12 min readConcept sheetSection A objective (2 marks) on units and Reynolds number; Section B very short answer (2 marks) and short answer (3 marks) on Stokes' law and terminal velocity; also a NEET favourite

Viscosity, types of flow and terminal velocity

Viscosity is fluid friction between layers. Flow stays streamlined below the critical velocity, and a falling sphere reaches a steady terminal velocity when the forces balance.

Viscous forceF = −ηA (dv/dx); A = area of the layer, dv/dx = velocity gradient, η = coefficient of viscosity

Unit of ηSI: N s m⁻² (Pa s); CGS: poise; 1 poise = 0.1 N s m⁻²; dimensions [ML⁻¹T⁻¹]

Reynolds numberR = vρd/η, with no unit. In your NIOS lesson: below 1000 laminar, 1000 to 2000 unsteady, above 2000 turbulent

Stokes' law and terminal velocityF = 6πηrv; v₀ = 2r²(ρ − σ)g / 9η, where ρ = density of the sphere, σ = density of the fluid

What viscosity is

Pour water and honey through two funnels. The water runs out fast; the honey crawls. When a fluid flows, each layer rubs on the next and slows it, like friction. Viscosity is the property of a fluid that opposes relative motion between its layers. In a pipe, the layer touching the wall is almost at rest, and layers nearer the middle move faster. The change of speed with distance across the flow, dv/dx, is the velocity gradient. The viscous force is F = −ηA(dv/dx). Here η (eta) is the coefficient of viscosity, and the minus sign shows that the force opposes the motion. Liquids become less viscous when heated, which is why warm honey flows easily. Gases become more viscous when heated.

Streamline flow vs turbulent flow

Streamline (laminar) flowEvery particle passing a point follows the same path as the one before it. The fluid moves in smooth layers (laminae) sliding over one another. It happens when the speed is below the critical velocity. Example: water flowing gently from a tap.

Turbulent flowThe paths become zig-zag, with swirls (eddies), and the layers mix. It happens when the speed is above the critical velocity. Example: a river in flood, or a tap opened fully.

Two streamlines never cross. At a crossing point the fluid would have two directions of velocity at once, which is impossible.

Critical velocity and Reynolds number

The critical velocity (vc) is the speed above which streamline flow turns turbulent. It is larger for a more viscous liquid, and smaller for a denser liquid or a wider tube: vc = Rη/(ρd). Here d is the diameter of the tube and R is the Reynolds number, a pure number with no unit. For any flow, R = vρd/η. In your NIOS lesson: R below 1000 means laminar flow, 1000 to 2000 unsteady flow, and above 2000 turbulent flow. Some other books use slightly different limits; use these in your exam.

Worked example: Reynolds number in a pipe

Problem

Water flows at 4.0 cm s⁻¹ through a pipe of diameter 2.0 cm. (a) Find the Reynolds number and the type of flow. (b) Above what speed will the flow become turbulent? (ρ = 1000 kg m⁻³, η = 1.0 × 10⁻³ N s m⁻²)

Working

v = 0.040 m s⁻¹, d = 0.020 m.
(a) R = vρd/η = (0.040 × 1000 × 0.020) ÷ (1.0 × 10⁻³) = 0.80 ÷ 0.0010 = 800.
R = 800 is below 1000, so the flow is laminar.
(b) Turbulence starts at R = 2000, so v = Rη/(ρd).
v = (2000 × 1.0 × 10⁻³) ÷ (1000 × 0.020) = 2.0 ÷ 20 = 0.10 m s⁻¹.

Answer

(a) R = 800, laminar flow; (b) above about 0.10 m s⁻¹ (10 cm s⁻¹)

Stokes' law and the derivation of terminal velocity

**Stokes' law:** a small smooth sphere moving slowly through a fluid feels a backward viscous force **F = 6πηrv**. Here r is its radius, v its speed and η the viscosity of the fluid. The method of dimensions gives F ∝ ηrv, and experiment gives the constant 6π.

1
Draw a sphere falling in a liquid. Mark its weight W downward, and the buoyant force B and the viscous force F upward.
2
Let the sphere have density ρ and the liquid density σ. Weight W = (4/3)πr³ρg; buoyant force B = (4/3)πr³σg.
3
As the sphere speeds up, F = 6πηrv grows, until the net force on it becomes zero.
4
It then falls at a steady speed v₀, the terminal velocity: 6πηrv₀ + (4/3)πr³σg = (4/3)πr³ρg.
5
So 6πηrv₀ = (4/3)πr³(ρ − σ)g.
6★
So, v₀ = 2r²(ρ − σ)g / 9η. Terminal velocity grows as r². If σ is more than ρ, as for an air bubble in water, v₀ is negative: the bubble rises.

★ key step

Worked example: a steel ball in glycerine

Problem

A steel ball of radius 1.0 mm falls through glycerine. Find its terminal velocity. (Density of steel ρ = 7800 kg m⁻³, density of glycerine σ = 1260 kg m⁻³, η = 1.5 N s m⁻², g = 9.8 m s⁻²)

Working

r = 1.0 × 10⁻³ m, so r² = 1.0 × 10⁻⁶ m²; ρ − σ = 6540 kg m⁻³.
v₀ = 2r²(ρ − σ)g / 9η.
Top line: 2 × 1.0 × 10⁻⁶ × 6540 × 9.8 = 0.128.
Bottom line: 9 × 1.5 = 13.5.
v₀ = 0.128 ÷ 13.5 = 9.5 × 10⁻³ m s⁻¹.

Answer

About 9.5 mm s⁻¹. A ball of twice the radius would fall 4 times as fast, about 38 mm s⁻¹

In real life

Monsoon raindrops reach a steady terminal velocity as air drag balances their weight, so they land gently, not at bullet speed. A paratrooper falls at a steady speed, and opening the parachute raises the drag for a slow, safe landing.

Exam angle

3 marks: “State Stokes' law. Derive an expression for the terminal velocity of a small sphere falling through a viscous liquid.” F = 6πηrv. Forces: weight (4/3)πr³ρg down; buoyancy (4/3)πr³σg and the viscous force up. At terminal velocity the net force is zero: 6πηrv₀ = (4/3)πr³(ρ − σ)g, so v₀ = 2r²(ρ − σ)g / 9η.

2 marks: “Define coefficient of viscosity. Write its SI unit and dimensions.” It is the viscous force per unit area of a layer for a unit velocity gradient: η = F ÷ (A dv/dx). SI unit N s m⁻² (1 poise = 0.1 N s m⁻²); dimensions [ML⁻¹T⁻¹].

2 marks: “What is Reynolds number? For what values is a flow laminar, unsteady and turbulent?” R = vρd/η, a pure number that tells the type of flow. Below 1000 laminar, 1000 to 2000 unsteady, above 2000 turbulent.

Remember it

Stokes' law is 6 π η r v: Six Pies Eaten Really Very slowly. Terminal velocity: two r squared, rho minus sigma, g, over nine eta.