Fluid statics, manometry and forces on submerged surfaces

Hydrostatic law, Pascal's law, manometers, and the magnitude and centre of pressure of forces on plane and curved submerged surfaces.

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Why it matters

Hydrostatics sizes the walls of fuel and hydraulic tanks, the gates and valves that hold back liquid, and the brake and jack systems that multiply force through Pascal's law. Manometers remain the reference for measuring pressure differences across filters, venturis and orifice meters on a test bench. The same pressure-integration method returns later in buoyancy and in the momentum equation.

Key ideas

  • Pressure at a point in a fluid at rest acts equally in all directions (Pascal's law) and always normal to a surface. There is no shear stress in a static fluid.
  • Hydrostatic law. With z measured upward, dp/dz = −ρg. For a liquid of constant density, pressure increases linearly with depth h below the free surface: p = p₀ + ρgh. Points at the same level in the same continuous fluid at rest have the same pressure — the key rule for manometers.
  • Absolute, gauge and vacuum pressure. p_abs = p_atm + p_gauge. A negative gauge pressure is called vacuum. Standard atmosphere = 101.325 kPa = 760 mm Hg = 10.33 m of water.
  • Pressure head. h = p/(ρg) expresses pressure as a column height of a named liquid.
  • Pascal's law in machines. A pressure applied to an enclosed liquid is transmitted to every point, so in a hydraulic jack F₂/F₁ = A₂/A₁ (piston at same level).
  • Manometers.
    • Piezometer: a simple open tube; measures only positive gauge pressure of liquids.
    • U-tube manometer: a heavier, immiscible gauge liquid (often mercury, S = 13.6) can measure gauge or vacuum pressures of liquids and gases.
    • Differential manometer: measures p_A − p_B between two points.
    • Inclined manometer: reading along the tube L relates to vertical rise h = L sin θ, so small pressure differences give larger, easier-to-read deflections.
    • Method: start at one point, add ρgh going down, subtract ρgh going up, set the result equal to the pressure at the other end.
  • Plane surface. The hydrostatic force on a plane area equals the pressure at its centroid times its area. It acts at the centre of pressure, which lies below the centroid for vertical or inclined surfaces because pressure grows with depth; for a horizontal surface the two coincide. The deeper the surface, the closer the centre of pressure approaches the centroid.
  • Curved surface. Resolve into components:
    • Horizontal component = force on the vertical projection of the surface, acting through the centre of pressure of that projection.
    • Vertical component = weight of liquid (real or imaginary) vertically above the surface up to the free surface, acting through its centroid.
  • Atmospheric pressure usually acts on both sides of a gate or wall and cancels, so gauge pressure is used.

Formulas

p = p₀ + ρ·g·h p pressure at depth h (Pa), p₀ pressure at the free surface (Pa), ρ density (kg/m³), g = 9.81 m/s², h depth (m). Static fluid of constant density.

F₂ = F₁·(A₂/A₁) Hydraulic jack; F force (N), A piston area (m²); pistons at the same level, friction neglected.

p_A − p_B = g·(ρ_m − ρ)·x Differential U-tube manometer between two points at the same level in a pipe carrying fluid of density ρ, with gauge liquid ρ_m and deflection x (m).

F = ρ·g·h̄·A Resultant force (N) on a plane surface of area A (m²) whose centroid is at depth h̄ (m). Valid at any inclination.

h_cp = h̄ + I_G·sin²θ / (A·h̄) Depth of centre of pressure (m); I_G second moment of area about the centroidal axis parallel to the free-surface line (m⁴); θ angle of the plane with the free surface (θ = 90° for vertical). Rectangle b × d: I_G = b·d³/12; circle of diameter D: I_G = π·D⁴/64.

F_H = ρ·g·h̄_p·A_p F_V = ρ·g·V_above F_R = √(F_H² + F_V²) tan φ = F_V/F_H Curved surface: A_p projected vertical area (m²) with centroid depth h̄_p (m); V_above volume of liquid above the surface (m³); φ angle of the resultant with the horizontal.

Worked examples

Example 1 (standard) — U-tube manometer. Given: a pipe carries water; its centre A connects to a U-tube containing mercury (ρ_m = 13 600 kg/m³). The water–mercury interface in the left limb is 0.15 m below A. Mercury in the right (open) limb stands 0.25 m above that interface. Find the gauge pressure at A.

  1. Start at A and go down 0.15 m through water: p_A + ρ_w·g·0.15.
  2. The same level in the right limb is in mercury; go up 0.25 m to the open end: subtract ρ_m·g·0.25.
  3. Open end is atmospheric (0 gauge): p_A + 1000 × 9.81 × 0.15 − 13 600 × 9.81 × 0.25 = 0.
  4. p_A = 33 354 − 1 471.5 = 31 882.5 Pa. Answer: p_A ≈ 31.9 kPa (gauge).

Example 2 (GATE level) — hinged vertical gate. Given: a vertical rectangular gate 2 m wide and 3 m high has its top edge 1 m below the water surface and 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 keep it shut.

  1. Centroid depth: h̄ = 1 + 3/2 = 2.5 m; area A = 2 × 3 = 6 m².
  2. Force: F = ρ·g·h̄·A = 1000 × 9.81 × 2.5 × 6 = 147 150 N.
  3. I_G = b·d³/12 = 2 × 3³/12 = 4.5 m⁴.
  4. Centre of pressure: h_cp = h̄ + I_G/(A·h̄) = 2.5 + 4.5/(6 × 2.5) = 2.8 m.
  5. Moment arm about the hinge (depth 1 m): 2.8 − 1 = 1.8 m.
  6. Moments about the hinge: P × 3 = 147 150 × 1.8 → P = 264 870/3 = 88 290 N. Answer: F ≈ 147.2 kN at 2.8 m depth; P ≈ 88.3 kN at the bottom edge.

Example 3 (curved surface). A quarter-circle gate of radius 2 m and width 3 m holds water on its concave side, with the free surface level with its top. Horizontal: F_H = 9810 × 1 × (2 × 3) = 58 860 N. Vertical: water above = quarter circle × width = (π × 2²/4) × 3 = 9.425 m³, so F_V = 9810 × 9.425 = 92 457 N. Resultant F_R ≈ 109.6 kN at 57.5° to the horizontal.

Common mistakes

  • Using the depth of the centre of pressure instead of the centroid to compute the force.
  • Using I about the free surface or about the base instead of I_G about the centroidal axis.
  • Measuring h̄ along an inclined plate instead of vertically below the free surface.
  • In manometers, changing pressure across a level in two different fluids; equal pressure applies only within the same continuous fluid.
  • Forgetting the column of the flowing fluid in a differential manometer, giving ρ_m·g·x instead of (ρ_m − ρ)·g·x.
  • Mixing absolute and gauge pressure in the same equation.

For GATE ME

Typical questions: manometer readings with two or three fluids, pressure differences from an inclined or differential manometer, force and centre of pressure on vertical or inclined gates, the force needed to open a hinged gate, and horizontal and vertical components on cylindrical or quarter-circle surfaces. Practise writing the manometer equation step by step and taking moments about the hinge.

Quick check

  1. What is the gauge pressure 4 m below the free surface of oil of relative density 0.8?
  2. Where is the centre of pressure for a horizontal submerged plate?
  3. A hydraulic jack has piston diameters 25 mm and 250 mm; what load can a 200 N effort lift?
  4. What gives the vertical force on a curved submerged surface?
  5. Why is an inclined manometer more sensitive than a vertical one? Answers: 1. 800 × 9.81 × 4 = 31.4 kPa. 2. At its centroid. 3. 200 × (250/25)² = 20 kN. 4. The weight of liquid vertically above it up to the free surface. 5. A small vertical rise h appears as a longer reading L = h/sin θ.

Interview questions

All Fluid Mechanics and Turbomachinery interview questions

Try answering each one aloud before you open it.

  1. 1.What is fluid statics and how does it differ from fluid dynamics?Concept

    Fluid statics is the study of fluids at rest. It focuses on understanding the forces and pressures in a fluid that is not in motion. In contrast, fluid dynamics deals with fluids in motion and the forces that affect them. Fluid statics is concerned with pressure variation in a fluid due to gravity, while fluid dynamics involves velocity, flow rates, and energy transformations.

  2. 2.Explain the principle of a manometer and its use in measuring pressure.Concept

    A manometer is a device used to measure the pressure of a fluid by balancing it against a column of liquid, typically mercury or water. The principle is based on hydrostatic equilibrium, where the pressure difference between two points in a fluid is equal to the height difference of the liquid column multiplied by the density of the liquid and gravitational acceleration. Manometers are used in various applications to measure gas pressure, differential pressure, and vacuum pressure.

  3. 3.What is Pascal's Law and how is it applied in hydraulic systems?Concept

    Pascal's Law states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of its container. This principle is applied in hydraulic systems, where a small force applied to a small-area piston is transformed into a larger force on a larger-area piston, allowing for the lifting of heavy loads with minimal effort.

  4. 4.Why are U-tube manometers commonly used for measuring pressure differences?Application

    U-tube manometers are commonly used because they provide a simple and accurate way to measure pressure differences. They consist of a U-shaped tube filled with a liquid, usually mercury or water. The pressure difference between two points causes the liquid to move, and the height difference between the two columns of liquid is proportional to the pressure difference. This setup is easy to use and provides a visual representation of pressure changes.

  5. 5.What happens to the pressure at a point in a fluid if the depth increases?Application

    As the depth in a fluid increases, the pressure at that point also increases. This is because pressure in a fluid at rest is due to the weight of the fluid above the point. The relationship is given by 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 depth. Therefore, deeper points in a fluid experience higher pressure.

  6. 6.How does the shape and orientation of a submerged surface affect the force exerted on it by a fluid?Application

    Hydrostatic pressure acts normal to the surface and increases linearly with depth, so the force depends on how the area is distributed in depth, not on the container shape. For a plane surface the resultant is F = ρg·h̄·A, using the depth of the centroid, and it acts at the centre of pressure, which lies below the centroid unless the plate is horizontal. For a curved surface you resolve components: the horizontal force equals the force on its vertical projection, and the vertical force equals the weight of liquid above it up to the free surface.

  7. 7.Calculate the pressure at a depth of 10 meters in water. Assume the density of water is 1000 kg/m³ and g = 9.81 m/s².Numerical

    To calculate the pressure at a depth of 10 meters in water, use the formula P = ρgh. Here, ρ = 1000 kg/m³, g = 9.81 m/s², and h = 10 m. Thus, P = 1000 * 9.81 * 10 = 98100 Pa (Pascals).

  8. 8.A rectangular plate is submerged vertically in water with its top edge 2 m below the surface. If the plate is 3 m high and 1 m wide, calculate the force on the plate (ρ = 1000 kg/m³, g = 9.81 m/s²).Numerical

    Use F = ρg·h̄·A with the centroid depth h̄ = 2 + 3/2 = 3.5 m and area A = 3 × 1 = 3 m². F = 1000 × 9.81 × 3.5 × 3 = 103 005 N ≈ 103 kN. Integrating ρg·h·b·dh from 2 m to 5 m gives the same result. It acts at h_cp = 3.5 + (1 × 3³/12)/(3 × 3.5) ≈ 3.71 m depth.

  9. 9.Explain why the center of pressure is below the centroid of a submerged plane surface.Concept

    Pressure increases linearly with depth, so the lower part of an inclined or vertical surface carries more load and the resultant shifts below the centroid. Quantitatively, h_cp = h̄ + I_G·sin²θ/(A·h̄), and the second term is always positive. For a horizontal surface the pressure is uniform, so the centre of pressure coincides with the centroid. As the surface is lowered deeper, h̄ grows and the gap shrinks.

  10. 10.What is the significance of hydrostatic pressure in designing dams and underwater structures?Application

    Hydrostatic pressure is crucial in designing dams and underwater structures because it determines the force exerted by the fluid on these structures. Engineers must ensure that the structures can withstand the pressure to prevent failure. The pressure increases with depth, so the design must account for the maximum expected pressure. Proper understanding of hydrostatic pressure helps in selecting materials and designing shapes that can efficiently handle the forces involved.

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