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Chapter 11: Problem 2

Gentamycin crystals are filtered through a small test filter medium with a negligible resistance. The following data were obtained: \(\begin{array}{lcccc}t(\mathrm{sec}) & 10 & 20 & 30 & 40 \\ V(1) & 0.6 & 0.78 & 0.95 & 1.1\end{array}\) The pressure drop in this test run was \(1.8\) times that when water was used with a filter area of \(100 \mathrm{~cm}^{2}\). The concentration of gentamycin in solution is \(5 \mathrm{~g} / \mathrm{l}\). How long would it take to filter \(5000 \mathrm{l}\) of gentamycin solution through a filter of \(1.5 \mathrm{~m}^{2}\), assuming the pressure drop is constant and \(\mu=1.2\) centipoise?

Short Answer

Expert verified
Calculate the average filtration rate, use the rate-area relationship, scale the volume and area, and apply the constants to find the time.

Step by step solution

01

- Calculate Filtration Rate

First, determine the average filtration rate for the given data. The filtration rate can be found by using the data provided in seconds and cubic meters.
02

- Determine the Relationship between Filtration Rate and Area

Next, use the relationship between filtration rate and filter area. Since the pressure drop and viscosity are given as constant, these will help determine the filtration capacity for the larger filter area.
03

- Calculate Time for Larger Volume

With the filtration rate known, scale it up to the larger volume (5000 liters) and the larger filter area (1.5 m²) to find the total time required.
04

- Use Provided Constants in Calculation

Apply the constant values (pressure drop and viscosity) in the final step to ensure consistency with the initial conditions.

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Key Concepts

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

gentamycin filtration
Gentamycin filtration is a crucial process in pharmaceuticals, where gentamycin crystals are separated from a solution using filtration. In the provided exercise, gentamycin's filtration was studied through a small test filter and the filtration rate was measured over several intervals. To understand this process better:
  • Filtration occurs when a liquid containing suspended particles is passed through a filter medium that captures these particles.
  • In this case, the gentamycin solution, with a concentration of 5 g/L, is filtered, and the data points collected show how quickly the volume of filtered solution grows over time.
  • The filtration rate is affected by factors such as the pressure drop across the filter medium and the viscosity of the solution.
Knowing these factors helps in planning larger-scale filtrations. Filtering large volumes efficiently is essential for manufacturing drugs like gentamycin while maintaining quality and cost-effectiveness.
filter area scaling
Scaling up the filtration area is vital when moving from small-scale experiments to large-scale production. In the exercise, we scale up the filtration process of gentamycin from a small test filter to a larger one. Here’s how we approach filter area scaling:
  • The filtration rate depends on the filter area. A larger filter area allows for faster filtration as more liquid can pass through the filter simultaneously.
  • Given the initial filter area of 100 cm², we calculate the filtration rate for this area and then scale it up to the large filter area of 1.5 m².
  • Mathematically, if the flow rate in the small filter is known, the same rate per unit area applies to the larger filter. Thus, we multiply the filtration rate by the ratio of the large filter area to the small filter area.
This scaling ensures that the larger filtering operation maintains the same efficiency as the smaller test run, assuming constant conditions like pressure and viscosity.
constant pressure drop
A constant pressure drop is a key assumption in filtration calculations, as it simplifies the relationship between the driving force for filtration and the flow rate. In this gentamycin filtration exercise, the pressure drop remains constant. Here’s why this is important:
  • The pressure drop (ΔP) influences the rate at which the liquid passes through the filter. A higher pressure drop generally increases the filtration rate.
  • When the pressure drop is constant, it means that the external forces driving the flow are steady, making it easier to predict how the system will behave at larger scales.
  • In our exercise, the pressure drop during the test run is 1.8 times that when water was used. By keeping this value constant, we make our calculations simpler and more reliable since the only variable changing significantly is the filter area.
This assumption helps ensure that the extrapolated filtration times and rates are accurate and consistent with the experimental data.

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Most popular questions from this chapter

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.

A solute protein is to be separated from a liquid phase in a chromatographic column. The adsorption isotherm is given by the following equation: $$ C_{S}=k C_{L}^{2} $$ where \(C_{S}\) is the solute concentration in solid phase (mg solute/mg adsorbent) and \(C_{L}\) is the liquid phase concentration of solute (mg solute/ml liquid). Use the following information: $$ \begin{aligned} &k=0.4, \quad \varepsilon=0.3, \quad A=25 \mathrm{~cm}^{2}, \\ &\mu=10 \mathrm{~g} \mathrm{ads} / 100 \mathrm{ml} \text { column }=100 \mathrm{mg} / \mathrm{ml} \end{aligned} $$ a. For \(V=400 \mathrm{ml}\) and \(X=25 \mathrm{~cm}\) determine the equilibrium solute concentrations in liquid and solid phases. b. Determine the ratio of travel distances of solute to solvent, \(R_{f}\).

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.

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?
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