Reactores

REACOTRES 1
EXAMEN PARCIAL FEBRERO 2012
FERNANDO ALONSO CUEVAS
PROBLEMA I
# Problema 1

Ts = 600 + 273 # Temperatura estándar
T1 = 600 + 273
T2 = 600 + 273
Pt = 2 # Presión de operación en bar
R = 8.314 # J/mol °K

# Generalizando para una reacción A + B ------> C + D
nA0 = 1 # Moles iniciales de C2H4
nB0 = 1 # Moles iniciales de H6O6
nC0 = 0 # Moles iniciales de C6H5H5
nD0 = 0nT = nA0 + nB0 + nC0 + nD0 # Moles totales

# Calores de formación a la temperatura estándar Hfi
DHfA = 52340 # Calor de formación de A
DHfB = 52210 # Calor de formación de B
DHfC = 29810 # Calor de formación de C
DHfD = 0

# Valores de las constantes del Cp/R de cada compuesto: ai, bi, ci, di
aA = 1.424
aB = -0.206
aC = 1.124
aD = 0

bA = 14.398e-3
bB = 39.064e-3
bC = 55.380e-3bD = 0

cA = -0.000004392
cB = -0.000013301
cC = -0.000018476
cD = 0

dA = 0
dB = 0
dC = 0
dD = 0

# Números estequiométricos
VA = -1
VB = -1
VC = 1
VD = 0
dV = VA + VB + VC + VD

# Delta de las constantes
DA = VA * aA + VB * aB + VC * aC + VD * aD
DB = VA * bA + VB * bB + VC * bC + VD * bD
DC = VA * cA + VB * cB + VC * cC + VD * cD
DD = VA * dA + VB * dB + VC * dC + VD *dD

dHr = VA * DHfA + VB * DHfB + VC * DHfC + VD * DHfD # Calor de reacción a la temperatura estándar
J = dHr - (DA * Ts + ((DB / 2) * (Ts ^ 2)) + ((DC / 3) * (Ts ^ 3)) + (DD / Ts)) * R
K = exp(-(J / R) * (1 / T2 - 1 / T1) + (DA * ln(T2 / T1)) + (DB / 2 * (T2 - T1)) + (DC / 6 * (T2 ^ 2 - T1 ^ 2)) + (DD / 2 * (1 / T2 ^ 2 - 1 / T1 ^ 2)))

f(e) = K * yA * yB * Pt - yC * yD # Extensión de lareacción
e(0) = 0
e(min) = 0
e(max) = 1

# Fracciones mol a la salida del reactor
yA = (nA0 + VA * e) / (nT + dV * e)
yB = (nB0 + VB * e) / (nT + dV * e)
yC = (nC0 + VC * e) / (nT + dV * e)
yD = (nD0 + VD * e) / (nT + dV * e)
sY = yA + yB + yC + yD

# Conversión de C2H4
CA = -(VA * e) / nA0

xA = -(VA * e) / nA0

Calculated values of NLE variables
| Variable | Value | f(x) |Initial Guess |
1 | e | 1. | 1.142E-15 | 0.5 ( 0 < e < 1. ) |

| Variable | Value |
1 | Ts | 873. |
2 | T1 | 873. |
3 | T2 | 873. |
4 | Pt | 2. |
5 | R | 8.314 |
6 | nA0 | 1. |
7 | nB0 | 1. |
8 | nC0 | 0 |
9 | nD0 | 0 |
10 | nT | 2. |
11 | DHfA | 5.234E+04 |
12 | DHfB | 5.221E+04 |
13 | DHfC | 2.981E+04 |
14 |DHfD | 0 |
15 | aA | 1.424 |
16 | aB | -0.206 |
17 | aC | 1.124 |
18 | aD | 0 |
19 | bA | 0.014398 |
20 | bB | 0.039064 |
21 | bC | 0.05538 |
22 | bD | 0 |
23 | cA | -4.392E-06 |
24 | cB | -1.33E-05 |
25 | cC | -1.848E-05 |
26 | cD | 0 |
27 | dA | 0 |
28 | dB | 0 |
29 | dC | 0 |
30 | dD | 0 |
31 | VA | -1. |
32| VB | -1. |
33 | VC | 1. |
34 | VD | 0 |
35 | dV | -1. |
36 | DA | -0.094 |
37 | DB | 0.001918 |
38 | DC | -7.83E-07 |
39 | DD | 0 |
40 | dHr | -7.474E+04 |
41 | J | -7.869E+04 |
42 | K | 1. |
43 | yA | 2.39E-08 |
44 | yB | 2.39E-08 |
45 | yC | 1. |
46 | yD | 0 |
47 | sY | 1. |
48 | CA | 1. |
49 | xA | 1. |Nonlinear equations
1 | f(e) = K * yA * yB * Pt - yC * yD = 0 |
| Extensión de la reacción |

Explicit equations
1 | Ts = 600 + 273 |
| Temperatura estándar |
2 | T1 = 600 + 273 |
3 | T2 = 600 + 273 |
4 | Pt = 2 |
| Presión de operación en bar |
5 | R = 8.314 |
| J/mol °K |
6 | nA0 = 1 |
| Moles iniciales de C2H4 |
7 | nB0 = 1 |
| Molesiniciales de H6O6 |
8 | nC0 = 0 |
| Moles iniciales de C6H5H5 |
9 | nD0 = 0 |
10 | nT = nA0 + nB0 + nC0 + nD0 |
| Moles totales |
11 | DHfA = 52340 |
| Calor de formación de A |
12 | DHfB = 52210 |
| Calor de formación de B |
13 | DHfC = 29810 |
| Calor de formación de C |
14 | DHfD = 0 |
15 | aA = 1.424 |
16 | aB = -0.206 |
17 | aC =...