BALANCES DE MATERIA Y ENERGIA
Examen No.5 28 de Mayo del 2009
Nombre del Alumno (a) :___________________________________Cal.___________
1. Determine el volumen molar del oxígeno a -50 °C y 10.5 atmósferas mediante la ecuación de Soave-Redlich-Kwong. Por favor muestre todos sus cálculos. Si usted emplea una grafica para determinar la raíz de la ecuación, grafique al menos tres puntos.
2. Una corriente de oxigeno a -65°C y 8.3 atm fluye a razon de 250 kg/h. Emplee la ecuación de estado SRK para estimar la velocidad de flujo volumetrico de esta corriente.
3. Se hidrogena propileno en un reactor intermitente:
C3H6(g) + H2(g) à C3H8 (g)
Se alimentan cantidades equimolares de propileno e hidrogeno al reactor a 25°C y una presion total absoluta de 32.0 atm. Se eleva la temperatura del reactor a 235°C y se mantiene constante hasta que se haya completado la reaccion. La conversión de propileno al principio del periodo isotermico es 53.2 %. Puede suponer comportamiento de gas ideal para este problema, aunque a las altas presiones a que se trabaja esta suposición constituye, en el mejor de los casos, una aproximación muy burda.
(a) ¿Cuál es la presion final del reactor?
(b) ¿Cuál es la fraccion de conversión del propileno cuando P=35.1 atm?
(c) Construya una grafica de presion contra fraccion de conversión del propileno que comprenda el perido de operación isotermico. Utilice la grafica para confirmar los resultados de los incisos (a) y (b).
4. El analisis de un aceite combustible Num. 4 es 86.47% por peso de carbono, 11.65% de hidrogeno, 1.35% de azufre y el resto productos inertes no combustibles. El aceite se quema en un horno para generar vapor con 15% de aire en exceso. El aire se precalienta a 175°C y entra a el horno a una presion manometrica de 180 mmHg. El azufre y el hidrogeno del combustible se oxida en su totalidad a SO2 y H2O, 5% del carbono se oxida a CO y el balance forma CO2
(a) Calcule la velocidad de alimentación (m3 aire)/(kg aceite)
(b) Calcule las fracciones molares (en base seca) y las ppm (partes por millon en base humeda o moles contenidos en 106 molesde gas de combustión humedo) de las especies de gas de combustión que podrian considerarse riesgos ambientales.
SOLUCION
SOLUCION PROBLEMA 5.38
POLYMATH Results
Example 2 - Gas volume from SRK equation of state 06-02-2009, Rev5.1.225
NLE Solution
Variable Value f(x) Ini Guess
V 1.7058415 7.42E-08 0.07
Pc 49.7
R 0.08206
T 223.15
omega 0.021
Tc 154.4
a 1.380724
b 0.0220872
m 0.5175971
alpha 0.801642
P 10.5
NLE Report (safenewt)
Nonlinear equations
[1] f(V) = R*T/(V-b)-alpha*a/(V*(V+b))-P = 0
Explicit equations
[1] Pc = 49.7
[2] R = 0.08206
[3] T = 223.15
[4] omega = 0.021
[5] Tc = 154.4
[6] a = 0.42747*R^2*Tc^2/Pc
[7] b = 0.08664*R*Tc/Pc
[8] m = 0.48508+1.55171*omega-0.1561*omega^2
[9] alpha = (1+m*(1-sqrt(T/Tc)))^2
[10] P = 10.5
Comments
[1] f(V) = R*T/(V-b)-alpha*a/(V*(V+b))-P
Gas volume from SRK equation of state (V in L/g-mol)
[2] a = 0.42747*R^2*Tc^2/Pc
SRK constant a
[3] b = 0.08664*R*Tc/Pc
SRK constant b
[4] alpha = (1+m*(1-sqrt(T/Tc)))^2
SRK constant alpha
[5] m = 0.48508+1.55171*omega-0.1561*omega^2
SRK constant m
[6] Tc = 154.4
Critical temperature (K)
[7] Pc = 49.7
Critical pressure (atm)
[8] R = 0.08206
Gas constant (L-atm/gmol-K)
[9] omega = 0.021
Acentric factor
[10] T = 223.15
Temperature (K)
[11] P = 10.5
Pressure (atm)
Settings
Max iterations = 150
Tolerance F = 0.0000001
Tolerance X = 0.0000001
Tolerance min = 0.0000001
General
Search range: 1.8 > V > 1.6
number of implicit equations: 1
number of explicit equations: 10
Data file: F:\BALANCE2009\ExaMen5.pol
PROB. 5.6 A
POLYMATH Results
Example 2 - Gas volume from SRK equation of state 05-27-2009, Rev5.1.225
NLE Solution
Variable Value f(x) Ini Guess
V 2.0120566 1.269E-08 0.07
Pc 49.7
R 0.08206
T 208.15
omega 0.021
Tc 154.4
a 1.380724
b 0.0220872
m 0.5175971
alpha 0.8401961
P 8.3
NLE Report (safenewt)
Nonlinear equations
[1] f(V) = R*T/(V-b)-alpha*a/(V*(V+b))-P = 0
Explicit equations
[1] Pc = 49.7
[2] R = 0.08206
[3] T = 208.15
[4] omega = 0.021
[5] Tc = 154.4
[6] a = 0.42747*R^2*Tc^2/Pc
[7] b = 0.08664*R*Tc/Pc
[8] m = 0.48508+1.55171*omega-0.1561*omega^2
[9] alpha = (1+m*(1-sqrt(T/Tc)))^2
[10] P = 8.3
Comments
[1] f(V) = R*T/(V-b)-alpha*a/(V*(V+b))-P
Gas volume from SRK equation of state (V in L/g-mol)
[2] a = 0.42747*R^2*Tc^2/Pc
SRK constant a
[3] b = 0.08664*R*Tc/Pc
SRK constant b
[4] alpha = (1+m*(1-sqrt(T/Tc)))^2
SRK constant alpha
[5] m = 0.48508+1.55171*omega-0.1561*omega^2
SRK constant m
[6] Tc = 154.4
Critical temperature (K)
[7] Pc = 49.7
Critical pressure (atm)
[8] R = 0.08206
Gas constant (L-atm/gmol-K)
[9] omega = 0.021
Acentric factor
[10] T = 208.15
Temperature (K)
[11] P = 8.3
Pressure (atm)
Settings
Max iterations = 150
Tolerance F = 0.0000001
Tolerance X = 0.0000001
Tolerance min = 0.0000001
General
Search range: 2.2 > V > 1.8
number of implicit equations: 1
number of explicit equations: 10
PROBLEMA 5.6 B
POLYMATH Results
Example 2 - Gas volume from SRK equation of state 05-27-2009, Rev5.1.225
NLE Solution
Variable Value f(x) Ini Guess
V 2.0120566 1.269E-08 0.07
Pc 49.7
R 0.08206
T 208.15
omega 0.021
Tc 154.4
a 1.380724
b 0.0220872
m 0.5175971
alpha 0.8401961
P 8.3
NLE Report (safenewt)
Nonlinear equations
[1] f(V) = R*T/(V-b)-alpha*a/(V*(V+b))-P = 0
Explicit equations
[1] Pc = 49.7
[2] R = 0.08206
[3] T = 208.15
[4] omega = 0.021
[5] Tc = 154.4
[6] a = 0.42747*R^2*Tc^2/Pc
[7] b = 0.08664*R*Tc/Pc
[8] m = 0.48508+1.55171*omega-0.1561*omega^2
[9] alpha = (1+m*(1-sqrt(T/Tc)))^2
[10] P = 8.3
Comments
[1] f(V) = R*T/(V-b)-alpha*a/(V*(V+b))-P
Gas volume from SRK equation of state (V in L/g-mol)
[2] a = 0.42747*R^2*Tc^2/Pc
SRK constant a
[3] b = 0.08664*R*Tc/Pc
SRK constant b
[4] alpha = (1+m*(1-sqrt(T/Tc)))^2
SRK constant alpha
[5] m = 0.48508+1.55171*omega-0.1561*omega^2
SRK constant m
[6] Tc = 154.4
Critical temperature (K)
[7] Pc = 49.7
Critical pressure (atm)
[8] R = 0.08206
Gas constant (L-atm/gmol-K)
[9] omega = 0.021
Acentric factor
[10] T = 208.15
Temperature (K)
[11] P = 8.3
Pressure (atm)
Settings
Max iterations = 150
Tolerance F = 0.0000001
Tolerance X = 0.0000001
Tolerance min = 0.0000001
General
Search range: 2.2 > V > 1.8
number of implicit equations: 1
number of explicit equations: 10
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