How to calculate water pressure at a depth
Water lower down has more water above it. Hydrostatic pressure tells you the extra pressure caused by that vertical column of still liquid.
What the formula is saying
A column with area A and depth h has volume Ah, mass ρAh and weight ρgAh. Dividing its weight by A leaves p = ρgh.
Read the symbols in plain language
- ρ
- Fluid densitykg/m³
- g
- Gravitym/s²
- h
- Depthm
Sort out the units first
Use ρ in kg/m³, g in m/s² and vertical depth h in m for Pa. The example assumes ρ = 1,000 kg/m³ and g = 9.81 m/s²; these are specified model inputs.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
- ρ · Fluid density
- 1000 kg/m³
- g · Gravity
- 9.81 m/s²
- h · Depth
- 3 m
Convert mass density to weight per volume
Multiply density by gravitational acceleration.
(1000) × (9.81) = 9810 N/m³Multiply by vertical depth
Depth measures the height of liquid above the point.
(9810) × (3) = 29430 Pa
Does this worked answer make sense?
The worked result is 29.43 kPa. At twice the depth, the pressure increase doubles; at the free surface it is zero.
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Use these new values. Work it out first, then check your answer.
- ρ · Fluid density
- 1000 kg/m³
- g · Gravity
- 9.81 m/s²
- h · Depth
- 5 m
Find: water pressure at a depth
A hint, not the answer
A column with area A and depth h has volume Ah, mass ρAh and weight ρgAh. Dividing its weight by A leaves p = ρgh.
Use ρ in kg/m³, g in m/s² and vertical depth h in m for Pa. The example assumes ρ = 1,000 kg/m³ and g = 9.81 m/s²; these are specified model inputs.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Convert mass density to weight per volume
Multiply density by gravitational acceleration.
(1000) × (9.81) = 9810 N/m³Multiply by vertical depth
Depth measures the height of liquid above the point.
(9810) × (5) = 49050 Pa
Avoid the common trap
Depth is vertical, not distance along a sloping wall. This gives pressure above the free-surface pressure, not absolute pressure by itself.
When this method applies — and when it does not
Stationary liquid of constant density with uniform gravity. Add the known free-surface pressure when absolute pressure is requested.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Lesson updated: · Worked examples checked against the implemented formula; not an independent engineering certification.