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TLS Online TPP Program

#Question id: 158

At the isoelectric pH of a tetrapeptide:

#Section 2: General Biology

  1. only the amino and carboxyl termini contribute charge.

  2. the amino and carboxyl termini are not charged.

  3. the total net charge is zero.

  4. there are four ionic charges

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TLS Online TPP Program

#Question id: 14265

#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 substrate concentration (S) in the reactor at steady state. -------------------------
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: 14270

#Section 5: Bioprocess Engineering and Process Biotechnology

Consider a well-mixed waste treatment system for a small-scale system. The system is operated with a reactor of 1000 l and flow rate of 100 l/h. The separator concentrates biomass by a factor of 2. The recycle ratio is 0.7. The kinetic parameters are µm = 0.5 h-1 , Ks = 0.2 g/l, Y M X/S = 0.5 g/g, and kd = 0.05 h-1 . What is the exit substrate concentration?