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# week9 - 4.9 The Joukowski mapping circles to ellipses A...

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4.9 The Joukowski mapping: circles to ellipses A particularly useful application of the mapping idea concerns the flow around bodies. We have solved the problem of the flow around a cylinder. Thus if we can find a conformal mapping between the unit circle and any given shape, we have solve the flow problem around this shape. z ζ R 2 c 2 d z = ζ + b 2 ζ Consider the (inverse) mapping z = ζ + b 2 ζ , (42) called the Joukowski mapping. Consider a circle of radius R , whose surface in the ζ -plane is described by the polar representation ζ = R e . Under the Joukowski mapping, z = R e + b 2 R e - = parenleftbigg R + b 2 R parenrightbigg cos θ + i parenleftbigg R b 2 R parenrightbigg sin θ = c cos θ + id sin θ is an ellipse with axes length 2 c and 2 d . The semiaxes come out to be c = R + b 2 R , d = R b 2 R .

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Now consider the uniform flow past an ellipse. To model an arbitrary angle α between the direction of the flow and the semi-major axis, we consider the flow around a cylinder that approaches the x -axis under an angle α : w 1 ( ζ ) = U parenleftbigg ζ e - + R 2 ζ e parenrightbigg . In principle, one can find w ( z ) using the inverse of the Joukowski mapping ζ = f ( z ) = 1 2 ( z + z 2 4 b 2 ) , so that w ( z ) = w 1 ( f ( z )). However, the resulting expressions are often not so useful. For example, to find the streamlines, it is much easier to find the streamlines of the w 1 ( ζ ) in the ζ -plane, and then to transform them using (42). This is how the pictures were produced. If one wants to calculate the velocity, one uses u iv = d w d z = d w 1 d ζ 1 d z/ d ζ = U (e - R 2 e 2 ) (1 b 2 2 ) . On the cylinder, ζ = R e , so u iv = U (e - e e - 2 ) (1 ( b 2 /R 2 )e - 2 ) = 2 iU sin( θ α ) (e ( b 2 /R 2 )e - ) . (43) This means there are stagnation points at θ = α and θ = π + α . This point is where a streamline leaves the surface. In other words, this streamline (plotted in red) has the same value of the streamfunction ψ then the surface of the ellipse. 4.10 Lift Now we want to fly! In principle, we know how to construct the flow around a wing of arbitrary shape, we only have to find the transformation, starting from a circle. We have seen already that the key ingredient is to have circulation around the wing. We have calculated the lift in the case of a cylindrical cross-section, but what is it for an arbitrary shape? 4.10.1 Blasius’ theorem C t = (cos χ, sin χ ) n = (sin χ, cos χ ) χ x y
We now calculate the force on a fixed body in a stream, using complex notation. This is a close relative of the calculation done in section (3.6), leading to (33). Suppose a fixed rigid body, boundary C is in a steady flow, generating a potential w ( z ). As we saw in 4.4, dw dz = u iv = q e - , (44) where | u | = q is the speed of the flow, and χ is the angle the flow direction makes to the horizontal. So, according to Bernoulli (no gravity): p = p 0 1 2 ρq 2 , where p 0 = p atm + ρU 2 / 2 is a constant.

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week9 - 4.9 The Joukowski mapping circles to ellipses A...

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