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,

Answer 1

1. (i) Let X_o is the cell concentration in the feed, X is the cell concentration at any time t, F is feed flow rate, V_r is the culture volume in the reactor, µ_g is the gross specific growth rate of bacteria cell, and k_d is the death rate. Now, if dX/dt is the rate of change of concentration of bacterial cells in the reactor, then the material balance on cell concentration can be given by-

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

• FX_o – FX + V_r µ_gX - V_rk_dX = V_rdX/dt …..(1)

Since it is given that the feed is sterile, and cell death rate is negligible, therefore, we have X_o and k_d both equal to zero. Hence eqn (1) can be written as

       V_r µ_gX - FX= V_rdX/dt or dX/dt = X(µ_g – F/ V_r)

We know F/ V_r = D, where D is dilution rate

Therefore, we have dX/dt = X(µ_g – D)……..(2)

Now since at steady state we have dX/dt = 0.

Therefore, eqn (2) will be reduced to µ_g = D

(ii) From above derivation, we know that for given chemostat conditions we have µ_g = D

And since the Monod kinetics fit the reactor operation, we have

µ_g = D = µ_mS/(K_s + S).......(3), where µ_m, S, and K_s have their usual meaning.

We can rewrite the above eqn as

1/D = K_s/µ_m S + 1/µ_m , Here you can see if we plot 1/D Vs 1/S, we will get K_s/µ_m as the slope of that plot and 1/µ_m as the intercept of that plot.

I’ve plotted the data for you in excel and calculated the above values

K_s/µ_m = 1537 and 1/µ_m = 9.392, which gives µ_m = 0.106/h and Ks = 163.64 microM.

(b) Eqn three from above can also be written for max conditions as

D_max = µ_mS_o/(K_s + S_o), Now we calculate Dmax for given conditions i.e. S_o = 100microM (given in question), K_s = 163.64 microM (we calculated above), and µ_m = 0.106/h (calculated above)

Therefore, D_max = 0.106*100/ (163.64 + 100) = 0.04/h.

Now we know D = F/V_r => D_max = F_max/V_r and therefore F_max = D_max * V_r . Now we will calculate crirical flow rate F_max by pluggin the values, V_r = 1L (given in question) and Dmax = 0.04/h (we just calculated).

Therefore, Fmax = 1* 0,04 = 0.04 L/h

Thanks!