Transcript CE319F Elementary Fluid Mechanics

FLUID PROPERTIES
Chapter 2
CE319F: Elementary Mechanics of Fluids
1
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 = 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
3
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
M = mega = 106
k = kilo = 103
c = centi = 10-2
m = milli = 10-3
m = 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
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 oC, 1 atm)
– Water
– Mercury
– Air
rwater = 1,000 kg/m3 (62.4 lbm/ft3)
rHg = 13,500 kg/m3
rair = 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
6
Example: Textbook Problem 2.8
•
Estimate the mass of 1 mi3 of air in slugs and kgs.
Assume rair = 0.00237 slugs/ft3, the value at sea level for standard conditions
7
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
g  rg
[ N / m3 ] or [lbf / ft 3 ]
• Weight per unit volume (e.g., @ 20 oC, 1 atm)
gwater
= (998 kg/m3)(9.807 m2/s)
= 9,790 N/m3
[= 62.4 lbf/ft3]
gair
= (1.205 kg/m3)(9.807 m2/s)
= 11.8 N/m3
[= 0.0752 lbf/ft3]
9
Specific Gravity
• Ratio of fluid density to density of water @
4 oC
rliquid
rliquid
SGliquid 

r water 1000 kg / m3
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 gasoline fills an
8 m3 tank on a transport truck, what is the weight of the gasoline in the
tank?
11
Ideal Gas Law (equation of state)
PV  nRuT
P = absolute (actual) pressure (Pa = N/m2)
V = volume (m3)
n = # moles
n
P  RuT
V
Ru = universal gas constant = 8.31 J/oK-mol
T = temperature (oK)
nM Ru
nM
P
T
RT  rRT
V M
V
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.
13
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.
14
Example: Textbook Problem 2.4
• 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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Viscosity
16
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
u( y) 
V
y
B
Fluid
Fixed plate
x
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u=0
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
  r 2 
u (r )  V 1    
  R  
V
Fluid
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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.
y
Moving plate
t0
u=V
t1 t2
Fluid
Fixed plate
x
u=0
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Fluid Deformation
For viscous fluid, shear stress is proportional
to deformation rate of the fluid (rate of strain)
da

dt
y
dy
t
dL
da 
dy
dL
dt 
dV
dL
Moving plate
da
t+dt
dx
da dV

dt dy
u=V+dV
Fluid
Fixed plate
dV

dy
x
u=V
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Viscosity
•
Proportionality constant = dynamic (absolute) viscosity
•
Newton’s Law of Viscosity
•
•
 m

Viscosity
m
Units
N / m2
dV / dy
m/s/m

dV
dy
V+d
v
V
N s
m2
m = 1x10-3 N-s/m2
•
Water (@ 20oC):
•
Air (@ 20oC): m = 1.8x10-5 N-s/m2
•
Kinematic viscosity  
m
r
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”
dV

dy
dV
 m
dy
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.
Liquids:
greater T = lower cohesive forces
between molecules = viscosity
down.
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24
Typical Viscosity Equations
3
Gas:
Liquid:
m  T  2 To  S
 
m To  T  S
m  Ce
b
T
T = Kelvin
S = Sutherland’s constant
Air = 111 oK
+/- 2% for T = 170 – 1900 oK
C and b = empirical constants
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Flow between 2 plates
Force is same on top
and bottom
F1  1 A1   2 A2  F2
A1  A2
1   2
y
1  m
du
du
m
2
dy 1
dy 2
Thus, slope of velocity
profile is constant and
velocity profile is a st. line
Moving plate
u=V
V
B
u( y) 
V
y
B
Fluid
Fixed plate
Force acting
ON the plate
x
u=026
Flow between 2 plates
Shear stress anywhere
between plates
 m
du
V
m
dy
B
m  0.1 N  s / m 2 ( SAE 30 @ 38o C )   (0.1 N  s / m 2 )( 3 m / s )
0.02 m
V  3 m/s
B  0.02 m
y
 15 N / m 2
Moving plate
V
B
u( y) 
V
y
B
u=V


Fixed plate
Shear
on fluid
x
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u=0
Flow between 2 plates
• 2 different coordinate systems
B
r
x
  r 2 
u (r )  V 1    
  B  
V
u ( y )  C  y  B  y 
y
x
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Example: Textbook Problem 2.33
Suppose 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

1 dp
u  
By
2m dx

y2
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
30
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)?
u
 1 dp
y
Hy  y 2  ut
2m ds
H


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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
dV  Vdp
1
dV  Vdp
Ev
Ev  
Ev = bulk modulus of elasticity
dp
dp

dV
dr
V
r
Small dV/V = large modulus of elasticity
How does second part of
equation come about?
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Example: Textbook Problem 2.45
• 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, s
•
surface tension = magnitude of
tension/length
Interface
water
air
Net force
inward
No net force
 s = 0.073 N/m (water @ 20oC)
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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
Fs
W
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Example: Textbook Problem 2.51
Find: Maximum capillary
rise between two vertical
glass plates 1 mm apart.
q
s
s
h
t
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Examples of Surface Tension
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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
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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.
120
Vapro Pressure (kPa)
100
80
60
40
20
0
0
10
20
30
40
50
60
70
80
90
100
Temperature (oC)
Vapor pressure of water (and other liquids) is a strong function of temperature.
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Vapor Pressure (Pvp) - continued
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
RH  100% x
PH 2O
Pvp , H 2O
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 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
4500
4000
3500
Pvp (Pa)
3000
2500
2000
1500
1000
500
0
0
5
10
15
20
degrees C
25
30
35
48
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