TLS Online TPP Program

#Question id: 14397


A purified hydrolytic enzyme (Mr 150 000) is available with the description: Gives a Vmax of 3 200 units/mg protein under optimal conditions. How many nmol of enzyme are there in 1 mg  

#Section 5: Bioprocess Engineering and Process Biotechnology
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TLS Online TPP Program

#Question id: 14266

#Section 5: Bioprocess Engineering and Process Biotechnology

For the activated-sludge, the specific growth rate of cells is given by 


 

The following parameter values are known: F = 500 l/h, α = 0.4, ɣ = 0.1, Xe = 0, V = 1500 l, Ks = 10 mg/l, µm = 1 h-1 , kd = 0.05 h^-1 , S0 = 1000 mg/l, Y M X/S = 0.5 g dw/g substrate.. Calculate the cell concentration in the reactor. 

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.