Solving Slurry Bed Adsorption: Calculating Residence Time & Isotherm Selection

[Will26040]

Summary:: I have an assignment on slurry bed adsorption and would like to know how to calculate residence time as it is not specified. I was also wondering if anyone had any suggestions for another isotherm to try and fit the data to (other than freundlich). Thanks

$\frac {dq} {dt} = \frac {k_la .t_{res}} {1+k_la.t_{res}}.\frac QS.[c_F-c^*]$

where:
$t_{res} =$ residence time $(s)$
$k_la =$ volumetric mass transfer coefficient $(s^{-1})$
$Q=$Volumetric flowrate
$S=$adsorbent mass
$c_F=$ feed concentration
$c^*=$Concentration in solution in equilibrium with the loading on the adsorbent (suitable isotherm used)

I know that I need to use the equation specified above and solve numerically to get the loading after 4 hours. However, I am not sure how to get the residence time, as this has been specified in previous examples that I have done. I was also wondering if anyone had any suggestions for another isotherm to try and fit the data to (other than freundlich). Thanks

 

[Chestermiller]

Please describe your understanding of the derivation of this equation. What is q defined as?

 

[Will26040]

Thanks for the reply! q is the loading on the adsorbent at time t

Here is the derivation, where $c_f$ is concentration of adsorbate in feed and $c_{out}$ is the concentration of adsorbent in outlet. I understand it but we are usually given residence time, however it is not given in this question.

 

[Chestermiller]

The parameter a is the total surface area of solid particles in the slurry divided by the total volume of slurry. So the total molar flow rate of solute to the solid surfaces is $kaV(c_{out}-c^*)$, and this must equal to the volumetric flow rate Q of solvent through the unit times the change in concentration of solvent between inlet and outlet of the unit: $$Q(c_{out}-c_{F})=-kaV(c_{out}-c^*)$$What does that tell you about the residence time?

 

[Will26040]

I'm not sure sorry. How would you get the total volume of slurry (V) in this example? as there is a continuous flow of liquid

 

[Eskimo18]

The parameter a is the total surface area of solid particles in the slurry divided by the total volume of slurry. So the total molar flow rate of solute to the solid surfaces is kaV(count−c∗), and this must equal to the volumetric flow rate Q of solvent through the unit times the change in concentration of solvent between inlet and outlet of the unit: Q(count−cF)=−kaV(count−c∗)What does that tell you about the residence time?
How would the parameter a be calculated? By using the the volume and the mass of the bed and researching particle size of activated carbon or by different means?

 

[Chestermiller]

I'm not sure sorry. How would you get the total volume of slurry (V) in this example? as there is a continuous flow of liquid

The problem statement states "the absorber bed ..., has a volume of 25 m^3." The value of V is 25 ^3, and the residence time is 12.5 seconds.

 

[Chestermiller]

Eskimo:

You. would have to know the average surface to volume ratio of the particles, the volume of the bed, and the volume fraction particles. The difficult parameter to obtain is k, or ka. In my judgment this would have to be determined experimentally.

 

[Will26040]

The problem statement states "the absorber bed ..., has a volume of 25 m^3." The value of V is 25 ^3, and the residence time is 12.5 seconds.

Is it really that simple? I thought the residence time was affected by how fast the adsorption equilibrium was reached

 

[Chestermiller]

Is it really that simple? I thought the residence time was affected by how fast the adsorption equilibrium was reached

No way. In a CSTR, the mean residence time is equal to the reactor volume divided by the volume rate of flow.

 

[Eskimo18]

You. would have to know the average surface to volume ratio of the particles, the volume of the bed, and the volume fraction particles. The difficult parameter to obtain is k, or ka. In my judgment this would have to be determined experimentally.

Can the value of k not be found using adsoprtion isotherms using the loading and pressure data given?
How would you know the volume fraction of the solvent, would it be 0.3%*2m^3s^-1?

 

[Chestermiller]

Can the value of k not be found using adsoprtion isotherms using the loading and pressure data given?

No. k has nothing to do with the equilibrium behavior. It is related to the rate of diffusion of solute (outside the surface of particles) across the solute concentration boundary layer between the bulk solvent at solute concentration $C_{out}$ and the solvent at the surface of the particles at solute concentration $C^*$. In terms of the solute concentration boundary layer thickness $\delta$, k is defined as $$k=\frac{D}{\delta}$$where D is the diffusivity of the solute in the solvent.

How would you know the volume fraction of the solvent, would it be 0.3%*2m^3s^-1?

Where did the "2" come from? Volume fraction in a gas is a synonym for mole fraction.

 

[Will26040]

Please could you point me in the direction of an equation used for desorption (part b), I have no idea what to use. Thanks

 

[Chestermiller]

Please could you point me in the direction of an equation used for desorption (part b), I have no idea what to use. Thanks

You use the same equation, except that now, C* exceeds count