Liquid pressure, or hydrostatic pressure, is the pressure caused by the weight of liquid above a point.
In a liquid at rest, pressure increases with depth.
1. Gauge pressure at depth
For a liquid of density \(\rho\), gravitational field strength \(g\), and depth \(h\), the gauge pressure is
\[
\begin{aligned}
P_{\mathrm{gauge}} &= \rho gh.
\end{aligned}
\]
Gauge pressure means pressure caused by the liquid itself, measured relative to the pressure at the free surface.
2. Absolute pressure
Absolute pressure includes the pressure already acting at the surface.
If the surface pressure is \(P_0\), then
\[
\begin{aligned}
P_{\mathrm{abs}}
&= P_0+P_{\mathrm{gauge}} \\
&= P_0+\rho gh.
\end{aligned}
\]
For an open container, \(P_0\) is usually atmospheric pressure:
\[
\begin{aligned}
P_0 &\approx 101325\ \mathrm{Pa}.
\end{aligned}
\]
3. Gauge pressure versus absolute pressure
| Pressure type |
Formula |
Meaning |
| Gauge pressure |
\(P_{\mathrm{gauge}}=\rho gh\) |
Extra pressure due to the liquid above the point |
| Absolute pressure |
\(P_{\mathrm{abs}}=P_0+\rho gh\) |
Total pressure measured relative to vacuum |
4. Where the formula comes from
Imagine a vertical column of liquid above a small horizontal area \(A\).
The column has height \(h\), so its volume is
\[
\begin{aligned}
V &= Ah.
\end{aligned}
\]
Its mass is
\[
\begin{aligned}
m &= \rho V \\
&= \rho Ah.
\end{aligned}
\]
Its weight is
\[
\begin{aligned}
W &= mg \\
&= \rho Ahg.
\end{aligned}
\]
Pressure is force divided by area:
\[
\begin{aligned}
P &= \frac{F}{A}.
\end{aligned}
\]
Using the weight of the liquid column as the force,
\[
\begin{aligned}
P_{\mathrm{gauge}}
&= \frac{W}{A} \\
&= \frac{\rho Ahg}{A} \\
&= \rho gh.
\end{aligned}
\]
5. Multiple liquid layers
If different liquids are stacked in layers, the total gauge pressure is found by adding the pressure from each layer above the point.
\[
\begin{aligned}
P_{\mathrm{gauge}}
&= \rho_1gh_1+\rho_2gh_2+\rho_3gh_3+\cdots.
\end{aligned}
\]
This can also be written as
\[
\begin{aligned}
P_{\mathrm{gauge}}
&= g(\rho_1h_1+\rho_2h_2+\rho_3h_3+\cdots).
\end{aligned}
\]
The top layer contributes only over its own thickness. Below that, the next layer begins contributing.
6. Solving for depth from pressure
If the gauge pressure is known, rearrange the hydrostatic formula:
\[
\begin{aligned}
P_{\mathrm{gauge}} &= \rho gh.
\end{aligned}
\]
\[
\begin{aligned}
h &= \frac{P_{\mathrm{gauge}}}{\rho g}.
\end{aligned}
\]
If absolute pressure is known, first subtract the surface pressure:
\[
\begin{aligned}
P_{\mathrm{gauge}}
&= P_{\mathrm{abs}}-P_0.
\end{aligned}
\]
Then use
\[
\begin{aligned}
h &= \frac{P_{\mathrm{abs}}-P_0}{\rho g}.
\end{aligned}
\]
7. Worked example: seawater at 35 m
Suppose seawater has density
\[
\begin{aligned}
\rho &= 1025\ \mathrm{kg/m^3}.
\end{aligned}
\]
At depth
\[
\begin{aligned}
h &= 35\ \mathrm{m},
\end{aligned}
\]
using \(g=9.81\ \mathrm{m/s^2}\), the gauge pressure is
\[
\begin{aligned}
P_{\mathrm{gauge}}
&= \rho gh \\
&= (1025)(9.81)(35) \\
&= 351934\ \mathrm{Pa} \\
&\approx 351.9\ \mathrm{kPa}.
\end{aligned}
\]
For an open container or the ocean surface, use atmospheric pressure:
\[
\begin{aligned}
P_0 &\approx 101.325\ \mathrm{kPa}.
\end{aligned}
\]
Then the absolute pressure is
\[
\begin{aligned}
P_{\mathrm{abs}}
&= P_0+P_{\mathrm{gauge}} \\
&\approx 101.325\ \mathrm{kPa}+351.9\ \mathrm{kPa} \\
&\approx 453.2\ \mathrm{kPa}.
\end{aligned}
\]
8. Important ideas
- Hydrostatic pressure depends on depth, density, and gravity.
- At the same depth, a denser liquid creates greater pressure.
- Gauge pressure is zero at the free surface.
- Absolute pressure at the surface equals \(P_0\).
- The shape of the container does not affect pressure at a given depth in a static liquid.
- Pressure acts equally in all directions at a point in a liquid at rest.
9. Formula summary
| Goal |
Formula |
Use |
| Gauge pressure |
\(P_{\mathrm{gauge}}=\rho gh\) |
Pressure due to liquid depth only |
| Absolute pressure |
\(P_{\mathrm{abs}}=P_0+\rho gh\) |
Total pressure including surface pressure |
| Depth from gauge pressure |
\(h=P_{\mathrm{gauge}}/(\rho g)\) |
Find depth from a gauge reading |
| Depth from absolute pressure |
\(h=(P_{\mathrm{abs}}-P_0)/(\rho g)\) |
Find depth from an absolute pressure reading |
| Layered liquid pressure |
\(P_{\mathrm{gauge}}=\sum \rho_i g h_i\) |
Add each layer’s pressure contribution |
| Force on a flat plate |
\(F=PA\) |
Estimate pressure force on a selected area |
10. Assumptions
- The liquid is at rest.
- Density is treated as constant within each layer.
- Gravity is treated as constant.
- The depth is measured vertically downward from the free surface.
- Dynamic effects such as waves, flow, and acceleration are ignored.
Key idea: the deeper you go, the taller the liquid column above you becomes, so the pressure increases.
