FLUID PROPERTIES Chapter 2 CE 319 F Elementary

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FLUID PROPERTIES Chapter 2 CE 319 F: Elementary Mechanics of Fluids 1

FLUID PROPERTIES Chapter 2 CE 319 F: Elementary Mechanics of Fluids 1

Fluid Properties • Define “characteristics” of a specific fluid • Properties expressed by basic

Fluid Properties • Define “characteristics” of a specific fluid • Properties expressed by basic “dimensions” – length, mass (or force), time, temperature • Dimensions quantified by basic “units” We will consider systems of units, important fluid properties (not all), and the dimensions associated with those properties. 2

Systeme International (SI) • • Length = meters (m) Mass = kilograms (kg) Time

Systeme International (SI) • • Length = meters (m) Mass = kilograms (kg) Time = second (s) Force = Newton (N) – Force required to accelerate 1 kg @ 1 m/s 2 – Acceleration due to gravity (g) = 9. 81 m/s 2 – Weight of 1 kg at earth’s surface = W = mg = 1 kg (9. 81 m/s 2) = 9. 81 kg-m/s 2 = 9. 81 N • Temperature = Kelvin (o. K) – 273. 15 o. K = freezing point of water – o. K = 273. 15 + o. C 3

Système International (SI) • Work and energy = Joule (J) J = N*m =

Système International (SI) • Work and energy = Joule (J) J = N*m = kg-m/s 2 * m = kg-m 2/s 2 • Power = watt (W) = J/s • SI prefixes: G = giga = 109 c = centi = 10 -2 M = mega = 106 m = milli = 10 -3 k = kilo = 103 m = micro = 10 -6 4

English (American) System • • Length = foot (ft) = 0. 3048 m Mass

English (American) System • • Length = foot (ft) = 0. 3048 m Mass = slug or lbm (1 slug = 32. 2 lbm = 14. 59 kg) Time = second (s) Force = pound-force (lbf) – Force required to accelerate 1 slug @ 1 ft/s 2 • Temperature = (o. F or o. R) – o. Rankine = o. R = 460 + o. F • Work or energy = ft-lbf • Power = ft-lbf/s Banana Slug Mascot of UC Santa Cruz – 1 horsepower = 1 hp = 550 ft-lbf/s = 746 W 5

Density • Mass per unit volume (e. g. , @ 20 o. C, 1

Density • Mass per unit volume (e. g. , @ 20 o. C, 1 atm) – Water – Mercury – Air rwater = 1, 000 kg/m 3 (62. 4 lbm/ft 3) r. Hg = 13, 500 kg/m 3 rair = 1. 205 kg/m 3 • Densities of gases = strong f (T, p) = compressible • Densities of liquids are nearly constant (incompressible) for constant temperature • Specific volume = 1/density = volume/mass 6

Example: Textbook Problem 2. 8 • Estimate the mass of 1 mi 3 of

Example: Textbook Problem 2. 8 • Estimate the mass of 1 mi 3 of air in slugs and kgs. Assume rair = 0. 00237 slugs/ft 3, the value at sea level for standard conditions 7

Example • A 5 -L bottle of carbon tetrachloride is accidentally spilled onto a

Example • A 5 -L bottle of carbon tetrachloride is accidentally spilled onto a laboratory floor. What is the mass of carbon tetrachloride that was spilled in lbm? 8

Specific Weight • Weight per unit volume (e. g. , @ 20 o. C,

Specific Weight • Weight per unit volume (e. g. , @ 20 o. C, 1 atm) gwater = (998 kg/m 3)(9. 807 m 2/s) = 9, 790 N/m 3 [= 62. 4 lbf/ft 3] gair = (1. 205 kg/m 3)(9. 807 m 2/s) = 11. 8 N/m 3 [= 0. 0752 lbf/ft 3] 9

Specific Gravity • Ratio of fluid density to density of water @ 4 o.

Specific Gravity • Ratio of fluid density to density of water @ 4 o. C Water Mercury SGwater = 1 SGHg = 13. 55 Note: SG is dimensionless and independent of system of units 10

Example • The specific gravity of a fresh gasoline is 0. 80. If the

Example • The specific gravity of a fresh gasoline is 0. 80. If the gasoline fills an 8 m 3 tank on a transport truck, what is the weight of the gasoline in the tank? 11

Ideal Gas Law (equation of state) P = absolute (actual) pressure (Pa = N/m

Ideal Gas Law (equation of state) P = absolute (actual) pressure (Pa = N/m 2) V = volume (m 3) n = # moles Ru = universal gas constant = 8. 31 J/o. K-mol T = temperature (o. K) R = gas-specific constant R(air) = 287 J/kg-o. K (show) 12

Example • Calculate the volume occupied by 1 mol of any ideal gas at

Example • Calculate the volume occupied by 1 mol of any ideal gas at a pressure of 1 atm (101, 000 Pa) and temperature of 20 o. C. 13

Example • The molecular weight of air is approximately 29 g/mol. Use this information

Example • The molecular weight of air is approximately 29 g/mol. Use this information to calculate the density of air near the earth’s surface (pressure = 1 atm = 101, 000 Pa) at 20 o. C. 14

Example: Textbook Problem 2. 4 • Given: Natural gas stored in a spherical tank

Example: Textbook Problem 2. 4 • Given: Natural gas stored in a spherical tank – Time 1: T 1=10 o. C, p 1=100 k. Pa – Time 2: T 2=10 o. C, p 2=200 k. Pa • Find: Ratio of mass at time 2 to that at time 1 • Note: Ideal gas law (p is absolute pressure) 15

Viscosity 16

Viscosity 16

Some Simple Flows • Flow between a fixed and a moving plate Fluid in

Some Simple Flows • Flow between a fixed and a moving plate Fluid in contact with plate has same velocity as plate (no slip condition) u = x-direction component of velocity y Moving plate u=V V B Fluid Fixed plate x 17 u=0

Some Simple Flows • Flow through a long, straight pipe Fluid in contact with

Some Simple Flows • Flow through a long, straight pipe Fluid in contact with pipe wall has same velocity as wall (no slip condition) u = x-direction component of velocity R r x V Fluid 18

Fluid Deformation • Flow between a fixed and a moving plate • Force causes

Fluid Deformation • Flow between a fixed and a moving plate • Force causes plate to move with velocity V and the fluid deforms continuously. y Moving plate t 0 u=V t 1 t 2 Fluid Fixed plate x u=0 19

Fluid Deformation For viscous fluid, shear stress is proportional to deformation rate of the

Fluid Deformation For viscous fluid, shear stress is proportional to deformation rate of the fluid (rate of strain) y dy t d. L Moving plate da t+dt dx u=V+d. V Fluid Fixed plate x u=V 20

Viscosity • Proportionality constant = dynamic (absolute) viscosity • Newton’s Law of Viscosity •

Viscosity • Proportionality constant = dynamic (absolute) viscosity • Newton’s Law of Viscosity • V+d v Viscosity V • Units • Water (@ 20 o. C): • Air (@ 20 o. C): m = 1. 8 x 10 -5 N-s/m 2 • Kinematic viscosity m = 1 x 10 -3 N-s/m 2 Kinematic viscosity: m 2/s 1 poise = 0. 1 N-s/m 2 1 centipoise = 10 -2 poise = 10 -3 N-s/m 2 21

Shear in Different Fluids • Shear-stress relations for different fluids • Newtonian fluids: linear

Shear in Different Fluids • Shear-stress relations for different fluids • Newtonian fluids: linear relationship • Slope of line = coefficient of proportionality) = “viscosity” Shear thinning fluids (ex): toothpaste, architectural coatings; Shear thickening fluids = water w/ a lot of particles, e. g. , sewage sludge; Bingham fluid = like solid at small shear, then liquid at greater shear, e. g. , flexible plastics 22

Effect of Temperature Gases: greater T = greater interaction between molecules = greater viscosity.

Effect of Temperature Gases: greater T = greater interaction between molecules = greater viscosity. Liquids: greater T = lower cohesive forces between molecules = viscosity down. 23

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Typical Viscosity Equations T = Kelvin Gas: S = Sutherland’s constant Air = 111

Typical Viscosity Equations T = Kelvin Gas: S = Sutherland’s constant Air = 111 o. K +/- 2% for T = 170 – 1900 o. K Liquid: C and b = empirical constants 25

Flow between 2 plates Force is same on top and bottom Thus, slope of

Flow between 2 plates Force is same on top and bottom Thus, slope of velocity profile is constant and velocity profile is a st. line y Moving plate u=V V B Fluid Fixed plate Force acting ON the plate x u=026

Flow between 2 plates Shear stress anywhere between plates y Moving plate V u=V

Flow between 2 plates Shear stress anywhere between plates y Moving plate V u=V t B t Fixed plate Shear on fluid x 27 u=0

Flow between 2 plates • 2 different coordinate systems B r x y V

Flow between 2 plates • 2 different coordinate systems B r x y V x 28

Example: Textbook Problem 2. 33 Suppose that glycerin is flowing (T = 20 o.

Example: Textbook Problem 2. 33 Suppose that glycerin is flowing (T = 20 o. C) and that the pressure gradient dp/dx = -1. 6 k. N/m 3. What are the velocity and shear stress at a distance of 12 mm from the wall if the space B between the walls is 5. 0 cm? What are the shear stress and velocity at the wall? The velocity distribution for viscous flow between stationary plates is 29

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Example: Textbook Problem 2. 34 A laminar flow occurs between two horizontal parallel plates

Example: Textbook Problem 2. 34 A laminar flow occurs between two horizontal parallel plates under a pressure gradient dp/ds (p decreases in the positive s direction). The upper plate moves left (negative) at velocity ut. The expression for local velocity is shown below. Is the magnitude of the shear stress greater at the moving plate (y = H) of at the stationary plate (y = 0)? 31

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Elasticity (Compressibility) • If pressure acting on mass of fluid increases: fluid contracts •

Elasticity (Compressibility) • If pressure acting on mass of fluid increases: fluid contracts • If pressure acting on mass of fluid decreases: fluid expands • Elasticity relates to amount of deformation for a given change in pressure Ev = bulk modulus of elasticity How does second part of equation come about? Small d. V/V = large modulus of elasticity 33

Example: Textbook Problem 2. 45 • Given: Pressure of 2 MPa is applied to

Example: Textbook Problem 2. 45 • Given: Pressure of 2 MPa is applied to a mass of water that initially filled 1000 -cm 3 (1 liter) volume. • • Find: Volume after the pressure is applied. • Ev = 2. 2 x 109 Pa (Table A. 5) 34

Example • Based on the definition of Ev and the equation of state, derive

Example • Based on the definition of Ev and the equation of state, derive an equation for the modulus of elasticity of an ideal gas. 35

Surface Tension • Below surface, forces act equal in all directions • At surface,

Surface Tension • Below surface, forces act equal in all directions • At surface, some forces are missing, pulls molecules down and together, like membrane exerting tension on the surface • Pressure increase is balanced by surface tension, s • surface tension = magnitude of tension/length Interface water air Net force inward No net force • s = 0. 073 N/m (water @ 20 o. C) 36

Surface Tension • Liquids have cohesion and adhesion, both involving molecular interactions – Cohesion:

Surface Tension • Liquids have cohesion and adhesion, both involving molecular interactions – Cohesion: enables liquid to resist tensile stress – Adhesion: enables liquid to adhere to other bodies • Capillarity = property of exerting forces on fluids by fine tubes or porous media – – – due to cohesion and adhesion If adhesion > cohesion, liquid wets solid surfaces at rises If adhesion < cohesion, liquid surface depresses at pt of contact water rises in glass tube (angle = 0 o) mercury depresses in glass tube (angle = 130 -140 o) • See attached information 37

Example: Capillary Rise • Given: Water @ 20 o. C, d = 1. 6

Example: Capillary Rise • Given: Water @ 20 o. C, d = 1. 6 mm • Find: Height of water W 38

Example: Textbook Problem 2. 51 Find: Maximum capillary rise between two vertical glass plates

Example: Textbook Problem 2. 51 Find: Maximum capillary rise between two vertical glass plates 1 mm apart. q s s h t 39

Examples of Surface Tension 40

Examples of Surface Tension 40

Example: Textbook Problem 2. 48 Given: Spherical soap bubble, inside radius r, film thickness

Example: Textbook Problem 2. 48 Given: Spherical soap bubble, inside radius r, film thickness t, and surface tension s. Find: Formula for pressure in the bubble relative to that outside. Pressure for a bubble with a 4 -mm radius? Should be soap bubble 41

Vapor Pressure (Pvp) • Vapor pressure of a pure liquid = equilibrium partial pressure

Vapor Pressure (Pvp) • Vapor pressure of a pure liquid = equilibrium partial pressure of the gas molecules of that species above a flat surface of the pure liquid – Concept on board – Very strong function of temperature (Pvp up as T up) – Very important parameter of liquids (highly variable – see attached page) • When vapor pressure exceeds total air pressure applied at surface, the liquid will boil. • Pressure at which a liquid will boil for a given temperature – At 10 o. C, vapor pressure of water = 0. 012 atm = 1200 Pa – If reduce pressure to this value can get boiling of water (can lead to “cavitation”) • If Pvp > 1 atm compound = gas • If Pvp < 1 atm compound = liquid or solid 42

Example • The vapor pressure of naphthalene at 25 o. C is 10. 6

Example • The vapor pressure of naphthalene at 25 o. C is 10. 6 Pa. What is the corresponding mass concentration of naphthalene in mg/m 3? (Hint: you can treat naphthalene vapor as an ideal gas). 43

Vapor Pressure (Pvp) - continued Vapor pressure of water (and other liquids) is a

Vapor Pressure (Pvp) - continued Vapor pressure of water (and other liquids) is a strong function of temperature. 44

Vapor Pressure (Pvp) - continued Pvp, H 2 O = Pexp(13. 3185 a –

Vapor Pressure (Pvp) - continued Pvp, H 2 O = Pexp(13. 3185 a – 1. 9760 a 2 – 0. 6445 a 3 – 0. 1299 a 4) P = 101, 325 Pa a = 1 – (373. 15/T) T = o. K valid to +/- 0. 1% accuracy for T in range of -50 to 140 o. C Equation for relative humidity of air = percentage to which air is “saturated” with water vapor. What is affect of RH on drying of building materials, and why? Implications? 45

Example: Relative Humidity The relative humidity of air in a room is 80% at

Example: Relative Humidity The relative humidity of air in a room is 80% at 25 o. C. (a) What is the concentration of water vapor in air on a volume percent basis? (b) If the air contacts a cold surface, water may condense (see effects on attached page). What temperature is required to cause water condensation? 46

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Saturation Vapor Pressure 48

Saturation Vapor Pressure 48

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