As a simplified means to explain how a flying aircraft’s weight is transmitted to the ground, consider two-dimensional ideal flow with density ρ and speed U past an ideal vortex of strength Γ a distance H above an infinite flat surface (see Figure 14.11). Integrate the pressure distribution on the

Question

As a simplified means to explain how a flying aircraft’s weight is transmitted to the ground, consider two-dimensional ideal flow with density ρ and speed U past an ideal vortex of strength Γ a distance H above an infinite flat surface (see Figure 14.11). Integrate the pressure distribution on the surface to show that it carries a load of ρUΓ when H → ∞.

Accepted Answer

Choose the flat surface at y = 0, and presume the uniform inflow is horizontal along the x-axis so that the stream function (see Sections 7.2 and 7.3) may be written:

[latex]\psi(x, y)=U y+\frac{\Gamma}{2 \pi} \ln \left(\sqrt{x^2+(y-H)^2} ight)-\frac{\Gamma}{2 \pi} \ln\left(\sqrt{x^2+(y+H)^2} ight)[/latex]

where Γ is positive in the clockwise direction, and the method of images (see Section 7.3) has been used to represent the surface. In Figure 14.11, the actual and image vortices appear as solid and open circles, respectively. With this stream function the velocity components are:

[latex]u(x, y)=\frac{\partial \psi}{\partial y}=U+\frac{\Gamma}{2 \pi}\left(\frac{y-H}{x^2+(y-H)^2}-\frac{y+H}{x^2+(y+H)^2} ight)[/latex], and

[latex]v(x, y)=-\frac{\partial \psi}{\partial x}=-\frac{\Gamma}{2 \pi}\left(\frac{x}{x^2+(y-H)^2}-\frac{x}{x^2+(y+H)^2} ight)[/latex].

For steady two-dimensional ideal flow in Cartesian coordinates, the simplest version of the Bernoulli equation applies along the surface:

[latex]p(x, y)+\frac{1}{2} ho\left(u^2(x, y)+v^2(x, y) ight)=p_{\infty}+\frac{1}{2} ho U^2, \quad ext { or for } \quad y=0: p(x, 0)-p_{\infty}=\frac{ ho}{2}\left(U^2-u^2(x, 0) ight)[/latex],

where the second form occurs because υ(x, 0) = 0. From above, the horizontal velocity on thesurface is:

[latex]u(x, 0)=U-\frac{\Gamma}{\pi}\left(\frac{H}{x^2+H^2} ight), \quad ext { so } \quad U^2-u^2(x,0)=2 U \frac{\Gamma}{\pi}\left(\frac{H}{x^2+H^2} ight)-\frac{\Gamma^2}{\pi^2}\left(\frac{H}{x^2+H^2} ight)^2[/latex]

Thus, pressure force (per unit length) is:

[latex]\overset{+\infty}{\underset{-\infty}{\int}}\left(p(x, 0)-p_{\infty} ight) d x=\frac{ ho U \Gamma}{\pi}\overset{+\infty}{\underset{-\infty}{\int}}\frac{H}{x^2+H^2} d x-\frac{ ho \Gamma^2}{2 \pi^2}\overset{+\infty}{\underset{-\infty}{\int}}\left(\frac{H}{x^2+H^2} ight)^2 d x[/latex].

Both integrals can be evaluated using the variable substitution x = Htanξ, to find:

[latex]\overset{+\infty}{\underset{-\infty}{\int}}\left(p(x, 0)-p_{\infty} ight) d x=\frac{ ho U \Gamma}{\pi}[\xi]_{-\pi / 2}^{+\pi / 2}-\frac{ ho \Gamma^2}{2 \pi^2 H}\overset{+\pi/2}{\underset{-\pi/2}{\int}}\cos ^2 \xi d \xi= ho U \Gamma- ho \frac{\Gamma^2}{4 \pi H}[/latex].

The first term of the final answer balances the lift force on the vortex and is independent of how far the vortex is from the surface. This lift force is transmitted to the surface through the combined effects of the vortex’s induced velocity and the free-stream flow. The second term of the final answer is an interference term that is negligible as H → ∞. For common aircraft geometries in three dimensions, this term is more complicated and includes contributions from multiple vortices. It leads to what pilots call ground effect when landing (see Exercise 14.18).

Question 14.3: As a simplified means to explain how a flying aircraft’s wei...

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