Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Biomass present in a fermentation broth is to be separated by vacuum filtration. Filter and broth characteristics are given below. $$ \begin{aligned} &A=50 \mathrm{~m}^{2}, \quad \Delta P=0.01 \mathrm{~N} / \mathrm{m}^{2}, \quad C=15 \mathrm{~kg} / \mathrm{m}^{3} \\ &\mu=0.003 \mathrm{~kg} / \mathrm{m}-\mathrm{s}, \quad \alpha=2 \mathrm{~m} / \mathrm{kg} \end{aligned} $$ a. If rate of filtration has a constant value of \(d V / d t=50 \mathrm{l} / \mathrm{min}\), determine the cake and filter resistances at \(t=30 \mathrm{~min}\). b. Determine the filter surface area \((A)\) required to filter 50001 broth within 60 min with the same pressure drop across the filter.

Short Answer

Expert verified
Cake resistance = 45 m^{-1} Filter resistance = 21.67 m^{-1} Filter surface area = 16.67 m^2

Step by step solution

01

Calculate the total volume filtered

Given the rate of filtration \(\frac{dV}{dt}=50 \text{ l/min}\), and the time \(t=30 \text{ min}\). The total volume filtered is \(V = \frac{dV}{dt} \times t = 50 \text{ l/min} \times 30 \text{ min} = 1500 \text{ l} = 1.5 \text{ m}^3\).
02

Calculate the cake resistance

The cake resistance \(R_c\) can be calculated using \(R_c = \alpha C V\), where \(\alpha = 2 \text{ m/kg}\), \(C = 15 \text{ kg/m}^3\), and \(V = 1.5 \text{ m}^3\). Substituting the values gives: \[R_c = 2 \text{ m/kg} \times 15 \text{ kg/m}^3 \times 1.5 \text{ m}^3 = 45 \text{ m}^{-1}\].
03

Calculate the filter resistance

Using the filtration equation which relates the pressure drop \(\Delta P = 0.01 \text{ N/m}^2\), viscosity \(\mu = 0.003 \text{ kg/(m-s)}\), and rate of filtration: \[\frac{dV}{dt} = \frac{A \Delta P}{\mu (R_f + R_c)}\]. Substitute values and solve for \(R_f\): \[50 \text{ l/min} \times \frac{1 \text{ m}^3}{1000 \text{ l}} = \frac{50 \text{ m}^2 \times 0.01 \text{ N/m}^2}{0.003 \text{ kg/(m-s)} \times (R_f + 45 \text{ m}^{-1})}\]. Simplifying, we get: \[\frac{1}{20} = \frac{1}{0.003(R_f + 45)}\], solving for \(R_f\) yields \[R_f = 21.67 \text{ m}^{-1}\].
04

Calculate filter surface area for different conditions

The filtration equation becomes \[\frac{83.33 \text{ l/min}}{1000} = \frac{A \times 0.01 \text{ N/m}^2}{0.003 \text{ kg/(m-s)} \times (21.67 + 45)\text{ m}^{-1}} \]. Simplifying and solving for \(A\) we get: \[\frac{1}{12} = \frac{A \times 0.01}{0.003 \times 66.67} \rightarrow A = 16.67 \text{ m}^2\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Filtration Resistance
When dealing with vacuum filtration, understanding filtration resistance is crucial. Filtration resistance is the opposition a filter offers to the flow of the liquid being filtered. It is divided into two parts: the resistance of the filter medium itself \((R_f)\) and the resistance of the filter cake \((R_c)\) that builds on the medium over time. In the given exercise, calculating these resistances allows us to determine how efficiently our filtration process will run. The formula to find the cake resistance \((R_c)\) is: \[ R_c = \alpha C V \] where \(\alpha\) is the specific cake resistance, \(C\) is the concentration of solids, and \(V\) is the volume of filtrate. The filter resistance \((R_f)\) can then be found using the overall filtration equation given by: \[ \frac{dV}{dt} = \frac{A \Delta P}{\mu (R_f + R_c)} \] where \(A\) is the filter area, \( \Delta P\) is the pressure drop, and \( \mu \) is the fluid viscosity. These calculations help us determine the effectiveness and efficiency of the filtration process.
Bioprocess Engineering Calculations
In bioprocess engineering, we utilize various calculations to optimize processes like vacuum filtration. This involves determining parameters like the rate of filtration, filter area, and resistances. In this exercise, we start with calculating the total volume filtered over a given time using the rate of filtration: \[ V = \frac{dV}{dt} \times t \] Here, \( \frac{dV}{dt}\) is the filtration rate and \(t\) is the time. From this volume, we can then calculate the cake resistance. Understanding these calculations is important for designing efficient filtration systems. Besides, the equations for filtration can guide engineers in modifying operational parameters to improve yield and reduce costs.
Cake Resistance
Cake resistance is a key concept in filtration processes. It refers to the resistance offered by the layer of filtered solids (the cake) that builds up on the filter medium. In the exercise, it is calculated using the formula: \[ R_c = \alpha C V \] The cake resistance directly impacts the efficiency of the filtration process. A higher cake resistance means more opposition to fluid flow, reducing the rate of filtration. Therefore, understanding and managing cake resistance is essential for optimal filter design and operation. Engineers can manipulate factors like pressure, filtration rate, and filter area according to the calculated cake and filter resistances to achieve desired filtration outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the use of gel chromatography to separate two proteins \(\mathrm{A}\) and \(\mathrm{B}\). The partition coefficient \(\left(K_{D}\right)\) for \(\mathrm{A}\) is \(0.5\) and for \(\mathrm{B}\) is \(0.15 . V_{o}\), the void volume in the column, is \(20 \mathrm{~cm}^{3}\). \(V_{i}\), the void volume within the gel particles, is \(30 \mathrm{~cm}^{3}\). The total volume of the column is 60 \(\mathrm{cm}^{3}\). The flow rate of elutant is \(100 \mathrm{~cm}^{3} / \mathrm{h}\). Ignoring dispersion and other effects, how long will it take for A to exit the column? How long for B?

Consider the scale-up of a chromatography column for purification of a protein. A column of \(40 \mathrm{~cm}\) length is used with a superficial velocity of \(40 \mathrm{~cm} / \mathrm{h}\). The peak concentration of the target protein exits at a time of \(100 \mathrm{~min}\). The standard deviation of the peak is \(14 \mathrm{~min}\). a. How long must you wait to collect \(90 \%\) of the protein? b. If the same column is used, but velocity is increased to \(60 \mathrm{~cm} / \mathrm{h}\) and external or Taylor dispersion controls, what will be the value of \(\sigma\) ? c. If the column is lengthened to \(60 \mathrm{~cm}\) while the velocity is at \(60 \mathrm{~cm} / \mathrm{h}\), how will \(\sigma\) change? Will the peak be sharper or broader than at \(40 \mathrm{~cm} / \mathrm{h}\) with a \(40-\mathrm{cm}\) column?

In a cross-flow ultrafiltration system used for filtration of proteins from a fermentation broth, gel resistance increases with protein concentration according to the following equation: \(R_{G}=0.5+0.01(C)\), where \(C\) is in \(\mathrm{mg} / \mathrm{l} .\) Pressure at the entrance of the system is \(P_{i}=6\) atm and at the exit is \(P_{0}=2\) atm. The shell side of the filter is open to the atmosphere, resulting in \(P_{f}=1 \mathrm{~atm}\). The membrane resistance is \(R_{M}=0.5 \mathrm{~atm} /\left(\mathrm{mg} / \mathrm{m}^{2} \cdot \mathrm{h}\right)\), and protein concentration in the broth is \(C=100 \mathrm{mg} / \mathrm{l}\). Determine: a. The pressure drop across the membrane. b. Filtration flux. c. Rejection coefficient of the membrane for effluent protein concentration of \(C_{f}=5 \mathrm{mg} / \mathrm{l}\).

In a cross-flow ultrafiltration unit, a protein of \(\mathrm{MW}=3 \times 10^{5}\) da is separated from the fermentation broth by using a UF membrane. The flow rate of liquid through a tube of diameter \(d=2 \mathrm{~cm}\) and length \(L=50 \mathrm{~cm}\) is \(Q=21 / \mathrm{min}\). The flow regime is turbulent, \(f=0.0005\), and \(C_{4}=2\left[\mathrm{~atm}(\mathrm{~s} / \mathrm{cm})^{2}\right] .\) The inlet pressure is \(P_{i}=2 \mathrm{~atm}\). Protein concentrations in the solution and on gel film are \(C_{B}=30 \mathrm{mg} / \mathrm{l}\) and \(C_{G}=100 \mathrm{~g} / \mathrm{l}\), respectively. a. Determine the exit pressure \(\left(P_{0}\right)\). b. Determine the transmembrane pressure drop \(\left(\Delta P_{M}\right)\). c. If the mass transfer coefficient \((k)\) for protein flux is \(k=5 \mathrm{~cm} / \mathrm{s}\), determine the flux of liquid through the UF membrane \((J)\). d. If the resistance of the filter is \(R_{M}=0.002 \mathrm{~atm} \cdot \mathrm{cm}^{2} \cdot \mathrm{s} / \mathrm{cm}^{3}\), determine the cake resistance, \(R_{G}\).

Components \(A\) and \(B\) of a binary mixture are to be separated in a chromatographic column. The adsorption isotherms of these compounds are given by the following equations: $$ \begin{aligned} &m_{A}=f_{A}(c)=\frac{k_{1} C_{A}}{k_{2}+C_{A}} \\ &m_{B}=f_{B}(c)=\frac{k_{1}^{\prime} C_{B}}{k_{2}^{\prime}+C_{B}} \end{aligned} $$ where \(k_{1}=0.2 \mathrm{mg}\) solute \(\mathrm{A}\) absorbed \(/ \mathrm{mg}\) adsorbent $$ \begin{aligned} &k_{2}=0.1 \mathrm{mg} \text { solute } / \mathrm{ml} \text { liquid } \\ &k_{1}^{\prime}=0.05 \mathrm{mg} \text { solute } \mathrm{B} \text { adsorbed } / \mathrm{mg} \text { adsorbent } \\ &k_{2}^{\prime}=0.02 \mathrm{mg} \text { solute/ml liquid } \end{aligned} $$ The bed contains \(3 \mathrm{~g}\) of very fine support particles. The bed volume is \(150 \mathrm{ml}\), bed porosity is \(\varepsilon=0.35\), and the cross-sectional area of the bed is \(A=6 \mathrm{~cm}^{2} .\) If the volume of the mixture added is \(\Delta V=50 \mathrm{ml}\), determine the following: a. Position of each band \(A\) and \(B\) in the column, \(L_{A}\) and \(L_{B}\) (or \(\Delta X_{A}\) and \(\Delta X_{B}\) ). b. \(L_{A} / L_{B} ; R_{f A}=L_{A} / L_{c} ; R_{f B}=L_{B} / L_{c}\) when \(C_{A}=10^{-1} \mathrm{mg} / \mathrm{ml}\) and \(C_{B}=0.05 \mathrm{mg} / \mathrm{ml}\) in liquid phase at equilibrium.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free