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Unformatted text preview: Fluid Dynamics 3 - Solutions to Sheet 9 1. (i) Use the transformation = e z , so the vortex is at = e ib in the upper half plane. Then from the method of images w ( ) = i 2 ( ln( ) ln( ) ) in the upper half plane and w ( z ) = i 2 ln bracketleftbigg e z e ib e z e 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 ) = i 2 ln( z ib )), has to be subtracted, giving w s ( z ) = w ( z ) w ( z ) = i 2 braceleftbigg ln bracketleftbigg e z e ib z ib bracketrightbigg ln(e z e 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 ib ) + e ib 2 ( z ib ) 2 + O ( z ib ) 3 , giving w s ( z ) = i 2 braceleftbigg ln bracketleftbigg e ib + e ib 2 ( z ib ) bracketrightbigg ln(e z e ib ) bracerightbigg . Then the velocity becomes u iv = dw s dz vextendsingle vextendsingle vextendsingle vextendsingle z = ib = i 2 bracketleftbigg 1 2 e ib e ib e ib bracketrightbigg . Now e ib e ib e ib = e ib 2 i sin b = i cos b 2 sin b + sin b 2 sin b , which gives the desired result u + iv = 4 cot b. 2. (i) The original vortex has images at z 1 = i and z 1 = i , and z 1 has images at z = 0 (upper wall) and z 2 = 2 i (lower wall). In general, the images of z n are at z n 1 and z 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 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 n 1 = a n +1 = a n . Thus the vortices are alternating in sign and a n = ( 1) n +1 i 2 ....
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This note was uploaded on 01/20/2012 for the course MATH 33200 taught by Professor Eggers during the Fall '11 term at University of Bristol.
- Fall '11