Question

4.12 Production of 1,3-propanediol

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 ((231-1803) to 1,3-propanedi01 (C3H802) and acetic acid (C2H402), with minimal formation of other fermentation products such as butyric acid, ethanol, and 1-12 gas. The fermentation and cell growth equation can be written as:

                 68 C3H803 + 3 NH3 = 3 C4H702N + 49 C3H802 + 15 C2H402 + 15 C02 + 40 H20

where C4H702N represents the biomass.

A continuous fermenter is set up for 1,3-propanediol production at 37^o c and atmospheric pressure. Anaerobic conditions are maintained by sparging the broth with nitrogen gas at a flow rate of 1000 litres min—I. The feed rate of medium into the fermenter is 1000 kg h—l; the medium contains ammonia and 14% w/ w glycerol. The yield of 1,3propanediol is strongly affected by glycerol concentration, which must be kept above a certain level to suppress the formation of undesired by-products. Accordingly, the fermentation process is designed so that the product stream contains an unreacted glycerol concentration of 3% w/ w.

A: What is the volumetric flow rate and composition of the off-gas?

B: What minimum concentration of NH3 is needed in the feed stream?

C: If the NH3 concentration from (b) is used, what is the concentration of 1,3-propanediol in the product stream?

Answer 1

The feed of glycerol is 14% w/w of glycerol which means that 100 Kg feed contains 14 Kg Glycerol. The flow rate of the medium is 1000 Kg/h. so it contains 140 Kg/h of glycerol.

The off gas from the reaction will be the CO2 produced and the Nitrogen blowing through the medium.

68 moles of glycerol produces 15 moles of CO2

140 Kg of glycerol is 140 x 10³ g/92.09382 g/mol

=1521.7 moles so this will produce 3% remains unreacted so 1476 moles will react

(1476/68) x 15 = 325.6 moles of CO₂/h

PV=nRT

1 atm x V = 325.6 moles x 0.08205 L atm K⁻¹ mol⁻¹ x 310

V = 8281 L/h = 138.0 L/min

A) So the off gas flow rate is 1000 L/min of nitrogen and 138 L/min of CO2 = 1138 L/min composition is 87.9% nitrogen and 12.1% CO2.

The feed stream contains 140 Kg/h of glycerol which is 1521.7 moles of glycerol of which 1476 moles will react

68 moles of glycerol requires 3 moles of ammonia so the minimum ammonia required is (1476/68)*3 = 65.11 moles of ammonia/h.

in 1000 Kg if we consider density to be 1 we have 65 moles/1000 Kg

The molality of a solution is calculated by taking the moles of solute and dividing by the kilograms of solvent.

B) 0.065 molality is the concentration of ammonia minimum required

C) 3 moles of ammonia will produce 49 moles of 1,3 propanediol so 65 moles will produce (65/3)*49 = 1061 moles of 1,3 propanediol

1061 moles in 1000 Kg will be 1.061 molality of 1,3 propanediol