TLS Online TPP Program

#Question id: 392


When an ionic compound such as sodium chloride (NaCl) is placed in water, the component atoms of the NaCl crystal dissociate into individual sodium ions (Na+) and chloride ions (Cl-). In contrast, the atoms of covalently bonded molecules (e.g., glucose, sucrose, glycerol) do not generally dissociate when placed in aqueous solution. Which of the following solutions would be expected to contain the greatest number of solute particles (molecules or ions)?

#Section 4: Fundamentals of Biological Engineering
  1. 1 liter of 0.5 M NaCl

  2. 1 liter of 1.0 M NaCl

  3. 1 liter of 1.0 M glucose

  4. 1 liter of 1.0 M NaCl and 1 liter of 1.0 M glucose will contain equal numbers of solute particles.

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TLS Online TPP Program

#Question id: 14267

#Section 5: Bioprocess Engineering and Process Biotechnology

In a trickling biological filter, the BOD value of the feed stream is S0i = 500 mg/l with a feed flow of F = 103 l/h. The effluent BOD value is desired to be S0 = 10 mg/l. The following kinetic parameters for the biocatalysts are known: rm = 20 mg/S/l .h and Ks = 200 mg S/l. The biofilm thickness is L = 0.1 mm. The cross-sectional area of the filter is A = 2 m2 , and the biofilm surface area per unit volume of the bed is a = 500 cm2 /cm3 . Assume that dissolved oxygen is the rate-limiting substrate and the diffusion coefficient of oxygen is DO2 = 2 × 10-5 cm2 /s. Determine the required height of the bed. You can assume first-order bioreaction kinetics?

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TLS Online TPP Program

#Question id: 14268

#Section 5: Bioprocess Engineering and Process Biotechnology

An activated-sludge waste treatment system is required to reduce the amount of BOD5 from 1000 mg/l to 20 mg/l at the exit. The sedimentation unit concentrates biomass by a factor of 3. Kinetic parameters are µm = 0.2 h-1 , Ks = 80 mg/l, kd = 0.01 h-1 , and Y M X/S = 0.5 g MLVSS/g BOD5. The flow of waste water is 10000 l/h and the size of the treatment basin is 50,000 l. What is the value of the solids residence time?

TLS Online TPP Program

#Question id: 14269

#Section 5: Bioprocess Engineering and Process Biotechnology

An activated-sludge waste treatment system is required to reduce the amount of BOD5 from 1000 mg/l to 20 mg/l at the exit. The sedimentation unit concentrates biomass by a factor of 3. Kinetic parameters are µm = 0.2 h-1 , Ks = 80 mg/l, kd = 0.01 h-1 , and Y M X/S = 0.5 g MLVSS/g BOD5. The flow of waste water is 10000 l/h and the size of the treatme nt basin is 50,000 l. What value of the recycle ratio must be used?

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TLS Online TPP Program

#Question id: 14271

#Section 5: Bioprocess Engineering and Process Biotechnology

Penicillin is produced by P. chrysogenum in a fed-batch culture with the intermittent addition of glucose solution to the culture medium. The initial culture volume at quasi-steady state is V0 = 500 l, and glucose-containing nutrient solution is added with a flow rate of F = 50 l/h. Glucose concentration in the feed solution and initial cell concentration are S0 = 300 g/l and X0 = 20 g/l, respectively. The kinetic and yield coefficients of the organism are mm = 0.2 h-1 , KS = 0.5 g/l, and YX/S = 0.3 g dw/g glucose. Determine the concentration of cells at quasi-steady state when t = 10 h.

TLS Online TPP Program

#Question id: 14272

#Section 5: Bioprocess Engineering and Process Biotechnology

Penicillin is produced by P. chrysogenum in a fed-batch culture with the intermittent addition of glucose solution to the culture medium. The initial culture volume at quasi-steady state is V0 = 500 l, and glucose-containing nutrient solution is added with a flow rate of F = 50 l/h. Glucose concentration in the feed solution and initial cell concentration are S0 = 300 g/l and X0 = 20 g/l, respectively. The kinetic and yield coefficients of the organism are mm = 0.2 h-1 , KS = 0.5 g/l, and YX/S = 0.3 g dw/g glucose. If qP = 0.05 g product/g cells h and P0 = 0.1 g/l, determine the product concentration in the vessel at t = 10 h