Fluid properties, Newtonian and non-Newtonian fluids

Why it matters

Every pump, pipeline, reactor and mixer in a chemical plant is sized from a handful of fluid properties: density sets pressure and inertia, viscosity sets friction and power, and the rheology (Newtonian or not) decides which design equations are even valid. Polymer melts, slurries, paints, food pastes and drilling muds are non-Newtonian, and treating them as water gives pressure drops and pump powers that can be wrong by a factor of several.

Key ideas

What a fluid is. A fluid is a substance that deforms continuously under any shear stress, however small. A solid resists shear by a finite deformation; a fluid resists only the rate of deformation. We treat fluids as a continuum, valid when the length scale of the problem is much larger than the molecular mean free path (Knudsen number Kn = λ/L ≪ 1).

Density and related quantities.

• Density ρ (kg/m³): water ≈ 1000, air at 20 °C and 1 atm ≈ 1.2.
• Specific weight γ = ρg (N/m³).
• Specific gravity SG = ρ/ρ_water (dimensionless).
• Specific volume v = 1/ρ (m³/kg).

Viscosity. Viscosity is the fluid's resistance to shear deformation, caused by momentum transfer between adjacent layers. In simple shear (flow between a fixed plate and a moving plate) the velocity varies linearly across the gap and the shear stress is proportional to the velocity gradient.

• Dynamic viscosity μ (Pa·s = N·s/m² = kg/(m·s)). 1 poise = 0.1 Pa·s; 1 cP = 1 mPa·s. Water at 20 °C ≈ 1.0 mPa·s.
• Kinematic viscosity ν = μ/ρ (m²/s). 1 stokes = 10⁻⁴ m²/s. Water ≈ 1.0 × 10⁻⁶ m²/s; air ≈ 1.5 × 10⁻⁵ m²/s, so air has the larger kinematic viscosity even though its μ is about 55 times smaller.
• Temperature effect: liquid viscosity falls with temperature (weaker intermolecular cohesion); gas viscosity rises with temperature (more molecular momentum exchange). Pressure has little effect on either at moderate pressures.

Newtonian fluids. Shear stress is linearly proportional to shear rate with zero intercept: a plot of τ against du/dy is a straight line through the origin with slope μ. Water, air, most gases, low-molecular-weight liquids and dilute solutions behave this way.

Non-Newtonian fluids (time-independent).

• Pseudoplastic (shear-thinning): apparent viscosity falls as shear rate rises; power-law index n < 1. Polymer solutions, paints, blood, many food products.
• Dilatant (shear-thickening): apparent viscosity rises with shear rate; n > 1. Concentrated suspensions such as cornstarch in water.
• Bingham plastic: no flow until a yield stress τ₀ is exceeded, then linear. Toothpaste, drilling mud, some slurries.
• Herschel–Bulkley: yield stress plus power-law behaviour.

Time-dependent fluids. Thixotropic fluids lose apparent viscosity with time at constant shear rate (yoghurt, some paints, ketchup); rheopectic fluids gain it. Viscoelastic fluids (polymer melts) show both viscous and elastic response (rod climbing, die swell).

Other properties.

• Surface tension σ (N/m): force per unit length at a liquid interface; gives capillary rise h = 4σcosθ/(ρgd) in a tube of diameter d. Water–air at 20 °C ≈ 0.073 N/m.
• Compressibility: bulk modulus K = −V·dp/dV (Pa). Water ≈ 2.2 GPa, so liquids are treated as incompressible. Gases are effectively incompressible only when the Mach number is below about 0.3.
• Vapour pressure p_v: the pressure at which a liquid boils at a given temperature; it decides cavitation in pumps (see NPSH).

Formulas

• τ = μ·(du/dy) — Newton's law of viscosity. τ shear stress (Pa), μ dynamic viscosity (Pa·s), du/dy velocity gradient (s⁻¹). Newtonian fluids, laminar flow.
• ν = μ/ρ — kinematic viscosity (m²/s), ρ density (kg/m³).
• τ = μ·U/h — simple shear (Couette) with a linear profile; U plate speed (m/s), h gap (m). Valid for a thin gap.
• τ = K·(du/dy)ⁿ — power-law (Ostwald–de Waele) model. K consistency index (Pa·sⁿ), n flow-behaviour index (–).
• μ_app = τ/(du/dy) = K·(du/dy)ⁿ⁻¹ — apparent viscosity (Pa·s).
• τ = τ₀ + μ_p·(du/dy) for τ > τ₀ — Bingham plastic. τ₀ yield stress (Pa), μ_p plastic viscosity (Pa·s). For τ ≤ τ₀ there is no shear flow.
• K_b = −V·Δp/ΔV — bulk modulus (Pa); ΔV/V = −Δp/K_b.
• h = 4σ·cosθ/(ρ·g·d) — capillary rise (m); σ surface tension (N/m), θ contact angle, d tube diameter (m).

Worked examples

Example 1 (standard). A flat plate of area 0.5 m² slides at 0.4 m/s on a 1 mm oil film over a fixed surface. Oil viscosity μ = 0.25 Pa·s. Find the shear stress, the drag force and the power needed.

1. Assume a linear velocity profile across the thin film: τ = μ·U/h.
2. τ = 0.25 Pa·s × 0.4 m/s / 0.001 m = 100 Pa.
3. Force F = τ·A = 100 Pa × 0.5 m² = 50 N.
4. Power P = F·U = 50 N × 0.4 m/s = 20 W. τ = 100 Pa, F = 50 N, P = 20 W.

Example 2 (GATE level). A polymer solution follows the power law with K = 0.8 Pa·sⁿ and n = 0.6. It is sheared between parallel plates 2 mm apart, the upper plate moving at 0.1 m/s. Find (a) the shear stress, (b) the apparent viscosity, and (c) the apparent viscosity if the plate speed is doubled.

1. Shear rate (linear profile): du/dy = U/h = 0.1 / 0.002 = 50 s⁻¹.
2. τ = K·(du/dy)ⁿ = 0.8 × 50^0.6 = 0.8 × 10.46 = 8.37 Pa.
3. μ_app = K·(du/dy)ⁿ⁻¹ = 0.8 × 50^(−0.4) = 0.167 Pa·s.
4. Doubling U doubles the shear rate to 100 s⁻¹: μ_app = 0.8 × 100^(−0.4) = 0.127 Pa·s. The fluid has thinned by a factor 2^0.4 = 1.32, as expected for n < 1. (a) τ ≈ 8.37 Pa, (b) μ_app ≈ 0.167 Pa·s, (c) μ_app ≈ 0.127 Pa·s.

Example 3 (compressibility). Water (K_b = 2.2 GPa) is pressurised by 10 MPa. Fractional volume change: ΔV/V = −Δp/K_b = −10 × 10⁶ / 2.2 × 10⁹ = −0.0045, i.e. a 0.45 % decrease — small enough to justify the incompressible assumption.

Common mistakes

• Saying "viscosity falls with temperature" for all fluids: true for liquids, false for gases.
• Mixing up μ and ν: Reynolds number uses ρVD/μ or VD/ν, never VD/μ.
• Unit slips: cP to Pa·s is ×10⁻³, not ×10⁻²; the stokes is 10⁻⁴ m²/s.
• Using τ = μ·U/h for a thick gap or for a non-Newtonian fluid with a non-linear profile.
• Treating a Bingham plastic as flowing at any stress: below τ₀ it moves as a rigid plug.
• Writing the apparent viscosity of a power-law fluid as K·(du/dy)ⁿ instead of K·(du/dy)ⁿ⁻¹.

For GATE CH

Expect short numericals on Newton's law of viscosity (plate on an oil film, shaft in a journal bearing), power-law and Bingham calculations of shear stress, shear rate or apparent viscosity, and conceptual MCQs identifying a rheogram (τ vs du/dy curve) as Newtonian, pseudoplastic, dilatant or Bingham. Practise unit conversion between poise, centipoise, stokes and SI, and the temperature trend for liquids versus gases.

Quick check

1. Give the SI unit of kinematic viscosity.
2. How does the viscosity of air change when it is heated?
3. A power-law fluid has n = 1.4. Is it shear-thinning or shear-thickening?
4. A Bingham fluid has τ₀ = 4 Pa and μ_p = 0.5 Pa·s. What shear rate results from τ = 6 Pa?

Answers: 1. m²/s. 2. It increases. 3. Shear-thickening (dilatant). 4. (6 − 4)/0.5 = 4 s⁻¹.