Question

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, \(R_{G}\).

Short answer

Exit pressure is 1.9924 atm. Transmembrane pressure is 1.9962 atm. Protein flux is -16661.67 cm/s. Cake resistance is -11.98 atm s/m^3.

Step-by-step solution

Calculate Exit Pressure (P0)

Use the Darcy-Weisbach equation for turbulent flow:\[\frac{\text{Pressure drop}}{L} = f \frac{\rho v^2}{2d}\]The density (\(\rho\)) can be assumed as that of water (\(1000\ kg/m^3\)) and the velocity (\(v\)) can be determined as the volumetric flow rate divided by the cross-sectional area. First, convert the flow rate from liters per minute to cubic meters per second: \(Q = 21 \text{ l/min} = 21 \times 10^{-3} / 60 \text{ m}^3/\text{s} = 3.5 \times 10^{-4} \text{ m}^3/\text{s}\). The diameter in meters is \(0.02 \text{ m}\), thus the cross-sectional area \(A = \pi (d/2)^2 = \pi (0.02 / 2)^2 = 3.14 \times 10^{-4} \text{ m}^2\).Hence, \(v = Q/A = 3.5 \times 10^{-4} / 3.14 \times 10^{-4} = 1.1 \text{ m/s}\).Now, using the Darcy-Weisbach equation,\[\frac{P_i - P_0}{L} = f \frac{\rho v^2}{2d}\]Plugging in the values,\[\frac{2 - P_0}{0.5} = 0.0005 \frac{1000 \times (1.1)^2}{2 \times 0.02}\ \rightarrow P_i - P_0 = 0.0005 \times \frac{1000 \times 1.21}{2 \times 0.02} \times 0.5 = 0.0076 \text{ atm}\ \rightarrow P_0 = 2 - 0.0076 = 1.9924 \text{ atm}\]

Determine Transmembrane Pressure Drop (ΔPM)

The transmembrane pressure drop (\(\Delta P_M\)) is the average pressure across the membrane.\[\Delta P_M = \frac{P_i + P_0}{2} = \frac{2 + 1.9924}{2} = 1.9962 \text{ atm}\]

Calculate Flux of Liquid through Membrane (J)

The protein flux (\(J\)) through the membrane can be found using the relationship:\[J = k \frac{C_B - C_G}{C_B} = 5 \frac{30 \text{ mg}/\text{l} - 100 \text{ g}/\text{l}}{30 \text{ mg}/\text{l}}\]Since \(100 \text{ g}/\text{l}\) is \(100000 \text{ mg}/\text{l}\), we have:\[J = 5 \times \frac{30 - 100000}{30} = 5 \times \frac{-99970}{30} = -16661.67 \text{ cm/s}\]

Determine Cake Resistance (RG)

The total resistance to the flow is given by the sum of the resistance of the filter and the cake:\[R_T = R_{M} + R_{G}\] and the flux equation is:\[J = \frac{\triangle P_{M}}{\mu \times R_{T}}\]Rearranging for \(R_G\), we get:\[R_{G} = \frac{\triangle P_{M}}{\mu \times J} - R_{M}\]Since \(\mu\) for water is about \(0.001 \text{ Pa} \text{ s} = 0.001 \text{ atm} \text{s}/\text{m}\), we convert \(J\) to m/s (assuming \(1 \text{ cm} = 0.01\; \text{m}\)), \(J = -166.62 \text{ m/s}\), so:\[R_{G} = \frac{1.9962}{0.001 \times -166.62} - 0.002 = -11.98 \text{ atm} \text{ s}/\text{m}^{3}\]

Key concepts

Protein Separation

Protein separation through cross-flow ultrafiltration is a highly efficient method to separate proteins from solutions such as fermentation broth. In this process, a membrane filters the liquid to retain larger molecules like proteins, while smaller molecules pass through. This is particularly useful in bioengineering and bioprocessing industries.
In the given exercise, a protein with a molecular weight (MW) of 300,000 Da is separated using a UF membrane. The flow rate and other parameters are used to determine various factors like flux and resistance. Understanding these parameters helps in optimizing the process for higher efficiency.

Darcy-Weisbach Equation

The Darcy-Weisbach equation computes the pressure drop due to friction in a pipe with turbulent flow. The formula is:\[ \Delta P = f \frac{L}{d} \frac{\rho v^2}{2} \]where:

• \( \Delta P \) is the pressure drop
• \( f \) is the friction factor
• \( L \) is the length of the pipe
• \( d \) is the diameter of the pipe
• \( \rho \) is the fluid density
• \( v \) is the velocity of the fluid

In the exercise, the pressure drop along the tube is calculated using the Darcy-Weisbach equation with given values and assumptions like the density of water.

Transmembrane Pressure Drop

Transmembrane Pressure Drop (TMPD) indicates the difference in pressure across the membrane. It's a crucial parameter in ultrafiltration as it drives the filtration process. Calculating the TMPD helps understand the efficiency of the separation.
In the exercise, TMPD is computed as the average of the inlet and outlet pressures:
\[ \Delta P_M = \frac{P_i + P_0}{2} \]The inlet pressure \(P_i\) is given, and the exit pressure \(P_0\) is determined using the Darcy-Weisbach equation.

Flux Calculation

Flux (J) refers to the flow rate of liquid through the membrane. It is important as it measures the performance of the ultrafiltration process. The flux formula used in the exercise is:\[ J = k \frac{C_B - C_G}{C_B} \]where:

• \( k \) is the mass transfer coefficient
• \(C_B\) is the solute concentration in the bulk solution
• \(C_G\) is the solute concentration on the membrane/gell film

In the problem, converting the units and substituting the given values result in the flux value. This calculation plays a crucial role in determining how effectively the membrane separates the proteins.

Cake Resistance

Cake resistance (\(R_G\)) is the additional resistance offered by the accumulated solid layer, or 'cake,' on the membrane. It affects the overall flow resistance significantly, often more than the membrane resistance itself. The total resistance to flow combines both the membrane resistance (\(R_M\)) and cake resistance according to:\[ R_T = R_M + R_G \]where the flux equation can be rearranged to solve for \(R_G\):\[ R_G = \frac{\Delta P_M}{\mu J} - R_M \]Here, \(\mu\) represents the viscosity of the fluid. Determining \(R_G\) helps in understanding the operational challenges and optimizing the process for minimal resistance and maximum efficiency of the ultrafiltration unit.