1 FLUID PROPERTIES Chapter 2 CE319F: Elementary Mechanics of Fluids.

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FLUID PROPERTIESFLUID PROPERTIESChapter 2

CE319F: Elementary Mechanics of Fluids

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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.

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Systeme International (SI)

• Length = meters (m)• Mass = kilograms (kg)• Time = second (s)• Force = Newton (N)

– Force required to accelerate 1 kg @ 1 m/s2

– Acceleration due to gravity (g) = 9.81 m/s2

– Weight of 1 kg at earth’s surface = W = mg = 1 kg (9.81 m/s2) = 9.81 kg-m/s2 = 9.81 N

• Temperature = Kelvin (oK)– 273.15 oK = freezing point of water – oK = 273.15 + oC

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Système International (SI)

• Work and energy = Joule (J)J = N*m = kg-m/s2 * m = kg-m2/s2

• Power = watt (W) = J/s

• SI prefixes:G = giga = 109 c = centi = 10-2

M = mega = 106 m = milli = 10-3

k = kilo = 103 = micro = 10-6

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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/s2

• Temperature = (oF or oR)– oRankine = oR = 460 + oF

• Work or energy = ft-lbf• Power = ft-lbf/s

– 1 horsepower = 1 hp = 550 ft-lbf/s = 746 W

Banana Slug

Mascot of UC Santa Cruz

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Density

• Mass per unit volume (e.g., @ 20 oC, 1 atm)– Water water = 1,000 kg/m3 (62.4 lbm/ft3)

– Mercury Hg = 13,500 kg/m3

– Air air = 1.205 kg/m3

• Densities of gases = strong f (T,p) = compressible• Densities of liquids are nearly constant

(incompressible) for constant temperature• Specific volume = 1/density = volume/mass

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Example: Textbook Problem 2.8

• Estimate the mass of 1 mi3 of air in slugs and kgs. Assume air = 0.00237 slugs/ft3, the value at sea level for standard conditions

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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?

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Specific Weight

• Weight per unit volume (e.g., @ 20 oC, 1 atm)

water = (998 kg/m3)(9.807 m2/s)

= 9,790 N/m3

[= 62.4 lbf/ft3]

air = (1.205 kg/m3)(9.807 m2/s)

= 11.8 N/m3

[= 0.0752 lbf/ft3]

]/[]/[ 33 ftlbformNg

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Specific Gravity• Ratio of fluid density to density of water @

4oC

3/1000 mkgSG liquid

water

liquidliquid

Water SGwater = 1

Mercury SGHg = 13.55

Note: SG is dimensionless and independent of system of units

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Example• The specific gravity of a fresh gasoline is 0.80. If the gasoline fills an

8 m3 tank on a transport truck, what is the weight of the gasoline in the tank?

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Ideal Gas Law (equation of state)

TnRPV u

TRV

nP u

RTRTV

nMT

M

R

V

nMP u

P = absolute (actual) pressure (Pa = N/m2)

V = volume (m3)

n = # moles

Ru = universal gas constant = 8.31 J/oK-mol

T = temperature (oK)

R = gas-specific constant

R(air) = 287 J/kg-oK (show)

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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 oC.

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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 oC.

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• Given: Natural gas stored in a spherical tank

– Time 1: T1=10oC, p1=100 kPa

– Time 2: T2=10oC, p2=200 kPa

• Find: Ratio of mass at time 2 to that at time 1

• Note: Ideal gas law (p is absolute pressure)

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ViscosityViscosity

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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

u=VMoving plate

Fixed plate

y

x

V

u=0

B yB

Vyu )( Fluid

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Some Simple Flows

• Flow through a long, straight pipeFluid in contact with pipe wall has same velocity as wall (no slip condition)

u = x-direction component of velocity

rx

R

21)(

R

rVru

VFluid

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Fluid Deformation

• Flow between a fixed and a moving plate

• Force causes plate to move with velocity V and the fluid deforms continuously.

u=VMoving plate

Fixed plate

y

xu=0

Fluid

t0t1 t2

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Fluid Deformation

u=V+VMoving plate

Fixed plate

y

xu=V

Fluid

t t+t

x

y

L

t

For viscous fluid, shear stress is proportional to deformation rate of the fluid (rate of strain)

V

Lt

y

L

y

V

t

y

V

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Viscosity• Proportionality constant = dynamic (absolute) viscosity

• Newton’s Law of Viscosity

• Viscosity

• Units

• Water (@ 20oC): = 1x10-3 N-s/m2

• Air (@ 20oC): = 1.8x10-5 N-s/m2

• Kinematic viscosity

dydV /

2

2

//

/

m

sN

msm

mN

V

V+dv

dy

dV

Kinematic viscosity: m2/s

1 poise = 0.1 N-s/m2

1 centipoise = 10-2 poise = 10-3 N-s/m2

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Shear in Different Fluids

• Shear-stress relations for different fluids

• Newtonian fluids: linear relationship

• Slope of line = coefficient of proportionality) = “viscosity”

dy

dV

dy

dV

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

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Effect of Temperature

Gases:

greater T = greater interaction between molecules = greater viscosity.

Liquids:

greater T = lower cohesive forces between molecules = viscosity down.

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Typical Viscosity Equations

ST

ST

T

T o

o

2

3

Liquid:

Gas:

T = Kelvin

S = Sutherland’s constant

Air = 111 oK

+/- 2% for T = 170 – 1900 oK

Tb

Ce C and b = empirical constants

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Flow between 2 plates

u=VMoving plate

Fixed plate

y

x

V

u=0

B yB

Vyu )( Fluid Force acting

ON the plate

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21

222111

AA

FAAF

221

1 dy

du

dy

du

Thus, slope of velocity profile is constant and velocity profile is a st. line

Force is same on topand bottom

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u=VMoving plate

Fixed plate

y

x

V

u=0

B yB

Vyu )(

B

V

dy

du Shear stress anywherebetween plates

Shearon fluid

mB

smV

CSAEmsN o

02.0

/3

)38@30(/1.0 2

2

2

/15

)02.0

/3)(/1.0(

mN

m

smmsN

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• 2 different coordinate systems

rx

B

21)(

B

rVru

Vy

x yByCyu )(

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Example: Textbook Problem 2.33Suppose that glycerin is flowing (T = 20 oC) and that the pressure gradient dp/dx = -1.6 kN/m3. 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

2

2

1yBy

dx

dpu

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H

yuyHy

ds

dpu t

2

2

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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)?

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Elasticity (Compressibility)

VdpdV VdpE

dVv

1

• 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

Small dV/V = large modulus of elasticity

ddp

VdV

dpEv How does second part of

equation come about?

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• Given: Pressure of 2 MPa is applied to a mass of water that initially filled 1000-cm3 (1 liter) volume.

•

• Find: Volume after the pressure is applied.

• Ev = 2.2x109 Pa (Table A.5)

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

equation for the modulus of elasticity of an ideal gas.

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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,

• surface tension = magnitude of tension/length

= 0.073 N/m (water @ 20oC)

water

air

No net force

Net forceinward

Interface

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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 = 0o)– mercury depresses in glass tube (angle = 130-140o)

• See attached information

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Example: Capillary Rise

• Given: Water @ 20oC, d = 1.6 mm• Find: Height of water

F

W

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Find: Maximum capillary rise between two vertical glass plates 1 mm apart.

t

h

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Examples of Surface Tension

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Given: Spherical soap bubble, inside radius r, film thickness t, and surface tension .

Find: Formula for pressure in the bubble relative to that outside. Pressure for a bubble with a 4-mm radius?

Should be soap bubble

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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 oC, 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

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Example• The vapor pressure of naphthalene at 25 oC is 10.6 Pa. What is the

corresponding mass concentration of naphthalene in mg/m3? (Hint: you can treat naphthalene vapor as an ideal gas).

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Vapor Pressure (Pvp) - continued

Vapor Press. vs. Temp.

0

20

40

60

80

100

120

0 10 20 30 40 50 60 70 80 90 100

Temperature (oC)

Vap

ro P

ress

ure

(kP

a)

Vapor pressure of water (and other liquids) is a strong function of temperature.

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Vapor Pressure (Pvp) - continued

OHvp

OH

P

PxRH

2

2

,

%100

Pvp,H2O = Pexp(13.3185a – 1.9760a2 – 0.6445a3 – 0.1299a4)

P = 101,325 Pa a = 1 – (373.15/T) T = oK

valid to +/- 0.1% accuracy for T in range of -50 to 140 oC

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?

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Example: Relative HumidityThe relative humidity of air in a room is 80% at 25 oC.

(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?

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Saturation Vapor Pressure

0

500

1000

1500

2000

2500

3000

3500

4000

4500

0 5 10 15 20 25 30 35

degrees C

Pv

p (

Pa)

1 FLUID PROPERTIES Chapter 2 CE319F: Elementary Mechanics of Fluids.

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