This set of Bioseparation Technology Multiple Choice Questions & Answers (MCQs) focuses on “Convective Mass Transfer”.
1. The connective mass transfer can be observed in ___________
a) Flowing fluids
b) Stagnant fluid
c) Fluid in bulk
d) Solids
View Answer
Explanation: The connective mass transfer is observed in the flowing fluids for example, transport of solute in a flowing fluid across the solid surface of the membrane. The driving force which helps in mass transfer is the difference of chemical potential through the thermodynamic gradient for the flow of mass and the flow is from the area of higher chemical potential to lower potential region.
2. What is the example of transport of solute in a liquid flowing through a solid surface?
a) Reverse osmosis
b) Haemodialysis
c) Liquid-liquid extraction
d) Diffusion
View Answer
Explanation: Haemodialysis is the example of transport of solute in a liquid flowing through a solid surface. In haemodialysis, the urea is transferred to the surface of dialysis membrane from blood.
3. What is the example of transport of a solute in a liquid flowing through another immiscible liquid?
a) Reverse osmosis
b) Haemodialysis
c) Liquid-liquid extraction
d) Diffusion
View Answer
Explanation: Liquid-liquid extraction is the example of transport of solutes in a liquid flowing through another immiscible membrane. In this, penicillin can be transferred within the filtered liquid medium which flows past another organic solvent to extract the liquids through extractor.
4. Which parameter allows the transfer of solute when the liquid flows through a solid surface?
a) Convective mass transfer
b) Heat transfer
c) Fluid flow
d) Molecular diffusion
View Answer
Explanation: Molecular diffusion allows the transfer of solute, when a liquid flows past a solid surface a stagnant boundary liquid layer is formed close to the surface and within these boundary layers the transfer of solutes takes place.
5. If the flow of liquid is turbulent, the transfer of solutes will take place due to molecular diffusion.
a) True
b) False
View Answer
Explanation: The turbulent flow of liquid and the mass transfer accompanied by it will be due to the combination of molecular diffusion as well as eddy diffusion and not just by the molecular diffusion. The laminar flow of liquid is the result of only molecular diffusion.
6. What is convective mass transfer?
a) Combination of molecular diffusion and eddy diffusion
b) Molecular diffusion
c) Eddy diffusion
d) Fluid flow
View Answer
Explanation: The convective mass transfer is the combination of molecular diffusion and the eddy diffusion. The mass transfer in convective mode is due to the movement of fluid in bulk, for example, flow of water will transport the molecules that are present or dissolved in the water.
7. What will be the flux equation for convective mass transfer?
a) NA = (D + E) \(\frac{dC_A}{dx}\)
b) NA = -(D – E) \(\frac{dC_A}{dx}\)
c) NA = -(D + E) \(\frac{dC_A}{dx}\)
d) NA = -(DE) \(\frac{dC_A}{dx}\)
View Answer
Explanation: The flux equation for convective mass transfer is NA = -(D + E) \(\frac{dC_A}{dx}\) where, D is the molecular diffusivity, E is eddy diffusivity, C is the concentration of A. The flux equation can be also written as NA = kA ΔcA where, kA is the mass transfer coefficient for the transfer of fluid. Flux measures the amount of molecules that can flow through unit area at unit interval time and it is proportional to negative gradient of concentration.
8. Calculate the eddy diffusivity for an aqueous solution of Ig A. The fluid is being pumped through a tube of diameter 1 mm and the mass transfer coefficient for the protein in the radial direction is 1 × 10-6 m/s.
a) 4.85 × 10-11 m2/s
b) -4.85 × 10-11 m2/s
c) 4.52 × 10-11 m2/s
d) 4.52 × 10-10 m2/s
View Answer
Explanation: The molecular weight of Ig G is 155000 kg/mole and the diffusivity of Ig G in water is DAB = (\(\frac{9.4 × 10^{-15} × 277}{0.001 × (155000)^{1/3}}\) = 4.85 × 10-11 m2/s and the radius of the tube is 0.5mm. The solute transfer is due to diffusion and the mass transfer coefficient is kD = \(\frac{4.85 × 10^{-11}}{0.5 × 10^{-3}}\) = 9.69 × 10-8 m/s and the eddy diffusion is E = (1 × 10-6 × 0.5 × 10-3) – 4.85 × 10-11 = 4.52 × 10-10 m2/s.
9. What will be the flux through the convective mass transfer when the mass transfer coefficient is 2 × 10-8 m/s and the change in the concentration is 25000 kg-mole/m3 and the eddy diffusion is 3 × 10-6 m2/s?
a) 0.0005 kg-moles/m2.s
b) 0.0020 kg-moles/m2.s
c) 0.0015 kg-moles/m2.s
d) 0.0025 kg-moles/m2.s
View Answer
Explanation: The flux NA = kA ΔcA where, kA is the mass transfer coefficient for the transfer of fluid. Therefore, NA = 2 × 10-8 × 25000 = 0.0005 kg-moles/m2.s.
10. Estimate the flux of the a solute when diffusivity of solute in water is 2.6 × 10-8 m2/s, the change is concentration of the solute is 5000 kg-mole/ m3, the length of the tube is 5m3 and the eddy diffusion is 4 × 10-6 m2/s.
a) -0.005 kg-moles/ m2.s
b) 0.005 kg-moles/ m2.s
c) -0.004 kg-moles/ m2.s
d) 0.004 kg-moles/ m2.s
View Answer
Explanation: Flux NA = -(D + E) \(\frac{dC_A}{dx}\) where, D is the molecular diffusivity, E is eddy diffusion, C is the change in the concentration of A, x is the length of the tube. Therefore, flux NA = -(2.6 × 10-8 + 4 × 10-6 \(\frac{5000}{5}\) = -0.004 kg – \(\frac{moles}{m^2}\).s.
Sanfoundry Global Education & Learning Series – Bioseparation Technology.
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