Hydrostatic Pressure

Consider a body of water which is stationary in a reference frame that is fixed with respect to the Earth's surface. In this chapter, such a frame is treated as approximately inertial. Let $z$ measure vertical height, and suppose that the region $z\leq 0$ is occupied by water, and the region $z>0$ by air. According to Equation (1.79), the air/water system remains in mechanical equilibrium (i.e., ${\bf v} = D{\bf v}/Dt= {\bf0}$ ) provided

$${\bf0} = \nabla p + \rho\,\nabla{\mit\Psi},$$ (2.1)

where $p$ is the static fluid pressure, $\rho$ the mass density, ${\mit\Psi}=g\,z$ the gravitational potential energy per unit mass, and $g$ the (approximately uniform) acceleration due to gravity. Now,

$$\rho(z) = \left\{ \begin{array}{ccc} 0&\mbox{\hspace{0.5cm}}&z>0\\ [0.5ex] \rho_0&&z\leq 0 \end{array}\right.,$$ (2.2)

where $\rho_0$ is the (approximately uniform) mass density of water. Here, the comparatively small mass density of air has been neglected. Because $\rho=\rho(z)$ and ${\mit\Psi}={\mit\Psi}(z)$ , it immediately follows, from Equation (2.1), that $p=p(z)$ , where

$$\frac{dp}{dz} = -\rho\,g.$$ (2.3)

We conclude that constant pressure surfaces in a stationary body of water take the form of horizontal planes. Making use of Equation (2.2), the previous equation can be integrated to give

$$p(z)=\left\{\begin{array}{llc} p_0&\mbox{\hspace{1cm}}&z>0\\ [0.5ex] p_0-\rho_0\,g\,z&&z\leq 0 \end{array}\right.,$$ (2.4)

where $p_0\simeq 10^5\,{\rm N\,m}^{-2}$ is atmospheric pressure at ground level (Batchelor 2000). According to this expression, pressure in stationary water increases linearly with increasing depth (i.e., with decreasing $z$ , for $z<0$ ). In fact, given that $g\simeq 9.8\,{\rm m\,s}^{-2}$ and $\rho_0\simeq 10^3\,{\rm kg\,m^{-3}}$ (Batchelor 2000), we deduce that hydrostatic pressure in water rises at the rate of 1 atmosphere (i.e., $10^5\,{\rm N\,m}^{-2}$ ) every $10.2\,{\rm m}$ increase in depth below the surface.