Umich反应工程 lec18.pptVIP

Lecture18 Thursday 3/13/08Solution to Tuesdays In-class Problem .User Friendly Energy Balance DerivationsAdiabatic (Tuesday’s lecture) .Heat Exchange Constant Ta .Heat Exchange Variable Ta Co-current .Heat Exchange Variable Ta Counter Current Adiabatic Operation CSTR Adiabatic Energy Balance – Adiabatic, DCP = 0 Irreversible for Parts (a) through (c) (b) (c)? Levenspiel Plot CSTR?????X = 0.6?????T = 360 PFR?????X = 0.6 Summary (d) At Equilibrium (e) Te = 358 Xe = 0.59 User Friendly Equations Relate T and X or Fi User Friendly Equations Relate T and X or Fi Heat Exchange Mole Balance (1) Energy Balance A. Constant Ta (17B) Ta = 300K * Elementary liquid phase reaction carried out in a CSTR The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1. (a) Assuming the reaction is irreversible, A ? B, (KC = 0) what reactor volume is necessary to achieve 80% conversion? (b) If the exiting temperature to the reactor is 360K, what is the corresponding reactor volume? (c) Make a Levenspiel Plot and then determine the PFR reactor volume for 60% conversion and 95% conversion. Compare with the CSTR volumes at these conversions (d) Now assume the reaction is reversible, make a plot of the equilibrium conversion as a function of temperature between 290K and 400K. Mole Balance Rate Law Stoichiometry (a) Given X = 0.8, find T and V CSTR?????X = 0.95?????T = 395 PFR?????X = 0.95 V = 5.28 dm3 Texit = 360 X = 0.6 PFR V = 6.62 dm3 Texit = 395 X = 0.95 PFR V = 7.59 dm3 T = 395 X = 0.95 CSTR V = 2.05 dm3 T = 360 X = 0.6 CSTR Elementary liquid phase reaction carried out in a PFR The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1. Rate Law (2) (3) (4) Stoichiometry (5)  (6) Parameters (7) – (15) Adiabatic and (16A) Additional Parameters (17A)??(17B) Heat