CHE 318 Lecture 21

In-Depth Analysis of Packed Bed Columns

Author

Dr. Tian Tian

Published

March 2, 2026

Note

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Learning outcomes

After this lecture, you will be able to:

Recall how packed-bed mass transfer coefficients compare with those of other geometries.
Describe mass-balance equations for packed-bed columns.
Identify concentration and pressure profiles inside a packed column.

Recall from last week: mass transfer correlations

Goal: calculate $k_c'$ from:
- geometry ($L_D, L, D_P$, etc)
- properties ($\mu, \rho, D_{AB}, v$, etc)
Calculate $N_{\text{Re}}$, $N_{\text{Sc}}$
Get $j_D$ and / or $N_{\text{Sh}}$ to back-calculate $k_c'$

A deeper look into packed bed

Recall from last week: 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 $\epsilon=0.35$

Case 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!

Why packed bed columns have better mass transfer?

Packed bed geometry provides large surface-to-volume ratio ($\epsilon$ and $D_p$)
The boundary layer length $\delta c$ is small due to continuous disruption of liquid/gas interface
Overall $k_c'$ is larger than single geometry
Pressure drop $\Delta p$ over the column can be larger than other types!

Compute mass transfer in packed bed columns

In a packed bed design, it is often desired to know the following quantities:

The outlet concentration $c_{A2}$
The total flux of mass transfer across interface

It is still desired to use [In] - [Out] + [Gen] = [Acc]
What are each of these quantities?

Mass balance in packed beds

Volumetric flow rate is $Q$ (unit m$^3$/s)
Inlet, outlet concentration: $c_{A1}$, $c_{A2}$
Cross sectional area $S$
Interfacial mass transfer flux $N_A$, effective area $A_{\text{eff}}$

\[\begin{align} \text{[In]} - \text{[Out]} + \text{[Gen]} &= \text{[Acc]} \\ Q (c_{A1} - c_{A2}) + A_{\text{eff}} \hat{N}_A = 0 \end{align}\]

The effective area $A_{\text{eff}}$

$A_{\text{eff}}$ related to the surface-to-volume ratio $a$ and total bed volume $V_b$

\[\begin{align} A_{\text{eff}} = a V_{b} \end{align}\]

For spheres, we can derive $a$ (unit m$^2$/m$^3$)

\[\begin{align} a = \frac{6(1 - \epsilon)}{D_p} \end{align}\]

How to get $\hat{N}_A$?

$\hat{N}_A$ is the average mass transfer flux across the interface.
To solve it, we still use a control volume from $z$ to $z+ dz$
Locally, driving force $c_{Ai} - c_{A}(z)$
Local mass transfer flux

\[\begin{align} N_A(z) = k_c \left[c_{Ai} - c_{A}(z)\right] \end{align}\]

Differential equation for $c_A(z)$

\[\begin{align} N_A a S dz &= Q d c_A \\ \frac{k_c a S}{Q} dz &= \frac{d c_A}{c_{Ai} - c_{A}(z)} \end{align}\]

Integrating over $z$ we get:

\[\begin{align} \ln\!\left( \frac{c_{Ai} - c_{A1}}{c_{Ai} - c_{A}(z)} \right) = \frac{k_c a S z}{Q} \end{align}\]

Concentration profile of packed bed (single phase)

When $c_{Ai}$ is constant (e.g. solid-gas interface), concentration follows:

\[\begin{align} c_{A}(z) = c_{Ai} - (c_{Ai} - c_{A1})\exp(- \frac{k_c a S}{Q} z) \end{align}\]

Analog: reactive wall in pipe system (unsteady-state mass transfer in Lecture 11)

\[\begin{align} c_{A}(z) = c_{Ai} - (c_{Ai} - c_{A1})\exp(- \frac{4 k_c z}{Dv_m}) \end{align}\]

The pipe wall problem corresponds to $a=4 / D$, similar to a packed bed problem.

Clearly, packed bed can achieve saturation much faster, because $D_p$ is usually a few mm to 1 inch.

Outlet concentration for packed bed

If a packed bed column has height $H$, outlet concentration $c_{A2}$ follows:

\[\begin{align} c_{A2} = c_{Ai} - (c_{Ai} - c_{A1})\exp(- \frac{k_c a S H}{Q}) \end{align}\]

Rearrange gives

\[\begin{align} \ln\!\left( \frac{c_{Ai} - c_{A1}}{c_{Ai} - c_{A2}} \right) &= \frac{k_c a S H}{Q} \\ &= \frac{k_c A_{\text{eff}}}{Q} \end{align}\]

Solving the average mass transfer flux $\hat{N}_A$

We can further use the mass balance equation $Q(c_{A1} - c_{A2}) + \hat{N}_A A_{\text{eff}} = 0$.
Use the fact $c_{A2} - c_{A1} = (c_{Ai} - c_{A1}) - (c_{Ai} - c_{A2})$

That gives us

\[ \hat{N}_{A} = k_c \frac{(c_{Ai} - c_{A1}) - (c_{Ai} - c_{A2})}{\ln\!\left( \frac{c_{Ai} - c_{A1}}{c_{Ai} - c_{A2}} \right)} \]

Average mass transfer flux by log-mean driving force

The previous result means the average driving force in a exponentially-changing concentration profile like in packed bed, should be expressed in log-mean driving force form.

It is an expression we will frequently see in other column setups with mixing interfaces

\[ \hat{N}_{A} = k_c \frac{(c_{Ai} - c_{A1}) - (c_{Ai} - c_{A2})}{\ln\!\left( \frac{c_{Ai} - c_{A1}}{c_{Ai} - c_{A2}} \right)} \]

Example 4: packed bed design

In this lecture, we show that the outlet concentration will be saturated when the tube is long enough.

\[\begin{align} c_{A2} = c_{Ai} - (c_{Ai} - c_{A1})\exp(- \frac{A k_c}{Q}) \end{align}\]

Consider all parameters in case 3, how high should we design the packed bed to ensure saturated outlet concentration?
$D_p=2.54$ cm, $\epsilon=0.35$, $k_c'=0.214$ m/s, $v_m=3.66$ m/s
You can consider effective saturation means $\frac{A k_c}{Q} = 5$

Example 4: packed bed design – results

Minimal height: $H_{\text{min}}$

\[ H_{\text{min}} = \frac{5 v_m D_p}{6 (1 - \epsilon) k_c'} \]

In case 3 setup, $H_{\text{min}}=0.556$ m

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

--- title: "CHE 318 Lecture 21" subtitle: "In-Depth Analysis of Packed Bed Columns" author: "Dr. Tian Tian" date: "2026-03-02" format: html: {} revealjs: output-file: slides.html pdf: output-file: L21.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L21.pdf) - Handwritten notes 👉 [Open in pdf](./public/L21_annotated.pdf) ::: ::: ## Learning outcomes {.center} After this lecture, you will be able to: - **Recall** how packed-bed mass transfer coefficients compare with those of other geometries. - **Describe** mass-balance equations for packed-bed columns. - **Identify** concentration and pressure profiles inside a packed column. ## Recall from last week: mass transfer correlations - Goal: calculate $k_c'$ from: - geometry ($L_D, L, D_P$, etc) - properties ($\mu, \rho, D_{AB}, v$, etc) - Calculate $N_{\text{Re}}$, $N_{\text{Sc}}$ - Get $j_D$ and / or $N_{\text{Sh}}$ to back-calculate $k_c'$ ## A deeper look into packed bed ![](./public/cheatsheet-packedbed-only.svg) ## Recall from last week: 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 $\epsilon=0.35$ ## Case 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! ## Why packed bed columns have better mass transfer? - Packed bed geometry provides large surface-to-volume ratio ($\epsilon$ and $D_p$) - The boundary layer length $\delta c$ is small due to continuous disruption of liquid/gas interface - Overall $k_c'$ is larger than single geometry - Pressure drop $\Delta p$ over the column can be larger than other types! ## Compute mass transfer in packed bed columns In a packed bed design, it is often desired to know the following quantities: 1) The outlet concentration $c_{A2}$ 2) The total flux of mass transfer across interface - It is still desired to use [In] - [Out] + [Gen] = [Acc] - What are each of these quantities? ## Mass balance in packed beds - Volumetric flow rate is $Q$ (unit m$^3$/s) - Inlet, outlet concentration: $c_{A1}$, $c_{A2}$ - Cross sectional area $S$ - Interfacial mass transfer flux $N_A$, effective area $A_{\text{eff}}$ ```{=tex} \begin{align} \text{[In]} - \text{[Out]} + \text{[Gen]} &= \text{[Acc]} \\ Q (c_{A1} - c_{A2}) + A_{\text{eff}} \hat{N}_A = 0 \end{align} ``` ## The effective area $A_{\text{eff}}$ - $A_{\text{eff}}$ related to the surface-to-volume ratio $a$ and total bed volume $V_b$ ```{=tex} \begin{align} A_{\text{eff}} = a V_{b} \end{align} ``` - For spheres, we can derive $a$ (unit m$^2$/m$^3$) ```{=tex} \begin{align} a = \frac{6(1 - \epsilon)}{D_p} \end{align} ``` ## How to get $\hat{N}_A$? - $\hat{N}_A$ is the **average** mass transfer flux across the interface. - To solve it, we still use a control volume from $z$ to $z+ dz$ - Locally, driving force $c_{Ai} - c_{A}(z)$ - Local mass transfer flux ```{=tex} \begin{align} N_A(z) = k_c \left[c_{Ai} - c_{A}(z)\right] \end{align} ``` ## Differential equation for $c_A(z)$ ```{=tex} \begin{align} N_A a S dz &= Q d c_A \\ \frac{k_c a S}{Q} dz &= \frac{d c_A}{c_{Ai} - c_{A}(z)} \end{align} ``` Integrating over $z$ we get: ```{=tex} \begin{align} \ln\!\left( \frac{c_{Ai} - c_{A1}}{c_{Ai} - c_{A}(z)} \right) = \frac{k_c a S z}{Q} \end{align} ``` ## Concentration profile of packed bed (single phase) - When $c_{Ai}$ is constant (e.g. solid-gas interface), concentration follows: ```{=tex} \begin{align} c_{A}(z) = c_{Ai} - (c_{Ai} - c_{A1})\exp(- \frac{k_c a S}{Q} z) \end{align} ``` - Analog: reactive wall in pipe system (unsteady-state mass transfer in [Lecture 11](../L11)) ```{=tex} \begin{align} c_{A}(z) = c_{Ai} - (c_{Ai} - c_{A1})\exp(- \frac{4 k_c z}{Dv_m}) \end{align} ``` The pipe wall problem corresponds to $a=4 / D$, similar to a packed bed problem. - Clearly, packed bed can achieve saturation much faster, because $D_p$ is usually a few mm to 1 inch. ## Outlet concentration for packed bed - If a packed bed column has height $H$, outlet concentration $c_{A2}$ follows: ```{=tex} \begin{align} c_{A2} = c_{Ai} - (c_{Ai} - c_{A1})\exp(- \frac{k_c a S H}{Q}) \end{align} ``` - Rearrange gives ```{=tex} \begin{align} \ln\!\left( \frac{c_{Ai} - c_{A1}}{c_{Ai} - c_{A2}} \right) &= \frac{k_c a S H}{Q} \\ &= \frac{k_c A_{\text{eff}}}{Q} \end{align} ``` ## Solving the average mass transfer flux $\hat{N}_A$ - We can further use the mass balance equation $Q(c_{A1} - c_{A2}) + \hat{N}_A A_{\text{eff}} = 0$. - Use the fact $c_{A2} - c_{A1} = (c_{Ai} - c_{A1}) - (c_{Ai} - c_{A2})$         That gives us $$ \hat{N}_{A} = k_c \frac{(c_{Ai} - c_{A1}) - (c_{Ai} - c_{A2})}{\ln\!\left( \frac{c_{Ai} - c_{A1}}{c_{Ai} - c_{A2}} \right)} $$ ## Average mass transfer flux by log-mean driving force The previous result means the average driving force in a exponentially-changing concentration profile like in packed bed, should be expressed in **log-mean driving force** form. It is an expression we will frequently see in other column setups with mixing interfaces $$ \hat{N}_{A} = k_c \frac{(c_{Ai} - c_{A1}) - (c_{Ai} - c_{A2})}{\ln\!\left( \frac{c_{Ai} - c_{A1}}{c_{Ai} - c_{A2}} \right)} $$ ![](./public/log-mean-df.png) ## Example 4: packed bed design In this lecture, we show that the outlet concentration will be saturated when the tube is long enough. ```{=tex} \begin{align} c_{A2} = c_{Ai} - (c_{Ai} - c_{A1})\exp(- \frac{A k_c}{Q}) \end{align} ``` - Consider all parameters in case 3, how high should we design the packed bed to ensure saturated outlet concentration? - $D_p=2.54$ cm, $\epsilon=0.35$, $k_c'=0.214$ m/s, $v_m=3.66$ m/s - You can consider effective saturation means $\frac{A k_c}{Q} = 5$ ## Example 4: packed bed design -- results - Minimal height: $H_{\text{min}}$ $$ H_{\text{min}} = \frac{5 v_m D_p}{6 (1 - \epsilon) k_c'} $$ - In case 3 setup, $H_{\text{min}}=0.556$ m ## 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

CHE 318 Lecture 21

Learning outcomes

Recall from last week: mass transfer correlations

A deeper look into packed bed

Recall from last week: comparison between geometries

Case 3: results

Why packed bed columns have better mass transfer?

Compute mass transfer in packed bed columns

Mass balance in packed beds

The effective area \(A_{\text{eff}}\)

How to get \(\hat{N}_A\)?

Differential equation for \(c_A(z)\)

Concentration profile of packed bed (single phase)

Outlet concentration for packed bed

Solving the average mass transfer flux \(\hat{N}_A\)

Average mass transfer flux by log-mean driving force

Example 4: packed bed design

Example 4: packed bed design – results

Summary