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  1. The  Zymomonas mobilis cells are used for chemostat culture in a 60 m 3  fermenter. The feed contains 12 g l -1  glucose; K  s  for the organism is 0.2 g l -1 . µ m = 0.3 h -1  and Y x/s = 0.06 gdw/g. a.   What flow rate is required for a steady-state substrate concentration of 1.5 g l -1  ?  b.   What is the corresponding steady state cell density and volumetric cell production rate? 1. Consider a continuous, aerobic bacterial culture in a chemostat with sterile feed. Three different dilution rates  D  are tested for a glucose feed concentration S   f   = 10 mM, and the biomass concentration  x  and glucose concentration S   in the exit stream are measured. The results are as follows:  D  (h - )  x  (g/L) S   (mM) 0.05 0.248 0.067 0.5 0.208 1.667 5 0 10 a. Estimate the glucose yield coefficient Y   X/S   (g biomass/mole glucose).  b. Assuming Monod growth kinetics, estimate the maximum specific growth rate μ max  (h -1 ) and the Monod constant  K  S   (mM). Ans. i. Between time t = 0.05 to 0.05     025.06.104.0067.0667.1 248.0208.0 /   S  X S  X  C C Y   Between time t = 0.5 to 5     025.0333.8208.0667.110 208.00.0 /   S  X S  X  C C Y   ii. Monod chemostat model equation 0 max       f   s  Dx x D s K  s    For sterile feed  x  f   = 0        D DK  sY  x  s f  S  X  max/feedsterile    and  D DK  s  s  maxfeedsterile    (any of this two equation could be use to solve the problem) D =   At 0.05 h -1   05.0K 0.050.067 maxs     (1) At 0.5 h -1   5.0 μ K 0.51.667 MAXS   (2) Solve equation (1) and (2) to get  MAX  (= 0.802) and K  S  (=1.007) 3. Consider a culture of bacteria making a valuable product in a chemostat operated at steady state. The liquid feed is sterile and contains 50 mM glucose (the limiting growth substrate S  ). In a series of steady state runs, the dilution rate  D  is increased incrementally, and the cell density exiting the chemostat is measured for each:  D  (h -1 )  x  (g/L) 0.05 1.58 0.10 2.17 0.15 2.47 0.20 2.65 0.25 2.76 0.30 2.83 0.35 2.86 0.40 2.84 0.45 2.74 0.50 2.25 0.52 1.32 0.53 0  a. Explain why the cell density can increase as the flow rate is increased, as shown here with  D  < 0.35 h -1 .  b. At what dilution rate should you operate the chemostat to optimize productivity of a strictly  growth-associated  product? Explain the basis for your answer. c. At what dilution rate should you operate the chemostat to optimize productivity of a strictly non-growth-associated  product? Explain the basis for your answer. 4. Consider cell growth in a chemostat at steady state, with sterile feed. In bench-scale experiments, it is found that the specific growth rate is inhibited at high glucose concentrations S  , modeled according to the phenomenological expression μ (h -1 ) = 0.6 S  /(0.2 + S   + 0.1 S  2 ), with S   in g/L. a. If the glucose concentration in the feed is 10 g/L, estimate the minimum dilution rate at which cell washout would occur.  b. Is the dilution rate estimated in part a the highest that will support cell growth in the  bioreactor?  Justify and briefly explain your answer.  5. A new strain of yeast is being considered for biomass production. The following data were obtained using a chemostat. An influent substrate concentration of 800 mg/l and an excess of oxygen were used at a pH of 5.5 and T  = 35 0 C. Using the following data, calculate µ m , K  s , Y X/SM, k  d  and m s  assuming µ net =µ m S/(K  s  + S)- k  d . Dilution rate (h -1 ) Carbon substrate concentration (mg/l) Cell concentration (mg/l) 0.1 16.7 366 0.2 33.5 407 0.3 59.4 408 0.4 101 404 0.5 169 371 0.6 298 299 0.7 702 59 6. The specific growth rate for unlimited growth in a chemostat is given by the following equation:                 where S 0  = 10 g/l K  s  = 1 g/l I= 0.05 g/l Y X/SM = 0.1 g cells/g subs X 0 =0 K  I =0.01 g/l µ m = 0.5 h -1  k  d =0 a.   Determine X and S as a function of D when I=0.  b.   With inhibitor added to a chemostat, determine the effluent substrate concentration and X as a function of D. c.   Determine the cell productivity, DX, as a function of dilution rate. 7 .    Pseudomonas sp.  has a mass doubling time of 2.4 h when grown on acetate. The saturation constant using this substrate is l  g   /3.1  (which is unusually high), and cell yield on acetate is  g  gcell  /46.0 acetate. If we operate a chemostat on a feed stream containing 38 g/l acetate, find the following: a.   Cell concentration when the dilution rate is one-half of the maximum  b.   Substrate concentration when the dilution rate is 8.0 D max  c.   Maximum dilution rate d. Cell productivity at 0.8 D max  8. Under substrate-limited conditions, a microorganism exhibits the following net specific growth rate, μ net , and yield coefficient, Y   X/S  : μ net  (h -1 ) = 0.70 S  /(0.1 + S  ), with S   in g/L; Y   X/S   (g DCW/g) = 0.40 The available growth medium contains 10 g/L substrate. a. When a batch bioreactor containing 100 L of the growth medium is inoculated with 1.0 g DCW of biomass, estimate the maximum cell density achieved, and the approximate time required to achieve it, after exponential growth is initiated.  b. You decide instead to culture the microorganism in a chemostat, using the same growth medium as the (sterile) feed. Estimate the dilution rate (h -1 ) at which the chemostat will achieve maximum steady-state productivity of biomass. c. Calculate and compare the overall biomass productivities (g DCW/L/h) of the two scenarios in  parts a & b. What other consideration will make the batch process even less productive compared to the chemostat?

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Jul 23, 2017
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