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Unformatted text preview: Fluid Dynamics 3  Solutions to Sheet 3 1. (i) Using the formula for u = ∇× A in cylin drical polars, one obtains u r = − 1 r ∂ Ψ ∂z , u z = 1 r ∂ Ψ ∂r and thus ∇· u = 1 r ∂ ( ru r ) ∂r + ∂u z ∂z = − 1 r ∂ 2 Ψ ∂r∂z + 1 r ∂ 2 Ψ ∂z∂r = 0 . (ii) Streamlines are given by d r u r = d z u z , so ∂ Ψ ∂r d r + ∂ Ψ ∂z d z = dΨ = 0 Therefore Ψ = const on streamlines (actually stream surfaces as 3D flow). (iii) The streamfunction has to obey the equa tions ∂ Ψ ∂z = 0 , ∂ Ψ ∂r = Ur, and thus Ψ is independent of r according to the first equation, and the second equation can be integrated to Ψ = Ur 2 2 . 2. Twodimensional flow u = ( u,v, 0), with u = ∂ψ ∂y , v = − ∂ψ ∂x (i) The streamline equations are v d x − u d y = 0, so d ψ = ∂ψ ∂x d x + ∂ψ ∂y d y = 0 and so ψ is constant along streamlines. (ii) Here  u  = ( u 2 + v 2 ) 1 2 = bracketleftBigg parenleftbigg ∂ψ ∂x parenrightbigg 2 + parenleftbigg ∂ψ ∂y parenrightbigg 2 bracketrightBigg = ∇ ψ  So if u increases then the gradient of ψ in creases and the streamlines thus get closer to gether. (iii) Volume flux, Q , across the curve C joining the points x and x 1 is Q = integraldisplay x 1 x u · n d s where d s is the arc length along the curve. But here, n = (d y, − d x ) / d s (because d s = (d x, d y ), so the unit tangent in the direction of the curve is (d x, d y ) / d s , and this is perpen dicular to n ). Now Q = integraldisplay x 1 x parenleftbigg ∂ψ ∂x , − ∂ψ ∂y parenrightbigg . (d y, − d x ) d s d s and so Q = integraldisplay x 1 x d ψ = ψ ( x 1 ) − ψ ( x ) 3. Here u = ( u,v, 0) with ∂u ∂x = − 2( y − b )( x − a ) [( x − a ) 2 + ( y − b ) 2 ] 2 and ∂v ∂y = 2( y − b )( x − a ) [( x − a ) 2 + ( y − b ) 2 ] 2 So mass conservation is satisfied: ∇· u = ∂u ∂x + ∂v ∂y = 0 We seek a ψ ( x,y ) such that u = ∂ψ ∂y , v = − ∂ψ ∂x Easy to see that ψ = 1 2 log[( x − a ) 2 + ( y − b ) 2 ] Streamlines are ψ = const which implies that ( x − a ) 2 + ( y − b ) 2 = const and these are circles centred at x = a , y = b : y x Figure 1: Streamlines are circles centred at ( a,b ). Flow direction clockwise (from observing signs of u and v in portions of the sketch) 4. Start with the Stokes streamfunction for a source of strength m . Ψ = m 4 π parenleftBigg 1 − z ( z 2 + r 2 ) 1 2 parenrightBigg (i) Define a source sink pair at z = 0, z = a : Ψ 1 = m 4 π parenleftBigg 1 − z ( z 2 + r 2 ) 1 2 parenrightBigg − m 4 π parenleftBigg 1 − z − a (( z − a ) 2 + r 2 ) 1 2 parenrightBigg = m 4 π parenleftBigg z − a (( z − a ) 2 + r 2 ) 1 2 − z ( z 2 + r 2 ) 1 2 parenrightBigg (ii) First we write Figure 2: Streamlines for part (i): A source/sink pair along the...
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 Fall '11
 Eggers
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