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Unformatted text preview: Fluid Dynamics 3  Solutions to Sheet 6 1. (i) (a) The potential of a uniform stream is = Uz , and the result follows from z/r = cos . (b) The potential of a dipole = z is 1 r = x r 3 = z r 3 = cos r 2 . (ii) The Laplacian gives (a) 4 = U cos r 2 r 2 r U r sin sin 2 = 2 U cos r 2 U cos r = 0 . and (b) 4 = 2 cos r 2 r 1 r + r 4 sin sin 2 = 2 cos r 4 + 2 cos r 4 = 0 . (iii) The velocity components are (a) u r = r = U cos , u = 1 r = Ur sin , and (b) u r = r = 2 cos r 3 , u = 1 r = sin r 3 . (iv) n = u r , and u r  r = R = U cos + 2 A cos R 3 , so A = UR 3 / 2 does the trick. 2. (i) In three dimensions, 4 (1 /r ) = 0. Thus 4 = A 4 1 r = 0 . Explicit calculation gives = A 1 r = A r r 3 . (ii) The boundary condition is u n = U n . Now u i = i = i A j r j r 3 = A i r 3 + 3 A j r j r i r 5 , and hence u n  r = R = A n R 3 + 3 A n R 3 = 2 A n R 3 . Thus using the boundary condition we have U = 2 A /R 3 and = R 3 2 r 2 U n . (iii) Since the centre of the sphere moves according to x = U t , the above potential has the form ( x ,t ) = ( x U t ), and thus t = U = R 3 2 r 3 3( U n ) 2 U 2 . This means one has to use the timedependent form of Bernoullis equation u 2 2 + p = p atm  t ....
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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
 Eggers

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