solnsheet4 - Fluid Dynamics 3 - Solutions to Sheet 4 1. (i)...

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Unformatted text preview: Fluid Dynamics 3 - Solutions to Sheet 4 1. (i) For the given flow field, u = f ( r ) in cylindrical polars. Thus, applying the formula from chapter 0, u = 1 r u = 0 . (ii) We have to solve the steady Euler equation ( u ) u = p/. According to chapter 0, the convective term on the left is ( u ) u = u 2 r r = f 2 r r . Thus p can only depend on r , so that p = p ( r ) r , and the required equation follows. (iii) According to chapter 0, u = z r rf r , which vanishes if f = A/r . From (ii), we have p ( r ) = A 2 r 3 , and so p ( r ) = A 2 2 r 2 + p atm . 2. (i) Conservation of mass is volume flux in equals volume flux out, so Ud = U 1 b 1 + U 2 b 2 (ii) Use the momentum integral theorem. Need to choose a control surface - this should be a boundary where we know stuff about the solu- tion. So follow example in lectures and take it to be around the flow cutting the inflow and outflows far away from the impingement. Let S be the contour ABCDEF in the figure: Deal with pressure term first: A B C D E F integraldisplay S p n d S = integraldisplay S ( p p atm ) n d S + integraldisplay S p atm n d S The last term is zero by the divergence theo- rem (see lectures). The first term is zero along all parts of S apart from the section DE . The other term is integraldisplay...
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solnsheet4 - Fluid Dynamics 3 - Solutions to Sheet 4 1. (i)...

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