101 102 chapter 4 applications of the navierstokes

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102 CHAPTER 4. APPLICATIONS OF THE NAVIER–STOKES EQUATIONS 4.1.1 Equations of fluid statics The equations of fluid statics are usually derived by a simple force balance under the assumption of mechanical equilibrium (pressure independent of time in a motionless— i.e. , static—fluid) for the fluid system. Here, however, we obtain them as a special case of the N.–S. equations. This is actually much easier; and now that we have these equations available, such an approach, among other things, emphasizes the unity of fluid mechanics in general as embodied in the N.–S. equations. Thus, we begin by writing the 3-D incompressible N.–S. equations in a Cartesian coordinate system with the positive z direction opposite that of gravitational acceleration. This gives u x + v y + w z = 0 , (4.1a) u t + uu x + vu y + wu z = 1 ρ p x + ν u , (4.1b) v t + uv x + vv y + wv z = 1 ρ p y + ν v , (4.1c) w t + uw x + vw y + ww z = 1 ρ p z + ν w g . (4.1d) Now because the fluid is assumed to be motionless, we have u = v = w 0, implying that all derivatives of velocity components also are zero. In particular, there can be no viscous forces in a static fluid. Clearly, the continuity equation is satisfied trivially, and the momentum equations collapse to p x = 0 , (4.2a) p y = 0 , (4.2b) p z = ρg = γ , (4.2c) where γ is the specific weight defined in Chap. 2. The first two of these equations imply that in a static fluid the pressure is constant throughout planes aligned perpendicular to the gravitational acceleration vector; i.e. , p x = 0 implies p is not a function of x , and similarly for y . We now integrate Eq. (4.2c) to formally obtain p ( x, y, z ) = γz + C ( x, y ) , where C ( x, y ) is an integration function . (When integrating a partial differential equation we usually obtain integration functions instead of integration constants.) But we have already noted that Eqs. (4.2a) and (4.2b) imply p cannot be a function of either x or y ; so in this special case C is a constant, and we then have p ( z ) = γz + C . (4.3) We remark here that the first liquid considered in this context was water, and as a result the contribution γz = ρgz to the pressure in a static fluid is often termed the hydrostatic pressure . The value of C must now be determined by assigning a value to p , say p 0 , at some reference height z = h 0 . This leads to p 0 = p ( h 0 ) = γh 0 + C C = p 0 + γh 0 , and from this it follows that p ( z ) = p 0 + γ ( h 0 z ) . (4.4) This actually represents the solution to a simple boundary-value problem consisting of the first- order differential equation (4.2c) and the boundary condition p ( h 0 ) = p 0 , and we will emphasize this viewpoint in the sequel as we introduce several examples to demonstrate use of the equations of fluid statics.
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4.1. FLUID STATICS 103 Pascal’s Law One of the first results learned about fluid statics in high school physics classes is Pascal’s law . We will state this, provide physical and mathematical explanations of it, and consider an important application.
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