Aerodynamic Forces

The net aerodynamic force acting on an three-dimensional airfoil of finite size can be written

$${\bf A} =- \int_S p\,{\bf n}\,dS,$$ (9.70)

where the integral is taken over the surface of the airfoil, $S$ . Here, ${\bf n}$ is an outward unit normal vector on $S$ , $dS$ is an element of $S$ , and $p$ is the air pressure. From Bernoulli's theorem (in an irrotational fluid), we can write $p=p_0-(1/2)\,\rho\,v^{\,2}$ , where $p_0$ is a constant pressure. Because a constant pressure exerts no net force on a closed surface, we get

$${\bf A} =\frac{1}{2}\,\rho\int_S U^{\,2}\,{\bf n}\,dS,$$ (9.71)

where ${\bf U}$ is the tangential air velocity just above the surface of the airfoil. Now,

$${\bf U}\times ({\bf n}\times {\bf U}) = U^{\,2}\,{\bf n}- ({\bf n}\cdot{\bf U})\,{\bf U} = U^{\,2}\,{\bf n},$$ (9.72)

because ${\bf n}\cdot{\bf U}=0$ on the surface. Hence,

$${\bf A} = \frac{1}{2}\,\rho\int_S {\bf U}\times({\bf n}\times {\bf U})\,dS.$$ (9.73)

Making use of Equations (9.64) and (9.69), the previous expression can be written

$${\bf A} = \rho\int_S({\bf V}+{\bf v}_{\mit\Sigma}+{\bf v}_S)\times \mbox{\boldmath$\Omega$} _S\,dS = {\bf L}+{\bf D}+{\bf F},$$ (9.74)

where

$${\bf L} = \rho\,{\bf V}\times \int_S \mbox{\boldmath$\Omega$} _S\,dS,$$ (9.75)

$${\bf D} = \rho\int_S {\bf v}_{\mit\Sigma}\times \mbox{\boldmath$\Omega$} _S\,dS,$$ (9.76)

$${\bf F} =\rho\int_S {\bf v}_S\times \mbox{\boldmath$\Omega$} _S\,dS.$$ (9.77)

Here, ${\bf V}$ , ${\bf v}_{\mit\Sigma}$ , and ${\bf v}_S$ are the incident wind velocity, the velocity induced by the free vortices in the wake, and the velocity induced by the bound vortices covering the surface of the airfoil, respectively. The forces ${\bf L}$ and ${\bf D}$ are called the lift and the induced drag, respectively. (Note, that ${\bf L}$ now represents a net force, rather than a force per unit length.) We shall presently demonstrate that the force ${\bf F}$ is negligible.

Let us assume that

$$\mbox{\boldmath$\Omega$} _S \simeq {\mit\Omega}_z\,{\bf e}_z:$$ (9.78)

that is, that the bound vortices covering the surface of the airfoil run parallel to the $z$ -axis. This assumption is exactly correct for an airfoil of infinite wingspan and constant cross-section. Moreover, it is a good approximation for an airfoil of finite wingspan, provided the airfoil's length greatly exceeds its width (i.e., $b\gg c$ ). Now, the incident wind velocity is written ${\bf V}= V\,{\bf e}_\parallel$ . Moreover, $dS = dl\,dz$ , where $dl$ is an element of length that runs parallel to the $x$ -$y$ plane whilst lying on the airfoil surface. Making use of the curl theorem, we can easily show that

$$\oint_C {\mit\Omega}_z\,dl = {\mit\Gamma}(z),$$ (9.79)

where the closed curve $C$ is the intersection of the airfoil surface with the plane $z=z$ , and ${\mit\Gamma}(z)$ is the air circulation about the airfoil in this plane. Thus, it follows from Equation (9.75) that

$${\bf L} = \rho\,V\int_{-b/2}^{b/2}{\mit\Gamma}(z)\,dz\,\,{\bf e}_\perp.$$ (9.80)

This expression is the generalization of Equation (9.31) for a three-dimensional airfoil of finite size. As before, the lift is at right-angles to the incident wind direction.

Let us make the further assumption--known as the lifting line approximation (because the lifting action of the wing is effectively concentrated onto a line)--that

$${\bf v}_{\mit\Sigma} = - w(z)\,{\bf e}_\perp$$ (9.81)

throughout $S$ , where $-w(z)\,{\bf e}_\perp$ is the induced velocity due to the free vortices in ${\mit\Sigma}$ , evaluated at the trailing edge of the airfoil. Here, the velocity $w(z)$ is called the downwash velocity. It follows from Equation (9.76) that

$${\bf D} = \rho\,\int_{-b/2}^{b/2} w(z)\,{\mit\Gamma}(z)\,dz\,\,{\bf e}_\parallel.$$ (9.82)

Note that the induced drag is parallel to the incident wind direction. The origin of induced drag is as follows. It takes energy to constantly resupply free vortices to the wake, as they are swept downstream by the wind (note that a vortex filament possesses energy by virtue of the kinetic energy of its induced flow pattern), and this energy is supplied by the work done in opposing the induced drag. The drag acting on a well-designed airfoil (i.e., an airfoil with an aerodynamic shape that minimizes form drag) situated in a high Reynolds number wind (which implies that the friction drag is negligible) is generally dominated by induced drag.

According to Equations (9.62) and (9.77), the force ${\bf F}$ is written

$${\bf F} = \frac{\rho}{4\pi}\int_S\int_{S'}\frac{[\mbox{\boldmath$... ...mbox{\boldmath$\Omega$}_S({\bf r})}{\vert{\bf r}-{\bf r}'\vert^{\,3}}\,dS\,dS'.$$ (9.83)

We can interchange primed and unprimed variables without changing the value of the integral. Hence,

$${\bf F} = \frac{\rho}{4\pi}\int_S\int_{S'}\frac{[\mbox{\boldmath$... ...box{\boldmath$\Omega$}_S({\bf r}')}{\vert{\bf r}'-{\bf r}\vert^{\,3}}\,dS'\,dS.$$ (9.84)

Taking the half the sum of the previous two equations, we obtain

$${\bf F} = \frac{\rho}{8\pi}\int_S\int_{S'}\frac{[\mbox{\boldmath$... ...box{\boldmath$\Omega$}_S({\bf r}')}{\vert{\bf r}-{\bf r}'\vert^{\,3}}\,dS\,dS'.$$ (9.85)

However, $({\bf a}\times{\bf b})\times {\bf c} + ({\bf b}\times{\bf c})\times{\bf a} + ({\bf c}\times {\bf a})\times {\bf b} = {\bf0}$ . Thus, the previous expression yields

$${\bf F} = \frac{\rho}{8\pi}\int_S\int_{S'}\frac{[\mbox{\boldmath$... ...{\bf r})]\times ({\bf r}-{\bf r}')}{\vert{\bf r}-{\bf r}'\vert^{\,3}}\,dS\,dS'.$$ (9.86)

But, the assumption (9.78) implies that $\Omega$ $_S({\bf r}')\times$$\Omega$ $_S({\bf r})\simeq {\bf0}$ . Hence, ${\bf F}$ is negligible, as was previously stated.

Consider a closed surface covering the small section of the airfoil lying between the parallel planes $z=z$ and $z=z+dz$ . The flux of vorticity into the surface due to bound vortices at $z$ is ${\mit\Gamma}(z)$ . The flux of vorticity out of the surface due to bound vortices at $z+dz$ is ${\mit\Gamma}(z+dz)$ . Finally, the flux of vorticity out of the surface due to the free vortices in the part of the wake lying between $z$ and $z+dz$ is $I(z)\,dz$ . However, the net flux of vorticity out of a closed surface is zero, because vorticity is divergence free. Hence,

$${\mit\Gamma}(z) = {\mit\Gamma}(z+dz) + I(z)\,dz,$$ (9.87)

which implies that

$$I(z) = - \frac{d{\mit\Gamma}}{dz}.$$ (9.88)

Figure 9.12: Semi-infinite vortex filament.

Finally, consider a semi-infinite straight vortex filament of vortex intensity $\Gamma$ $= -{\mit\Gamma}\,{\bf e}_x$ that terminates at the origin, $O$ , as shown in Figure 9.12. Let us calculate the flow velocity induced by this filament at the point $P=(0,\,0,\,z)$ . From the diagram $l=z\,\tan\phi$ , $dl= z\,\sec^2\phi\,d\phi$ , $\vert{\bf r}-{\bf r}'\vert= z\,\sec\phi$ , and $\Gamma$ $\times \vert{\bf r}-{\bf r}'\vert = {\mit\Gamma}\,z\,{\bf e}_y$ . Hence, from Equation (9.61), the induced velocity at $P$ is ${\bf v}= v_y\,{\bf e}_y$ , where

$$v_y = \frac{{\mit\Gamma}}{4\pi\,z}\int_0^{\pi/2}\cos\phi\,d\phi = \frac{{\mit\Gamma}}{4\pi\,z}.$$ (9.89)

This result allows us to calculate the downwash velocity, $w(z)=-v_y(z)$ , induced at the trailing edge of the airfoil by the semi-infinite free vortices in the wake. The vortex intensity in the small section of the wake lying between $z$ and $z+dz$ is $I(z)\,dz$ , so we obtain

$$w(z) =- \frac{1}{4\pi}\int_{-b/2}^{b/2} \frac{I(z')\,dz'}{z-z'}= \frac{1}{4\pi}\int\frac{d{\mit\Gamma}(z')}{z-z'},$$ (9.90)

where use has been made of Equation (9.88).