Mechanical Properties of Fluids

Introduction

This chapter explores the physical properties of liquids and gases, collectively known as fluids.
Fluids can flow, distinguishing them from solids.
Fluids are essential for life and mediate biological processes.

Fluids vs. Solids

Shape: Fluids lack a definite shape, unlike solids.
Volume: Solids and liquids have fixed volumes under atmospheric pressure, while gases expand to fill their container.
Compressibility: Solids and liquids have low compressibility compared to gases; their volume changes little under pressure.
Shear Stress: Fluids offer minimal resistance to shear stress, deforming easily, unlike solids.

Pressure

Pressure is the impact of a force on a given area.
Fluids at rest exert a normal force on submerged objects, perpendicular to the surface.
If a force had a component parallel to the surface, it would cause the fluid to flow, which contradicts the state of rest.
Average pressure ( $P{av}$ ) is defined as the normal force (F) per unit area (A): $P{av} = \frac{F}{A}$ (9.1)
Pressure at a point is the limiting value as the area approaches zero:
$P = \lim_{\Delta A \to 0} \frac{\Delta F}{\Delta A}$ (9.2)
Pressure is a scalar quantity with dimensions $[ML^{-1}T^{-2}]$ .
The SI unit of pressure is the pascal (Pa), equivalent to $N/m^2$ .
Atmosphere (atm) is a common unit: 1 atm = $1.013 \times 10^5$ Pa.

Density

Density ($\rho $) is mass (m) per unit volume (V): $ \rho = \frac{m}{V} $(9.3)</li><li>Density has dimensions$ [ML^{-3}] $and is measured in kg/$ m^3 $.</li><li>Density is a positive scalar quantity.</li><li>Liquids are nearly incompressible, maintaining constant density under pressure.</li><li>Gases exhibit significant density variation with pressure.</li><li>Water's density at 4°C (277 K) is$ 1.0 \times 10^3 $kg/$ m^3 $.</li><li>Relative density is the ratio of a substance's density to water's density at 4°C.<ul><li>It is dimensionless; for example, aluminum's relative density is 2.7, meaning its density is$ 2.7 \times 10^3 $kg/$ m^3 $.</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Example 9.1</h3><ul><li>Estimating pressure on thigh bones (femurs) supporting the upper body.</li><li>Given: Total mass = 40 kg, Area = 2 x 10$ cm^2 $= 20 x$ 10^{-4} m^2 $,$ g = 10 m/s^2 $.</li><li>Force acting on the femurs:$ F = 40 kg \times 10 m/s^2 = 400 N $.</li><li>$ P_{av} = \frac{F}{A} = \frac{400 N}{20 \times 10^{-4} m^2} = 2 \times 10^5 N/m^2 $</li></ul><h3 collapsed="false" seolevelmigrated="true">Pascal’s Law</h3><ul><li>Pascal's observation: Pressure in a fluid at rest is uniform at the same height.</li><li>Demonstration: A right-angled prism element ABC-DEF within a fluid at rest.</li><li>Pressures$ Pa $,$ Pb $, and$ Pc $exert normal forces$ Fa $,$ Fb $, and$ Fc $on faces BEFC, ADFC, and ADEB, with areas$ Aa $,$ Ab $, and$ A_c $, respectively.</li><li>Equilibrium conditions: $ Fb \sin{\theta} = Fc $ $ Fb \cos{\theta} = Fa $ $ Ab \sin{\theta} = Ac $ $ Ab \cos{\theta} = Aa $</li><li>Pressure equality: $ \frac{Fa}{Aa} = \frac{Fb}{Ab} = \frac{Fc}{Ac} \implies Pa = Pb = P_c $(9.4)</li><li>Pressure is not a vector; it has no direction.</li><li>In equilibrium, pressure is uniform in a horizontal plane; otherwise, flow would occur.</li></ul><h3 collapsed="false" seolevelmigrated="true">Variation of Pressure with Depth</h3><ul><li>Consider a fluid at rest with point 1 at height h above point 2; pressures are$ P1 $and$ P2 $, respectively.</li><li>For a cylindrical element of base area A and height h: $ (P2 - P1)A = mg $(9.5)</li><li>With fluid density$ \rho $, mass$ m = \rho V = \rho hA $: $ P2 - P1 = \rho gh $(9.6)</li><li>If point 1 is at the fluid surface open to the atmosphere (pressure$ Pa $), then:$ P = Pa + \rho gh $(9.7)</li><li>Gauge pressure is the excess pressure at depth h:$ P - P_a $.</li><li>Pressure depends on fluid column height, not container shape, illustrating hydrostatic paradox.</li></ul><h3 collapsed="false" seolevelmigrated="true">Hydrostatic Paradox</h3><ul><li>Vessels of different shapes connected at the bottom have the same water level because pressure is uniform at a given depth.</li></ul><h3 collapsed="false" seolevelmigrated="true">Example 9.2</h3><ul><li>Pressure on a swimmer 10 m below a lake's surface.</li><li>Given:$ h = 10 m $,$ \rho = 1000 kg/m^3 $,$ g = 10 m/s^2 $.</li><li>$ P = P_a + \rho gh = 1.01 \times 10^5 Pa + (1000 kg/m^3)(10 m/s^2)(10 m) = 2.01 \times 10^5 Pa \approx 2 atm $</li><li>Pressure increases by 100% at 10 m depth; at 1 km, it increases by 100 atm.</li></ul><h3 collapsed="false" seolevelmigrated="true">Atmospheric Pressure and Gauge Pressure</h3><ul><li>Atmospheric pressure is the weight of an air column per unit area.</li><li>At sea level:$ 1.013 \times 10^5 Pa $(1 atm).</li><li>Torricelli's mercury barometer measures atmospheric pressure using the height of a mercury column.</li><li>Pressure at point A (top of mercury column) is approximately zero.<ul><li>Point B inside the column at the mercury surface has the same pressure as atmospheric pressure (Pa) at point C outside.</li><li>$ P_a = \rho gh $(9.8), where$ \rho $is mercury's density and h is the column height.</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Pressure Measurement Units</h3><ul><li>Pressure is often expressed in cm or mm of mercury (Hg).<ul><li>1 mm of Hg is a torr (after Torricelli): 1 torr = 133 Pa.</li></ul></li><li>Meteorological units: 1 bar =$ 10^5 $Pa; 1 millibar = 100 Pa.</li><li>Open tube manometer: Measures pressure differences using liquid height in a U-tube.<ul><li>Gauge pressure ($ P - P_a $) is proportional to manometer height h as given by Eq. (9.8).</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Fluid Properties</h3><ul><li>Liquids are nearly incompressible with constant density.</li><li>Gases exhibit large density variations with pressure and temperature changes.</li></ul><h3 collapsed="false" seolevelmigrated="true">Example 9.3</h3><ul><li>Atmospheric height estimation assuming constant density.</li><li>Given:$ \rho = 1.29 kg/m^3 $,$ P_a = 1.01 \times 10^5 Pa $.</li><li>$ \rho gh = 1.01 \times 10^5 Pa \implies h = \frac{1.01 \times 10^5 Pa}{(1.29 kg/m^3)(9.8 m/s^2)} \approx 8 km $</li><li>Atmospheric density decreases with height; the atmosphere extends beyond 100 km.</li></ul><h3 collapsed="false" seolevelmigrated="true">Example 9.4</h3><ul><li>Pressure at 1000 m ocean depth:<ul><li>Given:$ h = 1000 m $,$ \rho = 1.03 \times 10^3 kg/m^3 $. $ P = P_a + \rho gh $ $ = 1.01 \times 10^5 Pa + (1.03 \times 10^3 kg/m^3)(10 m/s^2)(1000 m) $ $ = 104.01 \times 10^5 Pa \approx 104 atm $</li><li>Gauge pressure:$ P_g = \rho gh = 103 \times 10^5 Pa \approx 103 atm $.</li><li>Force on a submarine window (area$ A = 0.04 m^2 $): $ F = P_g A = (103 \times 10^5 Pa)(0.04 m^2) = 4.12 \times 10^5 N $</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Hydraulic Machines</h3><ul><li>Pascal's law application: Pressure applied to a fluid in a vessel is transmitted equally throughout.</li><li>Hydraulic lift: Force applied on a small piston (area$ A1 $) is transmitted to a larger piston (area$ A2 $).<ul><li>Pressure:$ P = \frac{F1}{A1} $.</li><li>Upward force on larger piston:$ F2 = PA2 = \frac{F1 A2}{A_1} $.</li><li>Mechanical advantage:$ \frac{A2}{A1} $.</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Example 9.5</h3><ul><li>Two syringes connected by a water-filled tube.<ul><li>Smaller piston diameter: 1.0 cm;<ul><li>Radius$ r_1 = 0.5 \times 10^{-2} m $</li></ul></li><li>Larger piston diameter: 3.0 cm<ul><li>Radius$ r_2 = 1.5 \times 10^{-2} m $</li></ul></li></ul></li><li>a) Force on larger piston with 10 N on smaller piston:<ul><li>$ \frac{F1}{A1} = \frac{F2}{A2} $</li><li>$ F2 = F1 \frac{A2}{A1} = 10N \frac{\pi (1.5 \times 10^{-2}m)^2}{\pi(0.5 \times 10^{-2}m)^2} = 90 N $</li></ul></li><li>b) Distance the larger piston moves when the smaller piston is pushed 6.0 cm:<ul><li>$ A1L1 = A2L2 $</li><li>$ L2 = L1 \frac{A1}{A2} = 6.0 cm \frac{\pi (0.5 \times 10^{-2}m)^2}{\pi (1.5 \times 10^{-2}m)^2} = 0.67 cm $</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Example 9.6</h3><ul><li>Car lift with compressed air.<ul><li>Small piston radius$ r_1 = 5.0 cm = 0.05 m $</li><li>Large piston radius$ r_2 = 15 cm = 0.15 m $</li><li>Car mass$ m = 1350 kg $</li></ul></li><li>a) Force$ F1 $on the small piston:$ P1 = P2\frac{F1}{A1} = \frac{mg}{A2} $ $ F1 = mg \frac{A1}{A_2} = 1350 kg \times 9.8 m/s^2 \times \frac{\pi (0.05m)^2}{\pi (0.15m)^2} = 1470 N $</li><li>b) Air pressure required: $ P = \frac{F1}{A1} = \frac{1470 N}{\pi (0.05m)^2} = 1.87 \times 10^5 Pa $</li></ul><h3 collapsed="false" seolevelmigrated="true">Hydraulic Brakes</h3><ul><li>Hydraulic brakes use Pascal's principle to amplify force.<ul><li>Small force on pedal moves master piston.</li><li>Pressure transmitted to larger pistons on wheels.</li><li>Equal braking effort on all wheels.</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Streamline Flow</h3><ul><li>Fluid dynamics studies fluids in motion.</li><li>Steady flow: Fluid velocity at a point remains constant over time.</li><li>Streamlines: Paths of fluid particles; they do not cross in steady flow.</li></ul><h3 collapsed="false" seolevelmigrated="true">Equation of Continuity</h3><ul><li>Mass conservation in incompressible fluids: $ \rhoP AP vP \Delta t = \rhoR AR vR \Delta t = \rhoQ AQ v_Q \Delta t $(9.9)</li><li>For incompressible fluids:$ AP vP = AR vR = AQ vQ $(9.10) $ Av = constant $</li><li>(9.11) (Equation of continuity)<ul><li>Av is the volume flux or flow rate.</li><li>Velocity increases at narrower pipe sections.</li></ul></li><li>Critical speed: Flow becomes turbulent above this limit.</li></ul><h3 collapsed="false" seolevelmigrated="true">Bernoulli’s Principle</h3><ul><li>Relates pressure, velocity, and height in steady flow.</li><li>Bernoulli's equation:$ P1 - P2 = \frac{1}{2} \rho (v2^2 - v1^2) + \rho g (h2 - h1)P1 + \frac{1}{2} \rho v1^2 + \rho g h1 = P2 + \frac{1}{2} \rho v2^2 + \rho g h2 $<ul><li>(9.12) $ P + \frac{1}{2} \rho v^2 + \rho g h = constant $</li><li>(9.13)</li></ul></li><li>Assumptions: Incompressible, non-viscous fluids, steady flow.</li><li>When fluid is at rest ($ v=0 $), Bernoulli’s equation simplifies to$ P1 + \rho g h1 = P2 + \rho g h2 $.</li></ul><h3 collapsed="false" seolevelmigrated="true">Speed of Efflux: Torricelli’s Law</h3><ul><li>Speed of fluid outflow from a tank.</li><li>$ v1 A1 = v2 A2 \implies v2 = v1\frac{A1}{A2} $(Equation of continuity)</li><li>Applying Bernoulli's equation: $ Pa + \frac{1}{2}\rho v1^2 + \rho g y1 = P + \rho g y2 $,$ v1 = \sqrt{\frac{2(P - Pa)}{\rho} + 2gh} $, if$ P >> P_a $.</li><li>If the tank is open to the atmosphere,$ v_1 = \sqrt{2gh} $(Torricelli's law).</li></ul><h3 collapsed="false" seolevelmigrated="true">Dynamic Lift</h3><ul><li>Force on a body due to its motion through a fluid.<ul><li>Spinning ball deviates due to pressure difference (Magnus effect).</li><li>Aerofoil generates lift due to higher flow speed above the wing.</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Example 9.7</h3><ul><li>Boeing aircraft dynamic lift calculation.<ul><li>Mass =$ 3.3 \times 10^5 kg $</li><li>Wing area = 500$ m^2 $</li><li>Speed = 960 km/h = 267 m/s</li></ul></li><li>a) Pressure difference: $ \Delta P \times A = mg $ $ \Delta P = \frac{mg}{A} = \frac{3.3 \times 10^5 kg \times 9.8 m/s^2}{500 m^2} = 6.5 \times 10^3 N/m^2 $</li><li>b) Fractional increase in speed: $ \Delta P = \frac{1}{2} \rho (v2^2 - v1^2) $ $ \frac{v2 - v1}{v{av}} \approx \frac{\Delta P}{\rho v{av}^2} =0.08

Viscosity

Resistance to fluid motion (internal friction).
Coefficient of viscosity ($\eta) defined as the ratio of shearing stress to the strain rate $\eta = \frac{F/A}{v/l}$ .
- SI unit: poiseuille (Pl) or N s/ $m^2$ or Pa s.
- Dimensions: $[ML^{-1}T^{-1}]$ .
Viscosity decreases with temperature in liquids, increases in gases.

Stokes’ Law

Viscous drag force $\vec{F} = -6 \pi \eta a \vec{v}$ .
Terminal velocity: $v_t = \frac{2a^2(\rho - \sigma)g}{9 \eta}$ , derived by setting viscous force plus buoyant force equal to gravitational force.

Example 9.9

Terminal velocity of a copper ball in oil.
Given:
Terminal velocity $v_t = 6.5\times 10^{-2} m/s$
Radius $a = 2\times 10^{-3} m$
Density of oil $\sigma= 1.5\times 10^{3} kg/m^{3}$
Density of copper $\rho = 8.9\times 10^{3} kg/m^{3}$
$\eta = \frac{2a^2(\rho - \sigma)g}{9v_t} = 0.99 Nsm^{-2}$

Surface Tension

Phenomenon related to free surfaces of liquids, where surfaces possess additional energy.
Liquids minimize surface area due to molecular attraction.

Surface Energy and Surface Tension

Surface tension (S) is surface energy per unit area, measured as force per unit length $S = \frac{F}{2l}$ .

Angle of Contact

Angle ($\theta $) between liquid surface tangent and solid surface.<ul><li>Determines liquid spreading or droplet formation.</li><li>Related to interfacial tensions:$ S{la}cos\theta + S{sl} = S_{sa} $.</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Drops and Bubbles</h3><ul><li>Free liquid drops and bubbles are spherical due to surface tension.<ul><li>Pressure inside a spherical liquid drop:$ Pi - Po = \frac{2S_{la}}{r} $.</li><li>Pressure inside a bubble:$ Pi - Po = \frac{4S_{la}}{r} $.</li></ul></li></ul><h3 collapsed="false" seolevelmigrated="true">Capillary Rise</h3><ul><li>Liquid rises in a narrow tube due to surface tension and pressure difference.<ul><li>Capillary rise height:$ h = \frac{2S cos \theta}{ \rho g a}$$.

Example 9.10

Pressure in a capillary tube to form a hemispherical bubble.
Po = (1.01 × 105 Pa + 0.08 m × 1000 kg m–3 × 9.80 m s–2)
= 1.01784 × 105 Pa
Pi = Po + 2S/r
= 1.01784 × 105 Pa+ (2 × 7.3 × 10-2 Pa m/10-3 m)
= (1.01784 + 0.00146) × 105 Pa= 1.02 × 105 Pa