{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# ch13 - -14 3 Simple Solutions In this chapter we consider...

This preview shows pages 1–3. Sign up to view the full content.

-14- 3. Simple Solutions In this chapter we consider several applications of the Navier-Stokes equation in which the difficult terms are identically zero. 3.1. Static Equilibrium The very simplest situation is when u = 0. The Navier-Stokes equation then reduces to p = − ρ g ˆ z Whose components are p x = 0 p y = 0 p z = − ρ g The first two equations imply that p = p ( z ). Assuming that g is constant (a very good approximation), the third then implies that only ρ = ρ ( z ) can be compatible with no motion. Alternatively, we can rewrite the hydrostatic equation using the gravitational potential. p = ρ φ The curl of a gradient is always zero. Thus the left side of this equation is simply ∇ × ∇ p = 0 The chain rule for the curl of a scalar times a vector applied to the right side gives ∇ × ρ φ = ∇ ρ × ∇ φ + ρ ∇ × ∇ φ Thus ρ × ∇ φ = 0 This equation implies that the surfaces of constant ρ must be parallel to the equipoten- tials. This is clearly the case when ρ = ρ ( z ) and g = g ˆ z = ∇ ( gz ) because the surfaces of constant density and the equipotentials are both horizontal and we have the same result deduced above. As a slightly more complicated illustration, consider the shape of the free surface of water in a vertical cylindrical container rotating about its axis (see Figure 9). After we have waited long enough that any transient motions associated with spinning up the fluid have died away, u in a reference frame rotating with the container is zero. If the water is of constant density, the only place where the the density gradient is non-zero is at the free surface. There it is perpendicular to the surface and points down. Thus the shape of the free surface must be an equipotential. In the rotating system, the total effective gravity potential is φ = − gz + 1 2 2 R 2 , where R is the horizontal distance from the rotation axis.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
-15- Choosing the equipotential φ = 0, we must have gz = 1 2 2 R 2 and hence the shape of the free surface must be parallel to the surface z = 2 2 g R 2 which is a paraboloid concave upwards. This result has recently been used to make very accurate large parabolic astronomical mirrors by freezing a rotating glass melt. Returning to the ˆ z component of the static Navier-Stokes equation, we have the ordinary differential equation dp dz = − ρ ( z ) g which can be directly integrated to give p hydrostatic ( z ) = p (0) z 0 ρ ( z ) g dz In the ocean, the constant of integration, p(0) is usually taken to be the mean atmospheric pressure. Because z is negative below the surface and ρ ( z ) is very nearly constant, the hydrostatic pressure increases downwards at the rate of about one atmosphere every 10 meters. Because hydrostatic pressure exists in the absence of motion and because we will generally be interested in moving fluids, its is useful to redefine the pressure in the the Navier-Stokes equation to be p = p hydrostatic + p It is the pressure perturbation p that is dynamically important. For the case when ρ = ρ 0 = constant
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern