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Lesson 5 of 8

Mechanical Properties of Fluids · Lesson 5 of 8

Viscosity and Motion Through Fluids

“Explain real-fluid shear, pipe resistance and terminal settling using models with explicit validity conditions.”

Learning Objectives

• Relate viscous shear stress to shear rate and identify viscosity units. • Explain no-slip velocity profiles and the effect of temperature. • Use pipe-flow resistance and Reynolds-number scaling where the chapter exercises require them. • Derive terminal speed from weight, buoyancy and Stokes drag, stating the low-speed assumptions.

Viscosity is resistance to relative motion inside a fluid

Slide a plate across a film of oil, then imagine replacing the oil by honey while keeping the speed and film thickness unchanged. A more viscous liquid needs a greater pulling force. Neighbouring fluid layers exchange momentum and exert tangential forces opposing their relative motion. This internal friction converts organised mechanical energy into internal energy. It explains why a real fluid can lose pressure along a pipe even when its average speed does not change.

A fluid at rest does not sustain a static shear stress like an elastic solid. Once shear stress is applied, however, the fluid deforms continuously. Its response depends on how quickly it is deforming. For the Newtonian fluids used in the simple model, shear stress is proportional to the rate of shear strain, not the accumulated strain. Some everyday materials are not Newtonian; the constant-viscosity formulas below are not universal descriptions of every paste or biological fluid.

From moving plates to the viscosity formula

Put a liquid film of thickness l between two broad parallel plates of area A. Hold the bottom plate fixed and move the upper plate to the right at constant speed v. With the no-slip boundary condition, the fluid touching the bottom is at rest and the fluid touching the top moves at v. In the simple steady, uniform, no-pressure-gradient arrangement, intermediate velocities increase linearly with distance from the bottom plate.

Velocity changes between plates and across a pipeTop plate speed vBottom plate fixedlPipe: fastest near axisNo slip: zero speed at wallsShear stress = η × velocity gradient
Viscous velocity profiles between plates and in a pipe— Schematic. The parallel-plate profile is linear in the stated arrangement; fully developed laminar pipe flow has a curved, approximately parabolic profile.

In time Δt the top moves Δx = vΔt. The shear strain acquired by the film is Δx/l. Divide this by Δt to obtain the shear rate v/l, with unit s⁻¹. If τ is shear stress, define the coefficient of viscosity η by τ = η(v/l). A layer is pulled forward by faster material above and pulled back by slower material below. The viscous force on the moving plate opposes its motion.

Newtonian viscous shearLaTeX
Force here is a magnitude. More generally τ is related to the local velocity gradient dv/dy; v/l applies to a linear profile across gap l. Strain rate, unlike strain, has unit s⁻¹.
Viscosity units and dimensionsLaTeX
The SI name poiseuille (Pl) also represents Pa·s; a millipoiseuille is 10⁻³ Pa·s. Do not confuse dynamic viscosity η with density or with the kinematic viscosity η/ρ.

In a stationary circular pipe the liquid touching the wall is approximately at rest, while liquid near the axis moves fastest. Different cylindrical layers move at different speeds. The cross-sectional mean speed is not the centreline speed. No slip at the wall does not imply that the entire pipe flow stops. A pressure difference maintains flow against the viscous resistance.

Measuring viscosity with a sliding block

Problem
A block of contact area 0.10 m² slides at constant speed 0.085 m/s on a liquid film 0.30 mm thick. A string over an ideal pulley connects it to a hanging 0.010 kg mass. Find the liquid’s viscosity, using g = 9.8 m/s².

  1. 1.Because the motion is at constant speed, the hanging mass has zero acceleration. Its weight is balanced by string tension T = mg = 0.098 N. The same tension balances viscous drag on the block.
  2. 2.Film thickness l = 0.30 × 10⁻³ m. Shear stress is F/A = 0.098/0.10 = 0.98 Pa.
  3. 3.Shear rate is v/l = 0.085/(0.30 × 10⁻³) = 283.3 s⁻¹.
  4. 4.Viscosity η = shear stress/shear rate = 0.98/283.3 = 3.46 × 10⁻³ Pa·s. Equivalently η = Fl/(Av).
  5. 5.This model neglects other friction, assumes a uniform film and a linear velocity profile, and uses an ideal string and pulley. If the mass were accelerating, T would not equal mg.

Temperature and representative viscosity values

FluidTemperature (°C)Viscosity (mPa·s)
Water201.0
Water1000.3
Blood372.7
Machine oil16113
Machine oil3834
Glycerine20830
HoneyNot specified in source200
Air00.017
Air400.019

These representative values show that glycerine is much more viscous than water, while air’s dynamic viscosity is much smaller. Real honey varies greatly with composition and temperature, and real blood has a more complex rheology than a single Newtonian value. The source describes blood as several times more viscous than water and states that their viscosity ratio remains approximately constant from 0°C to 37°C; treat that as its simplified comparison, not a universal biomedical calibration.

For ordinary liquids viscosity generally decreases as temperature rises: the molecules can rearrange more easily despite their attractions. In gases viscosity generally increases with temperature because faster random molecular motion transfers momentum more effectively between layers. These opposite trends concern viscosity, not density. Heating a liquid also usually reduces its surface tension, which is a different physical property.

Pipe resistance: the tool needed for the flow exercises

Consider fully developed, steady laminar flow of an incompressible Newtonian liquid through a straight circular tube of radius R and length L. Let ΔP be the inlet pressure minus outlet pressure at equal elevation. A coaxial cylindrical fluid core of radius r feels forward pressure force ΔP πr². The surrounding liquid exerts a resisting shear force τ(r)2πrL. Force balance gives τ(r) = ΔP r/(2L).

Speed decreases outward, so its radial derivative dv/dr is negative. Here a derivative means the local rate of speed change with radius. The positive shear magnitude is −η dv/dr. Substituting the force balance gives dv/dr = −ΔP r/(2ηL). Adding these small speed changes from the stationary wall, where v(R) = 0, inward gives a parabolic profile. This is a short mathematical bridge for the pipe-flow exercises rather than a new empirical assumption about ideal nonviscous flow.

Laminar pipe profile and volume flowLaTeX
Hagen–Poiseuille relation: straight circular tube, constant viscosity, no slip, fully developed laminar flow. Q is volume per second; use radius, not diameter, in R⁴.

To obtain Q, split the circular section into thin rings. A ring of radius r and width dr has area approximately 2πr dr and transports volume v(r)2πr dr each second. Summing these rings means integrating from 0 to R. The integrals of r and r³ are R²/2 and R⁴/4; inserting the profile gives πR⁴ΔP/(8ηL). The mean speed is half the centreline speed in this profile. These results do not require a separate ideal Bernoulli pressure drop; they explicitly describe viscous dissipation.

The fourth-power radius dependence is powerful. With viscosity, tube length and pressure difference fixed, halving the radius reduces Q by a factor of sixteen. Changing pressure changes Q only in direct proportion in the same model. This explains why syringe-needle bore can control flow much more strongly than modest changes in thumb pressure. A real injection also includes tissue resistance and possible non-Newtonian behaviour; the lesson is the strong geometric sensitivity.

Flow regime and the role of size

Reynolds numberLaTeX
For pipe flow D is the diameter. Re is dimensionless and compares inertial effects with viscous effects. The relevant length is chosen for the geometry; onset thresholds depend on the setup.

Small Reynolds number suggests strong viscous control and smooth laminar behaviour. In a long straight pipe the transition is commonly discussed around Re of a few thousand, with details depending on disturbances; this is not a universal boundary for all flows. For a similar shape and the same fluid, keeping the same Reynolds number makes characteristic speed inversely proportional to size. Thus a smaller wind-tunnel aircraft model generally requires a greater speed to reach a comparable transition condition. Other similarity requirements may matter in a real wind tunnel.

Glycerine through a tube

Problem
Glycerine of density 1.3 × 10³ kg/m³ and viscosity 0.83 Pa·s flows through a horizontal tube 1.5 m long and radius 1.0 cm. Collected mass per second is 4.0 × 10⁻³ kg/s. Find the pressure drop and check laminar consistency.

  1. 1.Convert mass flow to volume flow: Q = ṁ/ρ = 0.004/1300 = 3.077 × 10⁻⁶ m³/s. The radius is R = 0.010 m, so D = 0.020 m.
  2. 2.Mean speed = Q/(πR²) = 0.00979 m/s. Do not divide mass flow directly by area and call it a speed.
  3. 3.Rearrange Poiseuille: ΔP = 8ηLQ/(πR⁴) = 975.5 Pa, approximately 9.8 × 10² Pa.
  4. 4.Re = 1300 × 0.00979 × 0.020/0.83 = 0.307, approximately 0.3. This very small value is consistent with laminar flow, so the assumed regime is reasonable.
  5. 5.The pressure drop is positive in the direction of flow and supplies energy dissipated by viscosity. A loss-free horizontal Bernoulli calculation with equal end speeds would miss it.

Stokes’ drag on a slowly moving sphere

A sphere moving through a fluid drags the adjacent fluid, which disturbs neighbouring layers. For a small rigid sphere moving sufficiently slowly through an effectively unbounded Newtonian fluid, Stokes’ law gives a drag opposite the relative velocity. The source states this law without deriving the numerical factor 6π; dimensional analysis alone cannot supply that factor. This low-Reynolds-number law is not a general drag formula for every sphere at every speed.

Stokes’ lawLaTeX
a is sphere radius, v is speed relative to the distant fluid, and η is dynamic viscosity. Require creeping flow (sphere Reynolds number much less than 1), negligible wall effects and the appropriate no-slip sphere model.

A raindrop or a pendulum moving through air experiences resistance, but a typical large raindrop may deform and have a Reynolds number too large for Stokes’ law. Use the spherical low-speed model for sufficiently small droplets or suitable laboratory spheres. A force proportional to v gives a convenient model only within its regime.

Terminal speed is a force balance, not the disappearance of forces

Release a dense sphere in a stationary viscous liquid. Weight acts downward, while buoyancy and drag act upward once it falls. At first drag is small and the sphere accelerates. Increasing speed increases drag. Eventually the upward forces can equal weight and acceleration becomes zero even though the sphere keeps moving. That constant limiting speed is terminal speed.

Buoyancy equals the weight of displaced fluid, a prerequisite force used here: for sphere volume V = 4πa³/3, it is ρ_f Vg. If ρ_s is sphere density, weight is ρ_s Vg. At terminal speed, subtract buoyancy from weight and equate the remainder to Stokes drag. Keeping separate density symbols prevents using the fluid density twice.

Terminal-speed balance and resultLaTeX
This downward settling-speed formula assumes ρₛ > ρ_f and Stokes conditions. If the density difference reverses, buoyant rise replaces downward settling; equal densities give no gravity-driven settling in this model.
A settling sphere reaches a force balanceWeightBuoyancyViscous dragv downwardtvvₜAt vₜ: weight = buoyancy + drag
Settling and the approach to terminal speed— Schematic force arrows are qualitative, not a scale construction. The graph shows speed approaching a constant as drag builds up.

Terminal speed scales as a², the density difference and 1/η. Doubling radius gives four times the terminal speed only if the new motion still satisfies the Stokes assumptions. A neutral-density sphere does not settle simply because its mass is nonzero; buoyancy cancels its weight. Terminal motion has zero net force, not zero weight or zero drag.

Finding oil viscosity from a falling copper sphere

Problem
A copper sphere of radius 2.0 mm settles at 6.5 cm/s through oil at 20°C. Sphere density is 8.9 × 10³ kg/m³ and oil density is 1.5 × 10³ kg/m³. Infer viscosity using g = 9.8 m/s² and the Stokes model.

  1. 1.Convert a = 2.0 × 10⁻³ m and v_t = 6.5 × 10⁻² m/s. The density difference is 7400 kg/m³.
  2. 2.Solve for η: η = 2a²(ρ_s − ρ_f)g/(9v_t).
  3. 3.Substitution gives η = 2(0.0020)² × 7400 × 9.8/(9 × 0.065) = 0.9917 Pa·s ≈ 0.99 Pa·s.
  4. 4.A consistency estimate gives sphere Re = ρ_f v_t(2a)/η ≈ 0.39. This is below 1 but not much less than 1, so the strict creeping-flow assumption is marginal. The result is the source’s Stokes-model estimate; high-accuracy inference would need corrections and attention to tank-wall effects.

Quick Quiz

Quiz

Quick check

For a Newtonian fluid, viscous shear stress is proportional to what?

Explanation

τ = η times strain rate. An elastic solid instead relates shear stress to shear strain within its elastic range.

Quick check

What is the SI unit of dynamic viscosity?

Explanation

Viscosity is stress divided by rate, giving Pa/(s⁻¹) = Pa·s.

Quick check

At terminal settling speed in the Stokes model, which balance holds?

Explanation

Zero net force requires downward weight to equal both upward forces together.

Quick check

In laminar Poiseuille flow at fixed pressure drop, halving the tube radius makes Q what?

Explanation

Q is proportional to R⁴, so Q_new/Q_old = (1/2)⁴ = 1/16.

Quick check

Which ordinary temperature trend is correct?

Explanation

The mechanisms differ: liquid rearrangement becomes easier, while gas momentum transfer increases with molecular motion.

Practice Problems

Practice Problems
  1. A 0.20 m² plate moves at 0.10 m/s over a stationary plate with a uniform 0.50 mm liquid gap. If η = 0.020 Pa·s, calculate shear rate, shear stress, force and power needed to maintain the speed.
  2. Derive the laminar pipe speed profile from force balance on a cylindrical core and the no-slip condition. Explain how summing thin annular volume flows gives the R⁴ dependence of Q.
  3. A horizontal circular tube of radius 2.0 mm and length 0.50 m carries a liquid of viscosity 0.10 Pa·s at Q = 2.0 × 10⁻⁷ m³/s. Calculate the pressure drop using the laminar model. For density 1000 kg/m³, compute Re to check consistency.
  4. A sphere of radius 0.10 mm and density 2500 kg/m³ settles in a liquid of density 1000 kg/m³ and viscosity 0.10 Pa·s. With g = 9.8 m/s², calculate its Stokes terminal speed and sphere Reynolds number. State the force balance and whether the low-Re assumption is reasonable.
  5. For geometrically similar models in the same fluid, explain why a model one fifth the size of an aircraft would require five times the speed to match Reynolds number. State why this similarity argument alone does not prove that every real wind-tunnel behaviour is identical.

Key Takeaways

Key Takeaways

• Viscosity relates shear stress to strain rate and dissipates mechanical energy; its SI unit is Pa·s. • No slip produces different speeds across a pipe; the mean and centreline speeds are distinct. • Laminar tube flow is strongly radius dependent; check the regime before using the pipe formula. • Stokes terminal speed follows from weight minus buoyancy balancing drag, and requires sufficiently small Reynolds number.