CHE 318 Lecture 07

Steady State Mass Transfer – Other Geometries

Author

Dr. Tian Tian

Published

January 19, 2026

Note

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Recap

Examples of steady state diffusion
Applying steady state flux equations to solid and liquid systems
Examples for several special forms
- weight % and average concentration in liquid
- log mean form of concentration / molar fraction
- pressure - solubility - permeability

Learning Outcomes

(Continue from Lecture 06) After today’s lecture, you will be able to:

Identify the difference between diffusion in various geometries
Describe the approximations used for simplifying the soluton for $N_A$
Recall the reason to choose stagnant B solution for spherical diffusion
Solve pseudo-steady state diffusion problem
Analyze solutions of P.S.S solution and calculate diffusivity through experiment

Diffusion Through Varying Cross-Sectional Area

Steady state mass balance

Many industrial applications involde 1D transport with variable area $A(z)$, with mass balance:

\[\begin{align} [\mathrm{In}] - [\mathrm{Out}] &= 0 \\ N_{A1} A_1 - N_{A2} A_2 &= 0 \end{align}\]

Define the total molar flow rate of A:

\[ \overline{N}_A = N_A \, A(z) \]

At steady state: - $\overline{N}_A$ is constant - $N_A$ varies with position if $A(z)$ (or $A(r)$) varies

Applications of Variable-Area Diffusion

This framework can be used to solve diffusion through:

Sphere
Cylinder
Tube with varying diameter
Any system with a known area function $A = A(z)$ or $A=A(r)$

Key ideas

Do not solve as simple 1D Cartesian diffusion
Geometry enters through $A(z)$
Flux adjusts to maintain constant $\overline{N}_A$

Case Study 1: Diffusion Through a Sphere

Typical physical situations

Evaporation of a liquid droplet
Sublimation of a solid organic sphere

Conditions

Steady state
Stagnant B ($N_B = 0$)
Radial symmetry

Governing equation (stagnant B): use the ODE form, not the full 1D Cartesian solution:

\[\begin{align} N_A &= -\frac{D_{AB}}{RT}\frac{d p_A}{dr} + \frac{p_A}{p_T} N_A \\ N_A\!\left(1 - \frac{p_A}{p_T}\right) &= -\frac{D_{AB}}{RT}\frac{d p_A}{dr} \end{align}\]

Stagnant B in a Sphere: Reformulation

For spherical symmetry, $N_A(r)$ varies with $r$ and cannot appear alone in the balance equation L.H.S

Use the constant molar flow rate: \[ \overline{N}_A = N_A(r)\,4\pi r^2 = \text{constant} \]

\[\begin{align} \frac{\overline{N}_A}{4\pi r^2}\,dr &= -\frac{D_{AB}}{RT}\frac{1}{1-p_A/p_T}\,dp_A \end{align}\]

Stagnant B in a Sphere: Final Results

\[\begin{align} \boxed{ \frac{\overline{N}_A}{4\pi} \left(\frac{1}{r_1}-\frac{1}{r_2}\right) = \frac{D_{AB}p_T}{RT} \ln\!\left(\frac{p_T-p_{A2}}{p_T-p_{A1}}\right) } \end{align}\]

This looks very similar to the 1D stagnant B solution in cartesian coordinates. We can simplify it in several occations.

Stagnant B in a Sphere: Far-Field Limit ($r_2 \gg r_1$)

Special case: $r_2 \gg r_1$ (often $r_2 \to \infty$)

Typical examples: - Evaporating liquid droplet - Sublimation of a naphthalene (solid) sphere

From the general spherical stagnant-B result, use the log-mean pressure form:

\[\begin{align} \boxed{ N_{A1} = \frac{D_{AB} p_T}{RT\,p_{Bm}} \left(p_{A1} - p_{A2}\right) } \end{align}\]

Where $N_{A1} = N_A(r=r_1)$. For liquid, another form is often used (considering $p_{Bm}/p_T \approx 1$ and use $p = cRT$)

\[\begin{align} \boxed{ N_{A1} = \frac{D_{AB}}{r_1}\left(c_{A1} - c_{A2}\right) } \end{align}\]

Spherical Diffusion: Thin Shell Limit $r_1 \approx r_2$

Special case: thin membrane / shell
$\Delta r = r_2 - r_1 \ll r_1$

Approximation: \[ \frac{1}{r_1} - \frac{1}{r_2} \;\approx\; \frac{r_2 - r_1}{r_1^2} \;=\; \frac{\Delta r}{r_1^2} \]

Final result:

\[\begin{align} \boxed{ N_{A1} = \frac{D_{AB}}{RT}\frac{p_T}{p_{Bm}} \frac{(p_{A1}-p_{A2})}{\Delta r} } \end{align}\]

This acts like a 1D stagnant-B film with thickness $\Delta r$! (See lecture 3 wooclap question.)

Case 2: Diffusion Through a Non-uniform Area Conduit

Conditions:

Stagnant B
Steady state
Variable radius $r(z)$

Assume linear radius profile: \[ r(z) = r_1 + \frac{(r_2-r_1)}{(z_2-z_1)}(z-z_1) \]

Use the ODE form with $\dot N_A = N_A(z)\,A(z)$ and $A(z)=\pi r(z)^2$:

\[\begin{align} \frac{\dot N_A}{\pi}\int_{z_1}^{z_2}\frac{dz}{\left[r(z)\right]^2} &= -\frac{D_{AB}}{RT}\int_{p_{A1}}^{p_{A2}} \frac{dp_A}{1-p_A/p_T} \end{align}\]

Can you solve for $N_{A1}$?

Examples for Pseudo Steady State (P.S.S)

Example 1: Diffusion Through Stagnant B with Changing Path Length

Adapted from Geankoplis 6.2-3

Water vapor diffuses through a stagnant gas in a narrow vertical tube, dry air is constantly blown at the top of tube.
At time $t$, the liquid level is a distance $z$ from the tube top (i.e., the diffusion path length is $z$).
As diffusion proceeds, the liquid level drops slowly, so $z$ increases with time. The liquid has density $\rho_A$, and molecular weight $M_A$

Derive an expression for the time $t_F$ required for the level to drop such that the diffusion path length changes from $z=z_0$ at $t=0$ to $z=z_F$ at $t=t_F$.

Example 1: solutions

Tip

Use pseudo steady-state assumption
$N_A$ change over time!

Answer \[ t_F = \frac{\rho_A (z_F^2 - z_0^2) RT p_{Bm}}{2 D_{AB} M_A p_{T}} \frac{1}{(p_{A1} - p_{A2})} \]

Example 2: Determine $D_{AB}$ Through Evaporation

Adapted from Griskey 10-2

Sample setup as example 4, a vertical tube of diameter $D=0.01128$ m contains a liquid volatile species $A$ (chloropicrin, $CCl_3NO_2$) evaporating into stagnant air ($B$) at 1 atm. The gas-phase diffusion of $A$ occurs through the air column above the liquid surface.

At $t=0$, the distance from the tube top to the liquid surface is $z_0 = 0.0388$ m, after $t=1$ day, the distance is $z_1 = 0.0412$ m.

Vapor pressure at the interface: $p_{A1} = 3178.3$ Pa
Liquid density: $\rho_A = 1650$ kg/m$^3$
Molecular weight: $M_A = 164.39$ kg/kmol

Use your expression from example 4, determine the binary diffusivity $D_{AB}$ of $A$ in air.

Example 2: solutions

Tip

Pseudo steady state solution and assuming $N_A=\text{const}$ solution differ very little. Why?

Answer:

Pseudo-steady state: $D_{AB} = 8.56\times{}10^{-6}\ \text{m}^2/\text{s}$
$N_A$ constant: $D_{AB} = 8.75\times{}10^{-6}\ \text{m}^2/\text{s}$ ($+2.2\%$ error)

Example 3: P.S.S For Diffusion Through Sphere

Adapted from Geankoplis Ex 6.2-4

A sphere of naphthalene having a radius of 2.0 mm is suspended in a large volume of still air at 318 K and $1.01325 \times{}10^5$ Pa (1 atm). The surface temperature of the naphthalene can be assumed to be at 318 K and its vapor pressure at 318 K is 0.555 mm Hg. The $D_AB$ of naphthalene in air at 318 K is $6.92\times{}10^{−6}\ \text{m}^2/\text{s}$.

Calculate the rate of evaporation of naphthalene from the surface.
Write the expression for the time $t_F$ to evaporate a sphere from radius $r_0$ to $r_F$. The solid density for naphthalene is $\rho$ and molecular weight is $M_A$.
What is the $t_F$ value when the sphere is completely evaporated?

Example 3: solutions

Tip

Similar setup as example 5. $N_A$ is time-dependent

Answer:

$N_A = 9.68\times{}10^{-8}\ \text{kg mol}/\text{m}^2/\text{s}$
Expression for $t_F{r=r_F}$: \[ t_F(r=r_F) = \frac{\rho RT p_{Bm}}{2 M_A D_{AB} p_T} \frac{1}{(p_{A1} - p_{A2})} (r_0^2 - r_F^2) \]
Expression for $t_F{r=0}$: \[ t_F(r=0) = \frac{\rho RT p_{Bm} r_0^2}{2 M_A D_{AB} p_T} \frac{1}{(p_{A1} - p_{A2})} \]

Compare the solutions with Example 4. We can also measure the diffusivity of volatile organic molecules using the sphere evaporation methos!

Summary

Solving diffusion through varying cross-section
Pseudo-steady state solutions to diffusion-evaporation problems

--- title: "CHE 318 Lecture 07" subtitle: "Steady State Mass Transfer -- Other Geometries" author: "Dr. Tian Tian" date: "2026-01-19" format: html: {} revealjs: output-file: slides.html pdf: output-file: L07.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L07.pdf) - Handwritten notes 👉 [Open in pdf](./public/L07_annotated.pdf) ::: ::: ## Recap - Examples of steady state diffusion - Applying steady state flux equations to solid and liquid systems - Examples for several special forms - weight % and average concentration in liquid - log mean form of concentration / molar fraction - pressure - solubility - permeability ## Learning Outcomes {.center} (Continue from Lecture 06) After today's lecture, you will be able to: - **Identify** the difference between diffusion in various geometries - **Describe** the approximations used for simplifying the soluton for $N_A$ - **Recall** the reason to choose stagnant B solution for spherical diffusion - **Solve** pseudo-steady state diffusion problem - **Analyze** solutions of P.S.S solution and calculate diffusivity through experiment ## Diffusion Through Varying Cross-Sectional Area **Steady state mass balance** Many industrial applications involde 1D transport with variable area $A(z)$, with mass balance: ```{=tex} \begin{align} [\mathrm{In}] - [\mathrm{Out}] &= 0 \\ N_{A1} A_1 - N_{A2} A_2 &= 0 \end{align} ``` Define the **total molar flow rate** of A: $$ \overline{N}_A = N_A \, A(z) $$ At steady state: - $\overline{N}_A$ is constant - $N_A$ varies with position if $A(z)$ (or $A(r)$) varies ## Applications of Variable-Area Diffusion ![](./public/vary-area.png){width="75%"} :::{.columns} :::{.column width="50%"} This framework can be used to solve diffusion through: - Sphere - Cylinder - Tube with varying diameter - Any system with a known area function $A = A(z)$ or $A=A(r)$ ::: :::{.column width="50%"} **Key ideas** - Do **not** solve as simple 1D Cartesian diffusion - Geometry enters through $A(z)$ - Flux adjusts to maintain constant $\overline{N}_A$ ::: ::: ## Case Study 1: Diffusion Through a Sphere :::{.columns} :::{.column width="50%"} **Typical physical situations** - Evaporation of a liquid droplet - Sublimation of a solid organic sphere ::: :::{.column width="50%"} **Conditions** - Steady state - Stagnant B ($N_B = 0$) - Radial symmetry ::: ::: **Governing equation (stagnant B)**: use the **ODE form**, not the full 1D Cartesian solution: ```{=tex} \begin{align} N_A &= -\frac{D_{AB}}{RT}\frac{d p_A}{dr} + \frac{p_A}{p_T} N_A \\ N_A\!\left(1 - \frac{p_A}{p_T}\right) &= -\frac{D_{AB}}{RT}\frac{d p_A}{dr} \end{align} ``` ## Stagnant B in a Sphere: Reformulation For spherical symmetry, $N_A(r)$ varies with $r$ and **cannot** appear alone in the balance equation L.H.S Use the **constant molar flow rate**: $$ \overline{N}_A = N_A(r)\,4\pi r^2 = \text{constant} $$ ```{=tex} \begin{align} \frac{\overline{N}_A}{4\pi r^2}\,dr &= -\frac{D_{AB}}{RT}\frac{1}{1-p_A/p_T}\,dp_A \end{align} ``` ## Stagnant B in a Sphere: Final Results ```{=tex} \begin{align} \boxed{ \frac{\overline{N}_A}{4\pi} \left(\frac{1}{r_1}-\frac{1}{r_2}\right) = \frac{D_{AB}p_T}{RT} \ln\!\left(\frac{p_T-p_{A2}}{p_T-p_{A1}}\right) } \end{align} ``` This looks very similar to the 1D stagnant B solution in cartesian coordinates. We can simplify it in several occations. ## Stagnant B in a Sphere: Far-Field Limit ($r_2 \gg r_1$) **Special case:** $r_2 \gg r_1$ (often $r_2 \to \infty$) Typical examples: - Evaporating liquid droplet - Sublimation of a naphthalene (solid) sphere From the general spherical stagnant-B result, use the **log-mean pressure** form: ```{=tex} \begin{align} \boxed{ N_{A1} = \frac{D_{AB} p_T}{RT\,p_{Bm}} \left(p_{A1} - p_{A2}\right) } \end{align} ``` Where $N_{A1} = N_A(r=r_1)$. For liquid, another form is often used (considering $p_{Bm}/p_T \approx 1$ and use $p = cRT$) ```{=tex} \begin{align} \boxed{ N_{A1} = \frac{D_{AB}}{r_1}\left(c_{A1} - c_{A2}\right) } \end{align} ``` ## Spherical Diffusion: Thin Shell Limit $r_1 \approx r_2$ **Special case:** thin membrane / shell $\Delta r = r_2 - r_1 \ll r_1$ Approximation: $$ \frac{1}{r_1} - \frac{1}{r_2} \;\approx\; \frac{r_2 - r_1}{r_1^2} \;=\; \frac{\Delta r}{r_1^2} $$ Final result: ```{=tex} \begin{align} \boxed{ N_{A1} = \frac{D_{AB}}{RT}\frac{p_T}{p_{Bm}} \frac{(p_{A1}-p_{A2})}{\Delta r} } \end{align} ``` This acts like a 1D stagnant-B film with thickness $\Delta r$! (See lecture 3 wooclap question.) ## Case 2: Diffusion Through a Non-uniform Area Conduit Conditions: - Stagnant B - Steady state - Variable radius $r(z)$ Assume **linear radius profile**: $$ r(z) = r_1 + \frac{(r_2-r_1)}{(z_2-z_1)}(z-z_1) $$ Use the **ODE form** with $\dot N_A = N_A(z)\,A(z)$ and $A(z)=\pi r(z)^2$: ```{=tex} \begin{align} \frac{\dot N_A}{\pi}\int_{z_1}^{z_2}\frac{dz}{\left[r(z)\right]^2} &= -\frac{D_{AB}}{RT}\int_{p_{A1}}^{p_{A2}} \frac{dp_A}{1-p_A/p_T} \end{align} ``` _Can you solve for $N_{A1}$_? # Examples for Pseudo Steady State (P.S.S) ## Example 1: Diffusion Through Stagnant B with Changing Path Length _Adapted from Geankoplis 6.2-3_ Water vapor diffuses through a stagnant gas in a narrow vertical tube, dry air is constantly blown at the top of tube. At time $t$, the liquid level is a distance $z$ from the tube top (i.e., the diffusion path length is $z$). As diffusion proceeds, the liquid level drops slowly, so $z$ increases with time. The liquid has density $\rho_A$, and molecular weight $M_A$ 1) Derive an expression for the time $t_F$ required for the level to drop such that the diffusion path length changes from $z=z_0$ at $t=0$ to $z=z_F$ at $t=t_F$. ## Example 1: solutions :::{.callout-tip} 1. Use pseudo steady-state assumption 2. $N_A$ change over time! ::: Answer $$ t_F = \frac{\rho_A (z_F^2 - z_0^2) RT p_{Bm}}{2 D_{AB} M_A p_{T}} \frac{1}{(p_{A1} - p_{A2})} $$ ## Example 2: Determine $D_{AB}$ Through Evaporation _Adapted from Griskey 10-2_ Sample setup as example 4, a vertical tube of diameter $D=0.01128$ m contains a liquid volatile species $A$ (chloropicrin, $CCl_3NO_2$) evaporating into stagnant air ($B$) at 1 atm. The gas-phase diffusion of $A$ occurs through the air column above the liquid surface. At $t=0$, the distance from the tube top to the liquid surface is $z_0 = 0.0388$ m, after $t=1$ day, the distance is $z_1 = 0.0412$ m. - Vapor pressure at the interface: $p_{A1} = 3178.3$ Pa - Liquid density: $\rho_A = 1650$ kg/m$^3$ - Molecular weight: $M_A = 164.39$ kg/kmol 1) Use your expression from example 4, determine the binary diffusivity $D_{AB}$ of $A$ in air. ## Example 2: solutions :::{.callout-tip} Pseudo steady state solution and assuming $N_A=\text{const}$ solution differ very little. Why? ::: Answer: 1. Pseudo-steady state: $D_{AB} = 8.56\times{}10^{-6}\ \text{m}^2/\text{s}$ 1. $N_A$ constant: $D_{AB} = 8.75\times{}10^{-6}\ \text{m}^2/\text{s}$ ($+2.2\%$ error) ## Example 3: P.S.S For Diffusion Through Sphere _Adapted from Geankoplis Ex 6.2-4_ A sphere of naphthalene having a radius of 2.0 mm is suspended in a large volume of still air at 318 K and $1.01325 \times{}10^5$ Pa (1 atm). The surface temperature of the naphthalene can be assumed to be at 318 K and its vapor pressure at 318 K is 0.555 mm Hg. The $D_AB$ of naphthalene in air at 318 K is $6.92\times{}10^{−6}\ \text{m}^2/\text{s}$. 1. Calculate the rate of evaporation of naphthalene from the surface. 2. Write the expression for the time $t_F$ to evaporate a sphere from radius $r_0$ to $r_F$. The solid density for naphthalene is $\rho$ and molecular weight is $M_A$. 3. What is the $t_F$ value when the sphere is completely evaporated? ## Example 3: solutions :::{.callout-tip} Similar setup as example 5. $N_A$ is time-dependent ::: Answer: 1) $N_A = 9.68\times{}10^{-8}\ \text{kg mol}/\text{m}^2/\text{s}$ 2) Expression for $t_F{r=r_F}$: $$ t_F(r=r_F) = \frac{\rho RT p_{Bm}}{2 M_A D_{AB} p_T} \frac{1}{(p_{A1} - p_{A2})} (r_0^2 - r_F^2) $$ 3) Expression for $t_F{r=0}$: $$ t_F(r=0) = \frac{\rho RT p_{Bm} r_0^2}{2 M_A D_{AB} p_T} \frac{1}{(p_{A1} - p_{A2})} $$ Compare the solutions with Example 4. We can also measure the diffusivity of volatile organic molecules using the sphere evaporation methos! ## Summary - Solving diffusion through varying cross-section - Pseudo-steady state solutions to diffusion-evaporation problems

CHE 318 Lecture 07

Recap

Learning Outcomes

Diffusion Through Varying Cross-Sectional Area

Applications of Variable-Area Diffusion

Case Study 1: Diffusion Through a Sphere

Stagnant B in a Sphere: Reformulation

Stagnant B in a Sphere: Final Results

Stagnant B in a Sphere: Far-Field Limit (\(r_2 \gg r_1\))

Spherical Diffusion: Thin Shell Limit \(r_1 \approx r_2\)

Case 2: Diffusion Through a Non-uniform Area Conduit

Examples for Pseudo Steady State (P.S.S)

Example 1: Diffusion Through Stagnant B with Changing Path Length

Example 1: solutions

Example 2: Determine \(D_{AB}\) Through Evaporation

Example 2: solutions

Example 3: P.S.S For Diffusion Through Sphere

Example 3: solutions

Summary