Download - Examen de Reactores
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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
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8.16
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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
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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
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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.
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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