Question: (1)The space time required to achieve 80% conversion in a CSTR is 5 h. The entering volumetric flow rate and concentration of reactant A are

(-4, (a) Revisit Examples 21 through 23. How would your answers change if the flow rate, FA0. were cut in half? If it were doubled? (b) Example 2-5. How would your answers change if the two CSTRs (one 0.82m3 and the other 3.2m ) were placed in parallel with the flow: FA0. divided equally to each reactor. (c) Example 2-6. How would your answer change if the PFRs were placed in parallel with the flow, FA0. divided equally to each reactor? (d) Example 2-7. (1) What would be the reactor volumes if the two intermediate conversions were changed to 20% and 50%. respectively. (2) What would be the conversions, X1,X2, and X3, if all the reactors had the same volume of 100dm3 and were placed in the same order? (3) What is the worst possible way to arrange the two CSTRs and one PFR? (e) Example 2-8. The space time required to achieve 80% conversion in a CSTR is 5h. The cntering volumetric flow rate and concentration of reactant A are 1dm3/min and 2.5 molar. respectively. If possible. determine (1) the rate of reaction. rA= (2) the reactor volume, V= . (3) the exit concentration of A,CA= . and (4) the PFR space time for 80% conversion. You have two CSTRs and two PFRs each with a volume of 1.6m3. Use Figure 2-2 to calculate the conversion for each of the 1112 io 106 ig arrangements. (a) Two CSTRs in series. (b) Two PFRs in series. (c) Two CSTRs in parallel with the feed. FA0. divided equally between the two reactors. (d) Two PFRs in parallel with the feed divided equally between the two reactors. (e) A CSTR and a PFR in parallel with the flow equally divided. Also calculate the overall conversion. Xov Xov=FA0FA0FACSTRFAPFR,withFACSTR=2FA02FA0XCSTR.FAPFR=2FA0(1XPFR) (f) A PFR follow by a CSTR
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