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
- 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