Mass Transfer in a wetted-wall column

[Rogue]

Homework Statement

Ammonia in gas is being absorbed by water in a wtted-wall column. At one level of the column, the following data applies:
Gas phase mass transfer co-efficient:
5.22 x 10^-9 kmol m^-2s^-1Pa^-1

Liquid Phase mass transfer coefficient:
3.88 x 10^-5 m s^-1

Henrys constant 0.955kPa (kmol m^-3)^-1


Use the following additional information to find the mass transfer flux in the column:

mole fraction of ammonia in liquid* 1.351 x 10^-3
mole fraction of ammonia in gas* 0.065
total pressure of system 1.013 bar
mole mass of ammonia 17

Relevant Equations

Attempt at solution

Using Ca =55.51xa

Ammonia in liquid:
55.51 x 1.351 x 10^-3
=0.075kmol^-3

Ammonia in gas:
55.51 x 0.065
=3.608kmol^-3


I know my end equation needs to be in the order of:
Na = kc x (Ca1 - Ca2)


I attempted:

=(4.42 x 10^-6) x (3.608 - 0.075)

= 1.562 x 10^-5


But have since been advised that I need to be calculating partial pressure, molar mass transfer flux and equivalent molar concentration and to substitute these into my final equation.
At this point, I am finding myself to be very confused.

Please can someone assist me as this particular question seems to have me completely confounded?

 

[Chestermiller]

Let:

$k_L$ = mass transfer coefficient of ammonia in the liquid phase
$k_g$ = mass transfer coefficient of ammonia in the gas phase
$p_{\infty}$ = partial pressure of ammonia in the bulk of the gas phase
$p$ = partial pressure of ammonia in the gas phase at the gas-liquid interface
$C_{\infty}$ = concentration of ammonia in the bulk of the liquid phase
$C$ = concentration of ammonia in the liquid phase at the gas-liquid interface
H = Henry's law constant

What is the mass flux $\phi$ in terms of $k_L$, $C_{\infty}$, and C?
What is the same mass flux $\phi$ interns of $k_g$, $p_{\infty}$, and p?
What is the relationship between p, C, and H?

 

[Rogue]

Thanks for the response Chester.

Am I right in thinking?:

$\phi = k \iota (C - C \alpha)$

Answer = 1.371 x10^-4

$\phi = k \varrho (P - P \alpha)$

In terms of calculating partial pressure - I'll have to re-visit this to refresh my memory along with the relationship between p, C and H.

 

[Chestermiller]

Thanks for the response Chester.

Am I right in thinking?:

$\phi = k \iota (C - C \alpha)$

This equation is correct

$\phi = k \varrho (P - P \alpha)$

This equation is incorrect. It should read:$$\phi = k \varrho ( P \alpha-P)$$

So you have $$\phi = k_{\iota} (C - C _{\alpha})=k _{\rho} ( P _{\alpha}-P)$$

In terms of calculating partial pressure - I'll have to re-visit this to refresh my memory along with the relationship between p, C and H.

Henry's law relates C and P by: $$C = PH$$
Now, please eliminate C and P between these three equations, and express $\phi$ as a function of $C_{\alpha}$ and $P_{\alpha}$.