Question

15 points A pulse tracer experiment was conducted in a tubular reactor and RTD function e(t) was calculated and tabulated belowin Table 1. An irreversible, aqueous-phase, elementary reaction, 2A-B-C, with k = 0.5 Lmol-1 min -1 at 320K will be carried out isothermally at 320K in this reactor, whose mean residence time was found to be 15 min from RTD studies. The feed concentration of A is 2mol/dm3. Determine the conversion that can be attained in this reactor for the following 2 cases: a) Assuming that the reactor were an ideal PFR. (4 marks) b) Using the complete segregation model for its non-ideality, Compare and comment (11 marks) t, min 0 0 5 0.03 10 0.05 15 0.05 20 0.04 25 0.02 30 0.01 35 0 E Table 1 RTD data

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Answer 1

Given's Inrevessible aqueous-phase elementary reactions "-A=KCA ² 2A B+ c L 0.5 L [unit of k represent the order of reaction molimir is two T= 320K (Isothermal Mean residence time, tm = 15 min feed concentration, cao = 2 mol 2 mol dm3 a) convession achieved using Ideal PER. . It is observed that for constant volumetic flow rate and in the absent of dispersion, no matter what RTD exists for particular reacta, ideal on nou-ideal, T= tm Thus, I= tm = 15 min For and order constant volume reaction in PFR, the design equation is given as, CAOKE It Cook I substituting the value of all the variables an equ , me get

Er derivation, T= ChordXA 2 * 0.5*15 1 + 2 * 0.5*15 Cool OKA -NA = KCA2 = KCA0²(1-XA) ² I= CAO ARXA X = 6:9375 KCA0² (1-XA) 2 YA I KAO aXA The connession attained in this case is 93.757. (1-XA) 2 TKCAO = XA (1-KA XA = TKCAO HIKCAO

conversion using complete segregation model. Giren data between t & RTD function Elt) 30 25 35 २० 15 It, min o 1o OOL Elt) 003 0:05 O'02 0.05 0.04 To determine the mean conversion using segregation model, we must average the conversion of all the various globules en the exit stream. dx = Xit) * Elt) de ax at Xet) Elt) = ° xe) )4t XLt) for second ordee reaction is given as, Xet) = KECAO KECAO +1 X So keast L + Kaut Elt) at X(t) = 0.5 X 2 Xt If 0.5X2xt CY Xit) - t Itt

οος 0.04 Area under curve gives 0'03 XG) ett) 0'02+ 0.01 0 5 + LO IS 20 25 30 35 اسلم t, min

Irapezoidal sule formulla, Xe box [f(x0) + 2 (f(x) + f(x3) +-- ofery + f(xenti] Intewal -[ab] - [a,b] Ax = b-a n = subintewals. b=35 a = 0 AXs b-a 35-0 7 - 35 X= 5.0 xlt) € (t) dt [0,35) n= x = 35 x 22 0+2(0.028 +0.0 455 +0.0469 +0.0381+0.01927 91675X10-3) to] x = x x 2 x 01844 35 7 x = 0.9219 The conversion achieved by considering the segregation model is 92.19 % This decrease in conversion as is due to due to the non-ideality in the sesctor ie', turbulence, bypassing and dead gone formation compared to ideal PAR