Question

2. (10 pts) Mold spores are grown in a large 10,000 L continuous bioreactor (CSTR) at 35 °C to produce a vitamin. At steady state, the spore concentration is 10s cells/liter, and the vitamin concentration is 30 mg/L. Carbon and nitrogen nutrients are fed continuously through a liquid inlet, and oxygen bubbled through so that the OTR does not limit production. Because bacteria can be introduced along with the liquid feed containing carbon and nitrogen sources, the bioreactor is subjected to a daily sterilization step while the mold spores are producing the vitamin. It is heated to 121 °C for 15 minutes in an attempt to remove bacteria but not kill a large number of the mold spores. This works to preferentially kill bacteria since they are more susceptible to heat than mold. Unfortunately, the vitamin is also temperature sensitive, and some of it is inactivated in the heating step. Given the Arrhenius parameters for the mold and bacteria death constant (kd) and vitamin inactivation, and recalling that in the death phase,-0, In X/X,-.kat, and k.-α exp (-E04/RT ) Mold spores: a 10 min1 Eod 65 kcal/mol Bacteria cells: a 175 min1 Eo.d-5.2 kcal/mol Vitamin α-104 min-1 Eo.d 10 kcal/mol Assuming bacteria have reached a concentration of 10% cells/liter (matching the mold cell concentration) at the start of the sterilization, determine A. what 96 of mold cells remain alive after the thermal sterilization step B. What % of bacteria cells were killed by the thermal sterilization. C. What % of the vitamin product remains active after sterilization. D. How long would the reactorhave to be held at 121 °C to decrease the bacteria concentration by three orders of magnitude, to 100 cells/liter? R is the universal gas constant. It is 1.987 cal/mol K

Answer 1

This is a simple kinetics problem.

We have been given the 3 Arrhenius equations

thus we can find out the destruction rates of all 3 species

Let's evaluate the rate constant at T=121C

Km = 10^33*exp(-65000/(1.987*394)) = 10^33*exp(-83.02) = 10^33*8.81*10^-37 = 8.81*10^-4

Kb = 175*exp(-5200/(1.987*394)) = 175*exp(-6.64) = 175*0.00131 = 0.2285

Kv = 10000*exp(-10000/(1.987*394)) = 10000*exp(-12.77) = 10000*2.845*10^-6 = 0.02845

It is given that t=15

Thus

Ln(X/Xo) = -8.81*10^-4*15 ..............................for mold spores

Ln(X/Xo) = -0.01322

X = Xo*exp(-0.01322)

X = 10^5*0.987

X = 9.87*10^4 cells/L

Thus 98.7% of the mold spores survive

Ln(X/Xo) = -0.2285*15 ..............................for bacteria

Ln(X/Xo) = -3.4275

X = Xo*exp(-3.4275)

X = 10^5*0.0325

X = 3.25*10^3 cells/L

Thus 96.75% of bacteria die

Ln(X/Xo) = -0.02845*15 ..............................for vitamin

Ln(X/Xo) = -0.42675

X = Xo*exp(-0.42675)

X = 30*0.6526

X = 19.578 mg/L

Thus 34.75% of vitamin is deactivated

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Now we are given the expected (X/Xo) value

X/Xo = 0.001

Thus

0.001 = exp(-0.2285*t)

0.2285*t = Ln(1000)

0.2285*t = 6.9078

t = 30.231 min