problema 11
DESCRIPTION
Problema 11TRANSCRIPT
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