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In equimolar counter diffusion for every mole of one component diffusing in one direction there is , by definition, one mole of the other component

In equimolar counter diffusion for every mole of one component diffusing in one direction there is, by definition, one mole of the other component diffusing in the opposite direction.
In this problem we will consider mass transfer by diffusion in the vertical column of constant cross- section (shape of the cross-section is not significant) shown in Figure Q2 below. The column is open at both ends. Inside the column there are two components A and B. Gas streams containing both components A and B flow across each of the two open ends of the column at the top and the bottom. The concentration of A in the lower stream is greater than the concentration of A in the upper stream giving rise to a concentration difference, which drives the mass transfer of A from the bottom to the top of the column. For equimolar counter diffusion, we have NAy =-NBy in the column, where NAy and NBy are the molar fluxes of the components A and B respectively in the vertical direction in the column, relative to stationary coordinates.
The geometry has been chosen so that there is mass transfer in one direction only, the y-direction and Cartesian coordinates can be used. Use a differential control volume inside the column and
a) develop the mass balance equation in molar terms for components A and B; [16]
b) derive an expression for the molar fluxes NAy and NBy, relative to stationary coordinates; [16]
c) derive the concentration profiles of components A and B along the column. Plot the profiles in a graph with concentration in the x-axis and vertical distance in the y-axis; [12]
d) compare the molar flux of A in this case to that for diffusion through a stagnant film.[6]
You can assume that the gases are ideal.
Gas Stream B & A
y = y2
x y = y1 NAy
NBy xA2
xA xA1
Gas Stream A & B In equimolar counter diffusion for every mole of one component diffusing in one direction there is,
by definition, one mole of the other component diffusing in the opposite direction.
In this problem we will consider mass transfer by diffusion in the vertical column of constant cross-
section (shape of the cross-section is not significant) shown in Figure Q2 below. The column is
open at both ends. Inside the column there are two components A and B. Gas streams containing
both components A and B flow across each of the two open ends of the column at the top and the
bottom. The concentration of A in the lower stream is greater than the concentration of A in the
upper stream giving rise to a concentration difference, which drives the mass transfer of A from the
bottom to the top of the column. For equimolar counter diffusion, we have NAy=-NBy in the column,
where NAy and NBy are the molar fluxes of the components A and B respectively in the vertical
direction in the column, relative to stationary coordinates.
The geometry has been chosen so that there is mass transfer in one direction only, the y-direction
and Cartesian coordinates can be used. Use a differential control volume inside the column and
a) develop the mass balance equation in molar terms for components A and B;
b) derive an expression for the molar fluxes NAy and NBy, relative to stationary coordinates;
c) derive the concentration profiles of components A and B along the column. Plot the profiles in a
graph with concentration in the x-axis and vertical distance in the y-axis;
d) compare the molar flux of A in this case to that for diffusion through a stagnant film.
You can assume that the gases are ideal.
Figure Q2: Schematic of the vertical column where the equimolar counter diffusion of gases
A and B takes place.
image text in transcribed

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