Mass Transfer Mass Transfer Coefficients Notes 10-11-2015

ChE 330, Mass Transfer and Separa3on Processes

Mass Transfer Coefficients

Chapter 3 Coursepack Class Notes S&H 3.4-3.7

ObjecAves

Understand the moAvaAon for using Mass Transfer Coefficients. Solve problems involving mass transfer in Laminar Flow. Solve problems involving mass transfer in Turbulent Flow. Understand the difference between individual and overall mass

transfer coefficients. Use correlaAons to calculate mass transfer coefficients. Predict how mass transfer coefficients change, and understand

why they change, as a funcAon of system variables.

Mass Transfer in Laminar Flow


2uyz2

+ g = 0

uy f (y)uy = 0 at z = uyz

= 0 at z = 0

uy =g 2

21 z

%

& ' (

) *

2

NRe =4rH u y

=4u y

=4

NRe < 8-25, interface is flat Between 1200 & 8-25, interface may ripple


DABCAz

$

% &

'

( ) z

(yx) + uy CA( )y (zx) = DABCAz

$

% &

'

( ) z+z

(yx) + uy CA( )y+y (zx)

1. Flat interface. 2. Low Mass Transfer Rate. 3. No bulk flow in z. 4. No axial dispersion in y. 5. Mass Transfer in z, only by diffusion.

uy CA( )y+y uy CA( )yy

= DAB

CAz

%

& '

(

) * z+z

CAz

%

& '

(

) * z

z

uyCAy

= DAB 2CAz2


uyCAy

= DAB 2CAz2

CA = CAi at z = 0 for y > 0CA = CA 0 at y = 0 for 0 < z < CAz

= 0 at z = for 0 < y < L

CAi C ALCAi CA 0

= 0.7857e5.1213 + 0.09726e39.661 + 0.0361e106.25

C AL =1

u yu yCAydz

0

=2DAB L3u y

=8 /3

NReNSc ( /L)=

8 /3NPe ( /L)

Mass Transfer Coefficients

Q = hAT nA = kc ACA = kc A(CAi C A )

nA = u yWdC A = kc (CAi C A )Wdy

kc,avg =kC dy

0

L

L

=

u ydC A

(CAi C A )CA 0

CAL

L=

u yLln (CAi C A 0)(CAi C AL )

NA ,avg = kc,avg (CAi C A ) 0L = kc,avg (CAi C A )LM

NA ,avg = kc,avg(CAi C A 0) (CAi C AL )ln[(CAi C A 0) (CAi C AL )]

Mass Transfer Dimensionless Numbers

=2DAB L3u y

=8 /3

NReNSc ( /L)=

8 /3NPe ( /L)

NSc = /DAB

=momentum diffusivity

mass diffusivity

NRe =uL

=

intertial forceviscous force

NPe = NScNRe = /DAB

uL

=uLDAB

=bulk transport

molecular transport

=8 /3

NPe ( /L)=

molecular transportbulk transport

NSh,avg =kc,avgDAB

=convective transportmolecular transport

Sherwood Number

NSh,avg =kc,avgDAB

= 3.414

NSh,avg =4

Laminar Flow Across a Plate

x

=4.96NRe

0.5

Describes how the velocity boundary layer develops (3-129)

C

=1

NSc1/ 3

Describes how the concentraAon boundary layer develops (3-135) NSc ~ 1 (gases), NSc > 1 (liquid)

Laminar Flow in a Circular Tube

NSh,avg = 3.66 +0.0668[NPe /(x /D)]

1+ 0.04[NPe /(x /D)]2 / 3


1. Laminar flow allows momentum and mass transport to equaAons to be coupled.

2. Unfortunately, the boundary layer thickness is extremely difficult to determine.

3. The concept of mass transfer coefficient is sAll useful for determining the extent of mass transfer without invoking fluid mechanics directly.

4. CorrelaAons for mass transfer coefficients are specific to geometry (since the geometry influences the development of the momentum boundary layer).

Mass Transfer in Turbulent Flow

1. Streamlines no longer exist, except near a wall. 2. Eddies of fluid mix, in all direcAons. 3. Must account for mass transfer by molecular diffusion,

and eddy diffusion, superimposed on bulk flow.

NA = CT (DAB +D )xAz

The Reynolds Analogy

Applied at a solid boundary for mass, heat, and momentum

NA = (DAB +D )CAz z=0

= kc (CAi C A )

Use dimensionless velocity, temperature, and concentraAon:

=uxu x

=(Ti T)(Ti T )

=(CAi CA )(CAi C A )

Az z=0

=fu x

2( +M )=

hxCP ( + H )

=kc

(DAB + H )

For the flux equaAons:

If the three eddy diffusiviAes are equal, and the molecular diffusiviAes are negligible, then

f2

=hx

CP u x=

kcu x

= NSt ,M Only valid when NSc = 1.

The Chilton-Colburn Analogy Extends the Reynolds Analogy to Schmidt Numbers >1.

jM f2= jH =

hxCPux

NPr2/3 = jD =

kcuxNSc2/3 = NStNSc

2/3

Reynolds number dependence based on geometry: Eqs (3-166 through 3-171).

Mass Transfer Coefficient CorrelaAons

Mass Transfer in Fluids w/ Fluid-Fluid Interface + Turbulence

1. Previous topics covered fluids in contact with solids (plate, pipe) adsorpAon, membranes, drying.

2. DEAS involve fluid-fluid interfaces. 3. At fluid-fluid interfaces, turbulence may persist. 4. Three models will be developed to describe these

phenomena

Film Theory Assumes that flow field consists of two regions:

1. Outer Region Constant Velocity & ConcentraAon 2. Film Region Velocity & ConcentraAon vary. 3. Resistance to mass transfer is solely from Fickian Diffusion with the film. 4. Phase Equilibrium at the interface.

NA =CTDAB

(xAi xAb ) = kcCT (xAi xAb )

NA =CTDAB

(1 xA )LM(xAi xAb ) = k 'c (CAi CAb )

Subscript c denotes a molar concentraAon driving force. Superscript prime denotes the inclusion of bulk flow via the

log mean concentraAon term.

Two-Film Theory One fluid in film theory was pure.. So there is no resistance

to mass transfer in that phase.

What about two films?

Two-Film Theory Each phase has its own resistance to mass transfer each phase

has its own film. The fluid-fluid interface is always assumed to be at equilibrium,

so that there is no resistance at the interface.

The slope of the concentraAon curve tells you the direcAon of mass transport.

Two-Film Theory We can also write this in terms of mass transfer coefficients.

NA = kp (pAb pAi) = ky (yAb yAi)NA = kc (CAi CAb ) = kx (xAb xAi)

Flux to the interface = Flux away from the interface.

pAb

CAb

pAb

CAb

Two-Film Theory


The slope of the line connecAng the interfacial liquid & gas composiAon to the bulk liquid & gas composiAons is the negaAve of the raAo of the individual mass transfer coefficients

kC pAb

CAb

Solve for 3-212

Note the error in the text on 3-222.

NA =pAbHA CAb

(HA /kp ) + (1/kc )CAi = HApAb;

CAipAb

xAiyAi1KA

SubsAtute Henrys Law & Rearrange:


Try this with other forms of Henrys Law or Equilibrium Constant

Overall Mass Transfer Coefficient We can define mass transfer coefficients based on any

concentraAon driving force.

NA = KG (pAb pA* ) = Ky (yAb y A

* )

NA = KL (CA* CAb ) = Kx (x A

* xAb )

1KL

=HAkp

+1kc

1KG

=1kp

+1

HAkc

Denotes a composiAon in equilibrium with a bulk composiAon e.g., CA*=HApAi

Overall Mass Transfer Coefficient We can define mass transfer coefficients based on any

concentraAon driving force.

NA = KG (pAb pA* ) = Ky (yAb y A

* )

NA = KL (CA* CAb ) = Kx (x A

* xAb )

1KL

=HAkp

+1kc

1KG

=1kp

+1

HAkc

1Kx

=1

KAky+1kx

1Ky

=1ky

+KAkx

KA = Equilibrium Constant = yA/xA

RelaAonship between individual & overall mass transfer coefficients.

HA = Henrys Constant (defined in S&H) = CA/pA

pAb

CAb

CA,L pA,L

kC pA,G

CA,G

Overall Mass Transfer Coefficient

pAb

CAb

CA,L pA,L

kC pA,G

CA,G

Slope = -kc / kG would move toward zero; kc is much smaller. This means more of the overall mass transfer resistance resides in the liquid phase. Why is this important?

A wastewater stream is introduced to the top of a mass-transfer tower where it flows countercurrent to an air stream. At one point in the tower, the wastewater stream contains 1x10-6 kmol A/m3 and the air is essenAally free of any A. At the operaAng condiAons within the tower, the film mass-transfer coefficients are kC = 5x10-4 m/s, and kG = 0.01 kmol/m2-s-atm. The concentraAons are such that Henrys law is valid, and at the operaAng temperature, H = 10 atm/kmol/m3. Determine the overall mass flux of A. Determine the overall mass-transfer coefficients KL and KG. What fracAon of the total resistance is in the gas phase?

Mass Transfer Mass Transfer Coefficients Notes 10-11-2015

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