Chemical Engineering Chapter 7 Solutions For Reaction Mechanisms Pathways Bioreactions And Bioreactors Example The Graph

subject Type Homework Help
subject Pages 14
subject Words 1536
subject Textbook Elements of Chemical Reaction Engineering 4th Edition
subject Authors H. Scott Fogler
search this document

Unlock document.

This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
page-pf1
7-1
Solutions for Chapter 7 – Reaction Mechanisms,
Pathways, Bioreactions and Bioreactors
P7-1 (a) Example 7-1
The graph of Io/I will remain same if CS2 concentration changes. If concentration of M increases the slope
P7-1 (b) Example 7-2
For t = 0 to t = 0.35 sec, PSSH is not valid as steady state not reached.
And at low temperature PSSH results show greatest disparity.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 12 12
C1 0.1 2.109E-04 0.1 2.109E-04
C2 0 0 1.311E-09 1.311E-09
C6 0 0 3.602E-09 3.602E-09
C4 0 0 2.665E-07 1.276E-08
k5 3.98E+09 3.98E+09 3.98E+09 3.98E+09
k2 2.283E+06 2.283E+06 2.283E+06 2.283E+06
k4 9.53E+08 9.53E+08 9.53E+08 9.53E+08
k3 5.71E+04 5.71E+04 5.71E+04 5.71E+04
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(C1)/d(t) = -k1*C1-k2*C1*C2-k4*C1*C6
[6] d(C3)/d(t) = k2*C1*C2
[7] d(C5)/d(t) = k3*C4
page-pf2
7-2
P7-1 (c) Example 7-3
The inhibitor shows competitive inhibition.
P7-1 (d) Example 7-4
1)Now Curea = 0.001mol/dm3 and t = 10 min = 600 sec.
MAX
urea
MAX
M
V
XC
XV
k
t+
!
"
#
$
%
&
'
=1
1
ln
( )
3
3
3
3
/001.0
1
1
ln
/0266.0
sec600
Xdmmol
X
dmmol +
!
#
$
&
'
=
urea
urea
r
XC
!
=
"
ureaM
ureaMAX
urea CK
CV
r
+
=!
!
XC
rurea
urea ="
NLE Solution
Variable Value f(x) Ini Guess
X 0.6751896 -8.062E-10 0.5
NLE Report (safenewt)
Nonlinear equations
page-pf3
7-3
!"
=
O
ures
C
C
urea
r
dC
#
and
( )
XCC ureaurea !=1
0
!
urea
M
V
XC
XV
k+
!
#
$
&
'
=1
1
ln
(
P7-1 (e) Example 7-5
( ) gg
C
C
Y
P
S
PS /18.2
14.203.5
2457.238
/=
!
!!
=
"
"!
=
gg
Y
Y
PS
SP /46.0
18.2
11
/
/===
gg
Y
Y
PCS /87.1
46.0075.0
11
/=
+
==
+
P7-1 (f) Example 7-6
1) if we go for 24 hrs, fermentation will stop at 13.2 hrs as CP = CP
* .
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 13.2 13.2
Cc 1 1 16.558613 16.550651
Cs 250 39.292786 250 39.292786
rd 0.01 0.01 0.16559 0.1655065
Ysc 12.5 12.5 12.5 12.5
Ypc 5.6 5.6 5.6 5.6
umax 0.33 0.33 0.33 0.33
kobs 0.33 0.0039386 0.33 0.0039386
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd
[2] d(Cs)/d(t) = Ysc*(-rg)-rsm
page-pf4
7-4
Explicit equations as entered by the user
[1] rd = Cc*0.01
[2] Ysc = 1/0.08
[3] Ypc = 5.6
[4] Ks = 1.7
[5] m = 0.03
[6] umax = 0.33
2)Semi-Batch reactor:
See Polymath program P7-1-f-2.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 24 24
Cc 1.0E-04 1.0E-04 0.0474697 0.0474697
Cs 1.0E-04 1.0E-04 12.206266 12.206266
Ysc 12.5 12.5 12.5 12.5
Ks 1.7 1.7 1.7 1.7
m 0.03 0.03 0.03 0.03
kobs 0.33 0.3294925 0.33 0.3294925
Cso 5 5 5 5
vo 0.5 0.5 0.5 0.5
Vo 1 1 1 1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd
[2] d(Cs)/d(t) = vo*Cso/V + Ysc*(-rg)-rsm
Explicit equations as entered by the user
[1] rd = Cc*0.01
[2] Ysc = 1/0.08
[3] Ypc = 5.6
[6] umax = 0.33
[7] rsm = m*Cc
[8] kobs = (umax*(1-Cp/93)^0.52)
[11] vo = 0.5
[12] Vo = 1
page-pf5
7-5
3) Changes from part(2)
I
S
SS
SC
P
P
g
K
C
CK
CC
C
C
r2
52.0
*
max 1
++
!
!
"
#
$
$
%
&'=
µ
See Polymath program P7-1-f-3.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 24 24
Cc 1.0E-04 1.0E-04 1.514E-04 1.514E-04
Cp 0 0 4.669E-04 4.669E-04
rd 1.0E-06 1.0E-06 1.514E-06 1.514E-06
Ypc 5.6 5.6 5.6 5.6
Ks 1.7 1.7 1.7 1.7
m 0.03 0.03 0.03 0.03
rsm 3.0E-06 3.0E-06 4.541E-06 4.541E-06
kobs 0.33 0.3299991 0.33 0.3299991
Ki 0.7 0.7 0.7 0.7
Vo 1 1 1 1
page-pf6
7-6
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd
Explicit equations as entered by the user
[1] rd = Cc*0.01
[2] Ysc = 1/0.08
[3] Ypc = 5.6
[4] Ks = 1.7
[10] Cso = 5
[11] vo = 0.5
[12] Vo = 1
[13] V = Vo+vo*t
4) After 9.67 hrs, CS = 0.
page-pf7
7-7
P7-1 (g) Individualized solution
P7-1 (h) Individualized solution
P7-1 (i) Individualized solution
P7-1 (j) Individualized solution
P7-1 (k) Individualized solution
P7-1 (l) Individualized solution
P7-1 (m) Individualized solution
P7-1 (n) Individualized solution
P7-2 Solution is in the decoding algorithm given with the modules.
P7-3
Burning:
page-pf8
7-8
page-pf9
7-9
P7-3 (d) Individualized solution
P7-3 (e) Individualized solution
page-pfa
7-10
P7-4
page-pfb
7-11
P7-4 (d) Individualized solution
P7-4 (e) Individualized solution
P7-5 (a)
page-pfc
7-12
P7-5 (b)
Cl2
k1
" # "
k2
$ " " 2Cl
Cl +CO
k3
" # "
k4
$ " " COCl
5
2 2
k
COCl Cl COCl Cl+!+
( )( ) ( ) ( )( )
3 4 5 2
0
COCl
r k Cl CO k COCl k COCl Cl= = ! !
( ) ( )( )
( )
3
4 5 2
k Cl CO
COCl
k k Cl
=+
( )( ) ( )( )( )
( )
2
1 3 2
5 2
4 5 2
COCl
k k CO Cl Cl
r k COCl Cl
k k Cl
= = +
( ) ( ) ( )( ) ( ) ( )( )
2
1 2 2 3 4 5 2
0
Cl
r k Cl k Cl k Cl CO k COCl k COCl Cl= = ! ! + +
add rCOCl to rCl
( ) ( )2
1 2 2
0 0
Cl COCl
r r k Cl k Cl+ = + = !
( ) ( )
21
2
2
k
Cl Cl
k
=
( ) ( )
1
2
2
k
Cl Cl
k
=
( ) ( ) ( )
( )
( )( )
( )
2
3
0.5
1 1 2
1 3 2 2 1 3 2
2 2
4 5 2 4 5 2
COCl
k k
k k CO Cl Cl k k CO Cl
k k
r
k k Cl k k Cl
= =
+ +
page-pfd
7-13
( )
4 5 2
k k Cl>>
( )( )
2
23
1 3 2
2
2 4
COCl
k k
r CO Cl
k k
=
P7-5 (c) Individualized solution
P7-5 (d) Individualized solution
P7-6 (a)
NO2 + hv
1
K
!
NO + O
O2 + O + M
2
K
!
O3 + M
O3 + NO
3
K
!
NO2 + O2
Using PSSH,
MOONOO CCCkCkr
22 21
0!==
!
2
2
2
1
O
NO
MO Ck
Ck
CC =
NOOMOOO CCkCCCkr
323 32
0!==
!
NO
NO
NO
MOO
OCk
Ck
Ck
CCCk
C
3
1
3
222
3==
P7-6 (b)
NOOMOO
OCCkCCCk
dt
dC
32
3
32 !=
NOONO
NO CCkCk
dt
dC
32
2
31 +!=
NOONO
NO CCkCk
dt
dC
32 31 !=
NOOMOO
OCCkCCCk
dt
dC
32
2
32 +!=
P7-6 (c)
page-pfe
7-14
P7-6 (d) Individualized solution
P7-6 (e) Individualized solution
P7-7(a)
page-pff
7-15
P7-7(b)
Low temperatures with anti-oxidant
page-pf10
7-16
P7-7(c)
If the radicals are formed at a constant rate, then the differential equation for the concentration of the
radicals becomes:
[ ] [ ][ ]
00
i
d I
k k I RH
dt =!=
page-pf11
7-17
and
[ ] [ ]
0
i
k
I
k RH
=
The substitution in the differential equation for R· also changes. Now the equation is:
[ ] [ ][ ] [ ][ ] [ ][ ]
1 2 2 2 0
i P P
d R
k I RH k R O k RO RH
dt =!+ =
 
and solving and substituting gives:
[ ] [ ][ ]
0 2 2
P
k k RO RH
R
k O
+
=
Now we have to look at the balance for RO2·.
[ ] [ ][ ] [ ][ ] [ ]2
2
1 2 2 2 0
P P t
d RO
k R O k RO RH k RO
dt =! ! =
 
and if we substitute in our expression for [R·] we get
[ ]2
0 2
0t
k k RO=!
which we can solve for [RO2·].
[ ] 0
2
t
k
RO
k
=
Now we are ready to look at the equation for the motor oil.
[ ] [ ][ ] [ ][ ]
2 2i P
d RH
k I RH k RO RH
dt =! !
 
and making the necessary substitutions, the rate law for the degradation of the motor oil is:
[ ] [ ]
0
0 2RH P
t
d RH k
r k k RH
dt k
= = ! !
P7-7(d)
Without antioxidants With antioxidants
page-pf12
7-18
P7-7 (e) Individualized solution
P7-7 (f) Individualized solution
P7-7 (g) Individualized solution
P7-8 (a)
P7-8 (b)
P7-8 (c)
page-pf13
7-19
P7-8 (d)
See Polymath program P7-8-d.pol.
Everyone becomes ill rather quickly, and the rate at which an ill person recovers to a healthy person is
P7-8 (e) Individualized solution
P7-8 (f) Individualized solution
P7-9 (a)
Starting with the design equation for a batch reactor
page-pf14
7-20
Using the data at 40oF and 45oF the following graphs are made.
P7-9 (b)
P7-9 (c)
P7-9 (d)
The data appears that it may fit the Monod equation for substrate consumption at the stationary phase.
P7-9 (e) Individualized solution
P7-9 (f) Individualized solution

Trusted by Thousands of
Students

Here are what students say about us.

Resources
Company
Legal

Copyright ©2022 All rights reserved. | CoursePaper is not sponsored or endorsed by any college or university.