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solnsheet6

# solnsheet6 - Fluid Dynamics 3 Solutions to Sheet 6(ii The...

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Fluid Dynamics 3 - Solutions to Sheet 6 1. (i) (a) The potential of a uniform stream is φ = Uz , and the result follows from z/r = cos θ . (b) The potential of a dipole μ = μ ˆ z is μ · ∇ 1 r = - μ · x r 3 = - μz r 3 = - μ cos θ r 2 . (ii) The Laplacian gives (a) 4 φ = U cos θ r 2 ∂r 2 ∂r - U r sin θ sin 2 θ ∂θ = 2 U cos θ r - 2 U cos θ r = 0 . and (b) 4 φ = 2 μ cos θ r 2 ∂r - 1 ∂r + μ r 4 sin θ sin 2 θ ∂θ = - 2 μ cos θ r 4 + 2 μ cos θ r 4 = 0 . (iii) The velocity components are (a) u r = ∂φ ∂r = U cos θ, u θ = 1 r ∂φ ∂θ = - Ur sin θ, and (b) u r = ∂φ ∂r = 2 μ cos θ r 3 , u θ = 1 r ∂φ ∂θ = μ sin θ r 3 . (iv) n · ∇ φ = u r , and u r | r = R = U cos θ + 2 A cos θ R 3 , so A = UR 3 / 2 does the trick. 2. (i) In three dimensions, 4 (1 /r ) = 0. Thus 4 φ = A · ∇4 1 r = 0 . Explicit calculation gives φ = A · ∇ 1 r = - A · r r 3 . (ii) The boundary condition is u · n = U · n . Now u i = i φ = - 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 φ = - 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 φ ( x , t ) = φ ( x - U t ), and thus ∂φ ∂t = - U φ = - R 3 2 r 3 3( U · n ) 2 - U 2 . This means one has to use the time-dependent form of Bernoulli’s equation u 2 2 + p ρ = p atm ρ - ∂φ ∂t . Now u = φ = R 3 2 r 3 [3 n ( U · n ) - U ] , and consequently u 2 = R 6 4 r 6 3( U · n ) 2 + U 2 . Thus the pressure distribution around the sur- face is p - p atm | r = R = ρU 2 8 ( 9 cos 2 θ - 5 ) , where θ is the angle between U and n . This is the same as found in the lecture for a sphere 1

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solnsheet6 - Fluid Dynamics 3 Solutions to Sheet 6(ii The...

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