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