Question

The production of ethyl acetate occurs in a batch reactor with two raw materials, ethanol and acetic acid, as shown with the following reaction: ??2??5???? + ????3???????? ? ????3????????2??5 + ??2??

The reaction may be written simply in the form of: ?? + ?? ? ?? + ??

The reaction rate in the liquid phase is given by: ???? = ??1???????? At 100°C: ??1 = 7.93 × 10?6 m3kmol?1 s ?1

A feed of 38.4% by weight of acid, 50% by weight of alcohol, and no ester is used to promote a 35% conversion of the acid. The density of the mixture can be assumed to remain constant at 1020 kg m?3 . The daily production rate for the production of ethyl acetate is 50 metric tonnes and the reactor will be operated 24 hours a day with the time for filling, emptying, cleaning is 1 hour total for each batch.

Determine the following:

a) Initial and final concentration of each species (don’t forget about the water) [6 marks]

b) Batch reaction time [6 marks]

c) Number of batches per day [4 marks]

d) Reactor volume for required daily production [4 marks]

DATA:

Molecular Weight of Component

A 46 kg kmol?1

B 60 kg kmol?1

R 88 kg kmol?1

S 18 kg kmol?1

Answer 1

Given reaction is of the form A + B = R + S

Concentration of feed is given in wt % for calculation wte have to find out the concentration in kmol/m³

for this density of mixture and molecular weight of all components are given

a)

a1) Calculation of feed concentrations

Basis: 1 m³ of feed ( mass = 1020 kg)

  

Component	wt % in feed	Amount in 1m3 of feed (in kg)	molecular weight (kg/kmol)	Concentration in feed (kmol/L)
	(a)	(b) = 1020 *( (a)/100 )	(c)	(b)/(c)
A	50	510	46	11.087
B	38.4	391.68	60	6.528
R	0	0	88	0
S	11.6	118.32	18	6.57

a2) calculation of final concentration

Since the reaction mixture is constent density system final concentration are given by the equation

C_A = C_A0 ( 1 - X_A) = C_A0 - C_A0 X_A "-" sign for reactents and "+" sign for products

C_B = C_B0 ( 1 - X_B) = C_B0 - C_B0 X_B

C_R = C_R0 ( 1 + X_R) = C_R0 +C_R0 X_A

C_S = C_S0 ( 1 + X_A) = C_S0 +C_S0 X_{S ........................................(1)}

_{Also we are given}   X_B = 0.35 ( conversion of acid)

Since all stochiometric coefficents are 1 we can write

C_A0 X_A = C_B0 X_B = C_R0 X_R = C_S0 X_S

using this equation set (1) can be simplified as

  C_A = C_A0 - C_B0 X_B

C_B = C_B0 - C_B0 X_B

C_R = C_R0 +C_B0 X_B

C_S =C_S0 +C_B0 X_{B ........................................(2)}

C_B0 = 6.528 kmol/L

X_B = 0.35

using equation set (2) we can calculate final concentrations

Component (x)	Concentration in feed Cx0 (kmol/L)	Concentration in product Cx (kmol/ L)
A	11.087	8.8022
B	6.528	4.2432
R	0	2.2848
S	6.57	8.8548

b) For a reaction of form A+B = products we have

where M = C_B0/ C_A0 , M =0.589

X_A can be calculated using relation C_A0 X_A = C_B0 X_B

time can be calculated by re arranging this equation

t = 5537.862 sec = 1.538 hr ( approx 1.5 hr)

c) No of batches per day = 24 / time for one batch

time for one batch = reaction time + time for filling, emptying, cleaning

= 1.5 + 1

= 2.5 hr per batch

No. of batches per day = 24/ 2.5 = 9.6

No. of batches per day = 9 batches

d) Reactor volume for required daily production

Daily production rate of ester= 50 metric ton = 50000 kg = 568.18 kmol/day

production rate per batch = 568.18 / 9 = 63.13 kmol/ batch

Volume of reactor = production rate per batch/ ester concentration in final product

= 63.13/ 2.2848 =27.63 m³