Problema 2

4 A→R+6 S

a) Reactor Batch

−r A=k C A=k CAO (1−X A )

−r AV=−d N A

dt=¿

−d [N A 0−N A0X A ]dt

=N A0

d X A

dt

∫0

1

dt=N A0∫0

X A d X A

−r AV

t=N A 0∫0

x A d X A

−r AV

t=c A0∫0

xA d X A

−r A

t=C A0∫0

X A d X A

kC A0 (1−X A )

t=C A0∫0

X A d X A

1.052 x1019 e[−45707T ]

C A0 (1−X A )

t= 1

1.052x 1019∫0

X A d X A

e[−45707T ] (1−X A )

Aplicando un balance de materia:

N A0∆H R (T )4

d X A

dt+∑

i=1

n (N i0+v i4N A0 X A)Cpi dTdt =0

Se calcula el calor de reacción

∆ H R=∆ HR+∫TR

T

∆CpidT

∆Cpi=∑i=1

n

v iCpi¿productos−∑i=1

n

v iCpi ¿reactivos

Cálculo del Cp de reactivos:

CpP=4∗14.5cal

mol K=58 cal

mol K

Cálculo de Cp de productos

CpP=12.4cal

mol K+6 (7.2 ) cal

mol K=55.6 cal

mol K

∆Cpi=55.6−58=−2.4 calmolk

∆ H R (T )=5665−2.4 (T−298)

∆ H R (T )=5665−2.4 T+715.2

∆ H R (T )=6380.2−2.4T

dtN A0

[ N A 0∆H R (T )4

d X A

dt+∑

i=1

m

(N i0+v i4N A 0 X A)CpidT ]=0

∆ HR (T )4

d X A+∑i=1

m

( N A 0

N A 0

+v i4X A)CpidT=0

6380.2−2.4T4

d X A+[( N A 0

N A 0

−4i4X A)CpA+( N R0

N A0

+ 14X A)CpR+( N S0

N A0

+ 64X A)Cps]dT=0

(1595.05−0.6T )d X A+[ (1−X A )CpA+(0.25X A )CpR+( 64 X A)Cps]dT=0

(1595.05−0.6T )d X A+[ (1−X A )14.5+(0.25 X A )12.4+( 64 X A)7.2]dT=0

(1595.05−0.6T )d X A+ [ (1−X A )14.5+(0.25 X A )12.4+(1.5 X A )7.2 ]dT=0

(1595.05−0.6T )d X A+ [14.5−0.6 X A ] dT=0

(1595.05−0.6T )d X A=(0.6 X A−28.4 )dT

∫0

X A d X A

0.6 X A−14.5=−¿∫

T0

TdT

0.6T−1595.05¿

10.6∫0

X A 0.6d X A

0.6 X A−14.5=−¿ 1

0.6∫T 0

TdT

0.6T−1595.05¿

10.6ln (0.6 X A−14.5 )¿0

X A=¿− 10.6ln (0.6T−1595.06 ) ¿T 0

T ¿

ln (0.6 X A−14.5 )−ln (−28.4 )=−ln (0.6T−1595.06 )+ ln (0.6T0−1595.06 )

Aplicando leyes de los exponentes y la inversa del logaritmo natural:

0.6 X A−14.5−14.5

=0.6T 0−1595.050.6T−1595.05

0.6T−1595.5=(0.6T 0−1595.05 )( −14.50.6 X A−14.5 )

T= 10.6 [ (0.6T 0−1595.05 )( −14.5

0.6 X A−14.5 )+1595.05]T=(T 0+2658.42 )( −14.5

0.6 X A−14.5 )+2658.42Introduciendo la temperatura inicial

T=(945−2658.42 ) −14.50.6 X A−14.5

+2658.42

T= 24842.40.6 X A−14.5

+2658.42Línea deoperación .

Convirtiendo el gasto volumétrico y los tiempos muertos

vo=77.53Lmin ( 1m3

1000 L )(1min60 s )=0.001292m3/s

tm=0.763min( 60S1min )=45.78 s

t=Vvo

= 0.1m3

0.00129m3/s=77.39 s

t=t t−tm=77.39−45.78=31.6094 s

t= 1

1.052x 1019∫0

X A d X A

e[ −45707

24844.60.6 X A−14.5

+2658.42](1−X A )

31.6094= 1

1.052x 1019∫0

X A d X A

e[ −45707

24844.60.6 X A−14.5

+2658.42](1−X A )

Despejando la conversión:

X A=0.1985

T= 24842.40.6(0.2)−14.5

+2658.42=930.962K

b) Reactor PFR

Balance de materia

−r A=−dF A

dV=

−d [F A 0−X A F A0 ]dV

=F A 0dX A

dV

∫0

v

dv=FA 0∫0

X A dX A

−r A

V=F A 0∫0

X A dX A

−r A=¿ F A0∫

0

X A dX A

k C A0 (1−X A )(1+εA X A )

T 0T

=F A0

1.052x 1019∫0

X A (1+εA X A ) TT 0

dX A

e[−45707T ]

(1−X A)

¿

Balance de energía:

∑i=1

m

F A0∫T

T 0

CpdT=FA 0 X A∆ H R

4

F A0C p A (T 0−T )=FA 0 X A∆ H R(T )

4

14.5 (T 0−T )=X A (6380.2−2.4T )

4

14.5T 0−14.5T=1595.05 X A−0.6 X AT

T (0.6 X A−14.5 )=1595.05 X A−14.5T 0

T=1595.05 X A−14.5T 00.6 X A−14.5

líneade operación

Vv0

= 11.052x 1019

∫0

X A=0.3 (1+0.75 X A )

1595.05 X A−14.5 (945.15)0.6 X A−14.15

(945.15)dX A

e[−45707T ]

(1−X A)

=65.8 s

v0V

=1.5207 x10−21 /s

c) Reactor CSTR

F A0−F A+r AV=0

F A0−F A=−r AV

F A0−(F A0−X A F A0 )=−r AV

F A0−F A0+X AF A 0=−r AV

X A

−r A= VFA 0

VCA 0 v0

=X A (1+εA X A ) T

T 0KC A0 (1−X A )

Vv0

=X A (1+εA X A )TKC A0 (1−X A )T 0

Vv0

=X A (1+0.75 X A )T

KC A0 (1−X A )945.15

Vv0

=X A (1+0.75 X A )

1595.05 X A−14.15T 00.6 X A−14.15

1.052 x1019 e[ −457071595.05 X A−14.15T 0

0.6 X A−14.15](1−X A )

Utilizando una conversión de 0.23 y una temperatura de 945.15

t=77.49 s