University Physics I · Optional: Fluids · 15.8
Viscosity & Turbulence
Real fluids resist being sheared. That resistance drags on moving objects and dissipates energy in pipes; the Reynolds number decides whether the flow stays orderly or breaks into eddies.
Build the model
Connect the measurement to the mechanism.
Viscosity is a fluid's resistance to being sheared: neighbouring layers moving at different speeds exchange momentum, so a stress τ = η (dv/dy) is needed to keep them sliding. That one property does two jobs — it drags on anything moving through the fluid, and it dissipates the mechanical energy an ideal fluid would have conserved. Whether it actually controls a given flow depends on how it compares with inertia, and that comparison is the Reynolds number Re = ρvL/η.
When Re is small, viscosity wins: flow is laminar, drag on a sphere is Stokes' 6πηrv, and pipe flow follows Poiseuille's law. When Re is large, inertia wins: the flow breaks into eddies, drag becomes quadratic in speed, and the tidy linear results stop applying. Compute Re first, then choose the model.
- Simple definition
- Viscosity is a fluid's resistance to being sheared, and the Reynolds number compares inertial with viscous effects to predict whether a flow is laminar or turbulent.
- Example
- A steel ball dropped into thick oil stops accelerating within a millimetre and then sinks at a steady speed: viscous drag has grown to balance weight minus buoyancy.
Sliding layers resist: the stress needed grows with how fast speed changes across the gap.
η is the dynamic viscosity, in Pa s = kg m⁻¹ s⁻¹
Viscosity per unit density — how readily a fluid spreads momentum, regardless of how heavy it is.
Unit m² s⁻¹; water ≈ 1.0 × 10⁻⁶, air ≈ 1.5 × 10⁻⁵ at 20 °C
Inertia divided by viscosity. Small Re means viscosity governs; large Re means inertia does.
Dimensionless; L is the length that characterises the geometry
A small slow sphere feels drag set by radius, speed and viscosity — fluid density never appears.
Slow steady flow, Re ≲ 1 with Re taken on the diameter
Set viscous drag equal to weight minus buoyancy; settling speed then scales as radius squared.
From 6πηrvt = (4/3)πr³(ρs − ρf)g
Halving a pipe's radius cuts laminar flow to one sixteenth at the same pressure difference.
Steady laminar flow in a circular pipe; Q in m³ s⁻¹
No slip, and the stress it costs
A real fluid does not slide freely past a solid surface: the layer touching a wall moves with the wall — the no-slip condition — so the speed must change between wall and free stream, and a Newtonian fluid resists that shearing with a stress τ = η (dv/dy) proportional to the velocity gradient. The dynamic viscosity η is measured in Pa s, or kg m⁻¹ s⁻¹: water at 20 °C is 1.0 × 10⁻³ Pa s, air is 1.8 × 10⁻⁵ Pa s. Viscosity is not wall friction: the stress acts between every pair of layers — momentum carried sideways from faster fluid to slower, which is why heating a liquid lowers its viscosity (molecules escape each other's attraction more easily) while heating a gas raises it (molecules cross between layers faster).
Laminar flow and Poiseuille's law
In laminar flow the fluid moves in ordered layers that do not mix, and while it is steady each element traces the path of the one ahead. Drive a viscous fluid along a circular pipe and balance the pressure force on a cylindrical element against the shear on its surface: out comes a parabolic speed profile — zero at the wall, maximum on the axis — and a volume flow rate Q = πΔP r⁴/(8ηL). Q is only proportional to ΔP, so doubling the flow rate merely doubles the pressure difference; the striking part is the fourth power, which cuts the flow to one sixteenth when the radius is halved. That strict proportionality is also a test: a straight line through the origin on a ΔP against Q graph says the flow is still laminar.
Stokes drag on a small sphere
Solving the viscous equations for a sphere moving slowly through an unbounded fluid gives Fd = 6πηrv — linear in speed, and with no fluid density in it at all. Release the sphere and it accelerates until drag plus buoyancy balance weight, so vt = 2r²(ρs − ρf)g/(9η). A 1.0 mm-radius steel ball (ρs = 7800 kg m⁻³) in an oil with η = 1.0 Pa s and ρf = 900 kg m⁻³ settles at vt = 2(1.0 × 10⁻³)²(6900)(9.81)/(9 × 1.0) = 1.5 × 10⁻² m s⁻¹, about 1.5 cm s⁻¹. Falling-ball viscometers measure η exactly this way. Every step assumed the flow around the sphere stays viscous-dominated, so the result is only trustworthy once you check the Reynolds number.
The Reynolds number
Viscosity does not act alone; what matters is its size relative to inertia. For a flow of characteristic speed v across a characteristic length L, the inertial term in the equation of motion scales as ρv²/L and the viscous term as ηv/L², so their ratio is Re = ρvL/η = vL/ν. It is dimensionless, so it can be compared between systems of wildly different scale. A swimming bacterium (L ≈ 2 μm, v ≈ 30 μm s⁻¹ in water) has Re ≈ 6 × 10⁻⁵ and lives in pure viscosity — stop pushing and it stops instantly. Water at 0.10 m s⁻¹ in a 20 mm bore pipe has Re = (0.10)(0.020)/(1.0 × 10⁻⁶) = 2.0 × 10³. Two geometrically similar flows at the same Re behave identically, which is why scale models in wind tunnels tell you anything at all.
The transition to turbulence
Raise Re and small disturbances that viscosity used to damp out begin to grow instead. In a circular pipe, with L taken as the diameter, flow is reliably laminar below Re ≈ 2300, unstable through roughly 2300 to 4000, and fully turbulent above that. The same water in the same 20 mm pipe is laminar at 0.10 m s⁻¹ and turbulent at 0.50 m s⁻¹ (Re ≈ 1.0 × 10⁴). Turbulent flow is unsteady, three-dimensional and strongly mixing, with eddies of every size passing energy down to the smallest scales, where viscosity finally turns it into heat. The velocity profile flattens, wall shear rises, and the pressure drop stops being proportional to Q, climbing roughly as Q to the power 1.75 in smooth pipes and as Q² in rough ones.
Where the ideal-fluid model stops
Bernoulli's equation describes a fluid with no viscosity: no shear stress, no dissipation, so p + ½ρv² + ρgh is conserved along a streamline. Real flow always breaks that near a solid boundary, where a thin boundary layer carries the whole change from zero at the wall to the free-stream speed. At low Re that layer fills the pipe and Bernoulli is useless. At high Re it stays thin and Bernoulli works well across most of the cross-section — but the layer can separate, shed a turbulent wake, and dissipate energy anyway. Engineers restore the missing physics by adding a measured head loss to Bernoulli's equation. State the assumption, estimate Re, and you know in advance which model you are allowed to use.
Change one variable at a time
Make the relationship visible.
Hold the bore and the speed still and drag the viscosity down. The marker walks right across the 2300-4000 band and the parabola collapses into a flat core with a steep wall gradient: the centreline drops from 2.00 to 1.22 times the mean, while the mean itself never moves.
REYNOLDS NUMBER vd/ν2000
CENTRELINE ÷ MEAN SPEED2.00 ×
CENTRELINE SPEED0.200 m s⁻¹
SPEED THAT REACHES Re = 23000.115 m s⁻¹
Live interpretationREYNOLDS NUMBER vd/ν: 2000. CENTRELINE ÷ MEAN SPEED: 2.00 ×. CENTRELINE SPEED: 0.200 m s⁻¹. SPEED THAT REACHES Re = 2300: 0.115 m s⁻¹
Catch the common trap
Explain before calculating.
A small sphere settles at terminal speed through a viscous liquid at Re ≪ 1. A second sphere of the same material, in the same liquid, has twice the radius and is still in the Stokes regime. How does its terminal speed compare?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyWater at 20 °C (ρ = 1000 kg m⁻³, η = 1.0 × 10⁻³ Pa s) flows at a mean speed of 0.25 m s⁻¹ along a pipe of internal diameter 12 mm. Find the Reynolds number, name the regime, and find the fastest mean speed that would keep the flow laminar.
- Re = ρvL/η, and for a circular pipe the characteristic length L is the internal diameter: L = 12 mm = 0.012 m.
- Re = (1000 kg m⁻³)(0.25 m s⁻¹)(0.012 m)/(1.0 × 10⁻³ Pa s) = 3.0/(1.0 × 10⁻³) = 3.0 × 10³. Every unit cancels — Re is a pure number.
- Compare with the pipe thresholds: laminar below Re ≈ 2300, fully turbulent above Re ≈ 4000. Re = 3.0 × 10³ lands between them, in the unstable band, so neither Poiseuille's law nor a turbulent correlation can be trusted here.
- For laminar flow demand Re ≤ 2300, and use ν = η/ρ = 1.0 × 10⁻⁶ m² s⁻¹: v = Reν/L = (2300)(1.0 × 10⁻⁶ m² s⁻¹)/(0.012 m) = 0.19 m s⁻¹.
AnswerRe = 3.0 × 10³ — transitional, neither reliably laminar nor fully turbulent. Staying laminar needs v ≤ 0.19 m s⁻¹.
MediumA glass sphere of radius 0.50 mm (ρs = 2500 kg m⁻³) is released in glycerol (ρf = 1260 kg m⁻³, η = 1.41 Pa s). Find its terminal speed, then check that Stokes' law was allowed.
- At terminal speed the net force is zero: viscous drag plus upthrust balance weight, 6πηrvₜ = (4/3)πr³(ρs − ρf)g.
- Divide both sides by 6πηr: vₜ = 2r²(ρs − ρf)g/(9η). The density difference is ρs − ρf = 2500 − 1260 = 1240 kg m⁻³.
- With r = 0.50 mm = 5.0 × 10⁻⁴ m, so r² = 2.5 × 10⁻⁷ m²: vₜ = 2(2.5 × 10⁻⁷)(1240)(9.81)/(9 × 1.41) = (6.08 × 10⁻³)/(12.69) = 4.79 × 10⁻⁴ m s⁻¹. Units check: kg s⁻² ÷ kg m⁻¹ s⁻¹ = m s⁻¹.
- Stokes' law was assumed, so test it on the diameter D = 1.0 × 10⁻³ m: Re = ρf vₜ D/η = (1260)(4.79 × 10⁻⁴)(1.0 × 10⁻³)/(1.41) = 4.3 × 10⁻⁴.
- Re ≪ 1, so viscosity really does govern the flow round the sphere and the linear drag law stands. Had Re come out near 1 or above, the whole calculation would need redoing with a quadratic drag law.
Answervₜ = 4.8 × 10⁻⁴ m s⁻¹ (0.48 mm s⁻¹), at Re ≈ 4.3 × 10⁻⁴ — deep inside the Stokes regime.
HardOlive oil (ρ = 920 kg m⁻³, η = 0.081 Pa s) is driven through a horizontal tube of radius 2.0 mm and length 0.75 m by a pressure difference of 8.0 kPa. Find the volume flow rate and the mean speed, confirm the flow is laminar, then find the flow rate once the radius narrows to 1.6 mm at the same pressure difference.
- Poiseuille's law for steady laminar flow: Q = πΔP r⁴/(8ηL), with r⁴ = (2.0 × 10⁻³ m)⁴ = 1.6 × 10⁻¹¹ m⁴.
- Q = π(8.0 × 10³ Pa)(1.6 × 10⁻¹¹ m⁴)/(8 × 0.081 Pa s × 0.75 m) = (4.02 × 10⁻⁷)/(0.486) = 8.27 × 10⁻⁷ m³ s⁻¹, which is 0.83 mL s⁻¹.
- Mean speed is flow rate over cross-section: A = πr² = π(2.0 × 10⁻³)² = 1.257 × 10⁻⁵ m², so v̄ = (8.27 × 10⁻⁷ m³ s⁻¹)/(1.257 × 10⁻⁵ m²) = 6.58 × 10⁻² m s⁻¹.
- Poiseuille's law assumed laminar flow, so check it on the diameter: Re = ρv̄D/η = (920)(0.0658)(4.0 × 10⁻³)/(0.081) = 3.0 — far below 2300, so the assumption holds.
- At fixed ΔP, Q ∝ r⁴, so Q′/Q = (1.6/2.0)⁴ = 0.80⁴ = 0.4096 and Q′ = 0.4096 × 8.27 × 10⁻⁷ = 3.39 × 10⁻⁷ m³ s⁻¹.
- A 20% narrower tube passes 41% of the flow. Re follows too: Re ∝ Q/r ∝ r³, so it falls to 0.80³ = 0.512 of before, about 1.5 — the narrowed flow is even more firmly laminar.
AnswerQ = 8.3 × 10⁻⁷ m³ s⁻¹ (0.83 mL s⁻¹), v̄ = 6.6 × 10⁻² m s⁻¹, Re ≈ 3.0 — firmly laminar. At r = 1.6 mm, Q = 3.4 × 10⁻⁷ m³ s⁻¹, 41% of the original.