mass transfer boundary layer theory - eth z€¦ · h. schlichting, k. gersten, "boundary...

Boundary Layer Theory: Mass and Heat/Momentum Transfer

Mass Transfer

Lecture 11, 28.11.2019, Dr. M.Reza. Kholghy

Mass Transfer – Boundary Layer Theory 9-2

9.1 Fluid-Fluid Interfaces (lecture of 20.11.18)

9. Basic Theories for Mass Transfer Coefficients

9.2 Fluid-Solid Interfaces

Fluid-fluid interfaces are typically not fixed and are strongly affected

by the flow leading to heterogeneous systems that make it difficult

to development a general theory behind the MT correlations

(Cussler Table 8.3-2).

Solids typically have fixed and well-defined surfaces, allowing to

develop theoretical foundations for the empirical MT correlations

(Cussler Table 8.3-3). In contrast to fluid-fluid interfaces the models

are far more rigorous but, unfortunately, computationally

INTENSIVE.


In addition to this, fluid-solid interfaces have been investigated

intensely with respect to heat transfer. We can make use of this

due to the analogy between heat momentum and mass transfer.

9.2.1 MT from a plate (boundary layer theory)

MT correlation (Table 8.3-3):

3121

D

LU646.0

D

kLSh

Example: A sharp-edged, flat plate that is sparingly dissolvable is

immersed in a rapidly flowing solvent.

L: plate length; U∞: bulk fluid velocity


Prandtl first introduced the

concept of boundary layers.

Goal: Calculate the MTC for this fluid–solid interface

Literature:

R.B. Bird, W.E. Steward, E.N. Lightfoot, “Transport Phenomena”, 2nd ed. J. Wiley&Sons 2011.

H. Schlichting, K. Gersten, "Boundary Layer Theory", 8th ed., Springer 1999.

The transition from zero velocity at the plate to the velocity of the

surrounding free stream takes place in the boundary layer.


1. Calculate the velocity profile in the B.L.

2. Calculate the concentration profile in the B.L.

y 0

cj D |

y

and set it equal to kDc to obtain k.

3. Calculate the flux at the interface

Procedure:


The relative magnitude of the fluid flow (momentum) B.L. and

the concentration B.L. is given by

kinematic viscositySc

D diffusivity

Sc > 1: Momentum B.L. > Concentration B.L. (typical)

Sc = 1: Momentum B.L. = Concentration B.L.

Sc < 1: Momentum B.L. < Concentration B.L. (rare case)


Laminar B.L.

Continuity equation: 0z

w

y

v

x

u

(1)

Momentum Balance in x and y

2

2

2

2

2

2

2

2

y

v

x

v

y

p1

y

vv

x

vu

y

u

x

u

x

p1

y

uv

x

uu

at y = 0: u = 0 and v = 0

y = : u = U

= 0 (2 dim.)


For flow over a flat plate 0y

p

x

p

as we are far from the entrance point (Prandtl assumption).

Then,

2 2

2 2

smallbutnot negligiblenegligible

u u u uu v ( )

x y x y(2)

2

20

u uconst

x x

Shear stress on plate


(Negligible pressure changes in y (thin layer)

y = 0: u = 0 and v = 0

y = : u = U

Momentum equation in y (for steady flows):

2

2

2

2

y

v

x

v

y

p1

y

vv

x

vu

As v typically is small in the BL (for lam. case here =0), this reduces to:

0y

p

(3)

Boundary conditions:

(fluid adheres to plate)


Digression: How did Prandtl come up with that?

The B.L. thickness, , is related to Re by

Equations (1) and (2) are solved by introducing the variable

U

x

y

xURe~

x

21

U

x

yRe

x

y~

y 21


The two T-shirts are similar.

𝑙1

𝑙2

BL 1 and 2 are similar. They look

the same after normalization

𝛿1 𝛿2

X

Y

U

x

y


A stream function (with regard to 2 Dimensional Incompressible fluid flow) is

an arbitrary function whose derivatives give velocity components of a

particular flow situation.

The stream function is mathematically defined in such a way that lines of

constant stream function are streamlines (hence the name stream

function).

In other words, iso-stream function lines are streamlines in a fluid flow.

The stream function makes a lot of calculations and understanding of 2D

Incompressible fluid flows easy.

Reminder:What is a stream function?


The corresponding stream function is: )(fxU

Eq. (2) can be simplified by formulating it in terms of a stream

function Ψ. In order to satisfy the continuity equation (1), the velocity

components must be:

yu

xv

2

2

y

u

y

uv

x

uu

3

3

2

22

yyxyxy

Thus,

where f(η) is the dimensionless stream function.

yyu

fU

Ux

1fxU


1 2 1 21 1

2 2

U Uν f f

x x

fx

U

2

1f

U

x

yU

2

1

x

u3

fx

UU

y

u21

fx

U

y

u 2

2

2

1 2

1 23

1 1

2 2

U yν f U x f

x x x

U

and

ffx

U

2

121


Now the equation of motion (2) becomes:

0fff2

at = 0: f = 0 and f’ = 0

= ∞: f’ = 1 (see Table)

This is a third order nonlinear ordinary differential equation. A

solution in form of Blasius series expansion is available.

See attached Table from H. Schlichting (“Boundary Layer Theory”),

which gives the complete values of u and v for every x and y. The

velocity profiles have been calculated.


Source: H. Schlichting, “Boundary Layer Theory”


Blasius similarity solution of the velocity distribution in a laminar

boundary layer on a flat plate.P.K. Kundu, I.M. Cohen, “Fluid Mechanics” 2nd ed. 2002, Academic Press

U

u

x

Uy


Mass Transfer in Boundary Layers

The diffusion equation in the B.L. is:

21

2

21

211

y

c

x

cD

y

cv

x

cu

B.C.’s: y = 0: c1 = c10

y = : c1 = c1

Introduce dimensionless variables:

U

vv,

U

uu,

cc

ccc **

101

101*1

21

2

negligibleusually

21

211

y

c

x

c

U

D

y

cv

x

cu

B.C.s:

y = 0: c1*=0

y= : c1*=1

(4)

(5)


Compare to the momentum equation (2):

2

2

negligibleusually

2

2

y

u

x

u

Uy

uv

x

uu

B.C.’s: y = 0: u*=0 and v*=0

y= : u*=1

The two equations have the same form!

Sc = 1: In this special case where

U

D

U

the solution to the concentration profile is given by c1*=u*

or = D

The boundary layers for flow and concentration are IDENTICAL.


For this case we write Fick’s first law 0y1 |

y

cDj

101

0y

1 ccy

cDj

101

0y

101

0y

ccy

u

Ucc

y

uj

101

0

2

221

ccf

x

UU

Uj

101

Tablefromf

21

cc332.0x

Uj

with c1*:


As j = k (c10 - c1)21

x

U332.0k

21

x

U

D

x332.0

D

xkSh

2121

ReSc332.0xU

D332.0

Here, Sc = 1: 21Re332.0Sh

This is the Local Mass Transfer Coefficient to be distinguished from

the average MTC over a plate of length L (Cussler Table 8.3-3):

3121 ScRe646.0Sh

,


A more realistic case is with Sc >> 1 which is typical in liquids

and particle / gas systems.

Sc >> 1: (S.K. Friedlander, “Smoke, Dust and Haze”, 1st ed., 1977)

Examples: Dissolution of solid compounds in flowing liquid streams.

Flow (deposition) of pollutants over lakes or leaves


More specifically, if the B.L. (displacement) thickness is:

1/2

x* 1.72

U

For a wind blowing at a speed of 10 km/h, 0.56 mm.

E.g., at 2 cm from the leading edge of the flat surface:

Goal: To obtain the mass transfer coefficient

mm56.0m1033.072.13600sm10000

mkg1

s Pa1015m02.0

72.1 33

6

*

(Blasius, Schlichting)


Again at steady-state 2

*1

2*1*

*1*

y

c

U

D

y

cv

x

cu

Expand and

near the wall (η→0) into Taylor series:

fU

uu* )(f)(f

xU2

1

U

vv

21

*

a332.00...2

)0(f)0(f)0(fU

uu

2*

disregard higher order terms because

2

21

2

21

32

21

*

xU4

a

2

332.00

xU2

1

...)0(f3

1)0(f

2

1)0(f

xU2

1

U

vv

(5)

from table

1


Substitution in the above differential equation (5) gives:

Assume that c1* is a function only of η: )(gc*1

Then:

gx2

1

x

c*1

2

*

1

2*

12

21*

1

y

c

U

D

y

c

Ux4

a

x

ca

gx

U

y

g

y

c21*

1

gx

U

y

c2

*

1

2

1/2U

η = yx

with

Why?

Remember

similarity

solution


Introducing the above variables in the diffusion equation:

or

0

1g

0g

Solution: Set

0

0

gP

where

gx

U

U

Dg

x

U

Ux4

ag

x2a

21

2

212

0gD4

ag 2

with

0PSc4

aP gP 2

Integrate: Sc12

a

P

Pln 3

0

Reduce it to a

boundary value

problemSeperate

variables and

integrate


From the B.C’s.: at η = 0, g = 0:

From the B.C’s.: at η = , g = 1, hence:

Sc

12

aexpPP

g 3

0

dr rSc12

aexpPg

0

3

0

1

0

m

3

0 dr rSc12

aexpP

1/2U

η = yx

)(gc*1

* 1 101

1 10

,c c

cc c

=𝜕𝑔

𝜕𝜂𝜂=0

𝑚3 𝑑𝑚 =𝑎

12𝑆𝑐

13𝑑η


and

1 3

3η=0

0

1 3Γ(1 3)

aSc

g 12=

ηexp -m dm

1 3a

Sc12

=0.89

Remember (Gamma function): 1

0

( ) ( 1) ( 1) exp zz z z x x dx

𝑥 = 𝑚3 → 𝑑𝑥 = 3𝑚2𝑑𝑚

3( 1) 3 2 (3 1) 3

0 0

( ) exp 3 3 expz zz m m m dm m m dm


)(gc*1 dr rSc

12

aexpPg

0

3

0

0

0

gP

1 3a

Sc12

=0.89


0

*1

1 1

y=0

c=D c -c

y

1 2

10 1

η=0

U g=D c -c

x η

1 2 1 3

10 1

UD a= Sc c -c

U x 0.89 12

1 3-2 3 -1 2

10 1

k

U 0.332= Sc Re c -c

0.89 12

0y

1

y

cDj


2/13/2 ReScU34.0k So

This is the local mass transfer coefficient while the overall is found by:

L

1Av

y=00

-2/3 1/210 1

c1j = - D dx

L y

= 0.68 U Sc Re (c - c )


Let’s calculate Sh to compare it with our Table of correlations

(Table 8.3-3)2/3 1/2

2/3 1/2

2/3 1/2 1/2

1/3 1/2

1/3 1/2

kL L0.68 Sc Re U

D D

L D0.68 u

D u L

uL0.68

D

0.68 Sc Re

So we obtained the mass transfer coefficient from theory! Very rare.

Another example is MTC for laminar flow through a short circular tube

(look at section 9.4.1 of Cussler book )


In general2/13/2 ReScconst

U

k

cU

j

D

Applicable for flow around spheres, cylinders etc.

Mass transfer correlation over a wide range of Re:

3/12/1 ScRe6.02Sh


9.3 Theories for concentrated solutions

The MTC’s are based typically on dilute solutions but they are VERY

successful even in concentrated ones as, typically, the volume

average velocity normal to the fluid-fluid interface is rather small.

When the MTC’s fail, it is observed typically that “the k depends

strongly on concentration” especially when mass transfer is fast.

The flux is written as 0i11i11 vccckN

2211i11i1 NVNVccck


However, this is difficult and we also have the previous unknowns

again (film thickness, contact or residence times).

Instead we are looking for corrections to the dilute mass transfer

coefficient by the film theory:

0

k MTC in concentrated solution=

MTC in dilute solutionk

This framework can be used to build new film, penetration and

surface-renewal theories.


Dilute Concentrated


The total flux for a concentrated solution across a film:

01

11 vc

dz

dcDn with B.C.’s: z = 0: c1 = c1i

z = L c1 = c1

For concentrated solutions, n1 and v0 are constant, so the above

equation can be integrated from z=0 to L to give

D/Lvexpv/nc

v/nc 0

01i1

011

Rearranging:

Compare: 0i11i11 vccckN

0

i11i10

0

10z1 vccc1D/Lvexp

vNn


Then the mass transfer coefficient is

1D/Lvexp

vk

0

0

For dilute solutions:L

Dk0

Taking the ratio k/k0, we eliminate the dependency in L:

0 0

0 0 0

/

exp / 1

k v k

k v k


k can be smaller or larger than k0 depending the v0 direction

mass transfer boundary layer theory - eth z€¦ · h. schlichting, k. gersten, "boundary...

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