This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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. Twodimensional 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...
View
Full
Document
 Fall '11
 Eggers

Click to edit the document details