Problema 11 DATOS:

A↔B

Q̇=0

Fase líquida∴V=cte

PRF

D=50 cm=0.5m

T 0=57 °C=330K

AI=9∴F A0=9F I 0

C A0=9.3molL

C A=4.65molL

FB=20.375mols

Costo: $50,000/m

BALANCE DE MATERIA DEL PFR:E−S+G=A

E−S+G=0

F A∨V−F A∨(V +∆V )+rA∆V=0

F A∨V−F A∨(V +∆V )=−rA ∆V

−rA=−FA∨(V +∆V )−F A∨V

∆V

−r A=−[ lim∆V→0

F A∨(V +∆ V )−F A∨V

∆V ]−rA=

−dF A

dV

−r A=−d [F A0−X A F A0 ]

dV

−r A=FA 0dX A

dV

V=F A0∫0

X A dX A

−rA

−r A=k (CA−CB

keq )V=F A0∫

0

X A dX A

k (C A−CB

keq )

V=F A0∫0

X A dX A

k (C A0 (1−X A )−CB0−

ba C A0 X A

keq)

V=F A0∫0

X A dX A

k (C A0 (1−X A )−C A0 X A

keq)

Sabemos

C A=C A0 (1−X A )

C A=C A0−CA 0 X A

C A0 X A=CA 0−C A

X A=1−C A

C A0

X A=1−4.65mol

L

9.3 molL

=0.5

CB=CB 0−baC A0 X A=C A0 X A=(9.3 molL )(0.5 )=4.65 mol

L

v0=FB

CB=20.375 mol

s

4.65 molL

=4.38172 Ls=4.38172

Ls∗60 s

1min∗60min

1h=15,774.2 L

h

F A0=C A0 v0=(9.3molL )(15,774.2 Lh )=146,700molhV=146,700∫

0

0.5 dX A

31.1 e[7906( 1360− 1T )](9.3 (1−X A )−

9.3 X A

3.03e[−830( 1333− 1T )] )BALANCE DE ENERGÍA PARA EL PFR:

∑i=1

m

F i0∫T 1

T 0

Cpi dT−FA 0X A∆HR (T )

a=0

∑i=1

m

F i0∫T 1

T 0

Cpi dT=FA 0X A∆ HR (T )

a

Calor de reacción

∆ H R (T )=∆H R (T R )+∫T R

T

∆CpidT

∆Cpi=∑i=1

n

v iCpi¿productos−∑i=1

n

viCpi ¿reactivos

∆Cpi= (1 )(141 Jmol K )−(1 )(141 J

mol K )=0∆H R (T )=∆ H R (T R )+0

∆ H R (T )=∆ H R (T R )

∆H R (T )=−6900 Jmol

[FA 0CpA+F A0

aCp I ](T0−T )=

F A0 X A (−6900)1

F A0[CpA+CpI

9 ] (T 0−T )=F A0 X A(−6900)

[141+ 1619 ](T0−T )=−6900 X A

158.889 (T 0−T )=−6900 X A

158.889T0−158.889T=−6900 X A

T=158.889T 0+6900 X A

158.889

T=T 0+43.4265 X A

Sustituyendo la línea de operación en el balance de materia

V=146700∫0

0.5 dX A

31.1 e[7906(1360

− 1330+43.4265 X A

)](9.3 (1−X A )−9.3 X A

3.03e[−830( 1333− 1

300+43.4265 X A )] )V=1407.3 L

V=1.4073m3

V=π (D2 )2

L

L= V

π ( D2 )2=1.40732

π ( 0.52 )2=7.167m

CT=L( $50000m )=7.167m( $50000m )CT=$358,350