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# week7 - 4 Two-dimensional ows If the ow is in the plane and...

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4 Two-dimensional flows If the flow is in the plane, and there are only two independent variables x,y , the flow prob- lem is much simplified. Moreover, we will see that some of the physics is fundamentally different from three dimensions. One fact we know of already is that no vorticity is created in two dimensions, ω is simply convected with the flow. Of course, real flows are never truly two-dimensional; however if the geometry extends very far in one direction (think of the wing of an aeroplane), a two-dimensional description is a good approximation. 4.1 Flow past a cylinder Find the flow around a stationary cylinder in a steady stream U = U ˆ x . This is very closely analogous to the flow around a sphere; the effect of the sphere is modelled by a (two-dimensional) dipole, which is the derivative of a source, whose velocity field behaves like u r 1 /r (section 1.10). Thus the potential is φ = ln r , and a dipole in the μ -direction has potential φ = ( μ · ) ln r = μ · x r 2 , The ansatz for the velocity potential is thus φ = Ux + μ · x r 2 , and so u = U ˆ x + μ r 2 2( μ · x ) x r 4 . Now the boundary condition for r = R is u · n = u · x /r = 0, and from the velocity field we find u · n = U x r μ · x r 3 . Thus the boundary condition is satisfied if we choose μ = R 2 U . Thus the final answer for the potential is φ = U bracketleftbigg x + R 2 x r 2 bracketrightbigg , (35) and for the velocity field u = R 2 r 2 [ U 2 n ( U · n )] + U . (36)

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Now we plug this into Bernoulli’s equation to compute the pressure: u 2 2 + p ρ = U 2 2 + p atm ρ . On the surface u 2 vextendsingle vextendsingle r = R = (2 U 2 n ( U · n )) 2 = 4 U 2 4( U · n ) 2 = 4 U 2 (1 cos 2 θ ) = 4 U 2 sin 2 θ, so in other words p = p atm + ρU 2 2 (1 4 sin 2 θ ) . The pressure is once more symmetric about θ = π/ 2, i.e. about the midsection of the cylinder. It follows that the total force on the cylinder is again zero. 4.2 Non-uniqueness of the potential One particularly important aspect of two-dimensional flow is that any solid body placed in the flow domain will create a domain that is no longer simply connected. Defn : A closed curve C is reducible in a domain D if it can be shrunk to a point without ever leaving D . If every closed curve is reducible then D is simply connected . E.g. (i) if D is the interior of a circle, then it is simply connected. (ii) if D is the exterior of a circle then it is not simply connected. D i ii D Example : Let D be the domain a<r<b, 0 <θ< 2 π in cylindrical polars. D is not simply connected. We now place a point vortex (introduced at the beginning of chapter 3) at the centre, which creates a flow u = (0 , Γ / 2 πr, 0) . The strength of the vortex is measured in terms of its circulation Γ. Since the flow is irrotational, there exists a velocity potential s.t. u r = ∂φ ∂r = 0 , u θ = 1 r ∂φ ∂θ = Γ / (2 πr ) .
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week7 - 4 Two-dimensional ows If the ow is in the plane and...

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