EJERCICIOS DE DISEO DE REACTORESCICLO 2014-I

Ejercicio 1: La reaccin en fase gaseosa A B + C est siendo llevada a cabo en un reactor batch de 1000 dm3 a 500 K y 5.0 atm. El reactor es inicialmente cargado con una mezcla de gases equimolar de A y de un inerte.(a) Construya una tabla estequiometria mostrando las concentraciones molares al final de de la reaccin.(b) Determine el tiempo requerido para alcanzar 75% de conversin si la reaccin es elemental y la constante de velocidad es 0.0231 hr-1.(c) Encontrar la concentracin de B y la fraccin molar de inertes para 75% de conversin.Solucin:EspecieInicialCambioFinalConc(Ni/V)

ANAo-XNAoNAo(1-X)CA=CAo(1-X)

BX(b/a)NAoNAoXCB=CAoX

CX(c/a)NAoNAoXCC=CAoX

IINAoNAoCI=CAo

Total2NAoNT=NAo(2+X)

b) V=1000dm3 T= 500 K P= 5 atm X= 0.75 K= 0.0231 hr-1Reaccin de 1er orden: rA= dCA/dt (Reactor Batch) rA= -kCA

c)

Ejercicio 2:Calcular la masa de catalizador necesaria para convertir 90% de una alimentacin de 150 dm3/min de cantidades equimolares de reactante e inerte para la reaccin en fase gaseosa 2AB llevada a cabo en un PBR a 3 atm y 50C(k=0.1 dm6mol-1kg-1s-1).Solucin:Balance molar:

Orden de la reaccin:

Ejercicio propuesto:The gas-phase reaction A + 2B 2D is to be carried out in an isothermal plug flow reactor at 5.0 atm. The mole fractions of the feed stream are A = 0.20, B = 0.50, and inerts = 0.30.(a) What is the steady-state volumetric flow rate at any point in the reactor? (ignore pressure drop)(b) What are the stoichiometric expressions for the concentrations of A, B, and D as a function of conversion at any point along the reactor? In Out CA = FA/vA FAo -X FAo FAo (1-X) CAo (1-X)/(1-0.2X)B FBo = B FAo = (5/2) FAo -2 X FAo FAo (2.5-2X) CAo (2.5-2X)/(1-0.2X)D FDo =D FAo = 0 +2 X FAo FAo (2X) CAo (2X)/(1-0.2X)I FIo =I FAo = (3/2) FAo 0 FAo (1.5) CAo (1.5)/(1-0.2X)Total 5 FAo (5-X) FAo CAo (5-X)/(1-0.2X)(c) What is the feed concentration (mol/ dm3) of A if the feed temperature is 55C?(d) Determine how large the plug-flow reactor must be to achieve a conversion (based on A) of 0.70 if the temperature in the reactor is uniform (55C), volumetric feed rate is 50 dm3/min, and the rate law at 55C is = 1/2 , where k is 2.5 L0.5 / mol0.5 /min(e) Plot concentrations, volumetric flow rate, and conversion as a function of reactor length (L = 7.6 cm).(f) How large would a CSTR have to be to take the effluent from the PFR in part (d) and achieve a conversion of 0.85 if the temperature of the CSTR is 55C?(g) How many 1-inch diameter tubes, 20 ft in length, packed with a catalyst, are necessary to achieve 95% conversion of A using the original feed stream? Plot the pressure and conversion as a function of reactor length. The particles are 0.5 mm in diameter and the bed porosity is 45%.(h) Assuming the reaction is reversible, calculate the PFR volume required to achieve 70% of the equilibrium conversion, and the CSTR size necessary to raise the conversion of the PFR effluent to 85% of the equilibrium conversion if temperatures is constant at 100C. The activation energy for the reaction is 30 kJ/mol, and the reaction equilibrium constant at 100C is 10 (m 3 / kmol) 1/2.