Mechanical Properties of Fluids

A fluid at rest exerts a force perpendicular to any surface in contact with it. Pressure at a point is defined as the normal force (thrust) exerted per unit area:

P = F/A, measured in pascal (1 Pa = 1 N/m²).

A fluid at rest cannot sustain a shearing (tangential) force, so at a given point the pressure acts equally in every direction and depends only on the depth, the density of the fluid and g, not on the shape or the total amount of fluid present.

Variation of pressure with depth: Consider a thin horizontal fluid layer of thickness dh at depth h below the free surface. Balancing the weight of this layer against the net upward force on it gives dP = ρg dh. Integrating from the free surface, where the pressure is P₀ (usually atmospheric pressure), down to depth h:

P = P₀ + ρgh

Pressure therefore increases linearly with depth for a fluid of uniform density.

Pascal's law states that pressure applied at any point of an enclosed fluid at rest is transmitted undiminished to every point of the fluid and to the walls of the container. This is the working principle of hydraulic machines. In a hydraulic lift or a hydraulic brake, a small force F₁ applied on a piston of area A₁ creates a pressure P = F₁/A₁ in the enclosed liquid. Since this pressure acts equally on a second, larger piston of area A₂, the force obtained there is F₂ = P × A₂ = F₁(A₂/A₁). Because A₂ is much larger than A₁, a small applied force is converted into a large output force — this is how a hydraulic lift raises a car, and how a light push on a brake pedal produces a large clamping force at the wheel brake shoes, with the same pressure reaching every brake equally.

Worked example.

Given: In a hydraulic lift, the smaller piston has area A₁ = 10 cm² and the larger piston has area A₂ = 1000 cm². A force F₁ = 50 N is applied on the smaller piston.

Formula: F₂ = F₁ × (A₂/A₁)

Substitution: F₂ = 50 × (1000/10) = 50 × 100

Result: F₂ = 5000 N = 5 × 10³ N (5 kN), the load this lift can support.

Archimedes' principle states that whenever a body is partially or wholly immersed in a fluid, it experiences an upward buoyant force (upthrust) equal to the weight of the fluid displaced by the immersed part of the body:

F_b = ρ_fluid V_displaced g

This upthrust arises because the pressure at the bottom of the immersed body is greater than at the top (pressure increases with depth), so the net upward force due to the pressure difference equals the weight of the displaced fluid.

Law of floatation: A body floats in a fluid if it can displace a weight of fluid equal to its own weight before being fully submerged. For a body of density ρ_body floating in a fluid of density ρ_fluid, equating weight to upthrust gives the fraction of its volume that remains submerged:

V_submerged/V_body = ρ_body/ρ_fluid

If ρ_body > ρ_fluid, the body cannot displace enough fluid to balance its weight while floating, so it sinks.

Since the apparent weight of a submerged body equals its actual weight minus the buoyant force, measuring the loss of weight of a solid on immersion in water lets its density (or relative density) be found experimentally.

Worked example.

Given: A solid weighs 500 g (in air) and an apparent 400 g when fully immersed in water (density of water = 1000 kg/m³ = 1 g/cm³).

Formula: Loss of weight on immersion = weight of water displaced, so mass of water displaced = loss in the reading. Volume displaced = mass of water displaced / density of water. Density of solid = mass of solid / volume displaced.

Substitution: Loss of weight = 500 − 400 = 100 g, so mass of water displaced = 100 g. Volume displaced = 100 g / (1 g/cm³) = 100 cm³. Density of solid = 500 g / 100 cm³.

Result: Density of solid = 5 g/cm³ = 5000 kg/m³ (relative density 5).

In streamline (laminar) flow, every fluid particle passing through a given point follows exactly the same path as the particle before it, and the velocity at each point of the path is steady, though it may vary from one point to another. Streamlines can never cross one another, because a crossing point would mean the fluid there has two different velocities at once, which is not possible. Streamline flow occurs only up to a certain (critical) speed; beyond this speed the motion becomes irregular, with eddies and vortices, and is called turbulent flow.

Equation of continuity: Consider a tube of flow (bounded by streamlines) carrying an incompressible fluid in steady flow. Since fluid cannot accumulate anywhere inside the tube, the mass of fluid entering per second at any cross-section must equal the mass leaving per second at any other cross-section. For a fluid of constant density flowing through a pipe whose cross-sectional area changes from A₁ (speed v₁) to A₂ (speed v₂), this conservation of mass gives:

A₁v₁ = A₂v₂, i.e. Av = constant, along the tube.

This means the fluid must speed up where the tube narrows and slow down where it widens — exactly as water speeds up when a garden hose nozzle is narrowed.

Worked example.

Given: Water flows through a pipe of radius r₁ = 2 cm with speed v₁ = 3 m/s. The pipe narrows to a section of radius r₂ = 1 cm.

Formula: A₁v₁ = A₂v₂, and since A = πr², this gives v₂ = v₁(r₁/r₂)².

Substitution: v₂ = 3 × (2/1)² = 3 × 4

Result: v₂ = 12 m/s.

Bernoulli's principle applies the work-energy theorem to the steady, streamline flow of an ideal fluid (incompressible and non-viscous). Along any streamline, the sum of pressure energy, kinetic energy and gravitational potential energy per unit volume remains constant:

P + (1/2)ρv² + ρgh = constant

where P is the pressure, v is the flow speed, h is the height above a reference level, and ρ is the fluid density. This equation shows that, at the same height, a region of higher flow speed must have lower pressure, and vice versa.

Applications:

• Speed of efflux (Torricelli's law): For a small hole at depth h below the free surface of a liquid in a wide, open tank, applying Bernoulli's equation between the free surface (open to the atmosphere, speed ≈ 0) and the hole (also open to the atmosphere) gives the speed with which liquid leaves the hole: v = √(2gh).
• Venturi meter: A horizontal pipe is given a constriction; by the equation of continuity the flow speed is higher in the constriction, so by Bernoulli's principle the pressure there is lower. The measured pressure difference between the wide and narrow sections is used to calculate the flow rate.
• Dynamic lift: When the fluid (air) moves faster over one surface of a body (such as an aerofoil, or a spinning ball) than over the other, Bernoulli's principle predicts lower pressure on the faster side. This pressure difference produces a net force perpendicular to the flow, called dynamic lift.

Worked example.

Given: A large tank of water has a small hole at a depth h = 5 m below the free surface. Take g = 9.8 m/s².

Formula: v = √(2gh)

Substitution: v = √(2 × 9.8 × 5) = √98

Result: v ≈ 9.90 m/s. Notice this efflux speed is the same as the speed a body would attain falling freely through the same height h, and it does not depend on the direction in which the hole faces.

Viscosity is the property of a real fluid by which it opposes relative motion between its adjacent layers, acting like internal friction. For laminar flow between two fluid layers separated by a small distance dx, with a velocity difference dv between them, Newton's law of viscosity gives the tangential (backward-dragging) viscous force on an area A of the layer:

F = ηA(dv/dx)

where η (eta) is the coefficient of viscosity of the fluid, with SI unit Pa·s (equivalently N·s/m², also called the poiseuille).

Stokes' law: When a small sphere of radius r moves with speed v through a viscous fluid of coefficient of viscosity η (with the flow around it remaining streamline), the fluid exerts a viscous drag force on the sphere, opposing its motion:

F = 6πηrv

Terminal velocity: A small sphere (density ρ) falling through a viscous fluid (density σ) experiences three forces: its weight (downward), the buoyant upthrust (upward), and the viscous drag (upward, increasing with speed). As the sphere speeds up, the drag increases until the net force becomes zero; thereafter it falls with a constant terminal velocity v_t. Equating weight to the sum of upthrust and viscous force:

(4/3)πr³ρg = (4/3)πr³σg + 6πηrv_t

Solving for v_t:

v_t = 2r²(ρ − σ)g / (9η)

Worked example.

Given: A steel ball of radius r = 1 mm = 1 × 10⁻³ m and density ρ = 7800 kg/m³ falls through glycerine of density σ = 1260 kg/m³ and coefficient of viscosity η = 0.83 Pa·s. Take g = 9.8 m/s².

Formula: v_t = 2r²(ρ − σ)g / (9η)

Substitution: v_t = [2 × (1 × 10⁻³)² × (7800 − 1260) × 9.8] / (9 × 0.83) = [2 × 10⁻⁶ × 6540 × 9.8] / 7.47 = 0.128184 / 7.47

Result: v_t ≈ 1.72 × 10⁻² m/s = 1.72 cm/s.

Surface tension arises because molecules at the free surface of a liquid experience a net inward cohesive force (unlike molecules in the bulk, which are pulled equally in all directions by neighbouring molecules), so the surface behaves like a stretched elastic membrane that tends to minimise its area. It is defined as the force per unit length acting along the surface, tending to pull it inward, at right angles to an imaginary line drawn on the surface:

T = F/l, SI unit N/m.

Surface energy: Work must be done against this inward pull to increase the area of a surface, so surface tension is also equal to the surface energy per unit area:

T = W/(ΔA), SI unit J/m² (numerically equal to N/m).

Excess pressure: A curved liquid surface, due to surface tension, produces a pressure inside greater than outside. For a spherical liquid drop (one free surface), the excess pressure is P = 2T/r. For a soap bubble (which has two free surfaces, an inner and an outer film), the excess pressure is P = 4T/r.

Angle of contact (θ) is the angle between the tangent to the liquid surface at the point of contact and the solid surface, measured inside the liquid. It depends on the relative strengths of the cohesive force (between liquid molecules) and the adhesive force (between liquid and solid molecules). For pure water on clean glass, θ is small (nearly 0°, since adhesion dominates and water wets glass); for mercury on glass, θ is obtuse (greater than 90°, since cohesion dominates and mercury does not wet glass).

Capillary rise: When a narrow (capillary) tube is dipped in a liquid that wets it, the liquid surface inside the tube curves (meniscus), and surface tension pulls the liquid up the tube until the weight of the risen liquid column balances the net upward pull. The height of rise is:

h = 2T cosθ / (ρgr)

where r is the radius of the capillary tube and ρ is the density of the liquid. If θ is obtuse (as for mercury), cosθ is negative, and the formula correctly predicts a capillary depression (the level inside the tube falls below the level outside) instead of a rise.

Worked example.

Given: A capillary tube of radius r = 0.2 mm = 2 × 10⁻⁴ m is dipped in water of surface tension T = 7.2 × 10⁻² N/m and density ρ = 1000 kg/m³. The angle of contact θ ≈ 0°, so cosθ = 1. Take g = 9.8 m/s².

Formula: h = 2T cosθ / (ρgr)

Substitution: h = (2 × 7.2 × 10⁻² × 1) / (1000 × 9.8 × 2 × 10⁻⁴) = 0.144 / 1.96

Result: h ≈ 7.35 × 10⁻² m = 7.35 cm.