Hydrostatic pressure, manometers and forces on submerged surfaces
Hydrostatic equation, gauge and absolute pressure, the ISA atmosphere, manometer chains, and resultant force and centre of pressure on plane and curved submerged surfaces.
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Why it matters
Fluid statics gives the pressure field when nothing moves relative to the fluid. It is how an altimeter converts static pressure into altitude, how manometers in a wind tunnel read pressure differences, and how designers size fuel-tank walls, dam gates and hatches that hold back liquid. The same hydrostatic balance underlies the standard atmosphere used in every aircraft performance calculation.
Key ideas
Pressure at a point in a fluid at rest is the same in every direction (Pascal's law), because a static fluid carries no shear stress. Pressure is a scalar; the force it produces on a surface acts normal to that surface.
Hydrostatic equation. Balancing pressure and weight on a small fluid element with z measured vertically upward gives dp/dz = −ρg. Pressure therefore increases with depth. For a liquid of constant density, p = p₀ + ρgh, where h is depth below the free surface. Consequences:
- Pressure depends only on depth, not on the shape of the container (the hydrostatic paradox).
- Points at the same level in the same continuous, connected fluid at rest have the same pressure. This is the rule that solves every manometer problem.
- Pressure applied to a confined liquid is transmitted to every point (basis of the hydraulic press and aircraft hydraulic actuators).
Absolute and gauge pressure. Absolute pressure is measured from a perfect vacuum; gauge pressure is measured from local atmospheric pressure (p_gauge = p_abs − p_atm). Vacuum pressure is a negative gauge pressure. Gas laws need absolute pressure; hydrostatic forces on walls with air on the other side use gauge pressure.
Atmosphere. Air density varies with height, so dp/dz = −ρg is integrated with ρ = p/(RT). In the ISA troposphere (0–11 km) temperature falls linearly with lapse rate L = 6.5 K/km; above it, up to 20 km, the atmosphere is isothermal at 216.65 K and pressure decays exponentially.
Manometers. A piezometer tube measures gauge pressure of a liquid; a U-tube manometer with a heavier liquid measures larger or gas pressures; a differential manometer measures p₁ − p₂ between two points; an inclined manometer stretches a small vertical height h into a longer reading L = h/sinθ for better resolution. Method: start at one point, add ρgh when going down, subtract ρgh when going up, and end at the other point.
Forces on plane surfaces. The resultant pressure force on a plane surface equals the pressure at its centroid times its area. It acts at the centre of pressure, which lies below the centroid because pressure grows with depth. The gap shrinks as the surface goes deeper.
Forces on curved surfaces. Split the force into components. The horizontal component equals the force on the vertical projection of the surface. The vertical component equals the weight of fluid (real or imaginary) vertically above the surface up to the free surface. For a circular surface the resultant passes through the centre of curvature.
Buoyancy and stability, which follow from the same ideas, are treated in the next topic.
Formulas
dp/dz = −ρ·g
- p: pressure (Pa); z: height, positive upward (m); ρ: density (kg/m³); g = 9.81 m/s².
p = p₀ + ρ·g·h
- p₀: pressure at the free surface (Pa); h: depth below the free surface (m). Constant-density fluid.
p_gauge = p_abs − p_atm
T = T₀ − L·z and p = p₀·(T/T₀)^(g/(L·R))
- ISA troposphere: T₀ = 288.15 K, p₀ = 101 325 Pa, L = 0.0065 K/m, R = 287 J/(kg·K); g/(L·R) ≈ 5.26.
p = p₁·exp(−g·(z − z₁)/(R·T))
- Isothermal layer at temperature T (K), starting at height z₁ with pressure p₁.
p_A − p_B = (ρ_m − ρ)·g·h
- Differential U-tube: two points at the same level in pipes of fluid ρ, manometer fluid ρ_m, deflection h (m).
F = ρ·g·h_c·A
- F: resultant force on a plane surface (N); h_c: vertical depth of the centroid (m); A: area (m²). Gauge pressure, free surface at atmospheric pressure.
h_cp = h_c + I_G·sin²θ / (A·h_c)
- h_cp: vertical depth of the centre of pressure (m); I_G: second moment of area about the centroidal axis parallel to the surface line (m⁴); θ: angle of the plane with the free surface (θ = 90° for vertical). For a rectangle of width b and height d, I_G = b·d³/12.
F_H = ρ·g·h_c,proj·A_proj and F_V = ρ·g·V_above
- Curved surface: A_proj is the vertical projected area and h_c,proj its centroid depth; V_above is the volume of fluid above the surface up to the free surface (m³).
Worked examples
Example 1 (standard): vertical gate. Given: rectangular gate 2 m wide and 3 m high, vertical, top edge 1 m below the water surface; ρ = 1000 kg/m³.
- Centroid depth: h_c = 1 + 3/2 = 2.5 m. Area A = 2 × 3 = 6 m².
F = ρ·g·h_c·A= 1000 × 9.81 × 2.5 × 6 = 147 150 N.- I_G = b·d³/12 = 2 × 3³/12 = 4.5 m⁴.
h_cp = h_c + I_G/(A·h_c)= 2.5 + 4.5/(6 × 2.5) = 2.5 + 0.3 = 2.8 m. Answer: F ≈ 147.2 kN acting 2.8 m below the surface.
Example 2 (standard): differential manometer. Given: two water pipes A and B at the same level joined by a mercury U-tube (ρ_m = 13 600 kg/m³); deflection h = 0.2 m.
p_A − p_B = (ρ_m − ρ)·g·h= (13 600 − 1000) × 9.81 × 0.2.- = 12 600 × 1.962 = 24 721 Pa. Answer: p_A − p_B ≈ 24.7 kPa.
Example 3 (GATE level): curved gate. Given: a quarter-circle gate of radius R = 2 m and width w = 3 m holds water; the water fills the space above the curved surface, with the top of the gate at the free surface.
- Horizontal: projection is 2 m high × 3 m wide, centroid depth 1 m.
F_H = ρ·g·h_c·A= 1000 × 9.81 × 1 × 6 = 58 860 N. - Vertical: weight of water above = ρ·g·(πR²/4)·w = 9810 × π × 3 = 92 457 N.
- Resultant = √(58 860² + 92 457²) = 109 603 N, at tan⁻¹(92 457/58 860) = 57.5° below the horizontal, through the centre of the circle. Answer: F ≈ 109.6 kN at 57.5°.
Example 4 (GATE level): altimeter. ISA pressure at 5 km:
- T = 288.15 − 0.0065 × 5000 = 255.65 K.
p = 101 325 × (255.65/288.15)^5.2586= 54.0 kPa, and ρ = p/(RT) = 0.736 kg/m³.
Common mistakes
- Using the depth of the centroid as the line of action. The force acts at the centre of pressure, below the centroid.
- Mixing gauge and absolute pressure, especially when atmospheric pressure acts on both sides of a gate (it cancels).
- In manometers, equating pressures across two different fluids at the same level. The rule works only within one continuous fluid.
- Forgetting sin²θ for inclined surfaces, or using slant distance where vertical depth is needed.
- Applying p = ρgh to the atmosphere over kilometres. Air density changes with height.
- Using h (vertical) instead of the scale reading L in an inclined manometer, or the reverse.
For GATE AE
Common questions: multi-fluid manometer chains, inclined-manometer sensitivity, resultant force and centre of pressure on vertical, inclined or curved gates, and pressure, temperature and density in the standard atmosphere at a given altitude (troposphere and the isothermal layer). Practise writing the manometer equation one leg at a time and remembering the ISA constants, because atmosphere questions are routine in the aerospace paper.
Quick check
- Gauge pressure at 10 m depth in water (g = 9.81 m/s²)?
- A differential mercury–water manometer reads 0.1 m. What is the pressure difference?
- Where does the resultant force on a vertical rectangular gate with its top at the free surface act?
- Does an inclined manometer increase or decrease the reading for a given pressure difference?
Answers: 1. 98.1 kPa 2. 12 600 × 9.81 × 0.1 ≈ 12.4 kPa 3. At two-thirds of the gate height below the surface 4. Increases it (L = h/sinθ)
Interview questions
All Fluid Mechanics interview questionsTry answering each one aloud before you open it.
1.What is hydrostatic pressure and how is it calculated?Concept
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It is calculated using the formula P = ρgh, where P is the pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid column above the point in question.
2.Explain the working principle of a manometer.Concept
A manometer balances an unknown pressure against a column of liquid of known density using the hydrostatic law p = p₀ + ρgh. Points at the same level in the same continuous liquid at rest are at equal pressure, so you write the pressure from one end to the other, adding ρgh going down and subtracting going up. For a differential U-tube between two points at the same level, p₁ − p₂ = (ρ_m − ρ)gh; a heavier manometer liquid suits larger pressures and an inclined limb improves resolution for small ones.
3.What are the forces acting on a submerged surface?Concept
A fluid at rest exerts only normal pressure on a surface, so the force is the integral of p·dA normal to it. For a plane surface the resultant is F = ρg·h_c·A, the centroid pressure times the area, acting at the centre of pressure, which lies below the centroid. For a curved surface the horizontal component equals the force on its vertical projection and the vertical component equals the weight of fluid above it; for a fully immersed closed body the net of these pressure forces is the buoyant force.
4.Why are U-tube manometers commonly used in laboratories?Application
They are simple, cheap, have no moving parts and are a primary standard: the reading depends only on the liquid's density, g and a measured height, so they are used to calibrate electronic transducers. They can measure gauge, vacuum or differential pressure. Their limits are slow response, the need for a known liquid density (temperature dependent), and limited range, which is why wind tunnels use multitube or inclined manometers for small differences and transducers for unsteady pressures.
5.What happens to the hydrostatic pressure if the density of the fluid increases?Application
If the density of the fluid increases, the hydrostatic pressure at a given depth also increases. This is because hydrostatic pressure is directly proportional to the fluid density, as given by the formula P = ρgh.
6.How does the shape of a submerged object affect the buoyant force acting on it?Application
The shape of a submerged object does not affect the buoyant force acting on it. The buoyant force depends only on the volume of fluid displaced by the object, not its shape. This is a consequence of Archimedes' principle.
7.What is the significance of the center of pressure on a submerged surface?Concept
The center of pressure is the point on a submerged surface where the total hydrostatic pressure force is considered to act. It is significant because it determines the moment arm for calculating the torque due to the pressure force, which is important for stability and structural analysis.
8.Calculate the hydrostatic pressure at a depth of 5 meters in water. Assume the density of water is 1000 kg/m³ and g = 9.81 m/s².Numerical
Using the formula P = ρgh, where ρ = 1000 kg/m³, g = 9.81 m/s², and h = 5 m, the hydrostatic pressure P = 1000 * 9.81 * 5 = 49050 Pa.
9.A rectangular plate is submerged vertically in water with its top edge 2 meters below the surface. If the plate is 3 meters high and 1 meter wide, calculate the total hydrostatic force on the plate. Assume water density is 1000 kg/m³ and g = 9.81 m/s².Numerical
F = ρg·h_c·A. A = 3 × 1 = 3 m² and the centroid is 1.5 m below the top edge, so h_c = 2 + 1.5 = 3.5 m. F = 1000 × 9.81 × 3.5 × 3 = 103 005 N ≈ 103 kN. It acts at the centre of pressure, h_cp = h_c + I_G/(A·h_c) = 3.5 + 2.25/(3 × 3.5) ≈ 3.71 m below the surface.
10.Explain why hydrostatic pressure is independent of the shape of the container.Concept
In a fluid at rest there is no shear, so a force balance on a small element gives dp/dz = −ρg with no horizontal pressure gradient. Integrating, pressure depends only on vertical depth below the free surface: p = p₀ + ρgh. Walls that slope or overhang carry the difference in fluid weight as reaction forces, so the bottom pressure of a narrow and a wide vessel filled to the same height is identical (the hydrostatic paradox).
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