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Unformatted text preview: Fluid Dynamics 3 - Solutions to Sheet 9 1. (i) Use the transformation ΞΆ = e z , so the vortex is at ΞΆ = e ib in the upper half plane. Then from the method of images w ( ΞΆ ) = β i Ξ 2 Ο ( ln( ΞΆ β ΞΆ ) β ln( ΞΆ β ΞΆ ) ) in the upper half plane and w ( z ) = β i Ξ 2 Ο ln bracketleftbigg e z β e ib e z β e β ib bracketrightbigg in the channel. (ii) To find the velocity field convecting the vortex, the piece coming from the vortex at z = ib (which is w ( z ) = β i Ξ 2 Ο ln( z β ib )), has to be subtracted, giving w s ( z ) = w ( z ) β w ( z ) = β i Ξ 2 Ο braceleftbigg ln bracketleftbigg e z β e ib z β ib bracketrightbigg β ln(e z β e β ib ) bracerightbigg Since we want to know the velocity field at z = ib , we expand about this point: e z = e ib + e ib ( z β ib ) + e ib 2 ( z β ib ) 2 + O ( z β ib ) 3 , giving w s ( z ) = β i Ξ 2 Ο braceleftbigg ln bracketleftbigg e ib + e ib 2 ( z β ib ) bracketrightbigg β ln(e z β e ib ) bracerightbigg . Then the velocity becomes u β iv = dw s dz vextendsingle vextendsingle vextendsingle vextendsingle z = ib = β i Ξ 2 Ο bracketleftbigg 1 2 β e ib e ib β e β ib bracketrightbigg . Now e ib e ib β e β ib = e ib 2 i sin b = β i cos b 2 sin b + sin b 2 sin b , which gives the desired result u + iv = Ξ 4 Ο cot b. 2. (i) The original vortex has images at z 1 = i and z β 1 = β i , and z 1 has images at z = 0 (upper wall) and z β 2 = β 2 i (lower wall). In general, the images of z n are at z β n β 1 and z β n +1 . Evidently, this generates an infinity of images at all integer values. (ii) Let the contribution of each vortex to the complex poten- tial be ia n ln( z β z n ). To satisfy the boundary conditions at each wall, the amplitude of each images must be of op- posite sign and equal strength: a β n β 1 = a β n +1 = β a n . Thus the vortices are alternating in sign and a n = ( β 1) n +1 i Ξ 2 Ο ....
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- Fall '11
- Sin, Hyperbolic function, EIB