Mechanical Properties of Fluids · TOPIC LESSON
Count interfaces, track curvature and choose the right viscous-flow model.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
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No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
A liquid interface has a surface free-energy cost. For a uniform interface at fixed temperature and composition, a small reversible area increase requires work dW = γ dA, where γ is surface tension. Its units J m⁻² equal N m⁻¹. The same γ describes a tangential pulling force per unit length along a cut in the interface, perpendicular to that cut and tangent to the surface.
The surface is not literally an elastic rubber sheet with the same force–extension law. Its molecular environment differs from the bulk, and creating more interface requires rearranging molecules against that free-energy cost. The simple constant-γ model works when temperature and composition remain effectively fixed and surface gradients are absent.
A soap film has two liquid–air interfaces, front and back. If a straight movable wire of length L bounds the film, each surface pulls with γL, giving total F = 2γL. Moving the wire by distance x increases total area by 2Lx and requires quasistatic work 2γLx. A single liquid surface bounded by a line contributes only one interface unless the geometry says otherwise.
Surface tension tends to reduce area subject to volume and boundary constraints. A small isolated drop becomes approximately spherical when gravity and external forcing are weak because a sphere minimizes area at fixed volume. Larger drops can flatten under gravity; “all liquid surfaces must be spherical” is not the principle.
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ORIGINAL WORKED EXAMPLE
A soap film has surface tension 0.030 N m⁻¹ on each interface. Its movable wire is 0.10 m long and is pulled quasistatically by 0.020 m. Find required force and work, ignoring weight and friction.
Answer 6.0 mN; 1.2 × 10⁻⁴ J.
02 / UNDERSTAND & APPLY
For a spherical liquid drop of radius r with one interface, force balance across a hemisphere gives Δp πr² = γ(2πr), so internal excess pressure is Δp = 2γ/r. Equivalently, pressure work Δp dV balances γ dA during a small radius increase. Count the interface and use radius, not diameter.
A gas bubble inside a liquid also has one liquid–gas interface and has excess internal pressure 2γ/r under the same spherical constant-tension model. A thin soap bubble in air has two closely spaced interfaces, giving approximately 4γ/r when their radii are nearly equal. A thick shell would require separate curvatures and the appropriate pressure steps.
The general Laplace pressure jump is γ(1/R₁ + 1/R₂) with a consistent signed-curvature and normal convention. A spherical interface has both principal radii equal to r; a long cylindrical interface has one radius r and the other infinite, giving magnitude γ/r. The spherical formula is not automatically valid for every curved surface.
For two soap bubbles with the same constant γ, the smaller has larger excess pressure. If connected appropriately, this can drive gas from the smaller to the larger. Real rubber balloons have a tension that depends strongly on stretching and material behavior, so the constant-liquid-surface-tension conclusion should not be transferred blindly to balloons.
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ORIGINAL WORKED EXAMPLE
A spherical liquid drop and a thin soap bubble each have radius 2.0 mm and interface tension 0.040 N m⁻¹. Find their excess internal pressures.
Answer Drop 40 Pa; soap bubble 80 Pa above their respective surroundings.
03 / UNDERSTAND & APPLY
When n identical spherical drops of radius r merge into one sphere without volume loss, volume conservation gives R³ = nr³, or R = n^(1/3)r. Initial area is n4πr² and final area is 4πn^(2/3)r². The final surface area is smaller for n > 1, so constant-γ surface energy decreases.
The energy released is γ4πr²[n − n^(2/3)] for one-interface drops. It can become fluid motion, sound or thermal energy; it does not vanish. The reverse process of breaking a drop into smaller drops requires surface-energy input, even if total liquid volume stays unchanged.
For a soap bubble, count both surfaces and check whether the gas amount, gas temperature and surrounding pressure constrain volume. Merging gas bubbles is not automatically identical to merging incompressible liquid drops because gas can compress or expand as pressure changes. A problem using fixed volume must state or justify that approximation.
Surface energy scales with area, not volume. This makes surface effects relatively more important for small objects, whose area-to-volume ratio is large. A useful comparison of gravitational and surface effects is the dimensionless Bond number ρgr²/γ, using an appropriate density difference for two fluids. Small values indicate surface tension can dominate shape over gravity.
ORIGINAL WORKED EXAMPLE
Eight identical spherical liquid drops of radius r coalesce into one sphere with unchanged total volume and constant surface tension γ. Find final radius and the fraction of initial surface energy released.
Answer Final radius 2r; 50% of the initial surface energy released.
04 / UNDERSTAND & APPLY
The contact angle θ is measured through the liquid between the solid surface and the tangent to the liquid–gas interface. It depends on the three interfaces and on surface condition. A wetting liquid with θ < 90° commonly forms a concave meniscus in a vertical tube and rises; a nonwetting case with θ > 90° gives a depressed level.
For a circular capillary of radius r, the vertical component of surface-tension force around the contact line is 2πrγ cosθ. Balancing this against the weight of the raised liquid column, approximately ρgπr²h, gives h = 2γ cosθ/(ρgr). This neglects meniscus-volume corrections, air density, evaporation and other small-scale complications.
The sign of cosθ predicts rise or depression, and the magnitude scales inversely with tube radius. A tube with half the radius produces twice the ideal height for the same liquid, contact angle and gravity. The capillary pressure viewpoint gives the same result by comparing the curved meniscus pressure jump with the hydrostatic column.
The formula assumes a sufficiently long tube to accommodate the equilibrium meniscus. If a tube is cut shorter than the predicted rise, the liquid can reach the top and the interface geometry changes; it is not correct to insist that a full column with the original contact-angle curvature exists beyond the physical tube. Surface contamination and contact-angle hysteresis can also modify real observations.
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ORIGINAL WORKED EXAMPLE
A liquid of density 1000 kg m⁻³ and surface tension 0.072 N m⁻¹ wets a capillary of radius 0.50 mm with contact angle 60°. Use g = 10 m s⁻². Find ideal rise.
Answer Approximately 1.4 cm upward.
05 / UNDERSTAND & APPLY
A Newtonian fluid has shear stress proportional to the rate at which adjacent layers move relative to one another. For simple parallel flow with velocity v_x(y), the magnitude relation is τ = η|dv_x/dy|, with sign determined by which surface and force direction is considered. Dynamic viscosity η has SI units Pa s = kg m⁻¹ s⁻¹.
Between two large parallel plates separated by h, if the lower plate is fixed and the upper moves at constant speed U, the ideal steady no-slip Couette profile is linear when pressure gradient and edge effects are negligible. Shear force magnitude is F = ηAU/h. The fluid resists relative sliding, and mechanical energy supplied by the moving plate is dissipated as internal energy.
Viscosity is not density. A dense fluid need not be highly viscous, and a low-density fluid can still resist shear. Kinematic viscosity is ν = η/ρ, with units m² s⁻¹. Be careful because the same Greek letter ν is also used for frequency in other contexts; the definition and dimensions decide its meaning.
For many ordinary liquids, viscosity decreases as temperature rises, while dilute-gas viscosity often increases with temperature over familiar ranges. Those are material-dependent trends, not a universal single law for every fluid. Non-Newtonian fluids do not have one constant η independent of shear rate, so the simple linear model must be specified.
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ORIGINAL WORKED EXAMPLE
A Newtonian liquid of viscosity 0.50 Pa s fills a 2.0 mm gap between large plates. The upper plate has area 0.020 m² and moves at 0.10 m s⁻¹; the lower is fixed. Neglect pressure gradient and edge effects. Find required steady shear force and mechanical power.
Answer 0.50 N; 0.050 W.
06 / UNDERSTAND & APPLY
For steady, fully developed laminar flow of an incompressible Newtonian fluid through a long straight circular pipe of radius R, no-slip walls and negligible entrance effects give Poiseuille flow. The speed profile is v(r) = Δp(R² − r²)/(4ηL), where L is pipe length and Δp is the driving pressure difference for a horizontal pipe.
The profile is maximum on the axis and zero at the wall. Integrating the circular rings gives volume flow rate Q = πR⁴Δp/(8ηL). Mean speed Q/(πR²) is half the centreline speed. A drawing with uniform speed across the pipe is not the developed laminar profile, even though continuity can still use the cross-sectional mean.
The fourth-power radius dependence is especially important: doubling radius at fixed pressure difference, viscosity and length increases the ideal flow rate sixteenfold. Hydraulic resistance Δp/Q = 8ηL/(πR⁴) adds in series for segments carrying the same steady flow under the model. Parallel branches add conductances when they share the same pressure difference.
Reynolds number Re = ρv_mean D/η compares inertial and viscous effects for pipe flow with diameter D. Small Re favors laminar behavior, while sufficiently large Re permits turbulence and stronger sensitivity to disturbances and roughness. The familiar transition values are approximate, geometry-dependent guides rather than universal constants defining all flows.
Poiseuille’s relation includes viscous mechanical-energy loss, so its pressure drop is not the same phenomenon as the reversible pressure change in an ideal Bernoulli constriction. In a horizontal uniform pipe, steady mean speed can remain constant while pressure falls along the pipe to supply dissipation. That is why speed alone does not determine pressure everywhere.
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07 / UNDERSTAND & APPLY
For a rigid sphere of radius r moving slowly through an effectively unbounded Newtonian fluid with no slip and very small Reynolds number, Stokes drag magnitude is 6πηrv and opposes relative velocity. This is a creeping-flow result, not a universal drag law for every ball, raindrop or high-speed object. Container walls and nearby objects can modify drag.
A denser sphere falling through fluid has downward weight ρ_sVg and upward buoyancy ρ_fVg. At terminal speed, viscous drag balances their difference: 6πηrv_t = (ρ_s − ρ_f)(4πr³/3)g. Thus v_t = 2r²(ρ_s − ρ_f)g/(9η). The radius-squared dependence holds only while the small-Re assumptions remain valid.
If the sphere is less dense than the fluid, it rises rather than falls, and the magnitude uses the density-difference magnitude while direction follows the sign. Terminal speed means acceleration has become zero, not that forces have disappeared. Before reaching terminal motion, net force is nonzero and the speed evolves.
A simple particle-inertia model neglecting added-mass and history forces gives an exponential approach to terminal speed with characteristic time m/(6πηr). This transient shortcut is more restrictive than the steady Stokes terminal balance, so it should not be used as an exact full fluid-dynamics solution without its extra assumptions.
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ORIGINAL WORKED EXAMPLE
A sphere of radius 0.10 mm and density 2500 kg m⁻³ falls through a liquid of density 1000 kg m⁻³ and viscosity 0.10 Pa s. Use g = 10 m s⁻². Estimate terminal speed and the diameter-based Reynolds number.
Answer Approximately 3.3 × 10⁻⁴ m s⁻¹ downward; Re approximately 6.7 × 10⁻⁴.
08 / UNDERSTAND & APPLY
Count interfaces before choosing a pressure or film-force formula. For viscous flow, name the fluid model and check whether the assumed velocity gradient, laminar profile or low-Re drag law applies.
ORIGINAL PRACTICE · 1
N m⁻¹ and J m⁻².
Force per boundary length equals reversible energy change per interface area.
ORIGINAL PRACTICE · 2
2γL.
Each interface contributes γL.
ORIGINAL PRACTICE · 3
It halves.
Δp = 2γ/r.
ORIGINAL PRACTICE · 4
2γ/r for its single interface.
A thin soap bubble in air has two interfaces and gives approximately 4γ/r.
ORIGINAL PRACTICE · 5
Three times larger.
R³ = 27r³.
ORIGINAL PRACTICE · 6
Zero rise or depression.
cos90° = 0, so the vertical surface-tension contribution vanishes in this model.
ORIGINAL PRACTICE · 7
It triples.
h ∝ 1/r.
ORIGINAL PRACTICE · 8
Negative, indicating depression.
cosθ is negative when measured through the liquid.
ORIGINAL PRACTICE · 9
Pa s.
Shear stress divided by velocity gradient has units Pa/(s⁻¹).
ORIGINAL PRACTICE · 10
It halves.
F = ηAU/h.
ORIGINAL PRACTICE · 11
The flow rate increases sixteenfold.
Q ∝ R⁴.
ORIGINAL PRACTICE · 12
2.
Integrating the parabolic profile gives mean speed one half of its maximum.
ORIGINAL PRACTICE · 13
The speed increases fourfold.
v_t ∝ r² at fixed densities, viscosity and gravity.
ORIGINAL PRACTICE · 14
No.
Their vector sum is zero; the individual forces remain nonzero.
09 / UNDERSTAND & APPLY
Surface tension acts along an interface and is linked to its area and curvature. Viscosity describes stresses produced by relative fluid motion and velocity gradients. A liquid can exhibit both simultaneously, but their units, equations and model assumptions differ. A meniscus pressure jump is not a viscous pressure loss.
Check radii against diameters, single surfaces against double films, and contact-angle signs before doing arithmetic. For a flow prediction, verify whether the geometry supports a Couette, Poiseuille or Stokes approximation. If changing size makes Reynolds number large, the original scaling law may no longer apply.
KEEP THE BIG IDEAS
READ WITH CONTEXT
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