If the complex potential in the ζ-plane represents uniform horizontal flow past a cylinder with radius a and clockwise circulation Γ (see Figure 7.12a):

$w(\zeta)=U\left(\zeta+a^2 / \zeta\right)+\frac{i \Gamma}{2 \pi} \ln (\zeta / a)$ ,

set |b| = a and use the transformation from Example 14.4, $z=\zeta+\left(a e^{i \alpha}\right)^2 / \zeta$ , and the Kutta condition to show that $C_L=2\pi\sin\alpha$ for ideal flow past a flat plate at angle of attack α.

Question

If the complex potential in the ζ-plane represents uniform horizontal flow past a cylinder with radius a and clockwise circulation Γ (see Figure 7.12a):

[latex]w(\zeta)=U\left(\zeta+a^2 / \zeta\right)+\frac{i \Gamma}{2 \pi} \ln (\zeta / a)[/latex],

set |b| = a and use the transformation from Example 14.4, [latex]z=\zeta+\left(a e^{i \alpha}\right)^2 / \zeta[/latex], and the Kutta condition to show that [latex]C_L=2\pi\sin\alpha[/latex] for ideal flow past a flat plate at angle of attack α.

Accepted Answer

The complex velocity in the ζ-plane is:

[latex]\frac{d w}{d \zeta}=U\left(1-a^2 / \zeta^2 ight)+\frac{i \Gamma}{2 \pi \zeta}[/latex].

To find the angle(s), [latex] heta_s[/latex], of the cylinder-surface stagnation points, set dw/dζ = 0 and substitute in [latex]\zeta=a \exp \left(i heta_s ight)[/latex] to find:

[latex]0=U\left(1-a^2 / a^2 \exp \left(2 i heta_s ight) ight)+\frac{i \Gamma}{2 \pi a \exp \left(i heta_s ight)} ext {, which implies } \sin heta_s=-\Gamma / 4 \pi U a[/latex].

To satisfy the Kutta condition in the z-plane, choose Γ so that the downstream cylinder-surface stagnation point (the S on the right in Figure 7.12a) maps into the flat-plate’s trailing edge. From Example 4.4 with |b| = a, the (x,y)-location of the plate’s trailing edge is (2acosα, −2asinα) and it is mapped from the cylinder-surface stagnation point defined by [latex] heta_s+\alpha=0 ext {, or } heta_s=-\alpha[/latex]. Combine this result with that for the stagnation point location in the ζ-plane to find: [latex]\sin heta_s=-\sin\alpha=-\Gamma/4\pi Ua[/latex] or Γ = 4πUa sinα. Thus, the plate’s coefficient of lift is:

[latex]C_L=\frac{L}{(1 / 2) ho U^2(4 a)}=\frac{ ho U \Gamma}{(1 / 2) ho U^2(4 a)}=\frac{ ho U\cdot 4 \pi U a \sin \alpha}{(1 / 2) ho U^2(4 a)}=2 \pi \sin \alpha[/latex],

where the chord length of the plate is 4a.
While this result is readily anticipated from the formal results for Zhukhovsky airfoils, it does not require any approximations beyond those inherent to ideal flow, and is provided here to show that the body in the z-plane can be rotated by the transformation so that the free stream can be horizontal in both the ζ- and z-planes.

Question 14.5: If the complex potential in the ζ-plane represents uniform h...

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