Unsteady State Mass Transfer
2026-01-28
After today’s lecture, you will be able to:
Question: A gas mixture of CO\(_2\) (A) in argon (B) is flown through a cylindrical pipe of diameter \(D\) and length \(L\) with constant velocity \(v_m\). The pipe is covered with a porous material to absorb CO\(_2\). The porous material is coated uniformed on the inner wall of the pipe. While A is flowing through the pipe, it is being absorbed by a first-order reaction
\[ r_A = k(c_{A,s} - c_A) \]
Where \(c_{A, s}\) is the surface concentration on the absorbing material and \(c_A\) is the conentration of A in the gas mixture. We assume the concentration \(c_A(z)\) in the gas phase is uniform in the radial direction. The total pressure \(p_T\) and temperature \(T\) are both kept constant.
Can you estimate the concentration of CO\(_2\) at the end of the pipe?
Consider a differential gas-phase control volume of thickness \(\Delta z\) that intersects the catalytic wall.
\[\begin{align} \text{In} - \text{Out} + \text{Generation} &= \text{Accumulation} \\ \\ \frac{\pi D^2}{4} \left( N_A\big|_{z} - N_A\big|_{z+\Delta z} \right) + \pi D\,\Delta z\,k\,(c_{A,s}-c_A) &= \frac{\pi D^2}{4}\, \frac{\partial C_A}{\partial t} \\ -\frac{\partial N_A}{\partial z} + \frac{4k}{D}\,(c_{A,s}-c_A) = \frac{\partial c_A}{\partial t} \end{align}\]Use the convection–diffusion flux (constant \(v_m\)):
\[\begin{align} N_A = -\,D_{AB}\,\frac{\partial c_A}{\partial z} + c_A\,v_m \end{align}\]When \(v_m\) is constant, differentiate over \(N_A\) becomes:
\[\begin{align} \frac{\partial N_A}{\partial z} = -\,D_{AB}\,\frac{\partial^2 c_A}{\partial z^2} + v_m\,\frac{\partial c_A}{\partial z} \qquad (v_m=\text{const}) \end{align}\]Need:
Solve:
We can make some assumptions that do not have significant impact on the results:
After removing the \(\partial c_A/\partial t\) and \(D_{AB} \frac{\partial^2 c_A}{\partial z^2}\) terms, we have
\[\begin{align} \frac{4k}{D}\left(c_{A,s}-c_A\right) &= v_m \frac{dc_A}{dz} \\ \int_0^z \frac{4k}{D v_m} dz &= \int_{c_{Ai}}^{c_A} \frac{dc_A}{c_{A,s}-c_A} \\ \frac{4k}{D v_m} z &= -\ln\!\left(\frac{c_{A,s}-c_A(z)}{c_{A,s}-c_{A0}}\right) \end{align}\]We get \(c_A(z)\) profile, where \(c_{A0}=c_{A}(z=0)\)
\[\begin{align} c_A(z) = c_{A,s} -\left(c_{A,s} - c_{A0}\right) \exp\!\left(-\frac{4k}{D v_m} z\right) \end{align}\]For this problem, because the surface reaction occurs, we no longer have a linear \(c_A\) profile. It decays with a length scale of \(D v_m / 4k\).
We will investigate these questions in the second half of the course!
Question: A binary gas mixture containing species D and E occupies a stagnant gas film of thickness \(L\) in the \(z\)-direction. At \(z=0\), the gas is in contact with a solid catalyst surface that instantaneously and completely converts D to E. At \(z=L\), the gas composition is maintained at known values. Assumpt constant \(T\) and total pressure \(p_T\) and 100% conversion rate. The reaction between D and E follows: \[ m \text{D} \rightarrow n \text{E} \]
Develop the governing differential equation for \(N_D\) and \(c_D\) the molar flux of species D in the gas phase at steady state.
At anywhere outside the boundary, we have:
\[\begin{align} \text{[In]} - \text{[Out]} + \text{[Gen]} &= \text{[Acc]} \\ -\frac{\partial N_D}{\partial z} + 0 &= \frac{\partial c_D}{\partial t} \end{align}\]The reaction at the bottom catalyst is very fast, we can use the “general case” flux equation
\[\begin{align} N_D = -D_{DE} \frac{\partial c_D}{\partial z} + \frac{c_D}{c_T} (N_D + N_E) \end{align}\]In this system we have two conditions instantaneous and complete reaction. They have distinct meanings
We have in general \(N_D / m = - N_E / n\)
\[\begin{align} -\frac{\partial }{\partial z} \left[ -D_{DE} \frac{\partial c_D}{\partial z} + \frac{c_D}{c_T} (N_D + N_E) \right] &= \frac{\partial c_D}{\partial t} \\ -\frac{\partial }{\partial z} \left[ -D_{DE} \frac{\partial c_D}{\partial z} + \frac{c_D}{c_T} N_D (1 - \frac{n}{m}) \right] &= \frac{\partial c_D}{\partial t} \end{align}\]At steady state, we have
N_D = \left[-\frac{D_{DE}}{1 - \frac{c_D}{c_T} (1 - \frac{n}{m})}\right]{d c_D}{dz}The steady state solution can be easily integrated
A slighted adapted demo from lecture 3 (\(x_A\) profile). Can we modify the molar flux by interface reaction stoichiometry?
We will cover until this point in the mid-term exam!
To solve unsteady and steady state problem, use the following steps: