Fluid properties: viscosity, surface tension, compressibility

Why it matters

Every fluid-power and flow calculation starts from a few material properties. Viscosity sets the friction loss in a pipe, the drag torque in a journal bearing and the leakage past a spool valve; surface tension decides capillary rise in a manometer tube and the size of droplets from a nozzle; compressibility (bulk modulus) decides how "springy" a hydraulic actuator feels and how fast a pressure wave travels through oil.

Key ideas

Fluid and continuum. A fluid deforms continuously under any shear stress, however small. We treat it as a continuum, so density, velocity and pressure are smooth functions of position. This holds when the smallest length of interest is much larger than the molecular mean free path (true for liquids and for gases at ordinary pressures).

Density and specific weight. Density ρ is mass per unit volume (water ≈ 1000 kg/m³, mercury ≈ 13 600 kg/m³). Specific weight γ = ρg is weight per unit volume. Relative density (specific gravity) is ρ divided by the density of water at 4 °C, and has no unit.

Viscosity. Viscosity is a fluid's resistance to shear deformation, the internal friction between adjacent layers moving at different speeds.

• Newton's law of viscosity: shear stress is proportional to the rate of shear strain, τ = μ·du/dy. Fluids that obey it (water, air, most mineral oils) are Newtonian.
• Dynamic viscosity μ (Pa·s = N·s/m²; 1 poise = 0.1 Pa·s). Water at 20 °C: about 1.0 × 10⁻³ Pa·s.
• Kinematic viscosity ν = μ/ρ (m²/s; 1 stoke = 10⁻⁴ m²/s, 1 cSt = 10⁻⁶ m²/s). Hydraulic oils are graded by kinematic viscosity in cSt at 40 °C.
• No-slip condition: fluid in contact with a wall moves with the wall. This is why velocity gradients, and hence viscous stresses, appear near surfaces.
• Temperature effect: liquid viscosity falls as temperature rises (weaker cohesion between molecules); gas viscosity rises with temperature (more molecular momentum exchange). Pressure has only a small effect except at very high pressure.
• Non-Newtonian fluids: shear-thinning (pseudoplastic — paints, blood), shear-thickening (dilatant — cornflour slurry), and Bingham plastics (toothpaste, which need a yield stress before flowing). Their apparent viscosity depends on shear rate.
• For a thin film between a fixed and a moving surface, with gap h small, the velocity profile is nearly linear, so du/dy ≈ U/h. This is the basis of bearing friction and dashpot calculations.

Surface tension. At a liquid–gas interface, molecules have unbalanced cohesive forces, so the surface behaves like a stretched membrane. Surface tension σ is the force per unit length along a line on the surface (N/m), equal to the surface energy per unit area (J/m²). Water–air at 20 °C ≈ 0.073 N/m; mercury–air ≈ 0.48 N/m. σ decreases as temperature rises and is strongly affected by contaminants (soap lowers it).

• Pressure inside a curved surface is higher than outside: a droplet has one surface, a soap bubble has two.
• Capillarity: in a narrow tube the liquid rises if it wets the wall (contact angle θ < 90°, water on clean glass θ ≈ 0°) and is depressed if it does not (mercury on glass, θ ≈ 130–140°). This is why manometer tubes should be wider than about 10 mm.

Compressibility and bulk modulus. Bulk modulus K is the pressure rise needed per unit fractional decrease in volume. Compressibility β = 1/K. Water has K ≈ 2.2 GPa; mineral hydraulic oil about 1.4–1.8 GPa, and much less if air is entrained. For an ideal gas, K = p (isothermal) or K = γp (isentropic), so gases are roughly 10⁴ times more compressible than liquids at atmospheric pressure. Liquids are treated as incompressible in most flow problems; gases can be too, when the Mach number is below about 0.3. Compressibility still matters for liquids in water hammer and in the stiffness of hydraulic actuators.

Vapour pressure. The pressure at which a liquid boils at a given temperature. When local pressure in a pump inlet or valve falls below it, vapour cavities form and collapse — cavitation (covered in the pump topic).

Formulas

τ = μ · du/dy

• τ = shear stress (Pa), μ = dynamic viscosity (Pa·s), du/dy = velocity gradient normal to the flow (s⁻¹). Newtonian fluids, laminar shear.

τ ≈ μ · U / h

• U = relative speed of the surfaces (m/s), h = film thickness (m). Thin film with linear velocity profile.

ν = μ / ρ

• ν = kinematic viscosity (m²/s), ρ = density (kg/m³).

T = μ · π² · D³ · N · L / (120 · h) (journal bearing, N in rpm)

• T = viscous torque (N·m), D = shaft diameter (m), L = bearing length (m), h = radial clearance (m). Concentric shaft, thin film.

Δp = 4σ / d (droplet), Δp = 8σ / d (soap bubble), Δp = 2σ / d (liquid jet)

• Δp = inside minus outside pressure (Pa), σ = surface tension (N/m), d = diameter (m).

h = 4σ · cosθ / (ρ · g · d)

• h = capillary rise (m; negative means depression), θ = contact angle, d = tube diameter (m), g = 9.81 m/s².

K = −dp / (dV/V) = dp / (dρ/ρ)

• K = bulk modulus (Pa), dV/V = volumetric strain (–). The minus sign makes K positive because volume falls as pressure rises.

β = 1 / K

• β = compressibility (Pa⁻¹).

c = √(K / ρ)

• c = speed of a pressure (sound) wave in the fluid (m/s).

Worked examples

Example 1 — plate on an oil film (standard). A plate of area 0.5 m² slides at 2 m/s over a fixed surface, separated by an oil film 0.5 mm thick with μ = 0.1 Pa·s. Find the force and power needed.

1. Shear stress: τ = μ·U/h = 0.1 × 2 / 0.0005 = 400 Pa.
2. Force: F = τ·A = 400 × 0.5 = 200 N.
3. Power: P = F·U = 200 × 2 = 400 W. Answer: F = 200 N, P = 400 W.

Example 2 — journal bearing (GATE level). A 100 mm diameter shaft runs at 600 rpm in a 150 mm long bearing with a uniform radial clearance of 0.25 mm. Oil viscosity μ = 0.08 Pa·s. Find the power lost in viscous friction.

1. Surface speed: U = π·D·N/60 = π × 0.1 × 600/60 = 3.1416 m/s.
2. Shear stress: τ = μ·U/h = 0.08 × 3.1416 / 0.00025 = 1005.3 Pa.
3. Wetted area: A = π·D·L = π × 0.1 × 0.15 = 0.047124 m².
4. Friction force: F = 1005.3 × 0.047124 = 47.37 N.
5. Torque: T = F·D/2 = 47.37 × 0.05 = 2.369 N·m.
6. Power: P = T·ω, ω = 2π × 600/60 = 62.83 rad/s, so P = 2.369 × 62.83 = 148.8 W. Answer: about 149 W.

Example 3 — capillarity and compressibility. (a) Water (σ = 0.073 N/m, θ = 0°) in a clean glass tube of 1 mm bore. (b) Pressure needed to reduce a volume of water (K = 2.2 GPa) by 0.5 %.

1. (a) h = 4σ·cosθ/(ρ·g·d) = 4 × 0.073 × 1 / (1000 × 9.81 × 0.001) = 0.0298 m.
2. (b) Δp = K·(ΔV/V) = 2.2 × 10⁹ × 0.005 = 1.1 × 10⁷ Pa. Answer: (a) about 29.8 mm rise; (b) 11 MPa.

Common mistakes

• Using kinematic viscosity in τ = μ·du/dy. Shear stress needs dynamic viscosity; convert with μ = ρν.
• Forgetting unit conversions: poise to Pa·s (divide by 10), centistokes to m²/s (multiply by 10⁻⁶), film thickness in mm to m.
• Assuming viscosity of gases falls with temperature, like liquids. It rises.
• Using 4σ/d for a soap bubble. A bubble has two surfaces, so Δp = 8σ/d.
• Using the radius in place of the diameter in h = 4σcosθ/(ρgd), which doubles or halves the answer.
• Writing K = −V·dp/dV and then dropping the sign so that K comes out negative.
• Saying "water is incompressible" in a water-hammer or actuator-stiffness problem, where its finite bulk modulus is the whole point.

For GATE ME

Expect short numericals: force or power on a plate sliding over a film, torque and power lost in a journal bearing or a rotating disc/cone viscometer, capillary rise or depression, excess pressure inside drops and bubbles, and bulk modulus or wave-speed estimates. Conceptual questions test Newtonian versus non-Newtonian behaviour (identify a Bingham plastic or shear-thinning fluid from a τ versus du/dy graph), the opposite temperature trends of liquid and gas viscosity, and units and dimensions of μ, ν, σ and K. Practise converting units quickly and drawing the linear velocity profile in thin films.

Quick check

1. What are the SI units of dynamic and kinematic viscosity?
2. How does the viscosity of air change when it is heated?
3. What is the excess pressure inside a soap bubble of diameter 40 mm if σ = 0.03 N/m?
4. Which property of oil decides how much a trapped column compresses under load?
5. Does mercury rise or fall in a narrow glass tube, and why?

Answers: 1. Pa·s and m²/s; 2. it increases; 3. Δp = 8σ/d = 8 × 0.03/0.04 = 6 Pa; 4. bulk modulus; 5. it is depressed, because its contact angle with glass exceeds 90° (it does not wet glass).