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TP MT Problem-2

The document describes the absorption of a gas component A into a liquid containing a reactant B. It provides the following: 1) The concentration of A (CA) through the liquid film depends on the depth (x) and can be expressed as a function involving cosh, where the rate constant is defined. 2) The rate of mass transfer (NA) of A through the surface can be written as a function of the diffusivity (D), the initial concentration of A (CAi), and the thickness of the liquid film (δ), as shown. 3) The document derives these expressions for CA and NA by applying mass balance equations and boundary conditions to the diffusion-reaction process

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Gowri Shankar
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51 views4 pages

TP MT Problem-2

The document describes the absorption of a gas component A into a liquid containing a reactant B. It provides the following: 1) The concentration of A (CA) through the liquid film depends on the depth (x) and can be expressed as a function involving cosh, where the rate constant is defined. 2) The rate of mass transfer (NA) of A through the surface can be written as a function of the diffusivity (D), the initial concentration of A (CAi), and the thickness of the liquid film (δ), as shown. 3) The document derives these expressions for CA and NA by applying mass balance equations and boundary conditions to the diffusion-reaction process

Uploaded by

Gowri Shankar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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2.

A component ‘A’ is absorbed from a gas phase by a liquid


containing a constituent ‘B’ with which it reacts chemically by a pseudo first
order reaction. The rate of absorption can be assumed to depend on both the
rate of chemical reaction and the rate of molecular diffusion through the
laminar film of liquid of thickness ‘  ’,

The total liquid depth is  . Then show the concentration profile of ‘A’ through
the film is
cos h     x   K
C A  c Ai 
cos h    D
where
K

Where CA is the concentration at depth ‘x’ and D . Show also that the rate

of mass transfer through the surface is

DcAi   
 NA   
  cos h    
Solution:
At any distance x from the inter face, the concentration of comp. A is =C.
dc
D
Rate of input by absorption per unit area = dx

dc
 C  . x
At 
x   x dx
the conc. of component A =
Rate of output from this within thickness‘  x ’ is by two ways:

By molecular diffusion from 


x   x
(i) position
(ii) Consumption of ‘A’ by chemical reaction in  x thickness.

Rate of output by the molecular diffusion at 


 x   x
distance =

d  dc 
  c  . x 
dx  dx 
dc
. x KC dx
Rate of output by chemical reaction in  x thickness - dx = A
d  dc 
 c  . x   KCdx

 The net rate of output = dx  dx 

m2
Where D is the diffusivity, s , k- reaction rate constant, s-1

But Accumulation =0
 Rate of Input - rate of output = rate of accumulation

dc  d  dc  
 D   D  c  . x   KC x   0
dx  dx  dx  
dc dc d 2c
 D  D  D 2  x  KC x  0
dx dx dx
d 2c K
  C 0
dx 2 D
d 2c
  m 2C  0
where m 
2 K
2
Or dx D

Then the solution of this equation is:


C  c1e  mx  c2e  mx …(1)
Where c1 and c2 are the constants.

To find out these two constant values we must know the two boundary
conditions.
The boundary conditions are:
dc
x  0, C A  c Ai and at x   , 0
At dx , it is closed. That is at x=L, it is insulated.

BC  1: at x  0, c A  c Ai …(2)
dc A
BC  2 : at x   , 0
dx …(3)
 at x   , it is closed
From (1) and (2)
C A  c1e  m.0  c2e m.0

 c1  c2  c Ai …(4)
 mx
But  C A  c1e  c2e
mx

dcA
  m1c1e  mx  mc2 e mx
dx
dcA
0
But at x    dx
 0  mc1e  m  mc2e m

 m c2 e m  c1e  m 
…(5)

From (4) c2  c Ai  c1
 0  m[ c Ai  c1  e m  c1e  m ]

 m cAi em  c1  e m  e  m  

e m
 c1  c Ai
e m  e  m …(6)

But c2  c Ai  c1
e m
c Ai  c Ai
= e m  e  m

 e m  e  m  e m 
 c Ai  m  m 
 e e 

e  m
 c2  c Ai
e m  e  m …(7)
From equation (6) and (7)
e m  mx e  m
c A  c Ai m  m
e  c Ai m  m
e mx
e e e e
m   x 
e 
m  x
e
 c Ai
e m  e  m
e x  e x
cos h x 
But 2
cos h .m    x  K
 c A  c Ai m
cos h m where D

Then the rate of mass transfer per unit area through the surface is:
dcA
N  D x 0
dx
But
dc A sin h .m    x 
 c Ai  m 
dx cos h  m

dc A sin h .m
 x0  c Ai  m 
dx cos h m
sin h .m
 N A    D    m  c Ai
cos h m

Dc Ai  m 
 NA   
  cot h  m  
----------------

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