CHE 318 Lecture 26

Designing Packed Beds For Concentrated Systems

Author

Dr. Tian Tian

Published

March 13, 2026

Note

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Learning outcomes

After this lecture, you will be able to:

Recall the general procedure for designing a packed tower in concentrated systems.
Describe the concepts of transfer unit and number of transfer units in packed-bed design.
Apply realistic concentration profiles to estimate tower height.

Cheatsheet for packed bed design

Recall: packed tower height equation (individual phase)

General equation for gas. What the meanings for each term?

\[\begin{align} Z &= \int_{y_2}^{y_1} \frac{V'}{k_y'aS} \cdot \frac{(1-y)_{im}}{(1-y)^2(y-y_{i})} \, dy \\ &= \int_{y_2}^{y_1} \frac{V}{k_y'aS} \cdot \frac{(1-y)_{im}}{(1-y)(y-y_{i})} \, dy \end{align}\]

Recall: height equation for diluted system

If $x<0.1$ and $y<0.1$, typically use (for gas phase):

\[ Z = \left[ \frac{V}{k'_y a S} \frac{(1 - y)_{im}}{1 - y} \right] \int_{y_2}^{y_1} \frac{dy}{y - y_i} \]

or even simpler

\[ Z = \left[ \frac{V}{k'_y a S} \right] \int_{y_2}^{y_1} \frac{dy}{y - y_i} \]

what insights can we get?

Motivation: transfer units for packed beds

From unit analysis, the term $\frac{V}{k'_y a S}$ characterizes the height. We often use the concept of transfer unit to determine the performance of mass transfer in each phase. For gas phase, we have the “height” of the transfer unit $H_G$ as

\[\begin{align} H_G = \frac{V}{k_y' a S} \end{align}\]

$H_G$ is controlled by ratio of flow rate and mass transfer in gas phase
smaller $H_G$ means more efficient packing (thus needing shorter tower)

Define the number of transfer units

The rest of the equation for $Z$ can be rewritten as “number of transfer units”, similar to the theoretical number of trays in a tray tower. For gas phase general case, we have

\[\begin{align} N_G = \int_{y_2}^{y_1} \frac{(1-y)_{im}}{(1-y)(y-y_i)} \,dy \end{align}\]

For dilute system, we can approximate $N_G$ by

\[\begin{align} N_G &= \int_{y_2}^{y_1} \frac{dy}{y-y_i} \\ &= \frac{y_1 - y_2}{(y - y_i)_m} \end{align}\]

Transfer unit view of packed bed

For dilute system, we can see that

\[\begin{align} Z &= H_G N_G \\ &= H_G \frac{y_1 - y_2}{(y - y_i)_m} \end{align}\]

Total height $Z$ is a stack of “transfer units” with height $H_G$
Number of the units is governed by the total absorption $(y_1 - y_2)$ and driving force $(y - y_i)_m$
Need to absorb more (larger $(y_1 - y_2)$)? 👉 higher tower
Operation line close to equilibrium? 👉 higher tower

Example 1: packed bed height estimation using transfer units

Similar example in Lecture 24, Acetone is being absorbed by water in a packed bed column having a cross sectional area of 0.186 m$^2$ at 293 K and 1 atm. The inlet air contains 2.6 mol% acetone and outlet contains 0.5%. The gas flow is 13.65 kg mol inert air per hour. The pure water inlet flow is 45.36 kg mol water per hour. The coefficient $k_y' a$ is estimated to be $3.78 \times 10^{-2}$ kg mol/(s·m$^3$) and $k_x' a$ is $6.16 \times 10^{-2}$ kg mol/(s·m$^3$). The equilibrium line can be approximated by $y = 1.186 x$.

Now calculate the height using the dilute regime equation
Now calculate the height using $H_G$ and $N_G$ statements

Example 1 diagram

Example 1 solution

Key numbers during calculation:

Liquid outlet $x_1 = 0.00648$
$(1 - y)_{im} \approx (1 - x)_{im} \approx 1$ 👉 use dilute regime is justified
$(y - y_i)_{m} = 0.00602$
Average $V = (V_1 + V_2)/2 =3.852 \times 10^{-3}$ kg mol/s

Final results:

$Z=1.911$ m
$H_G=0.548$ m, $N_G=3.488$

Practical estimation of $k_y’ a$ or $H_G$

In real packed beds, coefficients are often obtained from correlations (using dimensionless numbers etc), which may depend on:

$N_{Re}$, $N_{Sc}$
Weight flow rates $G_y$, $G_x$
Packing type (shape? diameter?)

Design of packed bed in realistic situations

In many industrial applications, engineers are interested in one of the approaches to estimate the desired tower height:

Estimate the correlation of $k_y' a$ and $k_x' a$ as functions of $G_x$ and $G_y$

\[ k_y'a = f_y(G_x, G_y, \text{pack type});\quad k_x'a = f_x(G_x, G_y, \text{pack type}) \]

Estimate $H_G$ (and / or $H_L$) from the correlation

\[ H_G = f_G(G_x, G_y, N_{Sc}, \text{pack type}) \]

Either way, the flow rates are locally dependent, so integration should be used. Let’s see a detailed example from the textbook.

Example 2: concentrated packed bed design

(Example 10.7-1) A tower packed with 25.3 mm ceramic rings is used to absorb SO$_2$ from air using pure water at 293 K and 1 atm. The entering gas contains 20 mol% SO$_2$ and leaving gas has 2 mol%. The inert air flow is $6.53\times10^{-4}$ kg mol/s and inlet water flow is $4.20\times10^{-2}$ kg mol/s. Cross sectional area of the tower is 0.0929 m$^2$. From literature, the film mass transfer coefficients are estimated as

\[ k_y' a = 0.0594 G_y^{0.7} G_x^{0.25} \qquad k_x' a = 0.152 G_x^{0.82} \]

where $G_x$ and $G_y$ are total weight of liquid / gas flow per second per area, respectively. Estimate the tower height needed.

Example 2: steps?

We have seen $y_2 = 0.2$, so in general the dilute solution is not accurate. Use the following steps instead:

At each point on the operating line: determine $(x,y)$
Use local flow rates $V_y$, $L_x$ to find
- $G_y = (M_{\text{air}} V' + M_{A}V_y)/S$
- $G_x = (M_{\text{water}} L' + M_{A}V_x)/S$
Calculate $k_y' a$ and $k_x' a$ for each point
Use trial-and-error / numerical method to determine $(x_i,y_i)$
Integrate through the bed to get $Z$

Notes on the integration

The height equation is basically integration of a function $f(y)$ over the column. Manual integration using trapezoidal rule usually require at least 5 points along $y$

\[\begin{align} Z &= \int_{y_2}^{y_1} \frac{V dy}{\frac{k'_y a S}{(1 - y)_{im}} (1 - y)(y - y_i)} \\ &= \int_{y_2}^{y_1} f(y) dy \\ &= \sum_{j=0}^{n-1} [f(y_j) + f(y_{j + 1})] \frac{y_{j+1} - y_{j}}{2} \end{align}\]

Example 2: operating line

Example 2: textbook solutions (1)

Example 2: textbook solutions (2)

Final integration result: $Z = 1.588$ m

Summary

In this lecture we have seen how the height of a realistic packed bed tower is calculated and how to obtain relative statements using the transfer unit concepts.

In next weeks, we will talk about the final topic: mass transfer coupled with heat transfer. One useful example will be the humidification tower.

--- title: "CHE 318 Lecture 26" subtitle: "Designing Packed Beds For Concentrated Systems" author: "Dr. Tian Tian" date: "2026-03-13" format: html: {} revealjs: output-file: slides.html pdf: output-file: L26.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L26.pdf) - Handwritten notes 👉 [Open in pdf](./public/L26_annotated.pdf) ::: ::: ## Learning outcomes {.center} After this lecture, you will be able to: - **Recall** the general procedure for designing a packed tower in concentrated systems. - **Describe** the concepts of transfer unit and number of transfer units in packed-bed design. - **Apply** realistic concentration profiles to estimate tower height. ## Cheatsheet for packed bed design ![Physical copies distributed in class](../L25/public/cheatsheet-2phase.svg) ## Recall: packed tower height equation (individual phase) General equation for gas. What the meanings for each term? ```{=tex} \begin{align} Z &= \int_{y_2}^{y_1} \frac{V'}{k_y'aS} \cdot \frac{(1-y)_{im}}{(1-y)^2(y-y_{i})} \, dy \\ &= \int_{y_2}^{y_1} \frac{V}{k_y'aS} \cdot \frac{(1-y)_{im}}{(1-y)(y-y_{i})} \, dy \end{align} ``` ## Recall: height equation for diluted system If $x<0.1$ and $y<0.1$, typically use (for gas phase): $$ Z = \left[ \frac{V}{k'_y a S} \frac{(1 - y)_{im}}{1 - y} \right] \int_{y_2}^{y_1} \frac{dy}{y - y_i} $$ or even simpler $$ Z = \left[ \frac{V}{k'_y a S} \right] \int_{y_2}^{y_1} \frac{dy}{y - y_i} $$ what insights can we get? ## Motivation: transfer units for packed beds From unit analysis, the term $\frac{V}{k'_y a S}$ characterizes the height. We often use the concept of **transfer unit** to determine the performance of mass transfer in each phase. For gas phase, we have the "height" of the transfer unit $H_G$ as ```{=tex} \begin{align} H_G = \frac{V}{k_y' a S} \end{align} ``` - $H_G$ is controlled by ratio of flow rate and mass transfer in gas phase - smaller $H_G$ means more efficient packing (thus needing shorter tower) ## Define the number of transfer units The rest of the equation for $Z$ can be rewritten as "number of transfer units", similar to the theoretical number of trays in a tray tower. For gas phase general case, we have ```{=tex} \begin{align} N_G = \int_{y_2}^{y_1} \frac{(1-y)_{im}}{(1-y)(y-y_i)} \,dy \end{align} ``` For dilute system, we can approximate $N_G$ by ```{=tex} \begin{align} N_G &= \int_{y_2}^{y_1} \frac{dy}{y-y_i} \\ &= \frac{y_1 - y_2}{(y - y_i)_m} \end{align} ``` ## Transfer unit view of packed bed For dilute system, we can see that ```{=tex} \begin{align} Z &= H_G N_G \\ &= H_G \frac{y_1 - y_2}{(y - y_i)_m} \end{align} ``` - Total height $Z$ is a stack of "transfer units" with height $H_G$ - Number of the units is governed by the total absorption $(y_1 - y_2)$ and driving force $(y - y_i)_m$ - Need to absorb more (larger $(y_1 - y_2)$)? 👉 higher tower - Operation line close to equilibrium? 👉 higher tower ## Example 1: packed bed height estimation using transfer units Similar example in [Lecture 24](../L24), Acetone is being absorbed by water in a packed bed column having a cross sectional area of 0.186 m$^2$ at 293 K and 1 atm. The inlet air contains 2.6 mol% acetone and outlet contains 0.5%. The gas flow is 13.65 kg mol inert air per hour. The pure water inlet flow is 45.36 kg mol water per hour. The coefficient $k_y' a$ is estimated to be $3.78 \times 10^{-2}$ kg mol/(s·m$^3$) and $k_x' a$ is $6.16 \times 10^{-2}$ kg mol/(s·m$^3$). The equilibrium line can be approximated by $y = 1.186 x$. a) Now calculate the height using the dilute regime equation b) Now calculate the height using $H_G$ and $N_G$ statements ## Example 1 diagram ![](./public/ex1-solution.png) ## Example 1 solution Key numbers during calculation: 1. Liquid outlet $x_1 = 0.00648$ 2. $(1 - y)_{im} \approx (1 - x)_{im} \approx 1$ 👉 use dilute regime is justified 3. $(y - y_i)_{m} = 0.00602$ 4. Average $V = (V_1 + V_2)/2 =3.852 \times 10^{-3}$ kg mol/s Final results: 1. $Z=1.911$ m 2. $H_G=0.548$ m, $N_G=3.488$ ## Practical estimation of $k_y’ a$ or $H_G$ In real packed beds, coefficients are often obtained from correlations (using dimensionless numbers etc), which may depend on: - $N_{Re}$, $N_{Sc}$ - Weight flow rates $G_y$, $G_x$ - Packing type (shape? diameter?) ![](./public/packed-geom.png) ## Design of packed bed in realistic situations In many industrial applications, engineers are interested in one of the approaches to estimate the desired tower height: 1. Estimate the correlation of $k_y' a$ and $k_x' a$ as functions of $G_x$ and $G_y$ $$ k_y'a = f_y(G_x, G_y, \text{pack type});\quad k_x'a = f_x(G_x, G_y, \text{pack type}) $$ 2. Estimate $H_G$ (and / or $H_L$) from the correlation $$ H_G = f_G(G_x, G_y, N_{Sc}, \text{pack type}) $$ Either way, the flow rates are locally dependent, so integration should be used. Let's see a detailed example from the textbook. ## Example 2: concentrated packed bed design (Example 10.7-1) A tower packed with 25.3 mm ceramic rings is used to absorb SO$_2$ from air using pure water at 293 K and 1 atm. The entering gas contains 20 mol% SO$_2$ and leaving gas has 2 mol%. The inert air flow is $6.53\times10^{-4}$ kg mol/s and inlet water flow is $4.20\times10^{-2}$ kg mol/s. Cross sectional area of the tower is 0.0929 m$^2$. From literature, the film mass transfer coefficients are estimated as $$ k_y' a = 0.0594 G_y^{0.7} G_x^{0.25} \qquad k_x' a = 0.152 G_x^{0.82} $$ where $G_x$ and $G_y$ are total weight of liquid / gas flow per second per area, respectively. Estimate the tower height needed. ## Example 2: steps? We have seen $y_2 = 0.2$, so in general the dilute solution is not accurate. Use the following steps instead: 1. **At each point** on the operating line: determine $(x,y)$ 2. Use local flow rates $V_y$, $L_x$ to find - $G_y = (M_{\text{air}} V' + M_{A}V_y)/S$ - $G_x = (M_{\text{water}} L' + M_{A}V_x)/S$ 3. Calculate $k_y' a$ and $k_x' a$ for each point 4. Use trial-and-error / numerical method to determine $(x_i,y_i)$ 5. Integrate through the bed to get $Z$ ## Notes on the integration The height equation is basically integration of a function $f(y)$ over the column. Manual integration using trapezoidal rule usually require at least 5 points along $y$ ```{=tex} \begin{align} Z &= \int_{y_2}^{y_1} \frac{V dy}{\frac{k'_y a S}{(1 - y)_{im}} (1 - y)(y - y_i)} \\ &= \int_{y_2}^{y_1} f(y) dy \\ &= \sum_{j=0}^{n-1} [f(y_j) + f(y_{j + 1})] \frac{y_{j+1} - y_{j}}{2} \end{align} ``` ![](./public/non-linear-inte.png) ## Example 2: operating line ![](./public/ex2-operating.png) ## Example 2: textbook solutions (1) ![](./public/ex2-step1.png) ## Example 2: textbook solutions (2) ![](./public/ex2-step2.png) Final integration result: $Z = 1.588$ m ## Summary In this lecture we have seen how the height of a realistic packed bed tower is calculated and how to obtain relative statements using the transfer unit concepts. In next weeks, we will talk about the final topic: mass transfer coupled with heat transfer. One useful example will be the humidification tower.