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TLS Online TPP Program

#Question id: 8895

What is one characteristic that separates chordates from all other animals?

#SCPH06 I Botany
  1. true coelom
  2. post-anal tail
  3. blastopore, which becomes the anus
  4. bilateral symmetry
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TLS Online TPP Program

#Question id: 14265

#SCPH05 I 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

#SCPH05 I 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

#SCPH05 I 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

#SCPH05 I 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

#SCPH05 I 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

#SCPH05 I 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?