高等輸送二 — 質傳 Lecture 8 Forced convection 郭修伯 助理教授. Forced convection...

高等輸送二 — 質傳

Lecture 8Forced convection

郭修伯 助理教授

Forced convection

• Forced convection– The flow is determined by factors other than

diffusion, factors like pressure gradients and wetted area

– exist whether or not diffusion occurs

• Analyzing tools– simple physical models– elaborate analytical mathematics

The film theory (for interfacial mass transfer)

• Assuming a stagnant film exists near interface, a solute present at high dilution is slowly diffusing across this film.

• At steady-state:

p1

c1iGas Bulk liquid

c1

z = 0 z = l

Liquid film)( 111 cckN i or

1101011 ccl

DjnN izz

l

Dk

D

klnumberSherwood

D

klnumberSherwood

tcoefficiendiffusion

lengthsticcharacteritcoefficientransfermassnumberSherwood

))((

?

)( svariablesystemotherflengthsticcharacteri

factorcorrectionlengthsticcharacteri

tcoefficiendiffusiontcoefficientransfermass

Example

Carbon dioxide is being scrubbed out of a gas using water flowing through a packed bed of 1 cm Berl saddles. The carbon dioxide is absorbed at a rate of 2.3 x 10-6 mol/cm2 sec. The carbon dioxide is present at a partial pressure of 10 atm, the Henry’s law coefficient H is 600 atm, and the diffusion coefficient of carbon dioxide in water is 1.9 x 10-5 cm2/sec. Find the film thickness.

The interfacial concentration:

1

111 c

cHHxp i

10 atm 600 atm

1/18 mol/cm3

9.3 x 10-4 mol/cm3

)( 111 cckN i Mass transfer k = 2.5 x 10-3 cm/sec

cmk

Dl 0076.0 typical order: 10-2 cm

Penetration theory (Higbie, 1935)

• The falling film is very thick. In the z direction, diffusion is much more important than convection, and in the x direction, diffusion is much less important then convection.

• Flux at the interface:

)( 11max

01011 ccxDvjnN izz

dxdynWL

N z

L W

010 01

1 )(2 11

max1 ccL

DvN i

The flux averaged over x is:

p1

c1iGas Bulk liquid

c1

z = 0 z =

)( 111 cckN i

)(2 11max

1 ccLDvN i

LDvk

max2

Dk

2/1

max

2/14

D

Lv

D

kL

max0

3

2vv

2/102/16

D

Lv

D

kL

Sherwood number Peclet number

2/12/102/16

D

v

v

Lv

D

kL

Reynolds number Schmidt number

Surface-renewal theory (Dankwerts, 1951)

• It consists of two regions: – Interfacial region: mass transfer occurs by penetration

theory.

– Renewal region: constantly exchanged with new elements from a second bulk region.

The length of time that small fluid elements spend in the interfacial region is the key.

Residence time distribution

p1

c1iGas

Bulk liquid

c1

z = 0

In the interfacial region, the flux is that for diffusion into a infinite slab:

)( 1101 cct

Dn iz

Residence time distribution

0

1)( dttE

The probability that a given surface element will be at the surface for time t E(t)dt =

The fraction of surface elements remaining at time t :

t

t dttEe )(/

The residence time distribution of surface element :

/

)(te

tE

The average flux is:

)()( 110101 ccD

dtntEN iz

Dk Dk

Comparison of the three theories

• The film theory

• The penetration theory

• The surface-renewal theory

Dk Dk

LDvk

max2 Dk

l

Dk Dk lunknown :

maxv

Lunknown

:unknown

Contact time

Film thickness

Surface residence time

Boundary layer theory

• A more complete description of mass transfer• Based on parallel with earlier studies of fluid

mechanics and heat transfer

laminar region

turbulent region • The sharp-edged plate made of a sparingly soluble solute is immersed in a rapid flowing solvent.• A boundary layer is formed.• The boundary layer is usually defined as the locus of distance over which 99% of the disruptive effect occurs.• When the flow pattern develops, the solute dissolves off the plate.

turbulent region

Layers caused by flow and by mass transfer

• The distance that the solute penetrates produces a new concentration boundary layer c , but this layer is not the same as that observed for flow . The two layers influence each other.

• When the dissolving solute is only sparingly soluble, the boundary layer caused by the flow is unaffected.

Assuming flow varies as a power series in the boundary layer thickness

find c ?

Find the boundary layer for flow first:

Assuming the fluid flowing parallel to the flat plate follows:

...33

2210 yayayaavx

The boundary conditions are:

0,0,02

2

y

vvy x

xThe fluid sticks to the plate.The plate is solid and the stress on it is constant.

Far from the plate, the plate has no effect.0,, 0

y

vvvy x

x

0,, 0

y

vvvy x

x

3

0 2

1

2

3

yy

v

vx

turbulent region

Mass balance on the control volume of the width W, the thickness x and the height l:

lyyyxx

l

xx

l

x xWvdyvWdyvW

000

)0(~0

Dividing Wxx 0

l

xy dyvdx

dv

0

x-momentum balance gives:

)()0(0 00

00xWxWvvdyvvWdyvvW y

xx

l

xxx

l

xx

Dividing Wxx 0

y

l

x vvdyvdx

d 0

0

20

0

00 )( dyvvv

dx

d

y

vxxy

x

0

00 )( dyvvv

dx

d

y

vxxy

x

3

0 2

1

2

3

yy

v

vx

013

140

vdx

d

0,0 x

2/102/1

13

280

vx

x

We have now and use to find vx.3

0 2

1

2

3

yy

v

vx

How about c ?

Find the boundary layer for concentration:

Assuming the concentration profile parallel to the flat plate follows:

...33

22101 yayayaac

The boundary conditions are:

0,,021

2

11

y

cccy i

The concentration and flux are constant at the plate.

Far from the plate, the plate has no effect.0,0, 11

y

ccy

0,, 101

y

cvcy c

3

1

1

2

1

2

31

cci

yy

c

c

turbulent region

Mass balance on the control volume of the width W, the thickness x and the height l:

)0(0 010 10 1 lyyxx

l

xx

l

x xWNdyvWcdyvWc

Dividing Wxx 0

l

xy dyvcdx

dnN

0 1011

c

dyvcdx

d

y

cD xy

0 101

3

1

1

2

1

2

31

cci

yy

c

c

2/102/1

13

280

vx

x

3

0 2

1

2

3

yy

v

vx

D

dx

dx cc

33

3

4 c

D

dx

dx cc

33

3

4

0,0 cx

Dc

3

2/102/1

13

280

vx

x

3/12/1

064.4

D

vxxc

Schmidt number:

D

Mass transfer coefficient?

)( 111 cckN i

01

011

yy y

cDnN

3

1

1

2

1

2

31

cci

yy

c

c

c

iDcN

2

3 11

Similar to film theory

3/12/1

01

1 323.0

D

vxx

DcN i

3/12/1

0323.0

D

vxD

kx

3/12/1

0323.0

D

vxD

kx

Averaged over length L

3/12/1

0646.0

D

vxD

Lk

• Valid for a flat plate when the boundary layer is laminar (I.e., Re < 300,000)

• ,between the prediction of the film theory and the penetration/surface-renewal theories.

3/2Dk

Water flows at 10 cm/sec over a sharp-edged plate of benzoic acid. The dissolution of benzoic acid is diffusion-controlled, with a diffusion coefficient of 1.0 x 10-5 cm2/sec. Find (a) the distance at which the laminar boundary layer ends, (b) the thickness of the flow and concentration boundary layers at that point, and (c) the local mass transfer coefficients at the leading edge and at the position of transition, as well as the average mass transfer coefficient over this length.

(a) the length before the turbulent region begins:

000,300

sec01.0

sec101 30

cmg

cmcm

gxvx

x = 300 cm

2/102/1

13

280

vx

x

x = 300 cmcm5.2

Dc

3

cmc 25.0

3/12/1

0323.0

D

vxD

kxsec/109.5 5 cmk

Graetz-Nusselt problem

• Mass transfer across the walls of a pipe containing fluid in laminar flow.

• Find the dissolution rate as a function of quantities like Reynolds and Schmidt numbers

Flow conditions Diffusion conditions

zr

Sparing soluble solute

Fixed solute concentration at the wall of a short tube

zr

Mass balance for the solute in a constant-density fluid on a washer-shaped region:

Solute accumulation by convection

Solute accumulation by diffusion

Solute accumulation =

z

cv

z

c

r

cr

rrD z

121

211

0

z

c

R

rv

r

cr

rr

D

1

201 120

z

c

R

rv

r

cr

rr

D

1

201 120

rRs

z

c

R

sv

s

cD

10

21

2 40

0,,0

,0,0

0,,0

1

11

1

csz

ccsz

csallz

i

34

3

11

d

i

ecc

3/10

9

4

DRz

vs

incomplete gamma function

How to find the mass transfer coefficient?

)( 111 cckN i

RrRr r

cDnN

1

11

)0(

34

94

1

3/10

1

icDRzv

DN

3/10

9

4

34

DRz

vDk

Averaged over length L:3/13/13/103/1

34

3

L

d

D

v

v

dv

D

dk

Sherwood

1.62

Reynolds

Schmidt

diameter length

Water is flowing at 6.1 cm/sec through a pipe of 2.3 cm in diameter. The walls of a 14-cm section of this pipe are made of benzoic acid, whose diffusion coefficient in water is 1.0 x 10-5 cm2/sec. Find the average mass transfer coefficient over this section.

3/13/13/103/1

34

3

L

d

D

v

v

dv

D

dk

3/13/1

25

3/13/1

25 14

3.2

sec/101

sec)/1.6)(3.2(

34

3

sec/100.1

3.2

cm

v

v

cmcm

cm

cmk

sec/103.4 4 cmk Check before ending the question!

21001400Re Laminar flow?

cmRpenetratedhasdiffusioncmv

DLs 15.1)(01.0

40

Short pipe?

OK!

Concentrated solutions

• In most application, correlations for dilute solutions can also be applied to concentrated solutions.

• In a few cases, k is a function of the driving force:

11 ckN ),,( 1cScRefk

11 ckN ),( ScRefk

2211111

01111

)(

)(

NVNVccck

vccckN

ii

ii

mass transfer coefficient for rapid mass transfer ? dilute system

p1

c1i

Gas

Bulk liquid

c1

z = 0 z = l

A mass balance on a thin film shell z thick shows that the total flux is a constant:

dz

dn10

221111

01

11

nVnVcdz

dcD

vcdz

dcDn

11

11

,

,0

cclz

ccz i

Dlv

i

e

vnc

vnc 0

01

1

01

1

0111

0

011

10 vccc

e

vnN ii

Dlvz

...

)(12

)(

21

20

20

0

00

k

v

k

vkk

The relation between the dilute mass transfer coefficient k0 and the concentrated mass transfer coefficient k.

or

10

0

0

0

0

kv

e

kv

k

k

Benzene is evaporating from a flat porous plate into pure flowing air. Using the film theory, find N1/k0c1i and k/k0 as a function of the concentration of benzene at the surface of the plate.

The benzene evaporates off the plate into air flowing parallel to the plate : 02 n

221111

01

11

nVnVcdz

dcD

vcdz

dcDn

11

cV

cnv 10

i

z cc

c

l

DcnN

1011 ln

In dilute solution: )0( 1,1 id cl

DN

)1ln(1

111

01

iii

xxck

N

)1ln(1

111

1

10

10

0

00

10

0 ii

i

ckN

kv

xx

x

e

ckN

e

kv

k

k

Summary of forced convection

• Table 13.6-1• all predictions cluster around experimentally

observed values• In most cases. Sh Re1/2 and Sc1/3

• Recommend: film and penetration theories