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Prediction of the Break-Point Time in Fixed-Bed Adsorption At low concentrations, the equilibrium for the antibiotic novobiocin and Dowex \(21 \mathrm{~K}\) anion exchange resin is linear, $$ q_{i}=125 c_{i}^{*} $$ for \(q_{i}\) and \(c_{i}^{*}\) in units of milligrams per milliliter. For the range of concentrations where this isotherm is valid, the mass transfer coefficient \(K_{a}\) averages \(82 \mathrm{~h}^{-1}\). Assuming a linear isotherm, estimate the break-point time (where \(c_{i} / c_{i 0}=0.05\) ) in a fixed-bed adsorber with a bed length of \(20 \mathrm{~cm}\) and superficial velocity of \(40 \mathrm{~cm} / \mathrm{h}\). (Data from P. A. Belter, F. L. Cunningham, and J. W. Chen, "Development of a recovery process for novobiocin," Biotechnol. Bioeng., vol. 15, p. 533,1973 .)

Short Answer

Expert verified
The computed break-point time \(t_{B}\) provides the estimated time for the adsorption process to reach a point where \(c_{i} / c_{i 0}=0.05\).

Step by step solution

01

Calculate the volume rate per cross-sectional area

The superficial velocity is defined as the volume rate per cross-sectional area of the bed. Thus, keeping in mind that the given superficial velocity is \(40 \mathrm{~cm} / \mathrm{h}\), the volume flow rate \(V\) can be calculated as \(V = vA\).
02

Apply the definition of the mass transfer coefficient

The mass transfer coefficient \(K_{a}\) is given as \(82 \mathrm{~h}^{-1}\). Applying it to the equation \(J = K_{a}(C_{s} - C)\) and considering that at the start of the adsorption process, \(C = C_{0}\), the initial flux \(J_{0}\) becomes \(J_0 = K_{a}C_{0}\).
03

Define the initial solid phase concentration

Since we are at the start of the adsorption process, the solid phase has no solute. Thus, the initial solid phase concentration \(q_{0}\) is zero for \(q = q_{i} = 125 c_{i}^{*}\).
04

Determine the break-point time

The break-point time \(t_{B}\) is when \(c_{i} / c_{i 0}=0.05\). Thus, using the definition of \(q_{i}\) with \(c_{i}^{*} = c_{i 0}*0.05\), and keeping in mind that the initial solid phase concentration \(q_{0}\) is zero, we can find the bed length \(L\) and the volume flow rate \(V\), then compute the break-point time \(t_{B}\) as \(t_{B} = q_{i}*L/V\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Adsorption Isotherms
Adsorption isotherms are crucial in understanding how solutes interact with adsorbents in processes like water purification or air filtration. When we talk about isotherms in the context of a fixed-bed adsorber, we're referring to the relationship between the concentration of the adsorbate on the adsorbent (\( q_i \) and its concentration in the fluid phase (\( c_i^* \). The simplest type of isotherm is the linear isotherm, as shown in the given exercise.

For a linear isotherm, the formula is straight forward: \( q_i = k c_i^* \), where \( k \) is the proportionality constant. In this case, that constant is 125 mg/mL, indicating that the adsorption capacity increases linearly with the concentration of novobiocin in the solution. This direct proportionality simplifies calculations and predictions about the adsorber's performance but only applies within certain concentration ranges where the linearity holds true.
Mass Transfer Coefficient
The mass transfer coefficient, denoted as \( K_a \) in our exercise, is a measure that indicates the rate at which a solute moves from the fluid phase to the adsorbent surface. Think of it like the speed of a runner; the higher the \( K_a \), the faster the adsorption happens.

In technical terms, it represents how quickly the equilibrium is reached between the solid and liquid phases. We apply it using the equation \( J = K_a(C_s - C) \), where \( J \) is the flux of adsorbate, \( C_s \) is the concentration at the surface, and \( C \) is the concentration in the bulk fluid. Understanding and calculating the mass transfer coefficient is vital; it influences the design and operation of the adsorber by affecting the break-point time and the overall efficiency of the adsorption process.
Break-Point Time Calculation
One of the most important parameters in the design and operation of a fixed-bed adsorber is the break-point time, denoted as \( t_B \) in the exercise. This is the moment when the solute concentration in the effluent reaches a defined percentage of its initial value, which in our case is 5%. Break-point time is essential because it indicates when the adsorbent is near saturation and can no longer purify the incoming fluid effectively.

To calculate the break-point time, we incorporate our linear isotherm relation \( q_i = 125 c_i^* \) and the definition of \( q_i \) at the break-point concentration 0.05 \( c_{i0} \). By knowing the bed length (\( L \) and the volume flow rate (\( V \), we solve for \( t_B \) using the relation \( t_B = q_i * L / V \), which gives us the time at which the effluent concentration is no longer acceptable, and the adsorption bed thus needs to be regenerated or replaced. It's a critical calculation for ensuring the continuous treatment of the fluid without risking breakthrough of the adsorbate.

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Most popular questions from this chapter

Design of a Protein Purification Process (Mini-Case Study) Propose a purification process for the protein described. Assume a bacterial process. Design of the process will involve estimation of protein properties, for which a spreadsheet is provided at the textbook website (http://www.biosep.ou.edu). (Protein adapted from H. Zou, T. J. McGarry, T. Bernal, and M. W. Kirschner, "Identification of a vertebrate sister-chromatid separation inhibitor involved in transformation and tumorigenesis," Science, vol. 285, pp. \(418-422,1999 .)\) Sequence 1 MATLIYVDKENGEPGTRVVAKDGLKLGSGPSIKALDGRSQVSTPRFGKTFD 52 APPACLPKATRKALGTVNRATEKSVKT-1\ldots.....S-S \(\ldots \ldots+1\) KGPLKQKQPSCFSAKKMTEKTCVKAKS 106 SVPASDDAYPEIEKFFPFNPLDFESFDLPEEHQIAHLPLSGVPLMILDEER 157 ELEKLFQLGPPSPVKMPSPPWESNLLQSPSSILSTLDVELPPVCCDIDI Research Production Process A PCR-amplified gene was cloned into a modified pET28a vector to express recombinant protein in \(E\). coli. Protein more than \(90 \%\) pure was obtained by affinity purification with Ni-nitrilotriacetic acid (NTA) beads followed by Resource Q column chromatography. In addition to the traditional process and host-cell-related impurities, the following substance related impurities are present in the process: Impurity 1: Des-Met (N-terminal methionine deleted for the desired protein) Impurity 2: C-terminal cleavage product (cleavage between lysine 9 and glutamic acid 10) 1 ENGEPGTRVVAKDGLKLGSGPSIKALDGRSQVSTPRFGKTFD 43 APPACLPKATRKALGTVNRATEKSVK \(=1=\cdots \cdots\) TKGPLKQKQPSCFSAKKMTEKTCVKAKS 97 SVPASDDAYPEIEKFFPFNPLDFESFDLPEEHQIAHLPLSGVPLMILDEER 148 ELEKLFQLGPPSPVKMPSPPWESNLLQSPSSILSTLDVELPPVCCDIDI Impurity 3: N-terminal cleavage product (cleavage between lysine 9 and glutamic acid 10 ) 1 MATLIYVDK Impurity 4: Misfolded isoform 1 MATLIYVDKENGEPGTRVVAKDGLKLGSGPSIKALDGRSQVSTPRFGKTFD 52 APPACLPKATRKALGTVNRATEKSVK- TKGPLKQKQPSCFSAKKMTEKTCVKAKS TKGPLKQKQPSCFSAKKMTEKTCVKAKS 106 SVPASDDAYPEIEKFFPFNPLDFESFDLPEEHQIAHLPLSGVPLM 157 ELEKLFQLGPPSPVKMPSPPWESNLLQSPSSILSTLDVELPPVCCDIDE Impurity 5: Covalent dimer between cysteine 56 on adjacent molecules Information about chromatography adsorbents can be found through links to vari- ous companies at the textbook website (www.biosep.ou.edu). There is also a link to a spreadsheet to approximate protein charge as a function of pH at the textbook website.

Three Binding Solutes Derive the isotherms for three binding solutes that all compete for the same sites on the resin. If \(K_{\text {eq2 }}>K_{\text {eq } 1}\) (c) \(c_{2}\) is low initially, and increases throughout the elution process.

Estimation of the Number of Theoretical Plates for Ion Exchange Chromatography of a Protein Following a Langmuir Isotherm A \(20 \mathrm{~cm}\) long ion exchange column is loaded with a protein solution at a concentration of \(3 \mathrm{mg} / \mathrm{ml}\) and is eluted isocratically at a superficial velocity of \(30 \mathrm{~cm} / \mathrm{h}\). The void fraction is \(0.3\). The protein has a \(K_{\text {eq }}\) (Langmuir isotherm) of \(1.0 \mathrm{ml} / \mathrm{mg}\), and the column's resin is saturated with the protein at a resin-phase concentration of \(10 \mathrm{mg} / \mathrm{ml}\). If the protein peak has a standard deviation of \(4.0 \mathrm{~min}\), estimate the number of theoretical plates in the column.

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Langmuir Isotherm For the Langmuir isotherm [Equation (7.2.6)], determine the relationship between the equilibrium constant \(K_{\text {eq }}\) and the mobile phase concentration [C] when the adsorbent is half saturated.

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