CHEMICAL ENGINEERING

MASS TRANSFER 219

If the pressure is doubled to 200 kN/m^2 , the diffusivity is halved to 0.5D (from
equation 10.18) and:

PA 2 D 0. 1 ð 200

 D20 kN/m^2 andPA 1 D0kN/m^2

PB 2 D 200  20

 D180 kN/m^2 andPB 1 D200 kN/m^2

∴ PBMD 200  180
/ln 200 / 180
D 189 .82 kN/m^2

P/PBMD 200 / 189. 82

 D 1 .054 i.e. unchanged

Hence: NAD 0. 5 D/RTL

1. 054  20  0
   D 10. 54 D/RTL, that is the rate is
   unchanged

(b) If the absorbing solution now exerts a partial vapour pressure of ammonia of
5kN/m^2 , then at a total pressure of 100 kN/m^2 :

PA 2 D10 kN/m^2 andPA 1 D5kN/m^2

PB 2 D90 kN/m^2 andPB 1 D95 kN/m^2

PBMD 95  90 

/ln 95 / 90

 D 92 .48 kN/m^2

∴ P/PBMD 100 / 92. 48
D 1. 081

NADD/RTL

ð 1. 081  10  5

 D 5. 406 D/RTL

At 200 kN/m^2 , the diffusivityD 0. 5 Dand:

PA 2 D20 kN/m^2 andPA 1 D5kN/m^2

PB 2 D180 kN/m^2 andPB 1 D195 kN/m^2

∴ PBMD 195  180
/ln 195 / 180
D 187 .4kN/m^2

P/PBMD 1. 067

NAD 0. 5 D/RTL

 1. 067  20  5

 D 8. 0 D/RTL

Thus the rate of diffusion has been increased by 100 8  5. 406 

/ 5. 406 D48%.

PROBLEM 10.4

In the Danckwerts’ model of mass transfer it is assumed that the fractional rate of surface
renewalsis constant and independent of surface age. Under such conditions the expression
for the surface age distribution function issest.
If the fractional rate of surface renewal were proportional to surface age (saysDbt,
wherebis a constant), show that the surface age distribution function would then assume
the form:
 2 b/
^1 /^2 ebt

(^2) / 2
Solution
From equation 10.117: f^0 t
Dsft
D 0