Question 8.2: Consider a container of fluid at rest with a Cartesian coord...

From Eqs. (8.36), (8.40), and (8.42), with g=+\rho g\hat{j}, given the downward oriented coordinate direction, we have

                 -\frac{\partial p}{\partial x}+\mu \triangledown ^{2}\nu _{x}+\rho g_{x}=\rho a_{x}.            (8.36)

              -\frac{\partial p}{\partial y}+\mu \triangledown ^{2}\nu _{y}+\rho g_{y}=\rho a_{y}.              (8.40)

                -\frac{\partial p}{\partial z}+\mu \triangledown ^{2}\nu _{z}+\rho g_{z}=\rho a_{z}.                  (8.42)

From the first and third equations, p=p( y) at most, and the partial derivative becomes an ordinary derivative. Solving by integration,

and, consequently, we have FIGURE 8.3 Determination of the pressure as a function of depth in a static fluid. Fundamental Balance Relations

The integration constant c is found from the boundary conditions. Here, note that the so-called gauge pressure is defined as the absolute pressure minus atmospheric pressure. If we assume an atmospheric pressure at the surface, then p(y=0)=0(gauge) and c=0. Thus, p(y)=\rho gy. At y=h, therefore, we obtain the well-known result that p=\rho gh at depth h (that pressure increases with depth is easily appreciated as we swim deeper in a pool). In a sense, then, this is a solution of the Navier–Stokes equation. Because of the importance and utility of the Navier–Stokes equation, much of Chap. 9 is devoted to its solution.