Static Fluids

A fluid in static equilibrium must support its own weight. If one considers a small circular box of fluid with area $\Delta A$ perpendicular to gravity and sides of thickness $dx$, the force on the sides cancels due to symmetry. The force pushing down at the top of the box is $P_0 \Delta A$. The force pushing up from the bottom of the box is $(P_0 + dP)\Delta A$. The weight of the fluid in the box is $w = \rho g \Delta A dx$. Thus:

$$\left\{(P_0 + dP) - P_0\right\}\Delta A = \rho g \Delta A dx$$ (8.7)

or

$$dP = \rho g dx$$ (8.8)

If $\rho$ is itself a function of pressure/depth (as occurs with e.g. air) this can be a complicated expression to evaluate. For an incompressible fluid such as water, $\rho$ is constant over rather large variations in pressure. In that case, integration yields:

$$P = P_0 + \rho g D$$ (8.9)