lecture in transport in arbitrary continua

Transport in Arbitrary Continua (Application of General Transport Equations)

Chapters 3, 11, and 19 of BSL (2nd ed)

Equations of change for isothermal, nonisothermal, and mixture systems

What we have discussed so far…

• Introduction – what is transport? Examples; Why transport?

• Molecular transport – flux, driving force, and resistance; Newton’s law of viscosity; Fourier’s law of heat conduction; Fick’s law of binary diffusion

• Shell balances – profiles of τ and v; q and T; NA (nA, JA, jA); transport in simple systems; transport w/ generation

What we should cover for the next 3-5 lectures…

• Chapter 3 – Equations of change for isothermal systems (equation of continuity; equation of motion; application to problems)

• Chapter 11 – Equations of change for nonisothermal systems (energy equation)

• Chapter 19 – Equations of change for multicomponent systems (equation of continuity for a multicomponent mixture)

THE EQUATIONS OF CHANGE FOR ISOTHERMAL SYSTEMS

Chapter 3 of BSL 2nd ed

It is tedious to set up a shell balance for each problem that we want to solve.

• General mass balance and general momentum balance that can be applied to any problem

• Derivations of “The equation of continuity” and “The equation of motion” for isothermal systems

• The derived equations in terms of substantial derivative

• Use the equations to solve flow problems

The equation of continuity - 1 Mass balance over a volume element:

(Rate of increase of mass) = (rate of mass in) – (rate of mass out)

Rate of mass in and mass out in x direction:

xxx

xx

vzy

vzy

|x at x out

| at xin

Rate of increase of mass within the element:

tzyx

zzzzz

yyyyy

xxxxx

vvyx

vvxz

vvzyt

zyx

zyx v

zv

yv

xt

v

t

Time rate of change of the fluid density at a fixed point in space

The equation of continuity – 2

v

t

Rate increase of mass per unit volume

Net rate of mass addition per unit volume by convection

Divergence of ρv v

For a fluid of constant density (incompressible fluid):

0 v

The equation of motion - 1 Momentum balance over a volume element:

(Rate of increase of momentum) = (rate of momentum in) – (rate of momentum out) + (external force on the fluid) This is a VECTOR equation

Rates at which x-component of momentum enters and leaves the volume element:

zzzxzzxyyyxyyxxxxxxxx yxxzzy

ij

Meaning of terms:

-Total momentum flux (convective + molecular transport) - Subscript j tells us what velocity component is influencing the momentum transport at direction i

External force (usually gravity) acting on fluid element; x component is: zyxgx

The equation of motion - 2 Rate of increase of x momentum within the volume element: tvzyx x

xzzzxzzx

yyyxyyx

xxxxxxxx

gxyxyx

xz

zyt

vzyx

x component momentum balance over volume element:

xzxyxxxx gzyx

vt

y and z components of momentum balance:

yzyyyxyy gzyx

vt

zzzyzxzz gzyx

vt

The equation of motion - 3

xzxyxxxx gzyx

vt

yzyyyxyy gzyx

vt

zzzyzxzz gzyx

vt

gv

t

If the total momentum flux is: τvv p

gvvv

p

t

Rate of increase of momentum per unit volume

Rate of momentum addition by convection per unit volume

Rate of momentum addition by molecular transport per unit volume

External force on fluid per unit volume

The equations of change in terms of the substantial derivative - 1

The Partial Time Derivative ∂/ ∂t – time rate of change of a property (e.g. concentration) at a fixed location (see example on the right) zyxt

c

,,

The Total Time Derivative d/ dt – time rate of change of a property (e.g. concentration) when the observer is moving

tyxtzxtzyzyx z

c

dt

dz

y

c

dt

dy

x

c

dt

dx

t

c

dt

dc

,,,,,,,,

The Substantial Time Derivative D/ Dt – time rate of change of a property (e.g. concentration) when the observer is moving with the same velocity as the stream, fluid or substance of velocity v (derivative following the motion)

cvvvv zyx

t

c

z

c

y

c

x

c

t

c

Dt

Dc

The equations of change in terms of the substantial derivative - 2

zyx v

zv

yv

xt

v

t

zyxzyx

zyxzyx

vvvvvv

vvvvvv

zyxzyxt

zyxzyxt

Equation of continuity

v

Dt

D

Equation of motion gτv

pDt

D

Equation of continuity

Three most common simplifications of the equation of motion

gτv

pDt

D

1. Constant density ρ and viscosity μ gvv

2pDt

D

Popularly known as Navier-Stokes equation; starting point for describing isothermal flows of gases and liquids

2. Acceleration term neglected in Navier-Stokes equation gv 20 p

Also called Stokes flow equation/creeping flow equation; when flow is extremely slow - Used in lubrication theory, particle motion in suspension, flow through porous media; swimming of microbes, etc

3. Viscous forces are neglected gv

pDt

D

Also called Euler equation for “inviscid” fluids (ideal fluids with no viscosity) - Used to describe flow around airplane wings (except near solid boundary); flow of ocean currents; etc

How to use the equations of change to solve flow problems

To describe the flow of a Newtonian fluid (determine completely the pressure, density, and velocity distributions in the fluid) at constant temperature, you’ll need:

Equation of continuity v

t

Equation of motion gvvv

p

t

Usual procedure is substitute expression for τ (Newton’s law of viscosity) to equation of motion

The components of τ will discuss later the general form of Newton’s Law of Viscosity

The equations of state ChE 122; p = p(ρ)

The equations of the viscosities as function of density

Boundary and initial conditions

Equations of change are also written in polar and spherical coordinates

Equation of motion in terms of τ (Appendix B.5) and for a Newtonian fluid with constant density ρ and viscosity μ (Appendix B.6)

Equations of change are also written in polar and spherical coordinates

Generalization of Newton’s Law of Viscosity - 1

Newton’s law of viscosity (simplest form)

dy

dvxyx

Assumptions: -vx is a function of y alone - vy and vz are zero

What if the other two velocity components are not negligible? The velocity components depend on all three coordinates and time? What is τ?

tzyx ,,,xx vv

tzyx ,,,yy vv

tzyx ,,,zz vv

Generalization of Newton’s Law of Viscosity - 2

The viscous stress tensor τ has nine components (three normal stresses τxx, τyy, and τzz and six shear stresses τxy = τyx, τxz = τzx, τyz = τzy); in terms of viscosity μ and dilatational viscosity κ

ijzyx

j

i

i

j

ijz

v

y

v

x

v

x

v

x

v

3

2

δvvvτ

3

2T

Example of usage:

z

v

y

v

x

v

z

v

y

v

x

v

x

v

z

v

y

v

x

v

x

v

x

v

x

v

y

v

z

v

y

v

x

v

x

v

y

v

zyx

zyxxxx

zyxxxxx

yxyx

zyxyxyx

3

2

3

4

3

22

3

2

3

2

Generalized Newton’s Law of Viscosity in Cartesian Coordinates (B.1)

Generalized Newton’s Law of Viscosity in Cylindrical Coordinates (B.1)

Generalized Newton’s Law of Viscosity in Spherical Coordinates (B.1)

Example: Flow of a Falling Film

Find shear stress and velocity distribution profiles using equations of change.

First step: Write down your postulates.

Example: Flow of a Falling Film - 1

Postulates:

Steady-state flow with constant density ρ and viscosity μ

x and y components of velocity are zero (meaning of δz); vz doesn’t depend on y (film infinitely wide in y axis)

zxvzz ,δv

Next, simplify the tabulated equations of Generalized Newton’s Law of Viscosity, continuity, and motion given in Appendix B.


0

z

vz0

v

z

v

y

v

x

v zyx

xvzzδv


Tabulated Newton’s Law of Viscosity in Cartesian coordinates

0

0

0

0

0

0

x

vzxz


0

0

0

cos0

0

sin0

gxz

p

y

p

gx

p

xz

Example: Flow of a Falling Film - 5 Solve the simplified equations using initial and boundary conditions.

cos0

0

sin0

gxz

p

y

p

gx

p

xz

constantsin gxp

Constant is an arbitrary function in terms of y and z Constant is not a function of y (since dp/dy = 0)

If pressure in the gas phase is very nearly constant at the prevailing atmospheric pressure patm, then at the gas liquid interface x = 0, p = patm

atmpgxp sin

Pressure distribution in the falling film


atmpgxp sin 0

z

p

cos0

0

sin0

gxz

p

y

p

gx

p

xz

cosgx

xz

Same differential equation that you get from shell momentum balance!

BC1: x = 0 τxz = 0 cosgxxz

x

vzxz

Derived earlier from Generalized Newton’s Law of Viscosity

xdxg

dvz

cos

BC2: x = δ vz = 0

22

12

cos

xgvz

Example: Flow of a Falling Film - 7 In solving the velocity profile in the falling film, we can also use the Navier-Stokes equation (a simplification of Equation of Motion):

cos02

2

gx

vz

21

2

1

2

cos

cos

CxCxg

v

Cxg

x

v

z

z

0

0

0

Example: Flow Through a Circular Tube

Find shear stress and velocity distribution profiles using equations of change.

Example: Flow Through a Circular Tube - 1

Postulates:

Steady-state flow with constant density ρ and viscosity μ

no radial and tangential flows (vr and vθ = 0); vz doesn’t depend on θ

zrvzz ,δv


0

z

vz

Equation of continuity 0

0

0

0

0

0

r

vzrz

Generalized Newton’s Law of Viscosity


rzrzz rrrz

Pr

rrzgp

z

Pp

r

r

P

r

p

110

10

0 Equation of motion

Modified pressure P = p + ρgh h – upward height opposite to gravity

Since flow is downward, h = -z


rzrzz rrrz

Pr

rrzgp

z

Pp

r

r

P

r

p

110

10

0Modified pressure P is only a function of z orz Cr

dr

d

rdz

dP

1

1CzCP o BC1: z = 0 P = P0

BC2: z = L P = PL

00

01

00

PzL

PPP

PC

L

PPC

L

L

L

PPr

dr

d

r

Lrz

01

32

20

20

ln4

2

CrCrL

PPv

r

Cr

L

PP

Lz

Lrz

Generalized Newton’s Law of

Viscosity

r

vzrz

Distribution of modified pressure P inside the

circular tube

We solved these already. Review the BCs to solve the shear

stress and velocity distribution.

Final notes before moving to the next topic…

• Choose the proper coordinates (Cartesian, polar, or spherical?)

• Write reasonable postulates. Is the problem 1D, 2D, or 3D? Steady or transient? Is the mass density and/or viscosity constant?

• Equation of continuity Generalized Newton’s Law of Viscosity Equations of motion Solve the differential equations

THE EQUATIONS OF CHANGE FOR NONISOTHERMAL SYSTEMS


What we have discussed so far…

• Derivation of equation of continuity (conservation of mass) and equation of motion (conservation of momentum)

• Generalized Newton’s Law of Viscosity

• Use of equations of change to determine the shear stress and velocity profiles (also pressure!) of fluid in isothermal systems only

How about nonisothermal systems?

The equation of change for energy Energy balance (kinetic + internal) over a volume element:

{Rate of increase of kinetic and internal energy} = {net rate of kinetic and internal energy addition by convective transport} + {net rate of heat addition by molecular transport (conduction)} + {rate of work done on system by molecular mechanisms} + {rate of work done on system by external forces}

Rate of increase of kinetic and internal energy:

Uvt

zyx 2

2

1

Net rate of kinetic and internal energy addition by convection and conduction; rate of work done on system by molecular transport (shear stress and pressure forces):

zzzzzyyyyyxxxxx eeyxeezxeezy

e is the combined energy flux:

qvπve

Uv 2

2

1

π is the total molecular momentum flux:

τδπ p

The equation of change for energy - 1 Rate of work done on system by external forces (gravity):

zzyyxx gvgvgvzyx

zzyyxx

zzzzzyyyyyxxxxx

gvgvgvzyx

eeyxeezxeezyUvt

zyx

2

2

1

zzyyxxzyx gvgvgv

z

e

y

e

x

eUv

t

2

2

1

gve

Uvt

2

2

1

The equation of change for energy - 2

gve

Uvt

2

2

1

qvπve

Uv 2

2

1

τδπ p

gvvτvqv

pUvUvt

22

2

1

2

1

rate of increase

of energy per unit volume

rate of energy

addition per unit

volume by convective transport

rate of energy

addition per unit

volume by heat

conduction

rate of work done on fluid per

unit volume by pressure

forces

rate of work done

on fluid per unit

volume by viscous forces

rate of work done on fluid per

unit volume by

external forces

Where’s temperature T

in the equation??

We need the form of the energy equation in which temperature appears.

gvvv

p

t

Equation of motion

v

t

Perform a dot product (scalar product) with velocity vector v; some lengthy rearrangements;

apply equation of continuity

gvvτvτvvv

:

2

1

2

1 2 ppvvt

2

rate of increase of kinetic

energy per unit volume

rate of addition of kinetic energy by convection per

unit volume

rate of work done by

pressure of surroundings on the fluid

rate of reversible

conversion of kinetic

energy into internal energy

rate of work done by

viscous forces on the fluid

rate of irreversible conversion

from kinetic to internal

energy

rate of work by external force on the

fluid

The equation of change for kinetic energy


gvvτvτvvv

gvvτvqv

:2

1

2

1

2

1

2

1

2

22

ppvvt

pUvUvt

2

Equation of change for energy – Equation of change for kinetic energy =

Equation of change for internal energy vτvqv

:pUUt

rate of increase in internal

energy per unit volume

net rate of addition of

internal energy by convective transport, per unit volume

rate of internal energy

addition by heat

conduction, per unit volume

reversible rate of

internal energy

increase per unit volume

by compression

irreversible rate of

internal energy

increase per unit volume by viscous dissipation


vτvqv

:pUUt

in terms of substantial derivatives

vτvq

:pDt

UD

in terms of enthalpy and

using continuity equation

pHVpHU

Dt

D

Dt

HD

vτq :

Equation of change for internal energy


dpT

VTVdTCdp

p

HdT

T

HHd

p

p

Tp

Dt

D

Dt

HD

vτq : Equation of change for internal energy in terms of enthalpy

From classic thermodynamics (ChE 122):

Dt

Dp

T

VTV

Dt

DTC

Dt

HD

p

p

Dt

Dp

TT

Dt

DTC

Dt

HD

p

p

11

Dt

Dp

TTDt

DTC

Dt

HD

p

p

1

11

Dt

Dp

TDt

DTC

Dt

HD

p

p

ln

ln1

Dt

Dp

TDt

DTC

p

p

ln

ln:

vτq Equation of change for

temperature (our goal! )

How to use the equation of change for temperature

Dt

Dp

TDt

DTC

p

p

ln

ln:

vτq

Tk

Tk

2

q when Fourier’s Law is used

when Fourier’s Law is used and k is constant

vv vτ : when Generalized Newton’s Law of Viscosity is used

describes the degradation of mechanical energy to thermal energy that occurs in all flow systems (viscous dissipation heating) such as lubrication, rapid extrusion, rapid flight

22

3

2

2

1:: vvvτ-vτ

i j

ij

i

j

j

i

x

v

x

v

How to evaluate viscous dissipation heating term

vv

i j

ij

i

j

j

i

x

v

x

v

2

2

3

2

2

1:: vvvτ-vτ

See Table A.7 for this

Special restricted versions of the equation of change for temperature

Fourier’s law with constant k; omit the viscous dissipation term (only important in flows with enormous velocity gradients)

1. For an ideal gas, (∂ ln ρ / ∂ ln T)p = - 1 Dt

DpTk

Dt

DTCp

2

If Cp – Cv = R, and the equation of state is pM = ρRT, and using the equation of continuity:

v

TkDt

DTCv

2

2. For a fluid flowing in a constant pressure system, Dp / Dt = 0 TkDt

DTCp

2

3. For a fluid with constant density, (∂ ln ρ / ∂ ln T)p = 0 TkDt

DTCp

2

4. For a stationary solid, v is zero Tkt

TCp

2

The equation of change for temperature in terms of q

The equation of change for temperature for pure Newtonian fluids w/ constant ρ and k

Example: Steady-State Forced Convection Heat Transfer in Laminar Flow in a Circular Tube

A viscous fluid with physical properties (μ, k, ρ, Cp) assumed constant is in laminar flow in a circular tube of radius R. For z < 0 the fluid temperature is uniform at the inlet temperature T1. For z > 0 there is a constant radial heat flux qr = -q0. Set up the differential equations that will solve for the temperature profile.

Example: Steady-State Forced Convection Heat Transfer in Laminar Flow in a Circular Tube - 1

Earlier postulates and results from shell balance and simplification of equations of continuity and motion:

rvzzδv zPP pressure modified

Equation of continuity: 0

v

z

v

y

v

x

v zyx 0

z

vz

Equation of motion: 011

dr

dvr

dr

d

rdz

dPr

dr

d

rdz

dP zrz

00 Pz

L

PPP L

22

0 14 R

r

L

RPPv L

z

pressure profile

velocity profile


How to simplify equation of change for temperature? Use the equation of change for temperature for pure Newtonian fluids w/ constant ρ and k. Assume T = T(r,z).

Equation of energy:

2

2

21

r

v

z

T

r

Tr

rrk

z

TvC z

zp


2

2

21

r

v

z

T

r

Tr

rrk

z

TvC z

zp Equation of energy:

Two more assumptions: 1) In the z direction, heat conduction is much smaller than heat convection. 2) Flow is not sufficiently fast that viscous heating is significant.

heat convection in z direction

heat conduction in r and z directions

viscous dissipation heating term

r

Tr

rrk

z

T

R

r

L

RPPC L

p

11

4

22

0

Boundary conditions:

1

00

0:BC3

or :BC2

finite is0 :BC1

TTz

qr

TkqqRr

Tr

r

Analytical or numerical solution


We can put the problem in dimensionless form to minimize the number of parameters in the final problem formulation.

kR

L

RPPC

z

R

r

kRq

TT

Lp

22

00

1

4

11 2

00:BC3

11:BC2

finite is0 :BC1

Example: Steady flow in a nonisothermal film

A liquid is flowing downward in steady laminar flow along an inclined plane surface. The free liquid surface is maintained at temperature T0, and the solid surface at x = δ is maintained at Tδ. At these temperatures the liquid viscosity has values µ0 and µδ, respectively, and the liquid density and thermal conductivity may be assumed constant. Find the velocity distribution in this nonisothermal flow system. Neglect end effects and viscous heating. Assume the temperature dependence of viscosity is in the form of µ = Aexp(B/T), where A and B are constants.

Example: Steady flow in a nonisothermal film - 1

Postulates: T = T(x) and v = δzvz(x); steady-state and laminar flow; constant k and no viscous heating

2

2

0x

T

Equation of change for temperature (equation of energy):

211 CxCTCx

T

TTx

TTx

:BC2

0 :BC1 0

00

0102

TxTT

T

TTCTC

Temperature distribution in the nonisothermal film


The temperature dependence of viscosity can be written as:

00

11exp

TTB

T

To get viscosity µ as function of position x and temperature T:

00 Tx

TTT

x

TT

TTB

Tx

0

0

0

exp,

To approximate viscosity µ as function of position x alone, we can say T ≈ Tδ in the above equation if temperature does not change greatly through the film:

x

TT

TTB

xx

TT

TTB

Tx

0

0

00

0

0

expexp,

x

TTB

x

TT

TTB

x

00

0

0

11expexp

x

x

00


cosgxxz

xdxg

dvz

cos

Our results earlier (via shell momentum balance and

equations of change (continuity and motion))

x

x

0

0

0 :BC

cos00

z

x

z

vx

xdxg

dv

This is the differential equation that will give the

velocity profile in the flowing nonisothermal film.


A similar problem in Example 2.2-2 Falling Film with Variable Viscosity

(p. 47, BSL 2nd Ed.) 0 :BC

cos0

z

x

z

vx

xdxeg

dv

22

0

2 111cos

x

eeg

v

x

z

If α = -ln (μδ/ μ0)

0 :BC

cos00

z

x

z

vx

xdxg

dv

0

0

0

0

2

00

ln1ln1

ln

cos

xz

xgv

Final notes before going to the next topic…

• The equations of continuity and motion allow us to determine the pressure, shear stress, and velocity profiles.

• The equation of energy (equation of change for temperature) gives us the temperature profile.

• The equation of change for temperature can also be used to solve heat problems that we have discussed in the shell heat balances.

• Complete solution to a problem shows dependency of solution to space and time. What we usually offer as a solution is not complete but simplified based on our postulates.

EQUATIONS OF CHANGE FOR MULTICOMPONENT SYSTEMS


What we have covered so far…

• Derivation of equation of continuity (conservation of mass) and equation of motion (conservation of momentum) for isothermal, single component fluid systems; Generalized Newton’s Law of Viscosity; Simplification and application to problems

• Derivation of equation of energy in terms of temperature (conservation of energy) for nonisothermal, single component fluid systems; Viscous dissipation heating term; Simplification and set up of problems

• Equations of change for isothermal and nonisothermal, multicomponent systems??

Review of variables for mass transport

dy

dDj A

ABA

dy

dxcDJ A

ABA *

AABA Dj

AABA xcDJ *

molecular mass flux

molecular molar flux

BBAA vvv

BBAA vxvxv *

vA

*vcA

convective mass flux

convective molar flux

vDvjvn AAABAAAAA

*** vcxcDvcJvcN AAABAAAAA

combined mass flux

combined molar flux

The Equation of Continuity for a Multicomponent Mixture

Apply conservation of mass to each species α in a mixture of N components (α = 1, 2,… N). Let rα be the rate of production of species α (mass/volume-time).

Rate of increase of mass of α in the volume

element:

tzyx

Rate of addition of mass of α across face at x

xxnzy Rate of removal of mass

of α across face at x+∆x

xxxnzy

Rate of production of mass of α by chemical

reactions zryx

zryxnnyx

nnxz

nnzyt

zyx

zzzzz

yyyyy

xxxxx

rz

n

y

n

x

n

t

zyx

The Equation of Continuity for a Multicomponent Mixture - 2

rz

n

y

n

x

n

t

zyx

Equation of continuity for species α = 1, 2, … N

rt

n

vjn

rt

jv

Rate of increase of mass of α per unit volume

Net rate of addition of mass of α per unit

volume by convection

Net rate of addition of mass of α per unit

volume by diffusion

Net rate of production of mass of α per unit volume by reaction

If we add all equations from α = 1, 2, …N, we’ll get:

v

tEquation of continuity for the mixture

For a fluid mixture of constant mass density:

0 v

The Equation of Continuity for a Multicomponent Mixture – from mass units to molar units

rt

n

Rt

c

N

vJN c

Rct

c

Jv

Rate of increase of moles of α

per unit volume

Net rate of addition of moles of α

per unit volume by convection

Net rate of addition of moles of α

per unit volume by diffusion

Net rate of production

of moles of α per

unit volume by reaction

If we add all equations from α = 1, 2, …N, we’ll get:

N

Rct

c

1

v

Equation of continuity for the mixture

For a fluid mixture of constant molar density:

N

Rc 1

1

v

The Equation of Continuity for a Multicomponent Mixture – mass and molar fractions

rt

jv

Rct

c

Jv

Equation of continuity for species α = 1, 2,… N in mass units

Equation of continuity for species α = 1, 2,… N in molar units

r

t

jv

N

RxRxt

xc

1

Jv

Equation of continuity for species α = 1, 2,… N in mass fraction

Equation of continuity for species α = 1, 2,… N in molar fraction

Simplifications of the Equation of Continuity for a Multicomponent Mixture

Binary systems with constant ρDAB

r

t

jv AAABA

A rDt

2

v

AABA D j Diffusion in dilute liquid solutions at constant temperature and pressure

N

RxRxt

xc

1

Jv

Binary systems with constant cDAB

BAABAABAA RxRxxcDxt

xc

2v

Diffusion in low-density gases at constant temperature and pressure

Binary systems with constant cDAB ,with zero velocity, and no reaction

AAB xcD

AJ

BAABAABAA RxRxxcDxt

xc

2v

AABA cDt

c 2

Fick’s 2nd Law of Diffusion Diffusion equation

Diffusion in solids or stationary liquids

See Appendix B for full list of equation of continuity for a multicomponent mixture

Notes before solving some examples…

• The derived equation of continuity for a multicomponent mixture is applicable for isothermal/nonisothermal problems.

• We can use it to solve mass transport problems instead of using the shell mass balance approach.

• We still have the equation of motion and the equation of energy for multicomponent mixture systems.

Example: Diffusion Through a Stagnant Gas Film

Steady-state diffusion of A through stagnant gas B with

the liquid-vapor interface maintained at a fixed

position

Find the distribution profile of combined molar flux NAz

and molar fraction concentration xA in the

problem using the equation of continuity for

multicomponent mixture.

Example: Diffusion Through a Stagnant Gas Film - 2

• Write your postulates.

- The combined molar flux NAz and molar fraction concentration xA are only a function of z (NA = δzNA(z) and xA = xA(z)); 1D problem

- A has velocity components in the z direction only; B is stagnant (v* = vA = vz )

- No reaction

- Steady-state

- Constant molar density and DAB


2

1

RxRz

J

y

J

x

J

z

xv

y

xv

x

xv

t

xc AA

AzAyAxAz

Ay

Ax

A


2

1

RxRz

J

y

J

x

J

z

xv

y

xv

x

xv

t

xc AA

AzAyAxAz

Ay

Ax

A

z

J

z

xcv AzA

z

z

Jcx

z

AzA

v 0

AzA Jcz

v

convection

molecular diffusion

0

AzN

z

We arrived to this equation

during our discussion in

shell mass balances!

1CNAz

Constant distribution of

molar flux

dz

dx

x

cD

xcDNNx

JcN

A

A

AB

AABBzAzA

AzAAz

1

v

01

dz

dx

x

cD

dz

d A

A

ABWe solved this last time.

Review the boundary conditions and the technique

to solve this.

Example: Diffusion With a Homogeneous Chemical Reaction

A dissolves in liquid B in a beaker and diffuses

isothermally into the liquid phase. A undergoes an irreversible first order

homogeneous reaction: A + B AB

Rate of chemical decomposition of A is k1cA

Set up the differential equation that would determine the concentration profile of A in the given problem.

Example: Diffusion With a Homogeneous Chemical Reaction - 2

• Write your postulates.

- Binary solution of A and B; ignore small amount of AB that is present (pseudobinary assumption)

- Rate of decomposition of B and production of AB negligible

- Concentration of A is small; function of z

- Total molar concentration c and diffusion coefficient DAB are uniform and constant

- Diffusion of A in stationary liquid B


3

12

2

2

2

2

2

RxRz

x

y

x

x

xcD

z

xv

y

xv

x

xv

t

xc AA

AAAAB

Az

Ay

Ax

A

ABBAAAAAA

ABA

zA

yA

xA RRRxR

z

x

y

x

x

xcD

z

xv

y

xv

x

xv

t

xc

2

2

2

2

2

2


ABBAAAAAA

ABA

zA

yA

xA RRRxR

z

x

y

x

x

xcD

z

xv

y

xv

x

xv

t

xc

2

2

2

2

2

2

AA

AB Rz

xcD

2

2

0

0or0:BC2

0:BC1

0

0

12

2

dzdcNLz

ccz

ckz

cD

AAz

AA

AA

AB

We arrived to this same mathematical problem

formulation last time in our discussion of shell mass balances. Review your notes/textbook for the

solution of this.

Summary of the Multicomponent Equations of Change

Equation of continuity: α = 1, 2, …, N Equation of motion: 3 components (e.g. x, y, and z)

Equation of energy: single equation

Summary of the Multicomponent Equations of Change - 2

These are the expressions need to be substituted to multicomponent equations of change in Table 19.2-1.

Summary of the Multicomponent Equations of Change - 3

For example on nonisothermal binary mixtures, see Example 19.4-1 Simultaneous Heat and Mass Transport.

lecture in transport in arbitrary continua

Documents

momentum transport

x momentum

volume element

general momentum balance

xin rate of increase

general mass balance

ed equations of change

unit volume external