Factor:

a' = ω / (2 Ω)Tip Speed Ratio:

λ = (Ω · R) / V_{0}Local Tip Speed Ratio:

λ_{r}= λ · r/R

The ‘angular’ induction factor *a’* is defined as being proportional to *ω*. It is apparent that in the special case of * ω = 0* the angular induction factor becomes zero, and the rotor disk model reduces to the original actuator disk model. The next parameter, i.e. Tip Speed Ratio

*λ*, involves the rotor speed

*Ω*and describes the ratio of the blade tip speed

*Ω*·

*R*to the wind speed

*V*

_{0}*.*We will see later in this course that typical values for

*λ*range between 4 and 7. The third parameter shown is the Local Tip Speed Ratio

*λ*

_{r}, which is simply a fraction of

*λ*based on the local blade position

*r/R*.

Using these newly defined parameters and performing some algebra on the incremental power *dP* we obtain

...which gives us the incremental power *dP* as a function of wind speed *V _{0}*, blade radius

*R*, the rotational parameters

*λ*and

*λ*, and the dimensionless axial- and angular induction factors

_{r}*a*and

*a’*.

*Click here to view the derivation for dP*

Hence, the differential power coefficient *dC _{P}* becomes

*= dP / (½ ρ A V*

_{0}^{3}) = dP / (½ ρ πR^{2}V_{0}^{3})*Click for video transcript.*

#### Transcript: differential power coefficient dCp

The incremental power coefficient of an annulus on the rotor disk is a function of the tip speed ratio, the angular induction factor, the axial induction factor, and the local tip speed ratio, which is simply a product of the actual tip speed ratio times you local relative location along the blade.

The power coefficient *C _{P}* of the wind turbine is obtained by integrating the previous equation along the entire rotor disk, specifically for

*λ*ranging between 0 and the tip speed ratio

_{r}*λ*.

As the integral for the power coefficient *C _{P}* is performed over the local tip speed ratio

*λ*, our next task is to find two relations between

_{r}*λ*and the axial- and angular induction factors

_{r }*a*and

*a’*, which will enable us to compute the integral in (2a.27). We will do so by considering the following:

First, let us remember equation (2a.9) from actuator disk theory that related the pressure jump *Δp _{a}* across the actuator disk to the axial induction factor

*a*.

Actuator Disk: * Δp _{a} = ½ ρ (V_{0}^{2} - u_{1}^{2}) = *2

*ρ V*(1 -

_{0}^{2}a*a*)

What approach would we take for the Rotor Disk?

*Think about this, then click here for the answer.*

Our approach consists of having the same pressure jump cause the wake rotation *ω*. Hence, we are looking for *Δpa’ = Δpa*. In actuator disk theory, we used the Bernoulli equation applied to the axial velocity component through the streamtube.

We will use a similar approach here, however considering only the angular (or rotational) velocity. We write the Bernoulli equation as:

And substitute ω using the definition of the angular induction factor, i.e. *a’ = ω / (2 Ω).* Thus, we find for the pressure jump *Δp _{a’}:*

Equating both formulations for the pressure drop across the rotor disk, i.e. *Δp _{a} = Δp_{a’}* , we obtain the following after some algebra:

$$\u25b3{p}_{a}=\u25b3{p}_{{a}^{\prime}}\Rightarrow 2\text{\hspace{0.17em}}\rho \text{\hspace{0.17em}}{V}_{0}{}^{2}a\left(1-a\right)=2\rho {\Omega}^{2}{r}^{2}{a}^{\prime}(1+{a}^{\prime})$$

$$\Rightarrow a\left(1-a\right)={\left(\Omega \cdot \frac{r}{{V}_{0}}\right)}^{2}{a}^{\prime}(1+{a}^{\prime})$$

*Click here for a video transcript.*

#### Transcript: First relationship between a and a'

And this omega times r/V0 square is again the square of the local tip speed ratio lambda r. So we isolate this on the right hand side and so on the left hand side you get the ratio of a(1-a) divided by a'(1+a').

The above equation constitutes a first relation between *a, a’, *and* λ _{r}* . In order to evaluate the integral for the power coefficient

*C*in (2a.27), we must find a second relation.

_{P}We mentioned earlier that the addition of wake rotation is likely to reduce the maximum power coefficient *C _{P,max}=0.59* due to Betz in actuator disk theory. One of our objectives in rotor disk theory is to understand to what extent do the newly introduced parameters

*a’, λ,*and

*λ*reduce the Betz limit. By inspecting the integrand in (2a.27) we realize that the maximum power coefficient occurs for the integrand factor

_{r}*a’(1-a)*attaining a maximum for each

*λ*We therefore want to maximize the function

_{r}.*f(a, a’) = a’(1-a)*. As a first step, let us compute the first and second derivatives of

*f(a,a’)*using the product rule:

$$\frac{df}{da}=\frac{d{a}^{\prime}}{da}\left(1-a\right)-{a}^{\prime}$$

$$\frac{{d}^{2}f}{d{a}^{2}}=\frac{{d}^{2}{a}^{\prime}}{d{a}^{2}}\left(1-a\right)-2\frac{d{a}^{\prime}}{da}$$

Next, we must perform the necessary algebra find the second relation between *a* and *a'.*

*Click here to view the algebra.*

$$Eqn.\text{\hspace{0.17em}}\left(1\right)\text{\hspace{0.17em}}\Rightarrow \text{\hspace{0.17em}}a\left(1-a\right)={\lambda}_{r}^{2}{a}^{\prime}\left(1+{a}^{\prime}\right)$$

$$\frac{d}{da}\text{\hspace{0.17em}}\left(Eqn.\text{\hspace{0.17em}}1\right)\text{\hspace{0.17em}}\Rightarrow \text{\hspace{0.17em}}1-2a={\lambda}_{r}^{2}\left(\frac{d{a}^{\prime}}{da}\left(1+{a}^{\prime}\right)+{a}^{\prime}\frac{d{a}^{\prime}}{da}\right)$$

$$\Rightarrow \text{\hspace{0.17em}}1-2a={\lambda}_{r}^{2}\frac{d{a}^{\prime}}{da}\left(1+2{a}^{\prime}\right)$$

$$\Rightarrow \text{\hspace{0.17em}}1-2a=\frac{{a}^{\prime}}{\left(1-a\right){\lambda}_{r}^{2}\left(1+2{a}^{\prime}\right)}$$

$$\Rightarrow \text{\hspace{0.17em}}1-2a=\frac{{a}^{\prime}}{\left(1-a\right)}\cdot \frac{a\left(1-a\right)}{\left({a}^{\prime}\left(1+{a}^{\prime}\right)\right)}\cdot \left(1+2{a}^{\prime}\right)$$

$$\Rightarrow \text{\hspace{0.17em}}1-2a=\frac{a}{\left(1+{a}^{\prime}\right)}\cdot \left(1+2{a}^{\prime}\right)$$

$$\Rightarrow \text{\hspace{0.17em}}\left(1-2a\right)\cdot \left(1+{a}^{\prime}\right)=a\cdot \left(1+2{a}^{\prime}\right)$$

$$\Rightarrow \text{\hspace{0.17em}}1+{a}^{\prime}-2a-2a{a}^{\prime}=a+2a{a}^{\prime}$$

$$\Rightarrow \text{\hspace{0.17em}}{a}^{\prime}-4a{a}^{\prime}=a\left(1-4a\right)=3a-1$$

$$\Rightarrow \text{\hspace{0.17em}}{a}^{\prime}=\frac{\left(3a-1\right)}{\left(1-4a\right)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(2\right)$$