CHE 318 Lecture 20

Case Studies With Mass Transfer Coefficient (II)

Author

Dr. Tian Tian

Published

February 27, 2026

Note

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Learning outcomes

After this lecture, you will be able to:

Recall characteristic features of convective mass transfer over flat plates, spheres, and packed beds.
Identify suitable empirical correlations for each geometry and flow regime.
Apply equations to calculate $k_c'$, concentration changes, and fluxes in representative case studies.

Cheatsheet for mass transfer coefficient

Cheatsheet for using dimensionless numbers with $k_c'$. Printed version distributed in class

Case 2: flow parallel to flat plates

Can be used for gases or evaporation of liquid
Distinguished between laminar & turbulent flow
$N_{Re}$ regime cutoff different in gas & liquid!
Characteristic length $L$: length of plate in flow direction
Flow parallel to flat plate: video

Flow past parallel plates: regimes

Laminar flow ($N_{Re} < 15,000$)

\[\begin{align} j_D &= 0.664 N_{Re, L}^{-0.5} \\ \frac{k_c' L}{D_{AB}} &= 0.664 N_{Re, L}^{0.5} N_{Sc}^{1/3} \end{align}\]

This follows our derivation of boundary layer theory

Turbulent flow

Gas: $15,000 < N_{Re}< 300,000$

\[\begin{align} j_D = 0.036 N_{Re, L}^{-0.2} \end{align}\]

Liquid: $600 < N_{Re}< 50,000$

\[\begin{align} j_D = 0.99 N_{Re, L}^{-0.5} \end{align}\]

Example for flat plate

Example 7.3-2: A large volume of pure water at 26.1 $^\circ$C is flowing parallel to a flat plate of solid benzoic acid, where $L=0.244$ m in the direction of flow. The water velocity is 0.061 m/s. The solubility of benzoic acid in water is 0.02948 kg mol/m$^3$, and the diffusivity of benzoic acid in water is $1.245 \times 10^{-9}$ m$^2$/s. For water, $\mu=8.71\times 10^-4$ Pa$\cdot$s and $\rho=996$ kg/m$^3$ (basically the same conditions as in case 1-2).

Calculate the mass-transfer coefficient $k_c$ and $N_A$.

Flat surface: solution steps

Step 1: calculate $N_{Re}$, $N_{Sc}$. Regime?
Step 2: use flat-surface equations to calculate $j_D$
Step 3: convert $j_D$ to $k_c'$. Use $k_c = k_c'/x_{BM}$
Step 4: use master equation to get $N_A$

Flat surface example: results

Tip

We can use $x_{BM}=1$ in this case

$N_{\text{Re}}=1.700\times10^4$, $N_{\text{Sc}}=702$
Use liquid turbulent flow: $j_D = 0.99 N_{\text{Re}}^{-0.5}$ = 0.00758
$k_c' = j_D v N_{\text{Sc}}^{-2/3} = 5.85\times 10^{-6}$ m/s
$N_A = 1.726 \times 10^{-7}$ kg mol/m$^2$/s

Case 3: flow past single sphere

Frequent geometry in particle solutions
Low Reynolds regime 👉 solution for stagnant diffusion on spherical surface
High Reynolds regime 👉 correct $N_{Sh}$ and back calculate $k_c'$

Flow past single sphere: results

Low Reynolds ($N_{Re} < 2$)

\[\begin{align} N_A &= \boxed{\frac{2 D_{AB}}{D_p}} (c_{A1} - c_{A2}) \\ &= k_c (c_{A1} - c_{A2}) \\ &= \frac{k_c'}{x_{Bm}}(c_{A1} - c_{A2}) \\ \end{align}\]

For $x_{Bm} \approx 1$, we have:

\[ k_c' = \frac{2D_{AB}}{D_p} \]

Sherwood number: $N_{Sh} = 2$

High Reynolds ($N_{Re} > 2$)

Gas:

\[\begin{align} &N_{Sh} = 2 + 0.552 N_{Re}^{0.53} N_{Sc}^{1/3} \\ &0.6 < N_{Sc} < 2.7\;\ N_{Re} < 48000 \end{align}\]

Liquid:

\[\begin{align} N_{Sh} &= 2 + 0.95 N_{Re}^{0.5} N_{Sc}^{1/3};\ N_{Re} < 2000\\ N_{Sh} &= 0.347 N_{Re}^{0.62} N_{Sc}^{1/3};\ 2000 < N_{Re} < 17000 \end{align}\]

Back calculate $k_c' = N_{Sh} \frac{D_{AB}}{D_p}$

Case 4: mass transfer for packed beds

Video for fluidized packed bed

Packed bed: geometries

Geometry characteristics: void fraction $\varepsilon$:

\[ \varepsilon = \frac{\text{void space}}{\text{total space}} = \frac{\text{void space}}{\text{void space} + \text{solid space}} \]
- Typically $0.3 < \varepsilon < 0.5$
- Void fraction is difficult to measure experimentally
Total Effective Area for spherical particles:

\[ A = \frac{6(1 - \epsilon)}{D_p} V_b \]

Correlation equations in packed bed

Correlation 1, applicable to:

gase with $10 < N_{Re} <10,000$
liquid with $10 < N_{Re} < 1500$

\[\begin{align} j_D &= j_H = \frac{0.4548}{\varepsilon} N_{Re}^{-0.4069} \\ N_{Re} &= \frac{D_p v' \rho}{\mu} \end{align}\]

$D_p$: (average) particle diameter
$v’$: superficial velocity in the tube without packing

Correlation equations in packed bed (II)

Correlation 2, applicable to:

liquid with $0.0016 < N_{Re} < 55$, $165 < N_{Sc} < 70000$

\[\begin{align} j_D = \frac{1.09}{\varepsilon} N_{Re}^{-2/3} \end{align}\]

liquid with $55 < N_{Re} < 1500$, $165 < N_{Sc} < 10690$

\[\begin{align} j_D = \frac{0.250}{\varepsilon} N_{Re}^{-0.31} \end{align}\]

Correlation equations in packed bed (III)

Correlation 3, applicable to fluidized beds

$10 < N_{Re} < 4000$ (gas & liquid)

\[\begin{align} j_D = \frac{0.4548}{\varepsilon} N_{Re}^{-0.4069} \end{align}\]

$1 < N_{Re} < 10$ (liquid only)

\[\begin{align} j_D = \frac{1.1068}{\varepsilon} N_{Re}^{-0.72} \end{align}\]

Example 3: comparison between geometries

Adapted from Problem 7.3-3. Let’s estimate the gas-phase mass transfer coefficient $k_G$ (kg mol/(m$^2$ s Pa)) for mass transfer of water vapour to solids with different shapes. Consider a water vapour (A) in air (B) at 338.6 K and 101.32 Pa flowing through a big duct containing solids with various geometries. The flow velocity is 3.66 m/s. The water vapour concentration is small, so property of air is used ($\mu=2.03\times 10^-5$ Pa$\cdot$s, $\rho=1.043$ kg/m$^3$). From the table, $D_{AB} = 2.88 \times 10^{-5}$ m$^2$/s at 315 K. Compare the values for following geometries, which case?

Flow parallel to flat plate with length $L=2.54$ cm
A single sphere with diameter $D=2.54$ cm
Packed beds using spheres of average diameter $D_p=2.54$ cm and $\epsilon=0.35$

Example 3: results

Tip

Do not forget to correct $D_{AB}$ for temperature!
Choose the right $N_{\text{Sh}}$ or $j_D$ formula according to $N_{\text{Re}}$ and $N_{\text{Sc}}$
$k_G \approx k_G' = k_c' / (RT)$

Flat surface: $k_G = 1.738\times 10^{-8}$ kg mol/(m$^2$ s Pa). $N_{\text{Sh}} = 38.03$
Single sphere: $k_G = 1.984\times 10^{-8}$ kg mol/(m$^2$ s Pa). $N_{\text{Sh}} = 43.40$
Packed bed: $k_G = 7.60\times 10^{-8}$ kg mol/(m$^2$ s Pa). $N_{\text{Sh}} = 166.32$

Packed bed clearly wins!

Summary

Dimensionless numbers can be used to correlate mass transfer problems in different flow rate, dimension etc
Typically, start with a known geometry (pipe? parallel plate? sphere? packed bed?)
Find the correlation with dimensionless numbers $N_{Re}$, $N_{Sc}$
Calculate the final mass transfer rate
Geometry plays an important role in determining $k_c'$!

--- title: "CHE 318 Lecture 20" subtitle: "Case Studies With Mass Transfer Coefficient (II)" author: "Dr. Tian Tian" date: "2026-02-27" format: html: {} revealjs: output-file: slides.html pdf: output-file: L20.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L20.pdf) - Handwritten notes 👉 [Open in pdf](./public/L20_annotated.pdf) ::: ::: ## Learning outcomes {.center} After this lecture, you will be able to: - **Recall** characteristic features of convective mass transfer over flat plates, spheres, and packed beds. - **Identify** suitable empirical correlations for each geometry and flow regime. - **Apply** equations to calculate $k_c'$, concentration changes, and fluxes in representative case studies. ## Cheatsheet for mass transfer coefficient ![Cheatsheet for using dimensionless numbers with $k_c'$. Printed version distributed in class](../L19/public/cheatsheet.svg) ## Case 2: flow parallel to flat plates - Can be used for gases or evaporation of liquid - Distinguished between laminar & turbulent flow - $N_{Re}$ regime cutoff different in gas & liquid! - Characteristic length $L$: length of plate in flow direction - [Flow parallel to flat plate: video](https://www.youtube.com/watch?v=zsO5BQA_CZk) ## Flow past parallel plates: regimes :::{.columns} :::{.column width="50%"} **Laminar flow** ($N_{Re} < 15,000$) ```{=tex} \begin{align} j_D &= 0.664 N_{Re, L}^{-0.5} \\ \frac{k_c' L}{D_{AB}} &= 0.664 N_{Re, L}^{0.5} N_{Sc}^{1/3} \end{align} ``` - This follows our derivation of boundary layer theory ::: :::{.column width="50%"} **Turbulent flow** - Gas: $15,000 < N_{Re}< 300,000$ ```{=tex} \begin{align} j_D = 0.036 N_{Re, L}^{-0.2} \end{align} ``` - Liquid: $600 < N_{Re}< 50,000$ ```{=tex} \begin{align} j_D = 0.99 N_{Re, L}^{-0.5} \end{align} ``` ::: ::: ## Example for flat plate Example 7.3-2: A large volume of pure water at 26.1 $^\circ$C is flowing parallel to a flat plate of solid benzoic acid, where $L=0.244$ m in the direction of flow. The water velocity is 0.061 m/s. The solubility of benzoic acid in water is 0.02948 kg mol/m$^3$, and the diffusivity of benzoic acid in water is $1.245 \times 10^{-9}$ m$^2$/s. For water, $\mu=8.71\times 10^-4$ Pa$\cdot$s and $\rho=996$ kg/m$^3$ (basically the same conditions as in case 1-2). Calculate the mass-transfer coefficient $k_c$ and $N_A$. ## Flat surface: solution steps - Step 1: calculate $N_{Re}$, $N_{Sc}$. Regime? - Step 2: use flat-surface equations to calculate $j_D$ - Step 3: convert $j_D$ to $k_c'$. Use $k_c = k_c'/x_{BM}$ - Step 4: use master equation to get $N_A$ ## Flat surface example: results :::{.callout-tip} We can use $x_{BM}=1$ in this case ::: - $N_{\text{Re}}=1.700\times10^4$, $N_{\text{Sc}}=702$ - Use **liquid turbulent flow**: $j_D = 0.99 N_{\text{Re}}^{-0.5}$ = 0.00758 - $k_c' = j_D v N_{\text{Sc}}^{-2/3} = 5.85\times 10^{-6}$ m/s - $N_A = 1.726 \times 10^{-7}$ kg mol/m$^2$/s ## Case 3: flow past single sphere - Frequent geometry in particle solutions - Low Reynolds regime 👉 solution for stagnant diffusion on spherical surface - High Reynolds regime 👉 correct $N_{Sh}$ and back calculate $k_c'$ ## Flow past single sphere: results :::{.columns} :::{.column width="35%"} **Low Reynolds** ($N_{Re} < 2$) ```{=tex} \begin{align} N_A &= \boxed{\frac{2 D_{AB}}{D_p}} (c_{A1} - c_{A2}) \\ &= k_c (c_{A1} - c_{A2}) \\ &= \frac{k_c'}{x_{Bm}}(c_{A1} - c_{A2}) \\ \end{align} ``` - For $x_{Bm} \approx 1$, we have: $$ k_c' = \frac{2D_{AB}}{D_p} $$ - Sherwood number: $N_{Sh} = 2$ ::: :::{.column width="65%"} **High Reynolds** ($N_{Re} > 2$) - Gas: ```{=tex} \begin{align} &N_{Sh} = 2 + 0.552 N_{Re}^{0.53} N_{Sc}^{1/3} \\ &0.6 < N_{Sc} < 2.7\;\ N_{Re} < 48000 \end{align} ``` - Liquid: ```{=tex} \begin{align} N_{Sh} &= 2 + 0.95 N_{Re}^{0.5} N_{Sc}^{1/3};\ N_{Re} < 2000\\ N_{Sh} &= 0.347 N_{Re}^{0.62} N_{Sc}^{1/3};\ 2000 < N_{Re} < 17000 \end{align} ``` - Back calculate $k_c' = N_{Sh} \frac{D_{AB}}{D_p}$ ::: ::: ## Case 4: mass transfer for packed beds    - Video for [fluidized packed bed](https://www.youtube.com/watch?v=lFhrpSJZzck) ![Packed bed structure](./public/packed-bed.png) ## Packed bed: geometries - Geometry characteristics: void fraction $\varepsilon$: $$ \varepsilon = \frac{\text{void space}}{\text{total space}} = \frac{\text{void space}}{\text{void space} + \text{solid space}} $$ - Typically $0.3 < \varepsilon < 0.5$ - Void fraction is difficult to measure experimentally - Total Effective Area for spherical particles: $$ A = \frac{6(1 - \epsilon)}{D_p} V_b $$ ## Correlation equations in packed bed Correlation 1, applicable to: - gase with $10 < N_{Re} <10,000$ - liquid with $10 < N_{Re} < 1500$ ```{=tex} \begin{align} j_D &= j_H = \frac{0.4548}{\varepsilon} N_{Re}^{-0.4069} \\ N_{Re} &= \frac{D_p v' \rho}{\mu} \end{align} ``` - $D_p$: (average) particle diameter - $v’$: superficial velocity in the tube without packing ## Correlation equations in packed bed (II) Correlation 2, applicable to: - liquid with $0.0016 < N_{Re} < 55$, $165 < N_{Sc} < 70000$ ```{=tex} \begin{align} j_D = \frac{1.09}{\varepsilon} N_{Re}^{-2/3} \end{align} ``` - liquid with $55 < N_{Re} < 1500$, $165 < N_{Sc} < 10690$ ```{=tex} \begin{align} j_D = \frac{0.250}{\varepsilon} N_{Re}^{-0.31} \end{align} ``` ## Correlation equations in packed bed (III) Correlation 3, applicable to fluidized beds - $10 < N_{Re} < 4000$ (gas & liquid) ```{=tex} \begin{align} j_D = \frac{0.4548}{\varepsilon} N_{Re}^{-0.4069} \end{align} ``` - $1 < N_{Re} < 10$ (liquid only) ```{=tex} \begin{align} j_D = \frac{1.1068}{\varepsilon} N_{Re}^{-0.72} \end{align} ``` ## Example 3: comparison between geometries _Adapted from Problem 7.3-3_. Let's estimate the gas-phase mass transfer coefficient $k_G$ (kg mol/(m$^2$ s Pa)) for mass transfer of water vapour to solids with different shapes. Consider a water vapour (A) in air (B) at 338.6 K and 101.32 Pa flowing through a big duct containing solids with various geometries. The flow velocity is 3.66 m/s. The water vapour concentration is small, so property of air is used ($\mu=2.03\times 10^-5$ Pa$\cdot$s, $\rho=1.043$ kg/m$^3$). From the table, $D_{AB} = 2.88 \times 10^{-5}$ m$^2$/s at 315 K. Compare the values for following geometries, which case? a) Flow parallel to flat plate with length $L=2.54$ cm b) A single sphere with diameter $D=2.54$ cm c) Packed beds using spheres of average diameter $D_p=2.54$ cm and $\epsilon=0.35$ ## Example 3: results ::: {.callout-tip} - Do not forget to correct $D_{AB}$ for temperature! - Choose the right $N_{\text{Sh}}$ or $j_D$ formula according to $N_{\text{Re}}$ and $N_{\text{Sc}}$ - $k_G \approx k_G' = k_c' / (RT)$ ::: - Flat surface: $k_G = 1.738\times 10^{-8}$ kg mol/(m$^2$ s Pa). $N_{\text{Sh}} = 38.03$ - Single sphere: $k_G = 1.984\times 10^{-8}$ kg mol/(m$^2$ s Pa). $N_{\text{Sh}} = 43.40$ - Packed bed: $k_G = 7.60\times 10^{-8}$ kg mol/(m$^2$ s Pa). $N_{\text{Sh}} = 166.32$ Packed bed clearly wins! ## Summary - Dimensionless numbers can be used to correlate mass transfer problems in different flow rate, dimension etc - Typically, start with a known geometry (pipe? parallel plate? sphere? packed bed?) - Find the correlation with dimensionless numbers $N_{Re}$, $N_{Sc}$ - Calculate the final mass transfer rate - Geometry plays an important role in determining $k_c'$!