Please help solve this You are operating a steady state chem

Please help solve this

You are operating a steady state chemstat for bacteria production. From previous experiments, you have determined the doubling time for the bacteria is 30 minutes. You want to use 95% of the maximum dilution rate for steady state operation to avoid the washout of the bacteria. You can assume a perfect mixing in the reactor, no cell death occurs, and Monod growth kinetics. (The saturation constant for the bacteria is 0.2g/1) What is your operating dilution rate? What is your substrate concentration in the outlet What inlet substrate concentration do you need to achieve cell mass concentration of 10g/l assuming the reactor is operating at the maximum cell yield (0.7g cells/gram of substrate.)

Solution

The operating dilution rate in steady state can be calculated by the formula t_d = ln(2)/ D. Where t_d = doubling time and D = dilution rate. (Since the doubling time becomes the function of dilution time in steady state). Hence the operatin dilution rate = D = .30/ 30 ( .30 = value of log2, 30 = doubling time) = 10^-2

The substrate concentration can be calculated by the formula D = S/ K_s + S * maximum dilution rate (since highest growth rate = critical dilution rate). Here D = dilution rate, Ks = half of the saturation constant and S = concentration of substrate. The maximum dilution rate will be = 95 * 10^-4 (95 percent of dilution rate). Hence the concentration of substrate = 10.52 g/l.