Benemrita universidad de Guadalajara.

Centro Universitario de Ciencias Exactas e Ingenierias.

Materia: Analisis y diseo de reactores

Seccion: D03

Profesor: Macas Balleza Emma Rebeca

Trabajo: Examen III reactores (problemas)

Nombre:

Guevara Snchez Sandra Berenice

Naez Delgadillo Antonio

Guadalajara Jalisco 13/05 /2015

8.16

POLYMATH Results No Title 05-13-2015, Rev5.1.233

Calculated values of the DEQ variables Variable initial value minimal value maximal value final value T 358 358 450 450 G 0 -8992.4471 0 -8992.4471 R 1 1 7.452E+05 7.452E+05 E 4.0E+04 4.0E+04 4.0E+04 4.0E+04 tao 0.0848547 0.0066002 0.0848547 0.0066002 Cp0 50 50 50 50 k 2 2 2 2 Tc 350 350 350 350 deltaH -7500 -7500 -7500 -7500 X 0.1450868 0.0130285 0.1450868 0.0130285

ODE Report (RKF45) Differential equations as entered by the user

[1] d(G)/d(T) = deltaH*X

[2] d(R)/d(T) = Cp0*(1+k)*(T-Tc)

Explicit equations as entered by the user [1] E = 40000

[2] tao = 0.0066*exp((E/R)*((1/350)-(1/T)))

[3] Cp0 = 50

[4] k = 8000/(50*80)

[5] Tc = 350

[6] deltaH = -7500

[7] X = (tao*k)/(1+tao*k)

Independent variable variable name : T initial value : 358 final value : 450 Precision Step size guess. h = 0.000001 Truncation error tolerance. eps = 0.000001 General number of differential equations: 2 number of explicit equations: 7 Data file: C:\Users\andla_000\Desktop\reactores examen.pol

9.2

Como el problema nos da a elegir 1 problema de los ejemplos de este captulo y realizarlo

en polymat nosotros elegimos el ejemplo 9.1

Para nuestro caso donde nos pide el tiempo que se requiere para alcanzar el 90% con una

temperatura inicial de 20F = 497R por lo que en nuestro caso E9.-1.6 quedaria de laforma

= 497 + 89.8

y analizndolo en polymat el resultado es el siguiente.

POLYMATH Results 05-12-2015, Rev5.1.233

Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 10000 10000 X 0 0 0.9979573 0.9979573 T 497 497 586.61657 586.61657 K 2.655E-05 2.655E-05 0.003989 0.003989

ODE Report (RKF45) Differential equations as entered by the user

[1] d(X)/d(t) = K*(1-X)

Explicit equations as entered by the user [1] T = 497+89.8*X

[2] K = 0.000273*exp(16306*((1/535)-(1/T)))

Independent variable variable name : t initial value : 0 final value : 10000 Precision Step size guess. h = 0.000001 Truncation error tolerance. eps = 0.000001 General number of differential equations: 1 number of explicit equations: 2

y como vemos el tiempo que se requiere para llevar al 90% de conversin a esta temperatura es

aproximadamente 8300 segundos o 2 horas 18 minuto