4 Mass Transfer Coefficients.pdf

MASS TRANSFER THEORIES The equations developed so far can only predicts steady state

mass transfer provided the film thickness (zT) is known. However, in most cases, the value of zT

is not known as turbulent flow is desired to increase the rate of transfer per unit area or to create more interfacial area. Thus in turbulent flow, a mass transfer coefficient (k) is used,

which is defined as: Rate of mass transfer k Area x concentration difference

Unit 

mole/s  cm/s 2 3 cm x mole/cm

------ (1.59) 1

Since: Rate of mass transfer  molar flux, J Area

------ (1.60)

Therefore:

JA kc  c Ai  c A

in terms of concentration

------ (1.61)

JA ky  y Ai  y A

in terms of molar fraction in vapor phase

------ (1.62)

JA kx  x Ai  x A

in terms of molar fraction in ------ (1.63) liquid phase

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 Note that only the unit of kc is in cm/s, while the unit of kx and ky is similar to JA as molar fraction is dimensionless.  However, kc can be correlated with kx and ky by the molar density:

kc P k y  kc  M  RT kc  x k x  kc  M  M

------ (1.64)

------ (1.65)

where:

M x M

= molar density of fluid (mole/m3) = normal density of fluid (kg/m3) = average molecular weight of fluid (kg/mole)

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 The significance of kc is brought out by combining Eq. (1.61) with

Eq. (1.23) for steady state equimolar counterdiffusion in a stagnant film. Combination of both equations give:

DAB (c Ai  c A ) JA 1 kc   c Ai  c A zT (c Ai  c A )

------ (1.66)

DAB kc  zT

------ (1.67)

 Thus, the mass transfer coefficient is the molecular diffusivity

divided by the thickness of the diffusion path.  When dealing with unsteady state diffusion or diffusion in flowing streams, Eq. (1.67) still can be used to give an effective film thickness from known value of kc and DAB .

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EXERCISE: SO2 is absorbed into water from air in a absorption tower. At a specific location, SO2 is transferred at the rate of 0.0270 kmol SO2/m2.h and the liquid phase mole fraction are 0.0025 and 0.0003 respectively, at the two-phase interphase and in the bulk liquid. If the diffusivity of SO2 in water is 1.7x10-5 cm2/s, determine the mass-transfer coefficient, kc and the film thickness, zT neglecting the bulk flow effect.

cDAB NA  ( x Ai  x Ab ) zT

i.e. i = interphase, b = bulk flow

Ans: zT = 0.0028 cm 5

TYPE OF MT COEFFICIENTS A) Definition of MT coefficients: For turbulent MT with constant c :

J A  DAB

dc A  M  dz

------ (1.68)

where: DAB = molecular diffusivity (m2/s) M = mass eddy diffusivity (m2/s)

M is variable, near to zero at the interface or surface and increases as the distance from the wall increases.

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Since the variation of M is not generally known, use the average value of M. Integrating Eq. (1.68) between points 1 and 2 gives:

DAB   M J A1  (c A1  c A2 ) z2  z1

------ (1.69)

The flux JA1 is based on the surface area A1 since the cross sectional area may vary. The value of z2 - z1, the distance of the path is often not known. Hence, Eq. (1.69) is simplified and is written using kc :

J A1  kc (cA1  cA2 ) where:

DAB   M kc  z2  z1

------ (1.70) 7

B) MT coefficient for equimolar counterdiffusion Similar to the total molar flux, NA in molecular diffusion with the term M added:

dx A N A  c( DAB   M )  xA ( N A  N B ) dz

------ (1.71)

For the case of equimolar counterdiffusion, where NA = -NB, and

DAB   M integrating at steady state and k  gives: z2  z1 ' c

N A  k (c A1  c A2 ) ' c

------ (1.72) 8

MT coefficients can also be defined in several ways. Consider: yA = mole fraction in vapor phase xA = mole fraction in liquid phase Therefore, Eq. (1.72) can be written as follows for equimolar counterdiffusion: Gases:

N A  kc' (cA1  cA2 )  kG' ( pA1  pA2 )  k y' ( y A1  y A2 )

------ (1.73)

Liquids:

N A  kc' (cA1  cA2 )  kL' (cA1  cA2 )  k x' ( xA1  xA2 )

------ (1.74) 9

All of these MT coefficients can be related to each other. For example, using Eq. (1.74) and substituting yA1 = cA1/c and yA2 = cA2/c into the equation:

k y'

c A1 c A2 N A  k (c A1  c A2 )  k ( y A1  y A2 )  k (  )  (c A1  c A2 ) c c c ' c

' y

' y

------ (1.75) Hence,

kc' 

k y'

------ (1.76)

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C) MT coefficient for A diffusing through stagnant, nondiffusing B For A diffusing through stagnant, nondiffusing B where NB = 0, Eq. (1.71) gives for steady state:

kc' NA  (c A1 c A2 )  kc (c A1  c A2 ) xBM ' x

k  ( x A1  x A2 )  k x ( x A1  x A2 ) xBM

------ (1.77)

where: kc = MT coefficient for A diffusing through stagnant B. xBM and yBM are similar to the equations defined in the previous lectures.

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Rewriting Eq. (1.77) using other units: Gases:

N A  kc (cA1  cA2 ) k G( pA1  pA2 )  k y ( y A1  y A2 )

------ (1.78)

Liquids:

N A  kc (cA1  cA2 )  kL (cA1  cA2 )  k x ( xA1  xA2 )

------ (1.79)

All of these MT coefficients can be related to each other. For example, setting Eq. (1.77) equal to Eq. (1.79) gives:

kc'  c A1 c A2  NA  (c A1c A2 )  k x ( x A1  x A2 )  k x    xBM c   c Hence:

kc' kx  xBM c

------ (1.80) 12

------ (1.81)

SUMMARY:

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Example 8: A large volume of pure gas B at 2 atm pressure is flowing over a surface from which pure A is vaporizing. The liquid A completely wets the surface, which is the blotting paper. Hence, the partial pressure of A at the surface is the vapor pressure of A at 298 K, which is 0.20 atm. The k’y has been estimated to be 6.78 x 10-5 kg mole/m2·s·mole frac. Calculate NA, the vaporization rate, and also the value of ky and kG. Solution: This is the case of A diffusing through B, where the flux of B normal to the surface is zero, since B is not soluble in liquid A. pA1 = 0.20 atm pA2 = 0 (pure gas B)

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yA1 = pA1/PT = 0.20/2.0 = 0.10

From Eq. (1.78):

yA2 = 0

N A  k y ( y A1  y A2 )

However we have a value of k’y which is related to ky by:

k y yBM  k y'

------ (1.82)

Find ky by first calculating yBM: yB1 = 1.0 - 0.10 = 0.90

yBM

yB2 = 1.0 - 0 = 1.0

yB 2  yB1 1.0  0.90    0.95 ln( yB 2 / yB1 ) ln(1.0 / 0.90)

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Then, from Eq. (1.82), ky 

k y' yBM

6.78 x 105   7.138 x 105 kgmole/m 2  s  mole frac. 0.95

Similarly, calculate kG :

kG yBM P  k y yBM

------ (1.83)

7.138 x 105 10 2 kG    3 . 522 x 10 kgmole/m  s  Pa 5 PT 2 x 1.01325 x 10 Pa ky

7.138 x 105 kG    3.569 x 105 kgmole/m 2  s  atm PT 2.0 atm ky

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Using Eq. (1.78) to calculate the flux NA: N A  k y ( y A1  y A2 )  7.138 x 10-5 (0.10 - 0)  7.138 x 10-6 kgmole/m 2  s

Also: pA1 = 0.20 atm = 0.20(1.01325 x 105) = 2.026 x 104 Pa Using Eq. (1.78) again to calculate the flux NA: N A  kG ( p A1  p A2 )

N A  kG ( p A1  p A2 )

 3.522 x 10-10 (2.026 x 10 4 - 0)

 3.569 x 10-5 (0.20 - 0)

 7.138 x 10-6 kgmole/m 2  s

 7.138 x 10-6 kgmole/m 2  s

Note that in this case, since the concentration were diluted, yBM is close to 1.0 and ky and k’y differ very little.

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