Pressure in Fluids

In a fluid at rest, pressure varies with depth due to the weight of the fluid column above. This is the foundation of hydrostatics.

Concept

Pressure is normal force per unit area: $P = \frac{F_\perp}{A}.$

Properties of fluid pressure at rest:

• Acts equally in all directions (isotropic).
• Acts perpendicular to any surface element.
• Depends only on depth (not on shape of vessel) for a fluid at rest in a uniform gravitational field.

The hydrostatic equation: $P(h) = P_0 + \rho g h.$

Derivation

Consider a thin horizontal slab of fluid at depth $h$, area $A$, thickness $dh$. Equilibrium of vertical forces: $P(h+dh)\,A - P(h)\,A - (\rho A\,dh)\,g = 0.$

Dividing by $A\,dh$: $\frac{dP}{dh} = \rho g.$

Integrating with $P(0) = P_0$ (atmospheric): $P(h) = P_0 + \rho g h.$

For a fluid of variable density (e.g. atmosphere), $\rho$ depends on $P$ and integration becomes more involved.

Worked Example

A diver descends to depth $30$ m in seawater ($\rho = 1030$ kg/m$^3$). Find the absolute and gauge pressure. Take $P_0 = 1.01 \times 10^5$ Pa, $g = 9.8$ m/s$^2$.

Gauge: $P_g = \rho g h = 1030 \times 9.8 \times 30 \approx 3.03 \times 10^5$ Pa $\approx 3$ atm.

Absolute: $P = P_0 + P_g \approx 4.04 \times 10^5$ Pa $\approx 4$ atm.

Common Confusions

• Pressure does not depend on the cross-sectional area of the vessel — the hydrostatic paradox.
• Gauge pressure = absolute pressure − atmospheric; tyres and BP readings are gauge values.
• $P=\rho g h$ assumes incompressible fluid and constant $g$; not directly applicable to gases over large heights.

Key Takeaways

• $P(h) = P_0 + \rho g h$ for incompressible fluids in uniform gravity.
• Pressure at any horizontal level in a connected static fluid is the same.
• Hydrostatic paradox: shape of vessel does not affect pressure at the bottom.