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Reading Materials: Chapter 6 . Fluid Flow. LECTURES 16-18. What is a Fluid. Material that continually deforms under a shear stress Divided into two groups Liquid Gases. GASES. - PowerPoint PPT Presentation
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CHEM ENG 1007 1
Reading Materials: Chapter 6
LECTURES 16-18
CHEM ENG 1007 2
What is a Fluid
Material that continually deforms under a shear stress
Divided into two groupsLiquid
Gases
CHEM ENG 1007 3
Characterized as loosely-associated molecules which are normally not close together and which travel through space for long distances before colliding with each other. The velocity of their travel depends on the temperature of the gas.Characteristic of gases:
Readily compressibleExpand quickly and fill a container
To convert from volume to mass/mole use PV=nRT
GASES
CHEM ENG 1007 4
Characterized by molecules which are very close together and which are in collision with each other very frequently as they move around each other. The velocity of that motion and the rate of that collision depend on the temperature of the liquid.Characteristic of liquids:
Slightly compressibleTakes shape of container sides and bottom
To convert from volume to mass/mole use
Density = mass/volume
LIQUIDS
CHEM ENG 1007 5
Fluid-Like Systems
Solids present in fluids Slurries
fine particles suspended in liquid
Solids in a fluidized bed particles moving with the
fluid in a tall reactor.
CHEM ENG 1007 6
Variables associated with fluids
Density
Flow rate
Pressure
Viscosity
Surface tension
Others
Thermal conductivity
Electrical conductivity
Boiling point
Freezing point
Heat capacity
Enthalpy
Chapter 7 CHEM ENG 1007 7
Flow Rate
Rate at which a material is transported through a process line.
Mass flow rate:
Molar flow rate:
Volume flow rate:
where:
mass mm
time t
mole nn
time t
Volume VV
time t
avem V MW nρ: average density
aveMW : average molecular weight
Chapter 7 CHEM ENG 1007 8
Pressure
The pressure of the fluid is defined as the total force (exerted on the boundary by the fluid molecules) divided by the surface area of the boundary it is acting upon.
{F} {A} {P}
SI N m2 Nm-2 (Pa)
cgs dyne cm2 dynecm-2
American lbf in2 lbfin-2 (psi)
Chapter 7 CHEM ENG 1007 9
A woman’s high heels sink
into the soft ground,
but the larger shoes of the
much bigger man do not.
Pressure = force/area
Puzzle
Chapter 7 CHEM ENG 1007 10
PressureWe express pressure in two ways: absolute and gauge pressures
Gauge Pressure: fluid pressure that is measured relative to the atmospheric pressure. Absolute Pressure: the total magnitude of force exerted per unit area
In process calculations:Pabs = Pgauge + Patmosphere
Pabs = 0 in complete vacuumLetter “a” or “g” is added to designate absolute or gauge, thus psia or psig
Chapter 7 CHEM ENG 1007 11
Standard Atmospheric Pressure
At sea level, 0oC and 45o latitude:
Patm = 1 atm
= 101,325 Pa
= 14.7 psi
= 1.01325 bars
= 760 mmHg
= 10.333 m H2O = 33. 9 ft H2O
Chapter 7 CHEM ENG 1007 12
Torricelli filled a tube with mercury and inverted it into an open
container of mercury. Air pressure acting on the mercury in the dish
can support a column of mercury 76 cm in height.
How much does the atmosphere heigh?
Answer: the same as 76 cm of mercury. How?
Standard Atmospheric Pressure
Chapter 7 CHEM ENG 1007 13
Example 7.1
A man pumps his automobile tire until the tire gauge reads 34.0 psi. If the atmosphere in his community is 14.2 psia, what is the absolute pressure of the air in the tire?
Solution:
Pg = 34.0 psig
Patm = 14.2 psia
Pabs = 34.0 + 14.2 = 48.2 psia
Chapter 7 CHEM ENG 1007 14
Hydrostatic Pressure
It is the pressure (P) of the fluid at the base of the column. That is the force (F) exerted on the base divided by the base area (A). F thus equals the force on the top surface plus the weight of the fluid in the column.P = P0+ gh
h = height of a hypothetical column of the fluidA pressure may also be expressedas a head of a particular fluid (Ph)
Ph = P0 + h
Chapter 7 CHEM ENG 1007 15
Example 7.2
For the tank depicted in Fig. 7.2, if the NaOH solution is 8 ft high, what is the pressure at the bottom of the tank? Assume that the density of the NaOH solution is the same as that of water. Perform the calculation in metric units.
Solution:2 1
2 1 3
1000f
f
P P p ghkg
P P p ghm
9.80 m2
8 ft
s
0.3048 m1ft
2 1 223,896 23,896 23.896 kkg
msP P Pa Pa
P1 = 0 Pa (gauge), so P2 = 23,896 Pa (gauge)
or P2 = 23,896 + 101,325 = 125,221 Pa (absolute)
Chapter 7 CHEM ENG 1007 16
Quick Quiz 1
What is the pressure 30.0 m below the surface of a lake? Assume the atmospheric pressure (the pressure at the surface) is 10.4 m H2O? Express your answer in atm.
Solution:
Ph = P0 + h
= 10.4 + 30.0 = 40.4 m H2O
2
2
40.4 1 3.91
10.333 mH O atm
atmmH O
Chapter 7 CHEM ENG 1007 17
Figure 3.4-3 (p. 57) of Felder and RousseauBourdon gauge. It is used to measured fluid pressure from nearly perfect vacuums to about 7000 atm.
Fluid Pressure Measurement
A hollow tube closed at one end and bent into a C configuration. The open end of the tube is exposed to the fluid whose pressure is to be measured. As the pressure increases, the tube tends to straighten, causing a pointer attached to the tube to rotate.
Chapter 7 CHEM ENG 1007 18
Figure 3.4-4 (p. 58)Manometers. It is used for more accurate measurements of pressure (below about 3 atm)
Fluid Pressure MeasurementGauge pressure Absolute pressurePressure difference
Chapter 7 CHEM ENG 100719
Figure 3.4-5 (p. 58)Differential Manometer variables.
Fluid Pressure Measurement
1 2
1 2
1 2 1 2
f f
f
if p p p
P P p p gh p gh
P P p p g d d
1 1 1 2 2 2
( ) ( )
f
P a P b
P p g d P p g d p g h
If is gas then may be neglected
Chapter 7 CHEM ENG 1007 20
Illustration 1
A manometer reading gives 100 mmHg, calculate the absolute pressure.
Solution:
Measured gauge pressure:
Pabs = 13,328 + 101,325 = 114,653 Pa
= 115 kPa
3 2
13,600 9.80 0.1 13,328 ag
kg mP gh m P
m s
Chapter 7 CHEM ENG 1007 21
How does pressure relate to flow?
In the absence of other forces, fluids tend to flow from regions of high pressure to regions where the pressure is lower. Therefore, pressure differences provide a driving force for fluid flow.
Example: when a tire is punctured, air flows out of the high pressure tire to the atmosphere, which is a low pressure.
Chapter 7 CHEM ENG 1007 22
Viscosity ()
Characterises its resistance to flow A measure of “stickiness” of a fluid A “frictional force”
Less energy required for mixing
More energy required for mixing
Low viscosity fluid High viscosity fluid
Chapter 7 CHEM ENG 1007 23
Measurement - “Shear stress/shear rate”
Viscosity ()
dv
dyviscosity
Units SI: kg/(m.s) = N.s/m2 = Pa.s cgs: cp (centipoise) 1 cP = 10-2 poise = 10-3 Pa.s = 1 mPa.s
Chapter 7 CHEM ENG 1007 24
Viscosity of various fluids
Fluid Temperature (oC) ViscosityPa.s
Water 15.6 1.1x10-3
Gasoline 15.6 0.3x10-3
SAE 30 oil 15.6 383x10-3
Air 15 1.8x10-5
Methane 20 1.1x10-5
Chapter 7 CHEM ENG 1007 25
Substance (25°C) Viscosity (Pa.s)
Water 1 x 10-3
Mercury 1.5 x 10-3
Air 1.8 x 10-5
Castor oil 0.99
Viscosity of various fluids
Chapter 7 CHEM ENG 1007 26
Kinematic viscosity
2 -1 m .s
The property viscosity may also be combined with the fluid’s density to give the property kinematic viscosity
Chapter 7 CHEM ENG 1007 27
Types of viscous fluids
1
2
3
4
She
ar s
tres
s
Velocity gradient dv
dy
dv
dyviscosity
Chapter 7 CHEM ENG 1007 28
1. Newtonian fluid
e.g. water, air, other gases.
2-4 Non-Newtonian fluids
2. Bingham-plastic.
e.g. toothpaste, margarine, soap
3. Pseudo-plastic (shear thinning)
e.g. mayonnaise, polymer melts, paints.
4. Dilatant (shear thickening)
e.g. wet beach sand, starch in water
1
2
3
4
She
ar s
tres
s
Velocity gradient dv
dy
Types of viscous fluids
Chapter 7 CHEM ENG 1007 29
Newtonian fluids
With Newtonian fluids, shear stress increases proportionately with the shear rate
Question:
How would you determine the viscosity from a plot of shear stress against shear rate?
Shear rateVi
scos
ity
ExamplesWaterGlycerineAlcoholAir
Shear rate
Shea
r stre
ss
Chapter 7 CHEM ENG 1007 30
Other types of Non-Newtonian fluids
Pseudoplastic
Shear rate
Vis
cosi
ty
Dilatant
Shear rateV
isco
sity
Viscosity changes with power input
Chapter 7 CHEM ENG 1007 31
Thixotropic
Time
Vis
cosi
ty
Rheopetic
Time
Vis
cosi
tyViscosity changes with time at constant
shear rate
Other types of Non-Newtonian fluids
Chapter 7 CHEM ENG 1007 32
Non-Newtonian fluids can have more than one property
Example:
Damp gypsum Pseudoplastic, Thixotropic and Viscoelastic
Cream Dilatant, Rheopectic and not Viscoelastic
Viscoelastic: Fluid returns to original viscosity after power input ceases
Non-Newtonian fluids
Chapter 7 CHEM ENG 1007 33
Ideal / Inviscid Fluid
Hypothetical fluid which is incompressible and has zero viscosity.
Chapter 7 CHEM ENG 1007 34
Fluid Flow Principles
Flow patterns vary with:velocitygeometry of surface, andfluid properties such as viscosity, density.
Classic experiment by Osborne Reynolds (1883) observed two types of fluid flow:
Laminar flow – low flow ratesTurbulent flow – higher flow rates
Chapter 7 CHEM ENG 1007 35
Reynolds’ experiment
(i) Low flow rates: fluid moves in parallel layers.
(ii) High flow rates: cross currents (eddies) develop
Chapter 7 CHEM ENG 1007 36
Reynolds Number
Reynolds’ key variables
pipe diameter (D)mean velocity (v)fluid density ( )fluid viscosity ( )
Arrange into single “dimensionless group”:
vD
Re
Chapter 7 CHEM ENG 1007 37
Reynolds Number
For pipe flow:
Re < 2,100 - laminar flow
2,100 < Re < 10,000 - transition region
Re > 10,000 - turbulent flow
Pipe-flow systems with the same Re are said to be dynamically similar.
Chapter 7 CHEM ENG 1007 38
Fluid flow through a pipe
(i) Is the flow laminar flow or turbulent?
(ii) What is the effect of reducing the velocity by a factor of 10?
Water at 25oC flows through a pipe of internal diameter 0.1 m at a velocity 0.2 m/s.
Illustration 2
Chapter 7 CHEM ENG 1007 39
Solution
3
1000 0.2 0.1vD(i) Re 20,000
1x10
Hence flow is turbulent
(ii) Re v20,000
Re = 2,00010
Hence flow is laminar
Chapter 7 CHEM ENG 1007 40
Velocity Profiles: Laminar Flow
(i) Velocity profile is parabolic with the maximum velocity occurring in the centre of the pipe (r = 0)
2
max
2
x max 2
R pv
4 L
At other radial positions:
rv v 1
R
(i) Laminar Flow
Chapter 7 CHEM ENG 1007 41
Rr
L
(i) Velocities of a fluid in laminar flow through a circular pipe
2
x max 2
rv v 1
R
Velocity Profiles: Laminar Flow
Chapter 7 CHEM ENG 1007 42
(ii) Volumetric flow rate:
4R p
V8 L
Hagen Poiseuille equation.
(iii) Mean velocity:
maxV 1
v vA 2
Velocity Profiles: Laminar Flow
22
where
A cross-sectional area
DA R
4
Chapter 7 CHEM ENG 1007 43
Velocity Profiles: Turbulent Flow
Difficult to mathematically model due to its complex and rapidly changing flow patterns.
Experimental measurements show for time-averaged velocity and mean velocity
1/7
max
max
rv v 1
Rv 0.8v (ii) Turbulent Flow
Chapter 7 CHEM ENG 1007 44
(iii) Plug Flow
Velocity Profiles: Plug Flow
Common assumption for highly turbulent flow is that velocity does not vary over cross-section:
v v
Chapter 7 CHEM ENG 1007 45
Mass Conservation in Fluid Flow
Consider steady-state, one-dimensional fluid through pipe
A1
v
2
2
1
1
1
A
v2
2
Flow In Flow out
1 2
1 1 2 2
1 1 1 2 2 2
1 2
2 21
1
1 2
m m m
V V
A v A v
for the same fluid
A vv
A
v v
22
where
A cross-sectional area
DA R
4
Chapter 7 CHEM ENG 1007 46
Mass Conservation in Fluid Flow
Is v2 in (a) less than v2 in (b)? No, they are both the same.
1 1 1 2 2 2
1 2 1 2
1 2
A v A v
for the same fluid and A A
v v
Chapter 7 CHEM ENG 1007 47
Fluid Friction
Frictional force cause pressure drop during fluid flow through a constant diameter horizontal pipe.
Consider flow situation in pipe below.
L
v
1 2
D
p
w
w
p1 2
Chapter 7 CHEM ENG 1007 48
Fluid FrictionMomentum balance:
2
w
w
2
D0 p DL
4
L p 4
DL
p 2f vD
f = friction factor; L = length of pipe;
D = diameter of pipe;
Chapter 7 CHEM ENG 1007 49
Fluid Friction
2f
2f
2f
p L J2f v e kgD
Lfrictional energy e 2f v
D
e 2f Lfrictional head loss v
g g D
Chapter 7 CHEM ENG 1007 50
Example 7.7What value of the friction per mass of fluid (ef) is necessary to cause a decrease in pressure equal to
a) 10 psi (answer in Btu/lbm)?
f
3 2 2f
2 2m f
m
pe
10 lb ft 1 Btu 12 in
62.4 lb 778.1 ft lbin 1 ft
Btu0.030
lb
Chapter 7 CHEM ENG 1007 51
Example 7.7What value of the friction per mass of fluid (ef) is necessary to cause a decrease in pressure equal to
b) 68,900 Pa (answer in J/kg)?
f
pe
68,900 Pa 21 kg/ms
1 Pa
3 2
2
m 1 J s
1000 kg 1 kg m
J68.9
kg
Chapter 7 CHEM ENG 1007 52
Laminar Flow
f determined analytically from velocity profile.
16
i.e. fRe
Experimental data confirmed for Re < 2,100 (see Figure 2.10-3)
Chapter 7 CHEM ENG 1007 53
Figure 2.10-3: Friction Factor Chart
Chapter 7 CHEM ENG 1007 54
Turbulent Flowf cannot be determined analytically.
Experimental curves have been devised.
f also depends on surface roughness factor,
Pipe (mm)
Concrete 0.3-3.0
Cast iron 0.26
Galvanized iron 0.15
Plastic 0.0 (smooth)
Chapter 7 CHEM ENG 1007 55
0.25 f 0.079Re
Turbulent Flow
For smooth pipe & 2,100 < Re < 105
Blassius equation may be used:
Chapter 7 CHEM ENG 1007 56
Frictional losses in a pipe
Pipeline 5 km long & 30 cm internal diameter
Conveys water at 25oC at a rate of 180 kg/s.
Roughness factor /D = 0.001
Estimate the pressure drop across the pipe due to friction.
Illustration 3
Chapter 7 CHEM ENG 1007 57
Solution
For water
3
3
24
53
1,000 kg/m
10 Pa.s
m V vA
m 180v 2.55 m / s
A 1,000 0.3
1,000 2.55 0.3vDRe 7.65x10
10
Chapter 7 CHEM ENG 1007 58
From Figure 2.10-3, f = 0.005
Hence,
Solution
2 2L 5000p 2f v 2 0.005 1000 2.55
D 0.3
1,083,750 Pa
1084 kPa
Chapter 7 CHEM ENG 1007 59