Chapter 2 Fluid Properties taught by professor

1
FLUID PROPERTIES
Chapter 2
CE319F: Elementary Mechanics of Fluids

2
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,
and the dimensions associated with those properties.

3
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
In Chem Lab you were weighing Newtons, reported as N/g = kg
50 g = 0.05 kg is mass with 0.4905 Newtons weight
Temperature = Kelvin (o
K)
273.15 o
K = freezing point of water
o
K = 273.15 + o
C

4
Système International (SI)
• Force = Mass * Acceleration = kg * m/s2
= kg-m/s2
• Newton (N)
• Work and energy = Force * displacement = Joule
(J) = N*m = kg-m/s2
* m = kg-m2
/s2
• Power = Work/time = Watt (W) = J/s = N-m/s =
kg-m2
/s3
• SI prefixes:
G = giga = 109
c = centi = 10-2
M = mega = 106
m = milli = 10-3
k = kilo = 103
m = micro = 10-6

5
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 = (o
F or o
R)
– o
Rankine = o
R = 460 + o
F
• Work or energy = ft-lbf
• Power = ft-lbf/s
– 1 horsepower = 1 hp = 550 ft-lbf/s = 746 W
• Volume in cubic feet or (American) gallons or barrels or
acre-ft (= 325851 cubic feet)
Banana Slug
Mascot of UC Santa Cruz

6
Density (r) – Greek symbol rho
• Mass per unit volume (e.g., @ 20 o
C, 1 atm)
– Water rwater = 1,000 kg/m3
(62.4 lbm/ft3
)
– Mercury rHg = 13,550 kg/m3
– Air 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

7
Example:
• Estimate the mass of 1 mile3
of air in slugs and kgs.
Assume rair = 0.00237 slugs/ft3
, the value at sea level for standard
conditions or rair = 1.205 kg/m3
Assumption: density is constant, that is, the volume is not tall and T and P are
constant.
Given r = mass/volume
Therefore mass = r * volume
Mass = 0.00237 slugs/ft3
* 1 mi3
* (5280 ft/ mi)3
Mass = 1.205 kg/m3
* 1 mi3
* (1609.34 m/mile)3
= 5.023 x 109
kg

8
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 kg?
See Appendix / Google for density of CCl4
rcarbon tetrachloride = 1.59 g/cm³
mass = r * volume = (1.59 g/cm³ ) * (1 kg /1000 g) * (1000 cm³ /L) * 5 L

9
Specific Weight (g ) – Greek
symbol gamma
• Weight per unit volume (e.g., @ 20 o
C, 1 atm)
gwater = (998 kg/m3
)(9.807 m/s2
)
= 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
]
Note about 1000X difference
]
/
[
]
/
[ 3
3 ft
lbf
or
m
N
g

 

10
Specific Gravity
• Ratio of fluid density to density of water @
4o
C – Why 4o
C? Most dense at 4o
C.
3
/
1000 m
kg
SG
liquid
water
liquid
liquid





Water SGwater = 1
Mercury SGHg = 13.55
Note: SG is dimensionless and independent of system of units

12
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?
What does “fresh” mean? Volatile components have not
evaporated over time leaving the higher molecular weight compounds
behind for a more dense mixture.
SG = 0.8 = ggasoline / gwater @ 4o
C = ggasoline / (g * 1,000 kg/m3
)
SG = 0.8 = ggasoline / (9.81 m/s2
* 1000 kg/m3
) = ggasoline / 9810 N/m3
Weight = ggasoline * Volume = (0.8 * 9810 N/m3
) * 8 m3
Note, from Chem Lab, density (r) of water is 1 g / cm3
(1 g / cm3
) * ( 1 kg/1000 g ) * (1 m / 100 cm ) 3
= 1000 kg/m3

13
Ideal Gas Law (equation of state)
T
nR
PV u

T
R
V
n
P u

RT
RT
V
nM
T
M
R
V
nM
P u




P = absolute (actual) pressure (Pa = N/m2
)
V = volume (m3
)
n = # moles
Ru = universal gas constant = 8.31 J/o
K-mol
= 8.31 N-m/o
K-mol
T = absolute temperature (o
K)
M = molecular weight (g/mol)
R = gas-specific constant
R(air) = 287 J/kg-o
K [from text]
R(air) = Ru / M = 8.31 J/o
K-mol / 28.966 g/mol
R(air) = (8.31 J/o
K-mol) / ( 28.966 g/mol)
R(air) = (0.28689 J/ g - o
K) * ( 1000 g/ kg)
R(air) = 286.89 J/ ( kg - o
K) [same as text]

14
Example
• Calculate the volume occupied by 1 mol of any ideal gas at an
absolute pressure of 1 atm (101,000 Pa) and temperature of 20
o
C.
Tabsolute = 20 o
C + 273.15 = 293.15 o
K
P V = n R T
(101000 N/m2
) * V = (1 mol) * (8.31 J/o
K-mol) * (293.15 o
K)
V = 0.02412 m3
= 24.1 liters

15
Example
• Calculate the density of air at volume occupied by 1 mol of any
ideal gas at an absolute pressure of 1 atm (101,000 Pa) and
temperature of 20 o
C.
P V = n R T
(101,000 N/m2
) * V = (1 mol) * (8.31 J/o
K-mol) * (293.15 o
K)
V = 0.02412 m3
= 24.1 liters
Note: Sometime 1 bar (100,000 Pa = 105
N/m2
) is considered
standard atmosphere for convenience (1% difference)

16
Example
• The molecular weight of (dry) 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 = 101000 N/m2
)
at 20 o
C.
Pabs = r Rair Tabs
From couple of slides above R(air) = 286.89 J/ ( kg - o
K)
101,000 N/m2
= r (287 J/(kg-o
K )) * (293.15 o
K ) * (1 N-m / J)
1.20 kg/m3
= r

17
Example: Textbook Problem 2.4
• Given: Natural gas stored in a spherical tank of constant
volume V
– Time 1: T1=10o
C, p1=100
kPa gage
– Time 2 (after addition of gas): T2=10o
C, p2=200 kPa gage
– Local atmospheric pressure =100 kPa absolute
• Find: Ratio of mass at time 2 to that at time 1
• Note: Ideal gas law (P is absolute pressure)
Pabs,2 = r2 Rnatural gas Tabs,2
Pabs,1 = r1 Rnatural gas Tabs,1
Divide equations
Pabs,2 / Pabs,1 = r2 / r1
300 kPa / 200 kPa = (mass/V)2 / (mass/V)1

19
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=V
Moving plate
Fixed plate
y
x
V
u=0
B y
B
V
y
u 
)
( Fluid

20
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
x
R
















2
1
)
(
R
r
V
r
u
V
Fluid

21
Fluid Deformation
• Flow between a fixed and a moving plate
• Force causes plate to move with velocity V
and the fluid deforms continuously.
u=V
Moving plate
Fixed plate
y
x
u=0
Fluid
t0
t1 t2

22
Fluid Deformation
u=V+dV
Moving plate
Fixed plate
y
x
u=V
Fluid
t t+dt
dx
dy
da
dL
t


 
For viscous fluid, shear stress (force / area) is proportional
to deformation rate of the fluid (rate of strain)
V
L
t


 
y
L


 
y
V
t 




y
V


 

23
Viscosity
• Proportionality constant = dynamic (absolute) viscosity
• Newton’s Law of Viscosity
• Viscosity Greek symbol mu m
• Units
• Water (@ 20o
C): m = 1x10-3
N-s/m2
• Air (@ 20o
C): m = 1.8x10-5
N-s/m2
• Kinematic viscosity
Greek symbol nu n
dy
dV /

 
2
2
/
/
/
m
s
N
m
s
m
m
N 



 
V
V+d
v
dy
dV

 
Kinematic viscosity: m2
/s
1 poise = 0.1 N-s/m2 =
0.1 kg (m/s2
) - s/m2 =
0.1 kg /(m– s)
1 centipoise = 10-2
poise = 10-3
N-s/m2

24
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 or jello

25
Effect of Temperature
Gases:
greater T = greater interaction
between molecules = greater
viscosity. More energetic collisions
between molecules.
Liquids:
greater T = lower cohesive forces
between molecules = viscosity
down.

27
Typical Viscosity Equations
S
T
S
T
T
T o
o 








2
3



Liquid:
Gas:
T = Kelvin
S = Sutherland’s constant
Air = 111 o
K
+/- 2% for T = 170 – 1900 o
K
T
b
Ce

 C and b = empirical constants

28
Flow between 2 plates
u=V
Moving plate
Fixed plate
y
x
V
u=0
B y
B
V
y
u 
)
( Fluid Force acting
ON the plate
2
1
2
1
2
2
2
1
1
1









A
A
F
A
A
F
2
2
1
1 


 


dy
du
dy
du
Thus, slope of velocity
profile is constant and
velocity profile is a straight line
Force is same on top
and bottom

29
u=V
Moving plate
Fixed plate
y
x
V
u=0
B y
B
V
y
u 
)
(
B
V
dy
du


 

Shear stress anywhere
between plates
t
t
Shear
on fluid
m
B
s
m
V
C
SAE
m
s
N o
02
.
0
/
3
)
38
@
30
(
/
1
.
0 2





2
2
/
15
)
02
.
0
/
3
)(
/
1
.
0
(
m
N
m
s
m
m
s
N





30
• 2 different coordinate systems (radial for
cylinder versus linear for 2 flat plates)
r
x
B
















2
1
)
(
B
r
V
r
u
V
y
x
 
 
y
B
y
C
y
u 

)
(

31
Suppose that glycerin is flowing (T = 20 o
C) 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
1
y
By
dx
dp
u 




32
m glycerin (T = 20 o
C) = 1.41 N-s /m2
dp/dx = -1.6 kN/m3
= -1600 N/m3
u is velocity
t shear stress = m du/dy
y1 = 0.012 m and B = 0.05 m
y2 = 0.0 m (at the wall) and B = 0.05 m
u1 = (-1/(2* 1.41 N-s/m2
)) (-1600 N/m3
) (0.05 m * .012 m – (0.012m)2
)
u2 = (-1/(2* 1.41 N-s/m2
)) (-1600 N/m3
) (0.05 m * 0.0 m – (0.0 m)2
)
t1 = m*(-1/(2 m) )(dp/dy)(B – 2 y) = (-1/2)(-1600 N/m3
)(0.05m– 2(0.012m))
t2 = (-1/2) (-1600 N/m3
) (0.05 m – 2 (0.0 m))
Note at the wall ( y=0 ) the velocity goes to zero (no slip condition) but the
shear stress is the greatest
 
2
2
1
y
By
dx
dp
u 




33
  H
y
u
y
Hy
ds
dp
u t



 2
2
1

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)?

34
Elasticity (Compressibility)
Vdp
dV  Vdp
E
dV
v
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 (what are its units?)
Ev = bulk modulus of elasticity
Small dV/V = large modulus of elasticity


d
dp
V
dV
dp
Ev 

 How does second part of
equation come about?

35
• Given: Pressure of 2 MPa is applied to a mass of water that initially
filled 1000-cm3
(1 liter) volume. (dP = 2 x106
Pa)
• Find: Volume after the pressure is applied.
• Ev = 2.2x109
Pa (Table A.5)
• dV = (-1/ Ev) V dP
• dV = (-1/ 2.2x109
Pa ) (1 liter ) ( 2 x106
Pa)
• dV = -0.909 x10-3
liter = -0.909 cm3
Or about 1% smaller at a depth of about 200 m of freshwater

36
Example
• Based on the definition of Ev and the equation of state, derive an
equation for the modulus of elasticity of an ideal gas.
• Definition Ev = dP / (dr/r) = r (dP/dr)
• Ideal Gas Law
• P = rRT
• dP /dr = RT
• Plug in
• Ev = r (RT) = P

37
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
• s = 0.073 N/m (water @ 20o
C)
water
air
No net force
Net force
inward
Interface

38
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 and rises
– If adhesion < cohesion, liquid surface depresses at point of contact
– water rises in glass tube (angle = 0o
)
– mercury depresses in a glass tube (angle = 130-140o
)

40
Example: Capillary Rise
• Given: Water @ 10o
C, d = 1.6 mm
• Find: Height of water
• Do a force balance
• Weight = Surface Tension *
Contact length* cos(contact angle)
• Volume * g = s * p D cos (q)
Dh p D2
/4 * g = s * p D cos (q)
Dh = 4 s cos (q) / (g * D)
Dh = 4 (0.073 N/m) cos (0) / (9810 N/m3
* 0.0016 m)
Dh = 0.0186 m = 1.86 cm

F
W

41
Find: Maximum capillary rise
between two vertical glass plates 1 mm
apart.
• Do a force balance
• Weight = Surface Tension *
Contact length* cos(contact angle)
• Volume * g = s * 2L cos (q)
h * W * L * g = s * 2L cos (q)
h = 2 * s * cos (q) / (W * g )
h = 2 (0.073 N/m) cos(0) / (0.001 m * 9810 N/m3
)
Dh = 0.0149 m = 1.49 cm
t
s
s
q
h

42
Examples of Surface Tension

43
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

44
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

45
Example
• The vapor pressure of naphthalene at 25 o
C 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).
• Pabs V = n Runiversal Tabs
• 10.6 N/m2
= (n/V) (8.31 N-m/o
K-mol ) (25 + 273.15 o
K)
• 0.00428 mol/m3
= n/V
• Molecular weight of naphthalene (C10H8) = 128.1705 g/mol
• 0.00428 mol/m3
* 128.1705 g/mol * 1000 mg/g = Mass/V
• 548 mg/m3
= Mass Concentration

46
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)
Vapro
Pressure
(kPa)
Vapor pressure of water (and other liquids) is a strong function of temperature.

47
Saturation Vapor Pressure
0
500
1000
1500
2000
2500
3000
3500
4000
4500
0 5 10 15 20 25 30 35
degrees C
Pvp
(Pa)

48
Vapor Pressure (Pvp) - continued
O
H
vp
O
H
P
P
x
RH
2
2
,
%
100

Pvp,H2O = P exp(13.3185a – 1.9760a2
– 0.6445a3
– 0.1299a4
)
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?

49
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?
Psaturation vapor pressure = 3.17 kPa (look it up)
Relative Humidity Pwater actual / Psaturation vapor pressure = 0.8
Pwater / Ptotal = Vwater / Vtotal
0.8 * 3.17 kPa / 101 kPa = 0.0251 = Vwater / Vtotal

50
Saturation Vapor Pressure
0
500
1000
1500
2000
2500
3000
3500
4000
4500
0 5 10 15 20 25 30 35
degrees C
Pvp
(Pa)

Chapter 2 Fluid Properties taught by professor

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