A daily production of 50,000 kg (50 tons metric) of ethyl acetate is to be produ

Question

A daily production of 50,000 kg (50 tons metric) of ethyl acetate is to be produced in a batch reactor from ethanol and acetic acid:

C2H5OH + CH3COOH = CH3COOC2H5 + H2O

(A) (B) (R) (S)

The reaction rate in the liquid phase is given by:

rA = k(CACB - CRCS/K)

At 100C:

k = 7.93 x 10-6 m3/kmol sec

K = 2.93

2. (Froment and Bischoff, 2011) A daily production of 50,000 kg (50 tons metric of ethyl acetate is to be conducted in a batch reactor from ethanol and acetic acid: C2H5OH CH3COOH CH3COOC2H5 H20 (B) (C) (D) (A) The reaction rate in the liquid phase is given by: k (CACR CR, Cs/K) At 100°C, k-7.23 x 10 -6 ms/kmol/s and K-2.93 A feed of 23 percent by weight of acid, 46 percent of alcohol, and no ester is to be used with a 35 percent conversion of acid. The density is essentially constant at 1020 kg/m The reactor will be operated 24h per day, and the time for filling, emptying, and the like is 1 h total for reactors in the contemplated size range. What reactor volume is required?

Explanation / Answer

Feed composition : CB= 23% CH3COOH

CA= 46% C2H5OH

Conversion XB= 35 %

Production of ethyl acetate in 24 hrs = 50000 Kg/day

so, 35% of acid CR = 8.05 % product ethyl acetate.

So, rA = 7.23 * 10^(-6) [0.46 *0.23 - 0.0805 *0.0805 / 2.93 ]

= 7.23 * 10^(-6) * 0.0508

= 0.367 *10^(-6)

so , here v = 50000 / 1020 = 49.019 m3

Now,

V= CA * XA * v / rA = 0.46 * 0.35 * 49.019 / 0.367 * 10^(-6)

= 21.50 * 10^6

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