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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 z , u z = 1 r r and thus u = 1 r ( ru r ) r + u z z = 1 r 2 rz + 1 r 2 zr = 0 . (ii) Streamlines are given by d r u r = d z u z , so r d r + z d z = d = 0 Therefore = const on streamlines (actually stream surfaces as 3D flow). (iii) The streamfunction has to obey the equa- tions z = 0 , r = Ur, and thus is independent of r according to the first equation, and the second equation can be integrated to = Ur 2 2 . 2. Two-dimensional flow u = ( u,v, 0), with u = y , v = x (i) The streamline equations are v d x u d y = 0, so d = x d x + y d y = 0 and so is constant along streamlines. (ii) Here | u | = ( u 2 + v 2 ) 1 2 = bracketleftBigg parenleftbigg x parenrightbigg 2 + parenleftbigg y parenrightbigg 2 bracketrightBigg = | | So if u increases then the gradient of 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 x , y parenrightbigg . (d y, d x ) d s d s and so Q = integraldisplay x 1 x d = ( x 1 ) ( 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 ( x,y ) such that u = y , v = x Easy to see that = 1 2 log[( x a ) 2 + ( y b ) 2 ] Streamlines are = 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 . = m 4 parenleftBigg 1 z ( z 2 + r 2 ) 1 2 parenrightBigg (i) Define a source sink pair at z = 0, z = a : 1 = m 4 parenleftBigg 1 z ( z 2 + r 2 ) 1 2 parenrightBigg m 4 parenleftBigg 1 z a (( z a ) 2 + r 2 ) 1 2 parenrightBigg = m 4 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