CHE 318 Lecture 10

Unsteady State Mass Transfer (II)

Author

Dr. Tian Tian

Published

January 26, 2026

Note

Slides 👉 Open presentation🗒️
PDF version of course note 👉 Open in pdf
Handwritten notes 👉 Open in pdf

Midterm Exam Announcement

Date: Feb 09, 2026 (Monday)
Time: 50 min during class
Question types:
- Multichoice questions (conceptual, no derivation)
- Short-answer questions (conceptual, no derivation)
- Long-answer questions (derivation and / or calculation)
Formula sheet / calculator policy: refer to course syllabus

Midterm Exam Questions

Covers up to unsteady state mass transport
Sample questions to be released this week on Canvas
Use our AI helper wisely!

Recap

Unsteady state mass transfer: $\partial c_A/\partial t \neq 0$
- Mass balance equation: [In] - [Out] + [Gen] = [Acc]
- Flux equation: $N_A = J_{Az}^{*} + c_A/c_T * (N_A + N_B)$
General expression for U.S.S. M.T. in 3D vector
Boundary conditions and initial conditions

Learning Outcomes

After today’s lecture, you will be able to:

Formulate unsteady-state mass balances with appropriate assumptions and simplifications
Identify accumulation, transport, and generation terms in typical 1D mass transfer problems
Interpret time-dependent concentration profiles in unsteady-state mass transfer examples

B.C. Case 1. Concentration at surfaces

Interface can be gas|liquid, liquid|solid, gas|solid
Often assuming equilibrium

\[ c_{A}\vert_{\text{surf}} = c_{As}\qquad\text{eq. solubility} \]

B.C. Case 2: Chemical Reactions

\[\begin{align} N_A\big|_{\text{surf}} = \nu_A\, r_A \end{align}\]

Surface reaction couples mass transfer and kinetics
Molar flux at surface determined by reaction rate
Generally Neumann boundary
$\nu_A$: stoichiometric ratio

B.C. Case 3: Constant Flux

In many cases the flux $N_{A, \text{surf}}$ or $N_{B, \text{surf}}$ can be constant.
E.g. inpenetratable surface to stagnant gas $N_{B, \text{surf}} = 0$
Does not mean $N_{A}(z)$ or $N_B(z)$ elsewhere is constant!

U.S.S Example 1: Diffusion Through Stagnant B

We have seen in previous examples how to solve the molar flux of liquid evaporating into stagnant air. Let’s see the same system but in unsteady state.

Question: liquid (A) evaporates inside stagnant air (B) inside a vertical tube at constant temperature $T$ and pressure $p_T$. At the vent of the system dry air is continuous blown. Plot the molar fraction $x_A$ as a function of $z$ and time $t$. Assume the liquid level is $L$ away from the vent and does not change during the evaporation process.

Step 1: Species Mass Balance (Unsteady, 1D)

For a differential slice $A\,dz$, write the mass balance

\[\begin{align} \text{[IN]} - \text{[OUT]} &= \text{[ACC]} \\ A N_A \vert_{z=z} - A N_A \vert_{z=z+\Delta z} &= \frac{\partial}{\partial t}\left(A\,dz\,c_A\right) \\ -\frac{\partial N_A}{\partial z} = \frac{\partial c_A}{\partial t} \end{align}\]

We would have $\frac{\partial c_B}{\partial t} = -\frac{\partial N_B}{\partial z}$

Step 2: Couple With Flux Equation

This is a diffusion through stagnant B case, we can directly write

\[\begin{align} N_A(z, t) = -c_T D_{AB}\frac{\partial x_A(z, t)}{\partial z} + x_A \left[N_A(z, t) + N_B(z, t)\right] \end{align}\]

Can we use $N_B=0$ in this case?
No, $N_B$ changes by $z, t$!
$N_B=0$ only at $z=0$ (liquid interface)
Do not write the steady state $N_A$ solution!

Step 3: Conservation Equations

Generally, we still need to know the relation between $N_A$ and $N_B$ to solve the mass-balance-flux equations.

The total concentration $c_T=c_A + c_B$ is conserved, therefore we have constrains

\[\begin{align} \text{[In]}_{T} - \text{[Out]}_{T} &= 0 \\ \text{[In]}_A - \text{[Out]}_A &= -\text{[In]}_B + \text{[Out]}_B \\ \frac{\partial N_A(z, t)}{\partial z} &= -\frac{\partial N_B(z, t)}{\partial z} \end{align}\]

Step 4: Boundary Conditions

Boundary conditions (Left, Right, any time)

$x_A(0, t) = x_{A0}$ (equilibrium vapor fraction)
$x_A(L, t) = 0$ (dry air)
$N_B(0, t) = 0$ (No-flux boundary for B)

The last B.C for $N_B(0, t)$ gives:

\[\begin{align} N_A(0, t) = -\frac{c_T D_{AB}}{1 - x_{A0}}\frac{\partial x_A(0, t)}{\partial z} \end{align}\]

Step 5: Final solution

Unsteady state flux equation (stagnant B)

\[\begin{align} N_A = -c_T D_{AB} \frac{\partial x_A}{\partial z} + x_A \left[ \frac{-c_T D_{AB}}{1 - x_{A0}} \frac{\partial x_{A}}{\partial z}\big\vert_{z=0} \right] \end{align}\]

Governing equation for diffusion through stagnant B, unsteady state:

\[\begin{align} \frac{\partial x_A}{\partial t} = D_{AB} \frac{\partial^2 x_A}{\partial z^2} + \left[ \frac{D_{AB}}{1 - x_{A0}} \frac{\partial x_{A}}{\partial z}\big\vert_{z=0} \right] \frac{\partial x_A}{\partial z} \end{align}\]

Analytical solution exists, but numerical solution is more convenient

U.S.S Diffusion Through Stagnant B: Analytical Solution

$x_A{z, t}$ has an analytical solution if $L \rightarrow \infty$ (See Bird. Transport Phenomena Ch 20.1):

\[\begin{align} x_A(z, t) = x_{A0} \frac{1 - \operatorname{erf}\!(\frac{z}{\sqrt{4 D_{AB} t}} - \phi)}{1 + \operatorname{erf}\!\phi} \end{align}\]

$\operatorname{erf}$ is the error function:

\[\begin{align} \operatorname{erf}(\eta) = \frac{2}{\sqrt{\pi}} \int_{0}^{\eta} e^{-s^2}\,ds \end{align}\]

$\phi$: a dimensionless constant depending on $x_{A0}$, $t$ and $D_{AB}$
Scaling combining length and time: $\frac{z}{\sqrt{4D_{AB}t}}$
Higher $D_{AB}$ 👉 faster towards steady state! (Wooclap question 3, L09)
Penetration depth: $L_p \approx \sqrt{4D_{AB} t}$

Numerical Solutions To Evaporation Problem

The numerical solution using finite difference (FD) methods is beyond CHE 318, but we can briefly breakdown the process into:

Discretize space: $z \rightarrow z_i$
Approximate spatial derivatives with finite differences
Convert PDE into a system of ODEs in time
Integrate in time using standard ODE solvers

U.S.S Diffusion Through Stagnant B: Demo

U.S.S Example 2: Transport Through A Catalyst Wall

Question: A gas mixture containing species A flows through a cylindrical conduit of diameter $D$ with a constant mean velocity $v_m$. A porous catalytic wall of thickness $\Delta z$ is located at a fixed axial position inside the conduit. Inside the catalyst region, species A is consumed by a first-order surface reaction:

\[ r = k'(c_{A,s} - c_A) \]

Assume:

surface concentration on catalyst, $c_{A,s}$ is constant
uniform properties in the radial direction
constant $T$, $P$, and physical properties
no reaction outside the catalyst region

Step 1: Mass Balance

Consider a differential gas-phase control volume of thickness $\Delta z$ that intersects the catalytic wall.

\[\begin{align} \text{In} - \text{Out} + \text{Generation} &= \text{Accumulation} \\ \\ \frac{\pi D^2}{4} \left( N_A\big|_{z} - N_A\big|_{z+\Delta z} \right) + \pi D\,\Delta z\,k'\,(c_{A,s}-c_A) &= \frac{\pi D^2}{4}\, \frac{\partial C_A}{\partial t} \\ -\frac{\partial N_A}{\partial z} + \frac{4k'}{D}\,(c_{A,s}-c_A) = \frac{\partial c_A}{\partial t} \end{align}\]

Step 2: Coupling With Flux Equation

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:

\[\begin{align} \frac{\partial N_A}{\partial z} = -\,D_{AB}\,\frac{\partial^2 c_A}{\partial z^2} + v_m\,\frac{\partial c_A}{\partial z} \qquad (v_m=\text{const}) \end{align}\]

Step 3: General Equation for M.T + Surface Reaction

\[\begin{align} \frac{\partial c_A}{\partial t} &= -\left( -\,D_{AB}\,\frac{\partial^2 c_A}{\partial z^2} + v_m\,\frac{\partial c_A}{\partial z} \right) + \frac{4k'}{D}\,(c_{A,s}-c_A) \\ &= D_{AB}\,\frac{\partial^2 c_A}{\partial z^2} - v_m\,\frac{\partial c_A}{\partial z} + \frac{4k'}{D}\,(c_{A,s}-c_A) \end{align}\]

Need:

initial condition $c_A(z,0)$
boundary conditions at $z=0$ and $z=L$

Solve:

analytical (special cases)
numerical integration (finite difference)

Summary

Step-by-step solution to diffusion through stagnant B
Diffusion and reaction system setup

--- title: "CHE 318 Lecture 10" subtitle: "Unsteady State Mass Transfer (II)" author: "Dr. Tian Tian" date: "2026-01-26" format: html: {} revealjs: output-file: slides.html pdf: output-file: L10.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L10.pdf) - Handwritten notes 👉 [Open in pdf](./public/L10_annotated.pdf) ::: ::: ## Midterm Exam Announcement {.center} - Date: Feb 09, 2026 (Monday) - Time: 50 min during class - Question types: - Multichoice questions (conceptual, no derivation) - Short-answer questions (conceptual, no derivation) - Long-answer questions (derivation and / or calculation) - Formula sheet / calculator policy: refer to course syllabus ## Midterm Exam Questions {.center} - Covers up to unsteady state mass transport - Sample questions to be released this week on Canvas - Use our [AI helper](https://gemini.google.com/gem/f2d47200f0bf) wisely! ## Recap {.center} - Unsteady state mass transfer: $\partial c_A/\partial t \neq 0$ - **Mass balance equation**: [In] - [Out] + [Gen] = [Acc] - **Flux equation**: $N_A = J_{Az}^{*} + c_A/c_T * (N_A + N_B)$ - General expression for U.S.S. M.T. in 3D vector - Boundary conditions and initial conditions ## Learning Outcomes {.center} After today's lecture, you will be able to: - **Formulate** unsteady-state mass balances with appropriate assumptions and simplifications - **Identify** accumulation, transport, and generation terms in typical 1D mass transfer problems - **Interpret** time-dependent concentration profiles in unsteady-state mass transfer examples ## B.C. Case 1. Concentration at surfaces ![Concentration at phase interfaces](./public/bc-conc.svg){width="55%"} - Interface can be gas|liquid, liquid|solid, gas|solid - Often assuming equilibrium $$ c_{A}\vert_{\text{surf}} = c_{As}\qquad\text{eq. solubility} $$ ## B.C. Case 2: Chemical Reactions ![Boundary condition for reaction](./public/bc-reaction.svg){width="55%"} ```{=tex} \begin{align} N_A\big|_{\text{surf}} = \nu_A\, r_A \end{align} ``` - Surface reaction couples mass transfer and kinetics - Molar flux at surface determined by reaction rate - Generally Neumann boundary - $\nu_A$: stoichiometric ratio ## B.C. Case 3: Constant Flux ![Constant flux B.C.](./public/const-flux-bc.svg){width="55%"} - In many cases the flux $N_{A, \text{surf}}$ or $N_{B, \text{surf}}$ can be constant. - E.g. inpenetratable surface to stagnant gas $N_{B, \text{surf}} = 0$ - **Does not** mean $N_{A}(z)$ or $N_B(z)$ elsewhere is constant! ## U.S.S Example 1: Diffusion Through Stagnant B We have seen in previous examples how to solve the molar flux of liquid evaporating into stagnant air. Let's see the same system but in unsteady state. **Question**: liquid (A) evaporates inside stagnant air (B) inside a vertical tube at constant temperature $T$ and pressure $p_T$. At the vent of the system dry air is continuous blown. Plot the molar fraction $x_A$ as a function of $z$ and time $t$. Assume the liquid level is $L$ away from the vent and does not change during the evaporation process. ## Step 1: Species Mass Balance (Unsteady, 1D) For a differential slice $A\,dz$, write the mass balance ```{=tex} \begin{align} \text{[IN]} - \text{[OUT]} &= \text{[ACC]} \\ A N_A \vert_{z=z} - A N_A \vert_{z=z+\Delta z} &= \frac{\partial}{\partial t}\left(A\,dz\,c_A\right) \\ -\frac{\partial N_A}{\partial z} = \frac{\partial c_A}{\partial t} \end{align} ``` We would have $\frac{\partial c_B}{\partial t} = -\frac{\partial N_B}{\partial z}$ ## Step 2: Couple With Flux Equation This is a diffusion through stagnant B case, we can directly write ```{=tex} \begin{align} N_A(z, t) = -c_T D_{AB}\frac{\partial x_A(z, t)}{\partial z} + x_A \left[N_A(z, t) + N_B(z, t)\right] \end{align} ``` - Can we use $N_B=0$ in this case? - **No**, $N_B$ changes by $z, t$! - $N_B=0$ only at $z=0$ (liquid interface) - **Do not** write the steady state $N_A$ solution! ## Step 3: Conservation Equations Generally, we still need to know the relation between $N_A$ and $N_B$ to solve the mass-balance-flux equations. The total concentration $c_T=c_A + c_B$ is conserved, therefore we have constrains ```{=tex} \begin{align} \text{[In]}_{T} - \text{[Out]}_{T} &= 0 \\ \text{[In]}_A - \text{[Out]}_A &= -\text{[In]}_B + \text{[Out]}_B \\ \frac{\partial N_A(z, t)}{\partial z} &= -\frac{\partial N_B(z, t)}{\partial z} \end{align} ``` ## Step 4: Boundary Conditions Boundary conditions (Left, Right, any time) - $x_A(0, t) = x_{A0}$ (equilibrium vapor fraction) - $x_A(L, t) = 0$ (dry air) - $N_B(0, t) = 0$ (No-flux boundary for B) The last B.C for $N_B(0, t)$ gives: ```{=tex} \begin{align} N_A(0, t) = -\frac{c_T D_{AB}}{1 - x_{A0}}\frac{\partial x_A(0, t)}{\partial z} \end{align} ``` ## Step 5: Final solution - Unsteady state flux equation (stagnant B) ```{=tex} \begin{align} N_A = -c_T D_{AB} \frac{\partial x_A}{\partial z} + x_A \left[ \frac{-c_T D_{AB}}{1 - x_{A0}} \frac{\partial x_{A}}{\partial z}\big\vert_{z=0} \right] \end{align} ``` - Governing equation for diffusion through stagnant B, unsteady state: ```{=tex} \begin{align} \frac{\partial x_A}{\partial t} = D_{AB} \frac{\partial^2 x_A}{\partial z^2} + \left[ \frac{D_{AB}}{1 - x_{A0}} \frac{\partial x_{A}}{\partial z}\big\vert_{z=0} \right] \frac{\partial x_A}{\partial z} \end{align} ``` - Analytical solution exists, but numerical solution is more convenient ## U.S.S Diffusion Through Stagnant B: Analytical Solution $x_A{z, t}$ has an analytical solution if $L \rightarrow \infty$ (See Bird. _Transport Phenomena_ Ch 20.1): ```{=tex} \begin{align} x_A(z, t) = x_{A0} \frac{1 - \operatorname{erf}\!(\frac{z}{\sqrt{4 D_{AB} t}} - \phi)}{1 + \operatorname{erf}\!\phi} \end{align} ``` $\operatorname{erf}$ is the **error function**: ```{=tex} \begin{align} \operatorname{erf}(\eta) = \frac{2}{\sqrt{\pi}} \int_{0}^{\eta} e^{-s^2}\,ds \end{align} ``` - $\phi$: a dimensionless constant depending on $x_{A0}$, $t$ and $D_{AB}$ - Scaling combining length and time: $\frac{z}{\sqrt{4D_{AB}t}}$ - Higher $D_{AB}$ 👉 faster towards steady state! (Wooclap question 3, L09) - Penetration depth: $L_p \approx \sqrt{4D_{AB} t}$ ## Numerical Solutions To Evaporation Problem The numerical solution using finite difference (FD) methods is beyond CHE 318, but we can briefly breakdown the process into: - Discretize space: $z \rightarrow z_i$ - Approximate spatial derivatives with finite differences - Convert PDE into a system of ODEs in time - Integrate in time using standard ODE solvers ## U.S.S Diffusion Through Stagnant B: Demo ```{=html} <iframe width="100%" height="900" src="../../scripts/L10_uss_profile.html" title="Webpage example"></iframe> ``` ## U.S.S Example 2: Transport Through A Catalyst Wall **Question**: A gas mixture containing species A flows through a cylindrical conduit of diameter $D$ with a constant mean velocity $v_m$. A porous catalytic wall of thickness $\Delta z$ is located at a fixed axial position inside the conduit. Inside the catalyst region, species A is consumed by a first-order surface reaction: $$ r = k'(c_{A,s} - c_A) $$ Assume: - surface concentration on catalyst, $c_{A,s}$ is constant - uniform properties in the radial direction - constant $T$, $P$, and physical properties - no reaction outside the catalyst region ## Step 1: Mass Balance Consider a differential gas-phase control volume of thickness $\Delta z$ that intersects the catalytic wall. ```{=tex} \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} ``` ## Step 2: Coupling With Flux Equation Use the convection–diffusion flux (constant $v_m$): ```{=tex} \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: ```{=tex} \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} ``` ## Step 3: General Equation for M.T + Surface Reaction ```{=tex} \begin{align} \frac{\partial c_A}{\partial t} &= -\left( -\,D_{AB}\,\frac{\partial^2 c_A}{\partial z^2} + v_m\,\frac{\partial c_A}{\partial z} \right) + \frac{4k'}{D}\,(c_{A,s}-c_A) \\ &= D_{AB}\,\frac{\partial^2 c_A}{\partial z^2} - v_m\,\frac{\partial c_A}{\partial z} + \frac{4k'}{D}\,(c_{A,s}-c_A) \end{align} ``` Need: - initial condition $c_A(z,0)$ - boundary conditions at $z=0$ and $z=L$ Solve: - analytical (special cases) - numerical integration (finite difference) ## Summary - Step-by-step solution to diffusion through stagnant B - Diffusion and reaction system setup