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Unformatted text preview: 2.5 The momentum flux From § 2.2, equation (12), ∂ ∂t ( ρu i ) + ∂M ij ∂x j = F i (18) where M ij = ρu i u j + pδ ij is called the momentum flux , and ρu i is called the momentum density . Note that (18) has the same fundamental structure than (9), as is typical for conservation laws. For example, if I define the flux of xmomentum p x by f ( p x ) j = M 1 ,j , then it follows from (18) that ∂p x ∂t + ∇· f ( p x ) = 0 , which describes the conservation of xmomentum. The equations for the other components are analogous. For steady flows it is often useful to consider the momentum balance over a control volume V . Of course, this brings us back to the original balance the derivation of the Euler equation was based upon. Since the flow is steady, the left hand side of (18) is zero. We also assume that F = − ρ ∇ Φ. Then 0 = integraldisplay V ∂M ij ∂x j dV + ρ integraldisplay V ∂ Φ ∂x j dV = integraldisplay S ( M ij + ρ Φ) n j dS by the divergence theorem. So, back in vector form, integraldisplay S ρ u ( u · n ) + ( p + ρ Φ) n dS = 0 for any closed surface S in/bounding the fluid. This is the momentum integral theorem 2.6 Example: Axisymmetric jet impinging on a wall . What is the force exerted by the flow on the wall ? Ignore the effects of gravity. xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx z r film wall jet p = p atm p = p atm S s u = − U ˆ z n = ˆ z u = U r ˆ r n = ˆ r n = − ˆ z Take a control surface S = S s + S j + S f + S w of cylindrical shape that is composed of the free surface of the jet on one side, and the wall on the other. The surface is closed by a circular surface through the jet far from impact (marked “jet”), and a cylindrical surface where the fluid flows parallel to the wall (marked “film”). Assume uniform flow parallel to wall in the film region far from impact and across the cross section of the jet u = − U ˆ z . Also assume pressure on all boundaries apart from the wall is p atm . Then in the absence of body forces, integraldisplay S j + S s + S f + S w ( ρ u ( u · n ) + p n ) dS = 0 . Now integraldisplay S p n dS = integraldisplay S ( p − p atm ) n dS + integraldisplay S p atm n dS = − integraldisplay S w ( p − p atm )ˆ z dS since integraldisplay S p atm n dS = integraldisplay V ∇ p atm dV = 0 and p = p atm everywhere else. Also, integraldisplay S ρ u ( u · n ) dS = integraldisplay S j ρU 2 ˆ z dS + integraldisplay S f ρU 2 r ˆ r dS since u · n = 0 on the wall and on the surface of the liquid. Bringing it all together and equating terms in direction ˆ z gives integraldisplay S w ( p − p atm ) dS = ρU 2 A, where A is the cross section of the jet. So there is an extra pressure, and the force exerted by the fluid on the wall is ρU 2...
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This note was uploaded on 01/20/2012 for the course MATH 33200 taught by Professor Eggers during the Fall '11 term at University of Bristol.
 Fall '11
 Eggers

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