Hydrostatic Pressure and Forces on Submerged Surfaces
In a fluid at rest, pressure rises linearly with depth: p = p₀ + ρ·g·h. Integrating that pressure over a submerged wall, gate or hatch gives the resultant force F = ρ·g·h_c·A, which acts not at the centroid but lower, at the center of pressure.
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Pressure at depth
A fluid at rest carries no shear, so a free body diagram of a small column of fluid has only its weight and the pressures on its ends. Balancing them gives dp/dz = −ρ·g: pressure grows by ρ·g for every metre you go down. In a liquid of constant density, that integrates to p = p₀ + ρ·g·h, where h is depth below the free surface.
Pressure depends only on depth, not on the shape of the container. Points at the same level in one connected, continuous fluid are at the same pressure, which is the rule used to read manometers. With layered fluids, such as oil floating on water, add ρ·g·h for each layer in turn.
Force on a submerged plane surface
On a flat surface, the pressure grows linearly down its depth. The resultant is the pressure at the centroid times the area, F = ρ·g·h_c·A, using gauge pressure when atmosphere acts on both sides. It is the same idea as finding the resultant of a distributed load in shear and moment diagrams.
Because the deeper part pushes harder, the force acts below the centroid. Measured along the plane of the surface from the free-surface line, the center of pressure is at y_cp = y_c + I_xc/(y_c·A), where I_xc is the second moment of area about the centroidal axis. The centroids of composite areas lesson and the moment of inertia calculator give y_c and I_xc for awkward shapes.
The offset I_xc/(y_c·A) shrinks with depth, so a deep gate has its center of pressure close to its centroid. For curved surfaces, find the horizontal and vertical components separately: the horizontal force equals the force on the surface’s vertical projection, and the vertical force equals the weight of the fluid above it.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Hydrostatic pressure
p = p₀ + ρ·g·h
p₀ = surface pressure (Pa), ρ = density (kg/m³), h = depth (m)
Munson et al., Fundamentals of Fluid Mechanics
Resultant force on a plane surface
F_R = ρ·g·h_c·A
h_c = depth of the centroid (m), A = area (m²)
Munson et al., Fundamentals of Fluid Mechanics
Center of pressure
y_cp = y_c + I_xc / (y_c·A)
y measured along the plane from the free-surface line (m), I_xc = second moment of area about the centroid (m⁴)
White, Fluid Mechanics
Rectangle second moment of area
I_xc = b·h³ / 12
b = width (m), h = height in the plane (m)
White, Fluid Mechanics
Worked examples
Example: Pressure in a tank with an oil layer
Given
An open tank holds 2.0 m of water with a 1.2 m layer of oil floating on top.
Take the oil’s specific gravity as 0.87 (an assumed value), ρ_water = 998 kg/m³ and atmospheric pressure 101.3 kPa.
Find
The gauge pressure at the oil–water interface.
The gauge and absolute pressure at the bottom.
Solution
The oil density is SG × ρ_water ≈ 870 kg/m³. Pressure at the interface comes from the oil column alone.p_int = 870 × 9.81 × 1.2 = 10 240 Pa = 10.2 kPa gauge
Add the water column below the interface.p_bot = 10 240 + 998 × 9.81 × 2.0 = 29 820 Pa = 29.8 kPa gauge
Check: a gauge on the bottom reads as if the tank held 3.05 m of water, less than the 3.2 m total depth because the oil is lighter.h_eq = 29 820 / (998 × 9.81) = 3.05 m
Answer: The interface is at about 10.2 kPa gauge and the bottom at about 29.8 kPa gauge, or 131.1 kPa absolute. A level transmitter calibrated for water alone would read the total level about 0.15 m low.
Example: Force on a vertical access hatch
Given
A rectangular access hatch on the vertical wall of a water tank is 0.6 m wide and 0.9 m tall. Its top edge is 2.5 m below the water surface.
The hatch is hinged along its top edge and held by a latch at its bottom edge. Take ρ = 998 kg/m³; atmosphere acts on both sides.
Find
The resultant water force on the hatch.
Where it acts.
The latch force needed to hold the hatch shut.
Solution
Area and centroid depth.A = 0.6 × 0.9 = 0.54 m², h_c = 2.5 + 0.9/2 = 2.95 m
Resultant force, using gauge pressure.F_R = 998 × 9.81 × 2.95 × 0.54 = 15 600 N
Center of pressure. On a vertical wall, y equals depth.I_xc = 0.6 × 0.9³ / 12 = 0.0365 m⁴, y_cp = 2.95 + 0.0365 / (2.95 × 0.54) = 2.973 m
Moments about the top hinge. F_R acts 2.973 − 2.5 = 0.473 m below the hinge; the latch is 0.9 m below it.F_latch = 15 600 × 0.473 / 0.9 = 8200 N
Answer: The water pushes on the hatch with about 15.6 kN, acting 23 mm below its centroid at 2.97 m depth. The bottom latch must hold about 8.2 kN, and the hinge carries the remaining 7.4 kN.
Common mistakes
Using the depth of the top or bottom edge for the force. F_R uses the depth of the centroid, h_c. The edges give the minimum and maximum pressure, not the resultant.
Putting the resultant at the centroid. It acts lower, at y_cp. The difference matters for hinge and latch loads.
Measuring y vertically on an inclined surface. In y_cp = y_c + I_xc/(y_c·A), y runs along the plane. Use h = y·sin θ only for pressures.
Adding atmospheric pressure when it acts on both sides. Use gauge pressure when atmosphere is on both faces; it cancels.
Treating layered fluids as one density. Add ρ·g·h for each layer separately, starting at the free surface.
In industry
Frequently asked questions
Does the shape of a tank change the pressure at the bottom?
No. Pressure in a static fluid depends only on depth and density, so a narrow tube and a wide tank filled to the same height have the same bottom pressure. This is the hydrostatic paradox.
Why is the center of pressure below the centroid?
Pressure grows with depth, so the lower part of the surface carries more of the load. The resultant shifts toward it.
What changes for a horizontal surface?
The pressure is uniform over it, so the resultant acts at the centroid and equals ρ·g·h·A.
How do I handle a gas above the liquid?
Use the gas pressure as p₀ at the free surface. A pressurised tank adds p₀·A to the force on every wall.