Problem 2: A process for microbial synthesis of 1,3-propanediol is being develop

Question

Problem 2: A process for microbial synthesis of 1,3-propanediol is being developed for the manufacture of ‘green’ polyester fabric from renewable resources. Under anaerobic conditions, a selected strain of Klebsiella pneumoniae converts glycerol (C3H8O3) to 1,3- propanediol (C3H8O2) and acetic acid (C2H4O2), with minimal formation of other fermentation products such as butyric acid, ethanol, and H2 gas. The fermentation and cell growth equation can be written as:

where C4H7O2N represents the biomass. A continuous fermenter is set up for 1,3- propanediol production. Anaerobic conditions are maintained by sparging the broth with nitrogen gas at a flow rate of 70 kg/h. The feed rate of medium into the fermenter is 1000 kg/h; the medium contains water, ammonia and 20% w/w glycerol. The yield of 1,3- propanediol is strongly affected by glycerol concentration, which must be kept above a certain level to suppress the formation of undesired by-products. NH3 is completely consumed. If the desired concentration of 1,3-propanediol in the product stream is 10%:

(a) What is the concentration of glycerol in the final product stream?

(b) What minimum concentration of NH3 is needed in the feed stream?

Molecular weights:

Glucose = 180

Ethanol = 46

CO2 = 44

H2O = 18

Glycerol = 92

NH3 = 17

O2 =32

1,3-propanediol = 76

Explanation / Answer

ANS a)

Basis =1 litre of medium

concentration glycerol in medium = 100 g/L

molcular weight of glycerol= 92

that i.e. concentration in gmol/L = 100/92 = 1.0869

according to given reaction,

68 moles of glycerol= 3 moles of NH3

that i.e. 1.086956 moles of glycerol = X

so X= (3* 1.086956) / 68= 0.04795

so 100 g/L of glycorol require 0.04795 NH3 to achieve complete conversion

Ans b) Basis = 40 liters of medium

take concentration of glycerol=100

so :- glycerol content in 40 liter medium will be

=(40 * 100)=4000 g/40 liter

=4000 / 92= 43.478 gmols/40 liter

According to given reaction,

68 moles of glycerol= 3 moles of Biomass

43.478 gmole of Biomass=X

so:- X= (3 * 43.478) / 68 =1.9181 gmol

Moleculer weight of Biomass=101

Mass of C4H7O2N generated= (101 * 1.9181) = 193.733 grams

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