CHE 318 Lecture 11

Unsteady State Mass Transfer

Author

Dr. Tian Tian

Published

January 28, 2026

Note

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Recap

Step-by-step solution to U.S.S. evaporation through stagnant air
Mass balance & flux analysis for catalyst on wall system

Learning Outcomes

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

Derive mass balance and flux with reaction terms
Identify B.C. and I.C. for reaction-related problems
Apply simplifications for U.S.S. problems with generation terms

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

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?

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)

Simplifications for Solving U.S.S. Problems

We can make some assumptions that do not have significant impact on the results:

Assumption 1: convection $\gg$ diffusion
- We can write $D_{AB} \frac{\partial^2 c_A}{\partial z^2} \approx 0$
- Only need $v_m$ to solve the U.S.S. problem
- Typically tolerable error in industrial pipes
Asumption 2: system close to steady state
- U.S.S –> S.S.? Just let $\partial c_A/\partial t = 0$
- Can compare with other S.S. solutions!

Solutions to Simplified U.S.S. Governing Equations

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 $c_{A0}=c_{A}(z=0)$

\[\begin{align} c_A(z) = c_{A,s} -\left(c_{A,s} - c_{A0}\right) \exp\!\left(-\frac{4k}{D v_m} z\right) \end{align}\]

Interpretation of $c_A(z)$ Profile

For this problem, because the surface reaction occurs, we no longer have a linear $c_A$ profile. It decays with a length scale of $D v_m / 4k$.

At the exit of the pipe, concentration of $c_A(L)$ is no less than $c_{A, s}$!
Decay length $L_{d} \approx \frac{D v_m}{4k}$
Faster flushing 👉 incomplete absorption
Is $N_A$ constant inside the tube?
What is the unit of $k$ in this case?

We will investigate these questions in the second half of the course!

Case 3: Catalytic Conversion of D → E (1D, Steady State)

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.

Step 1: Mass Balance

At anywhere outside the boundary, we have:

\[\begin{align} \text{[In]} - \text{[Out]} + \text{[Gen]} &= \text{[Acc]} \\ -\frac{\partial N_D}{\partial z} + 0 &= \frac{\partial c_D}{\partial t} \end{align}\]

Catalyst at the bottom of the reactor 👉 [Gen] = 0
How do we get the boundary condition?

Step 2: Flux Equation

The reaction at the bottom catalyst is very fast, we can use the “general case” flux equation

\[\begin{align} N_D = -D_{DE} \frac{\partial c_D}{\partial z} + \frac{c_D}{c_T} (N_D + N_E) \end{align}\]

How do we get relation between $N_D$ and $N_E$?
What are the boundary conditions?

Diffusion-Controlled Reaction: Explanations

In this system we have two conditions instantaneous and complete reaction. They have distinct meanings

Instantaneous: the reaction rate at the boundary does not depend on the actual concentration, but rather what ever molar flux is at that interface (diffusion-controlled)
- Often we will have $c_{D, z=0} = 0$ (all D at interface consumed instantly)
Complete: the conversion between D and E occur 100%, meaning the fluxes of D and E follow stoichiometry
- Stoichiometry: $m D \rightarrow n E$
- Flux continuity $N_D / m + N_E / n = 0$

Step 3: Conditions And Solutions

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

Catalyst In Wall: Simulation Demo

A slighted adapted demo from lecture 3 ($x_A$ profile). Can we modify the molar flux by interface reaction stoichiometry?

Conclusion of Unsteady State Mass Transport

We will cover until this point in the mid-term exam!

To solve unsteady and steady state problem, use the following steps:

Draw scheme and list physical quantities / conditions
Write mass balance equation
Is it steady-state or unsteady-state?
If unsteady state, write flux equation in differential form
If steady state, use one of the solution examples
Do integration / calculation

Summary

Writing mass balance equation for unsteady state problems
Apply conditions for reaction
Apply conditions for fluxes

Summary

Unsteady state mass transfer governing equation
Step-by-step solution to diffusion through stagnant B
Diffusion and reaction system setup

--- title: "CHE 318 Lecture 11" subtitle: "Unsteady State Mass Transfer" author: "Dr. Tian Tian" date: "2026-01-28" format: html: {} revealjs: output-file: slides.html pdf: output-file: L11.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L11.pdf) - Handwritten notes 👉 [Open in pdf](./public/L11_annotated.pdf) ::: ::: ## Recap {.center} - Step-by-step solution to U.S.S. evaporation through stagnant air - Mass balance & flux analysis for catalyst on wall system ## Learning Outcomes {.center} After today's lecture, you will be able to: - **Derive** mass balance and flux with reaction terms - **Identify** B.C. and I.C. for reaction-related problems - **Apply** simplifications for U.S.S. problems with generation terms ## U.S.S Example 2: Transport Through A Reactive Wall **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? ## 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) ## Simplifications for Solving U.S.S. Problems We can make some assumptions that do not have significant impact on the results: - **Assumption 1**: convection $\gg$ diffusion - We can write $D_{AB} \frac{\partial^2 c_A}{\partial z^2} \approx 0$ - Only need $v_m$ to solve the U.S.S. problem - Typically tolerable error in industrial pipes - **Asumption 2**: system close to steady state - U.S.S --> S.S.? Just let $\partial c_A/\partial t = 0$ - Can compare with other S.S. solutions! ## Solutions to Simplified U.S.S. Governing Equations After removing the $\partial c_A/\partial t$ and $D_{AB} \frac{\partial^2 c_A}{\partial z^2}$ terms, we have ```{=tex} \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 $c_{A0}=c_{A}(z=0)$ ```{=tex} \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} ``` ## Interpretation of $c_A(z)$ Profile For this problem, because the surface reaction occurs, we no longer have a linear $c_A$ profile. It decays with a length scale of $D v_m / 4k$. - At the exit of the pipe, concentration of $c_A(L)$ is no less than $c_{A, s}$! - Decay length $L_{d} \approx \frac{D v_m}{4k}$ - Faster flushing 👉 incomplete absorption - Is $N_A$ constant inside the tube? - What is the unit of $k$ in this case? We will investigate these questions in the second half of the course! ## Case 3: Catalytic Conversion of D → E (1D, Steady State) **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. ## Step 1: Mass Balance At anywhere outside the boundary, we have: ```{=tex} \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} ``` - Catalyst at the bottom of the reactor 👉 [Gen] = 0 - How do we get the boundary condition? ## Step 2: Flux Equation The reaction at the bottom catalyst is very fast, we can use the "general case" flux equation ```{=tex} \begin{align} N_D = -D_{DE} \frac{\partial c_D}{\partial z} + \frac{c_D}{c_T} (N_D + N_E) \end{align} ``` - How do we get relation between $N_D$ and $N_E$? - What are the boundary conditions? ## Diffusion-Controlled Reaction: Explanations In this system we have two conditions **instantaneous** and **complete** reaction. They have distinct meanings - **Instantaneous**: the reaction rate at the boundary does not depend on the actual concentration, but rather what ever molar flux is at that interface (**diffusion-controlled**) - Often we will have $c_{D, z=0} = 0$ (all D at interface consumed instantly) - **Complete**: the conversion between D and E occur 100%, meaning the fluxes of D and E follow **stoichiometry** - Stoichiometry: $m D \rightarrow n E$ - Flux continuity $N_D / m + N_E / n = 0$ ## Step 3: Conditions And Solutions We have in general $N_D / m = - N_E / n$ ```{=tex} \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 ## Catalyst In Wall: Simulation Demo A slighted adapted demo from lecture 3 ($x_A$ profile). Can we modify the molar flux by interface reaction stoichiometry? ```{=html} <iframe width="100%" height="900" src="../../scripts/L11_stoichiometry.html" title="Webpage example"></iframe> ``` ## Conclusion of Unsteady State Mass Transport We will cover until **this point** in the mid-term exam! To solve unsteady and steady state problem, use the following steps: 1. Draw scheme and list physical quantities / conditions 2. Write mass balance equation 3. Is it steady-state or unsteady-state? 4. If unsteady state, write flux equation in differential form 5. If steady state, use one of the solution examples 6. Do integration / calculation ## Summary - Writing mass balance equation for unsteady state problems - Apply conditions for reaction - Apply conditions for fluxes ## Summary - Unsteady state mass transfer governing equation - Step-by-step solution to diffusion through stagnant B - Diffusion and reaction system setup

Recap

Learning Outcomes

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

Step 1: Mass Balance

Step 2: Coupling With Flux Equation

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

Simplifications for Solving U.S.S. Problems

Solutions to Simplified U.S.S. Governing Equations

Interpretation of \(c_A(z)\) Profile

Case 3: Catalytic Conversion of D → E (1D, Steady State)

Step 1: Mass Balance

Step 2: Flux Equation

Diffusion-Controlled Reaction: Explanations

Step 3: Conditions And Solutions

Catalyst In Wall: Simulation Demo

Conclusion of Unsteady State Mass Transport

Summary

Summary