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


Mass BC of 1st kind

BC of 2nd kind

BC of 3rd kind

BC of 4th kind


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


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


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 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


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


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


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


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


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

−

= = − =


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=


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/ ( ) /


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

++

==

+

= = − +

= +


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.


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

++ = ∆ = ∆ − +


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

/


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

=+

+


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:


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


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


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=


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


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)


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)


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


=ρ∆π ( ) ( )[ ] ( )

zzzzz z,rvz,rvrr2∆+=

ρ∆π

( ) ( )( ) ( )( ) βρ∆∆π

+∆+∆π−∆π

cosgzrr2

zzprr2

zprr2+ -

vz(r,z)


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


=ρ∆π ( ) ( )[ ] ( )

zzzzz z,rvz,rvrr2∆+=

ρ∆π

( ) ( )( ) ( )( ) βρ∆∆π

+∆+∆π−∆π

cosgzrr2

zzprr2

zprr2+ -

vz(r,z)

2π∆r∆zThickness of shell


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


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


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?


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


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


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??

momentum transfer - 1 (4009)

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