CHEMICAL ENGINEERING

226 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

approximately zero and:

NAD

D

L

CAiCAo

{

1 C

2 L^2

^2 Dte

∑^1

nD 1

[

^2

6



1

n^2

expn^2 ^2 Dte/L^2

]}

PROBLEM 10.9

According to the simple penetration theory the instantaneous mass flux:

NA (^) tDCAiCAo

(

D

t

) 0. 5

What is the equivalent expression for the instantaneous heat flux under analogous condi-
tions?
Pure sulphur dioxide is absorbed at 295 K and atmospheric pressure into a laminar
water jet. The solubility of SO 2 , assumed constant over a small temperature range, is
1 .54 kmol/m^3 under these conditions and the heat of solution is 28 kJ/kmol.
Calculate the resulting jet surface temperature if the Lewis number is 90. Neglect heat
transfer between the water and the gas.

Solution

The heat flux at any time,fDk∂)/∂x
wherekis the thermal conductivity,) the
temperature, andythe distance in the direction of transfer.
The flux satisfies the same differential equation as),thatis:

DH∂^2 f/∂y^2 D∂f/∂t

 y > 0 ,t> 0

whereDHDthermal diffusivityDk/+Cp.
This last equation is analogous to the mass transfer equation 10.66:
∂C/∂t
DD∂^2 C/∂y^2
The solution of the heat transfer equation withfDFo(constant) atyD0whent> 0
is:
fDFoerfc

y

2

p

DHt

The temperature rise is due to the heat of solutionHS. Heat is liberated at the jet
surface at a rateHt
DNoAHS,

or: Ht
DCAiCAo
HSD/t
^0.^5

The temperature rise,T, due to the heat fluxHt

into the surface is:

TD

1

+Cp

p

DH

∫L

0

Ht)

d)

p

)

and: TD

CAiCAo

Hs

p

D/DH

+Cp