1.2-
Mass Transfer Theories
1. The two film theory (Whitmaun,1923) 2. The penetration theory (Higbie, 1935) 3. The surface surface renewal renewal theory theory (Dankwert (Dankwerts,19 s,1951) 51) 4. The film-penetration theory (Toor and Marchello, 1958) (1) The two film theory It states that diffusion is a steady state process and the resistance to mass transfer lies in two films on both sides of the interface. It is described by Fick's first law, l aw, dx d Interface N A = − CDAB A = −DAB CA dy dy C P
i
i
C A
P A
X A
y A
L
Gas
1 ← →
L
2→ ←
Liquid
For Gas Phase − DAB dPA NA = RT dy Li
− D AB
0
RT
N A ∫ dy =
NA =
D AB L1RT
Pi
∫ dPA
PA
(PA − Pi )
(1.16)
For Liquid Phase
N A = −DAB
dCA dy
L2
CA
0
Ci
N A ∫ dy = − D AB ∫ dC A
(1.17)
2. The Penetration Theory It states that diffusion is unsteady state process and the molecules of the solute are in constant random motion, where clusters of these molecules arrive at the interface, remaining there for a fixed period of time, and some of them penetrate while the rest mixes back into the bulk of the phase. The process is described by Fick's second law with the boundary conditions: CAο =0 Gas
∂CA ∂ 2CA = DAB ∂t ∂y2 (i) At t=0; C A = C A = 0 °
Ci
y=0
Liquid
(ii) At y=0; C A = C i (iii) At y= ∞ ; C A = C A ° = 0
Putting y
η=
∂η = ∂y
4D AB t 1 4D AB.t
∂η η = ∂t 2t ∂C A ∂CA ∂η ∂C A − η = = . ( ) ∂t ∂η ∂t ∂η 2t ∂ 2C A ∂ ∂C A ∂ ∂η ∂C A ∂η ( )= . .( . ) = 2 ∂y ∂y ∂y ∂η ∂y ∂η ∂y ∂ 2C A ∂ 2 C A 1 = .( ) ∂y 2 ∂η2 4D AB.t Substitute
∂C A
∂ 2C A nd and in Fick's 2 law, we have; 2 ∂y ∂t
∂C A − η ∂ 2C A 1 .( ) = D AB ( ) ∂η 2t ∂η 2 4 D AB .t d 2C A
+ 2η
d η 2
put → P = And ,
∫
dP P
dP d η
dC A d η dC A d η
=0
⇒∴
d η
=
d C A 2
d η
+ 2η P = 0
+ ∫ 2η d η = 0
ln P + η2 − ln c1 = 0 ⇒ ln
∴ P = c1e −η ∴
2
dP
dCA dη
P c
= −η2
2
= c1e − η
2
2
−η ∫ dCA = c1 ∫ e .dη ⇒ C A = c1erf (η) + c 2
Where c1 , c 2 are constant B.C (i) or (iii) y erf =∞ 4DAB.t
erf (∞) = 1 0 = c1 (erf (∞)) + c2
∴ c1 = −c 2
The solution becomes:CA = c 2 − c 2erf (η) B.C(ii) Ci = c 2 − c 2 (erf 0)
0
c2 = Ci The solution is C A = C i (1 − erf η)
CA Ci CA Ci
= (1 − erf η) = erfc(η) = erfcη
N A = −D AB
dC A dy
= N A ° = − D AB
But fromC A = Ci (1 − erf dC A dy
−(
= 0 − Ci e
y 4D AB.t
y 4 D AB .t
dC A
dy y = 0
)
(1.18)
) 2
y =o
dCA
= −Ci dy y = 0
∴ N A = − D AB (−Ci )( NA = Ci 3.
.
1 4DAB .t
) = Ci
Surface Renewal Theory
D AB 4t (1.19)
This is similar to the Penetration theory except that molecules arriving at the interface remain, there, for different random periods of time. Furthermore in order to apply this we need to know the rate of renewal of the surface( , which is a very difficult task and may, only be achieved experimentally. Mass transfer rates across the interface between two phases (e.g., gas and liquid) will be enhanced by surface renewal process. Suppose that the surface contains( i) elements of the surface and a random distribution of ages would exist for each element. The rate of production of fresh surface is and that is independent of the state of the surface (age of the element of surface).
)
Suppose that the rate of production of fresh surface per unit total area of surface is, and that is independent of the age of the element in question. The area of surface of age between t and t+dt will be a function and may be written as V(t+dt).dt. This will be equal to the area passing in t ime, dt, from the age range [(tdt) to t ] to the age range [t to (t+dt)]. Further, this in turn will be equal to the area in the age group[t to (t+dt)], less that replaced by fresh surface in time dt:
i.e.
V(t+dt).dt = V(t).dt – [V(t).dt]. .dt
) ) = -.V(t) ∴ ∴ ) + .V(t)= 0 ∴ V (t) = C. e � Now the total area of surface considered was unity: ∴ V (t) dt = 1 ∴ C. e � dt = 1 C ) = 1, V(t) = . e � t
t
∴ C =
t
Substitute this result in the solution of the penetration theory (equation 1.19), which is: NA = Ci
.
NA = Ci
D4.AB� λ.� dt
to get,
λ �/ � dt Putting, .t = x . Then .dt = 2xdx. Substitute in above, . .2. � dt N = C ∴ N = C √ D.λ NA = Ci
2
A
i
A
i
(1.20)
4. The Film – Penetration Theory As the name suggests it is a combination of the Film and Penetration theories. The governing system of equations is,
∂CA ∂ 2CA = DAB ∂t ∂y2 (i) At t = 0; C A = C A ° = 0 (ii) At y = 0; C A = C i (iii) At y = ; C A = C A ° = 0
Giving the solution thus,
) - ∑ = ∑ ) ���� ���� . .
(1.21)
And the mean mass flux.
NA = CAi
. {1 + 2 ∑ ��� . )
(1.22)
1.3- Mass Transfer Coefficient
Interface
Gas side NA =
D AB L1RT
N A = k g (
(PA − Pi )
PA − Pi RT g
PA
)
Gas Pi
Ci
g
N A = k g (C A − C i ) N A = Ck g (y A − y i )
Liquid side D AB L2
NA = C
(Ci − C A )
D AB L2
L2
L1
Where y A and y i are mole fraction in gas phase
NA =
Liquid
(x i − x A )
N A = Ck l (x i − x A ) Where x i , x A are mole fractions in liquid phase. k G: Individual mass transfer coefficient in gas phase k L: Individual mass transfer coefficient in liquid phase In the above we replaced (D AB /L1RT) by k g and (cDAB /L2) by k l because L1and L2 can not be estimated theoretically or experimentally.
y=0
CA
1.3.1-Overall Mass Transfer Coefficient
In the previous section the equations obtained are not useful for practical purposes, since the concentrations at the interface (P i and Ci) can not be determined therefore we resort to concept of overall mass transfer coefficient. This is defined as, * For gas phase: N A = KG (PA – PA ) * For liquid phase: NA = KL (CA - CA) Where (*) refers to the equilibrium value. Or y A = mx A ⇒ (PA = HC A ) Where m is equilibrium constant. Also Pi = H.C i *
Or yi = mxi 1.3.2-Relation between overall and individual M.T.C N A = k g (y A − y i )
…….. (i) *
N A = K G (y A − y A )
……… (ii) ……… (iii)
N A = k l (x i − x A ) *
NA = KL (x A − xA )
……… (iv)
From equation (i) & (ii),
= ) )
Adding and subtracting yi to the numerator in LHS., *
1 KG 1 KG 1 KG
∴
= = =
1 KG
(yA − yA ) + yi − yi k g (yA − yi ) (yA − yi ) k g (yA − yi ) 1 k g
=
+
1 k g
*
+
(yi − y A ) k g (yA − yi )
m(xi − x A )
Using yA= mxA & yi= mxi and that NA=k g(yA-yi)=k L(xA-xi)
k l (x i − x A )
+
m k l
Excercise; Prove that
(1.23)
1 KL
=
1 mk g
+
1 k l
(1.24)
1.3.3- Volumetric mass transfer coefficient, K G.a and KL.a
When using KG or KL we need to find the interfacial (a), which is defined as.
a = 6* /dav. Where is the fractional hold ( fraction of bubbles or drops in the column) and d av.is the average bubble or diameter. Thus in order to estimate (a) we have to use correlations for and dav and, although there are many correlations in the literature, they are inadequate and unreliable. The best ones gives an error of nearly 70%, some, even, gives as much as 3000% error !. Therefore the use of the combined form ( K G.a or KL.a) is, clearly advantageous. And researchers should be working along these lines i.e., finding correlations for the volumetric mass transfer coefficient rather than the film or overall coefficients.
1.4-The Wetted Wall Column
This is simple apparatus to measure the average M.T.C it usually consists of a tube about 1 m in length and 2 cm in diameter Gas out y2
We have the equation, Liquid in
kmole
*
N A = K G (y A − y A )
m 2 .s * The mole rate (WA ) = N A * A = K G .A.(yA − y A ) kmole/s A = π.d.z WA = G1 (y1 − y 2 ) = L(x1 − x 2 )
x2
z (1.25)
Gas in y1
For the driving force ( yA − yA* ) we take the average value for the whole column and this is may be represented bythe log mean driving force:-
(y
A
− yA
*
) = (y L.m
A
− yA
*
) − (y 1
(y Ln (y
− yA
A
A − yA
A
− yA
*
*
) )
*
)
2
Liquid out x1
(1.26)
1 2
*
And
y1 = mx1 *
y2 = mx2
(1.27)
The final equation is,.
* * (y1 − y1 ) − (y 2 − y 2 ) = G1 (y1 − y 2 ) WA = K G (ππ.d.z) * − (y1 y1 ) Ln * (y 2 − y 2 )
(1.28)
Example 5: A wetted wall column is used to absorb Ammonia by using pure water from (6%) by volume mixture with air . The gas flow rate is (1.2 kmol ⁄ min) at 1atm 0 and 20 C. Calculate the overall mass transfer coefficient. Data given : The ratio of water to air flow rate = 1.4 The outlet ammonia concentration = 1.5% by volume Column diameter = 2 cm
Column height = 100 cm ;
and Henery s constant, m = 1.3
Solution: Find x1 using equation (1.25),
(x – x ) 0.06 – 0.015 = 1.4 (x – 0) y1 – y2 =
1
2
1
x1 = 0.032 From equation (1.26),
. .) . . . (y - ) = . ..) A
Lm
= 0.0238 Finally, from equation (1.28),
KG (3.14*2*100).(0.0238) = (1.2).(0.06 – 0.015)
∴ K = 3.61*10 kmol/cm .min -3
G
2
Calculation of Mass Transfer Coefficient 1
1
Sh = 2 + 0.6 Re .Sc 3 2
1
1
(1.29)
Sh = 0.023 Re 2 .Sc 3
. .. Re is the Reynolds number = � � Sc is the Schmidt number = .
Where Sh is the Sherwood number =
And k is the mass transfer coefficient.
1.5- Diffusion with Chemical Reaction
Consider gaseous (A) diffusing through liquid (B), which contains a solute (C). (A) reacts with( C) according to a first order irreversible reaction , with a reaction constant k 1. A
k 1
AC
Refer to the accompanying figure to perform material balance on (A) over elemental z,
In = Out
. z).S + k .S.z.C (S: cross sectional area) NA.S = ( N A +
1
= - k 1.CA
But NA = - DAB
Gas A A
NA
- k C = 0 ∴ D AB
1
z
A
– (k 1 /DAB ).CA = 0
NA+(dNA /dz). z
The auxiliary equation is, 2 m - k 1 /DAB = 0
∴ m = ��D
and The general solution is, 1
CA = .exp[
Liquid B m2 =
��D
��D .z] + .exp[ ��D .z]
Using the boundary conditions, B.C. 1: at z = 0, CA = CA0 B.C. 2:
⟶AC
A
at z = L,
The final solution is,
= � �
NA = 0
or
=0
