Blade element momentum theory

Suppose we are interested in finding the maximum power that can be extracted from the fluid. The power in the fluid is given by the following expression:

${\displaystyle \mathrm {Power} ={\frac {1}{2}}\rho Av^{3}}$

where ${\displaystyle \rho }$ is the fluid density as before, ${\displaystyle v}$ is the fluid velocity, and ${\displaystyle A}$ is the area of an imaginary surface through which the fluid is flowing. The power extracted from the fluid by a rotor in the scenario described above is some fraction of this power expression. We will call the fraction the power co-efficient, ${\displaystyle C_{p}}$. Thus the power extracted, ${\displaystyle \mathrm {Power} _{ext}}$ is given by the following expression:

${\displaystyle \mathrm {Power} _{ext}=\mathrm {Power} \times C_{p}}$

Our question is this: what is the maximum value of ${\displaystyle C_{p}}$ using the Betz model?

Let us return to our derived expression for the power transferred from the fluid to the rotor (${\displaystyle \mathrm {Power} _{ext}}$). We can see that the power extracted is dependent on the axial induction factor. If we differentiate ${\displaystyle \mathrm {Power} _{ext}}$ with respect to ${\displaystyle a}$, we get the following result:

${\displaystyle {\frac {\mathrm {d} \mathrm {Power} _{ext}}{\mathrm {d} a}}=2v_{\infty }^{3}\rho A_{D}\times \left((1-a)^{2}-2a(1-a)\right)}$

If we have maximised our power extraction, we can set the above to zero. This allows us to determine the value of ${\displaystyle a}$ which yields maximum power extraction. This value is a ${\displaystyle 1/3}$. Thus we are able to find that ${\displaystyle C_{P~max}=16/27}$. In other words, the rotor cannot extract more than 59 per cent of the power in the fluid.

Let us now go back to Bernoulli:

${\displaystyle {\frac {1}{2}}\rho v_{1}^{2}+P_{1}={\frac {1}{2}}\rho v_{2}^{2}+P_{2}}$

The velocity is the velocity of the fluid along a streamline. The streamline may not necessarily run parallel to a particular co-ordinate axis, such as the z-axis. Thus the velocity may consist of components in the axes that make up the co-ordinate system. For this analysis, we will use cylindrical polar co-ordinates ${\displaystyle (r,~\theta ,~z)}$. Thus ${\displaystyle v^{2}=v_{r}^{2}+v_{\theta }^{2}+v_{z}^{2}}$.

NOTE: We will in fact, be working in cylindrical co-ordinates for all aspects e.g. ${\displaystyle \mathbf {F} =F_{r}{\hat {\mathbf {r} }}+F_{\theta }{\hat {\theta }}+F_{z}{\hat {\mathbf {z} }}}$

Now consider the setup shown above. As before, we can break the setup into two components: upstream and downstream.

${\displaystyle P_{\infty }+{\frac {1}{2}}\rho v_{u}^{2}=P_{D+}+{\frac {1}{2}}\rho v_{D}^{2}}$

where ${\displaystyle v_{u}}$ is the velocity of the fluid along a streamline far upstream, and ${\displaystyle v_{D}}$ is the velocity of the fluid just prior to the rotor. Written in cylindrical polar co-ordinates, we have the following expression:

${\displaystyle P_{\infty }+{\frac {1}{2}}\rho v_{\infty }^{2}=P_{D+}+{\frac {1}{2}}\rho (v_{\infty }(1-a))^{2}}$

where ${\displaystyle v_{\infty }}$ and ${\displaystyle v_{\infty }(1-a)}$ are the z-components of the velocity far upstream and just prior to the rotor respectively. This is exactly the same as the upstream equation from the Betz model.

As can be seen from the figure above, the flow expands as it approaches the rotor, a consequence of the increase in static pressure and the conservation of mass. This would imply that ${\displaystyle v_{r}\neq 0}$ upstream. However, for the purpose of this analysis, that effect will be neglected.

${\displaystyle P_{D-}+{\frac {1}{2}}\rho v_{D}^{2}=P_{\infty }+{\frac {1}{2}}\rho v_{w}^{2}}$

where ${\displaystyle v_{D}}$ is the velocity of the fluid just after interacting with the rotor. This can be written as ${\displaystyle v_{D}^{2}=v_{D,~r}^{2}+v_{D,~\theta }^{2}+v_{D,~z}^{2}}$. The radial component of the velocity will be zero; this must be true if we are to use the annular ring approach; to assume otherwise would suggest interference between annular rings at some point downstream. Since we assume that there is no change in axial velocity across the disc, ${\displaystyle v_{D,~z}=(1-a)v_{\infty }}$. Angular momentum must be conserved in an isolated system. Thus the rotation of the wake must not die away. Thus ${\displaystyle v_{\theta }}$ in the downstream section is constant. Thus Bernoulli simplifies in the downstream section:

${\displaystyle P_{D-}+{\frac {1}{2}}\rho v_{D,~z}^{2}=P_{\infty }+{\frac {1}{2}}\rho v_{w,~z}^{2}=P_{D-}+{\frac {1}{2}}\rho (v_{\infty }(1-a))^{2}}$

In other words, the Bernoulli equations up and downstream of the rotor are the same as the Bernoulli expressions in the Betz model. Therefore, we can use results such as power extraction and wake speed that were derived in the Betz model i.e.

${\displaystyle v_{w,z}=(1-2a)v_{\infty }}$

${\displaystyle \mathrm {Power} =2a(1-a)^{2}v_{\infty }^{3}\rho A_{D}}$

This allows us to calculate maximum power extraction for a system that includes a rotating wake. This can be shown to give the same value as that of the Betz model i.e. 0.59. This method involves recognising that the torque generated in the rotor is given by the following expression:

${\displaystyle \delta \mathbf {Q} =2\pi r\delta r\times \rho U_{\infty }(1-a)\times 2a'r\omega }$

with the necessary terms defined immediately below.

Consider fluid flow around an airfoil. The flow of the fluid around the airfoil gives rise to lift and drag forces. By definition, lift is the force that acts on the airfoil normal to the apparent fluid flow speed seen by the airfoil. Drag is the forces that acts tangential to the apparent fluid flow speed seen by the airfoil. What do we mean by an apparent speed? Consider the diagram below:

The speed seen by the rotor blade is dependent on three things: the axial velocity of the fluid, ${\displaystyle v_{\infty }(1-a)}$; the tangential velocity of the fluid due to the acceleration round an airfoil, ${\displaystyle a'\omega r}$; and the rotor motion itself, ${\displaystyle \omega r}$. That is, the apparent fluid velocity is given as below:

${\displaystyle \mathbf {v} =\omega r(1+a'){\hat {\mathbf {\theta } }}+v_{\infty }(1-a){\hat {\mathbf {z} }}}$

Thus the apparent wind speed is just the magnitude of this vector i.e.:

${\displaystyle |\mathbf {v} |^{2}=(\omega r(1+a'))^{2}+(v_{\infty }(1-a))^{2}=W^{2}}$

We can also work out the angle ${\displaystyle \phi }$ from the above figure:

${\displaystyle \sin \phi ={\frac {v_{\infty }(1-a)}{W}}}$

Supposing we know the angle ${\displaystyle \beta }$, we can then work out ${\displaystyle \alpha }$ simply by using the relation ${\displaystyle \alpha =\phi -\beta }$; we can then work out the lift co-efficient, ${\displaystyle c_{L}}$, and the drag co-efficient ${\displaystyle c_{D}}$, from which we can work out the lift and drag forces acting on the blade.

Consider the annular ring, which is partially occupied by blade elements. The length of each blade section occupying the annular ring is ${\displaystyle \delta r}$ (see figure below).

The lift acting on those parts of the blades/airfoils each with chord ${\displaystyle c}$ is given by the following expression:

${\displaystyle \delta L={\frac {1}{2}}\rho NW^{2}c\times c_{L}(\alpha )\delta r}$

where ${\displaystyle c_{L}}$ is the lift co-efficient, which is a function of the angle of attack, and ${\displaystyle N}$ is the number of blades. Additionally, the drag acting on that part of the blades/airfoils with chord ${\displaystyle c}$ is given by the following expression:

${\displaystyle \delta D={\frac {1}{2}}\rho NW^{2}c\times c_{D}(\alpha )\delta r}$

Remember that these forces calculated are normal and tangential to the apparent speed. We are interested in forces in the ${\displaystyle {\hat {\mathbf {z} }}}$ and ${\displaystyle {\hat {\theta }}}$ axes. Thus we need to consider the diagram below:

Thus we can see the following:

${\displaystyle \delta F_{\theta }=\delta L\sin \phi -\delta D\cos \phi }$

${\displaystyle \delta F_{z}=\delta L\cos \phi +\delta D\sin \phi }$

${\displaystyle F_{\theta }}$ is the force that is responsible for the rotation of the rotor blades; ${\displaystyle F_{z}}$ is the force that is responsible for the bending of the blades.

Recall that for an isolated system the net angular momentum of the system is conserved. If the rotor acquired angular momentum, so must the fluid in the wake. Let us suppose that the fluid in the wake acquires a tangential velocity ${\displaystyle v_{\theta }=2a'\omega r}$. Thus the torque in the air is given by

${\displaystyle |\mathbf {\delta {Q}} |=\rho (2\pi r\delta r)U_{\infty }(1-a)\times (2\Omega a'r^{2})}$

By the conservation of angular momentum, this balances the torque in the blades of the rotor; thus,

${\displaystyle {\frac {1}{2}}\rho W^{2}Nc(c_{l}\sin \phi -c_{d}\cos \phi )r\delta r=\rho (2\pi r\delta r)U_{\infty }(1-a)\times (2\Omega a'r^{2})}$

${\displaystyle {\frac {1}{2}}W^{2}Nc(c_{l}\sin \phi -c_{d}\cos \phi )=4\pi U_{\infty }(1-a)\times \Omega a'r^{2}}$

Furthermore, the rate of change of linear momentum in the air is balanced by the out-of-plane bending force acting on the blades, ${\displaystyle \delta F_{z}}$. From momentum theory, the rate of change of linear momentum in the air is as follows:

${\displaystyle \delta F_{z}=\rho (2\pi r\delta r)U_{\infty }(1-a)\times (v_{\infty }-v_{w})}$

which may be expressed as

${\displaystyle \delta F_{z}=\rho (4\pi r\delta r)U_{\infty }^{2}a(1-a)}$

Balancing this with the out-of-plane bending force gives

${\displaystyle {\frac {1}{2}}W^{2}Nc(c_{l}\cos \phi +c_{d}\sin \phi )=\rho (4\pi r\delta r)U_{\infty }^{2}a(1-a)}$

Let us now make the following definitions:

${\displaystyle C_{y}=c_{l}\sin \phi -c_{d}\cos \phi }$

${\displaystyle C_{x}=c_{l}\cos \phi +c_{d}\sin \phi }$

So we have the following equations:

${\displaystyle {\frac {1}{2}}W^{2}NcC_{y}=4\pi U_{\infty }(1-a)\times \Omega a'r^{2}}$

(1)

${\displaystyle {\frac {1}{2}}\rho W^{2}NcC_{x}=4\pi \rho \left[(a'\Omega r)^{2}+U_{\infty }^{2}a(1-a)\right]r}$

(2)

Let us make reference to the following equation which can be seen from analysis of the above figure:

${\displaystyle \sin \phi ={\frac {U_{\infty }}{W}}(1-a)\rightarrow \sin ^{2}\phi =\left({\frac {U_{\infty }}{W}}(1-a)\right)^{2}}$

(3)

Thus, with these three equations, it is possible to get the following result through some algebraic manipulation:[1]

${\displaystyle {\frac {a}{1-a}}={\frac {C_{x}\sigma _{r}}{4\sin ^{2}\phi }}}$

We can derive an expression for ${\displaystyle a'}$ in a similar manner. This allows us to understand what is going on with the rotor and the fluid. Equations of this sort are then solved by iterative techniques.

• Assumes that each annular ring is independent of every other annular ring.[2]
• Does not account for wake expansion.
• Does not account for tip losses, though correction factors can be included.
• Does not account for yaw, though it can be made to do so.
• Based on steady flow (non-turbulent).