Compartment 1: 2.6 A closed stirred-tank reactor with two compartments is shown in Fig. E2.6. The basic idea is to feed the reactants continuously into the first compartment, where they will be preheated by energy liberated in the exothermic reaction, which is anticipated to occur primarily in the second compartment. The wall separating the two compartments is quite thin, thus allowing heat transfer; the outside of the reactor is well insulated; and a cooling coil is built into the second compartment to remove excess energy liberated in Over-all Balance: 0 = pq - pq. thus 91 = 4 Energy balance: V,pC = pqC(T; – T) – UA(T, – T2) the reaction. Tests are to be conducted initially with a single-component feed (i.e., no reaction) to evaluate the reactor's thermal characteristics. Compartment 2: Notes: Over-all Balance: U, A: Overall heat transfer coefficient and surface area between compartments. 0 = pq. – pg: thus 92 - 41 = 9 Energy balance: U, Ac: Overall heat transfer coefficient and surface area of cooling tube. = pqC(T1 – T2) + UA(T, – T2) – U.A.(T1 – T2 dT Vi: Volume of Compartment 1. V:: Volume of Compartment 2. Cooling medium Te 92 T2 T V2 Figure E2.6 Leeeee
Compartment 1: 2.6 A closed stirred-tank reactor with two compartments is shown in Fig. E2.6. The basic idea is to feed the reactants continuously into the first compartment, where they will be preheated by energy liberated in the exothermic reaction, which is anticipated to occur primarily in the second compartment. The wall separating the two compartments is quite thin, thus allowing heat transfer; the outside of the reactor is well insulated; and a cooling coil is built into the second compartment to remove excess energy liberated in Over-all Balance: 0 = pq - pq. thus 91 = 4 Energy balance: V,pC = pqC(T; – T) – UA(T, – T2) the reaction. Tests are to be conducted initially with a single-component feed (i.e., no reaction) to evaluate the reactor's thermal characteristics. Compartment 2: Notes: Over-all Balance: U, A: Overall heat transfer coefficient and surface area between compartments. 0 = pq. – pg: thus 92 - 41 = 9 Energy balance: U, Ac: Overall heat transfer coefficient and surface area of cooling tube. = pqC(T1 – T2) + UA(T, – T2) – U.A.(T1 – T2 dT Vi: Volume of Compartment 1. V:: Volume of Compartment 2. Cooling medium Te 92 T2 T V2 Figure E2.6 Leeeee
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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