Download - Momentum Transfer - 1 (4009)
CHEE4009Transport PhenomenaSemester 1, 2016
Lecturer: Prof. D. D. Do
What are Transport Phenomena?
Domain of Chemical Engineering
Process Design
Product Design
Transport Phenomena & Science
Objectives in CHEE4009 Unify the various transfers (momentum, heat and mass)
Formulate a shell balance to obtain an equation describing the transfer
Impose physical constraints (boundary conditions)
Solve differential equations for various examples by analytical means
Extrapolate solutions of simple systems to analyze complex systems
etc…
Assessments 3 Assignments (15%)
1 on momentum transfer (5% each)
1 on heat transfer (5% each)
1 on mass transfer (5% each)
1 Mid Semester Test (20%)
1 Final closed book examination (65%)
Text books Bird, Stewart and Lightfoot, “Transport
Phenomena”, Wiley, second edition, NY, 2002
This is a very good book in Transport Phenomena. However, there are many books on the same subject available in the library.
References R. G. Rice and D. D. Do, “Applied Mathematics for
Chemical Engineers”, Second Edition, Wiley, New York, 2012.
This book is a good source of various basic mathematical methods of solving various transport problems in chemical engineering
You also find many more applied mathematics books for engineers in the library
An Overview of Transport Principles Momentum Stress
Velocity gradient
Viscosity
Newton law
Heat Heat Flux
Temperature Gradient
Thermal Conductivity
Fourier law
Mass Mass Flux
Concentration Gradient
Diffusion Coefficient
Fick law
Quantity
Driving Force
Parameter Characterizing the Transport
Basic law
τ µyxxdv
dy= − q k
dT
dxx = − J DdC
dxx AB= −
An Overview of Transport Principles Momentum Balance around a
finite shell having surfaces perpendicular to the transport directions
1st order ODE in terms of shear stress
2nd order ODE in terms of velocity
Heat Balance around a
finite shell having surfaces perpendicular to the transport directions
1st order ODE in terms of heat flux
2nd order ODE in terms of temperature
Mass Balance around a
finite shell having surfaces perpendicularto the transport directions
1st order ODE in terms of mass flux
2nd order ODE in terms of concentration
Transport equation
Type of equation
An Overview of Transport Principles Momentum BC of 1st kind BC of 2nd kind
BC of 4th kind
Calculus, vector analysis, separation of variables, Laplace transform, combination of variables, numerical analysis
Heat BC of 1st kind
BC of 2nd kind
BC of 3rd kind
BC of 4th kind
Calculus, vector analysis, separation of variables, Laplace transform, combination of variables, numerical analysis
Mass BC of 1st kind
BC of 2nd kind
BC of 3rd kind
BC of 4th kind
Calculus, vector analysis, separation of variables, Laplace transform, combination of variables, numerical analysis
Boundary conditions (BC)
Methods of analysis
Overview contents Momentum transfer
Heat transfer
Mass transfer
Momentum Transfer – Part I
Contents of Momentum Transfer Macroscopic versus Microscopic
Newtonian fluids versus non-Newtonian fluids
Examples: 1: Flow on flat plate 2: Flow though circular tube (Newtonian fluids) 3: Flow of Bingham (non-Newtonian) fluid through tube Other examples ….
Molecular flow versus continuum flow
Surface tension
Things seem to be simple! Fluid flow (momentum), heat transfer and mass transfer have been learnt
before, but only at the process(macroscopic) level
Here we shall deal with at microscopic level
So what are these two levels? Illustrated with a few simple examples
Process vs MicroscopicExample 1: Adsorption of dye into carbon
Process level
Microscopic level
About few hours later
Process vs MicroscopicExample 2: Gas Absorber
Liquid
Gas
Process level Microscopic level
Process vs MicroscopicExample 3: Heating of a flowing fluid
Process level Microscopic level
The Beauty of Microscopic Analysis Better understanding of the transfer mechanism
Better description of the dependence of transfer on the system parameters
Mathematical treatments of all three transfers are identical For similar geometrical systems, solutions obtained for one type of
transfer can be directly used on the other transfers
Minimize the experimental efforts
Process versus MicroscopicExample 1: Adsorption of dye into carbon
Process level
Microscopic level
About few hours later
Diffusion coefficient, pore size, particle size
Overall mass transfer coefficient
Process versus MicroscopicExample 2: Gas Absorber
Liquid
Gas
Process level Microscopic levelOverall mass transfer coefficient Velocity, diffusion coefficient, solubility
Process versus MicroscopicExample 3: Heating of a flowing fluid
Process level Microscopic level
Overall heat transfer coefficient Velocity, thermal conductivity
Newton law of viscosity
Friction is felt only when you walk either slower or faster than other people
The extent of friction depends on the type of clothes (i.e. property of the medium)
RailwayStation
Exit
Example of Two Parallel Plates (1)
Shear force is due to the difference in the velocities
y
x
y
Shear force
V
Example of Two Parallel Plates (2)
Definition of Shear Stress: is the force exerted in the x-direction on the surface of constant y
by the fluid in the region of lesser y:
τ yxy
Shear force
x
y
Flow direction
Surface at which the force is acting
Example of Two Parallel Plates (3) Directionof shear stress transport:
The shear stress is moving in the y-direction because the bottom molecular layer exerts a shear stress on the next layer which in turn exerts a shear stress on the subsequent layer
y
Example of Two Parallel Plates (4) In this example, the shear stress is induced by the
motion of the bottom plate.
Shear stress can also be induced by the pressure gradient or a gravity force The pressure force is a force acting on a surface
The gravity force is the force acting on a volume
y
Newton law The shear stress is generally a function of the velocity gradient and the
properties of the fluid
If the fluid is called Newtonian, the relationship between the shear stress and the velocity gradient is:
where vx is the fluid velocity in the x-direction
µ is the viscosity of the fluid
τ µ∂∂yx
xv
y= −
V
y
x
NOTE!!Flow direction is xShear direction is y
Important reminder about the shear stress The shear stress τyx is the force exerted
in the x-direction on a surface of constant y by the fluid in the region of lesser y
Newtonian Fluids By definition, a fluid is called a Newtonian fluid if Newton
law is applicable to that fluid Air, water are Newtonian fluids
The viscosity of Newtonian fluids is constant for a given temperature and pressure (i.e. independent of velocity and its gradient).
Gases: At moderate pressure, the viscosity is independent of pressure and increase with T. The temperature dependence is between T0.6 and T.
Liquids: The viscosity decreases with temperature
Viscosity Units: Pa-s (g/cm/s = poise)
Magnitudes: Air @ 20 C: 0.00018 g/cm/s
Liquid water @ 20 C: 0.01 g/cm/s = 0.001 Pa-s
Non-Newtonian Fluids Fluids that do not satisfy Newton law are called non-Newtonian fluids
Many biochemical and pharmaceutical fluids are non-Newtonian fluids
There is no general relationship that relates shear stress, velocity and velocity gradient. Rather there are empirical correlations that take the following general functional form:
0propertiesfluid,v,y
v,f x
xyx =
∂∂
τ
This means that the shear stress is not only dependent on the fluid properties, but also on the velocity and its gradient (strain)
Non-Newtonian Fluids For any fluids, one can write
The coefficient ηηηη is regarded as the apparent viscosity. Generally it depends on the current state of the fluid
If η decreases with increasing rate of shear (-dvx/dy), the behavior is termed pseudoplastic.
If η increases with increasing rate of shear (-dvx/dy), the behavior is termed dilatant.
yxτ
y
vx
∂∂−
Pseudo-plastic Newtonian
Dilatant
τ η∂∂yx
xv
y= −
Corn starch in water
Latex paint
Non-Newtonian FluidsEmpirical Correlation Equations Correlation equations are empirical and the parameters do not bear any
physical meaning. These parameters are function of temperature, pressure and compositions of the fluid. These equations should not be used outside their range of validity
Bingham model Ostwald-de Waele model Eyring model Ellis model Reiner-Philippoff model
Non-Newtonian Fluids: Graphical representation
τyx
dy
dvx−
Newtonian Fluid
Bingham Fluid
Ostwald-de Waele (Pseudo-plastic)
Ostwald-de Waele (dilatant)
Non-Newtonian FluidsBingham model The model equations are
Two parameters: µ0 and τ0
Behavior: The fluid remains rigid when the shear stress is less than the yield
stress τ0
When the shear stress is greater than τ0, the Bingham fluid behaves somewhat like a Newtonian fluid
τ µ τ τ τ
τ τ
yxx
yx
xyx
dv
dyif
dv
dyif
= − ± >
= <
0 0 0
00
Non-Newtonian FluidsBingham model The model equations are
Two parameters: µ0 and τ0
Behavior: The fluid remains rigid when the shear stress is less than the yield
stress τ0
When the shear stress is greater than τ0, the Bingham fluid behaves somewhat like a Newtonian fluid
τ µ τ τ τ
τ τ
yxx
yx
xyx
dv
dyif
dv
dyif
= − ± >
= <
0 0 0
00
Non-Newtonian FluidsOstwald-de Waele model The model equation is (two parameters)
The apparent viscosity is a function of the velocity gradient
Behavior: If n < 1: pseudo-plastic
If n = 1: Newtonian
If n > 1: dilatant
τ yxx
n
xmdv
dy
dv
dy= −
− 1
τyx
-dvx/dy
n < 1
n = 1
n > 1
Theory of Viscosity of GasesMolecular Force Field theory (1) Developed by Chapman and Enskog
This theory is based on the pairwise potential energy between two molecules The most popular equation that describes this potential energy was developed
by Lennard-Jones
It is known as the Lennard-Jones 12-6 equation
The two molecular parameters are σ and ε
ϕ εσ σ
( )rr r
=
−
412 6
Theory of Viscosity of GasesMolecular Force Field theory (2) The famous LJ 12-6 equation
The parameter σ is called the collision diameter, and is defined as the distance at which the potential energy is zero
The parameter ε is called the well-depth of the interaction potential. It is a measure of the strength of interaction
ϕ εσ σ
( )rr r
=
−
412 6
- ε
ϕ(r)
rσ
Theory of Viscosity of GasesMolecular Force Field theory (3) Without going into details (beyond the level of this subject), the expression
for viscosity of monatomic gases derived from this theory is
The parameter Ω is called the collision integral and it is a function of temperature. Tabulation of this is given in Appendix E.2 (page 866).
µσ µ
= × −2 6693 1052
.MT
Ω
Viscosity of Mixtures Useful semi-empirical equation was proposed by Wilke
x is the mole fraction
µµ
mixi i
j ijj
ni
n x
x=
=
= ∑∑
Φ1
1 Φ iji
j
i
j
j
i
M
M
M
M= +
+
−1
81 1
1 2 1 2 1 42/ / /
µµ
Procedure of Transport Phenomena Analysis (1) First principles:
Draw a physical diagram Identify all transport mechanisms Choose a frame of coordinates
Draw a shell Such that its surfaces must be perpendicular to the transport directions
Carry out the momentum balance around the shell
In the limit of the shell shrinking to zero, this balance equation should be a first-order differential equation in terms of shear stress
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
Steady State
Procedure of Transport Phenomena Analysis (2) Use the appropriate equation that relates the shear stress to the velocity,
the balance equation of first-order differential equation in terms of shear stress will become a second order differential equation in terms of velocity If the fluid is Newtonian, that appropriate equation is the Newton law;
otherwise use the appropriate equation for non-Newtonian fluids
Impose physical constraint on the boundary of the system This should give the boundary conditions (BCs) for the differential equation
of the previous step
Solve the differential equation & BCs for the velocity distribution Knowing the velocity distribution, maximum velocity, average velocity,
volumetric flow rate and shear stress can be determined
Boundary conditions At solid-fluid interface, the fluid velocity equals to the velocity of the
solid surface. This is called boundary condition of the first kind (Dirichlet BC)
At gas-liquid interface, the shear stress is zero owing to the fact that the gas viscosity is low. Boundary condition of the second kind (Neumann BC)
At liquid-liquid interface, the shear stresses and the velocities are continuous across the interface. This is called boundary condition of the fourth kind.
What to follow? A number of fluid flow problems with simple geometry.
These are to show the application of the first principles.
Results of simple problems are useful in understanding the system, and very often they are used in solving related complex problems.
Example 1:Flow on Flat Plate (1) Draw the physical diagram
zx
L
Wββββ
x
x + ∆x
Direction of shear transport
Example 1:Flow on Flat Plate (2) Momentum balance
( )WL xz xτ ( )WL xz x x
τ+∆
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
(WL∆x)ρg cosβ- + = 0
∆x
This is the momentum balance equation for a FINITE shell of thickness ∆x
zx
ββββx
x + ∆x
What we need is an equation which is valid at ANY point!
Example 1:Flow on Flat Plate (3) Now let us make the shell as thin as possible (mathematically
this is equivalent to taking the limit of ∆x to zero)
Recall the definition of derivative
0cosgx
lim xxzxxxz
0x=βρ+
∆τ−τ
− ∆+→∆
df x
dx
f x x f x
xx
( )lim
( ) ( )=
+ −→∆
∆∆0
Example 1:Flow on Flat Plate (4) Rewrite the momentum balance equation:
The limit is:
This is the first-order differential equation in terms of shear stress. Upon deriving this equation, nothing has been said about the type of fluid. This means
that this equation is valid for both Newtonian fluid and non-Newtonian fluid.
0cosgx
lim xxzxxxz
0x=βρ+
∆τ−τ
− ∆+→∆
0cosgdx
d xz =βρ+τ−
Example 1:Flow on Flat Plate (5) Now this is the place where you have to specify the fluid. If the fluid is
Newtonian, the Newton law can be applied
Substitute this into the balance equation:
The final equation is a second-order differential equation in terms of velocity
τ µxzzdv
dx= −
0cosgdx
d xz =βρ+τ−d v
dx
gz2
2= −
ρ βµcos
Example 1:Flow on Flat Plate (6) Having the required balance equation in differential form, we need to
specify the boundary conditions. This is achieved by considering the constraint at the boundaries of the system
At Solid-Fluid interface, the velocity is zerox
x = δ; v = 0
At Gas-Liquid interface, the shear stress is zero
x = 0; τxz = 0
x = 0; -µ dvz/dx = 0
Example 1:Flow on Flat Plate (7) Thus the balance equation and its boundary conditions are:
Integrate the balance equation with respect to x once
Integrate it one more time
1z Cx
cosg
dx
dv +
µβρ−=
d v
dx
gz2
2= −
ρ βµcos
x = 0; dvz/dx = 0
x = δ; vz = 0
21
2
z CxC2
xcosgv ++
µβρ−=
The solution for velocitydistribution contains twoconstants of integration
They can be found fromthe two boundary conditions
Example 1:Flow on Flat Plate (8) The velocity distribution is
Apply the first boundary condition:
Apply the second boundary condition:
21
2
z CxC2
xcosgv ++
µβρ−= 1
z Cxcosg
dx
dv +
µβρ−=
x = 0; dvz/dx = 0 1C)0(cosg
0 +
µβρ−= 0C1 =
x = δ; vz = 0 2
2
C2
cosg0 +δ
µβρ−=
2
cosgC
2
2
δ
µβρ=
Example 1:Flow on Flat Plate (9) Now we know the two constants of integration, C1 and C2. The final
solution for the velocity distribution is:
Knowing the velocity distribution, you can obtain: The maximum velocity
The average velocity
The volumetric flow rate
The shear stress that the fluid exerts on the plate
v xg x
z ( )cos
=
−
ρ δ βµ δ
2 2
21 Fundamental variable
Example 1:Flow on Flat Plate (10) Maximum Velocity
occurs at the gas-liquid interface, x = 0 (can be easily proved with calculus)
µβδρ=
δ−
µβδρ==
==
2
cosgv
x1
2
cosg)x(vv
2
max
0x
22
0xzmax
Example 1:Flow on Flat Plate (11) Average velocity:
is defined as a velocity when multiplied by the cross-sectional area will give the volumetric flow rate.
Thus, to obtain the average velocity we must obtain the volumetric flow rate first.
Example 1:Flow on Flat Plate (12) Volumetric Flow Rate:
Have to start from the first principles (shell!!!)
The flow rate through the differential area of W∆x is
v xg x
z ( )cos
=
−
ρ δ βµ δ
2 2
21
x
x + ∆x
vz(x)
( ) ( )
δ−
µβδρ∆=×∆=
22
z
x1
2
cosgxW)x(vxWdQ
Example 1:Flow on Flat Plate (13) The flow rate through the differential area (W∆x) is
Thus the total flow rate through the cross section area of the film is simply the integration of the differential flow rate dQ with respect to x, from x = 0 to x = δ
( ) ( )
δ−
µβδρ∆=×∆=
22
z
x1
2
cosgxW)x(vxWdQ
( )
µβδρ=
δ−
µβδρ== ∫∫
δ
3
cosgWQ
x1
2
cosgWdxdQQ
3
0
22
Example 1:Flow on Flat Plate (14) Now we know the volumetric flow rate. The average velocity is simply the
volumetric flow rate divided by the cross-sectional area (which is Wδ)
Thus the average velocity is two third of the maximum velocity
µβδρ=
δ
µβδρ
=δ
=3
cosg
W
3cosgW
W
Qv
2
3
ave
µβδρ=
2
cosgv
2
max
Example 1:Flow on Flat Plate (15) The force that the liquid film exerts on the plate:
First we calculate the shear stress (force per unit area) at the plate (i.e. x = δ). The area of the plate is WL
Apply the Newton law
δ=δ=
µ−=τ=x
zxxz dx
dv
A
F
x
Example 1:Flow on Flat Plate (15) The force that the liquid film exerts on the plate:
First we calculate the shear stress (force per unit area) at the plate (i.e. x = δ). The area of the plate is WL
Apply the Newton law
δ=δ=
µ−=τ=x
zxxz dx
dv
A
F
v xg x
z ( )cos
=
−
ρ δ βµ δ
2 2
21
xcosg
dx
)x(dvz
µβρ−=
Differentiate
βδρ= cos)WL(gF
Example 1:Flow on Flat Plate (16) The force that the liquid film exerts on the plate:
Volume of the liquid above the plate
Gravity force of the film downward
Force acting along the z-direction (flow direction)
This is the expected result, which can be derived without solving transport problem
Example 1:Flow on Flat Plate (17) Finally we complete the analysis of the first example.
and …
you now become an expert in Transport Phenomena analysis
Let’s practice your newly acquired expertise to the same system, but the fluid is now a non-Newtonian fluid rather than the Newtonianfluid.
Let’s do this with the Ostwald-de Waele model for non-Newtonian fluid
Non-Newtonian FluidsOstwald-de Waele model The model equation is (two parameters)
The apparent viscosity is a function of the velocity gradient
Behavior: If n < 1: pseudo-plastic
If n = 1: Newtonian
If n > 1: dilatant
τ yxx
n
xmdv
dy
dv
dy= −
− 1
τyx
-dvx/dy
n < 1
n = 1
n > 1
Example 2:Flow of a non-Newtonian fluid on a Plate (1)
Draw the physical diagram (as we have done before)
zx
L
Wββββ
x
x + ∆x
Direction of shear transport
Example 2:Flow of a non-Newtonian fluid on a Plate (2)
Momentum balance
( )WL xz xτ ( )WL xz x x
τ+∆
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
(WL∆x)ρg cosβ- + = 0
∆x
This is the momentum balance equation for a FINITE shell of thickness ∆x
Example 2:Flow of a non-Newtonian fluid on a Plate (3)
Now let us make the shell as thin as possible (mathematically this is equivalent to taking the limit of ∆x to zero)
Recall the definition of derivative
0cosgx
lim xxzxxxz
0x=βρ+
∆τ−τ
− ∆+
→∆
df x
dx
f x x f x
xx
( )lim
( ) ( )=
+ −→∆
∆∆0
Why are we doing this?
Example 2:Flow of a non-Newtonian fluid on a Plate (4) Rewrite the momentum balance equation:
The limit is:
This is the first-order differential equation in terms of shear stress. Upon deriving this equation, nothing has been said about the type of fluid because the above equation is merely a force balance equation.
0cosgx
lim xxzxxxz
0x=βρ+
∆τ−τ
− ∆+→∆
0cosgdx
d xz =βρ+τ−This equation is exactly thesame as that obtained in Example 1
Example 2:Flow of a non-Newtonian fluid on a Plate (5) Now we have to specify the fluid. If the fluid is a Ostwald-de Waele non-
Newtonian fluid, the constitutive equation that relates the shear stress and the velocity gradient is
Substitute this into the balance equation:
0cosgdx
d xz =βρ+τ−
dx
dv
dx
dvm z
1n
zxz
−
−=τ
βρ=
cosg
dxdv
dxd
mn
z
τ µxzzdv
dx= −compared to
Differential equation involves absolute sign
Example 2:Flow of a non-Newtonian fluid on a Plate (6)
Having the required balance equation in differential form, we need to specify the boundary conditions. This is achieved by considering the constraint at the boundaries of the system
At Solid-Fluid interface, the velocity is zerox
x = δ; v = 0
At Gas-Liquid interface, the shear stress is zero
x = 0; τxz = 0
1
0; 0n
z zxz
dv dvx m
dx dxτ
−
= = − =
Example 2:Flow of a non-Newtonian fluid on a Plate (7)
Thus the balance equation and its boundary conditions are:
Integrate this equation once, we get:
It is clear from the boundary condition 1 that C1 = 0
( ) 1
n
z Cxcosgdx
dvm +βρ=
x = 0; dvz/dx = 0
x = δ; vz = 0
βρ=
cosg
dx
dv
dx
dm
n
z
( ) xcosgdx
dvm
n
z βρ=
Example 2:Flow of a non-Newtonian fluid on a Plate (8)
This equation involves an absolute sign and it must be removed before the differential equation could be integrated. Recognizing that dvz/dz is negative, we can rewrite the equation as follows:
To find C2, you need to apply the second boundary condition
( ) xcosgdx
dvm
n
z βρ=
( )[ ] n/1zn/1 xcosgdx
dvm βρ=− ( ) 2
1)n/1(n/1
zn/1 C
1)n/1(
xcosgvm +
+βρ=−
+
x = δ; vz = 0
Example 2:Flow of a non-Newtonian fluid on a Plate (9) The final solution for the velocity distribution is:
Knowing the velocity distribution, you can obtain: The maximum velocity The average velocity The volumetric flow rate The shear stress that the fluid exerts on the plate
vn
n
g
m
xz
n n n n
=+
−
+ +
11
1 1 1δ ρ βδ
cos/ ( ) /
Example 2:Flow of a non-Newtonian fluid on a Plate (10)
Maximum Velocity occurs at the gas-liquid interface, x = 0 (can be easily proved with calculus)
1/ ( 1)/1
max 0
0
1/1
max
cos( ) 1
1
cos
1
n n nn
z x
x
nn
n g xv v x
n m
n gv
n m
δ ρ βδ
δ ρ β
++
==
+
= = − +
= +
Example 2:Flow of a non-Newtonian fluid on a Plate (11)
Average velocity:
is defined as a velocity when multiplied by the cross-sectional area will give the volumetric flow rate.
Thus, to obtain the average velocity we must obtain the volumetric flow rate first.
Example 2:Flow of a non-Newtonian fluid on a Plate (12)
Volumetric Flow Rate: Have to start from the first principles
The flow rate through the differential area of W∆x is
x
x + ∆x
vz(x)
vn
n
g
m
xz
n n n n
=+
−
+ +
11
1 1 1δ ρ βδ
cos/ ( ) /
( ) ( )1/ ( 1)/1 cos
11
n n nn
z
n g xdQ W x v W x
n m
δ ρ βδ
++ = ∆ = ∆ − +
Example 2:Flow of a non-Newtonian fluid on a Plate (13)
The flow rate through the differential area (W∆x) is
Thus the total flow rate through the cross section area of the film is simply the integration of the differential flow rate dQ with respect to x, from x = 0 to δ
( ) ( )1/ ( 1)/1 cos
11
n n nn
z
n g xdQ W x v W x
n m
δ ρ βδ
++ = ∆ = ∆ − +
( )∫∫δ ++
δ−
βρδ+
==0
n/)1n(n/11n x1
m
cosg
1n
nWdxdQQ
QnW
n
g
m
n n
=+
+
2 1
2 1 1δ ρ βcos
/
Example 2:Flow of a non-Newtonian fluid on a Plate (14)
Now we know the volumetric flow rate. The average velocity is simply the volumetric flow rate divided by the cross-sectional area (which is Wδ)
Thus the average velocity is related to the maximum velocity as:
For n < 1 (pseudoplastic fluids), the average velocity is getting closer to the maximum velocity
1n2
1n
v
v
max
ave
++=
n/11n
n/11n2
ave m
cosg
1n2
n
W
mcosg
1n2nW
W
Qv
βρδ+
=δ
βρδ+
=δ
=+
+
Example 2:Flow of a non-Newtonian fluid on a Plate (15)
The force that the liquid film exerts on the plate: First we calculate the shear stress (force per unit area) at the plate (i.e. x = δ).
The area of the plate is WL
Apply the Ostwald-de Waele model
βδρ= cos)WL(gF
δ
−
δ
−=τ=
dx
dv
dx
dvm
A
F z
1n
zxz
( )[ ] n/1zn/1 xcosgdx
dvm βρ=−
The force that the liquid film exerts on the plate:
Example 2:Flow of a non-Newtonian fluid on a Plate (16)
Volume of the liquid above the plate
Gravity force of the film downward
Force acting along the z-direction (flow direction)
Simple Examples so far The two examples, thus far, are simple,
BUT …
Their results can be useful in solving more complex problems, such as Drainage of liquid film from a vertical plate
That will be done later, but next we will consider another simple example of flow through a circular tube.
Example 3:Flow of Liquid through a Circular Tube (1) Flat plate:
Laminar flow
No end effects
Rectangular
Gravityas the driving force
Circular tube: Laminar flow
No end effects
Cylindrical
Pressureforce & gravity force as the driving force
Example 3:Flow of Liquid through a Circular Tube (2)
Example 3:Flow of Liquid through a Circular Tube (2) Draw the physical diagram
rr+∆r
z z+∆z
0
r
z
L
R
Example 3:Flow of Liquid through a Circular Tube (3) Momentum balance
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
++
= 0
( ) ( )z,rzr2 rzτ∆π ( )[ ] ( )z,rrzrr2 rz ∆+τ∆∆+π
( ) ( )[ ] ( )z,rvz,rvrr2 zz ρ∆π ( ) ( )[ ] ( )zz,rvzz,rvrr2 zz ∆+ρ∆+∆π
( ) ( )( ) ( )( ) βρ∆∆π
+∆+∆π−∆π
cosgzrr2
zzprr2
zprr2+ -
vz(r,z)
z z+∆z
Example 3:Flow of Liquid through a Circular Tube (3) Momentum balance
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
( ) ( )( ) ( )( ) βρ∆∆π
+∆+∆π−∆π
cosgzrr2
zzprr2
zprr2
vz(r,z)
z z+∆z
Example 3:Flow of Liquid through a Circular Tube (3) Momentum balance
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
( ) ( )z,rzr2 rzτ∆π
( ) ( )[ ] ( )z,rvz,rvrr2 zz ρ∆π+
vz(r,z)
z z+∆z
( )d mvF
dt=
Example 3:Flow of Liquid through a Circular Tube (3) Momentum balance
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
( ) ( )2 ,rzr r z r r zπ τ + ∆ ∆ + ∆
( ) ( ) ( )2 , ,z zr r v r z z v r z zπ ρ ∆ + ∆ + ∆
+
vz(r,z)
z z+∆z
( )d mvF
dt=
Example 3:Flow of Liquid through a Circular Tube (4) Mathematical convention
( )[ ] ( ) ( ) ( )rrrrzrz z,rzr2z,rrzrr2
∆+=τ∆π=∆+τ∆∆+π
( ) ( )[ ] ( ) ( ) ( )[ ] ( )zzzzzzz z,rvz,rvrr2zz,rvzz,rvrr2
∆+=ρ∆π=∆+ρ∆+∆π
( ) ( )2 ,rzr r z r r zπ τ + ∆ ∆ + ∆
( ) ( ) ( )2 , ,z zr r v r z z v r z zπ ρ ∆ + ∆ + ∆
+
Rate of momentum out
Example 3:Flow of Liquid through a Circular Tube (5)
Momentum balance
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
++
= 0
( ) ( )z,rzr2 rzτ∆π ( )[ ] ( )z,rrzrr2 rz ∆+τ∆∆+π
( ) ( )[ ] ( )z,rvz,rvrr2 zz ρ∆π ( ) ( )[ ] ( )zz,rvzz,rvrr2 zz ∆+ρ∆+∆π
( ) ( )( ) ( )( ) βρ∆∆π
+∆+∆π−∆π
cosgzrr2
zzprr2
zprr2+ -
vz(r,z)
Example 3:Flow of Liquid through a Circular Tube (6)
Momentum balance
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
++
= 0
( ) ( )rrrz z,rzr2
=τ∆π ( )[ ] ( )z,rrzrr2 rz ∆+τ∆∆+π
( ) ( )[ ] ( )zzzz z,rvz,rvrr2
=ρ∆π ( ) ( )[ ] ( )zz,rvzz,rvrr2 zz ∆+ρ∆+∆π
( ) ( )( ) ( )( ) βρ∆∆π
+∆+∆π−∆π
cosgzrr2
zzprr2
zprr2+ -
vz(r,z)
Example 3:Flow of Liquid through a Circular Tube (7)
Momentum balance
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
++
= 0
( ) ( )rrrz z,rzr2
=τ∆π ( ) ( )
rrrrz z,rzr2∆+=
τ∆π
( ) ( )[ ] ( )zzzz z,rvz,rvrr2
=ρ∆π ( ) ( )[ ] ( )
zzzzz z,rvz,rvrr2∆+=
ρ∆π
( ) ( )( ) ( )( ) βρ∆∆π
+∆+∆π−∆π
cosgzrr2
zzprr2
zprr2+ -
vz(r,z)
Example 3:Flow of Liquid through a Circular Tube (8)
Momentum balance
Rate of
momentum in
Rate of
momentum out
All forces acting
on the system
−
+
= 0
++
= 0
( ) ( )rrrz z,rzr2
=τ∆π ( ) ( )
rrrrz z,rzr2∆+=
τ∆π
( ) ( )[ ] ( )zzzz z,rvz,rvrr2
=ρ∆π ( ) ( )[ ] ( )
zzzzz z,rvz,rvrr2∆+=
ρ∆π
( ) ( )( ) ( )( ) βρ∆∆π
+∆+∆π−∆π
cosgzrr2
zzprr2
zprr2+ -
vz(r,z)
2π∆r∆zThickness of shell
Example 3:Flow of Liquid through a Circular Tube (9)
Momentum balance
Now take the limit when the shell shrinks to zero, i.e.
= 0
( ) ( )r
z,rrz,rrrrrzrrrrz
∆τ−τ
− =∆+=( ) ( )
z
z,rvrz,rvrzz
2zzzz
2z
∆
ρ−ρ− =∆+=
( ) ( )βρ+
∆−
− =∆+= cosgrz
zprzprzzzzz
vz(r,z)
( ) ( )0cosgr
dz
dpr
z
vrr
r
2z
rz =βρ+−∂ρ∂−τ
∂∂−
∆r
means the equation is valid at any point
Example 3:Flow of Liquid through a Circular Tube (10) Momentum balance
Since the fluid is assumed incompressible, the velocity is independent of axial distance z
vz(r,z)
( ) ( )0cosgr
dz
dpr
z
vrr
r
2z
rz =βρ+−∂ρ∂−τ
∂∂−
( ) 0cosgrdz
dprr
r rz =βρ+−τ∂∂−
Shear transport(friction)
Pressure driving force
Gravity driving force
Example 3:Flow of Liquid through a Circular Tube (11)
Boundary conditions
At the surface of the tube, the velocity is zero
r = R; v = 0
At the center of the tube,shear is zero (i.e. the gradientof velocity is zero)
r = 0; dv/dr = 0
Example 3:Flow of Liquid through a Circular Tube (12) The momentum balance equation is
Integrate this equation once
Applying the boundary condition: we get C1 = 0
Therefore the equation is
( ) 0cosgrdz
dprr
r rz =βρ+−τ∂∂−
0Ccosg2
r
dz
dp
2
rr 1
22
rz =+βρ+−τ−
r = 0; dv/dr = 0
0cosg2
r
dz
dp
2
rr
22
rz =βρ+−τ− Is C1 always zero?
Example 3:Flow of Liquid through a Circular Tube (13)
the momentum balance equation is
To proceed further, you need to specify the type of fluid. Let us start with the Newtonian fluid
τ µxzzdv
dx= −
0cosg2
r
dz
dp
2
rr
22
rz =βρ+−τ− 0cosg2
r
dz
dp
2
rrz =βρ+−τ−
0cosg2
r
dz
dp
2
r
dr
dvz =βρ+−µ
0Ccosg4
r
dz
dp
4
rv 2
22
z =+βρ+−µ
Integrate
Example 3:Flow of Liquid through a Circular Tube (14) …the momentum balance equation is:
Since the boundary condition has been used before in determining the constant C1, you have to use the remaining boundary condition to determine the constant C2.
0Ccosg4
r
dz
dp
4
rv 2
22
z =+βρ+−µ
r = 0; dv/dr = 0
r = R; v = 0
( ) 0Ccosg4
R
dz
dp
4
R0 2
22
=+βρ+−µ
βρ−= cosg4
R
dz
dp
4
RC
22
2
Example 3:Flow of Liquid through a Circular Tube (15) ….the solution for the velocity is
Simplifying the equation
But the pressure gradient is
The solution for the velocity distribution is
βρ+−βρ−=µ cosg4
R
dz
dp
4
Rcosg
4
r
dz
dp
4
rv
2222
z
βρ+−
−=µ cosgdz
dp
4
rRv
22
z
L
PP
dz
dp 0L −=
βρ+−
µ−= cosg
L
PP
4
rR)r(v L0
22
z
Drivingforce
Pressureforce
Gravityforce
No driving force=no flow
Example 3:Flow of Liquid through a Circular Tube (16) So the final solution for the velocity distribution is
From this, you can calculate the maximum velocity the volumetric flow rate the average velocity the force that the fluid exerts on the tube’s surface
βρ+−
µ−= cosg
L
PP
4
rR)r(v L0
22
z
Example 3:Flow of Liquid through a Circular Tube (16) So the final solution for the velocity distribution is
To simplify the symbols, let’s define a new pressure variable to include the combined effect of static pressure and gravitational force
So
βρ+−
µ−= cosg
L
PP
4
rR)r(v L0
22
z
'zgP'P ρ+=
−
µ−=
L
'P'P
4
rR)r(v L0
22
z
( )0gP'P 00 ρ+=
( )β−ρ+= cosLgP'P LL
β
z’ = 0
z’ = - Lcosβ
Example 3:Flow of Liquid through a Circular Tube (17) The maximum velocity is the velocity at the center of the tube
βρ+−
µ−= cosg
L
PP
4
rRv L0
22
max
0
βρ+−
µ= cosg
L
PP
4
Rv L0
2
max
Example 3:Flow of Liquid through a Circular Tube (18)
Volumetric flow rate is obtained by applying the shell principle
The differential flow rate through a differential area at r with a thickness of dr is
Therefore the volumetric flow rate is
rr+∆r
vz(r)
( ) )r(vrdr2dQ zπ=
( )∫∫ π==R
0
z )r(vrdr2dQQ( )
L8
R'P'PQ
4L0
µ−π=
The most celebrated equation
Hagen-Poiseuille
Example 3:Flow of Liquid through a Circular Tube (19)
Now you know the volumetric flow rate. The average velocity is simply the ratio of the volumetric flow rate and the cross-sectional area of the tube
If you prefer to write the average velocity in terms of the differential pressure gradient, it is
( )( )
L8
R'P'P
RL8
R'P'P
R
Qv
2L0
2
4L0
2average µ−=
πµ
−π
=π
=
µ−=
8
R
dz
dPv
2
average
Example 3:Flow of Liquid through a Circular Tube (20) Force acting on the tube’s surface. This can be obtained as the product of
the shear stress and the surface area of the tube
The shear stress at the surface can be obtained from the Newton’s law
HenceRr
zRrrz dr
dv
==
µ−=τ
( )RrrzRrrz RL2AF
==τ×π=τ×=
βρ+−
µ−= cosg
L
PP
4
rR)r(v L0
22
z
( )L2
R'P'P L0−=
( )L02 'P'PRF −π=
02
0 PRF π=L
2L PRF π=
This is expected
Example 3:Flow of Liquid through a Circular Tube (21) Summary
Hagen-Poiseuille equation The most celebrated equation in fluid flow
( )L8
R'P'PQ
4L0
µ−π=
( )L8
R'P'Pv
2L0
average µ−=
Recap Flow over a flat plate
Newtonian fluids
Non-Newtonian fluids
Flow through a circular tube
Newtonian fluid
How about non-Newtonian fluid??