Liquid pressure calculations

Why liquid pressure changes with depth

A liquid has mass, so gravity gives it weight. At a point beneath the surface, the column of liquid above that point presses down. A deeper point has a taller column above it, so there is more liquid weight acting per unit area and the pressure is greater.

This can be seen using a water-filled pressure can with holes at different heights. Water from a lower hole usually travels farther than water from a higher hole because the pressure inside the can is greater at the lower hole.

Calculating the pressure difference

For a liquid of uniform density, the pressure difference caused by a vertical column of liquid is

$$\Delta p=h\rho g$$

Here:

• $\Delta p$ is the difference in pressure, in pascals (Pa);
• $h$ is the vertical height of liquid between the two depths, in metres (m);
• $\rho$ is the liquid’s density, in kilograms per cubic metre ();

Near the Earth’s surface, use $g=10\,\text{N/kg}$ for this OCR outcome. The symbol $g$ does not mean grams here: it describes the strength of the gravitational field.

The equation follows the physical pattern. Increasing $h$ means a taller and therefore heavier liquid column. Increasing $\rho$ means the same volume contains more mass and therefore has greater weight. A stronger gravitational field also gives the liquid more weight. Each change increases the pressure difference.

Choosing the correct height

The value of $h$ is the vertical separation between the two points. For points at depths $h_1$ and $h_2$ below the same surface,

$$\Delta h=|h_2-h_1|$$

and therefore

$$\Delta p=\Delta h\,\rho g.$$

For example, consider two points in water at depths of $0.50\,\text{m}$ and $1.70\,\text{m}$. Their vertical separation is $1.20\,\text{m}$. Taking the density of water as and gives

$$\Delta p=1.20\times1000\times10=12\,000\,\text{Pa}.$$

The pressure at the deeper point is therefore $12\,000\,\text{Pa}$ greater. This result is also $12\,\text{kPa}$ because $1000\,\text{Pa}=1\,\text{kPa}$.

Pressure difference rather than total pressure

The equation calculates the extra pressure produced by the liquid between two levels. If one point is the liquid surface, $h$ is the depth below that surface and $h\rho g$ is the pressure increase from the surface to that point.

In an open container, atmospheric pressure acts at the surface as well. It acts at both levels when their pressure difference is compared, so it cancels. Only add a surface pressure if a question supplies it and specifically asks for the total pressure at a depth.

A reliable calculation method

1. Identify the two levels and calculate their vertical separation in metres.
2. Write down the density in $\text{kg/m}^3$ and the value of $g$ in N/kg.

Equation

$$\Delta p=h\rho g$$

• $\Delta p$: pressure difference (Pa)
• $h$: vertical difference in depth (m)