Fluid properties and fluid statics

Why it matters

Every hydraulic press, clamping fixture, coolant system, die-casting machine and storage tank in a factory is designed with the ideas in this topic. Fluid properties decide how a lubricant, cutting fluid or molten metal behaves, and fluid statics gives the pressures and forces on tank walls, gates and manometer readings that a production engineer reads every day.

Key ideas

What a fluid is. A fluid deforms continuously under any shear stress, however small. A solid resists shear by a finite deformation; a fluid at rest therefore carries no shear stress at all, only normal (pressure) stress. We treat the fluid as a continuum: properties are smooth functions of position, valid as long as the length scales of interest are much larger than the molecular mean free path.

Density and specific weight. Density ρ = mass per unit volume (kg/m³). Specific weight w = ρg (N/m³). Specific gravity (relative density) S = ρ / ρ_water, with ρ_water ≈ 1000 kg/m³ at about 4 °C; mercury has S ≈ 13.6.

Viscosity. Newton's law of viscosity says the shear stress in a simple shear flow is proportional to the velocity gradient: τ = μ du/dy. μ is the dynamic viscosity (Pa·s); kinematic viscosity ν = μ/ρ (m²/s). Fluids that obey this linear law (water, air, most oils) are Newtonian; paints, slurries, polymer melts and blood are non-Newtonian. Liquid viscosity falls as temperature rises (weaker intermolecular cohesion), gas viscosity rises with temperature (more molecular momentum exchange). This is why a cold machine oil is sluggish and why lubricant grades are specified at a temperature.

Surface tension and capillarity. At a liquid–gas interface molecules are pulled inward, so the surface behaves like a stretched membrane with tension σ (N/m). It causes capillary rise or depression in narrow tubes (water wets glass and rises; mercury does not wet glass and is depressed) and an excess pressure inside drops and bubbles. It matters in wetting of solder and brazing filler, spreading of coolant and the shape of weld pools.

Compressibility. Bulk modulus K = −dp/(dV/V) measures resistance to volume change. For water K ≈ 2.2 GPa, so liquids are treated as incompressible in almost all statics and hydraulics problems. Gases are compressible, but low-speed gas flow (Mach number below about 0.3) can still be treated as incompressible.

Pressure at a point and Pascal's law. In a fluid at rest, pressure at a point is the same in all directions. Pascal's law: a pressure change applied at any point of an enclosed fluid at rest is transmitted undiminished to every other point. This is the principle of the hydraulic press and jack: a small force on a small piston produces a large force on a large piston, at the cost of a proportionally smaller stroke.

Hydrostatic law. For a fluid at rest under gravity, dp/dz = −ρg (z measured upward). For a liquid of constant density, pressure increases linearly with depth h below the free surface: p = p_atm + ρgh. Pressure depends only on depth, not on the shape of the vessel (the hydrostatic paradox). Points at the same level in the same continuous fluid at rest have the same pressure; this is the rule used to solve every manometer.

Absolute and gauge pressure. p_abs = p_atm + p_gauge. A vacuum is a negative gauge pressure. Standard atmosphere = 101.325 kPa = 760 mm of mercury = about 10.3 m of water.

Forces on submerged surfaces. The resultant hydrostatic 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. For curved surfaces, the horizontal component equals the force on the vertical projection, and the vertical component equals the weight of fluid directly above the surface (real or imaginary).

Buoyancy. Archimedes' principle: a body immersed in a fluid experiences an upward force equal to the weight of fluid displaced, acting through the centre of buoyancy. A floating body is stable if its metacentre lies above its centre of gravity (positive metacentric height).

Formulas

τ = μ · du/dy

• τ: shear stress (Pa); μ: dynamic viscosity (Pa·s); du/dy: velocity gradient (1/s). Newtonian fluids only.

ν = μ / ρ

• ν: kinematic viscosity (m²/s); ρ: density (kg/m³).

p = p_atm + ρ·g·h

• p: absolute pressure at depth h (Pa); g = 9.81 m/s²; h: depth below free surface (m). Incompressible fluid at rest.

F₂ / A₂ = F₁ / A₁

• Hydraulic press: F forces (N), A piston areas (m²); ideal, frictionless, same level.

h = 4·σ·cos θ / (ρ·g·d)

• Capillary rise h (m) in a tube of diameter d (m); σ: surface tension (N/m); θ: contact angle. Negative h means depression.

Δp = 4·σ / d (droplet), Δp = 8·σ / d (soap bubble with two surfaces)

• Excess pressure inside (Pa).

F = ρ·g·h_c·A

• Force on a plane surface (N); h_c: depth of the centroid (m); A: area (m²).

h_cp = h_c + I_G·sin²θ / (A·h_c)

• Depth of the 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 with the free surface (θ = 90° for a vertical plane).

F_B = ρ_fluid · g · V_displaced

• Buoyant force (N).

Worked examples

Example 1 (standard): U-tube mercury manometer. Given: water flows in a pipe; a mercury U-tube is connected to it. The pipe centre A is 0.15 m above the mercury surface in the left limb; the mercury in the right limb (open to atmosphere) stands 0.25 m above the left mercury surface. ρ_water = 1000 kg/m³, ρ_Hg = 13 600 kg/m³, g = 9.81 m/s². Find the gauge pressure at A.

1. Equate pressures at the level of the left mercury surface (same fluid, same level): p_A + ρ_w·g·h₁ = ρ_Hg·g·h₂ (gauge).
2. p_A = ρ_Hg·g·h₂ − ρ_w·g·h₁ = 13 600 × 9.81 × 0.25 − 1000 × 9.81 × 0.15
3. p_A = 33 354 − 1471.5 = 31 882.5 Pa Answer: p_A ≈ 31.9 kPa (gauge)

Example 2 (GATE level): force and centre of pressure on a vertical gate. Given: a vertical rectangular sluice gate 2 m wide and 3 m high has its top edge 1 m below the free water surface. Find the resultant hydrostatic force and the depth at which it acts.

1. Centroid depth: h_c = 1 + 3/2 = 2.5 m; area A = 2 × 3 = 6 m².
2. F = ρ·g·h_c·A = 1000 × 9.81 × 2.5 × 6 = 147 150 N
3. I_G = b·d³/12 = 2 × 3³ / 12 = 4.5 m⁴
4. Vertical plane, so 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 free surface

Example 3 (quick): capillary rise. Water (σ = 0.073 N/m, θ = 0°) in a clean glass tube of 1 mm bore: h = 4 × 0.073 × 1 / (1000 × 9.81 × 0.001) = 0.0298 m, i.e. about 29.8 mm.

Common mistakes

• Mixing absolute and gauge pressure, especially when a question gives a vacuum reading or asks for absolute pressure.
• Using the centroid depth as the point of action of the force; the centre of pressure is always deeper for a submerged inclined or vertical plane.
• Walking through a manometer and changing sign wrongly: going down a column adds ρgh, going up subtracts it.
• Using the density of the manometer fluid where the pipe fluid's density is needed (and forgetting the pipe-fluid column).
• Using mm or cm for h or d without converting to metres.
• Saying "viscosity always decreases with temperature" — true for liquids, false for gases.
• Using 4σ/d for a soap bubble, which has two surfaces (8σ/d).

For GATE PI

Expect short numericals on manometers (simple, differential, inclined), hydraulic press force ratios, capillary rise, pressure at depth with absolute/gauge conversion, and force and centre of pressure on vertical or inclined plane gates. Conceptual one-markers test Newtonian vs non-Newtonian behaviour, the temperature dependence of viscosity, and buoyancy and stability of floating bodies. Practise writing the manometer equation level by level and memorise I_G for a rectangle and a circle.

Quick check

1. What is the SI unit of kinematic viscosity?
2. A gauge reads −30 kPa where the atmosphere is 101 kPa. What is the absolute pressure?
3. Does the hydrostatic force on a vertical gate act above or below its centroid?
4. Why does mercury show a capillary depression in glass?
5. A hydraulic press has piston diameters 20 mm and 200 mm. What force ratio does it give?

Answers: 1. m²/s 2. 71 kPa (absolute) 3. Below the centroid 4. Mercury does not wet glass (contact angle above 90°), so cos θ is negative 5. 100 (area ratio = (200/20)²)