Bacteria \'B\' are being used for biological wastewater treatment. The kinetic p
ID: 262545 • Letter: B
Question
Bacteria 'B' are being used for biological wastewater treatment. The kinetic parameters for the Bacteria B' are determined using data obtained from chemostat culture. The growth of the bacteria is well fitted to Monod kinetics. A one-liter fermenter is used with a feed stream containing 100 HM of substrate. Table Q4 shows the dilution rate and substrate concentration obtained Table Q4: Dilution rate and substrate concentration 0.01 0.015 0.02 0.025 0.03 0.035 0.04 Substrate concentration (uM 17.4 25.1 39.8 46.8 69.4 80.1 100 Given Monod kinetics, (a)Steady state is achieved for the operation of chemostat without recycle. The death rate and product formation in the system are both negligible. And, the medium used is well sterilized (i)State the material balance on cell concentration (X) (ii) Plot the graph and determine the kinetic parameters ?max and K, for this microorganism (b)Estimate the critical operating flow rate of the chemostat. Given,Explanation / Answer
Cells coming in the reactor through feed – cells going out through effluent + cells getting produced in the reactor – cells dying in the reactor = change in cell concentration in the reactor
Since it is given that the feed is sterile, and cell death rate is negligible, therefore, we have Xo and kd both equal to zero. Hence eqn (1) can be written as
Vr µgX - FX= VrdX/dt or dX/dt = X(µg – F/ Vr)
We know F/ Vr = D, where D is dilution rate
Therefore, we have dX/dt = X(µg – D)……..(2)
Now since at steady state we have dX/dt = 0.
Therefore, eqn (2) will be reduced to µg = D
(ii) From above derivation, we know that for given chemostat conditions we have µg = D
And since the Monod kinetics fit the reactor operation, we have
µg = D = µmS/(Ks + S).......(3), where µm, S, and Ks have their usual meaning.
We can rewrite the above eqn as
1/D = Ks/µm S + 1/µm , Here you can see if we plot 1/D Vs 1/S, we will get Ks/µm as the slope of that plot and 1/µm as the intercept of that plot.
I’ve plotted the data for you in excel and calculated the above values
Ks/µm = 1537 and 1/µm = 9.392, which gives µm = 0.106/h and Ks = 163.64 microM.
(b) Eqn three from above can also be written for max conditions as
Dmax = µmSo/(Ks + So), Now we calculate Dmax for given conditions i.e. So = 100microM (given in question), Ks = 163.64 microM (we calculated above), and µm = 0.106/h (calculated above)
Therefore, Dmax = 0.106*100/ (163.64 + 100) = 0.04/h.
Now we know D = F/Vr => Dmax = Fmax/Vr and therefore Fmax = Dmax * Vr . Now we will calculate crirical flow rate Fmax by pluggin the values, Vr = 1L (given in question) and Dmax = 0.04/h (we just calculated).
Therefore, Fmax = 1* 0,04 = 0.04 L/h
Thanks!
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