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**Unformatted text preview: **◦ 15 ◦ 30 ◦ Figure 7.29. Fluid Flow Past a Tilted Plate. Note that ∇ ϕ = (1 , 0 ), and hence this flow satisfies the Neumann boundary conditions (7.95) on the horizontal segment D = ∂ Ω. The corresponding complex potential is χ ( z ) = z , with complex velocity f ( z ) = χ ′ ( z ) = 1. Example 7.41. Circular disk . Recall that the Joukowski conformal map ζ = g ( z ) = 1 2 parenleftbigg z + 1 z parenrightbigg (7 . 96) squashes the unit circle | z | = 1 down to the real line segment [ − 1 , 1] in the ζ plane. Therefore, it will map the fluid flow outside a unit disk to the fluid flow past the line segment, which, according to the previous example, has complex potential Θ( ζ ) = ζ . The resulting complex potential is χ ( z ) = Θ ◦ g ( z ) = g ( z ) = 1 2 parenleftbigg z + 1 z parenrightbigg . (7 . 97) Except for a factor of 1 2 , indicating that the corresponding flow past the disk is half as fast, this agrees with the potential we derived in Example 7.17. Example 7.42. Tilted plate . Let us next consider the case of a tilted plate in a uniformly horizontal fluid flow. Thus, the cross-section is the line segment z ( t ) = te i φ , − 1 ≤ t ≤ 1 , obtained by rotating the horizontal line segment [ − 1 , 1] through an angle φ , as in Figure 7.29. The goal is to construct a fluid flow past the tilted segment that is asymptotically horizontal at large distance....

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