Question

5) The most conventional method for producing propylene oxide is by the reaction of propylene with Hz and O2. However, propane is also produced from a competing reaction of propylene with Hz. If you mix a stream contains propylene, hydrogen and air (6, 18.6 and 27 mol/h, respectively) with an inert carbon dioxide (3000 mol/h) stream and fed them to a continuous reactor. The reactor effluents introduced to a condenser where the condensate liquids consisting propylene oxide, propylene and propane is received from the bottom while the gas stream is left from the top. If 15% fractional conversion of the limiting reactant is obtained and the fractional yield of reacted propylene to the produced propylene oxide is 0.75. (1) Based on the process description, draw the schematic flow chart and label all the information. (ii) Write the main and side reaction. Calculate the standard heat of reaction for both reactions. (iii) Perform DOF analysis for the overall process and for cach single unit based on the elemental (atomic balance method) (iv) Determine the flow rate of each component in the process.

Answer 1

Based on the supplied information, the required solution is provided below.

CO Hay (0 N2=c Og=d CO2 قا 3000 molsh Reactor GHG : GHO = x Hg = = y Condenser CO= 340 malla z N₂ = 81.33 malih H₂ = G46 = 6 mollh H₂ Air = 27 mol/ha 18. 6 mollh O2 = 4.995 molln GHGO=e H6. = f ç Hg = g Isa 오 N₂ 11) Hot H₂ + O₂ ť ho + Ho (mavi reaction) СИ. + Hy ç Hå (side on) Degree of freedom Analysis On Reactor AS 27 mol air is fed to reactor h and air contains 21 % Or and 79 % & fed = 0.24*24 mai 10l/h y fel a 0.79% 27 mo! 21:33 mol/h 5.67 mo

a AS ₃ 46 ofed a 6 mol/ in and a fed=5.64 mohh So; as per main reaction ; O₂ is limiting reactant. 15iny Fractional connession of O₂ = - o connected Orls x 5.67 mol/h 0.8505 mol/h Q umeacted Cleaving reactos)= (5.67-0.8505) = 4:8195 mol h. AS will be same My and cou are ineet. e) My and co. entering the and reactor leaving the reactor Atomic species (c) balance on Reactor excluding car. (cea xeract (36) 5.36 of 3% + 6= we werty H balance за 6X6 + 2x 18.6 60+ 6xt sy +22 o balance 8x. G+ = x + 2x4.8195

from Reaction stoichiometang ; 0.25 As giren efractional yield of reacted propylene to produced propilene oxide =0.75 = fractional yield of reacted propylece to Produced propane Bopylerie produced Propane produced. 3y no of unknowns = 4 [w,x,y, 2] O inceperolent equations = 4 Toto D.O.F. = no g independent Opleitory olos 0.75 0.25 X5 Noig unknowns D.OF. 4-4 = 0 (b) On condenser 5 6 waf Ne y = g @ 3900

z=b-① 1o C-81.33 d = 4.8195 00 NO. og 11 no unknowns = a, b, c, d, e, dig y ,2 Mony is dependent ens 7 I to ©] D:0. F. 11-7 = 4 (0) On orecall system c balance 3000 xlt 6x3 a + 3e +34 +39 H balance 6X6 + 2X18.6 26+60 t. 6ft 89 (13) O balarice 3000 x 2 of 5.6742 = 2 a tad t e H balance ax 21.33 gc g unknowns a [a,b,c,d, e, f, g long independent ears = 4 c6 to 0) D-OF- 7-4. - 3

(N) from ②, 11.34 2+9.639 X = 1.701 mol h from 4 70- 28 =) ys 0.567 mol ollh from ; 6 = 1.Tolto.567 3.732 mo ollh fome 364 37:2 = 6X3-732+ 6*1.701. + 8x0 567 +22 Z = 18-033 mol/h o/W] from ③, f = 3-732 molla from ©. e = 1 Tol mollal from g = 6.567 mol/ from 8 a = 3000 moſh from @ 6= 18-033 mol/h from 10 from ② ; E = 21:33 mol/h d = 4.8195 mol/h