4 Mass Transfer Coefficients.pdf

1

MASS TRANSFER THEORIES

difference ion concentrat x Area transfer mass of Rate  k

_{The equations developed so far can only predicts steady state}

mass transfer provided the film thickness (z_T) 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:

cm/s mole/cm x cm mole/s Unit  _{2 3}  --- (1.59)

2 Since:

--- (1.60)

Therefore:

in terms of concentration --- (1.61)

in terms of molar fraction in --- (1.62) vapor phase

in terms of molar fraction in --- (1.63) liquid phase J flux, molar Area transfer mass of Rate _ A Ai A c

c

c

J

k





y

y

J

k





x

x

J

k





3

 Note that only the unit of k_c is in cm/s, while the unit of k_x and k_y is similar to J_A as molar fraction is dimensionless.

 However, k_c can be correlated with k_x and k_y by the molar density: --- (1.64) --- (1.65) where:

_M = molar density of fluid (mole/m3₎ _x = normal density of fluid (kg/m3₎

M = average molecular weight of fluid (kg/mole)

RT

P

k

k

k







M

k

k

k









4

 The significance of k_c 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:

--- (1.66)

--- (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 k_c and D_AB .

)

(

1

)

(

c

c

z

c

c

D

c

c

J

k











z

D

k



5 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, k_c and the film

thickness, z_T neglecting the bulk flow effect.

Ans: z_T = 0.0028 cm

)

(

x

x

z

cD

TYPE OF MT COEFFICIENTS





dz

dc

D

J









For turbulent MT with constant c :

--- (1.68) where:

D_AB = 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.

7 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:

)

(

c

c

z

z

D

J











The flux J_A1 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 k_c :

)

(

k

c

c

J





z

z

D

k









8 B) MT coefficient for equimolar counterdiffusion

Similar to the total molar flux, NA in molecular diffusion with the term _M added:

)

(

)

(

x

N

N

dz

dx

D

c

N













For the case of equimolar counterdiffusion, where N_A = -N_B, and

integrating at steady state and gives:

1 2 '

z

z

D

k









)

(

k

c

c

N





9 MT coefficients can also be defined in several ways. Consider:

y_A = mole fraction in vapor phase

x_A = mole fraction in liquid phase

Therefore, Eq. (1.72) can be written as follows for equimolar counterdiffusion: Gases:

)

(

)

(

)

(

k

c

c

k

p

p

k

y

y

N













)

(

)

(

)

(

k

c

c

k

c

c

k

x

x

N













10 All of these MT coefficients can be related to each other. For

example, using Eq. (1.74) and substituting y_A1 = c_A1/c and

y_A2 = c_A2/c into the equation:

)

(

)

(

)

(

)

(

c

c

c

k

c

c

c

c

k

y

y

k

c

c

k

N

















c

k

k



11 C) MT coefficient for A diffusing through stagnant,

nondiffusing B

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

) ( ) ( ) ( ) ( 2 1 2 1 ' 2 1 2 1 ' A A x A A BM x A A c A A BM c A x x k x x x k c c k c c x k N         --- (1.77) where:

k_c = MT coefficient for A diffusing through stagnant B.

x_BM and y_BM are similar to the equations defined in the previous lectures.

12 Rewriting Eq. (1.77) using other units:

Gases:

)

(

)

(

)

(

k

c

c

k

p

p

k

y

y

N













)

(

)

(

)

(

k

c

c

k

c

c

k

x

x

N













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

      _      c c c c k x x k c c x k N _{A A x A A x} A A BM c A 2 1 2 1 2 1 ' ) ( ) ( Hence: --- (1.80) --- (1.81)

c

k

x

k



13

14 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 N}

A, the vaporization rate, and also the value of k_y and k_G.

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.

p_A1 = 0.20 atm

15 y_A1 = p_A1/P_T = 0.20/2.0 = 0.10 y_A2 = 0 From Eq. (1.78):

N



k

(

y



y

)

However we have a value of k’_y which is related to k_y by:

'

y BM

y

y

k

k



Find k_y by first calculating y_BM:

95

.

0

)

90

.

0

/

0

.

1

ln(

90

.

0

0

.

1

)

/

ln(











y

y

y

y

y

16 Then, from Eq. (1.82),

frac. mole s kgmole/m 10 x 138 . 7 95 . 0 10 x 78 . 6 5 _{5 2} '        BM y y y k k Similarly, calculate k_G : BM y BM G

y

P

k

y

k



17 Using Eq. (1.78) to calculate the flux N_A:

s kgmole/m 10 x 7.138 0) -(0.10 10 x 7.138 ) ( 2 6 -5 -2 1      _{y A A} A k y y N Also: p_A1 = 0.20 atm = 0.20(1.01325 x 105_{) = 2.026 x 10}4 _Pa

Using Eq. (1.78) again to calculate the flux N_A:

s kgmole/m 10 x 7.138 0) -10 x (2.026 10 x 3.522 ) ( 2 6 -4 10 -2 1      _{G A A} A k p p N s kgmole/m 10 x 7.138 0) -(0.20 10 x 3.569 ) ( 2 6 -5 -2 1      _{G A A} A k p p N

Note that in this case, since the concentration were diluted,