ch13 - -14-3. Simple SolutionsIn this chapter we consider...

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Unformatted text preview: -14-3. Simple SolutionsIn this chapter we consider several applications of the Navier-Stokes equation inwhich the difficult terms are identically zero.3.1. Static EquilibriumThe very simplest situation is whenu=0. The Navier-Stokes equation then reducestop= gzWhose components arepx=py=pz= gThe first two equations imply thatp=p(z). Assuming that g is constant (a very goodapproximation), the third then implies that only=(z) can be compatible with nomotion. Alternatively, we can rewrite the hydrostatic equation using the gravitationalpotential.p=The curl of a gradient is always zero. Thus the left side of this equation is simply p=The chain rule for the curl of a scalar times a vector applied to the right side gives = + Thus =This equation implies that the surfaces of constantmust be parallel to the equipoten-tials. This is clearly the case when=(z) andg=gz= (gz) because the surfaces ofconstant density and the equipotentials are both horizontal and we have the same resultdeduced above.As a slightly more complicated illustration, consider the shape of the free surfaceof water in a vertical cylindrical container rotating about its axis (see Figure 9). After wehave waited long enough that any transient motions associated with spinning up the fluidhave died away,uin a reference frame rotating with the container is zero. If the water isof constant density, the only place where the the density gradient is non-zero is at the freesurface. There it is perpendicular to the surface and points down. Thus the shape of thefree surface must be an equipotential. In the rotating system, the total effective gravitypotential is= gz+122R2, where R is the horizontal distance from the rotation axis.-15-Choosing the equipotential=0, we must havegz=122R2and hence the shape of thefree surface must be parallel to the surfacez=22gR2which is a paraboloid concave upwards. This result has recently been used to make veryaccurate large parabolic astronomical mirrors by freezing a rotating glass melt.Returning to the zcomponent of the static Navier-Stokes equation, we have theordinary differential equationdpdz= (z)gwhich can be directly integrated to givephydrostatic(z)=p(0)z(z)g dzIn the ocean, the constant of integration, p(0) is usually taken to be the mean atmosphericpressure. Because z is negative below the surface and(z) is very nearly constant, thehydrostatic pressure increases downwards at the rate of about one atmosphere every 10meters.Because hydrostatic pressure exists in the absence of motion and because we willgenerally be interested in moving fluids, its is useful to redefine the pressure in the theNavier-Stokes equation to bep=phydrostatic+pIt is the pressure perturbationpthat is dynamically important.For the case when==constant, substituting this into the Navier-Stokes equation and subtracting theequation=...
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This note was uploaded on 10/01/2011 for the course ME 509 taught by Professor Wereley during the Spring '11 term at Purdue University-West Lafayette.

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ch13 - -14-3. Simple SolutionsIn this chapter we consider...

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