Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced

Question

3 years ago

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Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced

chemical in the nation on a total pound basis. About one-half of the EG is used for antifreeze, while the other half is used in the manufacture of polyesters. In the polyester category, 88% was used for fibers and 12% for the manufacture of bottles and films. The 2013 selling price for EG was $0.60 per pound. It is desired to produce 200 million pounds per year of EG. The reactors are to be operated isothermally. A 16.1 mol/dm3 solution of ethylene oxide (EO) in water is mixed with an equal volumetric solution of water containing 0.9 wt% of the catalyst H2SO4 and fed to two 800-gal CSTRs in series at 7.24 dm3 /s. The specific reaction rate constant is 0.311 min-1 . Because water is present in such excess, the concentration of water at any time is virtually the same as its initial concentration, and the rate law is independent of the concentration of H2O. The reaction is first order in EO:
Catalyst
EO+ W â EG

a. What is the conversion exiting the first reactor?
b. What is the conversion exiting the second reactor?

Engineering

1 answer:

bekas [8.4K] · Answer 1 · 2022-05-18T09:10:58+03:00

Answer:

a) 0.684

b) 0.90

Explanation:

Catalyst

EO + W → EG

a) calculate the conversion exiting the first reactor

CAo = 16.1 / 2 mol/dm^3

Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst

Vo = 7.24 dm^3/s

Vm = 800 gal = 3028 dm^3

hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins

next determine the value of conversion exiting the reactor ( Xai ) using the relation below

KIm = $\frac{Xai}{1-Xai}$ ------ ( 1 )

make Xai subject of the relation

Xai = KIm / 1 + KIm --- ( 2 )

where : K = 0.311 , Im = 6.97 ( input values into equation 2 )

Xai = 0.684

B) calculate the conversion exiting the second reactor

CA1 = CA0 ( 1 - Xai )

therefore CA1 = 2.5438 mol/dm^3

Vo = 7.24 dm^3/s

To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below

XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )

where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )

XA2 = 0.90