Mechanical Properties of Fluids · Lesson 8 of 8

Chapter Summary and Practice

“Connect the full chapter through model choice, formula conditions, mixed worked examples and cumulative practice.”

Learning Objectives

• Select the correct static, ideal-flow, viscous or interfacial model for a fluid problem. • Connect mass, momentum and energy reasoning without mixing their roles. • Use the chapter’s formula set with correct units, radii, pressure references and assumptions. • Solve mixed problems and diagnose common errors across the entire chapter.

One chapter, four different physical questions

The chapter describes fluids at rest, fluids in organised motion, real-fluid resistance and liquid interfaces. These share quantities such as pressure and density but require different reasoning. A submerged window is a static force problem. A constricted ideal stream uses mass and energy conservation. A viscous tube needs a pressure drop even at constant mean speed. A capillary meniscus combines an interfacial pressure jump with hydrostatic balance. Choosing the model is the first part of solving the question.

Question	Main reasoning	Check before calculating
Pressure at depth or across a window	Hydrostatic force balance	Same connected fluid; density approximately constant; pressure on both sides
Hydraulic force and motion	Pressure transmission plus volume conservation	Piston areas, elevations, outside pressures and losses
Speed through a constriction	Continuity: conservation of mass	Constant density? Branches? Mean versus local speed?
Ideal pressure, speed and height relation	Bernoulli: mechanical energy	Same streamline, steady flow, negligible losses, no intervening pump
Viscous flow or slow sphere motion	Shear-rate relation; pipe law; Stokes drag	Newtonian model, geometry, no slip, flow regime and wall effects
Film force, drop pressure or capillary level	Interfacial energy and force balance	Number of interfaces, contact length, radius, contact angle and outside pressure

Formula reference with physical conditions

Static fluid and pressure referencesLaTeX

P = \frac{F _{⊥}}{A}, ρ = \frac{m}{V}, P = P_{0} + ρ g h, P_{g} = P_{abs} - P_{atm}

F/A is an average or uniform pressure. The depth formula assumes constant density and g. Pressure is in Pa, density in kg/m³, height in m.

Hydraulic force and displacementLaTeX

\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}, A_{1} x_{1} = A_{2} x_{2}, F_{1} x_{1} = F_{2} x_{2}

Ideal equal-level pistons under a common atmosphere; neglect piston weights, friction and leaks. Area ratios are squared diameter ratios.

Mass continuity and ideal energyLaTeX

ρ A \overset{v}{ˉ} = constant, Q = A \overset{v}{ˉ} (ρ constant), P + \frac{1}{2} ρ v^{2} + ρ g y = C

Continuity is mass conservation. Bernoulli applies along a streamline under the ideal steady incompressible assumptions; all its terms have unit Pa = J/m³.

Efflux and simplified liftLaTeX

v_{o} ≃ 2 g h + \frac{2 ( P _{s} - P _{atm} )}{ρ}, L ≃ A \frac{ρ}{2} (v_{upper}^{2} - v_{lower}^{2})

Efflux assumes a large tank and small outlet. The lift estimate uses comparable heights and an effective pressure difference over wing area, not an equal-transit-time rule.

Viscosity and laminar pipe flowLaTeX

τ = η \frac{d v}{d y}, Q = \frac{π R ^{4} Δ P}{8 η L}, Re = \frac{ρ v ˉ D}{η}

The shear-gradient expression gives magnitude with a suitable sign convention. η has unit Pa·s. The pipe formula requires fully developed Newtonian laminar flow in a straight circular tube.

Slow sphere drag and settlingLaTeX

F_{d} = 6 π η a v, v_{t} = \frac{2 a ^{2} ( ρ _{s} - ρ _{f} ) g}{9 η}

a is sphere radius. Use the Stokes creeping-flow model, negligible wall effects and the density difference, including buoyancy. Downward settling requires ρₛ > ρ_f.

Interfacial force and pressureLaTeX

Δ E = S Δ A, F_{film} = 2 S l, Δ P_{one interface} = \frac{2 S}{r}, Δ P_{soap} ≃ \frac{4 S}{r}

S has unit N/m = J/m². ΔA counts every new face; r is curvature radius. Add outside pressure to an excess when an absolute pressure is requested.

Wetting and capillary heightLaTeX

cos θ = \frac{S _{s a} - S _{s l}}{S _{l a}}, h = \frac{2 S cos θ}{ρ g a}

Measure θ through liquid; a is tube radius. h is signed relative to reservoir level. Narrow circular tube and negligible meniscus-volume correction are assumed.

Quantity	SI unit	Dimensions	Interpretation
Pressure P	Pa = N/m²	M L⁻¹ T⁻²	Scalar normal stress in a static fluid
Density ρ	kg/m³	M L⁻³	Mass per volume
Relative density	1	Dimensionless	Density divided by reference density
Volume flow Q	m³/s	L³ T⁻¹	Volume crossing per time
Dynamic viscosity η	Pa·s	M L⁻¹ T⁻¹	Shear stress divided by strain rate
Surface tension S	N/m = J/m²	M T⁻²	Interfacial force/length or energy/area
Reynolds number	1	Dimensionless	Inertial/viscous comparison for chosen size

Similar words do not imply the same quantity

Pressure, force and surface tension are different. Pressure multiplied by an appropriate area gives a normal force; surface tension multiplied by a contact length gives a tangential interfacial force. Viscosity describes resistance to relative motion and is not a static surface force. All three can appear together in a real situation, but their units and mechanisms remain distinct. At rest, ordinary fluid shear stress vanishes; pressure remains and varies with depth.

Likewise distinguish steady from uniform. A steady velocity field stays the same with time at each point, but particles may accelerate through spatial changes. A laminar viscous pipe has different speeds across its section and dissipates energy despite being orderly. Two streamlines cannot give two different velocity directions at a single ordinary point. A rapid’s visible turbulence makes the elementary steady ideal-flow energy formula an inadequate complete model.

Surface tension is approximately independent of the interface’s total area for unchanged conditions; total surface energy is not. A soap bubble has two interfaces, while a gas cavity in bulk liquid has one. A wetting capillary’s liquid pressure just below its raised meniscus is below atmospheric, not above. Use the radius specified by the geometry: sphere radius in Stokes drag, pipe radius to the fourth power in Poiseuille flow and tube radius in capillary rise.

A practical sequence for solving a mixed problem

Start by drawing the fluid regions and labelling their interfaces and connected paths. Choose one height datum and state the pressure reference. Convert centimetres, millimetres and squared areas before substituting. Identify whether a mass flow rate or a volume flow rate is given. Write the conservation or equilibrium relation before inserting numbers. Finally check units, direction, limiting behaviour and model assumptions. A correct-looking numerical answer cannot rescue a model whose assumptions contradict the physical setup.

Hydraulic force and energy in one calculation

Problem
An ideal lift has input area 2.0 cm² and output area 80 cm². It must raise an 800 N load by 2.0 cm. Find the input force, hydraulic gauge pressure, input movement and work.

1.The area ratio is 80/2 = 40, so input force is 800/40 = 20 N.
2.Convert input area to 2.0 × 10⁻⁴ m². Gauge pressure is 20/(2.0 × 10⁻⁴) = 1.0 × 10⁵ Pa.
3.Volume conservation gives input movement = 40 × 2.0 cm = 80 cm = 0.80 m.
4.Input work is 20 × 0.80 = 16 J. Output work is 800 × 0.020 = 16 J. The large output force is fully consistent with energy conservation.

A raised meniscus and a submerged bubble

Problem
For water with S = 0.072 N/m, ρ = 1000 kg/m³ and g = 10 m/s², compare (a) the capillary height in a completely wetting tube of radius 0.50 mm and (b) the pressure in a hemispherical air bubble at the end of the same-radius tube 10 cm underwater. Use P_atm = 1.00 × 10⁵ Pa.

1.For the wetting capillary, h = 2S/(ρga) = 0.144/(1000 × 10 × 0.00050) = 0.0288 m = 2.88 cm. Liquid just below the meniscus is below atmosphere by ρgh = 288 Pa.
2.For the submerged hemisphere, outside water is above atmosphere by ρgd = 1000 Pa. The gas bubble adds a one-interface excess 2S/r = 288 Pa, with r = a for this hemisphere.
3.Inside bubble pressure is 100000 + 1000 + 288 = 101288 Pa. Its gauge pressure is 1288 Pa.
4.The same numerical curvature term appears as a liquid pressure deficit under a wetting meniscus and as a gas pressure excess in a submerged cavity. The phases and the sign convention determine which side has the larger pressure.

A constriction versus a viscous pipe

Problem
Ideal water flow has mean speeds 1 m/s and 3 m/s at wide and narrow sections of a horizontal pipe. What pressure difference follows? Why does a constant-area viscous tube require a different treatment? Use water density 1000 kg/m³.

1.For the ideal two-section comparison, continuity requires the wide area to be three times the narrow area. Bernoulli gives P_wide − P_narrow = 1000(3² − 1²)/2 = 4000 Pa.
2.In a constant-area steady incompressible tube the mean speeds at both ends are equal, but viscous dissipation can still require positive inlet-to-outlet pressure drop. The ideal equal-speed Bernoulli result alone would miss that loss.
3.For the specified fully developed laminar circular-tube model, use ΔP = 8ηLQ/(πR⁴). A pressure difference may supply acceleration, gravitational energy change or dissipative losses; identify which is present.

What the source exercises collectively test

The conceptual exercises require explanations of blood-pressure height effects, the nonuniform atmosphere, scalar pressure, contact angle, wetting, surface-energy minimisation, temperature trends, steadiness and the roles of mass, energy and momentum conservation. Numerical questions require area conversion, barometer height, structural pressure estimates, hydraulic pressure, layered-liquid balances, viscous pipe flow, lift, multiple outlets, film force and curved-interface pressure. Each of these methods has been developed in the earlier lessons; this summary connects them rather than replacing those derivations.

Quick Quiz

Quick check

Which pairing is correct?

Explanation

Continuity relates mass flow. Bernoulli relates pressure work to kinetic and potential energy under ideal-flow assumptions.

Quick check

Which calculation directly requires including buoyancy?

Explanation

Terminal force balance contains weight, displaced-fluid buoyancy and viscous drag.

Quick check

Which statement about a steady laminar viscous pipe is correct?

Explanation

Laminar describes orderly motion; shear between layers still dissipates mechanical energy.

Quick check

A raised wetting meniscus and a submerged gas bubble are compared. Which is correct?

Explanation

Curvature and the phases on the two sides determine the pressure jump. Hydrostatic depth must also be included for the submerged bubble.

Quick check

A soap-film slider length doubles while S and the two-face geometry stay unchanged. What happens to its supporting force?

Explanation

F = 2Sl is proportional to slider contact length, not to total film area.

Practice Problems

A water tank is open at P_atm = 1.01 × 10⁵ Pa. A small horizontal window of area 50 cm² is 4.0 m below the free surface, with atmosphere on its dry side. Use ρ = 1000 kg/m³ and g = 10 m/s². Find absolute pressure and net force on the window. Then find the ideal efflux speed if it is replaced by a small hole at the same depth.
An ideal hydraulic system has piston diameters 2.0 cm and 10.0 cm. A 60 N input force moves the small piston inward by 12 cm. Find output force, output displacement and input/output work; state the pressure-reference and elevation assumptions.
Water enters a horizontal ideal pipe of area 10 cm² at 1.0 m/s and leaves a section of area 2.5 cm². Find volume flow rate and downstream speed. If upstream gauge pressure is 20 kPa, find downstream gauge pressure using ρ = 1000 kg/m³. Explain why the pressure reference cancels.
A sphere of radius 0.20 mm and density 3000 kg/m³ falls through liquid of density 1000 kg/m³ and viscosity 0.20 Pa·s. Using g = 10 m/s², calculate Stokes terminal speed and Re. Then explain how the speed changes if the sphere radius doubles, provided the model remains valid.
A soap bubble in air and a gas cavity in liquid both have radius 4.0 mm and surface tension 0.032 N/m. Find each excess pressure. If the gas cavity is 25 cm below a liquid surface with ρ = 1000 kg/m³, g = 10 m/s² and P_atm = 1.00 × 10⁵ Pa, find its absolute pressure.
A liquid has S = 0.040 N/m, density 800 kg/m³ and contact angle 60° in a tube of radius 0.20 mm. Use g = 10 m/s² to find its capillary rise and meniscus pressure deficit. Explain how the sign changes for contact angle 120° without changing the other data.
A Newtonian liquid of viscosity 0.10 Pa·s and density 1000 kg/m³ flows at 1.0 × 10⁻⁶ m³/s through a horizontal circular tube of radius 3.0 mm and length 0.80 m. Compute its pressure drop and Reynolds number. Explain why an ideal equal-speed Bernoulli comparison does not account for this pressure loss.
Give a reasoned explanation for each observation: a tall low-density barometer column, unequal mercury levels after adding different liquid columns, a spinning ball’s deflected path, the strong effect of needle bore on flow, and equal slider forces in differently shaped two-sided films with equal slider lengths.

Key Takeaways

• Choose the physical model before selecting a formula; static pressure, ideal flow, viscosity and interfaces have different roles. • Track units, pressure references, height datums, area ratios and the correct geometric radius. • Mass, energy and momentum conservation complement one another; no hydraulic device or jet creates energy. • Check assumptions and limiting cases, and keep excess pressure separate from large rounded absolute pressures.

Previous · Lesson 7

Drops, Bubbles and Capillary Action

End of chapter