Question

calculate the reactions conversion percent (by hand and solver in excel) to be able to find good result in aspen in producing Ethylene oxide. The given reactions are:

C2H4+ 0.5O2 --> C2H4O

C2H4+ 3O2 --> 2CO2+2H2O

C2H4O + 2.5O2 --> 2CO2+2H2O

Along with stream data (COMPONENT FLOWRATES (Kmol/h) related to first and second reactors as followed:

First reactor input

Ethylene    --> 1047.91                          

Ethylene oxide --> 6.47

Carbon dioxide--> 31.71

Oxygen --> 6331.12

Nitrogen --> 28,191.39

Water --> 30.98

First reactor output:

Ethylene    --> 838.67

Ethylene oxide --> 206.79

Carbon dioxide --> 49.56

Oxygen --> 6204.19

Nitrogen --> 28,191.39

Water --> 48.82

Second reactor output:

Ethylene --> 670.64                          

Ethylene oxide --> 175.83

Carbon dioxide --> 63.44

Oxygen --> 6101.72

Nitrogen --> 28,188.72

Water --> 77.13

Answer 1

% Conversion for any reactant is defined as = kmole reacted / kmole of feed. The below table indicates the kmol distribution of reactants and products from reactor 1 & 2.

Compo nent	Ist reactor Inlet ; N1 (kmol/hr)	Mole reacted/generated = Reactor inlet - Reactor outlet = N1-N2	Ist reactor Outlet OR 2nd Reactor Inlet ; N2 (kmol/hr)	Mole reacted/generated = Reactor inlet - Reactor outlet = N2-N3	2nd reactor Outlet; N3 (kmol/hr)
C2H4	1047.91	209.24	838.67	168.03	670.64
C2H4O	6.47	-200.32	206.79	30.96	175.83
CO2	31.71	-17.85	49.56	-13.88	63.44
O2	6331.12	126.93	6204.19	102.47	6101.72
N2	28191.39	0	28191.39	2.67	28188.72
H2O	30.98	-17.84	48.82	-28.31	77.13

Note : + ve sign indicates reactant being consumed in the reaction; - ve sign indicates product being formed in the reactor

First reactor
% Conversion with respect to C2H4 = 209.24 * 100 /1047.91 = 19.9 %
% Convsersion with respect to C2H4O is not feasible in first reactor as C2H4O product, is formed in the reaction.
Second reactor
% Conversion with respect to C2H4 = 168.03*100/ 838.67 = 20%
% Convsersion with respect to C2H4O =
30.96*100/206.79 = 15%