Fluid statics, manometry and forces on submerged surfaces
Hydrostatic pressure, absolute and gauge pressure, manometers, Pascal's law, and forces and centre of pressure on plane and curved submerged surfaces.
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Why it matters
Fluid statics sizes tank walls, sluice gates, dam faces and inspection hatches, and it is the physics behind every manometer, pressure gauge and hydraulic press. In fluid power, Pascal's law is why a small pump piston can drive a large cylinder, and manometer reasoning is how you read pressure taps on a test rig.
Key ideas
Pressure at a point. In a fluid at rest there is no shear stress, so the only surface force is normal pressure. Pressure at a point is the same in every direction (Pascal's law at a point), so it is a scalar.
Hydrostatic law. Balancing weight against pressure on a small fluid element gives dp/dz = −ρg, with z measured upward. For a liquid of constant density, pressure increases linearly with depth h below the free surface: p = p_atm + ρgh. Consequences:
- Points at the same level in the same continuous, connected static fluid are at the same pressure. This is the rule used to step through manometers.
- Pressure depends only on depth, not on container shape (the hydrostatic paradox).
- A pressure change applied anywhere in an enclosed liquid is transmitted to every point (Pascal's principle), which is the basis of the hydraulic press and jack.
Absolute, gauge and vacuum pressure. Absolute pressure is measured from perfect vacuum; gauge pressure from local atmosphere; vacuum (suction) is atmosphere minus absolute when below atmosphere. Standard atmosphere is 101.325 kPa ≈ 760 mm of mercury ≈ 10.33 m of water. A pressure can be expressed as a head h = p/(ρg) of a named liquid.
Manometers. A manometer balances an unknown pressure against liquid columns.
- Piezometer: a simple open tube; only for liquids at modest positive gauge pressure.
- U-tube manometer: uses a heavier, immiscible liquid (often mercury) for high pressure or a lighter one for small differences; can read negative gauge pressure.
- Differential manometer: connects two points and reads their pressure difference.
- Inclined manometer: tilting the reading limb at angle θ magnifies the reading length by 1/sinθ, increasing sensitivity for small differences.
- Micromanometer / inverted U-tube: for very small differences between two liquid points. Method: start at one point, add ρgh going down, subtract ρgh going up, jump across a level only within the same fluid, and end at the other point.
Force on a plane submerged surface. Pressure varies linearly with depth, so the resultant is the average (centroid) pressure times area: F = ρg·h̄·A, acting normal to the surface. Because pressure is larger lower down, the resultant acts at the centre of pressure, which is always below the centroid for an inclined or vertical surface (and coincides with it for a horizontal surface). The gap shrinks as the surface goes deeper. These results use gauge pressure; atmospheric pressure acts on both sides of most gates and cancels.
Force on a curved surface. Split the resultant into components:
- Horizontal component = force on the vertical projection of the curved surface, acting through that projection's centre of pressure.
- Vertical component = weight of the liquid directly above the curved surface up to the free surface (real or imaginary), acting through the centroid of that volume.
- Resultant = √(F_H² + F_V²); for a circular surface it passes through the centre of curvature.
Hydraulic press. Equal pressure on two pistons gives F₂/F₁ = A₂/A₁. Volume is conserved, so the large piston moves less: force is multiplied but work (F × stroke) is not.
Formulas
p = p_atm + ρ·g·h
- p = absolute pressure at depth h (Pa), ρ = liquid density (kg/m³), g = 9.81 m/s², h = depth below free surface (m). Constant-density fluid at rest.
p_gauge = p_abs − p_atm
- All in Pa. Vacuum = p_atm − p_abs.
p_A − p_B = (ρ_m − ρ)·g·x (differential U-tube, A and B at same level)
- ρ_m = manometer liquid density, ρ = flowing liquid density (kg/m³), x = manometer deflection (m).
F = ρ·g·h̄·A
- F = hydrostatic force on a plane surface (N), h̄ = vertical depth of centroid (m), A = area (m²).
h_cp = h̄ + I_G·sin²θ / (A·h̄)
- h_cp = vertical depth of centre of pressure (m), I_G = second moment of area about the centroidal axis parallel to the free surface (m⁴), θ = angle of the plane to the free surface (90° for vertical). For a rectangle of width b and height d, I_G = b·d³/12; for a circle of diameter D, I_G = π·D⁴/64.
F_H = ρ·g·h̄_proj·A_proj and F_V = ρ·g·V_above
- Components on a curved surface (N); A_proj = vertical projected area (m²), V_above = volume of liquid above the surface (m³).
F₂ / F₁ = A₂ / A₁
- Hydraulic press, friction neglected.
Worked examples
Example 1 — U-tube and differential manometers (standard). (a) A U-tube mercury manometer (ρ_m = 13 600 kg/m³) is connected to a water pipe. The pipe centre is 0.15 m above the mercury surface in the left limb, and the mercury in the open right limb stands 0.25 m higher than in the left limb. Find the gauge pressure at the pipe centre.
- Equal pressure at the level of the left mercury surface:
p_A + ρ_w·g·h₁ = ρ_m·g·h₂. - p_A = 9.81 × (13 600 × 0.25 − 1000 × 0.15) = 9.81 × 3250 = 31 882.5 Pa. Answer: p_A ≈ 31.9 kPa (gauge).
(b) A mercury differential manometer between two water pipes at the same level shows a deflection of 100 mm.
p_A − p_B = (ρ_m − ρ_w)·g·x= (13 600 − 1000) × 9.81 × 0.1 = 12 360.6 Pa. Answer: about 12.4 kPa. Note that using ρ_m alone would overstate it by 8 %.
Example 2 — vertical gate (GATE level). A vertical rectangular gate 2 m wide and 3 m high has its top edge 1.5 m below the water surface. It is hinged along the top edge. Find the hydrostatic force, the depth of the centre of pressure, and the horizontal force needed at the bottom edge to hold it shut.
- Centroid depth: h̄ = 1.5 + 3/2 = 3.0 m; area A = 2 × 3 = 6 m².
- Force:
F = ρ·g·h̄·A= 1000 × 9.81 × 3.0 × 6 = 176 580 N. - I_G = b·d³/12 = 2 × 27/12 = 4.5 m⁴.
h_cp = h̄ + I_G/(A·h̄)= 3.0 + 4.5/(6 × 3.0) = 3.25 m.- Lever arm of F about the hinge: 3.25 − 1.5 = 1.75 m. Moments about the hinge: P × 3 = 176 580 × 1.75, so P = 103 005 N. Answer: F ≈ 176.6 kN at 3.25 m depth; P ≈ 103.0 kN.
Example 3 — curved gate. A quarter-circle gate of radius 2 m and width 1 m retains water that fills the quarter-circle region bounded by the gate, with the free surface level with the top of the gate (so the quadrant of water sits directly above the curved surface).
- Horizontal: projection is 2 m × 1 m with centroid at 1 m: F_H = 9810 × 1 × 2 = 19 620 N.
- Vertical: weight of water above = 9810 × (π × 2²/4) × 1 = 30 819 N.
- Resultant = √(19 620² + 30 819²) = 36 534 N at tan⁻¹(30 819/19 620) = 57.5° to the horizontal, passing through the centre of the circle. Answer: about 36.5 kN at 57.5° below the horizontal.
Common mistakes
- Taking the force to act at the centroid. It acts at the centre of pressure, below the centroid.
- Using the inclined distance in F = ρg·h̄·A. h̄ is the vertical depth of the centroid.
- Using I about the free surface or the base instead of I_G about the centroidal axis.
- In a differential manometer, using ρ_m instead of (ρ_m − ρ) when the connecting pipes are filled with the working liquid.
- Equating pressures across a level that passes through two different fluids.
- Mixing gauge and absolute pressure — adding atmospheric pressure on one side of a gate only.
- For a curved surface, taking F_V as the weight of liquid below the surface when the liquid is actually above it (or forgetting the imaginary liquid column when it is below).
For GATE ME
Expect manometer chains with two or three liquids, pressure conversions between heads of different liquids, force and centre of pressure on vertical, inclined, rectangular, triangular and circular gates, the force to open a hinged gate, and horizontal and vertical components on cylindrical or quarter-circle gates. Questions on fluids in rigid-body motion (tanks under linear acceleration or rotation) also use the same hydrostatic reasoning. Practise writing one clean pressure balance from point to point and taking moments about the hinge.
Quick check
- What is the gauge pressure 10 m below a water surface?
- Why is the centre of pressure below the centroid for a vertical plate?
- A differential manometer uses mercury with water above it. What density goes into (ρ_m − ρ)?
- A hydraulic press has piston areas of 10 cm² and 500 cm². What load does 200 N lift?
- What is the vertical force on a curved surface equal to?
Answers: 1. 98.1 kPa; 2. pressure increases with depth, so the lower part carries more force; 3. 13 600 − 1000 = 12 600 kg/m³; 4. 10 000 N; 5. the weight of liquid (real or imaginary) vertically above the surface up to the free surface.
See it move
All Mechatronics animationsAdjust the depth of the fluid and the area of the submerged surface to see how the pressure and force change. Observe how pressure increases with depth and affects the force on the surface.
Equations used
- P = ρgh — P pressure, ρ density, g gravity, h depth
- F = P_avg · A — F force, P_avg average pressure, A area
Interview questions
All Fluid Mechanics and Fluid Power interview questionsTry answering each one aloud before you open it.
1.What is fluid statics and how does it differ from fluid dynamics?Concept
Fluid statics is the study of fluids at rest, focusing on the forces and pressures in a fluid that is not in motion. It differs from fluid dynamics, which deals with fluids in motion and the forces that affect them. In fluid statics, the primary concern is understanding how pressure varies with depth and how it acts on submerged surfaces.
2.Explain the principle of a manometer and its use in measuring pressure.Concept
A manometer balances an unknown pressure against the weight of one or more liquid columns, using the hydrostatic law: pressure rises by ρgh going down and falls by ρgh going up, and points at the same level in the same continuous static liquid are at equal pressure. Writing that balance from one point to the other gives gauge pressure (one limb open to atmosphere) or a pressure difference (differential manometer). Mercury is used for large pressures, a light liquid or an inclined limb for small ones. They need no calibration and are used on pipes, venturimeters, orifice meters and wind-tunnel taps.
3.What is Pascal's Law and how is it applied in hydraulic systems?Concept
Pascal's law states that pressure at a point in a fluid at rest is the same in all directions, and that a pressure change applied to an enclosed fluid is transmitted to every part of it. In a hydraulic press or jack, the same pressure acts on a small and a large piston, so F₂/F₁ = A₂/A₁ and force is multiplied. Volume is conserved, so the large piston moves A₁/A₂ times as far, and work is not multiplied. The same idea lets one pump drive several cylinders at the same system pressure.
4.How is a manometer made sensitive enough to measure small pressure differences?Application
The reading of a U-tube is x = Δp/((ρ_m − ρ)g), so a small Δp gives a readable deflection only if the density difference is small. Using a light manometric liquid (water, alcohol or paraffin for gas flows) or an inverted U-tube with air above a liquid magnifies the reading. An inclined manometer stretches the reading length by 1/sinθ for the same vertical rise, and micromanometers add optical or enlarged-reservoir arrangements. Wide tubes are used to keep capillary errors small.
5.What happens to the pressure at a point in a fluid if the depth is doubled?Application
In a constant-density liquid at rest, gauge pressure ρgh is proportional to depth, so doubling the depth doubles the gauge pressure. Absolute pressure is p_atm + ρgh, so it increases but does not double. For example, at 10 m in water gauge pressure is 98.1 kPa and absolute about 199 kPa; at 20 m they are 196.2 kPa and about 298 kPa.
6.How does the shape and orientation of a submerged surface affect the hydrostatic force on it?Application
For a plane surface, the force is ρg·h̄·A, so only the area and the depth of its centroid matter, and it acts normal to the surface at the centre of pressure, which lies below the centroid by I_G·sin²θ/(A·h̄) unless the surface is horizontal. For a curved surface, pressure acts normal to every element in different directions, so the resultant is found from components: the horizontal component equals the force on its vertical projection, and the vertical component equals the weight of liquid above the surface up to the free surface. For a circular surface the resultant passes through the centre of curvature.
7.Calculate the 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 hydrostatic law, gauge pressure p = ρgh = 1000 × 9.81 × 5 = 49 050 Pa ≈ 49.1 kPa. The absolute pressure is this plus atmospheric pressure, about 49.05 + 101.3 ≈ 150.4 kPa.
8.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 force on one side of the plate.Numerical
The force on the plate can be calculated by integrating the pressure over the area. The pressure at a depth h is P = ρgh. The average pressure on the plate is at the centroid, which is 3.5 meters below the surface. Thus, P_avg = 1000 * 9.81 * 3.5 = 34335 Pa. The area A = 3 * 1 = 3 m². The force F = P_avg * A = 34335 * 3 = 103005 N.
9.Explain why atmospheric pressure is considered when using a manometer.Concept
Atmospheric pressure is considered when using a manometer because it acts as a reference point for measuring the pressure of a fluid. The manometer measures the difference between the fluid pressure and atmospheric pressure, allowing for the determination of absolute pressure when atmospheric pressure is added to the gauge pressure.
10.What is the significance of the hydrostatic paradox in fluid statics?Concept
The hydrostatic paradox refers to the counterintuitive observation that the pressure at a given depth in a fluid is independent of the shape or volume of the container. This means that the pressure at a certain depth is determined solely by the height of the fluid column above it, not by the total amount of fluid or the shape of the container. This principle is crucial in understanding fluid behavior in various applications.
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