Solutions for Chapter 1 – Mole Balances
P1-1.
This problem helps the student understand the course goals and objectives. Part (d) gives
hints on how to solve problems when they get stuck.
P1-2.
Encourages students to get in the habit of writing down what they learned from each chapter.
It also gives tips on problem solving.
P1-3.
Helps the student understand critical thinking and creative thinking, which are two major
goals of the course.
P1-4.
Requires the student to at least look at the wide and wonderful resources available on the CDROM and the Web.
P1-5.
The ICMs have been found to be a great motivation for this material.
P1-6.
Uses Example 1-1 to calculate a CSTR volume. It is straight forward and gives the student an
idea of things to come in terms of sizing reactors in chapter 4. An alternative to P1-15.
P1-7.
Straight forward modification of Example 1-1.
P1-8.
Helps the student review and member assumption for each design equation.
P1-9.
The results of this problem will appear in later chapters. Straight forward application of chapter
1 principles.
P1-10.
Straight forward modification of the mole balance. Assigned for those who emphasize
bioreaction problems.
P1-11.
Will be useful when the table is completed and the students can refer back to it in later
chapters. Answers to this problem can be found on Professor Susan Montgomery’s
equipment module on the CD-ROM. See P1-14.
P1-12. Many students like this straight forward problem because they see how CRE principles can be
applied to an everyday example. It is often assigned as an in-class problem where parts (a)
through (f) are printed out from the web. Part (g) is usually omitted.
P1-13. Shows a bit of things to come in terms of reactor sizing. Can be rotated from year to year with
P1-6.
P1-14. I always assign this problem so that the students will learn how to use POLYMATH/MATLAB
before needing it for chemical reaction engineering problems.
P1-15 and P1-16. Help develop critical thinking and analysis.
CDP1-A
Similar to problems 3, 4, and 10.
1-2
Summary
Assigned
Alternates
Difficulty
Time (min)
 P1-1
AA
SF
60
P1-2
I
SF
30
 P1-3
O
SF
30
P1-4
O
SF
30
P1-5
AA
SF
30
P1-6
AA
SF
15
P1-7
I
SF
15
 P1-8
S
SF
15
P1-9
S
SF
15
P1-10
O
FSF
15
P1-11
I
SF
1
P1-12
O
FSF
30
P1-13
O
SF
60
 P1-14
AA
SF
60
P1-15
O
--
30
P1-16
O
FSF
15
CDP1-A
AA
FSF
30
1-13
Assigned
 = Always assigned, AA = Always assign one from the group of alternates,
O = Often, I = Infrequently, S = Seldom, G = Graduate level
1-3
Alternates
In problems that have a dot in conjunction with AA means that one of the problem, either the
problem with a dot or any one of the alternates are always assigned.
Time
Approximate time in minutes it would take a B/B+ student to solve the problem.
Difficulty
SF = Straight forward reinforcement of principles (plug and chug)
FSF = Fairly straight forward (requires some manipulation of equations or an intermediate
calculation).
IC = Intermediate calculation required
M = More difficult
OE = Some parts open-ended.
*
Note the letter problems are found on the CD-ROM. For example A
CDP1-A.
Summary Table Ch-1
Review of Definitions and Assumptions
1,5,6,7,8,9
Introduction to the CD-ROM
1,2,3,4
Make a calculation
6
Open-ended
8
1-4
P1-1 Individualized solution.
P1-2
(b)
The negative rate of formation of a species indicates that its concentration is decreasing as the
reaction proceeds ie. the species is being consumed in the course of the reaction.
A positive number indicates production of the particular compound.
(c)
The general equation for a CSTR is:
FA 0
V
FA
rA
Here rA is the rate of a first order reaction given by:
rA = - kCA
Given : CA = 0.1CA0 , k = 0.23 min-1, v0 = 10dm3 min-1
Substituting in the above equation we get:
V
CA0 v0 C Av0
kC A
C A0 v0 (1 0.1)
0.1kC A0
V = 391.304 m3
(d)
k = 0.23 min-1
From mole balance:
dNA
Rate law:
rA
k
dt
rA V
rA
k CA
NA
V
Combine:
dNA
dt
k NA
1-5
(10dm3 / min)(0.9)
(0.23 min 1 )(0.1)
at t
0 , NAO = 100 mol and t
t=
N
1
ln A0
k
NA
1
ln 100
0.23
t , NA = (0.01)NAO
min
t = 20 min
P1-3 Individualized solution.
P1-4 Individualized solution.
P1-5 Individualized solution.
P1-6 Individualized solution
P1-7 (a)
The assumptions made in deriving the design equation of a batch reactor are:
-
Closed system: no streams carrying mass enter or leave the system.
-
Well mixed, no spatial variation in system properties
-
Constant Volume or constant pressure.
1-6
P1- 7 (b)
The assumptions made in deriving the design equation of CSTR, are:
-
Steady state.
-
No spatial variation in concentration, temperature, or reaction rate throughout the vessel.
P1-7(c)
The assumptions made in deriving the design equation of PFR are:
-
Steady state.
-
No radial variation in properties of the system.
P1-7 (d)
The assumptions made in deriving the design equation of PBR are:
-
Steady state.
-
No radial variation in properties of the system.
P1-7 (e)
For a reaction,
A B
-rA is the number of moles of A reacting (disappearing) per unit time per unit volume [=]
moles/ (dm3.s).
-rA’ is the rate of disappearance of species A per unit mass (or area) of catalyst *=+ moles/
(time. mass of catalyst).
rA’ is the rate of formation (generation) of species A per unit mass (or area) of catalyst *=+
moles/ (time. mass catalyst).
-rA is an intensive property, that is, it is a function of concentration, temperature, pressure,
and the type of catalyst (if any), and is defined at any point (location) within the system. It
is independent of amount. On the other hand, an extensive property is obtained by
summing up the properties of individual subsystems within the total system; in this sense, rA is independent of the ‘extent’ of the system.
P 1-8
Rate of homogenous reaction rA is defined as the mole of A formed per unit volume of the reactor per
second. It is an Intensive property and the concentration, temperature and hence the rate varies with
spatial coordinates.
1-7
rA' on the other hand is defined as g mol of A reacted per gm. of the catalyst per second. Here mass of
catalyst is the basis as this is what is important in catalyst reactions and not the reactor volume.
Applying general mole balance we get:
dN j
dt
Fj0
Fj
r j dV
No accumulation and no spatial variation implies
0 F j0
Fj
r j dV
Also rj = ρb rj` and W = Vρb where ρb is the bulk density of the bed.
=>
0 ( Fj 0
Fj )
rj' ( b dV )
Hence the above equation becomes
Fj 0
W
Fj
r
'
j
We can also just apply the general mole balance as
dN j
dt
( Fj 0
Fj )
rj' (dW )
Assuming no accumulation and no spatial variation in rate, we get the same form
Fj 0
W
as above:
Fj
rj'
P1-9
Applying mole balance to Penicillin: Penicillin is produced in the cells stationary state (See Chapter 9), so
there is no cell growth and the nutrients are used in making product.
Let’s do part c first.
1-8
[Flowrate In (moles/time)] penicillin + [generation rate (moles/time)]penicillin – [ Flowrate Out(moles/time)]
penicillin =
[rate of accumulation (moles/time)]penicillin
dNp
dt
Fp,in + Gp – Fp,out =
Fp,in = 0 (because no penicillin inflow)
V
Gp =
rp .dV
Therefore,
V
rp .dV - Fp,out =
dNp
dt
Assuming steady state for the rate of production of penicillin in the cells stationary state,
dNp
=0
dt
And no variations
V
Fp ,in
Fp ,out
rp
Or,
V
Fp ,out
rp
Similarly, for Corn Steep Liquor with FC = 0
V
FC 0
FC
rC
FC 0
rC
Assume RNA concentration does not change in the stationary state and no RNA is generated or
destroyed.
P1-10
Given
A
2 * 1010 ft 2
TSTP
V
4 * 1013 ft 3
T = 534.7 R
R
0.7302
atm ft 3
lbmol R
491.69 R
H
2000 ft
PO = 1atm
CS
yA = 0.02
1-9
2.04 *10 10
lbmol
ft 3
C = 4*105 cars
FS = CO in Santa Ana winds
vA
3000
FA = CO emission from autos
ft 3
per car at STP
hr
P1-10 (a)
Total number of lb moles gas in the system:
N
N=
PV
0
RT
1atm (4 1013 ft 3 )
= 1.025 x 1011 lb mol
atm. ft 3
0.73
534.69 R
lbmol.R
P1-10 (b)
Molar flowrate of CO into L.A. Basin by cars.
FA
y A FT
FT
3000 ft 3
hr car
yA
v A C P0
R TSTP
1lbmol
400000 cars
359 ft 3
(See appendix B)
FA = 6.685 x 104 lb mol/hr
P1-10 (c)
Wind speed through corridor is v = 15mph
W = 20 miles
The volumetric flowrate in the corridor is
vO = v.W.H = (15x5280)(20x5280)(2000) ft3/hr = 1.673 x 1013 ft3/hr
P1-10 (d)
Molar flowrate of CO into basin from Sant Ana wind.
FS
v0 CS
= 1.673 x 1013 ft3/hr 2.04 10 10 lbmol/ft3
= 3.412 x 103lbmol/hr
P1-10 (e)
1-10
Rate of emission of CO by cars + Rate of CO by Wind - Rate of removal of CO =
FA
FS
vo C co
V
dC co
dt
(V=constant, N co
dN CO
dt
C coV )
P1-10 (f)
t = 0 , C co
C coO
t
Cco
dt
V
0
CcoO
dCco
FS voCco
FA
F FS vo C coO
V
ln A
vo
FA FS vo C co
t
P1-10 (g)
Time for concentration to reach 8 ppm.
CCO 0
2.04 10 8
lbmol
, CCO
ft 3
2.04
lbmol
10 8
4
ft 3
From (f),
F FS vO .CCO 0
V
ln A
vo
FA FS vO .CCO
t
3
lbmol
lbmol
3 lbmol
13 ft
3.4
10
1.673
10
2.04 10 8
3
4 ft
hr
hr
hr
ft 3
ln
ft 3
lbmol
lbmol
ft 3
lbmol
1.673 1013
6.7 104
3.4 103
1.673 1013
0.51 10 8
hr
hr
hr
hr
ft 3
6.7 104
t = 6.92 hr
P1-10 (h)
(1)
to
=
0
tf = 72 hrs
C co = 2.00E-10 lbmol/ft3
a = 3.50E+04 lbmol/hr
vo
= 1.67E+12 ft3 /hr
b = 3.00E+04 lbmol/hr
Fs
= 341.23 lbmol/hr
V = 4.0E+13 ft3
a
b sin
t
6
Fs
vo C co
V
dC co
dt
Now solving this equation using POLYMATH we get plot between Cco vs t
1-11
See Polymath program P1-10-h-1.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t
0
0
72
72
C
2.0E-10
2.0E-10
v0
1.67E+12
1.67E+12
a
3.5E+04
3.5E+04
3.5E+04
3.5E+04
b
3.0E+04
3.0E+04
3.0E+04
3.0E+04
F
341.23
341.23
341.23
V
4.0E+13
4.0E+13
4.0E+13
2.134E-08
1.877E-08
1.67E+12
1.67E+12
341.23
4.0E+13
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(C)/d(t) = (a+b*sin(3.14*t/6)+F-v0*C)/V
Explicit equations as entered by the user
[1] v0 = 1.67*10^12
[2] a = 35000
[3] b = 30000
[4] F = 341.23
[5] V = 4*10^13
1-12
(2) tf = 48 hrs
Fs = 0
a
t
6
b sin
vo C co
V
Now solving this equation using POLYMATH we get plot between Cco vs t
See Polymath program P1-10-h-2.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t
0
0
48
48
C
2.0E-10
2.0E-10
v0
1.67E+12
1.67E+12
a
3.5E+04
3.5E+04
3.5E+04
3.5E+04
b
3.0E+04
3.0E+04
3.0E+04
3.0E+04
V
4.0E+13
4.0E+13
4.0E+13
4.0E+13
1.904E-08
1.693E-08
1.67E+12
1.67E+12
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(C)/d(t) = (a+b*sin(3.14*t/6)-v0*C)/V
Explicit equations as entered by the user
[1] v0 = 1.67*10^12
[2] a = 35000
[3] b = 30000
[4] V = 4*10^13
1-13
dC co
dt
(3)
Changing a  Increasing ‘a’ reduces the amplitude of ripples in graph. It reduces the effect of
the sine function by adding to the baseline.
Changing b  The amplitude of ripples is directly proportional to ‘b’.
As b decreases amplitude decreases and graph becomes smooth.
Changing v0  As the value of v0 is increased the graph changes to a “shifted
sin-curve”. And as v0 is decreased graph changes to a smooth
increasing curve.
P1-11 (a)
– rA = k with k = 0.05 mol/h dm3
CSTR: The general equation is
FA 0
V
FA
rA
Here CA = 0.01CA0 , v0 = 10 dm3/min, FA = 5.0 mol/hr
Also we know that FA = CAv0 and FA0 = CA0v0, CA0 = FA0/ v0 = 0.5 mol/dm3
Substituting the values in the above equation we get,
V
C A0v0
C A v0
k
(0.5)10 0.01(0.5)10
0.05
1-14
 V = 99 dm3
FR: The general equation is
dFA
dV
rA
k , Now FA = CAv0 and FA0 = CA0v0 =>
dC A v0
dV
k
Integrating the above equation we get
v0 C A
dC A
k CA0
V
dV => V
0
v0
(C A0
k
CA )
Hence V = 99 dm3
Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of
concentration.
P1-11 (b)
- rA = kCA with k = 0.0001 s-1
CSTR:
We have already derived that
V
C A0v0
C A v0
v0 C A0 (1 0.01)
kC A
rA
k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1
V
(10dm 3 / hr )(0.5mol / dm 3 )(0.99)
=> V = 2750 dm3
1
3
(0.36hr )(0.01 * 0.5mol / dm )
PFR:
From above we already know that for a PFR
dC Av0
dV
rA
kC A
Integrating
v0 C A dC A
k CA0 C A
v0 C A0
ln
k
CA
V
dV
0
V
Again k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1
1-15
Substituing the values in above equation we get V = 127.9 dm3
P1-11 (c)
- rA = kCA2 with k = 3 dm3/mol.hr
CSTR:
CA0 v 0 CA v 0
rA
V
v 0CA0 (1 0.01)
kCA 2
Substituting all the values we get
(10dm 3 /hr)(0.5mol /dm 3 )(0.99)
(3dm 3 /hr)(0.01* 0.5mol /dm 3 ) 2
V
=> V = 66000 dm3
PFR:
dCA v 0
dV
rA
kCA 2
Integrating
C
v 0 A dCA
k C CA 2
A0
=> V
V
dV =>
0
v0 1
(
k CA
10dm 3 /hr
1
(
3
3dm /mol.hr 0.01CA0
1
) V
CA 0
1
) = 660 dm3
CA0
P1-11 (d)
CA = .001CA0
t
NA0
NA
dN
rAV
Constant Volume V=V0
t
C A 0 dC A
CA
rA
Zero order:
t
1
CA0 0.001CA0
k
.999CAo
0.05
9.99 h
First order:
1-16
t
1 CA0
ln
k
CA
1
1
ln
0.001 .001
6908 s
Second order:
t
1 1
k CA
1
CA0
1
1
3 0.0005
1
0.5
666 h
P1-12 (a)
Initial number of rabbits, x(0) = 500
Initial number of foxes, y(0) = 200
Number of days = 500
dx
dt
k1 x k2 xy …………………………….(1)
dy
dt
k3 xy k4 y ……………………………..(2)
Given,
k1
0.02day 1
k2
0.00004 /(day
k3
0.0004 /(day rabbits )
k4
0.04day 1
foxes )
See Polymath program P1-12-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t
0
0
500
500
x
500
2.9626929
519.40024
4.2199691
y
200
1.1285722
4099.517
117.62928
k1
0.02
0.02
k2
4.0E-05
4.0E-05
0.02
0.02
4.0E-05
4.0E-05
1-17
k3
4.0E-04
4.0E-04
k4
0.04
0.04
4.0E-04
0.04
4.0E-04
0.04
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(t) = (k1*x)-(k2*x*y)
[2] d(y)/d(t) = (k3*x*y)-(k4*y)
Explicit equations as entered by the user
[1] k1 = 0.02
[2] k2 = 0.00004
[3] k3 = 0.0004
[4] k4 = 0.04
When, tfinal = 800 and k3
0.00004 /(day rabbits )
1-18
Plotting rabbits Vs. foxes
P1-12 (b)
POLYMATH Results
See Polymath program P1-12-b.pol.
POLYMATH Results
NLES Solution
Variable
Value
f(x)
Ini Guess
x
2.3850387 2.53E-11
2
y
3.7970279 1.72E-12
2
NLES Report (safenewt)
Nonlinear equations
[1] f(x) = x^3*y-4*y^2+3*x-1 = 0
[2] f(y) = 6*y^2-9*x*y-5 = 0
P1-13 Enrico Fermi Problem – no definite solution
P1-14
Mole Balance:
V=
FA0
FA
rA
1-19
Rate Law :
rA
kC A2
Combine:
V=
FA0 FA
kC A2
FA0
v0C A
dm3 2molA
3
.
s
dm3
FA
v0C A
3
6molA
s
dm3 0.1molA
.
s
dm3
0.3molA
s
mol
s
V=
dm3
mol
(0.03
)(0.1 3 ) 2
mol.s
dm
(6 0.3)
19, 000dm3
1-20
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Solutions for Chapter 2 - Conversion and Reactor Sizing
P2-1.
This problem will keep students thinking about writing down what they learned every chapter.
P2-2.
This “forces” the students to determine their learning style so they can better use the
resources in the text and on the CDROM and the web.
P2-3.
ICMs have been found to motivate the students learning.
P2-4.
Introduces one of the new concepts of the 4th edition whereby the students “play” with the
example problems before going on to other solutions.
P2-5.
This is a reasonably challenging problem that reinforces Levenspiels plots.
P2-6.
Straight forward problem alternative to problems 7, 8, and 11.
P2-7.
To be used in those courses emphasizing bio reaction engineering.
P2-8.
The answer gives ridiculously large reactor volume. The point is to encourage the student to
question their numerical answers.
P2-9.
Helps the students get a feel of real reactor sizes.
P2-10.
Great motivating problem. Students remember this problem long after the course is over.
P2-11.
Alternative problem to P2-6 and P2-8.
P2-12.
Novel application of Levenspiel plots from an article by Professor Alice Gast at Massachusetts
Institute of Technology in CEE.
CDP2-A
Similar to 2-8
CDP2-B
Good problem to get groups started working together (e.g. cooperative learning).
CDP2-C
Similar to problems 2-7, 2-8, 2-11.
CDP2-D
Similar to problems 2-7, 2-8, 2-11.
Summary
P2-1
 P2-2
 P2-3
Assigned
O
A
A
Alternates
Difficulty
Time (min)
15
30
30
P2-4
P2-5
 P2-6
P2-7
P2-8
P2-9
 P2-10
P2-11
P2-12
CDP2-A
CDP2-B
CDP2-C
CDP2-D
O
O
AA
S
AA
S
AA
AA
S
O
O
O
O
7,8,11
6,8,11
6,7,8
8,B,C,D
8,B,C,D
8,B,C,D
8,B,C,D
M
FSF
FSF
SF
SF
SF
SF
M
FSF
FSF
FSF
FSF
75
75
45
45
45
15
1
60
60
5
30
30
45
Assigned
 = Always assigned, AA = Always assign one from the group of alternates,
O = Often, I = Infrequently, S = Seldom, G = Graduate level
Alternates
In problems that have a dot in conjunction with AA means that one of the problems, either the
problem with a dot or any one of the alternates are always assigned.
Time
Approximate time in minutes it would take a B/B+ student to solve the problem.
Difficulty
SF = Straight forward reinforcement of principles (plug and chug)
FSF = Fairly straight forward (requires some manipulation of equations or an intermediate
calculation).
IC = Intermediate calculation required
M = More difficult
OE = Some parts open-ended.
____________
*
Note the letter problems are found on the CD-ROM. For example A CDP1-A.
Summary Table Ch-2
Straight forward
1,2,3,4,9
Fairly straight forward
6,8,11
More difficult
5,7, 12
Open-ended
12
Comprehensive
4,5,6,7,8,11,12
Critical thinking
P2-8
P2-1 Individualized solution.
P2-2 (a) Example 2-1 through 2-3
If flow rate FAO is cut in half.
v1 = v/2 , F1= FAO/2 and CAO will remain same.
Therefore, volume of CSTR in example 2-3,
1 FA0 X
2 rA
F1 X
rA
V1
1
6.4
2
3.2
If the flow rate is doubled,
F2 = 2FAO and CAO will remain same,
Volume of CSTR in example 2-3,
V2 = F2X/-rA = 12.8 m3
P2-2 (b) Example 2-4
Levenspiel Plot
4.5
4
3.5
Fao/-ra
3
2.5
2
1.5
1
0.5
0
0
0.2
0.4
Now, FAO = 0.4/2 = 0.2 mol/s,
Table: Divide each term
X
[FAO/-rA](m3)
0.82
FA 0
rA
0.8
FA 0
in Table 2-3 by 2.
rA
0
0.445
Reactor 1
V1 = 0.82m3
V = (FAO/-rA)X
0.6
Conversion
0.1
0.545
0.2
0.665
0.4
1.025
0.6
1.77
0.7
2.53
0.8
4
Reactor 2
V2 = 3.2 m3
X1
3.2
X1
By trial and error we get:
X1 = 0.546
and
X2 = 0.8
Overall conversion XOverall = (1/2)X1 + (1/2)X2 = (0.546+0.8)/2 = 0.673
P2-2 (c) Example 2-5
(1) For first CSTR,
at X=0 ;
FA 0
rA
X2
X2
1
FA0
rA
1.28m3
at X=0.2 ;
FA0
rA
.94 m3
From previous example; V1 ( volume of first CSTR) = .188 m3
Also the next reactor is PFR, Its volume is calculated as follows
0.5
V2
0.2
FAO
dX
rA
0.247 m3
For next CSTR,
X3 = 0.65,
FAO
rA
2 m 3 , V3 =
FAO ( X 3 X 2 )
rA
(2)
Now the sequence of the reactors remain
unchanged.
But all reactors have same volume.
First CSTR remains unchanged
Vcstr = .1 = (FA0/-rA )*X1
=> X1 = .088
Now
For PFR:
X2
V
0.088
FAO
dX
rA
,
By estimation using the levenspiel plot
X2 = .183
For CSTR,
.3m3
VCSTR2 =
FAO X 3 X 2
rA
0.1m3
=> X3 = .316
(3) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the smaller
CSTR.
Conversion
X1 = 0.20
X2 = 0.60
X3 = 0.65
Original Reactor Volumes
V1 = 0.188 (CSTR)
V2 = 0.38 (PFR)
V3 = 0.10 (CSTR)
Worst Arrangement
V1 = 0.23 (PFR)
V2 = 0.53 (CSTR)
V3 = 0.10 (CSTR)
For PFR,
X1 = 0.2
X1
V1
0
FAO
dX
rA
Using trapezoidal rule,
XO = 0.1, X1 = 0.1
X1
V1
XO
rA
f XO
f X1
0.2
1.28 0.98 m3
2
0.23m3
For CSTR,
For X2 = 0.6,
FAO
1.32m3 ,
rA
V2 =
FAO
rA
V3 = 0.1 m3
FAO
X2
rA
X 1 = 1.32(0.6 – 0.2) = 0.53 m3
For 2nd CSTR,
For X3 = 0.65,
2m3 ,
P2-3 Individualized solution.
P2-4 Solution is in the decoding algorithm given with the modules.
P2-5
X
0
0.1 0.2 0.4 0.6 0.7 0.8
3
FAO/-rA (m ) 0.89 1.08 1.33 2.05 3.54 5.06 8.0
V = 1.6 m3
P2-5 (a) Two CSTRs in series
For first CSTR,
V = (FAo/-rAX1) X
=> X1 = 0.53
For second CSTR,
V = (FAo/-rAX2) (X2 – X1)
=> X2 = 0.76
P2-5 (b)
Two PFRs in series
X1
V
0
FAo
dX
rA
X2
X1
FAo
dX
rA
By extrapolating and solving, we get
X1 = 0.62
X2 = 0.84
P2-5 (c)
Two CSTRs in parallel with the feed, FAO, divided equally between two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1
V = (0.5FAO/-rAX1) X1
Solving we get, Xout = 0.68
P2-5 (d)
Two PFRs in parallel with the feed equally divided between the two reactors.
FANEW/-rAX1 = 0.5FAO/-rAX1
By extrapolating and solving as part (b), we get
Xout = 0.88
P2-5 (e)
A CSTR and a PFR are in parallel with flow equally divided
Since the flow is divided equally between the two reactors, the overall conversion is the average of the
CSTR conversion (part C) and the PFR conversion (part D)
Xo = (0.60 + 0.74) / 2 = 0.67
P2-5 (f)
A PFR followed by a CSTR,
XPFR = 0.50
(using part(b))
V = (FAo/-rA-XCSTR) (XCSTR – XPFR)
Solving we get, XCSTR = 0.70
P2-5 (g)
A CSTR followed by a PFR,
XCSTR = 0.44 (using part(a))
X PFR
V
FAO
dX
X CSTR rA
By extrapolating and solving, we get
XPFR = 0.72
P2-5 (h)
A 1 m3 PFR followed by two 0.5 m3 CSTRs,
For PFR,
XPFR = 0.50
(using part(b))
CSTR1: V = (FAo/-rA-XCSTR) (XCSTR – XPFR) = 0.5 m3
XCSTR = 0.63
CSTR2: V = (FAo/-rA-XCSTR2) (XCSTR2 – XCSTR1) = 0.5 m3
XCSTR2 = 0.72
P2-6
Exothermic reaction: A  B + C
X
0
0.20
0.40
0.45
0.50
0.60
0.80
0.90
P2-6 (a)
r(mol/dm3.min)
1
1.67
5
5
5
5
1.25
0.91
1/-r(dm3.min/mol)
1
0.6
0.2
0.2
0.2
0.2
0.8
1.1
To solve this problem, first plot 1/-rA vs. X from the chart above. Second, use mole balance as given
below.
CSTR:
Mole balance:
VCSTR
FA0 X
rA
300mol / min 0.4
=>
5mol / dm 3 . min
=>VCSTR = 24 dm3
PFR:
X
V PFR
FA 0
Mole balance:
dX
rA
0
= 300(area under the curve)
VPFR = 72 dm3
P2-6 (b)
For a feed stream that enters the reaction with a previous conversion of 0.40 and leaves at any
conversion up to 0.60, the volumes of the PFR and CSTR will be identical because of the rate is constant
over this conversion range.
.6
VPFR
FA 0
dX
r
A
.4
FA0 .6
dX
rA .4
P2-6 (c)
VCSTR = 105 dm3
Mole balance: VCSTR
FA0 X
rA
F A 0 .6
X
rA .4
105dm 3
300mol / min
X
rA
0.35dm 3 min/ mol
Use trial and error to find maximum conversion.
At X = 0.70,
1/-rA = 0.5, and X/-rA = 0.35 dm3.min/mol
Maximum conversion = 0.70
P2-6 (d)
From part (a) we know that X1 = 0.40.
Use trial and error to find X2.
FA 0 X 2 X 1
rA X
Mole balance: V
2
Rearranging, we get
X2
0.40
rA X
2
At X2 = 0.64,
V
FA 0
X2
0.008
0.40
rA X
0.008
2
Conversion = 0.64
P2-6 (e)
From part (a), we know that X1 = 0.40. Use trial and error to find X2.
X2
Mole balance: V PFR
72
FA 0
dX
rA
0.40
X2
300
At X2 = 0.908, V = 300 x (area under the curve)
=> V = 300(0.24) = 72dm3
Conversion = 0.908.
dX
rA
0.40
P2-6 (f)
See Polymath program P2-6-f.pol.
P2-7 (a)
V
FS 0 X
rS
FS0 = 1000 g/hr
At a conversion of 40%
Therefore V
1
rS
0.15
0.15 (1000)(0.40)
dm 3 hr
g
60 dm 3
P2-7 (b)
At a conversion of 80%,
1
rS
0.8
dm 3 hr
g
FS0 = 1000 g/hr
Therefore V
0.8 (1000)(0.80)
640 dm 3
P2-7 (c)
X
VPFR
FS 0
dX
rS
0
From the plot of 1/-rS Calculate the area under the curve such that the area is equal to V/FS0 = 80 / 1000
= 0.08
X = 12%
For the 80 dm3 CSTR, V
80 dm 3
FS 0 X
rS
X/-rs = 0.08. From guess and check we get X = 55%
P2-7 (d)
To achieve 80% conversion with a CSTR followed by a CSTR, the optimum arrangement is to have a CSTR
with a volume to achieve a conversion of about 45%, or the conversion that corresponds to the
minimum value of 1/-rs. Next is a PFR with the necessary volume to achieve the 80% conversion
following the CSTR. This arrangement has the smallest reactor volume to achieve 80% conversion.
For two CSTR’s in series, the optimum arrangement would still include a CSTR with the volume to
achieve a conversion of about 45%, or the conversion that corresponds to the minimum value of 1/-rs,
first. A second CSTR with a volume sufficient to reach 80% would follow the first CSTR.
P2-7 (e)
rs
kC S CC
and C C
K M CS
rs
kC S 0.1 C S 0 C S
K M CS
1
rs
K M CS
kC S 0.1 C S 0 C S
0.1 C S 0
CS
0.001
0.001
0.001
Let us first consider when CS is small.
CS0 is a constant and if we group together the constants and simplify then
1
rs
KM
k1C S2
CS
k 2 Cs
since CS < KM
1
rs
KM
which is consistent with the shape of the graph when X is large (if CS is small X is
k1C k 2 Cs
2
S
large and as CS grows X decreases).
Now consider when CS is large (X is small)
As CS gets larger CC approaches 0:
CC
0.1 C S 0
CS
0.001 and C S
CS 0
If
kC S CC
then
K M CS
rs
1
rs
K M CS
kC S CC
As CS grows larger, CS >> KM
And
1
rs
CS
kC S CC
1
kCC
And since CC is becoming very small and approaching 0 at X = 0, 1/-rs should be increasing with CS (or
decreasing X). This is what is observed at small values of X. At intermediate levels of CS and X, these
driving forces are competing and why the curve of 1/-rS has a minimum.
P2-8
Irreversible gas phase reaction
2A + B  2C
See Polymath program P2-8.pol.
P2-8 (a)
PFR volume necessary to achieve 50% conversion
Mole Balance
X2
V
FA0
dX
X 1 ( rA )
Volume = Geometric area under the curve of
(FA0/-rA) vs X)
V
1
400000 0.5
2
100000 0.5
V = 150000 m3
.
P2-8 (b)
CSTR Volume to achieve 50% conversion
Mole Balance
V
V
FA 0 X
( rA )
0.5 100000
V = 50000m3
P2-8 (c)
Volume of second CSTR added in series to achieve 80%
conversion
V2
FA0 ( X 2 X 1 )
( rA )
V2
500000 (0.8 0.5)
V2 = 150000m3
P2-8 (d)
Volume of PFR added in series to first CSTR to achieve
80% conversion
VPFR
(
1
400000 0.3) (100000 0.3)
2
VPFR = 90000m3
P2-8 (e)
For CSTR,
V = 60000 m3 (CSTR)
Mole Balance
V
FA0 X
( rA )
60000
( 800000 X
500000 ) X
X = 0.463
For PFR,
V = 60000 m3 (PFR)
Mole balance
X
V
FA0
dX
( rA )
0
X
60000
( 800000 X 100000)dX
0
X = 0.134
P2-8(f)
Real rates would not give that shape. The reactor volumes are absurdly large.
P2-9
Problem 2-9 involves estimating the volume of three reactors from a picture. The door on the side of the
building was used as a reference. It was assumed to be 8 ft high.
The following estimates were made:
CSTR
h = 56ft
d = 9 ft
V = πr2h = π(4.5 ft)2(56 ft) = 3562 ft3 = 100,865 L
PFR
Length of one segment = 23 ft
Length of entire reactor = (23 ft)(12)(11) = 3036 ft
D = 1 ft
V = πr2h = π(0.5 ft)2(3036 ft) = 2384 ft3 = 67,507 L
Answers will vary slightly for each individual.
P2-10 No solution necessary.
P2-11 (a)
The smallest amount of catalyst necessary to achieve 80 % conversion in a CSTR and PBR connected in
series and containing equal amounts of catalyst can be calculated from the figure below.
The lightly shaded area on the left denotes the CSTR while the darker shaded area denotes the PBR. This
figure shows that the smallest amount of catalyst is used when the CSTR is upstream of the PBR.
See Polymath program P2-11.pol.
P2-11 (b)
Calculate the necessary amount of catalyst to reach 80 % conversion using a single CSTR by determining
the area of the shaded region in the figure below.
The area of the rectangle is approximately 23.2 kg of catalyst.
P2-11 (c)
The CSTR catalyst weight necessary to achieve 40 % conversion can be obtained by calculating the area
of the shaded rectangle shown in the figure below.
The area of the rectangle is approximately 7.6 kg of catalyst.
P2-11 (d)
The catalyst weight necessary to achieve 80 % conversion in a PBR is found by calculating the area of the
shaded region in the figure below.
The necessary catalyst weight is approximately 22 kg.
P2-11 (e)
The amount of catalyst necessary to achieve 40 % conversion in a single PBR can be found from
calculating the area of the shaded region in the graph below.
The necessary catalyst weight is approximately 13 kg.
P2-11 (f)
P2-11 (g)
For different (-rA) vs. (X) curves, reactors should be arranged so that the smallest amount of catalyst is
needed to give the maximum conversion. One useful heuristic is that for curves with a negative slope, it
is generally better to use a CSTR. Similarly, when a curve has a positive slope, it is generally better to use
a PBR.
P2-12 (a) Individualized Solution
P2-12 (b) 1) In order to find the age of the baby hippo, we need to know the volume of the stomach. The
metabolic rate, -rA, is the same for mother and baby, so if the baby hippo eats one half of what the
mother eats then Fao (baby) = ½ Fao (mother). The Levenspiel Plot is shown:
Autocatalytic Reaction
5
4.5
4
mao/-raM1
3.5
3
Mother
2.5
Baby
2
1.5
1
0.5
0
0
0.2
0.4
Conversion
0.6
0.8
Vbaby
FaoX
rA
1.36
*0.34 0.23m3
2
Since the volume of the stomach is proportional to the age of the baby hippo, and the volume of the
baby’s stomach is half of an adult, then the baby hippo is half the age of a full grown hippo.
Age
4.5 years
2
2.25 years
2) If Vmax and mao are both one half of the mother’s then
1
mA0mother
2
rAM 2
mAo
rAM 2
and since
rAM 2
rAM 2baby
mAo
rAM 2
vmax C A
K M C A then
1
vmax C A
2
KM CA
baby
1
rAM 2mother
2
1
mAo
2
1
rAM 2
2
mAo
rAM 2
mother
mother
mAo
will be identical for both the baby and mother.
rAM 2
Assuming that like the stomach the intestine volume is proportional to age then the volume of the
intestine would be 0.75 m3 and the final conversion would be 0.40
P2-12 (c)
Vstomach = 0.2 m3
From the web module we see that if a polynomial is fit to the autocatalytic reaction we get:
mA0
= 127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499
rAM 1
And since Vstomach =
mA0
X,
rAM 1
solve V= 127X5 - 172.36X4 + 100.18X3 - 28.354X2 + 4.499X = 0.2 m3
Xstomach = .067.
For the intestine, the Levenspiel plot for the intestine is shown below. The outlet conversion is 0.178.
Since the hippo needs 30% conversion to survive but only achieves 17.8%, the hippo cannot survive.
P2-12 (d)
PFR CSTR
PFR:
Outlet conversion of PFR = 0.111
CSTR:
We must solve
V = 0.46 = (X-0.111)(127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499)
X=0.42
Since the hippo gets a conversion over 30% it will survive.
P2-13
For a CSTR we have :
V=X
FA0
rA | X X f
So the area under the
FA 0
versus X curve for a CSTR is a rectangle but the height of rectangle
rA
corresponds to the value of
FA 0
at X= Xf
rA
But in this case the value of
FA 0
is taken at X= Xi and the area is calculated.
rA
Hence the proposed solution is wrong.
CDP2-A (a)
Over what range of conversions are the plug-flow reactor and CSTR volumes identical?
We first plot the inverse of the reaction rate versus conversion.
Mole balance equations for a CSTR and a PFR:
CSTR: V
FA 0 X
rA
PFR: V
X dX
0
rA
Until the conversion (X) reaches 0.5, the reaction rate is independent of conversion and the
reactor volumes will be identical.
i.e. VPFR
0.5 dX
0
rA
FA0 0.5
dX
rA 0
FA0 X
rA
VCSTR
CDP2-A (b)
What conversion will be achieved in a CSTR that has a volume of 90 L?
For now, we will assume that conversion (X) will be less that 0.5.
CSTR mole balance:
V
FA0 X
rA
v0 C A 0 X
X
rA
V
v0 C A 0
rA
3
5
m
s
0.09m 3
3
mol
8 m .s
200 3 3 10
mol
m
3 10 13
CDP2-A (c)
This problem will be divided into two parts, as seen below:

The PFR volume required in reaching X=0.5 (reaction rate is independent of conversion).
V1

FA0 X
rA
v0 C A 0 X
rA
1.5 1011 m 3
The PFR volume required to go from X=0.5 to X=0.7 (reaction rate depends on conversion).
Finally, we add V2 to V1 and get:
Vtot = V1 + V2 = 2.3 x1011 m3
CDP2-A (d)
What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a CSTR to
raise the conversion to 90 %
We notice that the new inverse of the reaction rate (1/-rA) is 7*108. We insert this new value into our
CSTR mole balance equation:
VCSTR
FA0 X
rA
v0 C A 0 X
rA
1.4 1011 m 3
CDP2-A (e)
If the reaction is carried out in a constant-pressure batch reactor in which pure A is fed to the reactor,
what length of time is necessary to achieve 40% conversion?
Since there is no flow into or out of the system, mole balance can be written as:
Mole Balance: rAV
dN A
dt
Stoichiometry: N A
N A0 (1 X )
Combine: rAV
dX
dt
N A0
From the stoichiometry of the reaction we know that V = Vo(1+eX) and e is 1. We insert this into our
mole balance equation and solve for time (t):
rA
V0
(1 X )
N A0
dt
C A0
t
0
X
0
dX
dt
dX
rA (1 X )
After integration, we have:
t
1
C A0 ln(1 X )
rA
Inserting the values for our variables:
t = 2.02 x 1010 s
That is 640 years.
CDP2-A (f)
Plot the rate of reaction and conversion as a function of PFR volume.
The following graph plots the reaction rate (-rA) versus the PFR volume:
Below is a plot of conversion versus the PFR volume. Notice how the relation is linear until the
conversion exceeds 50%.
The volume required for 99% conversion exceeds 4*1011 m3.
CDP2-A (g)
Critique the answers to this problem.
The rate of reaction for this problem is extremely small, and the flow rate is quite large. To obtain the
desired conversion, it would require a reactor of geological proportions (a CSTR or PFR approximately
the size of the Los Angeles Basin), or as we saw in the case of the batch reactor, a very long time.
CDP2-B Individualized solution
CDP2-C (a)
For an intermediate conversion of 0.3, Figure above shows that a PFR yields the smallest volume, since
for the PFR we use the area under the curve. A minimum volume is also achieved by following the PFR
with a CSTR. In this case the area considered would be the rectangle bounded by X =0.3 and X = 0.7 with
a height equal to the CA0/-rA value at X = 0.7, which is less than the area under the curve.
CDP2-C (b)
CDP2-C (c)
CDP2-C (d)
For the PFR,
CDP2-C (e)
CDP2-D
CDP2-D (a)
CDP2-D (b)
CDP2-D (c)
CDP2-D (d)
CDP2-D (e)
CDP2-D (f)
CDP2-D (g)
CDP2-D (h)
CDP2-E
CDP2-F (a)
Find the conversion for the CSTR and PFR connected in series.
X
-rA
1/(-rA)
0
0.2
5
0.1
0.0167
59.9
0.4
0.00488
204.9
0.7
0.00286
349.65
0.9
0.00204
490.19
CDP2-F (b)
CDP2-F (c)
CDP2-F (d)
CDP2-F (e)
CDP2-F (f)
CDP2-F (g) Individualized solution
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3-1
Solutions for Chapter 3 – Rate Laws
P3-1 Individualized solution.
P3-2 (a) Example 3-1
For, E = 60kJ/mol
k 1.32 1016 exp
T (K)
For, E = 240kJ/mol
60000 J
RT
k (1/sec) 1/T
k 1.32 1016 exp
ln(k)
T (K)
240000 J
RT
k (1/sec) 1/T
ln(k)
310 1023100 0.003226 13.83918
310 4.78E-25 0.003226 -56.0003
315 1480488 0.003175
315
14.2087
2.1E-24 0.003175 -54.5222
320 2117757 0.003125 14.56667
320 8.77E-24 0.003125 -53.0903
325 2996152 0.003077 14.91363
325 3.51E-23 0.003077 -51.7025
330 4194548
330 1.35E-22
0.00303 15.25008
335 5813595 0.002985 15.57648
0.00303 -50.3567
335 4.98E-22 0.002985 -49.0511
3-2
P3-2 (b) No solution will be given
P3-2 (c)
A+
1
1
B C
2
2
Rate law: rA
rA
1
kC
rB
1/2
kB
2
k A CA CB and kA
rC
1/2
1 dm 3
12.5
s mol
1 dm 3
25
s mol
2
25CA CB
2
k B CA 2CB
1/2
kC CA 2CB
1/2
2
P3-3(a)
Refer to Fig 3-3
The fraction of molecular collisions having energies less than or equal to 50 Kcal is given by the area
under the curve, f(E,T)dE from EA = 0 to 50 Kcal.
P3-3(b)
The fraction of molecular collisions having energies between 10 and 20 Kcal is given by the area under
the curve f(E,T) from EA = 10 to 20 Kcal.
3-3
P3-3(c)
The fraction of molecular collisions having energies greater than the activation energy EA= 30 Kcal is
given by the area under the curve f(E,T) from EA =30 to 50 Kcal.
P3-4 (a)
Note: This problem can have many solutions as data fitting can be done in many ways.
Using Arrhenius Equation
For Fire flies:
T(in
K)
294
1/T
0.003401
Flashes/
min
9
ln(flashe
s/min)
2.197
298
0.003356
12.16
2.498
303
0.003300
16.2
2.785
Plotting ln (flashes/min) vs. 1/T,
We get a straight line.
See Polymath program P3-4-fireflies.pol.
For Crickets:
T(in K)
287.2
1/T
3
x10
3.482
chirps/
min
80
ln(chirps/
min)
4.382
293.3
300
3.409
3.333
126
200
4.836
5.298
Plotting ln (chirps/min) Vs 1/T,
We get a straight line.
Both, Fireflies and Crickets data
follow the Arrhenius Model.
ln y = A + B/T , and have the similar activation energy.
See Polymath program P3-4-crickets.pol.
P3-4 (b)
For Honeybee:
T(in K)
V(cm/s)
ln(V)
298
1/T
3
x10
3.356
0.7
-0.357
303
3.300
1.8
0.588
308
3.247
3
1.098
Plotting ln (V) vs. 1/T, almost straight line.
3-4
ln (V) = 44.6 – 1.33E4/T
At T = 40oC (313K)
At T = -5oC (268K)
V = 6.4cm/s
V = 0.005cm/s(But bee would not be alive at this temperature)
See Polymath program P3-4-bees.pol.
P3-4 (c)
For Ants:
T(in K)
1/T x10
283
3
V(cm/s)
ln(V)
3.53
0.5
-0.69
293
3.41
2
0.69
303
3.30
3.4
1.22
311
3.21
6.5
1.87
Plotting ln (V) vs. 1/T,
We get almost a straight line.
See Polymath program P3-4-ants.pol.
So activity of bees, ants, crickets and fireflies follow
Arrhenius model. So activity increases with an increase in temperature. Activation energies for fireflies
and crickets are almost the same.
Insect
Cricket
Firefly
Ant
Honeybee
Activation Energy
52150
54800
95570
141800
P3-4 (d)
There is a limit to temperature for which data for any one of he insect can be extrapolate. Data which
would be helpful is the maximum and the minimum temperature that these insects can endure before
death. Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperature it
could be useless.
P3-5
There are two competing effects that bring about the maximum in the corrosion rate: Temperature and
HCN-H2SO4 concentration. The corrosion rate increases with increasing temperature and increasing
concentration of HCN-H2SO4 complex. The temperature increases as we go from top to bottom of the
column and consequently the rate of corrosion should increase. However, the HCN concentrations (and
the HCN-H2SO4 complex) decrease as we go from top to bottom of the column. There is virtually no
HCN in the bottom of the column. These two opposing factors results in the maximum of the corrosion
rate somewhere around the middle of the column.
3-5
P3-6 Antidote did not dissolve from glass at low temperatures.
P3-7 (a)
If a reaction rate doubles for an increase in 10°C, at T = T1 let k = k1 and at T = T2 = T1+10, let k = k2 = 2k1.
Then with k = Ae-E/RT in general, k1
k2
k1
e
E 1 1
R T2 T1
E
R
or
Ae E / RT1 and k2
ln
k2
k1
1
T2
1
T1
Ae E / RT2 , or
k2
k1
T1 T2
T1T2
ln
Therefore:
ln
E
R
k2
k1
T1 T1 10
R
T2 T1
T1 T1 10
ln 2 T1 T1 10
10
10 E
R ln 2
which can be approximated by T
10 E 0.5
R ln 2
P3-7 (b)
Equation 3-18 is k
Ae
E
RT
From the data, at T1 = 0°C, k1
k
Dividing gives 2
k1
E
R ln
k2
k1
1
T2
1
T1
1.99
E
e
E 1 1
R T2 T1
Ae E / RT1 , and at T2 = 100°C, k2
, or
RT1T2
k
ln 2
T1 T2
k1
cal
273 K 373 K
mol K
.050
ln
100 K
.001
7960
3-6
cal
mol
Ae E / RT2
A k1e
E
RT1
7960
10 3 min 1 exp
1.99
cal
mol
cal
mol K
2100 min 1
273 K
P3-7 (c) Individualized solution
P3-8
From the
given data
k= rA(dm3/mol.s) T(K)
rA/(4*1.5)
0.002
300 0.00033333
0.046
320 0.00766667
0.72
340
0.12
8.33
360 1.38833333
1/T
ln (k)
0.003333 -8.00637
0.003125 -4.87087
0.002941 -2.12026
0.002778 0.328104
Plotting ln(k) vs (1/T), we have a straight line:
Since, ln k = ln A - (
, therefore ln A = 41.99 and E/R = 14999
3-7
(a) Activation energy (E),
E = 14999*8.314 =124700 J/mol = 124.7 kJ/mol
(b) Frequency Factor (A), ln A = 41.99
A = 1.72X1018
(c) k = 1.72X1018 exp(-
(1)
Given T0 = 300K
Therefore, putting T = 300K in (1), we get k(T0) = 3.33X10-4
Hence,
k(T) = k(T0)exp[
P3-9 (a)
From the web module we know that
dX
dt
k (1 x) and that k is a function of temperature, but not a
linear function. Therefore doubling the temperature will not necessarily double the reaction rate, and
therefore halve the cooking time.
P3-9 (b)
When you boil the potato in water, the heat transfer coefficient is much larger, but the temperature can
only be 100°C.
When you bake the potato, the heat transfer coefficient is smaller, but the temperature can be more
than double that of boiling water.
P3-10
1) C2H6 → C2H4 + H2
Rate law: -rA = kC C2 H 6
2) C2H4 + 1/2O2 → C2H4O
Rate law: -rA = kCC2 H 4 C 02
1/ 2
3) (CH3)3COOC(CH3)3 → C2H6 + 2CH3COCH3
A
→ B +
2C
Rate law: -rA = k[CA – CBCC2/KC]
4) n-C4H10 ↔ I- C4H10
Rate law: -rA = k[ C nC4 H10 – C iC4 H10 /Kc]
5) CH3COOC2H5 + C4H9OH ↔ CH3COOC4H9 + C2H5OH
A
+
B
↔
C
+
D
Rate law: -rA = k[CACB – CCCD/KC]
3-8
P3-11 (a)
2A + B → C
(1)
(2)
(3)
(4)
-rA = kCACB2
-rA = kCB
-rA = k
-rA = kCACB-1
P3-11 (b)
1/ 2
k1C H 2 C Br
2
Rate law: -rHBr =
C HBr
k2
C Br2
(1) H2 + Br2 → 2HBr
(2) H2 + I2 → 2HI
Rate law: -rA = k1CH 2 CI2
P3-12
a)
we need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium
condition. If we assume the rate law for the reverse reaction (B->A) is
=
then:
From Appendix C we know that for a reaction at equilibrium: KC
At equilibrium, rnet
0, so:
Solving for KC gives:
3-9
b)
If we assume the rate law for the reverse reaction
is
then:
From Appendix C we know that for a reaction at equilibrium: KC
At equilibrium, rnet
0, so:
Solving for KC gives:
c)
If we assume the rate law for the reverse reaction
is
then:
3-10
From Appendix C we know that for a reaction at equilibrium: KC
At equilibrium, rnet
0, so:
Solving for KC gives:
P3-13
The rate at which the beetle can push a ball of dung is directly proportional to its rate constant,
therefore
-rA =c*k, where c is a constant related to the mass of the beetle and the dung and k is the rate constant
k = A exp(
From the data given
-rA
6.5
13
18
T(K)
300
310
313
1/T
0.003333
0.003226
0.003195
ln k
1.871802
2.564949
2.890372
3-11
Refer to P3-8 (similar procedure)
Therefore, A = 1.299X1011
E = 59195.68 J/mol
k = 1.299X1011 exp(-7120/T)
Now at T = 41.5 C = 314.5 K
k = 19.12 cm/s
Therefore, beetle can push dung at 19.12 cm/s at 41.5 C
P3-14
Since the reaction is homogeneous, which means it involves only one phase.
Therefore,
So, option (4) is correct.
P3-15
Assuming the reactions to be elementary:
2 Anthracene -> Dimer

rA
2
k (C Anthracene
CDimer
)
KC
where, Kc = k+/k-
Similarly for the second reaction:
3-12
Norbornadiene  Quadricyclane

rNorbornadiene
k (CNorbornadiene
CQuadricyclane
KC
)
where, Kc = k+/k-
P3-16
The mistakes are as follows:
1. For any reaction, the rate law cannot be written on the basis of the stoichiometric equation. It
can only be found out using experimental data.
2. In the evaluation of the specific reaction rate constant at 100o C the gas constant that should
have been used was 8.314 J K-1 mol -1.
3. In the same equation, the temperatures used should have been in K rather than oC.
4. The units for calculated k(at 100oC) are incorrect.
5. The dimension of the reaction rate obtained is incorrect. This is due to the fact that the rate law
that has been taken is wrong.
CDP3-A
Polanyi equation: E = C – α(-ΔHR)
We have to calculate E for the reaction
CH3• + RBr  CH3Br + R•
Given: ΔHR = - 6 kcal/mol
From the given data table, we get
6.8 = C – α (17.5)
And,
6.0 = C – α (20)
=> C = 12.4 KJ/mol and α = 0.32
Using these values, and ΔHR = - 6 kcal/mol, we get E = 10.48 KJ/mol
3-13
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4-1
Solutions for Chapter 4 –Stoichiometry
P4-1
Individualized solution
P4-2 (a) Example 4-1 Yes, water is already considered inert.
P4-2 (b) Example 4-2
For 20% Conversion
CD = same as example & CB = CAo (ѲB – XA/3) = 10 (
-
) = 2.33 mol/dm3
–
) = 0 mol/dm3
For 90 % Conversion
CD = same as example & CB = CAo (ѲB – XA/3) = 10 (
The final concentration of glyceryl sterate is 0 instead of a negative concentration. Therefore 90 % of
caustic soda is possible.
P4-2 (c) Example 4-3
For the concentration of N2 to be constant, the volume of reactor must be constant. V = VO.
Plot:
1
rA
0.5(1 0.14 X )2
(1 X )(0.54 0.5 X )
1/(-ra) vs X
180
160
140
1/(-ra)
120
100
80
60
40
20
0
0
0.2
0.4
0.6
0.8
1
1.2
X
The rate of reaction decreases drastically with increase in conversion at higher conversions.
4-2
P4-2 (d) Example 4-4
POLYMATH Report
No Title
05-May-2009
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
epsilon
-0.14
-0.14
-0.14
-0.14
2
FA0
1000.
1000.
1000.
1000.
3
k
9.7
9.7
9.7
9.7
4
K
930.
930.
930.
930.
5
KO2
38.5
38.5
38.5
38.5
6
KSO2
42.5
42.5
42.5
42.5
7
PO2
2.214
0.19639
2.214
0.1963901
8
PSO2
4.1
0.0115348
4.1
0.011535
9
PSO20
4.1
4.1
4.1
4.1
10 PSO3
0
0
4.754029
4.754029
11 rA
-0.0017387
-0.0063092
5.915E-08
-1.659E-09
12 w
0
0
5.0E+05
5.0E+05
13 x
0
0
0.9975796
0.9975795
Differential equations
1 d(x)/d(w) = -rA/FA0
Explicit equations
1
FA0 = 1000
mol.hr-1
2
PSO20 = 4.1
3
K = 930
atm-1/2
4
KSO2 = 42.5
5
KO2 = 38.5
atm-1
6
k = 9.7
7
epsilon = -0.14
8
PSO3 = PSO20*x/(1+epsilon*x)
9
PO2 = PSO20*(1.08-x)/2/(1+epsilon*x)
10 PSO2 = PSO20*(1-x)/(1+epsilon*x)
11 rA = -k*(PSO2*PO2^0.5 - PSO3/K)/(1+(PO2*KO2)^0.5+PSO2*KSO2)^2
mol/hr.g.cat
General
4-3
Total number of equations
12
Number of differential equations 1
Number of explicit equations
11
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
4-4
Therefore, catalyst weight required for 30% conversion is 1.611x105 g = 161.1 kg
4-5
Therefore, catalyst weight required for 60% conversion is 2.945x105 g = 294.5 kg
Therefore, catalyst weight required for 99% conversion is 4.041x105 g = 404.1 kg
P4-2 (e) Example 4-5
For a given conversion, concentration of B is lower in flow reactor than a constant volume batch reactor.
Therefore the reverse reaction decreases.
CT0 = constant and inerts are varied.
N 2O4
2NO2
A ↔ 2B
Equilibrium rate constant is given by: KC
Stoichiometry:
y A0
y A0 (2 1)
CB,e 2
CA,e
y A0
Constant volume Batch:
CA
N A0 (1 X)
V0
CA0 (1 X)
and
CB
2N A 0 X
V0
2CA 0 X
Plug flow reactor:
CA
FA0 (1 X)
v 0 (1 X)
CA0 (1 X)
and CB
(1 X)
2FA0 X
v 0 (1 X)
4-6
2CA0 X
(1 X)
CAO
y AO PO
RTO
y AO 0.07176 mol /dm 3
Combining: For constant volume batch:
KC
CB,e 2
CA,e
2
4CAo
X2
CAO (1 X)
Xe
KC (1 X e )
4CA0
Xe
K C (1 X e )(1
4C A0
For flow reactor:
KC
C B ,e
2
C A,e
2
4C Ao
X2
C AO (1 X )(1
X)
See Polymath program P4-2-e.pol.
POLYMATH Results
NLES Report (safenewt)
Nonlinear equations
[1] f(Xeb) = Xeb - (kc*(1-Xeb)/(4*Cao))^0.5 = 0
[2] f(Xef) = Xef - (kc*(1-Xef)*(1+eps*Xef)/(4*Cao))^0.5 = 0
Explicit equations
[1] yao = 1
[2] kc = 0.1
[3] Cao = 0.07174*yao
[4] eps = yao
4-7
Xe)
Yinert
Yao
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.95
0.956
Xeb
Xef
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
0.044
0.44
0.458
0.4777
0.5
0.525
0.556
0.5944
0.6435
0.71
0.8112
0.887
0.893
0.508
0.5217
0.537
0.5547
0.576
0.601
0.633
0.6743
0.732
0.8212
0.89
0.896
P4-3 Solution is in the decoding algorithm available separately from the author.
P4-4
A
y A0
(a) C B0
C A0
(b) C A
C A0
CC
C A0
0.1
X
X
2
1
1
X
2
1 X
1
(c) C B
(d) C B
C A0
0.1
0.1
1 0.25
1
1
X
2
0.25
2
0.1
1
1
X
2
X
1
2
mol
dm 3
1 X
1
1
1
B
C
2
2
1 1 1
1
2 2 2
0.875
0.1 0.125
0.875
C A0 1
1
0.1 0.75
1
X
2
1
X
2
mol
dm 3
4-8
C A0
0.086
0.0143
0.1
mol
dm3
mol
dm 3
mol
dm 3
rA
4
(e)
rC
, rA
2
4 mol dm 3 min
2rC
P4-5 (a)
Liquid phase reaction,
O
CH2--OH
CH2 - CH2 + H2O → CH2--OH
A
+ B → C
CAO = 16.13mol/dm3
CBO = 55.5 mol/dm3
Stoichiometric Table:
Species
Ethylene
oxide
Water
Glycol
Symbol
A
Initial
CAO=16.13 mol/dm3
Change
- CAOX
B
CBO= 55.5 mol/dm3,
B =3.47
Remaining
CA= CAO(1-X)
= (1-X) lbmol/ft3
-CAOX
CB = CAO(
0
CAOX
=(3.47-X) lbmol/ft3
CC = CAOX
= X lbmol/ft3
C
Rate law:
-rA = kCACB
Therefore,
2
-rA = k C AO
(1-X) (
At 300K
E = 12500 cal/mol,
k = 0.1dm3/mol.s
CSTR
=
B
-X) = k (16.13)2(1-X) (3.47-X)
X = 0.9,
16.13 0.9
C AO X
=
2
rA
0.1 16.13 1 0.9 3.47 0.9
2.186sec
and, V = τ X FA0 = 2.186 X 200 liters = 437.2 liters
At 350K,
k2 = k exp((E/R)(1/T-1/T2))= 0.1exp((12500/1.987)(1/300-1/350))
= 19.99 dm3/mol.s
Therefore,
CSTR
=
16.13 0.9
C AO X
=
2
rA
19.99 16.13 1 0.9 3.47 0.9
and, V = τ X FA0 = 0.109 X 200 liters = 21.8 liters
4-9
0.109sec ,
B -X)
P4-5 (b)
Isothermal, isobaric gas-phase pyrolysis,
C2H6 C2H4 + H2
A  B + C
Stoichiometric table:
Species
C2H6
C2H4
H2
symbol
A
B
C
Entering
FAO
0
0
FTO=FAO
Change
-FAOX
+FAOX
+FAOX
Leaving
FA=FAO(1-X)
FB=FAOX
FC=FAOX
FT=FAO(1+X)
= yao = 1(1+1-1) = 1
v = vo(1+ X)
=> v = vo(1+X)
P
RT
1 6atm
CAO = yAO CTO = yAO
=
= 0.067 kmol/m3 = 0.067 mol/dm3
3
0.082
m atm
K .kmol
1100 K
CA =
FA
v
FAO (1 X )
vO (1 X )
C AO
1 X
1 X
CB =
FB
v
FAO ( X )
vO (1 X )
C AO
X
mol/dm3
1 X
CC =
FC
v
FAO ( X )
vO (1 X )
C AO
X
mol/dm3
1 X
mol/dm3
Rate law:
-rA = kCA= kCAO
1 X
1 X
=0.067 k
1 X
1 X
If the reaction is carried out in a constant volume batch reactor, =>( = 0)
CA = CAO(1-X) mol/dm3 CB = CAO X mol/dm3
CC = CAO X mol/dm3
P4-5 (c)
Isothermal, isobaric, catalytic gas phase oxidation,
1
O2  C2H4O
2
1
+
B 
C
2
C2H4 +
A
4-10
Stoichiometric table:
B
=
FBO
FAO
y AO
C AO
CA
1
FAO
2
FAO
CB
FB
v
CC
FC
v
Symbol
A
B
Entering
FAO
FBO
Change
-FAOX
C2H4O
C
0
+FAOX
1
2
2
1
1
1
3
2
y AO CTO
FA
v
Species
C2H4
O2
y AO
P
RT
FAO 1 X
vO 1 X
FAO
B
vO 1
X
2
X
FAO X
vO 1 X
y AO
FAO
FTO
-
Leaving
FA=FAO(1-X)
1
FB=FAO( B -X/2)
FAOX
2
FAO
FAO FBO
FC=FAOX
2
3
0.33
2
3
6atm
0.082
atm.dm
mol.K
CAO 1 X
1 0.33 X
0.092
3
533K
mol
dm3
0.092 1 X
1 0.33 X
0.046 1 X
1 0.33 X
0.092 X
1 0.33 X
If the reaction follow elementary rate law
Rate law: rA
kC AC
0.5
B
rA
0.092 1 X
k
1 0.33 X
0.046 1 X
1 0.33 X
Change
-FA0X
-2FA0X
FA0X
Leaving
FA=FA0(1-X)
FB=FA0(θB-2X)
Fc=FA0X
P4-5 (d)
Isothermal, isobaric, catalytic gas-phase reaction in a PBR
C6H6 + 2H2 → C6H10
A + 2B → C
Given:
0
50 dm 3 / min
Stoichiometric Table:
Species
C6H6
H2
C6H10
Symbol
A
B
C
Entering
FA0
FB0=2FA0
0
4-11
0.5
B
C A0
CA
CB
CC
FB 0
FA 0
2 FA 0
FA 0
2
P 1
RT 3
CT 0 y A0
FA 0
FT 0
y A0
FA 0
FA0 FB 0
1
3
6atm
1
0.0821
* (443K ) 3
0.055
dm3 atm
mol K
FA
FA0 (1 X )
X)
0 (1
FB
FA0 ( B 2 X )
X)
0 (1
FC
FA0 X
X)
0 (1
C A0
1
(1 2 1)
3
2
3
mol
dm3
(1 X )
(1 23 X )
C A0
C A0
y A0
(2 2 X )
(1 23 X )
2 * C A0
(1 X )
(1 23 X )
X
2
3
(1
X)
Rate Law:
NOTE: For gas-phase reactions, rate laws are sometimes written in terms of partial pressures instead of
concentrations. The units of the rate constant, k, will differ depending on whether partial pressure or
concentration units are used. See below for an example.
rA ' kPA PB
mol
kgcat min
2
mol
* atm * atm 2
3
kgcat min atm
Notice that if you use concentrations in this rate law, the units will not work out.
PA
y A0 P
( y A0 C ) * RT
rA
kPA PB
2
2
C A RT
kC AC B ( RT ) 3
4kC A0
3
(1 X ) 3
( RT ) 3
(1 23 X ) 3
Design Equation for a fluidized CSTR:
W
W
W
FA0 X
rA '
FA0 X (1
2
3
X )3
3
4kC A0 (1 X ) 3 ( RT ) 3
0
X (1
2
3
X )3
2
4kC A0 (1 X ) 3 ( RT ) 3
Evaluating the constants:
k
53
mol
at 300 K
kgcat min atm3
At 170°C (443K),
4-12
k 443
k 300
EA 1
R T300
1
T443
J
1
mol
53
J
300 K
8.314
mol K
80000
1
443K
1663000
mol
kgcat min atm
Plugging in all the constants into the design equation:
X = 0.8
3
W
3
2
50 dm
min 0.8 (1
3 0.8)
3
mol
mol 2
atm
4 1663000 kgat min
(0.055 dm
0.8) 3 (0.0821 dm
443K ) 3
3 ) (1
mol K
atm3
5.25 10 7 kgcat
At 270°C (543K),
k 543
k 300
EA 1
R T300
1
T543
J
1
mol
53
J
300 K
8.314
mol K
80000
1
543K
9079000
mol
Plu
kgcat min atm
gging in all the constants into the design equation:
X = 0.8
3
W
3
2
50 dm
min 0.8 (1
3 0.8)
3
3
mol
mol 2
atm
4 9079000 kgat min
(0.055 dm
0.8) 3 (0.0821 dm
3 ) (1
mol K 543K )
atm3
5.22 10 8 kgcat
P4-6 (a)
Let A = ONCB
B = NH3
A + 2B
-rA
C = Nibroanaline
D = Ammonium Chloride
C+D
kC ACB
P4-6 (b)
Species
A
B
C
D
Entering
FA0
FB0 = ΘBFA0
=6.6/1.8 FA0
0
0
Change
- FA0X
-2 FA0X
FA0X
FA0X
P4-6 (c)
For batch system,
CA=NA/V
-rA = kNANB/V2
P4-6 (d)
4-13
Leaving
FA0(1-X)
FB=
FA0(ΘB – 2X)
FC= FA0X
FD=FA0X
-rA
kC ACB
FA
NA
V
NA
V0
N A0
1 X
V0
FB
NB
V
NB
V0
N A0
V0
rA
kC A2 0 1 X
B
CB 0
C A0
6.6
1.8
C A0
1.8
kmol
m3
rA
k 1.8
2
B
B
FA
v
FA
v0
C A0 1 X
2 X , CB
FB
v0
C A0
C A0 1 X , C A
2X
C A0
B
B
2X
3.67
1 X
3.67 2 X
P4-6 (e)
1) At X = 0 and T = 188°C = 461 K
2
A0
rA0
kC
rA0
0.0202
m3
kmol
0.0017
1.8 3
kmol min
m
B
2
3.67
kmol
m3 min
2) At X = 0 and T = 25C = 298K
k
k O exp
E 1
R TO
1
T
cal
m
1
mol
k 0.0017
exp
cal
kmol. min
461
1.987
mol.K
m3
2.03 10 6
kmol. min
11273
3
1
298
-rAO = kCAOCBO = 2.41 X 10-5 kmol/m3min
3)
k
k0 exp
E 1
R T0
1
T
4-14
0.0202
kmol
m 3 min
2X
k
cal
m
1
mol
0.0017
exp
cal
kmol min
461 K
1.987
mol K
k
0.0152
11273
3
1
561 K
m3
kmol min
rA0
kC A0CB 0
rA
0.0152
m3
kmol
1.8 3
kmol min
m
rA
0.1806
kmol
m3 min
6.6
kmol
m3
P4-6 (f)
rA = kCAO2(1-X)(θB-2X)
At X = 0.90 and T = 188C = 461K
1) at T = 188 C = 461 K
rA
m3
0.0017
kmol. min
0.00103
kmol
1.8 3
m
2
1 0.9 3.67 2 0.9
kmol
m 3 min
2) at X = 0.90 and T = 25C = 298K
rA
2.03 10
1.23 10 6
6
m3
kmol. min
kmol
1.8 3
m
2
1 0.9 3.67 2 0.9
kmol
m 3 min
3) at X = 0.90 and T = 288C = 561K
rA
m3
0.0152
kmol. min
0.0092
kmol
1.8 3
m
2
1 0.9 3.67 2 0.9
kmol
m 3 min
P4-6 (g)
FAO = 2 mol/min
1) For CSTR at 25oC -rA
1.23 10 6
kmol
m 3 min
4-15
FAO 1 X
rA X 0.9
V
2mol / min 0.1
mol
1.23 10 3 3
m min
2)At 288oC, -rA
V
0.0092
162.60m 3
kmol
m 3 min
FAO 1 X
rA X 0.9
2mol / min 0.1
mol
0.0092 3
m min
21.739m 3
P4-7
C6H12O6 + aO2 + bNH3 → c(C4.4H7.3N0.86O1.2) + dH2O + eCO2
To calculate the yields of biomass, you must first balance the reaction equation by finding the
coefficients a, b, c, d, and e. This can be done with mass balances on each element involved in the
reaction. Once all the coefficients are found, you can then calculate the yield coefficients by simply
assuming the reaction proceeds to completion and calculating the ending mass of the cells.
P4-7 (a)
Apply mass balance
For C
6 = 4.4c + e
For N
b = 0.86c
For O
For H
6 + 2a = 1.2c + d + 2e
12 + 3b = 7.3c + 2d
Also for C, 6(2/3) = 4.4c which gives c = 0.909
Next we solve for e using the other carbon balance
6 = 4.4 (0.909) + e
e=2
We can solve for b using the nitrogen balance
b = 0.86c = 0.86* (0.909)
b = 0.78
Next we use the hydrogen balance to solve for d
12 + 3b = 7.3c + 2d
12 + 3(0.78) = 7.3(0.909) + 2d
4-16
d = 3.85
Finally we solve for a using the oxygen balance
6 + 2a = 1.2c + d + 2e
6 + 2a = 1.2(0.909) + 3.85 + 2(2)
a = 1.47
P4-7 (b)
Assume 1 mole of glucose (180 g) reacts:
Yc/s= mass of cells / mass of glucose = mass of cells / 180 g
mass of cells = c*(molecular weight) = 0.909 mol* (91.34g/mol)
mass of cells = 83.12 g
Yc/s = 83.12 g / 180 g
Yc/s = 0.46
Yc/o2 = mass of cells / mass of O2
If we assume 1 mole of glucose reacted, then 1.47 moles of O2 are needed and 83.12 g of cells are
produced.
mass of O2 = 1.47 mol * (32 g/mol)
mass of O2 = 47.04 g
Yc/o2 = 83.12 g /47.04 g
Yc/o2 =1.77
P4-8 (a)
Isothermal gas phase reaction.
1
N2
2
3
H2
2
NH 3
Making H2 as the basis of calculation:
H2
1
N2
3
A
1
B
3
2
NH 3
3
2
C
3
Stoichiometric table:
Species
H2
N2
Symbol
A
B
Initial
FAO
NH3
C
0
FBO=
B FAO
4-17
change
-FAOX
-FAOX/3
Leaving
FA=FAO(1-X)
+2FAOX/3
FC=(2/3)FAOX
FB=FAO(
B -X/3)
P4-8 (b)
C AO
2 1
1
3 3
2
3
y AO
2
3
0.5
1
3
16.4atm
0.5
0.082
atm.dm
mol.K
= 0.2 mol/dm3
3
500 K
CH 2
CA
C AO 1 X
1 X
0.2 1 X
X
1
3
CNH3
CC
2 C AO X
3 1 X
2
3
0.1mol / dm3
0.2 X
X
1
3
0.1mol / dm3
P4-8 (c)
kN2 = 40 dm3/mol.s
(1) For Flow system:
1
rN 2
k N2 CN2
2
3
CH 2
1
40 C AO
X
1
3
X
1
3
2
3
rN 2
1.6
2
3
2
2
1 X
X
1
3
2
1 X
X
1
3
(2) For batch system, constant volume.
4-18
1
rN2
k N2 CN2
CA
NA
V
CA
C A0 1 X
CB
NB
V
3
2
CH 2
2
N A0 1 X
V0
NA
V0
N A0
B2
X
3
V0
X
3
C H 2O 1
1
rN2
40 C A0
X
3
1.6 1
X
1
3
2
1
2
1 X
2
3
1 X
3
2
2
P4-9 (a)
Liquid phase reaction  assume constant volume
Rate Law (reversible reaction):
rA
k C AC B
CC
KC
Stoichiometry:
CA
C A 0 1 X , CB
C A0 1 X , CC
C A0 X
To find the equilibrium conversion, set -rA = 0, combine stoichiometry and the rate law, and solve for Xe.
C AC B K C
CC
C A2 0 1 X e
2
X e2
2
Xe
0.80
KC
C A0 X e
1
Xe 1 0
C A0 K C
To find the equilibrium concentrations, substitute the equilibrium conversion into the stiochiometric
relations.
CA
C A0 1 X
CB
C A0 1 X
mol
1 0.80
dm3
mol
2 3 1 0.80
dm
2
mol
dm3
mol
0.4 3
dm
0.4
4-19
CA
C A0 X
2
mol
mol
*0.80 1.6 3
3
dm
dm
P4-9 (b)
Stoichiometry:
y A0
1 3 1
2 and
CA
NA
V
N A0 1 X
V0 1
CC
NC
V
3 N A0 X
V0 1 X
C A0
X
C A0
0
C
1 X
1 2X
3X
1 2X
Combine and solve for Xe.
K C C A0
3
1 Xe
3X e
C A0
1 2Xe
1 2Xe
KC 1 X e 1 2 X e
4
Xe
2
27CA2 0 X e3
27C A2 0
X e3 3 X e 1 0
KC
0.58
Equilibrium concentrations:
C A0
P0
RT0
CA
0.305
CC
10 atm
dm3 atm
400 K 0.082
mol K
1 0.58
1 2 0.58
3 0.58 0.305
1 2 0.58
0.059
0.246
0.305
mol
dm3
mol
dm3
mol
dm3
P4-9 (c)
Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction.
Stoichiometry:
CA
NA
V
N A0 1 X
V0
C A0 1 X
4-20
CC
NC
V
3 N A0 X
V0
3C A0 X
Combine and solve for Xe
KC C A0 1 X e
Xe
3
3CA0 X e
0.39
Equilibrium concentrations
CA
0.305 1 0.39
CC
0.305 0.39
mol
dm3
mol
0.36 3
dm
0.19
P4-9 (d)
Gas phase reaction in a constant pressure, batch reactor
Rate law (reversible reaction):
rA
CC3
KC
k CA
Stoichiometry:
y A0
1 3 1
2 and
CA
NA
V
N A0 1 X
V0 1
CC
NC
V
3 N A0 X
V0 1 X
1 X
C A0
X
0
C
1 2X
3C A0
X
1 2X
Combine and solve for Xe:
K C C A0 1 X e
1 2Xe
Xe
3C A0 X e
1 2Xe
3
0.58
Equilibrium concentrations:
CA
0.305 1 0.58
1 2 0.58
0.059
mol
dm3
4-21
3 0.305 0.58
CC
0.246
1 2 0.58
mol
dm3
P4-10
Given: Gas phase reaction A + B  8C in a batch reactor fitted with a piston such that
V = 0.1P0
k 1.0
ft 3
2
lb mol 2 sec
kC A2 CB
rA
NA0 = NB0 at t = 0
V0 = 0.15 ft3
T = 140°C = 600°R = Constant
P4-10 (a)
N A0
N A0 N B 0
y A0
0.5
8 1 1 6
y A0
3
Now
V
Therefore
NA
rA
V0 P0
T
P
T0
1
10V02
1
10V
V
N A0 1 X
2
A
kC CB
kN A2 N B
V3
X
and
T
T0
1 , P0
10V0 , and P 10V
X
or
V2
V02 1
NB
N A0
kN A3 0 1 X
V03 1
X
B
X
X
B
3
N A0
3
2
Therefore
4-22
N B0
N A0
y A0 P0
V0
RT
1
1 X
y P
k A0 0
RT
rA
rA
5.03*10
3
3
1
X 2
1 X
9
1 3X
3
3
2
lb mol
ft 3 sec
P4-10 (b)
V2
V02 1
0.22
0.152 1
X
X
X
0.259
rA
8.63*10 10
lb mol
ft 3 sec
P4-11
(1) For any reaction ,we cannot write the rate law on the basis of the stochiometric
equation. The rate law is to be obtained from the experimental data.
It has been mentioned as an elementary reaction in the problem statement but in the
proposed solution the rate law is based on the reaction equation that has been divided
by stoichiometric coefficient of A.
(2) The value of calculated is incorrect.
= yA0δ = 0.6 (3+5-2-3) = 1.8 is the correct value.
(3) The expression for CA and CB will therefore be,
C A0 1 X
CA
1 X
C A0 1
CB
1
2
X
3
X
(4) According to the system of units being used in the calculations, R = 0.0821 atm. Liter/
mol & Temperature = 700 K should be used.
4-23
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these materials .
5-1
Solutions for Chapter 5: Isothermal Reactor Design- Conversion
P5-1 Individualized solution.
P5-2 (a) Example 5-1
There would be no error! The initial liquid phase concentration remains the same.
P5-2 (b) Example 5-2:
For n equal sized CSTRs (volume Vi each) with equal feed FAi0 = FA0/n
CSTR mole balance on any reactor:
Since FAi0 = FA0/n
rA= kCA= kCAO(1-X)
(1)
For single CSTR with volume nVI and feed rate FA0
CSTR mole balance:
rA= kCA= kCAO(1-X)
(2)
From equation (1) and (2), Xi = X.
Therefore, we can generalize that the conversion achieved in any one of the parallel reactors with volume V i and feed
rate of FA0/n will be identical to the conversion achieved in one large CSTR of volume nVi and feed rate FA0.
P5-2 (c) Example 5-3
For 50% conversion, X = 0.5 and k = 3.07 sec-1 at 1100 K (from Example 5-3)
5-2
FB = 200X106 / (365 X 24 X 3600 X 28) lbmol/sec = 0.226 lbmol/sec
F
0.026
FAO = B
0.452lbmol / sec Also, C AO 0.00415lbmol / ft 3
X
0.5
Now, we have from the example
V
FA0
kC A0
1
ln
1
1 X
X
0.452lbmol / sec
3.07sec 1 X 0.00415lbmol / ft 3
1 1 ln
1
1 0.5
1X 0.5
= 35.47 X 0.886 ft3
= 31.44 ft3
Now,
n = 31.44 ft3/0.0205 ft2 X40 ft = 38.33
So, we see that for lower conversion and required flow rate the volume of the reactor is reduced.
P5-2 (d) Example 5-4
New Dp = 3D0/4
Because the flow is turbulent
o
1
Dp
1
2
y
Now, 1
1
2 oL
Po
Dp 2
D p1
2 L
Po
0.0775
1
2
1
0.75
0.1033
atm
2 X 0.103
X 60 ft
ft
1
10atm
1
2
1
0.24 2
0 , so too much pressure drop P = 0 and the flow stops.
P5-2 (e) Example 5-5
For P = 60atm,
CAO = 0.0415 lbmol/ft3
( C AO
y AO PO
RTO
60
)
0.73 1980
5-3
Using equation E-4-3.6, for X = 0.8
We see that the only thing that changes is CA0 and it increases by a factor of 10, therby decreasing the
volume by a factor of 10.
1
P
V
P5-2 (f) Example 5-6
For turbulent flow
2
1
and
Dp
1
P0
DP1
DP 2
P01
P02
1
1
1
1
5
1
5
1
Therefore there is no change.
P5-2 (g) Example 5-3, Using ASPEN, we get (Refer to Aspen Program P5-2g from polymath CD)
(1) At 1000K, for the same PFR volume we get only 6.2% conversion. While at 1200K, we
get a conversion of nearly 100%. This is because the value of reaction constant ‘k’
varies rapidly with reaction temperature.
(2) Earlier for an activation energy of 82 kcal/mol we got approx. 81% conversion. For
activation energy of 74 kcal/mol keeping the PFR volume the same we get a
conversion of 71.1%. While for an activation energy of 90 kcal/mol we get a
conversion of 89.93%.
(3) On doubling both flow rate and pressure we find that the conversion remains the
same.
P5-2 (h) Individualized solution.
P5-2 (i) Individualized solution.
P5-3 Solution is in the decoding algorithm given with the modules.
5-4
P5-4
We have to find the time required to cook spaghetti in Cuzco, Peru.
Location
Ann Arbor
Boulder
Cuzco
Elevation (km)
0.21
1.63
3.416
Pressure (mm Hg)
739
625
493
Boiling Point (°C)
99.2
94.6
88.3
Time (min)
15
17
?
Assume reaction is zero order with respect to spaghetti conversion:
E
rA
dC A
dt
Ae RT
k
so that
CA
CC
Ae
E
RT
t
A0
For complete conversion (i.e.: well cooked) CA = 0 at time t.
Therefore
E
RT
C A0
tAe
C A0
A
te
E
RT
ln
C A0
A
ln k
ln t
E 1
R Tb
Now, plot the natural log of the cooking time versus 1/Tb and get a linear relationship. Extrapolation to
Tb = 88.3°C = 361.45 K yields t = 21 minutes.
5-5
P5-5 (a)
The blades makes two equal volumes zones of 500gal each rather than one ‘big’ mixing zone of 1000gal.
So, we get 0.57 as conversion instead of 0.5.
5-6
P5-5 (b)
A CSTR is been created at the bend due to back mixing, so the effective arrangement is a PFR is in series
with a CSTR.
CSTR zone due
to back mixing
A→B
k = 5 min-1
vo = 5 dm3/min.
Xexpected = 0.632 Xactual = 0.618
X
V
FAO dX
rA
0
vo
1
5
1
= ln
ln
k
1 X Expected
5
1 .632
1.0
Now,
For PFR,
VP
For CSTR,
Also,
ln
FAO ( X actual
rA
VC
VP
V
1
………………………………….(1)
1 X1
VC
V
X1 )
X actual X 1
……………(2)
1 X actual
1 ………………………………………..(3)
Solving 1, 2 and 3 by using polymath,
See Polymath program P5-5-b.pol.
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 Vc
0.3001282 -2.831E-15 1.
5-7
2 X1
0.503351 7.864E-11 0
Variable Value
1 V
1.
2 Vp
0.6998718
3 X2
0.618
Nonlinear equations
1 f(X1) = ln(1/(1-X1)) -Vp = 0
2 f(Vc) = (X2-X1)/(1-X2) - Vc = 0
Explicit equations
1 V=1
2 X2 = .618
3 Vp = V - Vc
General Settings
Total number of equations
5
Number of implicit equations 2
Number of explicit equations 3
Elapsed time
0.0000 sec
Solution method
SAFENEWT
Max iterations
150
Tolerance F
0.0000001
Tolerance X
0.0000001
Tolerance min
VC = 0.3 dm
0.0000001
3
; VP = 0.7 dm3 ;
X1 = 0.5
P5-5 (c)
CAO = 2 mol/dm3
A B
Assuming 1st order reaction,
C AO X
rA
For CSTR,
-rA = kCA = kCAO(1-X)
=>
X
1 X
k
0.4
0.6
X
For PFR,
V
FA0
=> X PFR
0.67
dX
, k
kC
1
X
A0
0
1 exp(
X
dX
1 X
0
k ) =1-exp(-0.67) = 0.486
Now assuming 2nd order reaction,
5-8
For CSTR,
Now, assuming 2nd order reaction,
C AO X
rA
For CSTR,
rA
2
kC A2 = kC AO
1 X
=> kC AO
X
1 X
2
0.4
0.62
X
1
dX
kC AO 0 1 X 2
For PFR,
=> X
1
1
1
kC AO
2
1.111
1
X
kC AO 1 X
1
1
.526
2.111
So, while calculating PFR conversion they considered reaction to be 1st order. But actually it is a second
order reaction.
P5-5 (d)
A graph between conversion and particle size is as follows: Originally we are at point A in graph, when
particle size is decreased by 15%, we move to point C, which have same conversion as particle size at A.
But when we decrease the particle size by 20%, we reach at point D, so a decrease in conversion is
noticed. Also when we increase the particle size from position A, we reach at point B, again there is a
decrease in the conversion.
5-9
P5-6
(a)
(3) 50% (Liquid phase reactions do not depend on pressure)
(b)
(2) <50% (Low concentration if we lower the pressure)
2
kC A0 1 X
rA
(c)
1
X
2
, C A02
2
1
C A01 , low rate
2
(1) >0.234
2
rA
kC A0 1 X
rA
ky 2
y2
y2
1
k true y 2
k true
2
k 0.234
0.234
0.234
y2
P5-7
(a)
Ans. (1) X > 0.5
~
1
D
Increase D, decrease , increase X.
(b) Ans. (3) Xe = 0.75
Xe
(c)
KC
1 KC
3
1 3
Ans. (3) remain the same
5-10
0.75
KC
CA
C Ae
C A0 1 X e y 1 X e
C A0 X e y
Xe
Xe is not a function of .
(d) Ans. (3) remain the same
Xe is not a function of .
(e)
Ans. (4) insufficient information to tell
Xe
dp
P5-8 (a)
X
V kC A2 0 1 X
2
10
5-11
dX
dV
X
1 X
rA
FA0
2
kC AO
(1 X )2
FA0
kC A2 0V
FA0
X=0.85
So, considering the above results, we will choose a CSTR.
P5-8 (b)
P5-8 (c)
P5-8 (d)
1) CSTR and PFR are connected in series:
5-12
(200dm3 )(0.07dm3 / mol.min)(1mol / dm3 ) 2 (1 X ) 2
10mol / min
X CSTR
Solving the quadratic equation, XCSTR = 0.44
For PFR,
FA0
dX
dV
rA
(0.07dm3 / mol / min)C AO (1 X )(1 X )2
dV
10 mole / min
dX
X
(0.07dm3 / mol / min)(1mol / dm3 )2 (800dm3 )
10mole / min
dX
(1 X )3
0.44
X
0.736
2) when CSTR and PFR are connected in parallel,
X CSTR
(200dm3 )(0.07dm3 / mol.min)(1mol / dm3 ) 2 (1 X ) 2
5mol / min .
XCSTR = 0.56
For PFR,
X
dX
(1 X )2
0
(0.07dm3 / mol.min)(1mol / dm3 )2 (800dm3 )
5mol / min
XPFR = 0.92
Hence, final conversion X =
0.56 0.92
=0.74
2
P5-8 (e)
To process the same amount of species A, the batch reactor must handle
5dm3
2M
min
60 min
hr
24h
day
14400
mol
day
If the reactants are in the same concentrations as in the flow reactors, then
V
14400
mol
day
1dm3
mol
14400
dm3
day
So the batch reactor must be able to process 14400 dm3 every 24 hours.
5-13
Now we find the time required to reach 90% conversion. Assume the reaction temperature is 300K.
rAV
N A0
dX
dt
kC A2 0 1 X V
N A0
tR
N
N A0
X
, and since A0
2
V
VkC A0 1 X
tR
1
X
kC A0 1 X
C A0
1
0.9
dm
mol 0.1
4.2
*1
mol hr
dm3
3
2.14hr
Assume that it takes three hours to fill, empty, and heat to the reaction temperature.
tf = 3 hours
ttotal = tR + tf
ttotal = 2.14hours + 3 hours = 5.14 hours.
Therefore, we can run 4 batches in a day and the necessary reactor volume is
14400dm3
4
3600dm3
Referring to Table 1-1 and noting that 3600 dm3 is about 1000 gallons, we see that the price would be
approximately $85,000 for the reactor.
P5-8 (f)
The points of the problem are:
1) To note the significant differences in processing times at different temperature (i.e. compare
part (b) and (c)).
2) That the reaction is so fast at 77°C that a batch reactor is not appropriate. One minute to
react and 180 to fill and empty.
3) Not to be confused by irrelevant information. It does not matter if the reactor is red or black.
P5-9
PFR
dX
dV
rA
FA0
kCA0 1 X
C A0 0 1 X
5-14
k 1 X
X
0 1
X 1
0
X
dX
1 X
1
k
1
ln
1
y A0
k
1
2
2s
11 1 1
1
1
1 1 ln
0.8
1 0.8
2 1.61 0.8
2
X
1 X
5 dm 3 s
1
k
0
10dm 3
k
dV
k
1.209 at 300K
CSTR
FA0 X
V
0C A0 X
rA
0X 1
1 X
kC A0
1
1
2
X
k1 X
X
1 1 1
0.5
k
1 X1
X
1 X
1 0.8 1 0.5 0.8
2
0.2
2.8
ln
ln
k2
E 1
1
k1
R T1
T2
2.8
E
1
1
E
1.21
R 300
320
R 320 300
E R
300 320
20
ln
2.8
1.21
E 8010
5-15
20
, R 1.987
cal
mol
cal
mol K
P5-10
A
kC A0 1 X
X
0C A0 1
dX
dV
kV
3B
1
ln
0
1 X
1
3 1
2
y A0
kV
V0
1
1 1 ln
FA0 X 2 X1
rA2
V
X
0C A0 X 2
V
1 X2
1 X2
kC A0
1
X 2 X1 1
1 X2
k
1
0.5
1 0.5
X 22
k 0.886
X2 1
X1
X2
X 2 X1 1 X 2
1 X2
k X1
X1
k
0
X 22 1.386X 2 1.386 0
X2
1.386X 2
1.386 2
4
1.386
2
X2
0.673
P5-11 (a)
POLYMATH Report
Ordinary Differential Equations
04-May-2009
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 w
0
0
2000.
2000.
2 x
0
0
0.9978192
0.9978192
3 FBuO
50.
50.
50.
50.
4 PBuO
10.
10.
10.
10.
5 PBu
10.
0.0109159
10.
0.0109159
6 k
0.054
0.054
0.054
0.054
7 K
0.32
0.32
0.32
0.32
8 rBu
-0.0306122
-0.0421872
-0.0005854
-0.0005854
5-16
Differential equations
1 d(x)/d(w) = -rBu / FBuO
Kgcat-1
Explicit equations
1 FBuO = 50
Kmol.hr-1
2 PBuO = 10
atm
3 PBu = PBuO * (1 - x) / (1 + x)
atm
4 k = 0.054
kmol.kgcat-1.hr-1.atm-1
5 K = 0.32
atm-1
6 rBu = -k * PBu / (1 + K * PBu) ^ 2
Kmol.kgcat-1.hr-1
General
Total number of equations
7
Number of differential equations 1
Number of explicit equations
6
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
5-17
5-18
Hence, weight of catalyst required for 80% conversion is 1054.1 kg .
P5-11 (b)
Differential Mole Balance:
FBuO
= -rBuʹ
PBu = PBuO
Rate equation
:
Here,
-rBuʹ = K*PBu
For Fluidized CSTR :
5-19
Therefore,
w = FBuo*X
Given,
X = 0.8
FBuO = 50 Kmol.hr-1
KBu = 0.32 atm-1
PBuO = 10 atm
Therefore, putting the values in the above equation
w = 50*0.8
= 1225 Kg
Therefore, 1225 Kg of fluidized CSTR catalyst weight is required to achieve 80% conversion.
P5-11 (c)
With pressure drop parameter alpha = 0.0006 kg-1
POLYMATH Report
Ordinary Differential Equations
04-May-2009
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
w
0
0
1100.
1100.
2
x
0
0
0.7750365
0.7750365
3
y
1.
0.2689385
1.
0.2689385
4
FBuO
50.
50.
50.
50.
5
f
1.
1.
6.600157
6.600157
6
PBuO
10.
10.
10.
10.
7
PBu
10.
0.3408456
10.
0.3408456
8
k
0.054
0.054
0.054
0.054
9
K
0.32
0.32
0.32
0.32
10 rBu
-0.0306122
-0.0421871
-0.0149635
-0.0149635
11 alpha
0.0006
0.0006
0.0006
0.0006
Differential equations
1 d(x)/d(w) = -rBu / FBuO
5-20
Kgcat-1
2 d(y)/d(w) = -alpha * (1 + x) / 2 / y
Kgcat-1
Explicit equations
1 FBuO = 50
Kmol.hr-1
2 f = (1 + x) / y
3 PBuO = 10
atm
4 PBu = PBuO * (1 - x) / (1 + x) * y
atm
5 k = 0.054
kmol.kgcat-1.hr-1.atm-1
6 K = 0.32
atm-1
7 rBu = -k * PBu / (1 + K * PBu) ^ 2
Kmol.kgcat-1.hr-1
8 alpha = 0.0006
Kg-1
General
Total number of equations
10
Number of differential equations 2
Number of explicit equations
8
Elapsed time
1.157 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
5-21
So, we see the maximum rate in case with pressure drop is at catalyst weight equal to around 600 Kg.
5-22
To achieve 70% conversion, catalyst weight required is 932.3 kg .
In case of (a), 915.5 kg of catalyst is required to achieve 70% conversion.
5-23
P5-11 (d) Individualized solution
P5-11 (e) Individualized solution
P5-12 (a)
P5-12 (b)
P5-12 (c)
For α = 0.001dm-3
See Polymath program P5-12-c.pol.
5-24
POLYMATH Results
Calculated values of the DEQ variables
Variable
v
x
y
Co
esp
alfa
C
k
r
fo
initial value
0
0
1
0.3
2
0.001
0.3
0.044
-0.0132
2.5
minimal value
0
0
0.1721111
0.3
2
0.001
0.0077768
0.044
-0.0132
2.5
maximal value
500
0.656431
1
0.3
2
0.001
0.3
0.044
-3.422E-04
2.5
Differential equations as entered by the user
[1] d(x)/d(v) = -r/fo
[2] d(y)/d(v) = -alfa*(1+esp*x)/(2*y)
Explicit equations as entered by the user
[1] Co = 0.3
[2] esp = 2
[3] alfa = 0.001
[4] C = Co*(1-x)*y/(1+esp*x)
[5] k = 0.044
[6] r = -k*C
[7] fo = 2.5
At V = 500,
x = 0.66, y = 0.17
P5-12 (d) Individualized solution
P5-12 (e)
A
B 2C
5-25
final value
500
0.656431
0.1721111
0.3
2
0.001
0.0077768
0.044
-3.422E-04
2.5
Using these equations in Polymath we get the volume to be 290 dm3.
For a batch reactor.
5-26
P5-12 (f)
A
B 2C
Using Polymath to solve the differential equation gives a volume of 290 dm3
See Polymath program P5-12-f.pol.
PFR with pressure drop: Alter the Polymath equations from part (c).
See Polymath program P5-12-f-pressure.pol.
POLYMATH Results
5-27
Calculated values of the DEQ variables
Variable
v
x
y
Kc
alfa
Cao
k
esp
fo
r
initial value
0
0
1
0.025
0.001
0.3
0.044
2
2.5
-0.0132
minimal value
0
0
0.3181585
0.025
0.001
0.3
0.044
2
2.5
-0.0132
maximal value
500
0.5077714
1
0.025
0.001
0.3
0.044
2
2.5
-1.846E-04
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(x)/d(v) = -r/fo
[2] d(y)/d(v) = -alfa*(1+esp*x)/(2*y)
Explicit equations as entered by the user
[1] Kc = .025
[2] alfa = 0.001
[3] Cao = 0.3
[4] k = 0.044
[5] esp = 2
[6] fo = 2.5
[7] r = -(k*Cao/(1+esp*x))*(4*Cao^2*x^3/((1+esp*x)^2*Kc))
5-28
final value
500
0.5077714
0.3181585
0.025
0.001
0.3
0.044
2
2.5
-1.846E-04
At V = 500 dm3 X = 0.507 and y = 0.381
P5-13
A
To = 300K
KCO (300K)= 3.0 V = 1000gal = 3785.4 dm3
FAO X
rA
V
Mole balance:
rA
Rate law:
Stoichiometry: C A
V
Now using:
where
B
kO C
and
C AO 1 X
FAO
2
kO C AO
1 X
2
V 1 X
Z
2902.2dm3 , Z
V
Z
z
KC
C BO X
X2
KC
2
FAO
kOC AO 2
k
kO
CB
X
Z
z
C B2
KC
2
A
X2
KC
3785.4
0.62
0.4
X
k
ko
exp
X
1 X
2
E 1
R TO
exp
K CO exp
2
X
KC
1
T
H RX 1
R
TO
E 1
R T0
1
T
f X
0 V
and
1
T
Solving using polymath to get a table of values of X Vs T.
See Polymath program P5-13.pol.
POLYMATH Results
NLE Solution
Variable
X
To
Value
0.4229453
300
f(x)
3.638E-12
Ini Guess
0.5
5-29
0.42
3
Z
z
X
1 X
2
X2
KC
T
z
V
E
R
y
Kco
Hrx
Kc
305.5
2902.2
3785.4
1.5E+04
2
1.5684405
3
-2.5E+04
1.4169064
NLE Report (safenewt)
Nonlinear equations
0.43
0.42
[1] f(X) = (z/y)*X/((1-X)^2 - X^2/Kc) -V = 0
Explicit equations
0.41
[1] To = 300
[2] T = 305.5
[3] z = 2902.2
[4] V = 3785.4
[5] E = 15000
[6] R = 2
[7] y = exp(E/R*(1/To-1/T))
[8] Kco = 3
[9] Hrx = -25000
[10] Kc = Kco*exp(Hrx/R*(1/To-1/T))
X
0.4
0.39
0.38
0.37
0.36
295
300
305
310
Temperature
T(in K)
X
300
0.40
301
0.4075
303
0.4182
304
0.4213
305
0.4228
305.5
0.4229
305.9
0.4227
307
0.421
310
0.4072
315
0.3635
We get maximum X = 0.4229 at T = 305.5 K.
P5-14
CH 3CO 2 O H 2O
A B
5-30
2CH 3COOH
2C
315
320
Part 1 CSTR
A
B
3.3 10 3 dm3 s
0
C A0 1M, C B 51.2M
3
V 1 dm
X= ?
(1)Mole Balance:
V
(2)Rate Law:
rA
(3)Stoichiometry:
FA0 X
rA
k CACB
Liquid
0
Stoichiometric Table
A
B
C
FA0
FB0
BFA0
0
FA0 X
FA
FA0 1 X
FA0 X
FB
FA0
2FA0 X
FC
Concentration
CA
CB
FA
FA0 1 X
0
0
FB
FA0
0
51.2M
B
1M
B
0
C A0 1 X
X
C A0
B
51.2
Substitute the value of X into CB.
5-31
X
B
2FA0 X
X
CB
CA0 51.2 X
CB0
(4)Combine
k
rA
k C AC B0
k C B0 C A0 1 X
dm3
k k C B0 1.95 10 4
V
mol s
FA0 X
0C A0 X
rA
kC A0 1 X
0
kC A0 1 X
51.2
mol
dm
3
0.01 s 1
X
k 1 X
X
k1 X
k
X
1
k
(5)Evaluate
V
1 dm3
0
3.3 10 3 dm3 s
k
X
303 s
3.03
3.03
1 3.03
0.75
Calculate the Conversion in a
0
3
3
3.3 10 dm s,
CA0 1M
CB 0
V
51.2M
PFR
5-32
0.311 dm3
X= ?
(1)Mole Balance:
(2)Rate Law:
(3)Stoichiometry:
dX
rA
dV
FA0
rA k CACB
Liquid
0
CA
C A0 1 X
CB
C B0
k
rA
k C B0 C A0 1 X
kC A0 1 X
(4)Combine
dX
rA
kC A0 1 X
dV
FA0
0C A0
Separate variables and Integrate with limits V = 0 then X = 0 and V = V then X = X
X
dX
0 1
ln
X
V k
0
1
kV
1 X
0
X 1 e
dV
0
k
k
(5)Evaluate
0.311 dm3
3.3 10 3 dm3 s
k
0.94
X
0.61
5-33
94 s
P5-15
Gaseous reactant in a tubular reactor: A  B
rA
kC A
k
0.0015min 1 at 80 F
E
25, 000
MWA
cal
g mol
MWB
58
lb
lb mol
P 132 psig 146.7 psia
lb
hr
lb
58
lb mol
1000
FB
X
0.90
MB
Dt
1 inch (I.D.)
L 10 ft
T
260 F
nt
lb mol
17.21
hr
720 R
FA0
FB
X
1000
lb
hr
number of tubes
lb mol
hr
0.9
17.21
19.1
For a plug flow reactor:
0.9
nt Dt2 L
4
V
FA0
dX
rA
0
1 1 0
rA
yA
kC A0 1 X
C A0
0.9
V
yA
1.0
FA0
dX
rA
0
0.9
FA0
dX
kC A0 1 X
0
PA
RT
FA0
1
ln
kC A0
1 0.9
0
P
RT
FA0 RT
ln10
kP
At T2 = 260°F = 720°R, with k1 = 0.0015 min-1 at T1 = 80°F = 540°R,
E 1
R T1
k2
k1 exp
k2
53.6 min 1
V
0.72 ft 3
V
nt Dt2 L
4
0.0015exp
25000 1
1.104 540
1
720
lb mol
hr
ft 3 psia
lb mol R
720 R
53.6 min 1
3219hr 1
FA0 RT
ln10
kP
V
1
T2
19.1
10.73
3219 hr 1 146.7 psia
5-34
ln10
lb mol
hr
nt
4V
Dt2 L
4 0.72 ft 3
1
ft
12
2
13.2
10 ft
Therefore 14 pipes are necessary.
P5-16(a)
A → B/2
Combining
2
VPFR kCA0
FA0
X
2
1 X
2
X 1
0
dX
2 1
ln 1 X
C AO
y AO CTO
2
2
X
1
X
1 X
(for the integration, refer to Appendix A)
from the Ideal Gas assumption.
Substituting Eqn. (5), X = 0.8 and = –1/4 to Eqn. (4) yields,
2
2
VPFR ky AO
CTO
FAO
2( 1/ 4)(1 1/ 4) ln(1 0.8) ( 1/ 4)2 0.8
5-35
(1 1/ 4)2 0.8
1 0.8
2.9 ……….(6)
2
VPFR ky '2AO C 'TO
F '2AO
2
2
CTO
8 VPFR ky AO
9
FAO
8
2.9
9
2.58
P5-16(b) Individualized solution.
P5-17(a)
A
Given: The metal catalyzed isomerization
rA
B liquid phase reaction
CB
with Keq = 5.8
K eq
k1 C A
For a plug flow reactor with yA = 1.0, X1 = 0.55
Case 1: an identical plug flow reactor connected in series with the original reactor.
B = 0. For a liquid phase reaction CA
Since yA = 1.0,
rA
kC A0
X
K eq
1 X
For the first reactor,
X1
V1
FA0
dX
rA
0
X1
FA0
0
dX
kC A0 1 X
or
X
K eq
5-36
CA0 1 X and C B
C A0 X
X1
kC A0V1
FA0
X1
dX
0
kC A0V1
FA0
1
1
X
K eq
1
1
1
1
K eq
1
ln 1
1
1
1
K eq
ln 1
1
1
X
K eq
0
1
X1
K eq
0.853ln .355
0.883
Take advantage of the fact that two PFR’s in series is the same as one PFR with the volume of the two
combined.
VF = V1 + V2 = 2V1 and at VF , X = X2
XF
XF
kC A0VF
FA0
0
dX
1
1
1
1
X
K eq
kC A0VF
FA0
2
kC A0V1
FA0
1
kC A0V1
FA0
2 0.883
1.766
2
1.766
1
1
1
5.8
ln 1
1
1
1
K eq
1
K eq
1
ln 1
1
ln 1
1
1
X
K eq
0
1
X2
K eq
1
X2
5.8
X2 = 0.74
P5-17(b)
Case 2: Products from 1st reactor are separated and pure A is fed to the second reactor,
5-37
The analysis for the first reactor is the same as for case 1.
kC A0V1
FA0
1
1
K eq
1
ln 1
1
X1
K eq
1
By performing a material balance on the separator, FA0,2 = FA0(1-X1)
Since pure A enters both the first and second reactor CA0,2 = CA0, CB0,2 = 0,
CA = CA0 1 - X
B=0
C B = C A0 X for the second reactor.
X2
FA0 1 X
dX
FA0,2
rA
0
V2
kC A0V2
FA0 1 X 1
kC A0
1
1
1
K eq
X2
0
ln 1
dX
1 X
1
X
K eq
1
X2
K eq
and since V1 = V2
kC A0V2
FA0
kC A0V1
FA0
or
1
1
1
K eq
ln 1
1
1
1
1 X1
ln 1
1
1
K eq
1
X1
K eq
1
X2
K eq
1
5-38
1
1
X1
K eq
1
1
1 X1
1
X2
K eq
1
1
X2
1
1
X1
K eq
1
1 X1
1
1
0.356 0.45
1
1
K eq
1.174
FA0
FA0 1 X1 1 X 2
0.766
Overall conversion for this scheme:
X
X
FA0
FA0,2 1 X 2
FA0
FA0
0.895
P5-18
Given: Meta- to ortho- and para- isomerization of xylene.
M
k1
P
M
k2
O
O
P (neglect)
Pressure = 300 psig
T = 750°F
V = 1000 ft3 cat.
Assume that the reactions are irreversible and first order.
Then:
rM
k
k1CM
k 2 CM
kCM
k1 k2
0
Check to see what type of reactor is being used.
Case 1:
v0
2500
gal
hr
X
0.37
gal
hr
X
0.50
Case 2:
v0
1667
Assume plug flow reactor conditions:
FM 0 dX
rM dV or
5-39
1
1 X1 1 X 2
X
V
FM 0
dX
rM
0
X
V
X
CM 0 v0 dX
rM
0
v0
v0
ln 1 X
k
dX
k 1 X
0
CM0, k, and V should be the same for Case 1 and Case 2.
Therefore,
kV Case1
v0 Case1 ln 1 X Case1
kV Case 2
v0 Case1 ln 1 X Case 2
gal
ln 1 0.37
hr
gal
1667
ln 1 0.50
hr
2500
gal
hr
gal
1155
hr
1155
The reactor appears to be plug flow since (kV)Case 1 = (kV)Case 2
As a check, assume the reactor is a CSTR.
FM 0 X
V
CM 0 v0 X
CM 0 X
v0
rM
rM V
v0 X
k 1 X
or kV
v0 X
1 X
Again kV should be the same for both Case 1 and Case 2.
kV Case1
kV Case 2
gal
0.37
hr
1 0.37
1468
gal
0.50
hr
1 0.50
1667
2500
v0 Case1 X Case1
1 X Case1
v0 Case 2 X Case 2
gal
hr
1667
1 X Case 2
gal
hr
kV is not the same for Case 1 and Case 2 using the CSTR assumption, therefore the reactor must be
modeled as a plug flow reactor.
kV
k
gal
hr
gal
1155
gal
hr
1.55
3
1000 ft cat.
hr ft 3 cat
1155
For the new plant, with v0 = 5500 gal / hr, XF = 0.46, the required catalyst volume is:
V
v0
ln 1 X F
k
gal
hr ln 1 0.46
gal
1.155
hr ft 3 cat
5500
2931 ft 3 cat
This assumes that the same hydrodynamic conditions are present in the new reactor as in the old.
5-40
P5-19
A→ B in a tubular reactor
Tube dimensions: L = 40 ft, D = 0.75 in.
nt = 50
50
nt D 2
L
4
V
FA0
0.75
12
4
lb
hr
lb
73
lb mol
2
40 6.14 ft 3
500
mA
MWA
6.86
lb mol
hr
X
V
FA0
rA
dX
rA
0
kC A0 1 X
1
X
V
FA0
dX
rA
0
X
FA0
dX
kC A0 1 X
0
y A0 P
RT
FA0 RT
1
ln
or k
ky A0 P
1 X
with C A0
V
kC A0 1 X
X
FA0
1
ln
kC A0
1 X
P
RT
FA0 RT
1
ln
Vy A0 P
1 X
Assume Arrhenius equation applies to the rate constant.
At T1 = 600°R, k1 = 0.00152
At T2 = 760°R, k2 = 0.0740
Ae
Ae
E
RT1
E
RT2
5-41
k2
k1
ln
E 1
R T2
exp
k2
k1
E 1
R T2
1
T1
E T2 T1
R T1T2
1
T1
660 760
TT
k
1 2
ln 2
T1 T2 k1
E
R
A k1 exp
so k
100
ln
0.740
19,500 R
0.00152
E
RT1
E 1
R T
k1 exp
1
T1
From above we have
k
FA0 RT
1
ln
Vy A0 P
1 X
so
FA0 RT
1
ln
Vy A0 P
1 X
E 1
R T
k1 exp
1
T1
Dividing both sides by T gives:
FA0 R
1
ln
Vy A0 P
1 X
6.86
.00152
sec
E 1
R T
k1 exp
lb mol
hr
1
T1
T
10.73
sec
3600
hr
psia ft 3
lb mol R
6.14 ft
3
exp
ln 5
114.7 psia
Evaluating and simplifying gives:
exp
0.0308 R 1
1
T
19500
1
660 R
T
Solving for T gives:
T = 738°R = 278°F
5-42
1
T
19500
T
1
660 R
P5-20
Rate law:
-rA= kCACB
(1)
Molar flow rate: FA0 =
FB0 =
Since FA0= FB0, therefore equimolar mixture.
CA0 = CB0 = FA0/ = 1.51/5.33 = 0.283 lbmol/ft3
From (1), -rA= kCACB = kCA0(1-X)CB0(1-X) = kCA02 (1-X)2
(2)
Now, for PFR
V = FA0
(3)
Putting -rA from (2) in (3) and then integrating, we get
VkCA02/ FA0 =
Now given, 50% conversion, therefore X = 0.5
So, VkCA02/ FA0 =
=1
5.33*k*0.2832/1.51 = 1
k = 3.54
Now we have CSTR in series
V = 100 gal = 13.37 ft3
Now, CA20 = CA0/2 = 0.1425 lbmol/ft3 , since we have 50% conversion if PFR
FA20 = FA0/2 = 1.51/2 = 0.755
CSTR mole balance: V = FA20
So, 13.37 = 0.755
= FA20
=
13.37 =
=> X = 0.42
So, CA = 0.1425(1-0.42) = 0.083
So, overall conversion X = (CA0- CA)/CA0 = (0.283 – 0.083)/(0.283) = .70
5-43
X = 0.7
P5-21
Five things wrong with the solution are as follows :
(i) The rate law is wrongly written
(ii) The integration is wrongly done
(iii) Value of Ɛ is wrongly calculated
(iv)The equation for calculation of k has been reaaranged wrongly.
(v) calculated value of k is therefore wrong
correct solution
For a first order reaction ;
-rA = -
1 dN A
V dt
kC A
By transforming into conversion units we have;
-rA =
C A0 dX A
1 A X A dt
kC A0 (1 X A )
1 AXA
Separating and integrating we obtain ;
ln(1 X A )
V
)
AV0
ln(1
kt
Now in this case ;
Ɛ = yA0δ = 1(3-1) = 2
Substituting
V
V0
4 2
1
2
we have;
-ln(1-0.5) = ln2 = kt
=> k = ln2/t = 0.693/2hr = 0.3465 hr-1
5-44
P5-22
Reversible isomerization reaction
m-Xylene → p-Xylene
Xe is the equilibrium conversion.
rm
Rate law:
k Cm
Cp
ke
At equilibrium,
-rm = 0 =>
Cm
Cp
ke
Cmo 1 X e
Xe
1 Xe
Ke
1
Ke
1
rm
1
1 Xe
Xe
kC A0 1
Cmo X e
Ke
Xe 1 Xe
Xe
1
Xe
X
Xe
P5-22 (a)
For batch reactor,
Mole balance:
dX
dt
rmV
N mO
kC A0
X
1
C A0
Xe
Xe
Xe
ln
k
Xe X
V
FAO
vo
k
For PFR,
PFR
PFR
dX
rm
dX
1
1
1
k
1
X
Ke
dX
1
1
1
X
Ke
Xe
Xe
ln
k
Xe X
5-45
P5-22 (b)
For CSTR,
V
Fmo X
rm
X
CSTR
k 1
1
1
X
Ke
Putting the value of Ke,
X
k
CSTR
Xe
Xe
X
P5-22 (c)
Volume efficiency =
PFR
V
CSTR
1
V
Xe
Xe
ln
k
Xe X
ln
Xe
X
k Xe X
Xe
Xe
X
Xe
X
X
Xe
X
X
Xe
1
ln
X
X
1
Xe
Xe
Following is the plot of volume efficiency as a function of the ratio (X/Xe),
See Polymath program P5-22-c.pol.
5-46
X
ln
Xe
Xe
X
P5-23(a)
5-47
P5-23 (b)
P5-23 (c)
P5-23 (d)
For turbulent flow
2
DP1
1
DP 2
Ac1
Ac 2
3
0.018kg
1
2
1
12
1.52
3
0.00316 kg 1
Pexit = P0(1- αW)0.5 = (20atm)(1-0.00316kg-1(50kg))1/2 = 18.35 atm
ln
1 0.5
1 X2
X2
0.81
3/ 2
10 3
2
20atm
1 1 0.00316kg 1 50kg
1
atm kg
3 0.00316kg
P5-24
Production of phosgene in a microreactor.
CO + Cl2
COCl2 (Gas phase reaction)
A + B
C
The euqtions that need to be solved are as follows :
d(X)/d(W) = -rA/FA0
d(y)/d(W) = -α*(1+Ɛ *X)/(2*y)
( from 4.30)
5-48
FB0 = FA0;
Fb = FB0-FA0*X;
Fc = FA0*X
See Polymath program P5-24.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
W
X
y
e
FA0
FB0
Fa
Fb
v0
v
Fc
Ca
Cb
a
k
rA
Cc
initial value
0
0
1
-0.5
2.0E-05
2.0E-05
2.0E-05
2.0E-05
2.83E-07
2.83E-07
0
70.671378
70.671378
3.55E+05
0.004
-19.977775
0
minimal value
0
0
0.3649802
-0.5
2.0E-05
2.0E-05
4.32E-06
4.32E-06
2.83E-07
2.444E-07
0
9.1638708
9.1638708
3.55E+05
0.004
-19.977775
0
maximal value
3.5E-06
0.7839904
1
-0.5
2.0E-05
2.0E-05
2.0E-05
2.0E-05
2.83E-07
4.714E-07
1.568E-05
70.671378
70.671378
3.55E+05
0.004
-0.3359061
53.532416
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -rA/FA0
[2] d(y)/d(W) = -a*(1+e*X)/(2*y)
Explicit equations as entered by the user
[1] e = -.5
[3] FB0 = FA0
[5] Fb = FB0-FA0*X
[7] v = v0*(1+e*X)/y
[9] Ca = Fa/v
[11] a = 3.55e5
[13] rA = -k*Ca*Cb
[2] FA0 = 2e-5
[4] Fa = FA0*(1-X)
[6] v0 = 2.83e-7
[8] Fc = FA0*X
[10] Cb = Fb/v
[12] k = .004
[14] Cc = Fc/v
5-49
final value
3.5E-06
0.7839904
0.3649802
-0.5
2.0E-05
2.0E-05
4.32E-06
4.32E-06
2.83E-07
4.714E-07
1.568E-05
9.1638708
9.1638708
3.55E+05
0.004
-0.3359061
33.259571
P5-24 (a)
P5-24 (b)
The outlet conversion of the reactor is 0.784
The yield is then MW*FA*X = 99 g/mol * 2 e-5 mol/s * 0.784 = .00155 g/s = 48.95 g/ year.
Therefore 10,000 kg/year / 48.95 kg/ year = 204 reactors are needed.
P5-24 (c)
Assuming laminar flow, α ~ Dp-2, therefore
2
DP21
1
DP2 2
3.55 105 kg 1 4 14.2 105 kg 1
P5-24 (d)
A lower conversion is reached due to equilibrium. Also, the reverse reaction begins to overtake the
forward reaction near the exit of the reactor.
P5-24 (e) Individualized solution
5-50
P5-24 (f) Individualized solution
P5-24 (g) Individualized solution
P5-25 No solution necessary
P5-26
A B
C D
P0
P
1. Mole Balance – Use Differential Form
dX
dW
rA
FA 0
2. Rate Law – Psuedo Zero Order in B
rA
3. Stoichiometry – Gas:
kC A C 0B
0 1
kC A
X
T P0
, Isothermal, therefore T
T0 P
5-51
T0
FA0 1 X P
FA
CA
0 1
X P0
CA
1
X
y , y
P
P0
0
C A0 1 X y
, where W
2y
1 X
1
1 1 1 1
2
y A0
dy
dW
C A0
0 then y 1 and X 0
Integrating
y
part (a)
1
W
12
Now since P/P0 >0.1 .
So y= .1 for minimum value of W
Now (1 – αW)1/2 = .1 => W = 99 kg
4. Combine
rA
kC A
kC A0 1 X y
kC A0 1 X 1
dX
dW
kC A0
1 X 1
FA0
1
W
Separate Variables
dX
1 X
k
12
dW
0
when W = 0, X = 0
ln
1
1 X
2k
1
3 0
1
W
32
5-52
W
12
W
12
2 1.2 dm 3 g min
2k
3
0
ln
3 0.01 g 25 dm 3 min
1
3.2 1 1
1 X
y
1
W
0
0
W
3.2
32
12
P0
P
1
y
5-53
W
y
1/y
X
P
0
1
1
0
10 atm
1
1
1
0.47
2.5
0.99
1.01
0.11
5
0.975
1.02
0.21
15
0.92
1.08
0.5
25
0.87
1.16
0.67
50
0.71
1.41
0.87
75
0.5
2.0
0.94
90
0.32
3.16
0.955
99
0.1
10
0.96
X = 0.9
2.3
1
3.2
1
1
W
32
W .429
W = 57g
for 5% W = 1.05g
for last 5% (85 to 90%) W = 57 – 45 = 12 g
Ratio =
12g
= 11.4
1.05g
Polymath
dX
dW
rA prime FA0
dy
dW
alpha 2 y
rA prime
CA
k CA
C A0
1 Xy
k 1.2 dm 3 g min
alpha 0.01g 1
5-54
1 atm
5-55
5-56
P5-27
0,
y
0
1
W
1
2
Mole Balance/Design Equation
FA 0
dX
dW
rA
Rate Law
rA
kC A
2
5-57
Stoichiometry
CA
C A0 1 X y
Combining
FA 0
dX
dW
FA 0
2
kC A0 1 X
dX
2
kC A0 1 X
FA 0
X
2
kC A0 1 X
10
2
1
2
W
2
y2
W dW
2
W2
100
99
2
101
2
19.8
dm 6
mol 2 min
k 300 K
0.1
k 300 K
2000
101
k 400 K
k 300 K exp
E 1
R 300
19.8 exp
10,000 1
1.987 300
1311.179
dm 6
mol 2 min
1
400
1
400
P5-28
A
rA
B C
kC A
Case1: 10 dm3 PFR with 80% conversion at 300K.
dV
FA0
dX A
rA
C A0
dX A
rA
Here,
= 10 dm3/5 dm3/sec = 2 sec &
C A0
dX A
1 XA
kC A0
1 XA
y A0
1(1 1 1) 1
dX A
1 XA
kC A0
1 XA
5-58
0.8
dX A
1 XA
1 XA
2k
0
Integrating, we get
k = 1.21 sec-1
Case2: 10 dm3 CSTR with 80% conversion at 320K.
V
FA0
XA
rA
C A0
XA
rA
Here,
= 10 dm3/5 dm3/sec = 2 sec &
C A0
XA
1 XA
kC A0
1 XA
y A0
1(1 1 1) 1
XA
1 XA
kC A0
1 XA
XA
1 XA
1 XA
2k
Putting XA=0.8, we get
k = 2.8 sec-1
Now,
k (T2 )
k (T1 )
exp
k (T
k (T
Ea 1
(
R T1
320 K )
300 K )
1
)
T2
exp
Ea
1
1
(
)
1.986 300 320
Ea
2.8
1
1
exp
(
)
1.21
1.986 300 320
Ea
8000cal / mol
P5-29
For the turbulent flow
0,
y
0
1
W
1
2
5-59
Mole Balance/Design Equation
FA 0
dX
dW
rA
Rate Law
rA
kC A
2
Stoichiometry
CA
C A0 1 X y
Combining
FA0
dX
dW
kC A0 2 1 X
FA0
dX
2
kC A0 1 X 2
FA0
X
2
kC A0 1 X
1
W
2
y2
W dW
2
W2
Combining
~
2
DP1
1
DP 2
X
1 X
1
2
1
DP
9.975 10 3 kg 1
2
13.47 0.244
10
2
4.99 10 3 100
100
2
0.0802 100 24.95
X
4.99 10 3 kg 1
2
6.02
0.86
P5-30 Individualized solution
5-60
CDP5-B
a)
Polymath solution (Ans CDP5-B-a)
b)
5-61
Polymath solution(Ans CDP5-B-b)
c)
Polymath solution(Ans CDP5-B-c)
d) This part is almost same as part(b) with minor changes:
V = 15000000 – 10000t
vo = 80000 and vout = 70000
The reason the graph looks so
different from(a) is that pure water is
evaporated, but water with atrazine is
coming in.
Polymath code:
5-62
d(ca)/dt=cao*vo/v)-(ca(kv+vout)/v)
ca(0)=4.5
#
vo = 80000
#
v = 15000000 - 10000*t
k =
0.0025
#
cao = 4.5
#
vout = 70000
#
t(0)=0
#
t(f)=1000
Polymath solution(Ans CDP5-B-d)
CDP5-C
5-63
CD5RD-1 No solution
5-64
CD5RD-2
5-65
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these materials .
6-1
Solutions for Chapter 6 : Isothermal Reactor Design- Molar
Flow rates
P6-1 Individualized solution.
P6-2 (a) Example 6-1
For pressure doubled and temperature decrease
CTO = 2*Po/RT and T = 688K
See Polymath program P6-2-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
V
Fa
Fb
Fc
E
T
Cto
Ft
Ca
k
ra
Fao
rb
vo
rc
X
Tau
rateA
initial value
0
2.26E-04
0
0
2.4E+04
688
0.573773
2.26E-04
0.573773
213.40078
-70.254837
2.26E-04
70.254837
3.939E-04
35.127419
0
0
70.254837
minimal value
0
1.363E-05
0
0
2.4E+04
688
0.573773
2.26E-04
0.0236075
213.40078
-70.254837
2.26E-04
0.1189309
3.939E-04
0.0594654
0
0
0.1189309
maximal value
1.0E-04
2.26E-04
2.124E-04
1.062E-04
2.4E+04
688
0.573773
3.322E-04
0.573773
213.40078
-0.1189309
2.26E-04
70.254837
3.939E-04
35.127419
0.9395277
0.253044
70.254837
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(V) = ra
[2] d(Fb)/d(V) = rb
[3] d(Fc)/d(V) = rc
6-2
final value
1.0E-04
1.363E-05
2.124E-04
1.062E-04
2.4E+04
688
0.573773
3.322E-04
0.0236075
213.40078
-0.1189309
2.26E-04
0.1189309
3.939E-04
0.0594654
0.9395277
0.253044
0.1189309
Explicit equations as entered by the user
[1] E = 24000
[2] T = 688
[3] Cto = 2*1641/8.314/T
[4] Ft = Fa+Fb+Fc
[5] Ca = Cto*Fa/Ft
[6] k = 0.29*exp(E/1.987*(1/500-1/T))
[7] ra = -k*Ca^2
[8] Fao = 0.000226
[9] rb = -ra
[10] vo = Fao/Cto
[11] rc = -ra/2
[12] X = 1-Fa/Fao
[13] Tau = V/vo
[14] rateA = -ra
P6-2 (b) Example 6-2 Individualized solution.
P6-2 (c) Example 6-3
Using trial and error, we get maximum feed rate of B = 0.0251dm3/s to keep concentration of B
0.01mol/dm3.
See Polymath program P6-2-c.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
ca
cb
cc
cd
k
v00
cb0
initial value
0
0.05
0
0
0
0.22
0.0251
0.025
minimal value
0
0.0063485
0
0
0
0.22
0.0251
0.025
maximal value
500
0.05
0.009981
0.0078965
0.0078965
0.22
0.0251
0.025
6-3
final value
500
0.0063485
0.009981
0.0078965
0.0078965
0.22
0.0251
0.025
v0
5
5
ca0
0.05
0.05
rate
0
0
v
5
5
x
0
0
Differential equations as entered by the user
[1] d(ca)/d(t) = -k*ca*cb-v00*ca/v
[2] d(cb)/d(t) = -k*ca*cb+v00*(cb0-cb)/v
[3] d(cc)/d(t) = k*ca*cb-v00*cc/v
[4] d(cd)/d(t) = k*ca*cb-v00*cd/v
5
0.05
3.91E-05
17.55
0.5543321
Explicit equations as entered by the user
[1] k = .22
[2] v00 = 0.0251
[3] cb0 = 0.025
[4] v0 = 5
[5] ca0 = 0.05
[6] rate = k*ca*cb
[7] v = v0+v00*t
[8] x = (ca0*v0-ca*v)/(ca0*v0)
If the concentration of A is tripled the
maximum feed rate becomes
0.064 dm3/s
P6-2 (d through g) Individualized solution.
P6-3 Solution is in the decoding algorithm given with the modules.
P6-4 (a)
A
B 2C
CA= CA0
CA0 = PA0/RT =
k = 10-4exp[
Next we express the rate law as:
6-4
5
0.05
1.394E-05
17.55
0.5543321
rA
k CA
CBCC2
KC
At equilibrium,
C F
k T0 A
FT
CT 0
FT
3
FB FC2
KC
(1)
= 0,
Therefore, from (1), k[CA-
]=0
CA0(1-Xe) – CA0Xe*(2CA0Xe)2/Kc = 0
(1-Xe) - Xe*22* CA02* Xe2/ Kc = 0
(1-Xe) - Xe*22* 0.32* Xe2/ 0.025 = 0
(1-Xe) – 14.4Xe3 = 0 => Xe = 0.355
So, the equilibrium conversion obtained in a PFR is 35.5 %.
P6-4 (b)
dFB
dV
dFC
dV
rB
RB
rC
Transport out the sides of the reactor:
RB = kcCB =
kB CT 0 FB
FT
Stoichiometery:
-rA = rB =1/2 rC
6-5
POLYMATH Report
No Title
06-May-2009
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Co
0.3
0.3
0.3
0.3
2
Fa
2.5
0.0323558
2.5
0.0323558
3
Fb
0
0
0.0197934
0.0002897
4
Fc
0
0
4.935288
4.935288
5
Ft
2.5
2.5
4.967934
4.967934
6
K
0.044
0.044
0.044
0.044
7
Kc
0.025
0.025
0.025
0.025
8
kc
4.8
4.8
4.8
4.8
9
ra
-0.0132
-0.0132
-8.324E-05
-8.324E-05
10 Rb
0
0
0.0099094
8.398E-05
11 V
0
0
1500.
1500.
12 X
0
0
0.9870577
0.9870577
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = -ra - Rb
3 d(Fc)/d(V) = -2*ra
Explicit equations
1 Ft = Fa+Fb+Fc
2 K = 0.044
3 kc = 4.8
min-1
4 Kc = 0.025
5 Co = 0.3
6 X = (2.5-Fa)/2.5
7 Rb = kc*Co*Fb/Ft
8 ra = - (K*Co/Ft)*(Fa- Co^2*Fb*Fc^2/(Kc*Ft^2))
General
Total number of equations
11
Number of differential equations 3
Number of explicit equations
8
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
6-6
6-7
So, if the reactor volume is 800 dm3, exit conversion will be 0.92.
6-8
We see that volume required for 80% conversion is 467 dm3, therefore plotting the flow rates till 467
dm3
6-9
P6-5 (a)
Assume constant volume batch reactor
Mole balance: C A0
dX
dt
rA
Rate law and stoichiometry:
rA
kC A
kC A0 1 X
6-10
Specific reaction rate: k 25 C
0.0022 weeks 1
Combine:
X
dX
kC A0 1 X
0
1
ln 1 X 52.2 weeks
k
1
ln 1 X
0.0022 weeks 1
t
C A0
X
0.108
CA
C A0 1 X but since volume and molecular weight are constant the equation can be written as:
mA
mA0 1 X
6500 IU
mA0
mA0 1 0.108
7287 IU
C A0 C A
*100
CA
%OU
7287 6500
*100 12.1%
6500
P6-5 (b)
10,000,000 lbs/yr = 4.58 * 109 g/yr of cereal
Serving size = 30g
Number of servings per year = 4.58 * 109 / 30 =1.51 * 109 servings/yr
Each serving uses an excess of 787 IU = 4.62 * 10-4 = 1.02 * 10-6 lb
Total excess per year = (1.51 * 108 servings/yr) * (1.02 * 10-6 lbs/serving) = 154.11 lb/yr
Total overuse cost = $100/lb * 154.11 lb/yr = $15411 / yr (trivial cost)
P6-5 (c)
If the nutrients are too expensive, it could be more economical to store the cereal at lower
temperatures where nutrients degrade more slowly, therefore lowering the amount of overuse. The
cost of this storage could prove to be the more expensive alternative. A cost analysis needs to be done
to determine which situation would be optimal.
P6-5 (d)
k 40 C
X
t
C A0
0.0048 weeks 1
dX
kC A0 1 X
0
26weeks
6 months = 26 weeks
1
ln 1 X
k
1
ln 1 X
0.0048 weeks 1
X
0.12
CA
C A0 1 X but since volume and molecular weight are constant the equation can be written as:
6-11
mA
mA0 1 X
6500 IU
mA0
%OU
mA0 1 0.12
7386 IU
C A0 C A
*100
CA
7386 6500
*100 13.6%
6500
P6-6
3
Suppose the volumetric flow rate could be increased to as much as 6,000 dm /h (9,000 mol/h) and the
total time to fill, heat, empty and clean is 4.5 hours. What is the maximum number of moles of ethylene
glycol (CH2OH)2 you can make in one 24 hour period? The feed rate of ethylene cholorhydrin will be
adjusted so that the volume of fluid at the end of the reaction time will be 2500 dm3. Now suppose CO2
leaves the reactor as fast as it is formed.
A B
C D CO2
Mole Balance
dN A
rAV
dt
dN B
FB0
dt
dNC
rC V
dt
N D NC
rBV
Overall Mass Balance
Accumulation = In – Out
dm
ÝCO 2
0 m
dt
m
V Assume constant density
mÝCO 2
dV
dt
0
The rate of formation of CO2 is equal to the rate of formation of ethylene glycol (C).
mÝCO 2
dV
dt
CO 2
FCO
2
0
MWCO2
FCO
2
FCO
2
MWCO2
0
CO 2
MWCO2
6-12
Rate Law and Relative Rates
rA
kCA CB
rB rA
rC
rA
rCO 2
rA
Stoichiometry
NA
V
NB
V
CA
CB
dC A
dt
dC B
dt
dCC
dt
CD CC
dN CO 2
dt
FCO 2
CO 2 C A
0
rA
V
0C B0
CO 2 CB
0
V
V
0
rA
CO 2 CC
V
0
FCO 2
rAV
rAV
6-13
P6-6(c)
FA0
0.15 mol min
vo Fao /Ca
3
9 mol hr
9 mol hr / 1.5 mol dm 3
6 dm 3 hr
3
1000 dm is needed to fill the reactor. At 6 dm /hr it will take 166.67 hours
Now solving using the code form part (a) with the changed equations:
See Polymath program
6-14
P6-7
C B0
0.25
mol
dm3
V
C A0
0.2
mol
V0
dm3
VT
VT
D 2L
1.5m
4
4.42m
2
2.5m
4
3
4, 420 dm 3
We don’t know V0 or v0. First try equal number of moles of A and B added to react.
NA
C A0 V0
V0
NB
C B0 V C B0 VT
VT
4420
C A0
0.8 1.0
1
C B0
V0
2456
V 1964
V
0
k0
tR
3
0.000052 dm mol s
0.187 dm3 mol h
For 1 batch tR = 24 – 3 = 21
VT
0
V0
21
(1) For one batch we see that only 198 moles of C are made so one batch will not work.
(2) For two batches, we have a down time of 2 3 = 6 h and therefore each batch has a reaction time of
18h/2 = 9 h. We see that 106 moles of C are made in one batch therefore 2 106 = 212 moles/day
are made in two batches.
6-15
P6-7
6-16
P6-7
P6-8 (a)
A
B 2C
To plot the flow rates down the reactor we need the differential mole balance for the three species,
noting that BOTH A and B diffuse through the membrane
dFA
dV
dFB
dV
dFC
dV
rA
RA
rB
RB
rC
Next we express the rate law:
First-order reversible reaction
rA
k CA
CBCC2
KC
C F
k T0 A
FT
CT 0
FT
3
FB FC2
KC
6-17
Transport out the sides of the reactor:
RA = kACA =
k ACT 0 FA
FT
RB = kBCB =
kB CT 0 FB
FT
Stoichiometery:
-rA = rB =1/2 rC
Combine and solve in Polymath code:
See Polymath program P6-9-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
v
Fa
Fb
Fc
Kc
Ft
Co
K
Kb
ra
Ka
Ra
Rb
Fao
X
initial value
0
100
0
0
0.01
100
1
10
40
-10
1
1
0
100
0
minimal value
0
57.210025
0
0
0.01
100
1
10
40
-10
1
0.472568
0
100
0
maximal value
20
100
9.0599877
61.916043
0.01
122.2435
1
10
40
-0.542836
1
1
2.9904791
100
0.4278998
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(v) = ra - Ra
[2] d(Fb)/d(v) = -ra - Rb
[3] d(Fc)/d(v) = -2*ra
6-18
final value
20
57.210025
1.935926
61.916043
0.01
121.06199
1
10
40
-0.542836
1
0.472568
0.6396478
100
0.4278998
Explicit equations as entered by the user
[1] Kc = 0.01
[2] Ft = Fa+ Fb+ Fc
[3] Co = 1
[4] K = 10
[5] Kb = 40
[6] ra = - (K*Co/Ft)*(Fa- Co^2*Fb*Fc^2/(Kc*Ft^2))
[7] Ka = 1
[8] Ra = Ka*Co*Fa/Ft
[9] Rb = Kb*Co*Fb/Ft
P6-8 (b)
The setup is the same as in part (a) except there is no transport out the sides of the reactor.
See Polymath program P6-9-b.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
v
Fa
Fb
Fc
Kc
Ft
Co
K
ra
Fao
X
initial value
0
100
0
0
0.01
100
1
10
-10
100
0
minimal value
0
84.652698
0
0
0.01
100
1
10
-10
100
0
maximal value
20
100
15.347302
30.694604
0.01
130.6946
1
10
-3.598E-09
100
0.153473
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(v) = ra
[2] d(Fb)/d(v) = -ra
[3] d(Fc)/d(v) = -2*ra
6-19
final value
20
84.652698
15.347302
30.694604
0.01
130.6946
1
10
-3.598E-09
100
0.153473
Explicit equations as entered by the user
[1] Kc = 0.01
[2] Ft = Fa+ Fb+ Fc
[3] Co = 1
[4] K = 10
[5] ra = - (K*Co/Ft)*(Fa- Co^2*Fb*Fc^2/(Kc*Ft^2))
P6-8 (c) Conversion would be greater if C were diffusing out.
P6-8 (d) Individualized solution
P6-9
(a)
CO + H 2O
CO2 + H 2
A + B
C
+ D
Assuming catalyst distributed uniformly over the whole volume
Mole balance:
dFA
dW
Rate law:
r
RH 2
rA
dFB
dW
r
rB
rC
r
rD
dFC
dW
k C AC B
K H2 CD
6-20
r
CC CD
K eq
dFD
dW
r
RH 2
Stoichiometry: C A
CTO
FA
FT
FT
FA
FB
CB
FC
CTO
FB
FT
CC
CTO
FC
FT
FD
Solving in polymath:
See Polymath program P6-10.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
W
Fa
Fb
Fc
Fd
Keq
Ft
Cto
Ca
Cb
Kh
Cc
Cd
Rh
k
r
initial value
0
2
2
0
0
1.44
4
0.4
0.2
0.2
0.1
0
0
0
1.37
-0.0548
minimal value
0
0.7750721
0.7750721
0
0
1.44
3.3287437
0.4
0.0931369
0.0931369
0.1
0
0
0
1.37
-0.0548
maximal value
100
2
2
1.2249279
0.7429617
1.44
4
0.4
0.2
0.2
0.1
0.147194
0.0796999
0.00797
1.37
-0.002567
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(W) = r
[2] d(Fb)/d(W) = r
[3] d(Fc)/d(W) = -r
[4] d(Fd)/d(W) = - r -Rh
Explicit equations as entered by the user
[1] Keq = 1.44
[2] Ft = Fa+Fb+Fc+Fd
[3] Cto = 0.4
[4] Ca = Cto*Fa/Ft
[5] Cb = Cto*Fb/Ft
[6] Kh = 0.1
[7] Cc = Cto*Fc/Ft
[8] Cd = Cto*Fd/Ft
[9] Rh = Kh*Cd
[10] k = 1.37
[11] r = -k*(Ca*Cb-Cc*Cd/Keq)
For 85% conversion, W = weight of catalyst = 430 kg
6-21
final value
100
0.7750721
0.7750721
1.2249279
0.5536716
1.44
3.3287437
0.4
0.0931369
0.0931369
0.1
0.147194
0.0665322
0.0066532
1.37
-0.002567
CD
CTO
FD
FT
(b)
In a PFR no hydrogen escapes and the equilibrium conversion is reached.
K eq
CC CD
C AC B
C A2 0 X 2
C A2 0 (1 X ) 2
X2
1 X
2
1.44
solve this for X,
X = .5454
This is the maximum conversion that can be achieved in a normal PFR.
(c)
If feed rate is doubled, then the initial values of Fa and Fb are doubled. This results in a conversion of
.459
P6-10 Individualized solution
P6-11 (a)
At equilibrium, r = 0 => C A C B
CC C D
KC
V = V0 + vot
CA
N AO 1 X
V
C AO 1 X
V
VO
C AO 1 X
1
vo
t
VO
6-22
C BO
CB
vO
t C AO X
VO
1
CC
vo
t
VO
C AO X
CD
1
vo
t
VO
v
C AO 1 X C BO O t C AO X
VO
t
VO C AO
X2
vO C BO K C 1 X
2
X
200 * 7.72
X2
0.05 *10.93 1.08 1 X
t
C AO X
KC
X
Solving in Polymath
POLYMATH Report
Nonlinear Equation
06-May-2009
Calculated values of NLE variables
Variable Value f(x) Initial Guess
1 x
0
0
0.495 ( 0 < x < 0.99 )
Nonlinear equations
1 f(x) = 2825.25*(x^2/1.08/(1-x)+x) = 0
General Settings
Total number of equations
1
Number of implicit equations 1
Number of explicit equations 0
Elapsed time
0.0000 sec
Solution method
SAFENEWT
Max iterations
150
Tolerance F
0.0000001
Tolerance X
0.0000001
Tolerance min
0.0000001
6-23
If we solve in Excel
X
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
0.99
t (sec)
0
148.1466374
311.591358
493.0338235
695.8486111
924.3101852
1183.914286
1481.84765
1827.692593
2234.515909
2720.611111
3312.40216
4049.525
4994.264683
6250.42963
8004.875
10631.31111
15001.7287
23732.1
49902.28611
259188.435
6-24
P6-11 (b)
See Polymath program P6-12-b.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
Cc
Cd
Kc
k
initial value
0
7.72
10.93
0
0
1.08
9.0E-05
minimal value
0
0.2074331
7.6422086
0
0
1.08
9.0E-05
maximal value
1.5E+04
7.72
10.93
3.2877914
3.2877914
1.08
9.0E-05
6-25
final value
1.5E+04
0.2074331
9.51217
1.41783
1.41783
1.08
9.0E-05
ra
vo
Vo
V
X
-0.0075942
0.05
200
200
0
-0.0075942
0.05
200
200
0
-1.006E-05
0.05
200
950
0.9731304
-1.006E-05
0.05
200
950
0.9731304
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra - Ca*vo/V
[2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93)
[3] d(Cc)/d(t) = -ra - vo*Cc/V
[4] d(Cd)/d(t) = -ra - vo*Cd/V
Explicit equations as entered by the user
[1] Kc = 1.08
[2] k = 0.00009
[3] ra = -k*(Ca*Cb - Cc*Cd/Kc)
[4] vo = 0.05
[5] Vo = 200
[6] V = Vo + vo*t
[7] X = 1 - Ca/7.72
Polymath solution
P6-11 (c)
Change the value of vo and CAO in the Polymath program to see the changes.
P6-11 (d)
As ethanol evaporates as fast as it forms:
Now using part (b) remaining equations,
Polymath code:
CD = 0
See Polymath program P6-12-d.pol.
6-26
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
k
ra
vo
Vo
V
X
initial value
0
7.72
10.93
9.0E-05
-0.0075942
0.05
200
200
0
minimal value
0
0.0519348
6.9932872
9.0E-05
-0.0075942
0.05
200
200
0
maximal value
6000
7.72
10.93
9.0E-05
-3.69E-05
0.05
200
500
0.9932727
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra - Ca*vo/V
[2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93)
Explicit equations as entered by the user
[1] k = 0.00009
[2] ra = -k*Ca*Cb
[3] vo = 0.05
[4] Vo = 200
[5] V = Vo + vo*t
[6] X = 1 - Ca/7.72
P6-11 (e) Individualized solution
P6-11 (f) Individualized solution
6-27
final value
6000
0.0519348
7.8939348
9.0E-05
-3.69E-05
0.05
200
500
0.9932727
P6-12
CA0,
CA0,
CSTR
A
B
CA0(1-X),
(rejected)
-rA = k[ CA – CB/Ke)
-rA = k[ CA0(1-X)- CA0X/Ke)
Given :
k = 0.4 h-1
Ke = 4
V = 60 m3
CA0= 100 kg/m3
6-28
= 12 m3/hr
FA0= 100*12 = 1200 kg/hr
FA20= FA0 + CA0
= FA0 (1+ f) = 1200(1+ f)
CSTR mole balance:
=
=
Putting the values, we have
=
(1)
Solving we get,
X=
Profit = Value or products – Operating Cost
Profit = $2*CA0*X
(1+f) - $50
(1+f)
Profit = $ 600*(1+f)*(4X-1)
(2)
Now, solving equation (1) in excel and corresponding finding out the vale or profit from
equation (2), we have
f
X
Profit($)
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.571429
0.56338
0.555556
0.547945
0.540541
0.533333
0.526316
0.519481
0.512821
0.506329
0.5
0.493827
771.4286
789.7183
806.6667
822.3288
836.7568
850
862.1053
873.1169
883.0769
892.0253
900
907.037
6-29
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
0.487805
0.481928
0.47619
0.470588
0.465116
0.45977
0.454545
0.449438
0.444444
913.1707
918.4337
922.8571
926.4706
929.3023
931.3793
932.7273
933.3708
933.3333
0.9
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
1
0.454545
0.453515
0.452489
0.451467
0.45045
0.449438
0.44843
0.447427
0.446429
0.445434
0.444444
932.7273
932.9116
933.0679
933.1964
933.2973
933.3708
933.417
933.4362
933.4286
933.3942
933.3333
0.96
0.961
0.962
0.963
0.964
0.965
0.966
0.967
0.968
0.969
0.97
0.971
0.972
0.44843
0.44833
0.448229
0.448129
0.448029
0.447928
0.447828
0.447728
0.447628
0.447527
0.447427
0.447327
0.447227
0.447127
0.447027
0.446927
0.446828
0.446728
0.446628
0.446528
0.446429
933.417
933.4202
933.423
933.4256
933.428
933.43
933.4318
933.4333
933.4346
933.4355
933.4362
933.4367
933.4369
933.4368
933.4364
933.4358
933.4349
933.4337
933.4322
933.4305
933.4286
0.973
0.974
0.975
0.976
0.977
0.978
0.979
0.98
6-30
So we see the maximum profit ($933.4369) occurs for f=0.972
Corresponding conversion of A, X = 0.4472
P6-13
A + B -> C
6-31
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 t
0
0
200.
200.
2 Cc
2.
2.
3.893617
3.893617
3 Cb
2.
0.106383
2.
0.106383
4 Ca
2.
0.106383
2.
0.106383
5 k
0.0445
0.0445
0.0445
0.0445
6 Cao
2.
2.
2.
2.
7 ra
-0.178
-0.178
-0.0005036
-0.0005036
8 rb
-0.178
-0.178
-0.0005036
-0.0005036
9 X
0
0
0.9468085
0.9468085
10 rc
0.178
0.0005036
0.178
0.0005036
Differential equations
1 d(Cc)/d(t) = rc
2 d(Cb)/d(t) = rb
3 d(Ca)/d(t) = ra
Explicit equations
1 k = 0.0445
6-32
2 Cao = 2
3 ra = -k * Ca * Cb
4 rb = ra
5 X = (Cao - Ca) / Cao
6 rc = -ra
P6-14
The 4 mistakes are as follows:
1) rb = - k * Ca2
2) CA0 = 0.4 and not CTO
3)
CTO = 0.8
4) The 5th equation should have been
alpha FT
2 y FT 0
dy
dW
CDGA 6-1 (a)
At equilibrium, r = 0 => C A C B
CC C D
KC
V = V0 + vot
CA
N AO 1 X
V
C BO
CB
C AO 1 X
V
VO
C AO 1 X
1
vo
t
VO
vO
t C AO X
VO
1
vo
t
VO
6-33
CC
CD
C AO X
1
vo
t
VO
v
C AO 1 X C BO O t C AO X
VO
t
VO C AO
X2
vO C BO K C 1 X
C AO X
KC
2
X
Now solving in polymath,
See Polymath program CDGA6-1-a.pol.
CDGA 6-1 (b)
See Polymath program CDGA6-1-b.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
Cc
Cd
Kc
k
ra
vo
Vo
V
X
initial value
0
7.72
10.93
0
0
1.08
9.0E-05
-0.0075942
0.05
200
200
0
minimal value
0
0.2074331
7.6422086
0
0
1.08
9.0E-05
-0.0075942
0.05
200
200
0
maximal value
1.5E+04
7.72
10.93
3.2877914
3.2877914
1.08
9.0E-05
-1.006E-05
0.05
200
950
0.9731304
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra - Ca*vo/V
[2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93)
[3] d(Cc)/d(t) = -ra - vo*Cc/V
[4] d(Cd)/d(t) = -ra - vo*Cd/V
Explicit equations as entered by the user
[1] Kc = 1.08
[2] k = 0.00009
[3] ra = -k*(Ca*Cb - Cc*Cd/Kc)
[4] vo = 0.05
[5] Vo = 200
6-34
final value
1.5E+04
0.2074331
9.51217
1.41783
1.41783
1.08
9.0E-05
-1.006E-05
0.05
200
950
0.9731304
[6] V = Vo + vo*t
[7] X = 1 - Ca/7.72
Polymath solution
CDGA 6-1 (c)
Change the value of vo and CAO in the Polymath program to see the changes.
CDGA 6-1 (d)
As ethanol evaporates as fast as it forms:
Now using part (b) remaining equations,
Polymath code:
CD = 0
See Polymath program CDGA6-1-d.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
k
ra
vo
Vo
V
X
initial value
0
7.72
10.93
9.0E-05
-0.0075942
0.05
200
200
0
minimal value
0
0.0519348
6.9932872
9.0E-05
-0.0075942
0.05
200
200
0
maximal value
6000
7.72
10.93
9.0E-05
-3.69E-05
0.05
200
500
0.9932727
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra - Ca*vo/V
[2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93)
Explicit equations as entered by the user
6-35
final value
6000
0.0519348
7.8939348
9.0E-05
-3.69E-05
0.05
200
500
0.9932727
[1] k = 0.00009
[2] ra = -k*Ca*Cb
[3] vo = 0.05
[4] Vo = 200
[5] V = Vo + vo*t
[6] X = 1 - Ca/7.72
CDGA 6-1 (e) Individualized solution
CDGA 6-1 (f) Individualized solution
CDGA 6-2 (a)
Mole balance on reactor 1:
C A0 v A0 C A1v rA1V
C A0
v0 C A1v rA1V
2
dN A1
with v A0
dt
dN A1
dt
1
v0
2
Liquid phase reaction so V and v are constant.
C A0
2
C A1
rA1
dC A1
dt
Mole balance on reactor 2:
C A1v0 C A 2 v0
C A1
C A2
rA 2
rA 2V
dN A 2
dt
dC A 2
dt
6-36
Mole balance for reactor 3 is similar to reactor 2:
C A 2 v0 C A3v0
C A2
C A3
rA3V
rA3
dN A3
dt
dC A3
dt
Rate law:
rAi
kC Ai CBi
Stoichiometry
For parts a, b, and c CAi = CBi
so that
rAi
kC Ai2
Combine:
C A0
2
C A1
C A1
kC A21
C A2
kC A2 2
C A2
C A3
kC A2 3
dC A1
dt
dC A 2
dt
dC A3
dt
See Polymath program CDGA6-2-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca1
Ca2
Ca3
k
Cao
tau
X
initial value
0
0
0
0
0.025
2
10
1
minimal value
0
0
0
0
0.025
2
10
0.3890413
maximal value
100
0.8284264
0.7043757
0.6109587
0.025
2
10
1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca1)/d(t) = (Cao/2 -Ca1)/tau -k*Ca1^2
[2] d(Ca2)/d(t) = (Ca1 - Ca2)/tau -k*Ca2^2
[3] d(Ca3)/d(t) = (Ca2 - Ca3)/tau -k*Ca3^2
6-37
final value
100
0.8284264
0.7043757
0.6109587
0.025
2
10
0.3890413
Explicit equations as entered by the user
[1] k = 0.025
[2] Cao = 2
[3] tau = 10
[4] X = 1 - 2*Ca3/Cao
From Polymath, the steady state conversion of A is approximately 0.39
CDGA 6-2 (b)
99% of the steady state concentration of A (the concentration of A leaving the third reactor) is:
(0.99)(0.611) = 0.605
This occurs at t =
CDGA 6-2 (c)
The plot was generated from the Polymath program given above.
CDGA 6-2 (d)
We must reexamine the mole balance used in parts a-c. The flow rates have changed and so the mole
balance on species A will change slightly. Because species B is added to two different reactors we will
also need a mole balance for species B.
Mole balance on reactor 1 species A:
dN A1
with v A0
dt
dN A1
C A1v rA1V
dt
C A0 v A0 C A1v rA1V
2C A0
v0
3
2
v0 and v0
3
Liquid phase reaction so V and v are constant.
6-38
200
15
C A0
2
C A1
rA1
dC A1
dt
Mole balance on reactor 1 species B:
dN B1
and vB 0
dt
CB 0 vB 0 CB1v rB1V
1
v0
3
Stoichiometry has not changed so that –rAi = -rBi and it is a liquid phase reaction with V and v constant.
CB 0
3
CB1
rA1
dCB1
dt
Mole balance on reactor 2 species A:
We are adding more of the feed of species B into this reactor such that v2 = v0 + vB0 = 20
C A1v0 C A 2 v2
C A1
1
C A2
rA2
2
rA 2V
dN A 2
dt
dC A2
where
dt
1
V
and
v0
2
V
v2
Mole balance on reactor 2 species B:
dN B 2
dt
CB1v0 CB 0 vB 0 CB 2 v rB 2V
CB1
1
C B 0 vB 0
V
CB 2
rA2
2
dCB 2
dt
Mole balance for reactor 3 species A:
C A 2 v2 C A3v2
C A2
C A3
2
2
rA3
rA3V
dN A3
dt
dC A3
dt
Mole balance for reactor 3 species B:
CB 2 v2 CB 3v2
CB 2
CB 3
2
2
rA3
rA3V
dN B 3
dt
dCB 3
dt
6-39
Rate law:
rAi
kC Ai CBi
See Polymath program CDGA6-2-d.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca1
Ca2
Ca3
Cb1
Cb2
Cb3
k
Cao
tau
X
tau2
V
vbo
initial value
0
0
0
0
0
0
0
0.025
2
13.333333
1
10
200
5
minimal value
0
0
0
0
0
0
0
0.025
2
13.333333
0.3721856
10
200
5
maximal value
100
1.1484734
0.7281523
0.6278144
0.4843801
0.7349863
0.6390576
0.025
2
13.333333
1
10
200
5
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca1)/d(t) = (2*Cao/3 -Ca1)/tau -k*Ca1*Cb1
[2] d(Ca2)/d(t) = Ca1/tau - Ca2/tau2 -k*Ca2*Cb2
[3] d(Ca3)/d(t) = (Ca2 - Ca3)/tau2 -k*Ca3*Cb3
[4] d(Cb1)/d(t) = (1*Cao/3-Cb1)/tau-k*Ca1*Cb1
[5] d(Cb2)/d(t) = Cb1/tau+Cao*vbo/V-Cb2/tau2-k*Ca2*Cb2
[6] d(Cb3)/d(t) = (Cb2-Cb3)/tau2-k*Ca3*Cb3
Explicit equations as entered by the user
[1] k = 0.025
[2] Cao = 2
[3] tau = 200/15
[4] X = 1 - 2*Ca3/Cao
[5] tau2 = 10
[6] V = 200
[7] vbo = 5
Equilibrium conversion is 0.372.
This conversion is reached at t = 85.3 minutes.
6-40
final value
100
1.1484734
0.7281523
0.6278144
0.4821755
0.7291677
0.6309679
0.025
2
13.333333
0.3721856
10
200
5
CDGA 6-2 (e) Individualized solution
6-41
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7-1
Solutions for Chapter 7 – Collection and Analysis of Rate Data
P7-1 (a) Individualized solution
P7-1 (b) Individualized solution
P7-1 (c) Individualized solution
P7-1 (d) Individualized solution
P7-2 (a) In Example 7-1, we have assumed CB as constant, but CB is varying and the error is due to the
term
and the maximum error occurs when this ratio is minimum.
Error = 1 CB = CA0(Θ - X)
Θ = 0.5/0.05 = 10
CB = CA0(10 - X)
So, we have
t(min)
0
50
100
150
200
250
300
CA
X
CB
error(%)
0.05
0.038
0.0306
0. 256
0.0222
0.0195
0.0174
0
0.024
0.0388
0.0488
0.0556
0.061
0.0652
0.5
0.4988
0.49806
0.49756
0.49722
0.49695
0.49674
0
0.24
0.388
0.488
0.556
0.61
0.652
So, we have maximum error of 0.652%, hence
k = 0.24
.652*0.24/100
k = 0.24
0.0015648
P7-2 (b) Example 7-3 because when α is set equal to 2, the best value of k must be found.
P7-2 (c) Example 7-4


rate law: rCH 4  kPCO PH 2
7-2
Regressing the data
r’(gmolCH4/gcat.min)
5.2e-3
13.2e-3
30e-3
4.95e-3
7.42e-3
5.25e-3
PCO (atm)
1
1.8
4.08
1
1
1
PH2 (atm)
1
1
1
0.1
0.5
4
See Polymath program P7-2-c.pol.
POLYMATH Results
Nonlinear regression (L-M)
Model: r = k*(PCO^alfa)*(PH2^beta)
Variable
Ini guess
Value
k
0.1
0.0060979
alfa
1
1.1381455
beta
1
0.0103839
Precision
R^2
= 0.9869709
R^2adj
= 0.9782849
Rmsd
= 4.176E-04
Variance = 2.093E-06
95% confidence
6.449E-04
0.0850634
0.1951217
Therefore order of reaction = 1.14
Again regressing the above data putting   1
POLYMATH Results
Nonlinear regression (L-M)
Model: r = k*(PCO^0.14)*(PH2)
Variable
Ini guess
Value
k
0.1
0.0040792
Precision
R^2
= -0.8194508
R^2adj
= -0.8194508
Rmsd
= 0.0049354
Variance = 1.754E-04
95% confidence
0.0076284
Therefore, k = 0.004 (gmolCH4/(gcat.min.atm1.14))
P7-3 Solution is in the decoding algorithm given with the modules.
P7-4 Individualized solution
7-3
P7-5 (a)
P7-5 (b)
First we fit a polynomial to the data. Using Polymath we use regression to find an expression for X(z)
See Polymath program P7-5-b.pol.
POLYMATH Results
Polynomial Regression Report
Model: X = a0 + a1*z + a2*z^2 + a3*z^3 + a4*z^4 + a5*z^5 + a6*z^6
Variable
Value
95% confidence
7-4
a0
a1
a2
a3
a4
a5
a6
2.918E-14
0.0040267
-6.14E-05
7.767E-06
-5.0E-07
1.467E-08
-1.6E-10
0
0
0
0
0
0
0
General
Order of polynomial = 6
Regression including free parameter
Number of observations = 7
Statistics
R^2 =
R^2adj =
Rmsd =
Variance =
1
0
1.669E-10
1.0E+99
Next we differentiate our expression of X(z) to find dX/dz and knowing that
 dX 
ln 
  ln a  n ln 1  X 
 dz 
n
kC Ao
Ac
where, a 
FAo
 dX 
 as a function of ln 1  X  gives us similar vaules of slope and intercept
 dz 
Linear regression of ln 
as in the finite differences.
POLYMATH Results
Linear Regression Report
Model: ln(dxdz) = a0 + a1*ln(1-X)
Variable
a0
a1
Value
-5.531947
1.2824279
95% confidence
0.0241574
0.3446187
General
Regression including free parameter
Number of observations = 7
Statistics
R^2 =
R^2adj =
Rmsd =
Variance =
0.9482059
0.9378471
0.0044015
1.899E-04
n = 1.28
ln a = -5.53, a = 0.00396
7-5
k
FAO a
45.7 106 moles / s

3.96 103 cm  4.0s 1
n
6
3
2 
CAO AC 2.3 10 moles / cm  0.0196cm
Hence rate law is,
 rA  4.0 C 1.28
A
mol
dm3 s
P7-6 (a)
Liquid phase irreversible reaction:
A  B + C ; CAO = 2 mole/dm3
C AO  C A
 kC A

 C  CA 
ln  AO
  ln k   ln C A



Space time (  )min. CA(mol/dm3) ln(CA)
15
38
100
300
1200
1.5
1.25
1.0
0.75
0.5
0.40546511
0.22314355
0
-0.28768207
-0.69314718
ln((CAO-CA)/  )
-3.4011974
-3.9252682
-4.6051702
-5.4806389
-6.6846117
By using linear regression in polymath:
See Polymath program P7-6-a.pol.
POLYMATH Results
Linear Regression Report
Model: y = a0 + a1*lnCa
 C  CA 
ln  AO
  ln k   ln C A



Variable
Value
95% confidence
7-6
a0
-4.6080579
0.0162119
a1
2.9998151
0.0411145
Statistics
R^2 =
0.9999443
R^2adj =
0.9999258
Rmsd =
0.003883
Variance =
1.256E-04
Hence,
  slope  3
ln(k) = intercept = -4.6
therefore, k = 0.01 mole-2min-1.
Rate law:

dC A
 0.01C A3 mol / dm3 min
dt
P7-6 (b) Individualized solution
P7-6 (c) Individualized solution
P7-7 (a)
Constant voume batch reactor:
Mole balance:

A B +C
dC A
 kC A
dt
Integrating with initial condition when t = 0 and CA = CAO for   1.0
t
(1 )
 C A(1 ) 1 (2)(1 )  C A(1 )
1 C AO

……………substituting for initial concentration CAO = 2 mol/dm3.
k
(1   )
k
(1   )
t (min.)
0
5
9
15
22
30
40
60
CA (mol/dm3)
2
1.6
1.35
1.1
0.87
0.70
0.53
0.35
See Polymath program P7-7-a.pol.
POLYMATH Results
Nonlinear regression (L-M)
Model: t = (1/k)*((2^(1-alfa))-(Ca^(1-alfa)))/(1-alfa)
Variable
Ini guess
Value
95% confidence
k
0.1
0.0329798
3.628E-04
7-7
alfa
2
Precision
R^2 = 0.9997773
R^2adj = 0.9997327
Rmsd = 0.1007934
Variance = 0.0995612
1.5151242
0.0433727
K= 0.03 (mol/dm3)-0.5.s-1 and   1.5
Hence , rate law is

dC A
3
 0.03C 1.5
A mol / dm .s
dt
P7-7 (b) Individualized solution
P7-7 (c) Individualized solution
P7-7 (d) Individualized solution
P7-8 (a)
Liquid phase reaction of methanol and triphenyl in a batch rector.
CH3OH + (C6H5)3CCl  (C6H5)3COCH3 + HCl
A +
B

C
+
D
Using second set of data, when CAO = 0.01 mol/dm3 and CBO = 0.1 mol/dm3
CA(mol/dm3)
0.1
0.0847
0.0735
0.0526
0.0357
t (h)
0
1
2
5
10
Rate law:
 rA  kC Am C Bn
n
For table 2 data: CAO  CBO =>  rA  k ' C Am where k '  kCBO
(1 m )
 C A(1 m ) 1 (0.01)(1m )  C A(1m )
1 C AO

k'
(1  m)
k'
(1  m)
See Polymath program P7-8-a-1.pol.
Using eqn E7-3.1, t 
POLYMATH Results
Nonlinear regression (L-M)
Model: t = (1/k)*((0.1^(1-m))-(Ca^(1-m)))/(1-m)
Variable
k
m
Ini guess
1
2
Value
1.815656
2.0027694
95% confidence
0.0109025
0.0021115
7-8
Nonlinear regression settings
Max # iterations = 64
Precision
R^2
R^2adj
Rmsd
Variance
= 1
= 0.9999999
= 3.268E-04
= 8.902E-07
Therefore, m = 2
For first set of data, equal molar feed => CA = CB
Hence, rate law becomes rA  kC A2CBn  kC A(2 n )
Observation table 2: for CA0 =0.01 and CB0 = 0.1
t (h)
0
0.278
1.389
2.78
8.33
16.66
t
CA(mol/dm3)
1.0
0.95
0.816
0.707
0.50
0.37
(1 (2  n ))
 C A(1(2 n )) 1 (0.1)( 1n )  C A( 1n )
1 C AO

k
(1  (2  n))
k
(1  n))
See Polymath program P7-8-a-2.pol.
POLYMATH Results
Nonlinear regression (L-M)
Model: t = (1^(-1-n)-Ca^(-1-n))/(k*(-1-n))
Variable
n
k
Ini guess
3
2
Value
0.8319298
0.1695108
95% confidence
0.0913065
0.0092096
Nonlinear regression settings
Max # iterations = 64
Precision
R^2
R^2adj
Rmsd
Variance
= 0.9999078
= 0.9998848
= 0.0233151
= 0.0048923
Therefore, n = 0.8
7-9
Hence rate law is:
 rA  0.17 C A2 CB0.8
mol
dm3 h
P7-8 (b) Individualized solution
P7-9 (a)
At t = 0, there is only (CH3)2O. At t = ∞, there is no(CH3)2O. Since for every mole of (CH3)2O consumed
there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure.
P(∞) = 3P0
931 = 3P0
P0 ≈ 310 mm Hg
P7-9 (b)
Constant volume reactor at T = 504°C = 777 K
Data for the decomposition of dimethylether in a gas phase:
Time
PT(mm Hg)
0
312
390
408
777
488
1195
562
(CH 3 ) 2 O  CH 4  H 2  CO
y A0  1
  3 1  2
   y A0  2
P 
V  V0  0  1   X   V0 because the volume is constant.
P
P  P0 1   X 
at t = ∞, X = XAF = 1
N dX
1 dN A
  A0
 rA
V dt
V0 dt
Assume  rA  kC A (i.e. 1st order)
C A  C A0 1  X  (V is constant)
dX
 kC A0 1  X 
dt
P  P0
and X 
 P0
Then: C A0
7-10
3155
799

931
Therefore:
dX
1 dP

dt  P0 dt
 P  P0 
1 dP
k
 k 1 
1   P0  P 

 P0 dt
 P0   P0

or
dP
 k 1   P0  P 
dt
P
t
dP
P 1   P0  P  0 kdt
0


 2 P0 
 P0
 624 
 kt
  ln 
  ln 
 936  P 
 3P0  P 
 1    P0  P 
Integrating gives: ln 
Therefore, if a plot of ln
624
versus time is linear, the reaction is first order. From the figure below,
936  P
we can see that the plot is linear with a slope of 0.00048.
Therefore the rate law is:
 rA  0.00048C A
1.6
y = 0.00048x - 0.02907
1.2
0.8
0.4
0
-0.4
0
1000
2000
3000
4000
P7-9 (c) Individualized solution
P7-9 (d) The rate constant would increase with an increase in temperature. This would result in the
pressure increasing faster and less time would be need to reach the end of the reaction. The opposite is
true for colder temperatures.
7-11
P7-10 (a)
Photochemical decay of bromine in bright sunlight:
t (min)
CA (ppm)
10
2.45
20
1.74
30
1.23
40
0.88
50
0.62
60
0.44
Mole balance: constant V
dC A
 rA  kC A
dt
 dC 
ln   A   ln k    ln C A 
 dt 
Differentiation
T (min)
Δt (min)
CA (ppm)
ΔCA (ppm)
C A  ppm 


t  min 
10
20
10
2.45
30
10
1.74
40
10
1.23
50
10
0.88
60
10
0.62
-0.71
-0.51
-0.35
-0.26
-0.18
-0.071
-0.051
-0.035
-0.026
-0.018
7-12
10
0.44
After plotting and differentiating by equal area
-dCA/dt
ln(-dCA/dt)
ln CA
0.082
-2.501
0.896
0.061
-2.797
0.554
0.042
-3.170
0.207
0.030
-3.507
-0.128
0.0215
-3.840
-0.478
0.014
-4.269
-0.821
Using linear regression: α = 1.0
ln k = -3.3864
k = 0.0344 min-1
P7-10 (b)
dN A
 VrA  FB  0
dt
ppm
mg
at CA = 1 ppm
rA  0.0344
 0.0344
min
l min


mg   min   1 g  3.7851l  1lbs 
lbs
FB  25000 gal  0.0344
  60


  0.426

l min  
hr   1000 mg  1 gal  453.6 g 
hr

P7-10 (c) Individualized solution
P7-11
For the reaction,
Oz + wall  loss of O3
k1
Oz + alkene  products
k2
rO z 
Rate law:
dCO z
dt
m n
 k 1CO z  k 2CO
C
z Bu
7-13

Using polymath nonlinear regression we can find the values of m, n, k1 and k2
See Polymath program P7-11.pol.
7-14
P7-12
Given: Plot of percent decomposition of NO2 vs V/FA0
% Decomposition of NO 2
100
Assume that  rA  kC An
X=
For a CSTR V 
or
FA0 X
rA
V
X
X


FA0 rA kC An
with n = 0, X  k
V
FA0
X has a linear relationship with
V
as
FA0
shown in the figure.
7-15
Therefore the reaction is zero order.
P7-13
SiO2  6 HF  H 2 SiF6  2 H 2O
NS = moles of SiO 2 =
A C ρS δ
MWS
AC = cross-sectional area
ρS = silicon dioxide density
MWS = molecular weight of silicon dioxide = 60.0
δ = depth of Si
N F = moles of HF=
w V
100 MWF
w = weight percentage of HF in solution
ρ = density of solution
V = volume of solution
MWF = molecular weight of HF = 20.0
Assume the rate law is rS  kCF
Mole balance:
dN S
 rSV
dt



A  d
wV
 C S
k
 V
MWS dt
 100V MWF 

kMWS   
d




 Vw

dt 100 AC  S  MWF 

kMWS   
d

  w where  

 V
dt
100 AC  S  MWF 
 d 
ln  
  ln    ln w
 dt 
 d 
ln  

 dt 
ln w
-16.629
-15.425
-14.32
-13.816
-13.479
2.079
2.996
3.497
3.689
3.871
7-16
m
 d 
 is in min
 dt 
where  
 d 
 and ln w we have:
 dt 
From linear regression between ln  
slope = α = 1.775
intercept = ln β = -20.462 or β = 1.2986 * 10-9
1.775
  
kMWS



1001.775 AC  S  MWF 

V

AC  10*106 m 10m 2sides 1000 wafers   0.2 m2
MWS  60
 S  2.32
g
gmol
g
g
 2.32*106 3
ml
m
(Handbook of Chemistry and Physics, 57th ed., p.B-155)
g
g
 106 3
ml
m
g
MWF  20
gmol
 1
V  0.5 dm3  0.0005 m3
  1.2986*109
1.775
 6 g 
 10 3 
9
m 
1.2986*10  
 20 g 
 gmol 



g 
k  60

 gmol 
0.0005 m3
g 
1.775

0.2 m2  2.3*106 3  100 
m 




7-17

 m3 
k  3.224*10 

 gmol 
0.775
7
Final concentration of HF 
min 1
5  2.316
0.2   0.107
5
weight fraction = 10.7%
Initial concentration of HF = 0.2 (given)
Mole balance for HF:
weight fraction = 20%
dN
dN F
6 S
dt
dt



dw
w

 6k 
 V
100MWF dt
 100 MWF 
V


dw

  1.775  6k 

w
 100MWF 
20
10.7
10.7
where α = 1.775
0.775 t
 dt
0
1  1 
106 
7 

6
3.224*10




0.775  w0.775  20
 20*100 


0.775
t
1 
1
1 
 0.775   2.389*104 t

0.775
0.775  10.7
20

t  331 min
P7-14
See Polymath program P7-14.pol
7-18
P7-15
The following are the mistakes in the given solution, followed by the corrected solution:-
AB+C
For a batch Reactor,
dC A
 rA  kC A
dt
dC
 A  kC A
dt
Taking ln both sides,
ln(
dC A
)  ln k   ln C A
dt
7-19
(1) The value of △CA/△t cannot be calculated for t=0. It can only be calculated for t=1 onwards.
(2) The values of △CA/△t calculated for t=1, 2, 3 etc. have been incorrectly calculated.
t
Ca
ln Ca
Delta t
delta Ca/ delta t ln ( -delta Ca/ delta t)
0
1
0
1
0.5
-0.7
1
-0.5
-0.7
2
0.3
-1.2
1
-0.2
-1.61
3
0.2
-1.61
1
-0.1
-2.3
4
0.15
-1.9
1
-0.05
-3
(3) The graph has to be plotted for ln (-△CA/△t) vs. ln (CA) and not ‘t’.
(4) So, the value of α from the graph = 1.88, not 0.5.
(5) ln k = 0.64 from the graph, which gives
k = 1.89
Therefore, we get
dC A
 rA  1.89C1.88
A
dt
7-20
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materials .
8-1
Solutions for Chapter 8 – Multiple Reactions
P8-1 Individualized solution
P8-2 (a) Example 8-1
For PFR (gas phase with no pressure drop or liquid phase),
dC A
d
dC B
d
k1
k 2C A
dC X
d
dCY
d
k 3C A2
k 2C A
k1
k 3 C A2
In PFR with V = 1566 dm3 we get
τ = V/v0 = 783
X = 0.958 , SB/XY (instanteneous) = 0.244, and SB/XY (overall) = 0.624
also at τ = 350, SB/XY (instanteneous) is at its maximum value of 0.84
See polymath problem P8-2-1.pol
Calculated values of the DEQ variables
Variable initial value
tau
0
Ca
0.4
Cx
1.0E-07
Cb
0
Cy
1.0E-06
Cao
0.4
X
0
k1
1.0E-04
k2
0.0015
k3
0.008
Sbxy_over
0
Sbxy_inst
0.4347826
minimal value
0
0.0166165
1.0E-07
0
1.0E-06
0.4
0
1.0E-04
0.0015
0.008
0
0.2438609
maximal value
783
0.4
0.0783001
0.1472919
0.1577926
0.4
0.9584587
1.0E-04
0.0015
0.008
0.6437001
0.8385161
8-2
final value
783
0.0166165
0.0783001
0.1472919
0.1577926
0.4
0.9584587
1.0E-04
0.0015
0.008
0.6238731
0.2438609
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(tau) = -k1-k2*Ca-k3*Ca^2
[2] d(Cx)/d(tau) = k1
[3] d(Cb)/d(tau) = k2*Ca
[4] d(Cy)/d(tau) = k3*Ca^2
Explicit equations as entered by the user
[1] Cao = 0.4
[2] X = 1-Ca/Cao
[3] k1 = 0.0001
[4] k2 = 0.0015
[5] k3 = 0.008
[6] Sbxy_overall = Cb/(Cx+Cy)
[7] Sbxy_instant = k2*Ca/(k1+k3*Ca^2)
Independent variable
variable name : tau
initial value : 0
final value : 783
(2) PFR - Pressure increased by a factor of 100.
(a) Liquid phase: No change, as pressure does not change the liquid volume appreciably.
(b) Gas Phase:
Now CA0 = P/RT = 100 (P0_initial)/RT = 0.4 × 100 = 40 mol/dm3
Running with CA0 = 40 mol/dm3, we get:
X = 0.998 , SB/XY (instanteneous) = 0.681, and SB/XY (overall) = 0.024
As a practice, students could also try similar problem with the CSTR.
P8-2 (b) Example 8-2
(a) CSTR: intense agitation is needed, good temperature control.
8-3
(b) PFR: High conversion attainable, temperature control is hard – non-exothermic reactions, selectivity
not an issue
(c) Batch: High conversion required, expensive products
(d) and (e) Semibatch: Highly exothermic reactions, selectivity i.e. to keep a reactant concentration low,
to control the conversion of a reactant.
(f) and (g) Tubular with side streams: selectivity i.e. to keep a reactant concentration high, to achieve
higher conversion of a reactant.
(h) Series of CSTR’s: To keep a reactant concentration high, easier temperature control than single CSTR.
(i) PFR with recycle: Low conversion to reuse reactants, gas reactants, for highly exothermic reactions
(j) CSTR with recycle: Low conversions are achieved to reuse reactants, temperature control, liquid
reactants , for highly exothermic reactions
(k) Membrane Reactor: yield i.e. series reactions that eliminate a desired product , used for
thermodynamically limited reactions where the equilibrium shifts on left side.
(l) Reactive Distillation: when one product is volatile and the other is not ; for thermodynamically limited
gas phase reactions.
P8-2 (c) Example 8-3
(1) For k1 = k2, we get
CA
k1 '
C AO exp
and
d
d C B exp k1
'
dCB
d '
Wopt
CB
0
'
opt
vo
k1
k1C B
k1C AO exp
k1 '
k1C AO
0 , CB = 0
Optimum yield:
'
'
d '
Now at
dC B
k1C AO ' exp
k1C AO exp
k1 '
k1 '
k1 ' exp
1 exp
k1 '
k1 '
1
k1
and
X opt
1 e 1 =0.632
P8-2 (d) Example 8-4
According to the plot of CB versus T the maximum temperature is T = 310.52 0C.
8-4
CB
k2
k1C A0
1 k1 1
POLYMATH Results
See polymath problem P8-2-1.pol
Calculated values of the DEQ variables
Variable
t
T
Cao
tau
k1o
k2o
E1
E2
R
k1
k2
Cb
initial value
0
300
5
0.5
0.4
0.01
10
20
0.001987
0.4
0.4
0.6944444
minimal value
0
300
5
0.5
0.4
0.01
10
20
0.001987
0.4
0.4
0.0052852
maximal value
100
400
5
0.5
0.4
0.01
10
20
0.001987
26.513034
1757.3524
0.8040226
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = 1
Explicit equations as entered by the user
[1] Cao = 5
[2] tau = .5
[3] k1o = .4
[4] k2o = .01
[5] E1 = 10
[6] E2 = 20
[7] R = .001987
[8] k1 = k1o*exp((-E1/R)*(1/T-1/300))
8-5
final value
100
400
5
0.5
0.4
0.01
10
20
0.001987
26.513034
1757.3524
0.0052852
[9] k2 = k1o*exp((-E2/R)*(1/T-1/300))
[10] Cb = (tau*k1*Cao)/(tau*k2+1)/(tau*k1+1)
P8-2 (e) Individualized solution
P8-2 (f) Individualized solution
P8-2 (g) Individualized solution
P8-2 (h) Example 8-8
(1)
Membrane Reactor
PFR
SD/U Original Problem
2.58
0.666
SD/U P8-2 h
1.01
0.208
Doubling the incoming flow rate of species B lowers the selectivity.
(2) The selectivity becomes 6.52 when the first reaction is changed to A+2B  D
P8-2 (i) CDROM Example
Original Case – CDROM example
P8-2 i
The reaction does not go as far to completion when the changes are made. The exiting concentration of
D, E, and F are lower, and A, B, and C are higher.
See Polymath program P8-2-i.pol.
P8-2 (j)
For equal molar feed in hydrogen and mesitylene.
CHO = yHOCTO = (0.5)(0.032)lbmol/ft3 =0.016 lbmol/ft3
CMO = 0.016 lbmol/ft3
Using equations from example, solving in Polymath, we get
8-6
opt
0.38 hr. At
0.5 hr all of the H2 is reacted and only the decomposition of X takes place.
XH
CH
CM
CX
~
S X /T
CD-ROM
example
0.50
0.0105
0.0027
0.00507
0.2hr
0.596
This question
0.99
0.00016
0.0042
0.0077
0.38hr
1.865
See Polymath program P8-2-j.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
tau
CH
CM
CX
k1
k2
r1M
r2T
r1H
r2H
r1X
r2X
initial value
0
0.016
0.016
0
55.2
30.2
-0.1117169
0
-0.1117169
0
0.1117169
0
minimal value
0
1.64E-06
0.0041405
0
55.2
30.2
-0.1117169
0
-0.1117169
-0.0159818
2.927E-04
-0.0159818
maximal value
0.43
0.016
0.016
0.0077216
55.2
30.2
-2.927E-04
0.0159818
-2.927E-04
0
0.1117169
0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(CH)/d(tau) = r1H+r2H
[2] d(CM)/d(tau) = r1M
[3] d(CX)/d(tau) = r1X+r2X
Explicit equations as entered by the user
[1] k1 = 55.2
[2] k2 = 30.2
[3] r1M = -k1*CM*(CH^.5)
[4] r2T = k2*CX*(CH^.5)
[5] r1H = r1M
[6] r2H = -r2T
[7] r1X = -r1M
[8] r2X = -r2T
~
Increasing θH decreases τopt and S X/T.
P8-2 (k)
8-7
final value
0.43
1.64E-06
0.0041405
0.0077207
55.2
30.2
-2.927E-04
2.986E-04
-2.927E-04
-2.986E-04
2.927E-04
-2.986E-04
From CD-ROM example
Polymath code:
See Polymath program P8-2-k.pol.
POLYMATH Results
NLES Solution
Variable
CH
CM
CX
tau
K1
K2
CHo
CMo
Value
4.783E-05
0.0134353
0.0023222
0.5
55.2
30.2
0.016
0.016
f(x)
-4.889E-11
-1.047E-11
-9.771E-12
Ini Guess
1.0E-04
0.013
0.002
NLES Report (safenewt)
Nonlinear equations
[1] f(CH) = CH-CHo+K1*(CM*CH^.5+K2*CX*CH^.5)*tau = 0
[2] f(CM) = CM-CMo+K1*CM*CH^.5*tau = 0
[3] f(CX) = (K1*CM*CH^.5-K2*CX*CH^0.5)*tau-CX = 0
Explicit equations
[1] tau = 0.5
[2] K1 = 55.2
[3] K2 = 30.2
[4] CHo = 0.016
[5] CMo = 0.016
A plot using different values of is given.
For =0.5, the exit concentration are
CH = 4.8 ×10-5 lbmol/ft3 CM =0.0134 lbmol/ft3
CX =0.00232 lbmol/ft3
The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for
YMX
FX
FMO
CX
FM
C MO
CM
0.00232
0.016 0.0134
0.89mole xylene produced
mole mesitylene reacted
The overall selectivity of xylene relative to toluene is:
FX
FT
8.3mole xylene produced
mole toluene produced
YMX
CD-ROM example
0.0089
0.0029
0.0033
0.5
0.41
~
S X /T
CH
CM
CX
This Question
4.8 x 10-5
0.0134
0.00232
0.5
0.89
8-8
=0.5:
SX/T
0.7
8.3
P8-2 (l)
At the beginning, the reactants that are used to create TF-VIIa and TF-VIIaX are in high concentration. As
the two components are created, the reactant concentration drops and equilibrium forces the
production to slow. At the same time the reactions that consume the two components begin to
accelerate and the concentration of TF-VIIa and TF-VIIaX decrease. As those reactions reach equilibrium,
the reactions that are still producing the two components are still going and the concentration rises
again. Finally the reactions that consume the two components lower the concentration as the products
of those reactions are used up in other reactions.
P8-2 (m) Individualized solution
P8-3 Solution is in the decoding algorithm given with the modules (ICM problem )
P8-4 (a)
Assume that all the bites will deliver the standard volume of venom. This means that the initial
concentration increases by 5e-9 M for every bite.
After 11 bites, no amount of antivenom can keep the number of free sites above 66.7% of total sites.
This means that the initial concentration of venom would be 5.5e-8 M. The best result occurs when a
dose of antivenom such that the initial concentration of antivenom in the body is 5.7e-8 M, will result in
a minimum of 66.48% free sites, which is below the allowable minimum.
P8-4 (b)
8-9
The victim was bitten by a harmless snake and antivenom was injected. This means that the initial
concentration of venom is 0. From the program below, we see that if an amount of antivenom such that
the initial concentration in the blood is 7e-9 M, the patient will die.
See Polymath program P8-4-b.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
fsv
fs
Cv
Ca
fsa
Cp
kv
ksv
ka
kia
Cso
ksa
kp
kov
koa
kop
g
h
m
j
initial value
0
0
1
0
7.0E-09
0
0
2.0E+08
6.0E+08
2.0E+08
1
5.0E-09
6.0E+08
1.2E+09
0
0.3
0.3
0
0
0
-2.1E-09
minimal value
0
0
0.6655661
0
4.503E-09
0
0
2.0E+08
6.0E+08
2.0E+08
1
5.0E-09
6.0E+08
1.2E+09
0
0.3
0.3
0
0
0
-2.1E-09
maximal value
0.5
0
1
0
7.0E-09
0.3344339
0
2.0E+08
6.0E+08
2.0E+08
1
5.0E-09
6.0E+08
1.2E+09
0
0.3
0.3
0
0
0
-1.351E-09
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(fsv)/d(t) = kv * fs * Cv - ksv * fsv * Ca
[2] d(fs)/d(t) = -kv*fs*Cv - ka * fs * Ca + kia * fsa + g
[3] d(Cv)/d(t) = Cso * (-kv * fs * Cv - ksa * fsa * Cv) + h
[4] d(Ca)/d(t) = Cso*(-ka * fs * Ca + kia * fsa) + j
[5] d(fsa)/d(t) = ka * fs * Ca - kia * fsa - ksa * fsa * Cv
[6] d(Cp)/d(t) = Cso * (ksv * fsv * Ca + ksa * fsa * Cv) + m
Explicit equations as entered by the user
[1] kv = 2e8
[2] ksv = 6e8
[3] ka = 2e8
[4] kia = 1
[5] Cso = 5e-9
[6] ksa = 6e8
8-10
final value
0.5
0
0.6655661
0
4.503E-09
0.3344339
0
2.0E+08
6.0E+08
2.0E+08
1
5.0E-09
6.0E+08
1.2E+09
0
0.3
0.3
0
0
0
-1.351E-09
[7] kp = 1.2e9
[8] kov = 0
[9] koa = 0.3
[10] kop = 0.3
[11] g = ksa * fsa * Cv + ksv * fsv * Ca
[12] h = -kp * Cv * Ca - kov * Cv
[13] m = kp * Cv * Ca - kop * Cp
[14] j = -Cso * ksv * fsv * Ca - kp * Cv * Ca - koa * Ca
P8-4 (c)
The latest time after being bitten that antivenom can successfully be administerd is 27.49 minutes. See
the cobra web module on the CDROM/website for a more detailed solution to this problem
P8-4 (d) Individualized Solution
P8-5 (a)
Plot of CA, CD and CU as a function of time (t):
See Polymath program P8-5-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
8-11
Variable
t
Ca
Cd
Cu
k1
k2
K1a
K2a
Cao
X
initial value
0
1
0
0
1
100
10
1.5
1
0
minimal value
0
0.0801802
0
0
1
100
10
1.5
1
0
maximal value
15
1
0.7995475
0.5302179
1
100
10
1.5
1
0.9198198
final value
15
0.0801802
0.7995475
0.1202723
1
100
10
1.5
1
0.9198198
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))
[2] d(Cd)/d(t) = k1*(Ca-Cd/K1a)
[3] d(Cu)/d(t) = k2*(Ca-Cu/K2a)
Explicit equations as entered by the user
[1] k1 = 1.0
[2] k2 = 100
[3] K1a = 10
[4] K2a = 1.5
[5] Cao = 1
[6] X = 1-Ca/Cao
To maximize CD stop the reaction after a long time. The concentration of D only increases with time
P8-5 (b)
Conc. Of U is maximum at t = 0.31 min.(CA = 0.53)
P8-5 (c)
Equilibrium concentrations:
CAe = 0.08 mol/dm3
CDe = 0.8 mol/dm3
CUe = 0.12 mol/dm3
8-12
P8-5 (d)
See Polymath program P8-5-d.pol.
POLYMATH Results
NLES Solution
Variable
Ca
Cd
Cu
Ca0
k1
k2
K1a
K2a
t
Value
0.0862762
0.7843289
0.1293949
1
1
100
10
1.5
100
f(x)
-3.844E-14
-2.631E-14
6.478E-14
Ini Guess
1
0
0
NLES Report (safenewt)
Nonlinear equations
[1] f(Ca) = Ca0-t*(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))-Ca = 0
[2] f(Cd) = t*k1*(Ca-Cd/K1a)-Cd = 0
[3] f(Cu) = t*(k2*(Ca-Cu/K2a))-Cu = 0
1 min
10 min
100min
CAexit
0.295
0.133
0.0862
CDexit
0.2684
0.666
0.784
CUexit
0.436
0.199
0.129
X
0.705
0.867
0.914
Explicit equations
[1] Ca0 = 1
[2] k1 = 1
[3] k2 = 100
[4] K1a = 10
[5] K2a = 1.5
[6] t = 100
[7] X = 1-Ca/Ca0
P8-6 (a)
A
X
rX
1/ 2
k1C A
A
B
rB
k 2C A
rY
2
A
Y
k 3C A
1/ 2
k1
mol
0.004
dm 3
k2
0.3 min 1
k3
dm 3
0.25
mol. min
Sketch SBX, SBY and SB/XY as a function of CA
8-13
min
See Polymath program P8-6-a.pol.
It shows the table of SBX, SBY and SB/XY in correspondence with values of CA .
1)
SB/X =
rB
rX
2)
SB/Y =
rB
rY
3)
SB/XY =
k 2C A
k1C A1/ 2
k 2C A
k 3C A 2
rB
rX
k2
C A1/ 2
k1
k2
k 3C A
k 2C A
rY
k1C A1/ 2
k 3C A 2
8-14
P8-6 (b)
Volume of first reactor can be found as follows
We have to maximize SB/XY
From the graph above, maximum value of SBXY = 10 occurs at CA* = 0.040 mol/dm3
So, a CSTR should be used with exit concentration CA*
Also, CA0 = PA/RT = 0.162 mol/dm3
And
rA
=> V
v0 (C A0 C A* )
rA
rX
rB
rY
(k1C A1/ 2
k 3C A 2 )
k 2C A
v0 (C A0
* 1/ 2
(k1 (C A )
C A* )
*
k 3 (C A ) )
CB*
0.11
k 2C A
* 2
P8-6 (c)
Effluent concentrations:
CB
rB
We know, τ = 9.24 min =>
Similarly: C X
*
0.007
mol
dm 3
and
CB
k 2C A
CY *
0.0037
P8-6 (d)
Conversion of A in the first reactor:
C A0
CA
C A0 X
X
0.74
P8-6 (e)
A CSTR followed by a PFR should be used.
Required conversion = 0.99
=> For PFR, Mole balance:
dV
dX
F A0
rA
8-15
mol
dm 3
mol
dm 3
92.4dm 3
0.99
V
10 0.162
dX
1/ 2
0.74 ( k1C A
k 2C A
2
92.8dm 3
k 3C A )
P8-6 (f)
If we notice that E2 is the smallest of the activation energies, we get a higher selectivity at lower
temperatures. However, the tradeoff is that the reaction rate of species B, and therefore production of
B, decrease as temperature drops. So we have to compromise between high selectivity and production.
To do this we need expressions for k1, k2, and k3 in terms of temperature. From the given data we know:
ki
Ei
1.98T
Ai exp
Since we have the constants given at T = 300 K, we can solve for Ai.
A1
A2
A3
.004
1.49e12
20000
exp
1.98 300
.3
5.79e6
10000
exp
1.98 300
.25
1.798e21
30000
exp
1.98 300
Now we use a mole balance on species A
V
V
FA0
FA
rA
v C A0 C A
rA
C A0 C A
rA
C A0 C A
k1C
k2C A k3C A2
0.5
A
A mole balance on the other species gives us:
Fi
vCi
Ci
ri
rV
i
8-16
Cb versus temp
0.072
0.07
0.068
0.066
Cb
0.064
0.062
0.06
285
290
295
300
305
310
315
Using these equations we can make a Polymath program and by varying the temperature, we can find a
maximum value for CB at T = 306 K. At this temperature the selectivity is only 5.9. This may result in too
much of X and Y, but we know that the optimal temperature is not above 306 K. The optimal
temperature will depend on the price of B and the cost of removing X and Y, but without actual data, we
can only state for certain that the optimal temperature will be equal to or less than 306 K.
See Polymath program P8-6-f.pol.
POLYMATH Results
NLE Solution
Variable
Ca
T
R
k1
k2
Cao
Cb
k3
tau
Cx
Cy
Sbxy
Value
0.0170239
306
1.987
0.0077215
0.4168076
0.1
0.070957
0.6707505
10
0.0100747
0.0019439
5.9039386
f(x)
3.663E-10
Ini Guess
0.05
NLE Report (safenewt)
Nonlinear equations
[1] f(Ca) = (Cao-Ca)/(k1*Ca^.5+k2*Ca+k3*Ca^2)-10 = 0
Explicit equations
[1] T = 306
8-17
[2] R = 1.987
[3] k1 = 1.49e12*exp(-20000/R/T)
[4] k2 = 5790000*exp(-10000/R/T)
[5] Cao = .1
[6] Cb = 10*k2*Ca
[7] k3 = 1.798e21*exp(-30000/R/T)
[8] tau = 10
[9] Cx = tau*k1*Ca^.5
[10] Cy = tau*k3*Ca^2
[11] Sbxy = Cb/(Cx+Cy)
P8-6 (g)
Concentration is proportional to pressure in a gas-phase system. Therefore:
PA
S B / XY ~
PA
PA2
which would suggest that a low pressure would be ideal. But as before the
tradeoff is lower production of B. A moderate pressure would probably be best.
We know that :
SB/XY =
rB
rX
k 2C A
rY
k1C A1/ 2
k 3C A 2
Substituting PA = CA R T ;
in the above co – relation we get
rB
SB/XY = rX
rY
k2 ( PA / RT )
k1 ( PA / RT )1/2 k3 ( PA / RT ) 2
See Polymath program P8-6-g.pol.
Thus we find that the maximum is obtained at P= 1 atm .
8-18
P8-7
US legal limit: 0.8 g/l
Sweden legal limit: 0.5 g/l
A
k1
B
k2
C
Where A is alcohol in the gastrointestinal tract and B is alcohol in the blood stream
dC A
k1C A
dt
dCB
k1C A k2
dt
k1 10 hr 1
k2
0.192
g
L hr
Two tall martinis = 80 g of ethanol
Body fluid = 40 L
C A0
80 g
40 L
2
g
L
Now we can put the equations into Polymath.
See Polymath program P8-7.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
k1
k2
initial value
0
2
0
10
0.192
minimal value
0
7.131E-44
0
10
0.192
maximal value
10
2
1.8901533
10
0.192
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -k1*Ca
[2] d(Cb)/d(t) = -k2+k1*Ca
Explicit equations as entered by the user
[1] k1 = 10
[2] k2 = 0.192
P8-7 (a)
In the US the legal limit it 0.8 g/L.
8-19
final value
10
7.131E-44
0.08
10
0.192
This occurs at t = 6.3 hours..
P8-7 (b)
In Sweden CB = 0.5 g/l , t = 7.8 hrs.
P8-7 (c) In Russia CB = 0.0 g/l, t = 10.5 hrs
P8-7 (d)
For this situation we will use the original Polymath code and change the initial concentration of A to 1
g/L. Then run the Program for 0.5 hours. This will give us the concentration of A and B at the time the
second martini is ingested. This means that 1 g/l will be added to the final concentration of A after a half
an hour.
At a half an hour CA = 0.00674 g/L and CB = 0.897 g/L. The Polymath code for after the second drink is
shown below.
See Polymath program P8-7-d.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
k1
k2
initial value
0.5
1.0067379
0.8972621
10
0.192
minimal value
0.5
5.394E-42
0.08
10
0.192
maximal value
10
1.0067379
1.8069769
10
0.192
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -k1*Ca
[2] d(Cb)/d(t) = -k2+k1*Ca
Explicit equations as entered by the user
[1] k1 = 10
[2] k2 = 0.192
for the US t = 6.2 hours
Sweden: t = 7.8 hours
Russia: t =10.3 hours.
P8-7 (e)
8-20
final value
10
5.394E-42
0.08
10
0.192
The mole balance on A changes if the drinks are consumed at a continuous rate for the first hour. 80 g
of ethanol are consumed in an hour so the mass flow rate in is 80 g/hr. Since volume is not changing the
rate of change in concentration due to the incoming ethanol is 2 g/L/hr.
For the first hour the differential equation for CA becomes:
dC A
dt
k1C A
2t after that it reverts back to the original equations.
See Polymath program P8-7-e.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
k1
k2
initial value
0
0
0
10
0.192
minimal value
0
0
-1.1120027
10
0.192
maximal value
11
0.1785514
0.7458176
10
0.192
final value
11
6.217E-45
-1.1120027
10
0.192
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = if(t<1)then(-k1*Ca+2*t)else(-k1*Ca)
[2] d(Cb)/d(t) = -k2+k1*Ca
Explicit equations as entered by the user
[1] k1 = 10
[2] k2 = 0.192
US: CB never rises above 0.8 g/L so there is no time that it would be illegal.
Sweden: t = 2.6 hours
Russia: t = 5.2 hours
P8-7 (f)
60 g of ethanol immediately  CA = 1.5 g/L
CB = 0.8 g/L at 0.0785 hours or 4.71 minutes.
So the person has about 4 minutes and 40 seconds to get to their destination.
P8-7 (g)
A heavy person will have more body fluid and so the initial concentration of CA would be lower. This
means a heavier person will reach the legal limit quicker. The opposite is true for a slimmer person. They
will take longer to reach the legal limit, as their initial concentration will be higher.
8-21
P8-8 (a)
See Polymath program P8-8-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
k1
k2
k4
k3
initial value
0
6.25
0
0.15
0.6
0.2
0.1
minimal value
0
0.3111692
0
0.15
0.6
0.2
0.1
maximal value
4
6.25
0.5977495
0.15
0.6
0.2
0.1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -k1*Ca-k2*Ca
[2] d(Cb)/d(t) = k1*Ca-k3-k4*Cb
Explicit equations as entered by the user
[1] k1 = 0.15
[2] k2 = 0.6
[3] k4 = 0.2
[4] k3 = if(Cb<0)then (k1*Ca-k4*Cb) else (0.1)
P8-8 (b)
8-22
final value
4
0.3111692
0.4057018
0.15
0.6
0.2
0.1
8-23
P8-8 (c)
If one takes initially two doses of Tarzlon, it is not recommended to take another dose within the first
six hours. Doing so will result in build up of the drug in the bloodstream that can cause harmful effects.
P8-8 (d)
If the drug is taken on a full stomach most of it will not reach the wall at all. The processed food can also
drag the drug to the intestines and may limit its effectiveness. This effect can be seen in the adsorption
constant k1 and elimination constant k2 values. If k1 decreases this means that the adsorption process is
slow and if k2 increases means that the rate of elimination of Tarzlon increases. The next graph shows
the concentration profiles for k1 = 0.10 h-1 and k2 = 0.8 h-1. Note that the maximum amount of the drug
in the bloodstream is reduced by two.
Concentration Profiles
Ca
(mg/dm3)
Cb
(mg/dm3)
0.6
CB (mg/dm3)
CA (mg/dm3)
8
6
0.4
4
0.2
2
0
0
0
2
4
6
8
10
12
Time (h)
Concentration profile for Tarzlon in the stomach (A) and bloodstream (B).
The maximum amount of Tarzlon in the bloodstream is 0.3 mg/dm3.
P8-9 (a)
Reactor selection
A B
D
rD
r1 A
r1 A 10 exp( 8000 K / T )C AC B
A B
U
rU
r2 A
r2 A
SDU =
rD
rU
10 exp( 8000 K / T )C A C B
1/ 2
100 exp( 1000 K / T )C A C B
1/ 2
100 exp( 1000 K / T )C A C B
exp( 8000 K / T )C A
3/ 2
10 exp( 1000 K / T )C B
At T = 300K
1/ 2
k1 = 2.62 x 10-11 &
k2 = 3.57
SD/U
1/ 2
7.35 10 12 C A
1/ 2
CB
At T = 1000K
8-24
1/ 2
3/ 2
-3
k1 = 3.35 x 10 &
k2 =36.78
SD/U
9.2 10 5 C A
CB
1/ 2
1/ 2
Hence In order to maximize SDU, use higher concentrations of A and lower concentrations of B. This can
be achieved using:
1) A semibatch reactor in which B is fed slowly into a large amount of A
2) A tubular reactor with side streams of B continually fed into the reactor
3) A series of small CSTR’s with A fed only to the first reactor and small amounts of B fed to each reactor.
Also, since ED > EU, so the specific reaction rate for D increases much more rapidly with temperature.
Consequently, the reaction system should be operated at highest possible temperature to maximize SDU.
1
CB
Note that the selectivity is extremely low, and the only way to increase it is to keep
CA
2
10 6 and
add B drop by drop.
P8-9 (b)
A B
D
and
r1 A 100 exp( 1000 K / T )C AC B
A B
U
and
r2 A 106 exp( 8000K / T )C ACB
SDU =
rD
rU
100 exp( 1000 K / T )C AC B
10 6 exp( 8000 K / T )C AC B
exp( 1000 K / T )
10 4 exp( 8000 K / T )
At T = 300K
k1 = 3.57 &
k2 = 2.623
SDU = 1.14 x106
At T = 1000K
k1 = 36.78 & k2 = 3354.6
SDU = 0.103
Hence temperature should be kept as low as possible to maximize SDU, but at the same time care should
be taken to have a significant reaction extent.
P8-9 (c)
A B D and
r1 A
10 exp( 8000 K / T )C ACB
B D
r2 A
109 exp( 10, 000 K / T )CB CD
U
and
8-25
r1 A
r2 A
S DU
10 exp
8000 K / T C ACB
109 exp
10000 K / T CBCD
exp
S DU
8000 / T C A
8
10 exp
10000 / T CD
Therefore the reaction should be run at a low temperature to maximize SDU, but not too low to limit the
production of desired product. The reaction should also take place in high concentration of A and the
concentration of D should be limited by removing through a membrane or reactive distillation.
P8-9 (d)
A D and
D U1 and
A
U 2 and
rD
S D / U 1U 2
rU 1
rU 2
At T = 300K
k1 = 1.18 x 10-14 &
r1 A
4280 exp( 12000 K / T )C A
r2 D
10,100 exp( 15000 K / T )C D
r3 A
26 exp( 10800 K / T )C A
4280 exp( 12000 K / T )C A 10,100 exp( 15000 K / T )C D
10,100 exp( 15000 K / T )C D 26 exp( 10800 K / T )C A
k2 = 1.94 X 10-18 & k3 = 6.03 x 10-15
If we keep CA > 1000CD
S D / U 1U 2
1.18 10 14 C A 1.94 10 18 C D
1.94 10 18 C D 6.03 10 15 C A
At T = 1000K
k1 = 0.026 &
1.18
1.96
.603
k2 = 3.1 X 10-3 & k3 = 5.3 x 10-4
If we keep CA > 1000CD
S D / U 1U 2
0.026C A 3.1 10 3 C D
3.1 10 3 C D 5.3 10 4 C A
.026
.00053
49
Here, in order to lower U1 use low temperature and high concentration of A
But low temperature and high concentration of A favours U2
So, it’s a optimization problem with the temperature and concentration of A as the variables .
Membrane reactor in which D is diffusing out can be used.
P8-9 (e)
A B
D
and
r1 A
10 9 exp( 10000 K / T )C A C B
8-26
D A B
A B U
S D /U
rD
rU
and
r2 D
20 exp( 2000 K / T )C D
and
r3 A
103 exp( 3000K / T )C AC B
10 9 exp( 10000 K / T )C AC B 20 exp( 2000 K / T )C D
10 3 exp( 3000 K / T )C A C B
At T = 300K
k1 = 3.34 x 10-6 &
k2 = 0.025
& k3 = 0.045
The desired reaction lies very far to the left and CD is probably present at very low concentrations so
that:
S D /U
3.34 10 6 C A C B 0.025C D
0.045C A C B
S D /U
0
At T = 1000K
k1 = 45399.9 &
0.000334
4.5
0.000074
k2 = 2.7 & k3 = 49.7
If we assume that CACB > 0.001CD then,
S D /U
45399.9C A C B 2.7C D
49.7C A C B
45399
49.7
913
Here we need a high temperature for a lower reverse reaction of D and lower formation of U
Also we need to remove D as soon as it is formed so as to avoid the decomposition.
P8-9 (f)
A B
D
and
r1 A
10 exp( 8000 K / T )C ACB
A
U
and
r2 A
26 exp( 10,800 K / T )C A
U
A
and
r3U
1000 exp( 15, 000 K / T )CU
SD /U
rD
rU
10exp( 8000 K / T )C ACB
26exp( 10,800 K / T )C A 1000exp 15, 000 K / T CU
We want high concentrations of B and U in the reactor. Also low temperatures will help keep the
selectivity high.
If we use pseudo equilibrium and set –rA = 0.
rA 10 exp
8000
C AC B
T
26 exp
10800
C A 1000 exp
T
8-27
15000
CU
T
0
10 exp
CA
CU
CA
CU
8000
CB
T
26 exp
1000 exp
15000
T
10800
T
1
7000
26
4200
exp
CB
exp
100
T
1000
T
P8-9 (g)
A B
D
and
r1 A
800 exp( 8000 K / T )C A0.5CB
A B
U1
and
r2 B
10 exp( 300 K / T )C ACB
D B
U2
and
r3 D
106 exp( 8000 K / T )CD CB
800 exp
8000 / T C A0.5CB
80 exp
10 exp
300 / T C ACB
exp
(1)
S D / U1
8000 / T
300 / T C A0.5
At T = 300
S D / U1
2.098*10 10
0.368C A0.5
At T = 1000
S D / U1
29.43
0.7408C A0.5
To keep this selectivity high, low concentrations of A, and high temperatures should be used.
S D /U 2
800 exp
8000 / T C A0.5CB
106 exp
8000 / T CD CB
800C A0.5
106 CD
To keep this selectivity high, high concentrations of A and low concentrations of D should be used. Try
to remove D with a membrane reactor or reactive distillation. The selectivity is not dependant on
temperature.
To keep optimize the reaction, run it at a low temperature to maximized SD/U1 in a membrane reactor
that allows only D to diffuse out.
(2)
S D / U 1U 2
S D / U 1U 2
800 exp
10 exp
300 / T C ACB 106 exp
800 exp
10 exp
8000 / T C A0.5CB
8000 / T CD CB
8000 / T C A0.5
300 / T C A 106 exp
8000 / T CD
8-28
At T = 300
S D /U 1U 2
2.09*10 9 C A0.5
3.67C A 2.62*10 6 CD
0
At T = 1000 and very low concentrations of D
S D /U 1U 2
0.268C A0.5
7.408C A 335CD
.03617
C A0.5
If temperature is the only parameter that can be varied, then the highest temperature possible will
result in the highest selectivity. Also removing D will help keep selectivity high.
P8-10 (a)
Species A :
dCa
dt
rA
;
-rA = k1 * Ca
Species B:
dCb
dt
rb
rB = k1 * Ca – k2 * Cb
Species C:
dCc
dt
rc
rC = k2*Cb
See polymath program P8-10-a.pol
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
1.6
6.797E-18
1.6
6.797E-18
2 Cb
0
0
1.455486
0.6036996
3 Cc
0
0
0.9963004
0.9963004
4 k1
0.4
0.4
0.4
0.4
5 k2
0.01
0.01
0.01
0.01
8-29
6 ra
-0.64
-0.64
-2.719E-18
-2.719E-18
7 rb
0.64
-0.0132415
0.64
-0.006037
8 rc
0
0
0.0145549
0.006037
9 t
0
0
100.
100.
Differential equations
1 d(Cc)/d(t) = rc
2 d(Cb)/d(t) = rb
3 d(Ca)/d(t) = ra
Explicit equations
1 k2 = 0.01
2 rc = k2*Cb
3 k1 = 0.4
4 ra = - k1*Ca
5 rb = k1*Ca - k2*Cb
P8-10 (b)
Now the rate laws will change ra = k1-1Cb – k1*Ca
rb =k1*Ca – k1-1Cb – k2*Cb
rc = k2*Cb
See polymath program P8-10-b.pol
Calculated values of DEQ variables
8-30
Variable Initial value Minimal value Maximal value Final value
1
Ca
1.6
0.3949584
1.6
0.3949584
2
Cb
0
0
0.8736951
0.5191343
3
Cc
0
0
0.6859073
0.6859073
4
k1
0.4
0.4
0.4
0.4
5
k1b
0.3
0.3
0.3
0.3
6
k2
0.01
0.01
0.01
0.01
7
ra
-0.64
-0.64
-0.0022431
-0.0022431
8
rb
0.64
-0.0047669
0.64
-0.0029483
9
rc
0
0
0.008737
0.0051913
10 t
0
0
100.
100.
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
Explicit equations
1 k1 = 0.4
2 k2 = 0.01
3 k1b = 0.3
4 ra = k1b*Cb - k1*Ca
5 rb = k1*Ca - k1b*Cb - k2*Cb
6 rc = k2*Cb
8-31
P8-10 (c)
rC = k2*CB – k-2*Cc
See polymath program P8-10-c.pol
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
1.6
0.4500092
1.6
0.4500092
2
Cb
0
0
0.8742432
0.5953978
3
Cc
0
0
0.554593
0.554593
4
k1
0.4
0.4
0.4
0.4
5
k1b
0.3
0.3
0.3
0.3
6
k2
0.01
0.01
0.01
0.01
7
k2c
0.005
0.005
0.005
0.005
8
ra
-0.64
-0.64
-0.0013843
-0.0013843
9
rb
0.64
-0.0044688
0.64
-0.0017967
10 rc
0
0
0.0085186
0.003181
11 t
0
0
100.
100.
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
Explicit equations
1 k1 = 0.4
2 k2 = 0.01
3 k1b = 0.3
4 k2c = 0.005
5 ra = k1b*Cb - k1*Ca
6 rb = k1*Ca - k1b*Cb - k2*Cb + k2c*Cc
7 rc = k2*Cb - k2c*Cc
8-32
P8-10 (d)
When k1 > 100 and k2 < 0.1 the concentration of B immediately shoots up to 1.5 and then slowly comes
back down , while CA drops off immediately and falls to zero . This is because the first reaction is very
fast and the second reaction is slower with no reverse reactions.
When k2 = 1 then the concentration of B spikes again and remains high, while very little of C is formed.
This is because after B is formed it will not further get converted to C because the reverse reaction is
fast.
When k-2 = 0.25 , B shoots up , but does not stay as high because for the second reaction in the reverse
direction is a slightly slower than seen before , but still faster than the reaction in the forward direction.
P8-11 (a)
Intermediates (primary K-phthalates) are formed from the dissociation of K-benzoate with a CdCl2
catalyst reacted with K-terephthalate in an autocatalytic reaction step:
A k1
R S
C AO
P
RT
R
k2
Series
S
k3
2S
110kPa
kPa.dm3
8.314
683K
mol.K
Autocatalytic
0.02mol / dm3
Maximum in R occurs at t = 880 sec.
See Polymath program P8-11-a.pol.
POLYMATH Results
Variable
t
A
initial value
0
0.02
minimal value
0
0.003958
maximal value
1500
0.02
8-33
final value
1500
0.003958
R
S
k1
k2
k3
0.00159
0
0
0.00108
0.00119
0.00159
0.00159
0
0
0.00108
0.00119
0.00159
0.0069892
0.0100382
0.00108
0.00119
0.005868
0.0100382
0.00108
0.00119
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(A)/d(t) = -k1*A
[2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S)
[3] d(S)/d(t) = (k2*R)-(k3*R*S)
Explicit equations as entered by the user
[1] k1 = 1.08e-3
[2] k2 = 1.19e-3
[3] k3 = 1.59e-3
P8-11 (b)
1) T = 703 K
CAO = 0.019 mol/dm3
k1'
k1'
k1 exp
E1 1
R T
1
T'
(1.08 10 3 s 1 ) exp
(42600cal / mol )
1
(1.987cal / mol.K ) 683K
1
703K
2.64 10 3 s 1
Similarly,
k2'
3.3 10 3 s 1
And, k3'
3.1 10 3 dm3 / mol.s
Maxima in R occurs at around t =320 sec.
See Polymath program P8-11-b1.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
A
R
S
k1
k2
k3
initial value
0
0.019
0
0
0.00264
0.0033
0.0031
minimal value
0
3.622E-04
0
0
0.00264
0.0033
0.0031
maximal value
1500
0.019
0.0062169
0.0174625
0.00264
0.0033
0.0031
8-34
final value
1500
3.622E-04
8.856E-04
0.0174625
0.00264
0.0033
0.0031
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(A)/d(t) = -k1*A
[2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S)
[3] d(S)/d(t) = (k2*R)-(k3*R*S)
Explicit equations as entered by the user
[1] k1 = 2.64e-3
[2] k2 = 3.3e-3
[3] k3 = 3.1e-3
2) T = 663 K
CAO = 0.19 mol/dm3
(42600cal / mol )
1
(1.987cal / mol.K ) 683K
k1'
(1.08 10 3 s 1 ) exp
k2'
0.4 10 3 s 1
k3'
0.78 10 3 dm3 / mol.s
1
663K
0.42 10 3 s 1
See Polymath program P8-11-b2.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
A
R
S
k1
k2
k3
initial value
0
0.019
0
0
4.2E-04
4.0E-04
7.8E-04
minimal value
0
2.849E-04
0
0
4.2E-04
4.0E-04
7.8E-04
maximal value
10000
0.019
0.0071414
0.016889
4.2E-04
4.0E-04
7.8E-04
final value
10000
2.849E-04
0.0012573
0.016889
4.2E-04
4.0E-04
7.8E-04
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(A)/d(t) = -k1*A
[2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S)
[3] d(S)/d(t) = (k2*R)-(k3*R*S)
Explicit equations as entered by the user
[1] k1 = 0.42e-3
[2] k2 = 0.4e-3
[3] k3 = 0.78e-3
Independent variable
variable name : t
initial value : 0
final value : 10000
Maxima in R occurs around t = 2500 sec.
P8-11 (c)
Use the Polymath program from part (a) and change the limits of integration to 0 to 1200. We get:
CAexit = 0.0055 mol/dm3
CRexit = 0.0066 mol/dm3
8-35
CSexit = 0.0078 mol/dm3
P8-12 (a)
P8-12 (b)
P8-12 (c)
P8-12 (d)
P8-12 (e)
P8-12 (f)
P8-12 (g)
P8-12 (h)
Mole balance: C AO
CA
rA
CC
rC
CD
rD
8-36
Rate law:
Solving in polymath:
rA
k1 A C A
1
k 2 D C A CC2
3
rC
1
k1 A C A
3
2
k 2 D C A CC2
3
rD
k 2 D C A CC2
4
k 3 E CC C D
3
CA
0.0069 M C B
k 3 E CC C D
0.96M CC
SB/D = rB/rD = 247
SB/C = 1.88
See Polymath program P8-12-h.pol.
POLYMATH Results
NLES Solution
Variable
Ca
Cb
Cc
Cd
Ce
kd
ka
rb
ra
ke
rc
rd
re
tau
Cao
Value
0.0068715
0.9620058
0.5097027
0.0038925
0.2380808
3
7
0.0160334
-0.0498855
2
0.008495
6.488E-05
0.003968
60
3
f(x)
-2.904E-10
-1.332E-15
-1.67E-08
-2.391E-08
1.728E-08
0.51M C D
Ini Guess
3
0
0
0
0
NLES Report (safenewt)
Nonlinear equations
[1] f(Ca) = Cao-Ca+ra*tau = 0
[2] f(Cb) = Cb - rb*tau = 0
[3] f(Cc) = Cc-rc*tau = 0
[4] f(Cd) = Cd-rd*tau = 0
[5] f(Ce) = Ce - re*tau = 0
Explicit equations
[1] kd = 3
[2] ka = 7
[3] rb = ka*Ca/3
[4] ra = -(ka*Ca+kd/3*Ca*Cc^2)
[5] ke = 2
[6] rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc
[7] rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc
8-37
0.004M
[8] re = ke*Cd*Cc
[9] tau = 60
[10] Cao = 3
P8-12 (i)
For PFR and gas phase:
dFC
dV
Mole balance:
dFA
dV
rA
dFB
rB
dV
1
k 2 D C A CC2
3
Rate law:
rA
k1 A C A
rB
1
k1 A C A
3
rC
1
k1 A C A
3
2
k 2 D C A CC2
3
rD
k 2 D C A CC2
4
k 3 E CC C D
3
rE
k 3E CC C D
Stoichiometry: C A
CTO
FA
y
FT
FT
FA
FB
dy
dV
FT
2 y FTO
FC
rC
dFD
dV
rD
CD
CTO
dFE
dV
K 3 E CC C D
CC
CTO
FD
FE
FC
y
FT
FD
y
FT
Plot of CB and CC are overlapping.
See Polymath program P8-12-i.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
V
Fa
Fb
Fc
Fd
Fe
y
Ft
Cto
Cc
ka
kd
ke
Ca
initial value
0
20
0
0
0
0
1
20
0.2
0
7
3
2
0.2
minimal value
0
9.147E-04
0
0
0
0
0.9964621
13.330407
0.2
0
7
3
2
1.367E-05
maximal value
100
20
6.6638171
6.6442656
0.0201258
0.0043322
1
20
0.2
0.0993605
7
3
2
0.2
8-38
final value
100
9.147E-04
6.6638171
6.6442167
0.0171261
0.0043322
0.9964621
13.330407
0.2
0.0993325
7
3
2
1.367E-05
rE
rb
ra
Cd
Fto
rc
rd
re
alfa
X
0.4666667
-1.4
0
20
0.4666667
0
0
1.0E-04
0
3.191E-05
-1.4
0
20
-1.923E-05
-7.012E-05
0
1.0E-04
0
0.4666667
-9.586E-05
3.0E-04
20
0.4666667
8.653E-04
5.908E-05
1.0E-04
0.9999543
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(V) = ra
[2] d(Fb)/d(V) = rb
[3] d(Fc)/d(V) = rc
[4] d(Fd)/d(V) = rd
[5] d(Fe)/d(V) = re
[6] d(y)/d(V) = -alfa*Ft/(2*y*Fto)
Explicit equations as entered by the user
[1] Ft = Fa+Fb+Fc+Fd+Fe
[2] Cto = 0.2
[3] Cc = Cto*Fc/Ft*y
[4] ka = 7
[5] kd = 3
[6] ke = 2
[7] Ca = Cto*Fa/Ft*y
[8] rb = ka*Ca/3
[9] ra = -(ka*Ca+kd/3*Ca*Cc^2)
[10] Cd = Cto*Fd/Ft*y
[11] Fto = 0.2*100
[12] rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc
[13] rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc
[14] re = ke*Cd*Cc
[15] alfa = 0.0001
[16] X = 1-Fa/20
8-39
3.191E-05
-9.586E-05
2.56E-04
20
-1.923E-05
-6.742E-05
5.087E-05
1.0E-04
0.9999543
P8-12 (j) Changes in equation from part (i):
dFC
dV
rC
RC
RC
k diffuse C C
k diffuse
2 min 1
See Polymath program P8-12-j.pol.
P8-13(a)
8-40
See Polymath program P8-13-a.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
1.5
0.3100061
1.5
0.3100061
2
Cb
2.
0.3384816
2.
0.3384816
3
Cc
0
0
0.1970387
0.105508
4
Cd
0
0
0.6751281
0.6516263
5
Ce
0
0
0.1801018
0.1801018
6
Cf
0
0
0.3621412
0.3621412
7
kd1
0.25
0.25
0.25
0.25
8
ke2
0.1
0.1
0.1
0.1
9
kf3
5.
5.
5.
5.
10 ra
-1.5
-1.5
-0.0694818
-0.0694818
11 rb
-3.
-3.
-0.0365984
-0.0365984
12 rc
1.5
-0.0490138
1.5
-0.0085994
13 rd
1.5
-0.0128609
1.5
-0.0126825
14 rd1
1.5
0.0088793
1.5
0.0088793
15 re
0
0
0.0523329
0.0202008
16 re2
0
0
0.0523329
0.0202008
17 rf
0
0
0.2571468
0.0188398
18 rf3
0
0
0.2571468
0.0188398
19 V
0
0
50.
50.
20 vo
10.
10.
10.
10.
Differential equations
1 d(Ca)/d(V) = ra/vo
2 d(Cb)/d(V) = rb/vo
3 d(Cc)/d(V) = rc/vo
4 d(Cd)/d(V) = rd/vo
5 d(Ce)/d(V) = re/vo
6 d(Cf)/d(V) = rf/vo
Explicit equations
1
vo = 10
2
kf3 = 5
8-41
3
ke2 = .1
4
kd1 = 0.25
5
rf3 = kf3*Cb*Cc^2
6
rd1 = kd1*Ca*Cb^2
7
re2 = ke2*Ca*Cd
8
rf = rf3
9
re = re2
10 rd = rd1-2*re2+rf3
11 ra = -rd1-3*re2
12 rb = -2*rd1-rf3
13 rc = rd1+re2-2*rf3
P8-13 (b)
8-42
P8-13 (c)
8-43
8-44
8-45
P8-13 (d)
As θB increases the outlet concentration of species D and F increase, while the outlet concentrations of
species A, C, and E decrease. When θB is large, reactions 1 and 3 are favored and when it is small the
rate of reaction 2 will increase.
P8-13 (e)
When the appropriate changes to the Polymath code from part (a) are made we get the following.
POLYMATH Results
Calculated values of the DEQ variables
Variable
V
Fa
Fb
Fc
Fd
Fe
Ff
vo
Ft
Cto
kd1
ke2
kf3
Cc
Cd
initial value
0
20
20
0
0
0
0
100
40
0.4
0.25
0.1
5
0
0
minimal value
0
18.946536
18.145647
0
0
0
0
100
38.931546
0.4
0.25
0.1
5
0
0
maximal value
500
20
20
0.9342961
0.8454829
0.0445942
0.0149897
100
40
0.4
0.25
0.1
5
0.0095994
0.0086869
8-46
final value
500
18.946536
18.145647
0.9342961
0.8454829
0.0445942
0.0149897
100
38.931546
0.4
0.25
0.1
5
0.0095994
0.0086869
Cb
Ca
rd1
re2
rf3
re
rf
rd
ra
rb
rc
Scd
Sef
0.2
0.2
0.002
0
0
0
0
0.002
-0.002
-0.004
0.002
1
0
0.1864364
0.1946651
0.0016916
0
0
0
0
0.0014393
-0.0021989
-0.004
0.0016889
1
0
0.2
0.2
0.002
1.691E-04
8.59E-05
1.691E-04
8.59E-05
0.002
-0.002
-0.003469
0.002
1.1734311
83.266916
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(V) = ra
[2] d(Fb)/d(V) = rb
[3] d(Fc)/d(V) = rc
[4] d(Fd)/d(V) = rd
[5] d(Fe)/d(V) = re
[6] d(Ff)/d(V) = rf
Explicit equations as entered by the user
[1] vo = 100
[2] Ft = Fa+Fb+Fc+Fd+Fe+Ff
[3] Cto = .4
[4] kd1 = 0.25
[5] ke2 = .1
[6] kf3 = 5
[7] Cc = Cto*Fc/Ft
[8] Cd = Cto*Fd/Ft
[9] Cb = Cto*Fb/Ft
[10] Ca = Cto*Fa/Ft
[11] rd1 = kd1*Ca*Cb^2
[12] re2 = ke2*Ca*Cd
[13] rf3 = kf3*Cb*Cc^2
[14] re = re2
[15] rf = rf3
[16] rd = rd1-2*re2+rf3
[17] ra = -rd1-3*re2
[18] rb = -2*rd1-rf3
[19] rc = rd1+re2-2*rf3
[20] Scd = rc/(rd+.0000000001)
[21] Sef = re/(rf+.00000000001)
8-47
0.1864364
0.1946651
0.0016916
1.691E-04
8.59E-05
1.691E-04
8.59E-05
0.0014393
-0.0021989
-0.003469
0.0016889
1.1734311
1.9686327
P8-13 (f)
The only change from part (e) is:
dFD
dV
rD
kcD CD
dFB
dV
rB
FB 0
where VT = 500 dm3 and FB0 = 20 mol/min
VT
P8-13 (g)
The only change from part (e) is:
8-48
P8-14 (a)
Equations:
Rate law:
-rA = k1CACB1/2 + k2CA2
rB = rA
rC= k1CACB1/2 - k3CC + k4CD
rD = k2CA2 – k4CD
rE = k3CC
rG = k3CC
rG = k4CD
Mole Balance:
See polymath problem P8-14-a.pol
Maximum values can be found out from this table :
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
8-49
1
Ca
0.098
0.0064314
0.098
0.0064314
2
Cb
0.049
0.0032157
0.049
0.0032157
3
Cc
0
0
0.0138769
0.0004265
4
Cd
0
0
0.0001148
6.459E-07
5
Ce
0
0
0.0642751
0.0642751
6
Cg
0
0
0.0083756
0.0083756
7
Cto
0.147
0.147
0.147
0.147
8
Cw
0
0
0.0642751
0.0642751
9
Fa
10.
0.9041349
10.
0.9041349
10 Fb
5.
0.4520674
5.
0.4520674
11 Fc
0
0
1.463177
0.0599575
12 Fd
0
0
0.0116183
9.08E-05
13 Fe
0
0
9.035817
9.035817
14 Fg
0
0
1.177438
1.177438
15 Ft
15.
14.87618
20.66532
20.66532
16 Fw
0
0
9.035817
9.035817
17 k1
0.014
0.014
0.014
0.014
18 k2
0.007
0.007
0.007
0.007
19 k3
0.014
0.014
0.014
0.014
20 k4
0.45
0.45
0.45
0.45
21 ra
-0.0003709
-0.0003709
-5.395E-06
-5.395E-06
22 rb
-0.0001855
-0.0001855
-2.698E-06
-2.698E-06
23 rc
0.0003037
-3.328E-05
0.0003037
-5.744E-07
24 rd
6.723E-05
-8.355E-07
6.723E-05
-1.1E-09
25 re
0
0
0.0001943
5.971E-06
26 rg
0
0
5.166E-05
2.906E-07
27 rw
0
0
0.0001943
5.971E-06
28 Sae
0
0
25.89133
0.1000612
29 Scd
0
0
660.3424
660.3424
30 Sdg
0
0
0.0501992
7.711E-05
31 V
0
0
2.0E+05
2.0E+05
32 Yc
0
0
0.2299208
0.0663148
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = rb
3 d(Fc)/d(V) = rc
4 d(Fd)/d(V) = rd
5 d(Fe)/d(V) = re
8-50
6 d(Fw)/d(V) = rw
7 d(Fg)/d(V) = rg
Explicit equations
1
Cto = 0.147
2
k1 = 0.014
3
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
4
k2 = 0.007
5
Cb = Cto*(Fb/Ft)
6
k3 = 0.014
7
k4 = 0.45
8
Ca = Cto*(Fa/Ft)
9
Cc = Cto*(Fc/Ft)
10 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
11 rb = ra/2
12 re = k3*Cc
13 Cd = Cto*(Fd/Ft)
14 Ce = Cto*(Fe/Ft)
15 Cg = Cto*(Fg/Ft)
16 Cw = Cto*(Fw/Ft)
17 rg = k4*Cd
18 rw = k3*Cc
19 rd = k2*(Ca)^2-k4*Cd
20 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
21 Yc = Fc/Fa
22 Scd = if (V>0.0001)then (Fc/Fd) else (0)
23 Sae = if (V>0.0001)then (Fa/Fe) else (0)
24 Sdg = if (V>0.0001)then (Fd/Fg) else (0)
8-51
8-52
8-53
8-54
P8-14(b)
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.098
0.0951293
0.098
0.0951293
2
Cb
0.049
0.0475647
0.049
0.0475647
3
Cc
0
0
0.0032152
0.0032152
4
Cd
0
0
0.0001412
0.0001411
5
Ce
0
0
0.0002229
0.0002229
6
Cg
0
0
0.0005038
0.0005038
7
Cto
0.147
0.147
0.147
0.147
8
Cw
0
0
0.0002229
0.0002229
9
Fa
10.
9.63739
10.
9.63739
10 Fb
5.
4.818695
5.
4.818695
11 Fc
0
0
0.3257277
0.3257277
12 Fd
0
0
0.0143026
0.014299
13 Fe
0
0
0.0225834
0.0225834
14 Fg
0
0
0.0510394
0.0510394
15 Ft
15.
14.89232
15.
14.89232
16 Fw
0
0
0.0225834
0.0225834
17 k1
0.014
0.014
0.014
0.014
18 k2
0.007
0.007
0.007
0.007
19 k3
0.014
0.014
0.014
0.014
20 k4
0.45
0.45
0.45
0.45
21 ra
-0.0003709
-0.0003709
-0.0003538
-0.0003538
22 rb
-0.0001855
-0.0001855
-0.0001769
-0.0001769
23 rc
0.0003037
0.0003037
0.000336
0.000309
8-55
24 rd
6.723E-05
-1.674E-07
6.723E-05
-1.674E-07
25 re
0
0
4.501E-05
4.501E-05
26 rg
0
0
6.352E-05
6.351E-05
27 rw
0
0
4.501E-05
4.501E-05
28 Sae
0
0
7.875E+05
426.7466
29 Scd
0
0
22.77975
22.77975
30 Sce
0
0
594.3536
14.42332
31 Sdg
0
0
18.10358
0.2801563
32 V
0
0
1000.
1000.
33 Yc
0
0
0.0337983
0.0337983
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = rb
3 d(Fc)/d(V) = rc
4 d(Fd)/d(V) = rd
5 d(Fe)/d(V) = re
6 d(Fw)/d(V) = rw
7 d(Fg)/d(V) = rg
Explicit equations
1
Cto = 0.147
2
k1 = 0.014
3
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
4
k2 = 0.007
5
Cb = Cto*(Fb/Ft)
6
k3 = 0.014
7
k4 = 0.45
8
Ca = Cto*(Fa/Ft)
9
Cc = Cto*(Fc/Ft)
10 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
11 rb = ra/2
12 re = k3*Cc
13 Cd = Cto*(Fd/Ft)
14 Ce = Cto*(Fe/Ft)
15 Cg = Cto*(Fg/Ft)
16 Cw = Cto*(Fw/Ft)
17 rg = k4*Cd
18 rw = k3*Cc
8-56
19 rd = k2*(Ca)^2-k4*Cd
20 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
21 Yc = Fc/Fa
22 Scd = if (V>0.0001)then (Fc/Fd) else (0)
23 Sae = if (V>0.0001)then (Fa/Fe) else (0)
24 Sdg = if (V>0.0001)then (Fd/Fg) else (0)
25 Sce = if (V>0.0001)then (Fc/Fe) else (0)
Then taking different values of FA0 and FB0, we find the following data:
FA0
FB0 Θ= FA0/FB0
YC
SC/E
0.01
1
5
10
100
1000
5
5
5
5
5
5
SD/G
SC/D
0.002
0.2
1
2
20
200
0.139038
4.424789
0.070678
3.48E+04
0.111343
5.379092
0.088876
321.5455
0.057955
9.342066
0.168248
52.61217
0.033798
14.42332
0.280156
22.77975
0.001968
107.1563
2.87001
1.931864
5.99E-05
1001.268
30.05269
0.413009
50
100
150
200
250
0.16
0.14
0.12
0.1
0.08
Yc
0.06
0.04
0.02
0
-50
-0.02 0
ΘO2
8-57
1200
1000
800
SC/E
600
400
200
0
-50
0
50
100
150
200
250
150
200
250
ΘO2
35
30
25
20
SD/G
15
10
5
0
-50
-5
0
50
100
ΘO2
8-58
25
20
15
SD/G
10
5
0
-500
0
500
-5
1000
1500
2000
2500
ΘO2
P8-14(c)
Now we have a pressure drop parameter α = 0.002, so we modify our polymath program as follows:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.002
0.002
0.002
0.002
2
Ca
0.098
0.0041647
0.098
0.0041647
3
Cb
0.049
0.0020823
0.049
0.0020823
4
Cc
0
0
0.0004581
3.917E-05
5
Cd
0
0
5.56E-05
3.278E-06
6
Ce
0
0
8.655E-06
9.533E-07
7
Cg
0
0
3.75E-05
3.755E-06
8
Cto
0.147
0.147
0.147
0.147
9
Cw
0
0
8.655E-06
9.533E-07
10 Fa
10.
9.896873
10.
9.896873
11 Fb
5.
4.948437
5.
4.948437
12 Fc
0
0
0.0930717
0.0930717
13 Fd
0
0
0.0082669
0.0077896
14 Fe
0
0
0.0022655
0.0022655
15 Fg
0
0
0.0089238
0.0089238
16 Ft
15.
14.95962
15.
14.95963
17 Ft0
15.
15.
15.
15.
18 Fw
0
0
0.0022655
0.0022655
19 k1
0.014
0.014
0.014
0.014
20 k2
0.007
0.007
0.007
0.007
21 k3
0.014
0.014
0.014
0.014
8-59
22 k4
0.45
0.45
0.45
0.45
23 ra
-0.0003709
-0.0003709
-2.782E-06
-2.782E-06
24 rb
-0.0001855
-0.0001855
-1.391E-06
-1.391E-06
25 rc
0.0003037
3.587E-06
0.0003037
3.587E-06
26 rd
6.723E-05
-4.638E-06
6.723E-05
-1.354E-06
27 re
0
0
6.413E-06
5.483E-07
28 rg
0
0
2.502E-05
1.475E-06
29 rw
0
0
6.413E-06
5.483E-07
30 Sae
0
0
3.344E+06
4368.521
31 Scd
0
0
11.94813
11.94813
32 Sce
0
0
1218.255
41.08224
33 Sdg
0
0
37.46977
0.8729093
34 V
0
0
500.
500.
35 y
1.
0.0428242
1.
0.0428242
36 Yc
0
0
0.0094042
0.0094042
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = rb
3 d(Fc)/d(V) = rc
4 d(Fd)/d(V) = rd
5 d(Fe)/d(V) = re
6 d(Fw)/d(V) = rw
7 d(Fg)/d(V) = rg
8 d(y)/d(V) = -alpha/2/y*(Ft/Ft0)
Explicit equations
1
alpha = 0.002
2
Ft0 = 15
3
Cto = 0.147
4
k1 = 0.014
5
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
6
k2 = 0.007
7
Cb = Cto*(Fb/Ft)*y
8
k3 = 0.014
9
k4 = 0.45
10 Ca = Cto*(Fa/Ft)*y
11 Cc = Cto*(Fc/Ft)*y
12 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
8-60
13 rb = ra/2
14 re = k3*Cc
15 Cd = Cto*(Fd/Ft)*y
16 Ce = Cto*(Fe/Ft)*y
17 Cg = Cto*(Fg/Ft)*y
18 Cw = Cto*(Fw/Ft)*y
19 rg = k4*Cd
20 rw = k3*Cc
21 rd = k2*(Ca)^2-k4*Cd
22 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
23 Yc = Fc/Fa
24 Scd = if (V>0.0001)then (Fc/Fd) else (0)
25 Sae = if (V>0.0001)then (Fa/Fe) else (0)
26 Sdg = if (V>0.0001)then (Fd/Fg) else (0)
27 Sce = if (V>0.0001)then (Fc/Fe) else (0)
8-61
8-62
8-63
P8-14(d) Since C(Formic acid) is our desired product, so temperature corresponding to the maximum
yield of C should be recommended.
8-64
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.098
3.689E-08
0.098
3.689E-08
2
Cb
0.049
1.844E-08
0.049
1.844E-08
3
Cc
0
0
0.0150519
1.497E-12
4
Cd
0
0
0.0089776
1.018E-10
5
Ce
0
0
0.0496431
0.0496431
6
Cg
0
0
0.0477138
0.0477138
7
Cto
0.147
0.147
0.147
0.147
8
Cw
0
0
0.0496431
0.0496431
9
Fa
10.
7.43E-06
10.
7.43E-06
10 Fb
5.
3.715E-06
5.
3.715E-06
11 Fc
0
0
1.854
3.016E-10
12 Fd
0
0
1.030757
2.05E-08
13 Fe
0
0
9.999993
9.999993
14 Fg
0
0
9.611354
9.611354
15 Ft
15.
14.91758
29.61135
29.61135
16 Fw
0
0
9.999993
9.999993
17 k1
0.014
0.014
30.64368
30.64368
18 k10
0.014
0.014
0.014
0.014
19 k2
0.007
0.007
7.341E+07
7.341E+07
20 k20
0.007
0.007
0.007
0.007
21 k3
0.014
0.014
6.707E+04
6.707E+04
22 k30
0.014
0.014
0.014
0.014
23 k4
0.45
0.45
984.9755
984.9755
24 k40
0.45
0.45
0.45
0.45
25 ra
-0.0003709
-0.0625305
-1.0E-07
-1.0E-07
26 rb
-0.0001855
-0.0312653
-5.002E-08
-5.002E-08
27 rc
0.0003037
-0.0238556
0.0236439
-5.93E-12
28 rd
6.723E-05
-0.0093009
0.0120515
-3.804E-10
29 re
0
0
0.0773459
1.004E-07
30 rg
0
0
0.0627128
1.003E-07
31 rw
0
0
0.0773459
1.004E-07
32 Sae
0
0
5.205E+05
7.43E-07
33 Scd
0
0
3.394658
0.0147083
34 Sdg
0
0
18.42613
2.133E-09
35 T
300.
300.
550.
550.
36 V
0
0
1000.
1000.
37 Yc
0
0
0.4483122
4.059E-05
8-65
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = rb
3 d(Fc)/d(V) = rc
4 d(Fd)/d(V) = rd
5 d(Fe)/d(V) = re
6 d(Fw)/d(V) = rw
7 d(Fg)/d(V) = rg
8 d(T)/d(V) = 0.25
Explicit equations
1
k20 = 0.007
2
k10 = 0.014
3
k1 = k10*exp((10000/1.97)*(1/300-1/T))
4
k2 = k20*exp((30000/1.97)*(1/300-1/T))
5
k30 = 0.014
6
k3 = k30*exp((20000/1.97)*(1/300-1/T))
7
k40 = 0.45
8
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
9
Cto = 0.147
10 Ca = Cto*(Fa/Ft)
11 Cb = Cto*(Fb/Ft)
12 Cc = Cto*(Fc/Ft)
13 Cd = Cto*(Fd/Ft)
14 Ce = Cto*(Fe/Ft)
15 Cg = Cto*(Fg/Ft)
16 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
17 rb = ra/2
18 re = k3*Cc
19 k4 = k40*exp((10000/1.97)*(1/300-1/T))
20 Cw = Cto*(Fw/Ft)
21 rg = k4*Cd
22 rw = k3*Cc
23 rd = k2*(Ca)^2-k4*Cd
24 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
25 Yc = Fc/Fa
26 Scd = if (V>0.0001)then (Fc/Fd) else (0)
27 Sae = if (V>0.0001)then (Fa/Fe) else (0)
8-66
28 Sdg = if (V>0.0001)then (Fd/Fg) else (0)
Temperature corresponding to maximum yield = 367.8 K
P8-15
(1)
C2H4 + 1/2O2 → C2H4O
E + 1/2O → D
(2)
E
C2H4 + 3O2 → 2CO2 + 2H2O
+ 3O → 2U1 + 2U2
8-67
FIO
0.82 FTO
0.007626
P8-15 (a)
Selectivity of D over CO2
S
FD
FU 1
See Polymath program P8-15-a.pol.
POLYMATH Results
Variable
W
Fe
Fo
Fd
Fu1
Fu2
Finert
Ft
K1
K2
Pto
Pe
Po
k1
k2
X
S
r1e
r2e
initial value
0
5.58E-04
0.001116
0
1.0E-07
0
0.007626
0.0093001
6.5
4.33
2
0.1199987
0.2399974
0.15
0.088
0
0
-0.0024829
-0.0029803
minimal value
0
1.752E-10
4.066E-05
0
1.0E-07
0
0.007626
0.0091804
6.5
4.33
2
3.817E-08
0.008858
0.15
0.088
0
0
-0.0024829
-0.0029803
maximal value
2
5.58E-04
0.001116
2.395E-04
6.372E-04
6.371E-04
0.007626
0.0093001
6.5
4.33
2
0.1199987
0.2399974
0.15
0.088
0.9999997
0.4101512
-3.692E-10
-8.136E-10
Differential equations as entered by the user
[1] d(Fe)/d(W) = r1e+r2e
[2] d(Fo)/d(W) = 1/2*r1e + 3*r2e
[3] d(Fd)/d(W) = -r1e
[4] d(Fu1)/d(W) = -2*r2e
[5] d(Fu2)/d(W) = -2*r2e
Explicit equations as entered by the user
[1] Finert = 0.007626
[2] Ft = Fe+Fo+Fd+Fu1+Fu2+Finert
[3] K1 = 6.5
[4] K2 = 4.33
[5] Pto = 2
[6] Pe = Pto*Fe/Ft
[7] Po = (Pto*Fo/Ft)
[8] k1 = 0.15
[9] k2 = 0.088
[10] X = 1 - Fe/0.000558
[11] S = Fd/Fu1
[12] r1e = -k1*Pe*Po^0.58/(1+K1*Pe)^2
[13] r2e = -k2*Pe*Po^0.3/(1+K2*Pe)^2
8-68
final value
2
1.752E-10
4.066E-05
2.395E-04
6.372E-04
6.371E-04
0.007626
0.0091804
6.5
4.33
2
3.817E-08
0.008858
0.15
0.088
0.9999997
0.3758225
-3.692E-10
-8.136E-10
X = 0.999 and S = 0.376(mol of ethylene oxide)/(mole of carbon dioxide)
P8-15(b)
Changes in equation from part (a):
dFO
dW
RO
1
r1E 3r 2 E RO
and FO o
2
0.12 0.0093 0.001116 mol
W
2
kg.s
From Polymath program:
0
X = 0.71
S = 0.04 (mol of ethylene oxide)/(mole of carbon dioxide)
P8-15 (c)
Changes in equation from part (a):
dFE
dW
r1E r 2 E RE
and FE o
RE
0.06 0.0093
W
0.000558 mol
2
kg.s
From Polymath program:
0
X = 0.96
S = 0.41(mol of ethylene oxide)/(mole of carbon dioxide)
P8-15 (d) No solution will be given.
P8-16
The reactions are
1. C + W  6 H2 + 6CO
2. L + W  13 H2 + 10 CO
Rate laws are :
- r1C = k1C CC CW
-r2L = k2L CL Cw2
CT0 = P0/RT = ( 1 atm )/(0.082)(1473) = 0.00828 mol/dm3
FC0 = 0.00411 mol/s
FL0 = 0.0185 mol/s
FW0 = 0.02 mol/s
P8-16 (a)
See polymath program P 8-16(a)
Calculated values of DEQ variables
8-69
Variable Initial value Minimal value Maximal value Final value
1
ConvC
0
0
0.8337198
0.8337198
2
ConvL
0
0
0.7379696
0.7379696
3
Ct0
0.0082791
0.0082791
0.0082791
0.0082791
4
Fc
0.0041152
0.0006843
0.0041152
0.0006843
5
Fc0
0.0041152
0.0041152
0.0041152
0.0041152
6
Fco
0
0
0.0342518
0.0342518
7
Fh
0
0
0.0383516
0.0383516
8
Fl
0.0018519
0.0004852
0.0018519
0.0004852
9
Fl0
0.0018519
0.0018519
0.0018519
0.0018519
10 Ft
0.0259671
0.0259671
0.0807757
0.0807757
11 Fw
0.02
0.0070028
0.02
0.0070028
12 k1c
3.0E+04
3.0E+04
3.0E+04
3.0E+04
13 k2l
1.4E+07
1.4E+07
1.4E+07
1.4E+07
14 Mc
0.6666673
0.1108536
0.6666673
0.1108536
15 Mco
0
0
0.9590499
0.9590499
16 Mh
0
0
0.0767032
0.0767032
17 Ml
0.333333
0.0873434
0.333333
0.0873434
18 Mw
0.36
0.1260502
0.36
0.1260502
19 P
1.
1.
1.
1.
20 R
0.082
0.082
0.082
0.082
21 r1c
-0.2509955
-0.2509955
-0.0015102
-0.0015102
22 r2l
-0.3361048
-0.3361048
-0.0003587
-0.0003587
23 T
1473.
1473.
1473.
1473.
24 V
0
0
0.417
0.417
Differential equations
1 d(Fc)/d(V) = r1c
2 d(Fl)/d(V) = r2l
3 d(Fw)/d(V) = r1c+7*r2l
4 d(Fh)/d(V) = -6*r1c-13*r2l
5 d(Fco)/d(V) = -6*r1c-10*r2l
Explicit equations
1
T = 1473
2
R = 0.082
3
P=1
4
k2l = 14000000
5
k1c = 30000
8-70
6
Ft = Fc+Fl+Fw+Fh+Fco
7
Ct0 = P/(R*T)
8
Fc0 = 0.00411523
9
Fl0 = 0.00185185
10 Mc = Fc*162
11 Ml = Fl*180
12 Mh = Fh*2
13 Mco = Fco*28
14 Mw = Fw*18
15 r1c = -k1c*(Ct0*(Fc)/(Ft))*(Ct0*(Fw)/(Ft))
16 r2l = -k2l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2)
17 ConvC = (Fc0-Fc)/Fc0
18 ConvL = (Fl0-Fl)/Fl0
Plot of Fc vs V
8-71
Plot of Fl vs V
Plot of Fw vs V
8-72
Plot of Fh vs V
plot of Fco vs V
(b) W is the key reactant
YC = (-dCC/-dCW) = -r1C/(-r1C -7*r2L)
8-73
The modification is made in the polymath code for Prob 8.16(a)
The plot of YC versus V is as follows ;
For poltting YW versus volume ;
since W is our key reactant then,
Yw = -dW/(-dW) = 1
Now YL = -dL/(-dW) = -r2L/(-r1C – 7*r2L)
The plot is as follows :
8-74
Overall selectivity
SCO / H 2 = NCO = exit molar flow rate of CO/exit molar flow rate of H2 =
N H2
CCO |exit /CH2 | exit at any point in the reactor.
P8-16 (c) Individualized solution
P8-17 (a)
The reactions are
(i) L + 3 W  3 H2 + 3 CO + char
(ii) Ch + 4 W  10 H2 + 7CO
8-75
rate laws at 1200 0C are :
-r1L = k1LCLCw2
-r2CH = k2ChCchCw2
; k1L = 3721 (dm3/mol)2/sec
; k2CH = 1000 ( dm3/mol)2/sec
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 ConL
0
0
0.99
0.99
2 Ct0
0.2
0.2
0.2
0.2
3 Fch
0
0
0.0076175
0.0047293
4 Fco
0
0
0.089117
0.089117
5 Fh
0
0
0.1115957
0.1115957
6 Fl
0.0123457 0.0001235
0.0123457
0.0001235
7 Fl0
0.0123457 0.0123457
0.0123457
0.0123457
8 Ft
0.1240226 0.1240226
0.2506041
0.2506041
9 Fw
0.1116769 0.0450387
0.1116769
0.0450387
10 k1l
3721.73
3721.73
3721.73
3721.73
11 k2ch
1000.
1000.
1000.
1000.
12 r1l
-2.403113
-2.403113
-0.0004738
-0.0004738
13 r2ch
0
-0.0992252
0
-0.0048763
14 V
0
0
0.417002
0.417002
Differential equations
1 d(Fl)/d(V) = r1l
2 d(Fch)/d(V) = r2ch-r1l
3 d(Fw)/d(V) = 3*r1l+4*r2ch
4 d(Fh)/d(V) = -3*r1l-10*r2ch
5 d(Fco)/d(V) = -3*r1l-7*r2ch
Explicit equations
1 k2ch = 1000
2 k1l = 3721.73
3 Ft = Fch+Fl+Fw+Fh+Fco
4 Ct0 = 0.2
5 r1l = -k1l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2)
8-76
6 r2ch = -k2ch*(Ct0*(Fch)/(Ft))*((Ct0*(Fw)/(Ft))^2)
7 Fl0 = 0.012345679
8 ConL = (Fl0-Fl)/(Fl0)
8-77
P8-17 (b)
(c) SCO/Ch
Let S1 = S~CO/Ch = Fco/Fch
The plot is as follows-
8-78
Considering W as the principle reacting species;
Yw = 1
and Yl = (rl)/(rw) = (r1l)/(3*r1l + 4*r2ch)
the plot is as follows:
(d)
The molar flow rate for char is maximum at about V= 0.02085 dm3.
8-79
P8-18 (a)
Isothermal gas phase reaction in a membrane reactor packed with catalyst.
A
r1'C
B C
k1C C A
AD
r2' D
k2 D C A
2C + D  2E
r3' E
k3 E CC2 CD
C AO
P
RT
CB CC
K1C
24.6atm
0.082dm atm / mol.K (500 K )
3
0.6mol / dm3
Fa(0) = 10 mol/min
Fb(0) = Fc(0) = Fd(0) = Fe(0) = 0
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alfa
0.008
0.008
0.008
0.008
2 Ca
0.6
0.0029024
0.6
0.0029024
3 Cb
0
0
0.1525032
0.0075281
4 Cc
0
0
0.1425644
0.0541764
5 Cd
0
0
0.0254033
0.0036956
6 Ce
0
0
0.274725
0.235079
7 Cto
0.6
0.6
0.6
0.6
8 Fa
10.
0.0702103
10.
0.0702103
9 Fb
0
0
3.391481
0.1821054
10 Fc
0
0
3.16936
1.310531
11 Fd
0
0
0.5564071
0.0893965
12 Fe
0
0
5.686575
5.686575
13 Ft
10.
7.338818
12.8179
7.338818
14 Fto
10.
10.
10.
10.
15 k1c
2.
2.
2.
2.
16 K1c
0.2
0.2
0.2
0.2
17 k2d
0.4
0.4
0.4
0.4
18 k3e
400.
400.
400.
400.
8-80
19 kb
1.
1.
1.
1.
20 r1c
1.2
0.0017264
1.2
0.0017264
21 r2d
0.24
0.001161
0.24
0.001161
22 r3e
0
0
0.200243
0.0043387
23 ra
-1.44
-1.44
-0.0028874
-0.0028874
24 rb
1.2
0.0017264
1.2
0.0017264
25 rc
1.2
-0.0492057
1.2
-0.0026123
26 rd
0.24
-0.0164816
0.24
-0.0010084
27 re
0
0
0.200243
0.0043387
28 W
0
0
100.
100.
29 y
1.
0.5056359
1.
0.5056359
Differential equations
1 d(Fa)/d(W) = ra
2 d(Fb)/d(W) = rb-(kb*Cb)
3 d(Fc)/d(W) = rc
4 d(Fd)/d(W) = rd
5 d(Fe)/d(W) = re
6 d(y)/d(W) = -alfa*Ft/(2*Fto*y)
Explicit equations
1 k2d = 0.4
2 K1c = 0.2
3 Ft = Fa+Fb+Fc+Fd+Fe
4 Cto = 0.6
5 Cb = Cto*(Fb/Ft)*y
6 Ca = Cto*(Fa/Ft)*y
7 Cd = Cto*(Fd/Ft)*y
8 Cc = Cto*(Fc/Ft)*y
9 kb = 1
10 k1c = 2
8-81
11 r2d = k2d*Ca
12 k3e = 400
13 r1c = k1c*(Ca-(Cb*Cc/K1c))
14 ra = -r1c-r2d
15 r3e = k3e*(Cc^2)*Cd
16 rd = r2d-(r3e/2)
17 rb = r1c
18 rc = r1c-r3e
19 re = r3e
20 Ce = Cto*(Fe/Ft)*y
21 alfa = 0.008
22 Fto = 10
P8-18 (b)
Species B, C, D, and E all go though a maximum. The concentration of species are affected by
two factors: reaction and pressure drop. We can look at species B for example:
Species B:
8-82
Production by reaction 1 => tends to increase concentration of B
Pressure drop => tends to decrease concentration of B
At W<10, concentration of B rises because the production of B by reaction outweighs the
dilution of B by pressure drop. Further down the reactor, the concentration of B drops because
the effect of pressure drop outweighs the production of B by reaction.
Species E:
The reasoning for species E is similar to species B.
Species C and D:
Species C and D are produced in reactions 1 and 2, respectively. However, species C and D are
also reactants in reaction 3; this contributes to their decline at later times (in addition to the
pressure drop).
P8-18 (c)
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alfa
0.008
0.008
0.008
0.008
2 Ca
0.6
0.0024106
0.6
0.0024106
3 Cb
0
0
0.2447833
0.0593839
4 Cc
0
0
0.1181101
0.0028718
5 Cd
0
0
0.0408517
0.0119119
6 Ce
0
0
0.1034882
0.0252849
7 Cto
0.6
0.6
0.6
0.6
8 Fa
10.
0.2791706
10.
0.2791706
9 Fb
0
0
6.877205
6.877205
10 Fc
0
0
2.623093
0.3325842
11 Fd
0
0
1.379513
1.379513
12 Fe
0
0
2.928221
2.928221
13 Ft
10.
10.
13.00145
11.79669
14 Fto
10.
10.
10.
10.
15 k1c
2.
2.
2.
2.
16 K1c
0.2
0.2
0.2
0.2
8-83
17 k2d
0.4
0.4
0.4
0.4
18 k3e
400.
400.
400.
400.
19 kb
1.
1.
1.
1.
20 r1c
1.2
0.0031158
1.2
0.0031158
21 r2d
0.24
0.0009642
0.24
0.0009642
22 r3e
0
0
0.1548711
3.93E-05
23 ra
-1.44
-1.44
-0.0040801
-0.0040801
24 rb
1.2
0.0031158
1.2
0.0031158
25 rc
1.2
0.001223
1.2
0.0030765
26 rd
0.24
0.0009446
0.24
0.0009446
27 re
0
0
0.1548711
3.93E-05
28 W
0
0
100.
100.
29 y
1.
0.169772
1.
0.169772
Differential equations
1 d(Fa)/d(W) = ra
2 d(Fb)/d(W) = rb
3 d(Fc)/d(W) = rc-(kb*Cc)
4 d(Fd)/d(W) = rd
5 d(Fe)/d(W) = re
6 d(y)/d(W) = -alfa*Ft/(2*Fto*y)
Explicit equations
1 k2d = 0.4
2 K1c = 0.2
3 Ft = Fa+Fb+Fc+Fd+Fe
4 Cto = 0.6
5 Cb = Cto*(Fb/Ft)*y
6 Ca = Cto*(Fa/Ft)*y
7 Cd = Cto*(Fd/Ft)*y
8 Cc = Cto*(Fc/Ft)*y
8-84
9 kb = 1
10 k1c = 2
11 r2d = k2d*Ca
12 k3e = 400
13 r1c = k1c*(Ca-(Cb*Cc/K1c))
14 ra = -r1c-r2d
15 r3e = k3e*(Cc^2)*Cd
16 rd = r2d-(r3e/2)
17 rb = r1c
18 rc = r1c-r3e
19 re = r3e
20 Ce = Cto*(Fe/Ft)*y
21 alfa = 0.008
22 Fto = 10
Major Differences:
8-85
(1) When C diffuses out instead of B, the concentration of C is lower and the concentration
of B is higher.
(2) The concentration of D is higher because reaction 3 is slower (C is a reactant in reaction
3)
(3) The concentration of E is lower because of the same reasoning as (2).
P8-18 (d) Individualized Solution
P8-20
The reactions are :
k3
2A
A
k1
D
B C
Volume of PFR = 100 dm3
flow rate =υ = 10 dm3/min
C0 = 3 mol / m3
K1 = 0.05 min
K3 = 0.015 (dm3/mol)/min
Kc = 0.5 dm3 /mol
The mistakes in the solution are as follows :
k1
(i)
k1 is the rate constant for the reaction for A
B C . But the rate constants for
(ii)
(iii)
the reactions have been reversed.
The molar flow rates have not been used when rate laws are written in the code.
The rate law expression for various species (A,B,C,D) is incorrect .
The mistakes are corrected hereby;
The rate laws are written as follows;
rC = rB = k1(CA – (1/KC )CCCB) = 0.05 *(CA – CB*CC/0.5)
rA = - k3*CA – k1(CA – (1/KC )CCCB) = - 0.05 *(CA – CA*CC/0.5) - 0.015*CA
rD = 1/2 *k3*CA = 0.0075
Now for plotting the concentrations with the volume we need the performance
equation for PFR;
Thus
8-86
-rA *dV = dFA
dFA/dV = -rA
dCa/dV = -1/υ *rA
This being a gaseous reaction the volumetric flow rate will change with volume as
we proceed in the PFR,
Assuming the extent of reactions to be XA1 and XA2 for the reactions 1 and 2 ;
then υ =υ0 *(1 + εA1XA1 +εA2XA2 )
the performance equations will be ;
dXa/dV =
CA0 * 0 (1
1
A1 X A1
A2 X A2 )
* Ca0*(k 3*(1 Xa) k1((1 Xa) (1/ KC )* B * C *(1 Xa1)*(1 Xa2))
Here ,
Xa = Xa1 + Xa2
Similarly performance equations for other species in the reactor can be written and the
equations will have to be solved simultaneously for getting the plots.
8-87
CDGA8-1
8-88
8-89
8-90
8-91
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9-1
Solutions for Chapter 9 – Reaction Mechanisms, Pathways,
Bioreactions and Bioreactors
P9-1 Individualized Solution
P9-2 (a) Example 9-1
The graph of Io/I will remain same if CS2 concentration changes. If concentration of M increases the
slope of line will decrease.
P9-2 (b) Example 9-2
The inhibitor shows competitive inhibition.
P9-2 (c) Example 9-3
1)Now Curea = 0.001mol/dm3 and t = 10 min = 600 sec.
t
kM
1
ln
VMAX
1 X
600 sec
Curea X
VMAX
0.0266mol / dm 3
1
ln
3
1 X
0.000266mol / s.dm
0.001mol / dm 3 X
0.000266mol / s.dm 3
Solving, we get X = 0.9974.
t 461.7 sec
2) For CSTR,
C urea X
rurea
rurea
rurea
Curea X
9-2
VMAX Curea
K M Curea
0.000266mol / s.dm 3 0.1mol / dm 3 1 X
0.0266mol / dm 3 0.1mol / dm 3 1 X
Solving, we get X = 0.675
See Polymath program P9-2-c.pol.
POLYMATH Results
NLE Solution
Variable Value
f(x)
Ini Guess
X
0.6751896 -8.062E-10 0.5
NLE Report (safenewt)
Nonlinear equations
[1] f(X) = 0.0000266*(1-X)/(0.0266+0.1*(1-X))-0.1*X/461.7 = 0
3) For PFR,
CO
dC urea
r
Cures
and C urea
kM
1
ln
VMAX
1 X
Curea X
VMAX
C urea0 1 X
Same as batch reactor, but t replaced by
X = 0.8
P9-2 (d) Example 9-4
YS / P
YP / S
YS / C P
CS
CP
1
238.7 245
5.03 2.14
1
2.18
YS / P
1
YC P / S
2.18 g / g
0.46 g / g
1
0.075 0.46
1.87 g / g
Yes there is disparity as substrate is also used in maintenance.
P9-2 (e) Example 9-5 See Polymath program P9-2-e.pol.
Part (1)
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t
0
0
24
24
Cc
2.0E-04
1.276E-04 0.002742 0.002742
9-3
0.1mol / dm 3 X
461.7 sec
Cs
Cp
rd
Ysc
Ypc
Ks
m
umax
rsm
kobs
rg
vo
Cso
Vo
V
mp
5.0E-04
5.0E-04
4.5794959 4.5794959
0
0
0.0159354 0.0159354
2.0E-06
1.277E-06 2.742E-05 2.742E-05
12.5
12.5
12.5
12.5
5.6
5.6
5.6
5.6
1.7
1.7
1.7
1.7
0.03
0.03
0.03
0.03
0.33
0.33
0.33
0.33
6.0E-06
3.83E-06
8.226E-05 8.226E-05
0.33
0.3299706 0.33
0.3299706
1.941E-08 1.941E-08 6.598E-04 6.598E-04
0.5
0.5
0.5
0.5
5
5
5
5
1
1
1
1
1
1
13
13
0
0
0.20716
0.20716
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd-vo*Cc/V
[2] d(Cs)/d(t) = Ysc*(-rg)-rsm+(Cso-Cs)*vo/V
[3] d(Cp)/d(t) = rg*Ypc-vo*Cp/V
Explicit equations as entered by the user
[1] rd = Cc*.01
[2] Ysc = 1/.08
[3] Ypc = 5.6
[4] Ks = 1.7
[5] m = .03
[6] umax = .33
[7] rsm = m*Cc
[8] kobs = (umax*(1-Cp/93)^.52)
[9] rg = kobs*Cc*Cs/(Ks+Cs)
[10] vo = 0.5
[11] Cso = 5
[12] Vo = 1
[13] V = Vo+vo*t
[14] mp = Cp*V
Comments
[2] d(Cs)/d(t) = Ysc*(-rg)-rsm+(Cso-Cs)*vo/V
Cso is concentration of substrate in feed stream, NOT initial concentration of substrate in reactor
[17] mp = Cp*V
mass of product in grams
9-4
Part (2)
Change the cell growth rate law:
rg
CC C S
k obs
2
KS
CS
CS
KI
All other equations are the same as in part (1).
9-5
Notice that uncompetitive inhibition by the substrate causes the concentration of cells to decrease with
time, whereas without uncompetitive inhibition, the concentration of cells increased with time.
Consequently, the concentration of product is very low compared to the case without uncompetitive
inhibition.
Part (3)
Change the observed reaction rate constant:
k obs
max 1
CP
10000
0.52
All other equations are the same as in part (1).
CP
Since CP is so small, the factor 1
*
CP
0.52
1 whether CP* = 93 g/dm3 or CP* = 10,000 g/dm3. Thus
the plots for this part are approximately the same as the plots in part (1).
P9-2 (f) Example on CDROM
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.
See Polymath program P9-2-f.pol.
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
C7
0
0
0.0979179 0.0979179
C3
0
0
0.0012475 0.0012475
9-6
C5
C8
CP5
CP1
k5
T
k1
k2
k4
k3
0
0
0.0979179 0.0979179
0
0
6.237E-04 6.237E-04
0
0
0.0979123 0.0979123
0.1
2.166E-04 0.1
2.166E-04
3.98E+09
3.98E+09
3.98E+09 3.98E+09
1000
1000
1000
1000
0.0014964 0.0014964 0.0014964 0.0014964
2.283E+06 2.283E+06 2.283E+06 2.283E+06
9.53E+08
9.53E+08
9.53E+08 9.53E+08
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
[2] d(C2)/d(t) = 2*k1*C1-k2*C1*C2
[3] d(C6)/d(t) = k3*C4-k4*C6*C1
[4] d(C4)/d(t) = k2*C1*C2-k3*C4+k4*C6*C1-k5*C4^2
[5] d(C7)/d(t) = k4*C1*C6
[6] d(C3)/d(t) = k2*C1*C2
[7] d(C5)/d(t) = k3*C4
[8] d(C8)/d(t) = 0.5*k5*C4^2
[9] d(CP5)/d(t) = k3*(2*k1/k5)^0.5*CP1^0.5
[10] d(CP1)/d(t) = -k1*CP1-2*k1*CP1(k3*(2*k1/k5)^0.5)*(CP1^0.5)
P9-2 (g) Individualized solution
P9-3
Burning:
9-7
9-8
9-9
P9-4
k1
CH 3 CHO
CH 3
CH 3
CH 3CHO
k2
CHO
CH 3CHO
k3
2CH 3
rAC
k4
r1
r1
CHO
CH 3
CO CH 4
r2
k 2 C AC CCH3
CH 3
2CO H 2
r3
k 3CCHO C AC
r4
C2 H 6
r2
r3
k1C AC
k 2 C AC CCH3
C AC k1 k 2 CCH3
k 3C AC CCHO
k 3CCHO
Active intermediates: CH 3 , CHO
rCHO
r1
r3
k1C AC
C AC
k1
CCHO
rCH3
0
k 3C AC CCHO
k 3CCHO
0
0
k1
k3
r1 r2 r2 r3 2r4
k1C AC
k1C AC
k3C AC CCHO
0
2k4CCH3 2
0
Plugging in expression for C CHO :
9-10
k 4 CCH3
2
k1C AC
2k4CCH3 2
k1C AC
0
k1C AC
k4
CCH3
Now, substitute expressions for CH 3 and CHO into equation for
rAC
C AC k1
k 2 CCH3
k 3CCHO
C AC k1 k2
k1C AC
k4
k1C AC 2 k2
C AC
k1k4
k1
P9-4 (b)
When k2
C AC
k1k4
1,
rAC
k2
3
k1
C AC 2
k4
3
kC AC 2
9-11
rAC :
P9-4 (c)
CH3CHO
k1
k4
C2H6
CH3•
CHO•
k2
k3
CH4
CO
H2
P9-5 (a)
9-12
P9-5 (b)
k1
Cl2
2Cl *
k2
k3
Cl * CO
COCl *
k4
k5
COCl * Cl2
rCOCl*
COCl 2 Cl *
0 k3 Cl * CO
k4 k5 Cl2
k5 k3 CO Cl * Cl2
k5 COCl * Cl2
rCl
k5 COCl * Cl2
k3 Cl * CO
COCl *
rCOCl2
k4 COCl *
0 k1 Cl2
k2 Cl *
k4 k5 Cl2
2
k3 Cl * CO
add rCOCl to rCl
rCl
rCOCl
Cl *
Cl *
2
0 0 k1 Cl2
k2 Cl
2
k1
Cl2
k2
k1
Cl2
k2
9-13
k4 COCl *
k5 COCl * Cl2
k1
k5 k3 CO
Cl2
k2
k4 k5 Cl2
rCOCl2
k4
0.5
Cl2
k5 k3
k1
k2
k4
1/ 2
3
CO Cl2 2
k5 Cl2
k5 Cl2
rCOCl2
3
k11/ 2 k3 k5
CO Cl2 2
k21/ 2 k4
P9-5 (c)
The proposed mechanism for this reaction is:
Through this mechanism, it may be deduced that the net rate of formation of HBr
(product) is given by:
where the concentrations of the intermediates may be determined by invoking the
steady state approximation, i.e.:
and:
i.e. from (ii) + (i) we obtain:
9-14
i.e.:
which substituted back in (i) gives:
i.e.:
i.e.:
or: dividing 'top' and 'bottom' by k`b :
Thus, the net rate of formation of HBr may be written solely in terms of reactants,
9-15
i.e.:
which simplifies to:
as predicted by the empirical rate law:
where:
9-16
P9-6(a)
9-17
P9-6(b)
Low temperatures with anti-oxidant
9-18
P9-6(c)
If the radicals are formed at a constant rate, then the differential equation for the concentration of the
radicals becomes:
d I
dt
k0
and I
ki I
RH
0
k0
k i RH
The substitution in the differential equation for R· also changes. Now the equation is:
d R
dt
ki I
RH
k P1 R
O2
and solving and substituting gives: R
k P 2 RO 2
k0
RH
0
k P 2 RO 2
k P1 O2
RH
Now we have to look at the balance for RO2·.
d RO 2
dt
k P1 R
O2
k P 2 RO 2
RH
k t RO 2
and if we substitute in our expression for *R·+ we get
0
k0
RO 2
k t RO 2
2
0 which we can solve for [RO2·+.
k0
kt
Now we are ready to look at the equation for the motor oil.
9-19
2
0
d RH
dt
ki I
RH
k P 2 RO 2
RH
and making the necessary substitutions, the rate law for the degradation of the motor oil is:
d RH
dt
r RH
k0
kP 2
k0
RH
kt
P9-6(d)
Without antioxidants
With antioxidants
P9-6 (e) Individualized solution
P9-7 (a)
9-20
P9-7 (b)
P9-7 (c)
P9-7 (d)
See Polymath program P9-7-d.pol.
9-21
Everyone becomes ill rather quickly, and the rate at which an ill person recovers to a healthy person is
much slower than the rate at which a healthy person becomes ill. Eventually everyone is ill and people
start dying.
P9-7 (e) Individualized solution
P9-7 (f) Individualized solution
P9-8 (a)
E S k1 E.S
E.S k2 E S
E.S k3 P E
P E k4 E.S
By applying PSSH for the complex [E.S], we have
k1 [ E ][S ] k 2 [ E.S ] k 3 [ E.S ] k 4 [ P][E ]
k1 [ S ] k 4 [ P ]
[E]
k 2 k3
[ E.S ]
rP
0
rS
k 3 [ E.S ] k 4 [ P][E ]
k 3 k1 [ E ][S ] k 3 k 4 [ P][E ]
k 4 [ P][E ]
k 2 k3
k 3 k1 [ E ][S ] k 3 k 4 [ P][E ] k 2 k 4 [ P][E ] k 3 k 4 [ P][E ]
k 2 k3
k 3 k1 [ S ] k 2 k 4 [ P ]
[E]
k 2 k3
9-22
k1 k 3 [ s ]
k2
[ P]
[E]
k1 k 2
k2k4
k3
Since E is not consumed, we have
where [ ET ] is a constant
[ ET ] [ E ] [ E.S ]
k1 [ S ] k 4 [ P ]
[E]
k 2 k3
[ ET ] [ E ]
k2
k3
k1 [ S ] k 4 [ P ]
[E]
k 2 k3
So,
k1 k 3 [ S ]
rP
k2
k3
[ P]
[ ET ]
ke
k1k 3
k2k4
where k e
k1 [ S ] k 4 [ P ]
P9-8 (b)
E
k1
S
E S
E P
E S
k3
E P
E P
k5
P
0 k3 E S
E P
k3 E S
k 4 k5
k1 E S
k4 E P
k2 E S
ET
k4
S
E S
k3
E S
k4
k5
k5 E P
k3 E S
since E is not consumed: ET
E
E
E
rE S
rE S
k2
E S
E
E S
k4 E P
E S E P or E
E
ET
E S
ET
1
E S E P
k3
k4
k5
Insert this into the equation for rE·S and solve for the concentration of the intermediate:
E S
k1 S E T
1
rP
k5 E P
k3
k4
k1 S
k2
k3
k 4 k3
k 4 k5
k5
k 3 k5 E S
k4 k5
9-23
rP
k1k3 k5 S ET
k4 k5 k1 S k2 k4
k3
k3k5
P9-8 (c)
k1
E S1
k2
E S1 S2
E S1
k3
E S1S2
k4
k5
E S1S2
P E
(1) rE S1
0 k1 E S1
(2) rE S1S2
0 k3 E S S2
k2 E S
k3 E S S2
k4 E S1S2
k4 E S1S2
k5 E S1S2
k2 E S
k5 E S1S2
If we add these two rates we get:
(3) rE S1
rE S1S2
0 k1 E S
k3 E S S2
From equation (2) we get E S1S2
k4 k5
Plug this into equation 3 and we get:
k1 E S1
k3 S 2
k2
k 4 k5
E S1
E S1S2
k3 S2 k1 E S1
k4 k5
rP
k5 E S1S2
ET
E
ET
E 1
E S
k1k3 E S1 S2
k1k2 k5 S1 S2
k2 k4
k2 k4 k5 k2 k3 S2
k3 S2
k4 k5
k2
E
k2 k5 k3 S2
E S1S2
k1 S1
k3 S 2
k2
k 4 k5
k1k3 S1 S 2
k2 k4
k5 k 2
k3 S 2
9-24
k1k2 k5 S1 S2
rP
ET
k1 S1
k3 S2
k2
k4 k5
k 2 k 4 k 2 k5 k3 S 2 1
k1k3 S1 S2
k2 k4 k5 k2 k3 S2
P9-8 (d)
k3
k1
E S
E S
k2
k4
E P
E P
k5
rE S
0 k1 E S
E S
rE P
k1 S
E
k2
k3
rP
rP
k5 E P
k5
k3 E S
k4 E P
k1k3 S E
k1k3 S
k2
k5 E P
k4 E P
k2 k3
k4 E P
E
k3
E
ET
E
ET
E 1
E S
k1 S
E P
E
k4 E P
k2 k3
k5
k1 S
k4 P
k2
k5
k3
k1k3 S ET
rP
k1 S
k2 k3
k2 k3 1
rP
k3 E S
k4 E P
ET
rP
k2 E S
0 k4 E P
E P
rP
P
k4 P
k5
k3 S ET
k2 k3
k1
S
k4 k2 k3
k1k5
P
Vmax S
KM
S
KP P
9-25
P9-8 (e) No solution will be given
P9-8 (f) No solution will be given
P9-9 (a)
The enzyme catalyzed reaction of the decomposition of hydrogen peroxide. For a batch reactor:
9-26
P9-9 (b)
P9-9 (c) Individualized solution
P9-9 (d) Individualized solution
9-27
P9-10 (a)
P9-10 (b)
9-28
9-29
X
CS 0 CS
CS 0
50 2.1
.958
50
P9-10 (c)
rS
kCS ET
1 K1Cs K 2CS2
If ET is reduced by 33%, -rS will also decrease by 33%. From the original plot, we see that if the curve –rS
is decreased by 33%, the straight line from the CSTR calculation will cross the curve only once at
approximately CS = 40 mmol/L
X
CS 0 CS
CS 0
50 40
50
0.2
P9-10 (d) Individualized solution
P9-10 (e) Individualized solution
P9-11 (a)
9-30
P9-11 (b)
9-31
P9-11 (c) Individualized solution
P9-11 (d) Individualized solution
P9-12
For No Inhibition, using regression,
Equation model:
1
rs
a0 = 0.008
a0 a1
1
S
a1 = 0.0266
For Maltose,
Equation model:
1
rs
a0 a1
1
S
9-32
a0 = 0.0098
a1 = 0.33
For α-dextran,
1
rs
Equation model:
a0 = 0.008
a0 a1
1
S
a1 = 0.0377
Maltose show non-competitive inhibition as slope and intercept, both changing compared to no
inhibition case.
α-dextran show competitive inhibition as intercept same but slope increases compared to no
inhibition case.
P9-13 (a)
rS
rEHS
rP
k EHS
EHS
rP
rP
K M EH
kK M EH
S
S
EH 2
K2 H
EH
EH
K1 H
E
E
ET
ET
EH
K1 H
E
EH
EH
K1 H
EH 2
EH
K2 H
EH
ET
EH
1
1
K1 H
K2 H
Now plug the value of (EH) into rP
9-33
rS
rP
kK M EH
kK M ET K1 H
S
1 K1 H
S
K1 K 2 H
2
At very low concentrations of H+ (high pH) rS approaches 0 and at very high concentrations of H+ (low
pH) rS also approaches 0. Only at moderate concentrations of H+ (and therefore pH) is the rate much
greater than zero. This explains the shape of the figure.
P9-13 (b) Individualized solution
P9-13 (c) Individualized solution
P9-14 (a)
Case1: Without drug inhibition:Reactions:
1. GLP-1(7-36)  Stimulates insulin releases ( Rate constant k1)
(a)
2. GLP-1(7-36) + DPP-4  GLP-1(9-36) (Rate constant k2)
(a)
(b)
Rates of reaction (assuming 1st order) are given by: -rGLP-1 =
dCa
= k1*Ca+ k2*Ca*Cb
dt
- rDPP-4 =
dCb
= k2*Ca*Cb
dt
Sub to: - (Assumed values)
at t=0, Ca0 = Cb0 =1 gmol/liter
Now,
For a 1st order reaction, half life period (t 1/2) = ln (2)/ rate constant
Therefore, assuming the half life period of GLP-1(7-36) to be very short (i.e. 10 sec), we get
 k1 = ln (2)/ 10 (sec-1)
 k1 = 0.0693 sec-1
9-34
Similarly, given that half life of GLP-1(9-36) is 0.5 to 1 min (i.e. average 0.75 min or 45 sec), we
get
 k2 = ln (2)/ 45 (sec-1)
 k2 = 0.0154 sec-1
Solving the differential equations using POLYMATH, we get
See Polymath program P9-14-a.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
1.
0.0002638
1.
0.0002638
2 Cb
1.
0.831363
1.
0.831363
3 k1
0.0693
0.0693
0.0693
0.0693
4 k2
0.0154
0.0154
0.0154
0.0154
5 t
0
0
100.
100.
6 X
0.0847
2.166E-05
0.0847
2.166E-05
Differential equations
1 d(Cb)/d(t) = -k2*Ca*Cb
2 d(Ca)/d(t) = -k1*Ca-k2*Ca*Cb
Explicit equations
1 k2 = 0.0154
2 k1 = 0.0693
3 X = k1*Ca+k2*Ca*Cb
9-35
Here, we get the rate of reaction (-rGLP-1) value at t = 10 sec to be 0.035973 (gmol/ liter sec).
Case2: With drug inhibition
Reactions:
1. GLP-1(7-36) + Drug  Stimulates insulin releases (Rate constant k1)
(a)
(e)
2. GLP-1(7-36) + DPP-4  GLP-1(9-36) (Rate constant k2)
(a) (b)
3. Drug + DPP-4  E.DPP-4 (Inactive) (Rate constant k3)
(e)
(b)
Rates of reaction are given by: -
Sub to: - (Assumed values)
9-36
-rGLP-1 =
dCa
= k1*Ca*Ce + k2*Ca*Cb
dt
- rDPP-4 =
dCb
= k2*Ca*Cb + k3*Ce*Cb
dt
- rDrug =
dCe
= k1*Ca*Ce + k3*Ce*Cb
dt
at t=0, Ca0 = Cb0 =Ce0=1 gmol/liter
Now, as derived earlier
k1 = 0.0693 sec-1
k2 = 0.0154 sec-1
k3 = rate constant for drug inhibition (to be varied)
Taking, k3 = 0.0693, 0.03465, 0.0231, 0.0173, 0.0154, 0.01390, 0.01155, 0.0099, 0.00866, 0.0077,
0.00693 respectively and solving the 3 differential equations using POLYMATH:
Example: For k3 = 0.03465
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
1.
0.1451749
1.
0.1451749
2 Cb
1.
0.3512064
1.
0.3512064
3 Ce
1.
0.0232788
1.
0.0232788
4 k1
0.0693
0.0693
0.0693
0.0693
5 k2
0.0154
0.0154
0.0154
0.0154
6 k3
0.03465
0.03465
0.03465
0.03465
7 t
0
0
100.
100.
8 X
0.0847
0.0010194
0.0847
0.0010194
Differential equations
9-37
1 d(Cb)/d(t) = -k3*Ce*Cb-k2*Ca*Cb
2 d(Ce)/d(t) = -k1*Ca*Ce-k3*Ce*Cb
3 d(Ca)/d(t) = -k1*Ca*Ce-k2*Ca*Cb
Explicit equations
1 k3 = 0.03465
2 k2 = 0.0154
3 k1 = 0.0693
4 X = k1*Ca*Ce+k2*Ca*Cb
Here, we get the rate of reaction (-rGLP-1) value at t = 10 sec to be 0.0234211 (gmol/ liter sec).
Similarly, we obtain the rate of reaction (-rGLP-1) value at t = 10 sec, for all values of k3:
9-38
Rate of reaction without
Rate constant for inhibition (-rGLP-1) at t=10 Rate of reaction(-rGLP-1) at
inhibition( k3)
sec
t=10sec
Ratio ( t= 10 sec)
0.0693
0.035973
0.0198914
1.808469992
0.03465
0.035973
0.0234211
1.535922736
0.0231
0.035973
0.0249
1.444698795
0.0173
0.035973
0.0257217
1.398546752
0.0154
0.035973
0.026
1.383576923
0.0139
0.035973
0.0262283
1.371533801
0.01155
0.035973
0.0265907
1.352841407
0.0099
0.035973
0.0268515
1.339701693
0.00866
0.035973
0.0270509
1.329826364
0.0077
0.035973
0.0272075
1.322172195
0.00693
0.035973
0.0273344
1.316034008
Plotting the ratio of reaction of –rGLP (without inhibition) to the rate – rGLP (with inhibition) as a
function of drug inhibition constant k3, we get
9-39
P9-14 (b)
The performance equation for mixed reactor is given by:-
V
vCAo
XA
rA
(1)
Where,
V = Volume of the reactor, v = volumetric flow rate of the reactant
XA = Conversion of the reactant, -rA = rate of reaction in term of conversion
CAo = Initial concentration of the reactant, τ = Residence time = (V/ v)
As derived earlier, in case of drug inhibition the rate of disappearance of GLP-1(7-36) is given by:-rA = k1*Ca*Ce + k2* Ca* Cb
Let the conversion be XA,
Therefore we get,
-rA = k1*CAo (1- XA)* CAo (M1 - XA) + k2* CAo (1- XA)* CAo (M2 - XA) --------- (2)
Where, M1 = (CEo / CAo) & M2 = (CBo / CAo)
Putting, equation (2) in (1), we get

V
=
vCAo k1CAo 2 1 X A M1

V
=
v k1CAo 1 X A M1

τ = k1C
XA
XA
k2 CAo 2 1
XA
M2
XA
XA
k2 C Ao 1
X A M2
XA
k2 C Ao 1
X A M2
XA
XA
Ao
1
XA
M1
XA
P9-15
-rs = (μmax*Cs*Cc)/(Km+Cs)
9-40
XA
μmax= 1hr-1
Km = 0.25 gm/dm3
Yc/s = 0.5g/g
(a)
Cc0 = 0.1g/dm3
Cs0 = 20 g/dm3
Cc=Cc0 + Yc/s(Cs0-Cs)
See polymath problem P9-15-a.pol
Calculated values of DEQ variables
Variable Initial value Minimal value
Maximal value
Final value
1
Cc
0.1
0.1
10.1
10.1
2
Cc0
0.1
0.1
0.1
0.1
3
Cs
20.
6.301E-11
20.
6.301E-11
4
Cs0
20.
20.
20.
20.
5
Km
0.25
0.25
0.25
0.25
6
rc
0.0493827
1.273E-09
4.043487
1.273E-09
7
rs
-0.0987654
-8.086974
-2.546E-09
-2.546E-09
8
t
0
0
10.
10.
9
umax
1.
1.
1.
1.
0.5
0.5
0.5
0.5
10 Ycs
Differential equations
1 d(Cs)/d(t) = rs
Explicit equations
1 Cc0 = 0.1
2 Ycs = 0.5
3 Cs0 = 20
4 umax = 1
5 Cc = Cc0+Ycs*(Cs0-Cs)
6 Km = 0.25
7 rs = -umax*Cs*Cc/(Km+Cs)
8 rc = -rs*Ycs
9-41
Plot of Cc and Cs versus time
Plot of rs and rc with time
(b) Change the polymath code to include
rg = μmax*(1-Cc/C∞)*Cc
C∞= 1 g/dm3
9-42
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cc
0.1
0.1
0.5751209
0.5751209
2 Cc0
0.1
0.1
0.1
0.1
3 Cinf
1.
1.
1.
1.
4 Cs
20.
19.04976
20.
19.04976
5 Cs0
20.
20.
20.
20.
6 Km
0.25
0.25
0.25
0.25
7 rc
0.0225
0.0225
0.0624999
0.0610892
8 rg
0.09
0.09
0.2499995
0.2443569
9 rs
-0.045
-0.1249997
-0.045
-0.1221784
10 t
0
0
10.
10.
11 umax
1.
1.
1.
1.
12 Ycs
0.5
0.5
0.5
0.5
Differential equations
1 d(Cs)/d(t) = rs
Explicit equations
1 Cc0 = 0.1
2 Ycs = 0.5
3 Cs0 = 20
4 umax = 1
5 Cinf = 1
6 Cc = Cc0+Ycs*(Cs0-Cs)
7 Km = 0.25
8 rg = umax * (1-Cc/Cinf) * Cc
9 rs = -Ycs*rg
10
rc = -rs*Ycs
9-43
Plot of Cc with time
(c) Cs0 = 20g/dm3
Cc0 = 0
The dilution rate at which wash-out occurs will be by setting Cc=0 in equation;
Cc = Ycs*(Cs0 – (DKs)/(umax – D))
Dmax =
 Dmax =
= 0.987 hr
-1
-1
Thus dilution rate at which washout occurs is 0.987hr .
(d) DC C
rg , D C SO
m CC vO
rgV
Divide by CCV,
CS
CS
rS
,
rg
CC
CCV
max
D
KmD / ( max
KM
Cs
Cs
D)
9-44
Cs
Km Cs
max
,
Now
rS
DC C
DYC / S C SO
CC
YS / C rg
d DC C
dD
 Dmax,prod = 0.88 hr
CS
CS
 DCc = D YC/S (CS0 - KmD / (
Now, for Dmax . prod ,
YC / S C SO
max
D) )
0
-1
Using this value of D we can find the value of Cs
Cs = CS
KmD / ( max
D)
3
= 1.83 g/dm
CC
YC / S C SO
CS
= 9.085g/dm
3
3
rs = D(Cs0 – Cs) = 15.98 g/dm /hr
(e) Cell death cannot be neglected.
Kd =0.02 hr-1
DCc = rg –rd
And D(CS0 –Cs)= rS
For steady state operation to obtain mass flow rate of cells out of the system, Fc
FC =CCv0 = (rg-rd)V= (µ-kd) CCV
After dividing by CcV;
D=µ-kd
Now since maintenance is neglected.
Substituting for µ in terms of substrate concentration;
Cs =
( D kd )k s
( D kd )
max
The stoichiometry equation can be written as :
-rs = rg YS/C
CC = Yc/s
CS 0
( D kd )ks
( D kd )
max
9-45
Now the dilution rate at can be found by substituting Cc =0;
Thus Dmax =
max
KS
CS 0
CS 0
kd = 0.96 hr-1
Similarly the expression for dilution rate for maximum production is given by solving the equation ;
(D+kd) Cc = (D+kd ) YC/S (CS0 - Km( D
Now for Dmax . prod ,
d DC C
dD
kd ) / ( max
( D kd )) )
0
Thus we obtain Dmax,prod = 0.86 hr-1
(f) In this case the maintenance cannot be neglected.
m = 0.2 g/hr/dm3
The correlation for steady substrate concentration will remain the same.
Cs =
DK S
max
D
But the cell maintenance cannot be neglected. Thus the stoichiometry equation will be changed. The
equation will be -rs = Ys/Crg + mCc
=> -rs = rg/YS/C + mCc
t
=> (Cs0 – Cs) = Cc/YC/S +
mCc dt
0
Also we know that by mass balance -
dCc / dt
(
D )CC
Using this relation, the stoichiometric equation for substrate consumption changes to -
(CS 0 CS ) CC (1/ YC / S
 Cc =
m/(
D))
D(CS 0 DK S / ( max D))
1/ YC / S m / ( max D))
Now for finding the dilution rate at which wash out occurs,
Cc =0;
9-46
So Dwashout = 0.98 hr
-1
Similarly to calculate Dmax . prod ,
d DC C
dD
0
Thus we obtain Dmax,prod = 0.74 hr-1
(g) Individualised soultion
(h) Individualised solution
P9-16
Tessier Equation,
rg
max
1
e CS / k CC
Redoing P9-15 part (a)
For batch reaction,
dC S
dt
rS ,
rS
YS / C rC ,
CC
C CO
YC / S C SO
CS
rg
See Polymath program P9-16-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t
0
0
7
7
Cs
20
0.0852675 20
0.0852675
Cco
0.1
0.1
0.1
0.1
Ycs
0.5
0.5
0.5
0.5
Cso
20
20
20
20
Cc
0.1
0.1
10.057366 10.057366
k
8
8
8
8
umax
1
1
1
1
Ysc
2
2
2
2
rg
0.0917915 0.0917915 3.8563479 0.1066265
rs
-0.183583 -7.7126957 -0.183583 -0.213253
Rates
0.183583
0.183583
7.7126957 0.213253
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cs)/d(t) = rs
9-47
max
1 e CS / k CC
Explicit equations as entered by the user
[1] Cco = 0.1
[2] Ycs = 0.5
[3] Cso = 20
[4] Cc = Cco+Ycs*(Cso-Cs)
[5] k = 8
[6] umax = 1
[7] Ysc = 2
[8] rg = umax*(1-exp(-Cs/k))*Cc
[9] rs = -Ysc*rg
[10] RateS = -rs
Redoing P9-15 part (b)
D C SO
DC C rg
m
C C vO
Divide by CCV,
CS
rg V
D
k ln 1
CS
rS
rg
CC
max
1 e CS / k
CCV
max
1 e CS / k
D
max
Now
rS
CC
YS / C rg
YC / S C SO
For dilution rate at which wash out occur,
k ln 1
CSO = CS
CS
CC = 0
D
max
DMAX
max
1 e CS 0 / k
3
1hr 1 1 e 20 g / dm / 8 g / dm
#
Redoing P9-15 part (c)
DC C
DYC / S C SO
CS
CS
k ln 1
D
max
9-48
0.918hr 1
DCC
DYC / S C SO
k ln 1
D
max
Now
d DC C
dD
For Dmax . prod ,
0
Dmax . prod = 0.628 hr-1
P9-16 (b) Individualized solution
P9-17 (a)
rg =
max
CC
For CSTR,
KM
DC C
CS
CS
CS
D C SO
rg
YC / S C SO
DC C
rg
CS
v0CC
dCC
dt
V
dCC
dt
rg
Substrate Balance:
rg
FC
dCS
dt
v0CS 0 v0CS
YS rg
C
CS CC
K M CS
max
9-49
YS / C rg
D
4.5 g / dm
0.8hr 1 1g / dm 3 4.5 g / dm 3
4 1 g / dm 3
6250dm 3
rgV
3
C S CC
K M CS
Flow of cells out = Flow of cells in
Cell Balance: FC
1g / dm 3
0.5(10 1) g / dm
P9-17 (b)
FC
rS
max
vO
4.5 g / dm 3
V
V
rS
10 g / dm 3 1 0.9
C SO 1 X
CC
CS
3
vO
V
This would result in the Cell concentration growing exponentially. This is not realistic as at some point
there will be too many cells to fit into a finite sized reactor. Either a cell death rate must be included or
the cells cannot be recycled.
P9-17 (c)
Two CSTR’s
For 1st CSTR,
V = 5000dm3 , DC C
rS
rg
D C SO
YS / C rg
rg =
CS
max
KM
CS
CC
CS
See Polymath program P9-17-c-1cstr.pol.
POLYMATH Results
NLES Solution
Variable Value
f(x)
Ini Guess
Cc
4.3333333 9.878E-12 4
Cs
1.3333333 1.976E-11 5
umax
0.8
Km
4
Csoo
10
Cso
10
Ysc
2
rg
0.8666667
rs
-1.7333333
V
5000
vo
1000
D
0.2
X
0.8666667
Cco
4.33
NLES Report (safenewt)
Nonlinear equations
[1] f(Cc) = D*(Cc)-rg = 0
[2] f(Cs) = D*(Cso-Cs)+rs = 0
Explicit equations
[1] umax = 0.8
[2] Km = 4
[3] Csoo = 10
[4] Cso = 10
[5] Ysc = 2
[6] rg = umax*Cs*Cc/(Km+Cs)
9-50
rS
D
vO
V
[7] rs = -Ysc*rg
[8] V = 5000
[9] vo = 1000
[10] D = vo/V
[11] X = 1-Cs/Csoo
[12] Cco = 4.33
CC1 = 4.33 g/dm3
CS1 = 1.33 g/dm3
For 2nd CSTR, D C C 2
X = 0.867
CP1 = YP/CCC1 =0.866 g/dm3
C C1
rg
D C S1
See Polymath program P9-17-c-2cstr.pol.
POLYMATH Results
NLES Solution
Variable Value
f(x)
Ini Guess
Cc
4.9334151 3.004E-10 4
Cs
0.1261699 6.008E-10 5
umax
0.8
Km
4
Csoo
10
Cs1
1.333
Ysc
2
rg
0.120683
rs
-0.241366
V
5000
vo
1000
D
0.2
X
0.987383
Cc1
4.33
NLES Report (safenewt)
Nonlinear equations
[1] f(Cc) = D*(Cc-Cc1)-rg = 0
[2] f(Cs) = D*(Cs1-Cs)+rs = 0
Explicit equations
[1] umax = 0.8
[2] Km = 4
[3] Csoo = 10
[4] Cs1 = 1.333
[5] Ysc = 2
[6] rg = umax*Cs*Cc/(Km+Cs)
[7] rs = -Ysc*rg
[8] V = 5000
[9] vo = 1000
9-51
CS
rS
[10] D = vo/V
[11] X = 1-Cs/Csoo
[12] Cc1 = 4.33
CC2 = 4.933 g/dm3
CS2 = 1.26 g/dm3
X = 0.987
CP1 = YP/CCC1 =0.9866 g/dm3
P9-17 (d)
For washout dilution rate,
Dmax
DMAX .PROD
max 1
CC = 0
max
KM
C SO
C SO
0.8hr 1 10 g / dm 3
4 g / dm 3 10 g / dm 3
KM
K M C SO
0.57hr 1
4 g / dm 3
4 g / dm 3 10 g / dm 3
0.8hr 1 1
0.37 hr 1
Production rate = CCvO(24hr) = 4.85 g / dm 3 x1000dm3/hrx24hr = 116472.56g/day
P9-17 (e)
For batch reactor,
V = 500dm3,
dC C
dt
rg
dC S
dt
rS
CCO = 0.5 g/dm3 CSO = 10g/dm3
rS
YS / C rg
rg =
max
KM
See Polymath program P9-17-e.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t
0
0
6
6
Cc
0.5
0.5
5.4291422 5.4291422
Cs
10
0.1417155 10
0.1417155
Km
4
4
4
4
Ysc
2
2
2
2
umax
0.8
0.8
0.8
0.8
rg
0.2857143 0.1486135 1.403203 0.1486135
rs
-0.5714286 -2.8064061 -0.2972271 -0.2972271
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg
[2] d(Cs)/d(t) = rs
9-52
CS
CC
CS
Explicit equations as entered by the user
[1] Km = 4
[2] Ysc = 2
[3] umax = 0.8
[4] rg = umax*Cs*Cc/(Km+Cs)
[5] rs = -Ysc*rg
For t = 6hrs,
CC = 5.43g/dm3.
So we will have 3 cycle of (6+2)hrs each in 2 batch reactors of V = 500dm3.
Product rate = CC x no. of cycle x no. of reactors x V = 5.43 g/dm3 x 3 x 2 x 500dm3
= 16290g/day.
P9-17 (f) Individualized solution
P9-17 (g) Individualized solution
P9-18 (a)
max
D
CS
D
KS
max
CS
CS
KS
KS
CS
max
max
is the intercept and
1
is the slope
max
9-53
CS gm /dm2
D(day-1)
CS/D
1
1
1
Slope = 0.5 day
max
3
1.5
2
4
1.6
2.5
intercept = 0.5 gm/dm3
= 2 KS = 0.5*2 = 1
P9-18 (b)
CC
DK S
YC S
CSO
max
D
Inserting values from dataset 4
4
1.8 1.0
2.0 1.8
YC S
50
YS C
1
YC / S
4
50 9.0
10.25
P9-18 (c) Individualized solution
P9-18 (d) Individualized solution
P9-19 (a)
See Polymath program P9-19-a.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cc
1.
1.
1.774022
1.774022
2 f
0
0
0.99985
0
3 t
0
0
48.
48.
Differential equations
1 d(Cc)/d(t) = f* Cc * 0.9 / 24
Explicit equations
9-54
10
1.8
5.6
1 f = (If (((t > 0) And (t < 12)) Or ((t > 24) And (t < 36))) Then (sin(3.14 * t / 12)) Else (0))
(b) Given the initial concentration as Co = 0.5 mg/litre we have,
=
ln
=
Cc =
exp
= 0.9 day -1 = 0.0375 hr -1
From 6am to 6pm , t= 12 hrs
 Cc =
exp
 Cc =
exp
 Cc =
exp
 Cc =
1.332
}
}
*0.0375*
}
Thus the concentration progresses as = 1.332(conc. of previous day)
9-55
Therefore the time taken to reach the concentration of 200 mg/dm 3 using MS Excel is 22 days.
(d) From part (b)
Concentration at the end of 1st day = 1.33Co
Concentration at the start of 2nd day =1.33Co/2 + 100
(mg/lit)
Concentration at the end of 2rd day= 1.33(1.33Co/2 + 100)
And so forth.
Thus the concentration progresses as = 1.332(conc. of previous day)+100
Solving for the progression on MS Excel we find the concentration to converge to
299.4011mg/lit
Thus, at the end of 365 days the conc. becomes = 299.4011mg/lit
(e) Assuming that dilution and removal of algae is done at the end of the day after the growth
period is over. Then,
Rate of decrease of conc. =
= - k Cc
where, k = 1 (day-1)
At t=0,
Co= 200mg/lit (i.e. maximum concentration allowed for maximum productivity at
the end of the day)
So, removal per day:Cc= Co exp (-kt) = 200 exp (-1 day-1 x 1 day) = 200 exp (-1)
 Cc= 73.5 mg/dm3/day
 Volume of the pond = 5000 gallons or 18940.5 dm3
 Total mass flow rate of algae = 1392.1 g/day
9-56
(f) Now, since CO2 will affect the rate of growth of algae too, then let the reaction be:CO2 + Algae  more algae
Hence,
A
+C

2C
Where, A: CO2
C: algae
Now, rate of growth of algae should be
rc = k sin(
CaCc
Earlier, Ca = 1.69gm/kg of water = 1.69 gm/dm3 of water (density of water = 1kg/lit)
We have rate law as
rc = sin (
Cc
( CO2 concentration was assumed to be a constant = 1.69g /dm3
of water )
Hence, = k 1.69 gm / dm3
k = (.9 per day)
(1.69 gm / dm3)
k = 0.5325 dm3/g day or 0.022 dm3/g hr
A
At t=0
+
Ca0
C
Cc0

2C
0
Let conversion with respect to C is X
Then,
At t=t
C
Ca0 – XCc0
Cc0 (1-X)
2Cc0X
Cc= Cc0 (1+ X)
Ca= Ca0 – XCc0 = Cc0 (M – X)
where, M= Ca0/Cc0
Hence,
= k sin (
CaCc
9-57
thus, total number of moles of
=>
= k sin (
=>
= Cc0 k sin (
=>
(1+X) (M-X)
=
=>
ln
= (M+1)k Cc0 {1 -
=>
=>
Cc0 (1+X) Cc0 (M-X)
}
= exp [(M+1) k Cc0 {1 X (1 + exp [k (M+1) Cc0 {1 -
{1 -
}
}
}
]
]/M) =
exp [k (M+1) Cc0
]-1
=> X =
Now, X= 1Hence
,
=1-
We have,
Ca0 = Ks = 2g/dm3 & Cc0 = 1mg/ dm3 (assuming initial seeding value of algae from (c))
M= Ca0/Cc0 = 2000
Substituting the values we have the concentration profile i.e.
(g)
Let A: the species of unwanted algae
C: the species of desired algae
Then,
Rate of growth of A = 2

= 2

for A = 2 x
for C
Now overall density of the medium at any time t
9-58
Vs. t
= Ca + Cc
Hence, from given conditions,
Ca=0.5(Ca + Cc)
So, at that time
Ca=Cc
Now,

=
Cc
 Cc = Cc0 exp(

=
Ca
 Ca = Ca0 exp (2
Assuming, Cco= initial seed concentration of desired algae= 1 mg/litre & Cao = 0.1 mg/litre
(given).
Now, since the concentration is very less assuming there is no constraint of sunlight.
 Ca = Cc
 Cc0 exp(
= Ca0 exp (2
 ln (Cco/Cao) = exp (
where, = 0.9 per day
Putting the values, we get
t = 2.56 days or 61.4 hrs.
Since the number of days is coming less than 4.347 days which was calculated in part (c), so the
effect of daylight can be neglected. Hence, the initial assumption is verified.
P9-20
The following errors are present in this solution1. For the Eadie-Hofstee model, we plot –rS as a function of [-rS/(S)] and for the HanesWoolf model, we plot [(S)/-rS] as a function of (S). Since the y-axis plot parameter here is
[(S)/-rS], it is a Hanes-Woolf plot and not the Eadie-Hofstee plot.\
2. For Competitive Inhibition, Hanes-Woolf form is:
For Uncompetitive Inhibition, Hanes-Woolf form is:
9-59
Since the intercept is fixed and slope is changing in the given plot, this is uncompetitive
inhibition rather than being competitive one.
3. The x-axis variable is written wrong as (S + S2). It should be (S) for Hanes-Woolf form.
4. The expression for intercept is correct but the slope is given wrong. The correct
expressions are -
5. As concentration of inhibitor (I) increases, slope increases. In the given plot, slope for
line 1 is more as compared to line 2 in spite of having lower concentrations. This implies
that the concentration values are switched. Also the numerically calculated slope values
are wrong. The correct values are as –
Line 1:
(I) = 0.05 M
Slope 1 = 35
Line 2:
(I) = 0.02 M
Slope 2 = 15.5
6. Calculations on the basis of correct values –
;
;
Solving these above equations, we get kinetic law parameters
Vmax = 0.4 ;
KM = 0.04
KI = 3.85e-3
;
9-60
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10-1
Solutions for Chapter 10 – Catalysis and Catalytic Reactors
P10-1 Individualized solution
P10-2 (a) Example 10-1
(1)
CT .S
CB . s
Cv KT PT
Cv K B PB
KT PT 0 (1 X )
K B PT 0 X
KB
1.39
KT
KT (1 X )
KB X
1.038
CT .S
Therefore, C = f(X) can be plotted.
B.s
(2)
Cv
Ct
1
1 KT PT K B PB
1
1 KT PT 0 (1 X ) K B PT 0 X
Here, PT0 = yA0PTotal = 0.3X40 atm = 12 atm & K B
1.39 KT
1.038
Cv
Therefore, C = f(X) can be plotted.
t
(3)
CB . S
Ct
Cv K B PB
Cv (1 KT PT K B PB )
K B PT 0 X
1 KT PT 0 (1 X ) K B PT 0 X
CB .S
Therefore, C = f(X) can be plotted.
t
P10-2 (b) Example 10-2
(1) See polymath problem P10-2-b.pol
Increasing the pressure will increase the conversion for the same catalyst weight.
At 40 atm, we had 68.2% conversion for a catalyst weight of 10000kg.
While, at 80 atm we have 85.34% conversion for the same weight of the catalyst.
10-2
At 1 atm, we get 0.95% conversion for 10000 kg of catalyst. However, it’s not practically
possible to operate at inlet pressure of 1 atm, because there will be no flow of feed into
the PFR, due to absence of pressure difference.
(2) If the flow rate is decreased the conversion will increase for two reasons: (1) smaller
pressure drop and (2) reactants spend more time in the reactor.
(3) From figure E10-3.1 we see that when X = 0.6, W = 5800 kg.
P10-2 (c) Example 10-3
With the new data, model (a) best fits the data
(a)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Kea*Pea+Ke*Pe)
Variable
Ini guess
Value
95% confidence
k
3
3.5798145
0.0026691
Kea
0.1
0.1176376
0.0014744
Ke
2
2.3630934
0.0024526
Precision
R^2
R^2adj
Rmsd
Variance
= 0.9969101
= 0.9960273
= 0.0259656
= 0.0096316
(b)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Ke*Pe)
Variable
Ini guess
Value
95% confidence
k
3
2.9497646
0.0058793
Ke
2
1.9118572
0.0054165
Precision
R^2
R^2adj
Rmsd
Variance
= 0.9735965
= 0.9702961
= 0.0759032
= 0.0720163
(c)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/((1+Ke*Pe)^2)
Variable
Ini guess
Value
95% confidence
10-3
k
Ke
Precision
R^2
R^2adj
Rmsd
Variance
3
2
1.9496445
0.3508639
0.319098
0.0756992
= 0.9620735
= 0.9573327
= 0.0909706
= 0.1034455
(d)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe^a*Ph^b
Variable
Ini guess
Value
95% confidence
k
3
0.7574196
0.2495415
a
1
0.2874239
0.0955031
b
1
1.1747643
0.2404971
Precision
R^2
R^2adj
Rmsd
Variance
= 0.965477
= 0.9556133
= 0.0867928
= 0.107614
Model (e) at first appears to work well but not as well as model (a). However, the 95%
confidence interval is larger than the actual value, which leads to a possible negative value for
Ka. This is not possible and the model should be discarded. Model (f) is the worst model of all.
In fact it should be thrown out as a possible model due to the negative R^2 values.
(e)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/((1+Ka*Pea+Ke*Pe)^2)
Variable
Ini guess
Value
95% confidence
k
3
2.113121
0.2375775
Ka
1
0.0245
0.030918
Ke
1
0.3713644
0.0489399
Precision
R^2
R^2adj
Rmsd
Variance
= 0.9787138
= 0.9726321
= 0.0681519
= 0.0663527
(f)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Ka*Pea)
10-4
Variable
k
Ka
Precision
R^2
R^2adj
Rmsd
Variance
Ini guess
3
1
Value
95% confidence
44.117481
7.1763989
101.99791
16.763192
= -0.343853
= -0.5118346
= 0.5415086
= 3.6653942
P10-2 (d) Individualized solution
P10-3 Solution is in the decoding algorithm given with the modules
P10-4
P10-4 (a)
10-5
P10-4 (b)
Adsorption of isobutene limited
P10-4 (c)
10-6
P10-4 (d)
P10-4 (e) Individualized solution
P10-5 (a)
H H2
E
Ethylene
A Ethane
H2
C 2H 4
H E
cat
C 2H 6
A
Because neither H2 or C2H6 are in the rate law they are either not adsorbed or weakly adsorbed.
Assume H2 in the gas phase reacts with C2H6 adsorbed on the surface and ethane goes directly into
the gas phase. Then check to see if this mechanism agrees with the rate law
Eley Rideal
E S
E S H
E S
A S
rE
k AD PE CV
rS
k SC E SPH 2
CE S
KE
10-7
Assume surface reaction
CE S
K E PE C V
rS
k S C E SPH
CT
CV
rA
k SK E C T
CE S
PE PH
1 K E PE
P10-5 (b) Individualized solution
P10-6
O2
2S
2O S
A2
C3 H 6 O S
C3 H 5OH S
C3 H 6OH S
C3 H 5OH
rB
rS
rAD
k A PA2 CV2
rAD
kA
0
CA S
CV K A PA
rB
rS
rC S
k D CC S
rC S
kD
0
CC S
K C PC CV
CT
CV
rB
rS
C S
S
C A2 S
KA
k3 PBCV K A PA
CA S
CC S
k D CC S
CV 1
K C PC CV
K A PA
K C PC
k3CT PB K A PA
1
K A PA
2A S
B A S
k3 PB C A S
PC CV
KD
2S
KC PC
10-8
C S
C S
P10-7 (a)
P10-7 (b)
10-9
P10-7 (c)
10-10
P10-7 (d) Individualized solution
P10-7 (e) Individualized solution
P10-8
10-11
P10-9
P10-9 (a)
10-12
P10-9 (b)
P10-10
10-13
P10-10 (a)
10-14
P10-10 (b)
10-15
Substituting the expressions for CV and CA·S into the equation for –r’A
P10-10 (c) Individualized solution
P10-10 (d)
First we need to calculate the rate constants involved in the equation for –r’A
in part (a). We can rearrange the equation to give the following
10-16
Thus from the slope and intercept data
See Polymath program P10-10-d.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
W
X
e
initial value
0
0
1
minimal value
0
0
1
maximal value
23
0.9991499
1
10-17
final value
23
0.9991499
1
Pao
Pa
k1
k2
Fao
ra
rate
10
10
560
2.04
600
-12.228142
12.228142
10
0.0042521
560
2.04
600
-68.5622
2.3403948
10
10
560
2.04
600
-2.3403948
68.5622
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -ra/Fao
Explicit equations as entered by the user
[1] e = 1
[2] Pao = 10
[3] Pa = Pao*(1-X)/(1+e*X)
[4] k1 = 560
[5] k2 = 2.04
[6] Fao = 600
[7] ra = -k1*Pa/((1+(k2*Pa))^2)
[8] rate = -ra
P10-10 (e) Individualized solution
P10-10 (f)
10-18
10
0.0042521
560
2.04
600
-2.3403948
2.3403948
Use these new equations in the Polymath program from part (d).
See Polymath program P10-10-f.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
W
X
y
e
Pao
Pa
k1
k2
Fao
ra
rate
alpha
initial value
0
0
1
1
10
10
560
2.04
600
-12.228142
12.228142
0.03
minimal value
0
0
0.0746953
1
10
7.771E-05
560
2.04
600
-68.584462
0.0435044
0.03
maximal value
23
0.9997919
1
1
10
10
560
2.04
600
-0.0435044
68.584462
0.03
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -ra/Fao
[2] d(y)/d(W) = -alpha*(1+X)/2/y
Explicit equations as entered by the user
[1] e = 1
[2] Pao = 10
[3] Pa = y*Pao*(1-X)/(1+e*X)
[4] k1 = 560
[5] k2 = 2.04
[6] Fao = 600
[7] ra = -k1*Pa/((1+(k2*Pa))^2)
[8] rate = -ra
[9] alpha = .03
10-19
final value
23
0.9997919
0.0746953
1
10
7.771E-05
560
2.04
600
-0.0435044
0.0435044
0.03
P10-11(a)
10-20
P10-11 (b)
P10-11 (c)
The estimates of the rate law parameters were given to simplify the search techniques to make
sure that it converged on a false minimum. In real life, one should make a number of guesses of
the rate law parameters and they should include a large range of possibilities
P10-11 (d)
A
ra
PA
B H
kPA
1 K A PA
K B PB
PA0 (1 X )
, PB
(1 X )
K H PH
2
PA0 X
, PH
(1 X )
PA0 X
here,
(1 X )
10-21
y A0
1(2 1) 1
Where,
k 0.00137
K A 4.77
KB
0.259
KH
0.424
PA0
15atm
FA0 10mol / sec
Now, the design equation for the PFR,
rA
dX
dW FA0
Solving the above equations using Polymath (Refer to P10-11-d.pol)
Calculated values of DEQ variables
Variable
Initial
value
Minimal
value
Maximal
value
Final
value
1
Fao
10.
10.
10.
10.
2
k
0.00137
0.00137
0.00137
0.00137
3
Ka
4.77
4.77
4.77
4.77
4
Kb
0.262
0.262
0.262
0.262
5
Kh
0.423
0.423
0.423
0.423
6
Pa
15.
3.064E-09
15.
3.064E-09
7
Pao
15.
15.
15.
15.
8
Pb
0
0
7.5
7.5
9
Ph
0
0
7.5
7.5
10 ra
-3.904E-06
-1.26E-05
-1.114E-13
-1.114E-13
11 W
0
0
2.0E+06
2.0E+06
12 X
0
0
1.
1.
10-22
Differential equations
1 d(X)/d(W) = -ra/Fao
Explicit equations
1
Pao = 15
2
k = 0.00137
3
Ka = 4.77
4
Kb = 0.262
5
Kh = 0.423
6
Fao = 10
7
Ph = Pao*X/(1+X)
8
Pa = Pao*(1-X)/(1+X)
9
Pb = Pao*X/(1+X)
10 ra = -k*Pa/((1+Ka*Pa+Kb*Pb+Kh*Ph)^2)
From the above graph, we get the weight of catalyst required for 85% conversion is
1.2083x106 kg.
The required catalyst weight is so high because of very low value of reaction constant ’k’.
10-23
P10-12(a)
Proposed Single site Mechanism:
S ' H 2O
S ' H 2O
S ' H 2O
S ' H2
S ' H2
S ' H2
S'
1
O2
2
1
O2
2
S
Rate of adsorption: rAD
Rate of surface reaction: rs
Rate of desorption: rD
CS ' H 2O
k AD (Cv ' PH 2O
K H 2O
)
ks CS ' H 2O ---------- (1)
k D (CS ' H 2
K H 2 PH 2 Cv ' )
Assuming, surface reaction to be rate limiting, we get
rAD rD
0
k AD k D
CS ' H 2 O
K H 2OCv ' PH 2O and CS ' H 2
K H 2 PH 2 Cv '
From total site balance, we get
Ct
Cv '
1 K H 2O PH 2O K H 2 PH 2
Putting all the values in equation (1), we get
k s K H 2O Ct PH 2O
rs
1 K H 2O PH 2O K H 2 PH 2
rs
kPH 2O
1 K H 2O PH 2O
K H 2 PH 2
Where, k
ks K H 2OCt
P10-12(b)
Proposed Mechanism:
S ' H 2O
S ' H 2O
S ' H 2O
S H2
S H2
S
H2
Rate of adsorption: rAD
k D (Cv ' PH 2O
CS ' H 2O
K H 2O
)
10-24
Rate of surface reaction: rs
Rate of desorption: rD
ks (CS ' H 2O CS H 2 ) ---------- (1)
k D (CS H 2
K H 2 PH 2 Cv )
Assuming, surface reaction to be rate limiting, we get
rAD rD
0
k AD k D
K H 2OCv ' PH 2O and CS H 2
CS ' H 2 O
K H 2 PH 2 Cv
From total site balance, we get
Ct '
Cv '
1 K H 2O PH 2O
Cv
Ct
1 K H 2 PH 2
Putting all the values in equation (1), we get
K H O PH O Ct ' K H 2 PH 2 Ct
rs k s ( 2 2
)
1 K H 2O PH 2O 1 K H 2 PH 2
P10-12(c)
Proposed Mechanism:
Step (1)
S h
S ' O2
S ' O2
S ' O2
S ' H 2O
Step (2) S ' H 2O
S H2
S ' H 2O
S H2
S
H2
For Step (1)
Rate of adsorption of O2: rAD O 2
Rate of Reaction: rSO 2
kD O 2 (Cv
kSO 2 (CS ' O2
CS ' O2
KS
)
KO2 PO2 Cv ' )
Rate of surface reaction of water is rate controlling, we get
rAD O 2 rS O 2
0
kD O 2 kS O 2
CS ' O2
Cv K S ----------- (2)
10-25
CS ' O2
KO2 PO2 Cv ' ------- (3)
Equating (2) and (3), we get
Cv
KO2 PO2 Cv '
KS
-------- (4)
Now, here the site balance gets modified because oxygen is also getting adsorbed.
Ct
Cv Cv ' CS ' O2
CS ' H 2O CS H 2
Putting, the values and replacing Cv using equation (4), we get
Cv '
(1 K O2 PO2
And, Cv
rs
K H 2O PH 2O
KO2 PO2 Cv '
KS
Ct
K O2 PO2
K O2 K H 2 PO2 PH 2
KS
KS
=
K S (1 KO2 PO2
)
KO2 PO2 Ct
KO2 PO2
K H 2O PH 2O
KS
Putting all the values in equation (1), we get
K H 2O PH 2OCt
ks (
KO2 PO2 K O2 K H 2 PO2 PH 2
(1 KO2 PO2 K H 2O PH 2O
)
KS
KS
P10-13 (a)
10-26
K O2 K H 2 PO2 PH 2
KS
K S (1 KO2 PO2
)
K H 2 PH 2 K O2 PO2 Ct
K O2 PO2
K H 2O PH 2O
KS
K O2 K H 2 PO2 PH 2
KS
)
)
P10-13 (b)
P10-14
Assume the rate law is of the form rDep
At high temperatures K
rDep
rDep
2
PVTIPO
Run 1
k
0.028
0.45
0.2
Run 5
2
as T and therefore KPVTIPO
2
kPVTIPO
0.05
Run 2
2
kPVTIPO
2
1 KPVTIPO
11.2
2
11.28
2
11.25
7.2
0.8
2
At low temperature and low pressure
rDep
2
kPVTIPO
10-27
1
rDep
k
2
PVTIPO
Run 1
0.004
0.1
Run 2
2
0.015
0.2
2
0.4
0.375
These fit the low pressure data
2
At high pressure KPVTIPO
2
kPVTIPO
2
KPVTIPO
rDep
1
k
K
This fits the high pressure data
At PVTIPO = 1.5, r = 0.095 and at PVTIPO = 2, r = 0.1
Now find the activation energy
At low pressure and high temperature k = 11.2
At low pressure and low temperature k = 0.4
ln
k2
k1
E 1
R T1
ln
11.2
0.4
E
R
E
R
7743
E 15330.65
1
T2
E T2 T1
R T1T2
473 393
473 393
cal
mol
10-28
P10-15
P10-16 (a)
Using Polymath non-linear regression few can find the parameters for all models:
See Polymath program P10-16.pol.
(1)
POLYMATH Results
Nonlinear regression (L-M)
Model: rT = k*PM^a*PH2^b
Variable
k
a
b
Ini guess
1
0.1
0.1
Value
1.1481487
0.1843053
-0.0308691
95% confidence
0.1078106
0.0873668
0.1311507
10-29
Precision
R^2
R^2adj
Rmsd
Variance
= 0.7852809
= 0.7375655
= 0.0372861
= 0.0222441
α = 0.184 β = -0.031 k = 1.148
(2)
POLYMATH Results
Nonlinear regression (L-M)
Model: rT = k*PM/(1+KM*PM)
Variable
k
KM
Precision
R^2
R^2adj
Rmsd
Variance
Ini guess
1
2
Value
12.256274
9.0251862
95% confidence
2.1574162
1.8060287
= 0.9800096
= 0.9780106
= 0.0113769
= 0.0018638
k = 12.26 KM = 9.025
(3)
POLYMATH Results
Nonlinear regression (L-M)
Model: rT = k*PM*PH2/((1+KM*PM)^2)
Variable
k
KM
Precision
R^2
R^2adj
Rmsd
Variance
Ini guess
1
2
Value
8.4090333
2.8306038
95% confidence
18.516752
4.2577098
= -4.3638352
= -4.9002187
= 0.1863588
= 0.5001061
k = 8.409 KM = 2.83
(4)
10-30
POLYMATH Results
Nonlinear regression (L-M)
Model: rT = k*PM*PH2/(1+KM*PM+KH2*PH2)
Variable
k
KM
KH2
Ini guess
1
2
2
Value
101.99929
83.608282
67.213622
95% confidence
4.614109
7.1561591
5.9343217
Nonlinear regression settings
Max # iterations = 300
Precision
R^2
R^2adj
Rmsd
Variance
= -3.2021716
= -4.1359875
= 0.1649487
= 0.4353294
k = 102 KM = 83.6
KH2 = 67.21
P10-16 (b)
We can see from the precision results from the Polymath regressions that rate law (2) best
describes the data.
P10-16 (c) Individualized solution.
P10-16 (d)
We have chosen rate law (2)
Proposed Mechanism
M
S
M S
M S
H 2( g ) T( g )
Rate of adsorption: rAD
k AD (Cv PM
Rate of surface reaction: rs
CM S
)
KM
k s CM S ---------- (1)
Assuming, surface reaction to be rate limiting:
rAD
0
k AD
CM S
Cv PM K M
10-31
Putting, the value of CM S in Equation (1)
rs
ks Cv PM K M ----------------- (2)
Now, applying site balance
Ct Cv CM S
Ct
Cv Cv PM K M
Cv
Ct
1 PM K M
Putting in Equation (2), we get
ks K M Ct PM
rs
1 K M PM
rs
kPM
Where, k = ks K M Ct
1 K M PM
P10-17
Mistakes in the solution:
1) In the CC.S expression, the constant should be KA2 instead of KDC
2) The overall site balance should include the product, as it too is getting adsorbed.
3) The final rate law expression derived is wrong. The correct expression has been derived below:
Assuming, desorption to be rate controlling, we get:
rS1
kS1
rAD
k AD
A S
C A.S
A.S
rS 2
kS 2
0
A.S
K A PACV
A.S
A2 .S S
C A2 .S
K S C A2.s / CV
C A2 .S
K S K A 2 PA 2CV
10-32
A2 .S B ( g )
rS2
C.S
CC .S
]
K A2
k S2 [ PB C A 2 .S
CC .S
K A2 PB C A2 .S
C A2. S
CC .S
K A2 PB K S
CC .S
K A2 PB K S K A 2 PA 2Cv
CV
now,
Ct
CV
C A.S
C A2 .S
CC .S
Putting, all the values in equation (1), we get
Ct
CV
K A PACV
K S C A2.S
CV
Ct
CV (1 K A PA
K S K A2 PA 2
C A2 . S
K A2 PB K S
K A2 K S K A2 PB PA2 )
2
CV
now,
rC
rC
rC
kc CC .S
kC K A2 PB K S
C A2.S
CV
kC K A2 K S K A2 PB PA2Cv
kC K A2 K S K A2 PB PA2
(1 K A PA
K S K A2 PA 2
(1 K A PA
kPB PA2
K S K A2 PA 2 K A2 K S K A2 PB PA2 )
Where, k
K A2 K S K A2 PB PA2 )
kC K A2 K S K A2
10-33
CD10GA-1
10-34
10-35
CD10GA-2
10-36
CD10GA-3
10-37
10-38
10-39
The authors and the publisher have taken care in the preparation of this book but make no expressed or
implied warranty of any kind and assume no responsibility for errors or omissions. No liability is
assumed for the incidental or consequential damage in connection with or arising out of the use of the
information or programs contained herein.
Visit us on the Web : www.prenhallprofessional.com
Copyright © 2011 Pearson Education,Inc .
This work is protected by United States copyright laws and is provided solely for the use of
the instructors in teaching their courses and assessing student learning. Dissemination or
sale of any part of this work (including the World Wide Web ) will destroy the integrity of the
work and is not permitted . The work and the materials from it should never be made
available to the students except by the instructors using the accompanying texts in the
classes. All the recipient of this work are expected to abide by these restrictions and to
honor the intended pedagogical purposes and the needs of the other instructors who rely on
these materials .
11-1
Solutions for Chapter 11: Non-isothermal Reactor Design-The Steady-state
energy balance and adiabatic PFR applications
P11-1 Individualized solution
P11-2 (a) Example 11-1
For CSTR
V
X
FA0 X
rA
X
k 1 X
k
Ae E RT
1 Ae E RT
1
k
T
T0
0
One equation, two unknowns
Adiabatic energy balance
H Rx X
CPA
In two equations and two unknowns
In Polymath form the solution
Ae E RT
X
1 Ae E RT
f X
f T
H Rx X
CPA
T0
Enter X, A, E, R, C PA , T0 and HRx to find τ and from that you can find V.
P11-2 (b) Example 11-2
Helium would have no effect on calculation
%Error
CP T TR
H Rx
CP T TR
1270
100 5.47%
23,210
11-2
P11-2 (c) Example 11-3
Set Vfinal = 0.8 m3
POLYMATH Results
Calculated values of the DEQ variables
Variable
V
X
Ca0
Fa0
T
Kc
k
Xe
ra
rate
initial value
0
0
9.3
146.7
330
3.099466
4.2238337
0.7560658
-39.281653
39.281653
minimal value
0
0
9.3
146.7
330
2.852278
4.2238337
0.7404133
-56.196156
39.281653
maximal value
0.8
0.2603491
9.3
146.7
341.27312
3.099466
9.3196276
0.7560658
-39.281653
56.196156
final value
0.8
0.2603491
9.3
146.7
341.27312
2.852278
9.3196276
0.7404133
-56.196156
56.196156
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fa0
Explicit equations as entered by the user
[1] Ca0 = 9.3
[2] Fa0 = .9*163
[3] T = 330+43.3*X
[4] Kc = 3.03*exp(-830.3*((T-333)/(T*333)))
[5] k = 31.1*exp(7906*(T-360)/(T*360))
[6] Xe = Kc/(1+Kc)
[7] ra = -k*Ca0*(1-(1+1/Kc)*X)
[8] rate = -ra
Change the entering temperature T0 in the polymath code above (highlighted in yellow), and
record the exiting conversion:
T
X
330
0.26
340
0.54
350
0.68
370
0.66
390
0.65
X
11-3
420
0.62
450
0.59
500
0.55
600
0.48
We see that there is a maximum in the exit conversion corresponding to T0 = 350K. Therefore, the inlet
temperature we would recommend is 350K.
P11-2(d) Aspen Problem
P 11-2(e)
(1)
(2)
At high To, the graph becomes asymptotic to the X-axis, that is the conversion approaches 0.
At low To, the conversion approaches 1.
(3)
On addition of equal molar inerts, the energy balance reduces to:
X EB
CPA T T0
CPI (T T0 )
H Rx
CPA CPI
11-4
 X EB
2CPA T T0
H Rx
(4)
It is observed that at the same inlet temperature, the equilibrium conversion on the addition of
inerts is greater. This also follows from the Le Chatelier’s principle.
P11-2(f)
.
.
Q
m CP T
.
Q
220kcal / s
220 103
m
18(400 270)
854.7 mol / s
.
For the2nd case:
Hot Stream: 460K  350K
Cold Stream: 270K  400K
LMTD = 69.52 oC
220 103
A
100 69.52
31.64m2
11-5
P11-3
A B
C
3
Since the feed is equimolar, CA0 = CB0 = .1 mols/dm
CA = CA0(1-X)
CB = CB0(1-X)
Adiabatic:
T
X[
H R (T0 )]
X CP
i CPi
T0
CP
C pC C pB C pA
H R (T )
HC
Ci
C pA
i
T
30 15 15 0
HB
H A = -41000-(-15000-(-20000) = -6000 cal/mol A
cal
15 15 30
B C pB
mol K
6000 X
300 200 X
30
k C A2 0 (1 X ) 2 .01 k (1 X )2
300
rA
P11-3 (a)
dX
rA
VPFR
FA0
VCSTR
FA0 X
rA
For the PFR, FA0 = CA0v0 = (.1)(2) = .2 mols/dm3
See Polymath program P11-3-a.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 X
0
0
0.85
0.85
2 V
0
0
308.2917
308.2917
3 Ca0
0.1
0.1
0.1
0.1
4 Fa0
0.2
0.2
0.2
0.2
5 T
300.
300.
470.
470.
6 k
0.01
0.01
4.150375
4.150375
7 ra
-0.0001
-0.0018941
-0.0001
-0.0009338
Differential equations
1 d(V)/d(X) = -Fa0 / ra
Explicit equations
1 Ca0 = .1
2 Fa0 = .2
11-6
3 T = 300 + 200 * X
4 k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
5 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)
V = 308.2917dm3
For the CSTR,
X = .85, T = 300+(200)(85) = 470 K.
k = 4.31 (Using T = 470K in the formula).
-rA = .000971 mol/dm3/s
FA0 X
V
rA
.1 2 .85
175 dm3
-4
9.71 10
The reason for this difference is that the temperature and hence the rate of reaction remains constant
throughout the entire CSTR (equal to the outlet conditions), while for a PFR, the rate increases gradually
with temperature from the inlet to the outlet, so the rate of increases with length.
P11-3 (b)
T
T0
X[
HR ]
i CPi
For boiling temp of 550 k,
550 = T0 + 200
T0 = 350K
P11-3 (c)
P11-3 (d)
FA0 X
rA
VCSTR
VCSTR
( rA )
FA0
X
For V = 500 dm3, FA0=.2
rA
T
k C A2 0 (1 X ) 2
300 200 X
.01 k (1 X )2
Now use Polymath to solve the non-linear equations.
11-7
See Polymath program P11-3-d-1.pol.
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
0
480.
1 T
484.4136
2 X
0.9220681 -2.041E-09 0.9
Variable Value
1 k
6.072856
2 ra
0.0003688
Nonlinear equations
1 f(T) = 300 + 200 * X - T = 0
2 f(X) = 500 - .2 * X / ra = 0
Explicit equations
1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))
2 ra = 0.01 * k * (1 - X) ^ 2
Hence, X = .922 and T = 484.41 K
For the conversion in two CSTR’s of 250 dm3 each,
For the first CSTR, using the earlier program and V = 250 dm3,
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 T
476.482 1.137E-13 480.
2 X
0.88241 -5.803E-09 0.9
Variable Value
1 k
5.105278
2 ra
0.0007059
Nonlinear equations
1 f(T) = 300 + 200 * X - T = 0
2 f(X) = 250 - .2 * X / ra = 0
Explicit equations
1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))
2 ra = 0.01 * k * (1 - X) ^ 2
T = 476.48 ad X = .8824
Hence, in the second reactor,
11-8
VCSTR
X
T
FA0 ( X X 1 )
rA
VCSTR
( rA ) X 1
FA0
Tout ,CSTR1 200( X
X1 )
See Polymath program P11-3-d-2.pol.
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
0
480.
1 T
493.8738
2 X
0.9693688 -1.359E-09 0.8824
Variable Value
1 k
7.415252
2 ra
6.958E-05
3 X1
0.8824
Nonlinear equations
1 f(T) = 476.48 + 200 * (X - X1) - T = 0
2 f(X) = 250 - .2 * (X - X1) / ra = 0
Explicit equations
1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))
2 ra = 0.01 * k * (1 - X) ^ 2
3 X1 = .8824
Hence, final X = .9694
P11-3 (e) Individualized solution
P11-3 (f) Individualized solution
11-9
P11-4 (a)
11-10
P11-4 (b)
P11-4 (c) Individualized solution
P11-4 (d)
11-11
11-12
11-13
P11-5 (a)
A
B C
CA
CT
I
FA
FT
FI
FA
CT
C A CI
FT
FA
FI
C A01
C A 0 CI 0
C A01
C A 0 CI 0
1
I
FA0
FA0 FI 0
P11-5 (b)
Mole balance:
dX
dV
rA
FA0
Rate law:
rA
kC A
Stoichiometry: C A
C A01
1 X T0
1 X T
y A0
1 1 1 1
11-14
FA0
FT 0
1
y A0
1
T
FA0
FA0
1
Fi 0
1
CPA
i
i
i
X H RX
CPA
i
CPi T0
CPi
Enter these equations into Polymath
See Polymath program P11-5-b.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
V
X
Cao
Cio
theta
Fao
Cao1
e
To
dHrx
Cpa
Cpi
T
k
ra
initial value
0
0
0.0221729
0.0221729
100
10
4.391E-04
0.009901
1100
8.0E+04
170
200
1100
25.256686
-0.0110894
minimal value
0
0
0.0221729
0.0221729
100
10
4.391E-04
0.009901
1100
8.0E+04
170
200
1098.3458
24.100568
-0.0110894
maximal value
500
0.417064
0.0221729
0.0221729
100
10
4.391E-04
0.009901
1100
8.0E+04
170
200
1100
25.256686
-0.0061524
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] Cao = 2/(.082*1100)
[2] Cio = Cao
[3] theta = 100
[4] Fao = 10
[5] Cao1 = (Cao+Cio)/(theta+1)
[6] e = 1/(1+theta)
[7] To = 1100
[8] dHrx = 80000
[9] Cpa = 170
[10] Cpi = 200
[11] T = (X*(-dHrx)+(Cpa+theta*Cpi)*To)/(Cpa+theta*Cpi)
[12] k = exp(34.34-34222/T)
[13] ra = -k*Cao1*(1-X)*To/(1+e*X)/T
11-15
final value
500
0.417064
0.0221729
0.0221729
100
10
4.391E-04
0.009901
1100
8.0E+04
170
200
1098.3458
24.100568
-0.0061524
P11-5 (c)
There is a maximum at θ = 8. This is because when θ is small, adding inerts keeps the temperature low to
favor the endothermic reaction. As θ increases beyond 8, there is so much more inert than reactants that
the rate law becomes the limiting factor.
P11-5 (d)
The only change to the Polymath code from part (b) is that the heat of reaction changes sign. The new
code is not shown, but the plots are below.
See Polymath program P11-5-d.pol.
11-16
The maximum conversion occurs at low values of theta (θ < 8) because the reaction is now exothermic.
This means heat is generated during the reaction and there is no advantage to adding inerts as there was in
the endothermic case.
P11-5 (e)
We need to alter the equations from part (c) such that rA kC A2 and CA0 = 1
A plot of conversion versus theta shows a maximum at about θ = 5.
See Polymath program P11-5-e.pol.
P11-5 (f)
We need to alter the equations from part (c) such that
We already know that C A
C A0
rA
k CA
CB CC
KC
1 X T0
. Now weed need expressions for CB and CC. From
1 X T
stoichiometry we can see that CB = CC. In terms of CA0 we find that:
CB
CC
C A0
X
1
T0
X T
We also need an equation for KC: K C
KC1 exp
H RX 1
R
T1
1
T
2 exp
80000 1
1
8.314 1100 T
When we enter these into Polymath we find that the maximum conversion is achieved at approximately θ
= 8.
See Polymath program P11-5-f.pol.
11-17
P11-5 (g)
See Polymath program P11-5-g.pol.
P11-6
(a)
Adiabatic
Mole Balance
(1)
Rate Law
Stoichiometry
dX
dW
W
rA FA0
bV
dX
dV
rA B
FA0
rA
FA0
(2)
rA
k CA
(3)
k
k 1 exp
(4)
KC
KC2 exp
(5)
CA
CA0 1 X y T0 T
(6)
CB
CA0Xy T0 T
CB
KC
E 1
R T1
11-18
1
T
H Rx 1
R
T2
1
T
dy
FT T
dW y FT 0 T0
W
V
dy
b T
dV
2y T0
Parameters
(7) – (15)
T
2y T0
FA0, k1, E, R, T1, KC2,
H Rx , T2, CA0 , T0 , , b
Energy Balance
CP
Adiabatic and
(16A)
0
H Rx X
iC Pi
T T0
Additional Parameters (17A) & (17B)
T0 ,
dT
dV
rA
FiCPi
FA0
Heat Exchange
(16B)
dT
dV
rA
i C Pi
C PA
I C PI
H Rx Ua T Ta
FiCPi
iCPi
H Rx
FA0
11-19
CP X , if CP
Ua T Ta
iC Pi
0 then
(a) Adiabatic
11-20
Adiabatic
11-21
P11-7
(a)
Mole Balance (1)
dX
dW
rA
rA
FA 0
k C AC B
Rate Law
(2)
Rate Law
(3)
k
Rate Law
(4)
k 1 0.0002
Rate Law
(5)
T1
Rate Law
(6)
E 25,000
k 1 exp
E 1
R T1
CC2
KC
1
T
310
11-22
KC
K C 2 exp
H Rx 1
R
T2
1
T
Rate Law
(7)
Rate Law
(8)
H Rx
Rate Law
(9)
T2
Rate Law
(10)
K C2 1,000 KC 1000exp
20,000
305
20,000 1
1.987 305
1
T
2 1 1 0
y A0
Stoichiometry
(11)
CA
0.5 2 1 1
CA0 1 X y
B
Stoichiometry
(12)
Stoichiometry
(13)
Stoichiometry
(14)
Stoichiometry
(15)
2
C B C A0 2 X
CC
T
2y T0
= 0.00015
11-23
T0
y
T
T
2CA 0X 0 y
T
dy
dW
(b)
T0
T
0
rA
k C A0 1 X y
T0
T
C A0 2 X y
rA
kC 2A0
KC
4C 2A0 X 2e
C 2A0 1 X e 2 X e
T
4X 2
1 X 2 X
2C A0 Xy
T0
KC
T0
T
KC
y
T0
2
T
(c)
Equilibrium Conversion
X 2e
1 Xe 2 Xe
KC
4
2
KC
4
3K C
Xe
4
KC
1 X 2e
4
Equilibrium Conversion (19)
Xe
3KC
4
3KC
4
2
Xe
X 2e
3K C
K
Xe 2 C
4
4
2
KC
4
Xe
K C X 2e
4
2KC
KC
4
0
1
1
T
X
W
W
Section 1.01 Case 1 Adiabatic
11-24
2
(d)
Energy Balance
(16)
Energy Balance
(17)
T
H Rx X
T0
iC Pi
T0 = 325
H Rx
CP
2C PC
HR
C PA
C P T TR
C PB
CP
i C Pi
C PA
20 2 20
Energy Balance
(e)
iC Pi
(18)
T T0
H Rx X
i C Pi
11-25
2 20
20 20 0
0
B C PB
C C PC
0 20
I C PI
Everything that enters
1 40
100 cal mol K
400
20,000X
100
400 200X
(f) Adiabatic
11-26
0
Adiabatic
Gas Phase Adiabatic
11-27
Gas Phase Adiabatic
11-28
P11-8
1 Xe
2
KC
Xe
T
X e2
CC CD
C AC B
KC
1
T0
KC
HR X
CPA CPB
30000
X
25 25
300
300 600 X
See Polymath program P11-8.pol.
Calculated values of NLE variables
Variable Value
1 Xe
f(x)
Initial Guess
0.9997644 3.518E-11 0.5 ( 0 < Xe < 1. )
Variable Value
1 T
300.
2 Kc
1.8E+07
Nonlinear equations
1 f(Xe) = Xe - (1 - Xe) * Kc ^ 0.5 = 0
Explicit equations
1 T = 300
2 Kc = 500000 * exp(-30000 / 1.987 * (1 / 323 - 1 / T))
T
X
300
1
320
0.999
340
0.995
360
0.984
380
0.935
400
0.887
420
0.76
440
0.585
460
0.4
480
0.26
500
0.1529
520
0.091
540
0.035
560
0.035
11-29
Xe
1
Xe (Eqm Conversion)
0.9
0.8
0.7
0.6
Xe
0.5
0.4
0.3
0.2
0.1
0
300
350
400
450
500
550
Temp (K)
P11-9
For first reactor,
KC
X e1
or X e1
1 X e1
KC
1 KC
For second reactor
KC
X e2
orX e 2
1 X e2
B2
KC
B2
1 KC
For 3rd reactor
KC
X e3
orX e3
1 X e3
B3
KC
B3
1 KC
1st reactor: in first reaction Xe = 0.3
So, FB = FA01(.3)
2nd reactor: Moles of A entering the 2nd reactor: FA02 = 2FA01 - FA01(.3) = 1.7FA01
B2
.2 FA01
1.7 FA0
.12
FA02 CPi i T T0
X
C pA
B
FA02 X
HR
0
C pB T T 0
HR
Slope is now negative
3rd reactor:
11-30
X e2 = 0.3
Say
B
(.2 FA01 ) .3FA02
FA01 (.2 .3 1.8)
FA03
FA01 FA02 (1 X e 2 )
FA03
2.26 FA01
FA01 1.8 FA01 (1 X e 2 ) 1 1.8(1 .3) FA01
Feed to the reactor 3:
(2 FA01 ) .3FA02
B3
FA01 (.2 .3 1.8) 0.7FA01
.74
2.26
Feed Temperature to the reactor 2 is (520+450)/2 = 485 K
Feed Temperature to reactor 3 is 480 K
Xfinal = .4
Moles of B = .2FA01 + .3FAo2 + .4FA03 = FA01(.2 + .54 + (.4)(2.26)) = 1.64 FA01
X = FB/3FA01 = .54
P11-9 (b)
The same setup and equations can be used as in part (a). The entering temperature for reactor 1 is now
450 K and the outlet is 520 K. When the two streams are joined prior to entering reactor 2 the temperature
is (520+450)/2 = 485 K
Say that the outlet temperature for reactor 2 is 510 K. Then the entering temperature for reactor 3 would
be (510+510+450)/3 = 490 K
For any reactor j,
FA0 j CPi i T T0
X
C pA
B
FA0 j X
HR
0
C pB T T 0
HR
and θB for reactor 1 = 0. For reactor 2, θB > 0. This means that the slope of the conversion line from the
energy balance is larger for reactor 2 than reactor 1. And similarly θB for reactor 3 > θB for reactor 2. So
the line for conversion in reactor 3 will be steeper than that of reactor 2. The mass balance equations are
the same as in part (b) and so the plot of equilibrium conversion will decrease from reactor 1 to reactor 2,
and, likewise, from reactor 2 to reactor 3.
11-31
P11-10
1) The standard heat of reaction should have been -10000 cal/mol instead of -5000 cal/mol
2) While calculating
CP
1
CP
2 B
CP for the reaction, the following expression should have been employed:
1
CP CPA
2 C
The Cp of inerts should not have appeared in the
CP .
3) In the equation for the variation of T with conversion terms , the following should have been
used,
T
0
X ( H Rx
(TR )) X CPTR
i
 T
 T
CPi
i
CPi T
X CP
X (10000) X ( 9)(298) 33*300
33 X ( 9)
7318 X 9900
33 9 X
11-32
The authors and the publisher have taken care in the preparation of this book but make no expressed or
implied warranty of any kind and assume no responsibility for errors or omissions. No liability is
assumed for the incidental or consequential damage in connection with or arising out of the use of the
information or programs contained herein.
Visit us on the Web : www.prenhallprofessional.com
Copyright © 2011 Pearson Education,Inc .
This work is protected by United States copyright laws and is provided solely for the use of
the instructors in teaching their courses and assessing student learning. Dissemination or
sale of any part of this work (including the World Wide Web ) will destroy the integrity of the
work and is not permitted . The work and the materials from it should never be made
available to the students except by the instructors using the accompanying texts in the
classes. All the recipient of this work are expected to abide by these restrictions and to
honor the intended pedagogical purposes and the needs of the other instructors who rely on
these materials .
12-1
Solutions for Chapter 12: Steady State Non-isothermal Reactor Design- Flow Reactors with
Heat Exchange
P12-1 Individualized solution
P12-2(a) (1)
Part (a) Co-current
See polymath program P12-2-a(1).pol
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca0
9.3
9.3
9.3
9.3
2 Cp0
159.
159.
159.
159.
3 Cpc
28.
28.
28.
28.
4 deltaH
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
5 Fa0
14.67
14.67
14.67
14.67
6 k
0.5927441 0.5927441
654.8375
3.449562
7 Kc
1817.59
215.5325
1817.59
1063.552
8 m
500.
500.
500.
500.
9 Qg
1.103E+05 0.1990791
1.448E+06
0.1990791
10 Qr
-2.5E+04
-2.5E+04
5.301E+05
5.445171
11 ra
-5.51252
-72.4152
-9.954E-06
-9.954E-06
12 rate
5.51252
9.954E-06
72.4152
9.954E-06
13 T
305.
305.
417.9988
327.2341
14 Ta
310.
309.9439
327.233
327.233
15 Ua
5000.
5000.
5000.
5000.
16 V
0
0
5.
5.
17 X
0
0
0.9990603
0.9990603
18 Xe
0.9994501 0.9953818
0.9994501
0.9990606
Differential equations
1 d(Ta)/d(V) = Ua*(T-Ta)/m/Cpc
2 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cp0/Fa0
3 d(X)/d(V) = -ra/Fa0
12-2
Explicit equations
1 Cpc = 28
2 Ua = 5000
3 k = 31.1*exp(7906*(T-360)/(T*360))
4 Cp0 = 159
5 deltaH = -20000
6 Ca0 = 9.3
7 Kc = 1000*exp(-(20000/8.314)*((T-330)/(T*330)))
8 ra = -k*Ca0*(1-(1+1/Kc)*X)
9 Xe = Kc/(1+Kc)
10 Fa0 = .9*163*.1
11 rate = -ra
12 m = 500
13 Qg = ra*deltaH
14 Qr = Ua*(T-Ta)
Plot of X versus V
Plot of Ta versus V
12-3
Plot of T versus V
Part (b) Counter current
First multiply the right hand side of equation (E12-1.2) with by minus one to obtain;
d(Ta)/d(V) = - Ua*(T-Ta)/m/Cpc
12-4
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca0
9.3
9.3
9.3
9.3
2 Cp0
159.
159.
159.
159.
3 Cpc
28.
28.
28.
28.
4 deltaH
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
5 Fa0
14.67
14.67
14.67
14.67
6 k
0.5927441 0.1050316
640.6661
0.1050316
7 Kc
1817.59
216.9721
3077.31
3077.31
8 m
500.
500.
500.
500.
9 ra
-5.51252
-73.92857
-0.000591
-0.000591
10 rate
5.51252
0.000591
73.92857
0.000591
11 T
305.
285.9125
417.5158
285.9125
12 Ta
310.
285.8822
310.0614
285.8822
13 Ua
5000.
5000.
5000.
5000.
14 V
0
0
5.
5.
15 X
0
0
0.9990703
0.9990703
16 Xe
0.9994501 0.9954123
0.9996751
0.9996751
Differential equations
1 d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc
2 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cp0/Fa0
3 d(X)/d(V) = -ra/Fa0
Explicit equations
1 Cpc = 28
2 Ua = 5000
3 k = 31.1*exp(7906*(T-360)/(T*360))
4 Cp0 = 159
5 deltaH = -20000
6 Ca0 = 9.3
7 Kc = 1000*exp(-(20000/8.314)*((T-330)/(T*330)))
8 ra = -k*Ca0*(1-(1+1/Kc)*X)
9 Xe = Kc/(1+Kc)
10 Fa0 = .9*163*.1
12-5
11 rate = -ra
12 m = 500
Plot of X versus V
Plot of T versus V
Plot of Ta versus V
12-6
Part (c) Constant Ta
For constant Ta, we have to use
Polymath program (a) but multiply the right hand side of E12-1.1 by 0 in the program;
i.e d(Ta)/d(V) = Ua*(T-Ta)/m/Cpc *0
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca0
9.3
9.3
9.3
9.3
2 Cp0
159.
159.
159.
159.
3 Cpc
28.
28.
28.
28.
4 deltaH
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
5 Fa0
14.67
14.67
14.67
14.67
6 k
0.5927441 0.5927441
648.9771
0.9009882
7 Kc
1817.59
216.1228
1817.59
1600.16
8 m
500.
500.
500.
500.
9 ra
-5.51252
-72.24709
-0.000498
-0.000498
10 rate
5.51252
0.000498
72.24709
0.000498
11 T
305.
305.
417.8002
310.0078
12 Ta
310.
310.
310.
310.
13 Ua
5000.
5000.
5000.
5000.
12-7
14 V
0
0
5.
5.
15 X
0
0
0.9993161
0.9993161
16 Xe
0.9994501 0.9953943
0.9994501
0.9993755
Differential equations
1 d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc *0
2 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cp0/Fa0
3 d(X)/d(V) = -ra/Fa0
Explicit equations
1 Cpc = 28
2 Ua = 5000
3 k = 31.1*exp(7906*(T-360)/(T*360))
4 Cp0 = 159
5 deltaH = -20000
6 Ca0 = 9.3
7 Kc = 1000*exp(-(20000/8.314)*((T-330)/(T*330)))
8 ra = -k*Ca0*(1-(1+1/Kc)*X)
9 Xe = Kc/(1+Kc)
10 Fa0 = .9*163*.1
11 rate = -ra
12 m = 500
Plot of X versus V
12-8
Plot of T versus V
Plot of Ta versus V
Ta will remain constant with V
Part (d) Adiabatic
Using the polymath program of part (a) and making the parameter Ua =0; we have;
12-9
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca0
9.3
9.3
9.3
9.3
2 Cp0
159.
159.
159.
159.
3 Cpc
28.
28.
28.
28.
4 deltaH
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
5 Fa0
14.67
14.67
14.67
14.67
6 k
0.5927441 0.5927441
1115.107
1115.105
7 Kc
1817.59
183.3032
1817.59
183.3033
8 m
500.
500.
500.
500.
9 ra
-5.51252
-265.5745
0.0023739
-4.937E-05
10 rate
5.51252
-0.0023739
265.5745
4.937E-05
11 T
305.
305.
430.1037
430.1037
12 Ta
310.
310.
310.
310.
13 Ua
0
0
0
0
14 V
0
0
5.
5.
15 X
0
0
0.9945744
0.9945742
16 Xe
0.9994501 0.9945742
0.9994501
0.9945742
Differential equations
1 d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc *0
2 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cp0/Fa0
3 d(X)/d(V) = -ra/Fa0
Explicit equations
1 Cpc = 28
2 Ua = 5000*0
3 k = 31.1*exp(7906*(T-360)/(T*360))
4 Cp0 = 159
5 deltaH = -20000
6 Ca0 = 9.3
7 Kc = 1000*exp(-(20000/8.314)*((T-330)/(T*330)))
8 ra = -k*Ca0*(1-(1+1/Kc)*X)
9 Xe = Kc/(1+Kc)
10 Fa0 = .9*163*.1
12-10
11 rate = -ra
12
m = 500
Plot of X versus V;
Plot of T versus V
Plot of Ta versus V;
12-11
Ta remains a constant
(2)
Since the temperature reaches a maximum which is above 400K in all of the above systems at
some point during the operation thus they cannot prevent reaching the 400 K temperatures
which will start of another exothermic reaction. Hence the heat exchanger needs to be changed
in order that the heat removed by the exchanger is such that the maximum temperature in the
reactor does not go above 400K at any point during the operation.
(3)
Making the necessary changes in the polymath code and adding variables Qg and Qr as asked
for in the problem we plot them for a co-current exchanger as follows:
12-12
For counter current heat exchanger we have
For constant Ta;
For adiabatic operation;
12-13
(4)
Varying the values of m
(i) when m =0; a plot of Qg and Qr versus volume is as follows, this seems to be an unsafe
operating condition because generated heat is not taken out at the same time.
(ii) when m =2000;
12-14
In this case the conditions are safer than for m=500 and the heat generated is simultaneously
removed.
Variation of entering feed temperature:
(i) Keeping other conditions constant we change value of T0 only now.
Let T0 = 273K then,
12-15
Plot of Qg and Qr versus V is as
follows;
(ii) T0 = 315K;
Plot of Qg and Qr versus V is as follows;
Since again the heat generation term, Qg is much larger than Qr thus an unsafe
operating condition can result in this condition.
12-16
(5) An unsafe condition will result if the entering coolant temperature is made Ta(0) = 300K in
a concurrent heat exchanger ;resulting in great amount of heat generation compared to the
rate of heat removal.
The plot of Qg and Qr with V is as shown;
Similarly other unsafe operating conditions can be found out keeping in view that the heat
generated during the course of the reaction, is not removed efficiently by the exchanger such
that there is a wide difference between the two terms at any position in the reactor.
P12-2(b)
(1) The terms Qg = ra*△HRx and Qr = Ua*(T-Ta) are added in the polymath code given in
example 12.2.
See polymath program P12.2-b(1)
Case1: adiabatic operation
Qr=0 in this case
12-17
Case 2 : For constant heat exchange conditions
Case 3: Co-current heat exchange
12-18
Case 4 : counter current heat exchange :
We need to guess a value of Ta such that at exit Ta = 1250K
If we take Ta(0) = 995.15K then this can be done.
(2)
Now, V = 0.5 m3
T0 = 1050 K
Ta0 = 1250K
Case 1: adiabatic operation;
Substitute; Ua = 0 in the polymath code
12-19
(3) Substitute, the volume V= 5 m3 in the polymath code and get the results.
(4) Plot of Qg for all 4 cases against volume
Plot of Qr versus volume for all
cases
Plot of -ra (rate) versus V
12-20
It can be seen that the rate for constant heat exchange fluid temperature Ta , is higher than the
rest of cases because the difference beween heat generated and heat removed in this case is
highest.
The rate of reaction for all cases is decreasing because the temperature of the system is
decreasing with volume.
The rate of reaction for counter-current heat exchanger system is a U shaped curve plotted
against volume. At the front of the reactor, the reaction takes place very rapidly, drawing
energy from the sensible heat of the gas causing the gas temperature to drop because the heat
exchanger cannot supply energy at an equal or greater rate. This drop in temperature, coupled
with the consumption of reactants, slows the reaction rate as we move down the reactor.
(5)
Introduction of inert will introduce a change in energy balance equation and the value of Ѳ1 as
well.
(6)
(i) ѲI = 0
All the plots will remain as for part(1)
( ii ) ѲI = 1.5
Instead of the energy equation which was used previously
12-21
d (T )
= (Ua*(Ta-T)+ra* H)/(Fa0*(CpA+X*delCp)) the equation will change.
d (V )
Now we have to change the value of
i
Cpi
Now the value will be ∑Cp = ѲI*CpI + CpA = 1.5×50 + 1×163 = 238
Value of Ѳ will change as well
FX 0 FX 1
(1.5C A0 2C A0 ) (1.5C A0
Ѳ=
=
FX 0
(1.5C A0 C A0 )
C A0 )
1
2.5
0.4
(iii) Ѳ I = 3
∑Cp = ѲI*CpI + CpA = 3×50 + 1×163 = 313J/molA/K
FX 0 FX 1
Ѳ=
FX 0
(3CA0 2C A0 ) (3C A0 C A0 ) 1
0.25
=
(3CA0 C A0 )
4
Incorporating these changes in the code and plotting X versus V for different cases.
See polymath code P12-2-b(6).
The analysis is as follows:
Case I: adiabatic operation:
12-22
Case2: Constant heat exchange fluid temperature Ta
Case 3: Co-current heat exchange
The polymath program for reference is for the case of co-current heat exchange with ѲI =3.
By changing values of ∑Cp and ε variables as shown we can change the program for various cases
and sub cases.
12-23
Case 4: Countercurrent heat exchange
Remarks:
Thus we can see that for all cases when inert gas concentration is more then the reaction
proceeds faster but then the overall yield is less as well. In the case of adiabatic operation this
phenomenon is very significant . In case of constant heat exchange fluid temperature the effect
of inert gas is negligible.
(7)
Here we will change the polymath program as entered in P12-2(b) part1.
The Ta value will be changed and the program will be tested for following values of Ta.
1000K, 1175K, 1350K
Case 1: adiabatic conditions
12-24
Case 2: Constant heat exchange fluid temperature Ta
Case 3: Co-current conditions
12-25
Case 4: Countercurrent conditions
We need to enter Ta (V =0) values such that at V=Vf, Taf = 1000K, 1175K and 1350K respectively
Ta ( V=0) (K)
983.75
992
999
Ta( V=V f) (K)
1000
1175
1350
(d) Example 12-3
(1)
CA0 will decrease but this will have no effect
12-26
(2)
Will decrease
401.1 ft 3
466.1 ft 3 s
(3)
In the energy balance the slope of the energy balance of X vs. T will be greater
i CPC
35
18.65 18
4
501
1.67 19.5
35 335.7 130.2
BTU
kmol R
XM
B
XM,
T
Base case
Change QM
12-27
Less Conversions
P12-2 (e) Example 12-4
Change CP = 29 and –ΔH = 38700
POLYMATH Results
NLES Solution
Variable Value
f(x)
Ini Guess
X
0.7109354 2.444E-11 0.367
T
593.6885 1.2E-09 564
tau
0.1229
A
1.696E+13
E
3.24E+04
R
1.987
k
20.01167
NLES Report (safenewt)
Nonlinear equations
[1] f(X) = X-(397.3*(T-535)+92.9*(T-545))/(38700+7*(T-528)) = 0
[2] f(T) = X-tau*k/(1+tau*k) = 0
Explicit equations
[1] tau = 0.1229
[2] A = 16.96*10^12
[3] E = 32400
[4] R = 1.987
[5] k = A*exp(-E/(R*T))
Vary the heat exchanger area to find the effect on conversion.
12-28
P12-2 (f) (1) Example 12-5
See Polymath program P12-2-f.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alpha
1.05
1.05
1.05
1.05
2 Ca
0.1
1.306E-07
0.1
1.306E-07
3 Cb
0
0
0.0208092
8.93E-07
4 Cc
0
0
0.0038445
1.739E-07
5 Cto
0.1
0.1
0.1
0.1
6 Fa
100.
9.521981
100.
9.521981
7 Fb
0
0
65.11707
65.11707
8 Fc
0
0
12.68047
12.68047
9 Ft
100.
87.31953
100.
87.31953
10 Fto
100.
100.
100.
100.
11 k1a
482.8247
482.8247
1.753E+04
1.706E+04
12 k2a
553.0557
553.0557
1.79E+06
1.683E+06
13 r1a
-48.28247
-136.5345
-0.0022276
-0.0022276
14 r2a
-5.530557
-90.93151
-2.87E-08
-2.87E-08
15 T
423.
423.
682.1122
678.9481
16 To
423.
423.
423.
423.
17 V
0
0
0.8
0.8
18 y
1.
1.922E-05
1.
1.922E-05
Differential equations
1 d(Fa)/d(V) = r1a+r2a
2 d(Fb)/d(V) = -r1a
3 d(Fc)/d(V) = -r2a/2
4 d(T)/d(V) = (4000*(373-T)+(-r1a)*20000+(-r2a)*60000)/(90*Fa+90*Fb+180*Fc)
5 d(y)/d(V) = -alpha/(2*y)*(Ft/Fto)*(T/To)
Explicit equations
1 k1a = 10*exp(4000*(1/300-1/T))
2 k2a = 0.09*exp(9000*(1/300-1/T))
3 Cto = 0.1
4 Ft = Fa+Fb+Fc
5 To = 423
6 Ca = Cto*(Fa/Ft)*(To/T)*y
7 Cb = Cto*(Fb/Ft)*(To/T)*y
8 Cc = Cto*(Fc/Ft)*(To/T)*y
12-29
9 r1a = -k1a*Ca
10 r2a = -k2a*Ca^2
11 Fto = 100
12 alpha = 1.05
(f) (2)
In the polymath program of part (f) (1), rate equation will be changed like –
r1a = -k1a*(Ca-Cb/Kc)
with
Kc= 10*exp(20000*(1/450-1/T))
12-30
P12-2(g)
See Polymath program P12-2-g.pol.
Entering Temperature (T0 in K)
Reactor temperatures (Ts in K)
1.
200
250
2.
220
265, 625
3.
244
275, 563, 644
4.
261
292, 409, 571, 661
12-31
5.
283
310, 363, 449, 558, 677
6.
292
373, 463, 562, 674
7.
304
475, 551, 686
8.
338
518, 713
9.
350
720
Plot of steady state reactor temperature; Ts(Y axis) versus T0 (X axis)
The extinction temperature is 220 K and the ignition temperature is 338K.
Vary UA
UA 4000 J m3 s K
No steady state occurs in this case and G (T) is always more than R (T).
UA 40, 000 J m3 s K
five steady states will exist here T = 310, 363K, 449K ,558K and 677 depending how you read
the intersection on the graph.
UA 4, 00, 000 J m3 s K
Only one steady states occurs at around T = 350K (about).
Vary
0.001 Only the lower steady state exists at about T = 316 K
0.01 ; A number of steady states occur; 5 steady states occur
12-32
0.1, Only an upper steady state occurs at 678K.
Thus we see that if we go on increasing the value of residence time, firstly the number of steady
states will increase and then it will start decreasing.
P12-2(h)
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 V
0
0
10.
10.
2 Fa
5.
0.3455356
5.
0.3455356
3 Fb
10.
0.7169561
10.
0.7169561
4 Fc
0
0
4.62858
4.62858
5 Fd
0
0
0.0258848
0.0258848
6 T
300.
300.
914.7896
518.2818
7 Ta
325.
322.7616
506.8902
506.8902
8 E2
1.2E+04
1.2E+04
1.2E+04
1.2E+04
9 y
1.
1.
1.
1.
10 R
1.987
1.987
1.987
1.987
11 Ft
15.
5.716956
15.
5.716956
12 To
310.
310.
310.
310.
13 k2c
2.
2.
1.511E+06
9619.448
14 E1
8000.
8000.
8000.
8000.
15 Cto
0.2
0.2
0.2
0.2
16 Ca
0.0688889 0.0072303
0.0688889
0.0072303
17 Cc
0
0
0.096852
0.096852
18 r2c
0
-0.0081056
0
-0.0004569
19 Cpco
10.
10.
10.
10.
20 m
50.
50.
50.
50.
21 Cb
0.1377778 0.0150021
0.1377778
0.0150021
22 k1a
40.
3.318E+05
1.14E+04
23 r1a
-0.0523079 -3.727241
-0.0185467
-0.0185467
24 r1b
-0.1046159 -7.454481
-0.0370935
-0.0370935
25 rb
-0.1046159 -7.454481
-0.0370935
-0.0370935
26 r2a
0
-0.0081056
0
-0.0004569
27 DH1b
-1.5E+04
-1.5E+04
-1.5E+04
-1.5E+04
28 DH2a
-10000.
-10000.
-10000.
-10000.
40.
12-33
29 r1c
0.0523079 0.0185467
3.727241
0.0185467
30 Ta55
325.
325.
325.
325.
31 Cpd
16.
16.
16.
16.
32 Cpa
10.
10.
10.
10.
33 Cpb
12.
12.
12.
12.
34 Cpc
14.
14.
14.
14.
77.2731
170.
77.2731
35 sumFiCpi 170.
36 rc
0.0523079 0.0180899
3.725968
0.0180899
37 Ua
80.
80.
80.
80.
38 r2d
0
0
0.0162112
0.0009137
39 ra
-0.0523079 -3.728513
-0.0190036
-0.0190036
40 rd
0
0
0.0162112
0.0009137
41 Qg
1569.238
560.9706
1.118E+05
560.9706
42 Qr
-2000.
-2000.
4.139E+04
911.3246
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = rb
3 d(Fc)/d(V) = rc
4 d(Fd)/d(V) = rd
5 d(T)/d(V) = (Qg - Qr) / sumFiCpi
6 d(Ta)/d(V) = Ua * (T - Ta) / m / Cpco
Explicit equations
1 E2 = 12000
2 y=1
3 R = 1.987
4 Ft = Fa + Fb + Fc + Fd
5 To = 310
6 k2c = 2 * exp((E2 / R) * (1 / 300 - 1 / T))
7 E1 = 8000
8 Cto = 0.2
9 Ca = Cto * (Fa / Ft) * (To / T) * y
10 Cc = Cto * (Fc / Ft) * (To / T) * y
11 r2c = -k2c * Ca ^ 2 * Cc ^ 3
12-34
12 Cpco = 10
13 m = 50
14 Cb = Cto * (Fb / Ft) * (To / T) * y
15 k1a = 40 * exp((E1 / R) * (1 / 300 - 1 / T))
16 r1a = -k1a * Ca * Cb ^ 2
17 r1b = 2 * r1a
18 rb = r1b
19 r2a = r2c
20 DH1b = -15000
21 DH2a = -10000
22 r1c = -r1a
23 Ta55 = 325
24 Cpd = 16
25 Cpa = 10
26 Cpb = 12
27 Cpc = 14
28 sumFiCpi = Cpa * Fa + Cpb * Fb + Cpc * Fc + Cpd * Fd
29 rc = r1c + r2c
30 Ua = 80
31 r2d = -2 * r2c
32 ra = r1a + r2a
33 rd = r2d
34 Qg = r1b * DH1b + r2a * DH2a
35 Qr = Ua * (T - Ta)
12-35
Addition of Inerts:
The adding of inerts will decrease the peak in the reactor temperature. By trial and error, an
inert flow rate of 6.2 mol/min is seen to be sufficient to keep the reactor temperature below
700K.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 V
0
0
10.
10.
2 Fa
5.
0.4822244
5.
0.4822244
3 Fb
10.
0.9811024
10.
0.9811024
4 Fc
0
0
4.501122
4.501122
5 Fd
0
0
0.0166535
0.0166535
6 T
300.
300.
697.1199
570.593
7 Ta
325.
322.0417
477.6044
477.6044
8 E2
1.2E+04
1.2E+04
1.2E+04
1.2E+04
9 y
1.
1.
1.
1.
10 R
1.987
1.987
1.987
1.987
11 Ft
15.
5.981102
15.
5.981102
12 To
310.
310.
310.
310.
12-36
13 k2c
2.
2.
1.914E+05
2.8E+04
14 E1
8000.
8000.
8000.
8000.
15 Cto
0.2
0.2
0.2
0.2
16 Ca
0.0688889 0.0087606
0.0688889
0.0087606
17 Cc
0
0
0.081772
0.081772
18 r2c
0
-0.0040086
0
-0.0011748
19 Cpco
10.
10.
10.
10.
20 m
50.
50.
50.
50.
21 Cb
0.1377778 0.0178237
0.1377778
0.0178237
22 k1a
40.
8.369E+04
2.323E+04
23 r1a
-0.0523079 -1.876998
-0.0523079
-0.0646601
24 r1b
-0.1046159 -3.753997
-0.1046159
-0.1293201
25 rb
-0.1046159 -3.753997
-0.1046159
-0.1293201
26 r2a
0
-0.0040086
0
-0.0011748
27 DH1b
-1.5E+04
-1.5E+04
-1.5E+04
-1.5E+04
28 DH2a
-10000.
-10000.
-10000.
-10000.
29 r1c
0.0523079 0.0523079
1.876998
0.0646601
30 Ta55
325.
325.
325.
325.
31 Fi
6.2
6.2
6.2
6.2
32 Cpa
10.
10.
10.
10.
33 Cpb
12.
12.
12.
12.
34 Cpc
14.
14.
14.
14.
35 Cpd
16.
16.
16.
16.
36 rc
0.0523079 0.0523079
1.876239
0.0634852
37 Ua
80.
80.
80.
80.
38 r2d
0
0
0.0080172
0.0023497
39 ra
-0.0523079 -1.877758
-0.0523079
-0.0658349
40 rd
0
0
0.0080172
0.0023497
41 Qg
1569.238
1569.238
5.632E+04
1951.55
42 Qr
-2000.
-2000.
2.358E+04
7439.087
43 Cpi
10.
10.
10.
10.
141.8776
232.
141.8776
44 sumFiCpi 232.
40.
12-37
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = rb
3 d(Fc)/d(V) = rc
4 d(Fd)/d(V) = rd
5 d(T)/d(V) = (Qg - Qr) / sumFiCpi
6 d(Ta)/d(V) = Ua * (T - Ta) / m / Cpco
Explicit equations
1 E2 = 12000
2 y=1
3 R = 1.987
4 Ft = Fa + Fb + Fc + Fd
5 To = 310
6 k2c = 2 * exp((E2 / R) * (1 / 300 - 1 / T))
7 E1 = 8000
8 Cto = 0.2
9 Ca = Cto * (Fa / Ft) * (To / T) * y
10 Cc = Cto * (Fc / Ft) * (To / T) * y
11 r2c = -k2c * Ca ^ 2 * Cc ^ 3
12 Cpco = 10
13 m = 50
14 Cb = Cto * (Fb / Ft) * (To / T) * y
15 k1a = 40 * exp((E1 / R) * (1 / 300 - 1 / T))
16 r1a = -k1a * Ca * Cb ^ 2
17 r1b = 2 * r1a
18 rb = r1b
19 r2a = r2c
20 DH1b = -15000
21 DH2a = -10000
22 r1c = -r1a
23 Ta55 = 325
24 Fi = 6.2
12-38
25 Cpa = 10
26 Cpb = 12
27 Cpc = 14
28 Cpd = 16
29 rc = r1c + r2c
30 Ua = 80
31 r2d = -2 * r2c
32 ra = r1a + r2a
33 rd = r2d
34 Qg = r1b * DH1b + r2a * DH2a
35 Qr = Ua * (T - Ta)
36 Cpi = 10
37 sumFiCpi = Cpa * Fa + Cpb * Fb + Cpc * Fc + Cpd * Fd + Cpi * Fi
P12-2 (i) CD Example CDR12.4-1
~
2
1
dP1 P01
dP2 P0 2
1
dP P0
1 1
12 2
1
12-39
No effect for turbulent flow if both dP and P changed at the same time.
P12-3
T (K)
(a)
390
380
370
360
350
340
330
320
310
Fao = 8 mol/s
Fao = 5 mol/s
Fao = 1 mol/s
0
1000
2000
3000
4000
5000
X
W (g)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Fao = 8 mol/s
Fao = 5 mol/s
Fao = 1 mol/s
0
1000
2000
3000
4000
5000
W (g)
1
Xe
0.8
Fao = 8 mol/s
0.6
Fao = 5 mol/s
0.4
Fao = 1 mol/s
0.2
0
0
1000
2000
3000
4000
5000
W (g)
When the flowrate of A is large (high Fao), the reactants spend less time in the
reactor. That is, the residence time is short, so the exit conversion is small. At small
Fao, the residence time is long, so the exit conversion is large. Notice that at low
Fao, the conversion reaches equilibrium conversion, while at high Fao, the
conversion does not reach equilibrium conversion.
12-40
(b)
420
T (K)
400
380
thetaI = 4
360
thetaI = 1
340
thetaI = 0.5
320
300
0
1000
2000
3000
4000
5000
W (g)
0.6
X
0.5
0.4
thetaI = 4
0.3
thetaI = 1
0.2
thetaI = 0.5
0.1
0
0
1000
2000
3000
4000
5000
W (g)
1
Xe
0.8
thetaI = 4
0.6
thetaI = 1
0.4
thetaI = 0.5
0.2
0
0
1000
2000
3000
4000
5000
W (g)
As the fraction of inerts in the entering stream increases, the concentration of A in
the entering stream decreases. Low concentration of A
slower reaction rate
lower conversion and less heat evolved by the reaction (that is, lower temperature
in the reactor).
(c)
12-41
T (K)
400
380
Ua/rho = 0.1
cal/kg/s/K
360
340
Ua/rho = 0.5
cal/kg/s/K
320
Ua/rho = 0.8
cal/kg/s/K
300
0
1000
2000
3000
4000
5000
W (g)
0.6
0.5
Ua/rho = 0.1
cal/kg/s/K
X
0.4
Ua/rho = 0.5
cal/kg/s/K
0.3
0.2
Ua/rho = 0.8
cal/kg/s/K
0.1
0
0
1000
2000
3000
4000
5000
W (g)
Xe
1
0.8
Ua/rho = 0.1
cal/kg/s/K
0.6
0.4
Ua/rho = 0.5
cal/kg/s/K
0.2
Ua/rho = 0.8
cal/kg/s/K
0
0
1000
2000
3000
4000
5000
W (g)
A higher heat transfer coefficient means there is a faster rate of heat transfer from
the reactor to the coolant. Thus, at high Ua the reactor temperature is low. The
lower reactor temperature causes reaction rate to be slow, and thus lower
conversion is observed.
12-42
(d)
420
T (K)
400
380
To = 310K
360
To = 330K
340
To = 350K
320
300
0
1000
2000
3000
4000
5000
W (g)
0.6
X
0.5
0.4
To = 310K
0.3
To = 330K
0.2
To = 350K
0.1
0
0
1000
2000
3000
4000
5000
W (g)
1
0.8
To = 310K
0.6
To = 330K
0.4
To = 350K
0.2
0
0
1000
2000
3000
4000
5000
A higher entrance temperature causes reaction rate to be fast, thus high conversion
is observed. Conversely, a low entrance temperature causes reaction rate to be
slow, thus a lower conversion is observed.
12-43
(e)
400
T (K)
380
Tao = 300K
360
Tao = 320K
340
Tao = 340K
320
300
0
1000
2000
3000
4000
5000
W (g)
0.6
X
0.5
0.4
Tao = 300K
0.3
Tao = 320K
0.2
Tao = 340K
0.1
0
0
1000
2000
3000
4000
5000
W (g)
1
Xe
0.8
Tao = 300K
0.6
Tao = 320K
0.4
Tao = 340K
0.2
0
0
1000
2000
3000
4000
5000
W (g)
A lower coolant temperature results in lower reactor temperatures. Lower reactor
temperatures result in lower reaction rate, thus lower conversion.
(f) T0 = 350K
12-44
420
400
T (K)
380
mc = 50 g/s
360
mc = 1000 g/s
340
320
300
0
1000
2000
3000
4000
5000
W (g)
0.6
0.5
X
0.4
mc = 50 g/s
0.3
mc = 1000 g/s
0.2
0.1
0
0
1000
2000
3000
4000
5000
W (g)
1
Xe
0.8
0.6
mc = 50 g/s
0.4
mc = 1000 g/s
0.2
0
0
1000
2000
3000
4000
5000
W (g)
With a higher coolant flowrate, the coolant can remove heat from the reactor at a
faster rate. Thus, the temperature of the reactor will be lower with a higher coolant
flowrate. Again, a lower reactor temperature causes slower reaction rate, which
results in smaller conversion.
(g) T0 = 350K
12-45
420
400
mc = 50 g/s
(countercurrent)
T (K)
380
360
mc = 1000 g/s
(countercurrent)
340
320
300
0
2000
4000
6000
W (g)
0.6
0.5
mc = 50 g/s
(countercurrent)
X
0.4
0.3
mc = 1000 g/s
(countercurrent)
0.2
0.1
0
0
2000
4000
6000
W (g)
1
Xe
0.8
0.6
mc = 50 g/s
(countercurrent)
0.4
mc = 1000 g/s
(countercurrent)
0.2
0
0
2000
4000
6000
W (g)
The counter-current case is similar to the co-current case, except when mc=50 g/s,
the temperature of the reactor is lower in the counter-current case. This is
because, in general, counter-current heat exchange is more efficient than cocurrent heat exchange.
(h) Fluidized bed CSTR
Parameters:
Weight of the bed : 5000 kg
UA
: 500 cal/s.k
Ta
: 320 K
Density of bed : 2 Kg/m3
Rest all parameters are as specified in the base case.
12-46
Mole Balance:
X MB
W
FAo
rA'
rA'
CC2
)
KC
k (C A C B
CA
CB
C A0 1 X MB
CC
2C A0 X MB
T0
T
T0
T
W k C A0 1 X MB
2
T0
T
2C Ao X MB
KC
2
X MB
FA0
Solving this implicit equation will give X MB T
Energy Balance:
C P 0 and B
I
mc C Pc (Ta
1,
C
T ) 1 exp
X EB T
0
UA
mc C Pc
0
H Rx
So we have X EB (T )
Equilibrium Conversion:
At equilibrium,
rA'
C AC B
CC2
KC
C A0 1 X e
Xe
0
2
2C A0 X e
KC
2
K C0.5
K C0.5
2
12-47
FA 0
i
C Pi (T Ti 0 )
1.2
1
X
0.8
XMB
0.6
XEB
0.4
0.2
39
3
38
9
38
5
38
1
37
7
37
3
36
9
36
5
36
1
35
7
35
3
34
9
0
T
Xe
0.7
0.6
X
0.5
0.4
Xe
0.3
0.2
0.1
39
3
38
9
38
5
38
1
37
7
37
3
36
9
36
5
36
1
35
7
35
3
34
9
0
T
The intersection of mole balance and energy balance curve gives the conversion
achieved in the reactor.
Conversion achieved X = 0.438
Temperature
= 369.7 K (Xe = 0.441)
T (K)
(i) Endothermic, varying Fao
332
330
328
326
324
322
320
318
316
Fao = 8 mol/s
Fao = 5 mol/s
Fao = 1 mol/s
0
1000
2000
3000
4000
5000
W (g)
12-48
0.12
X
0.1
0.08
Fao = 8 mol/s
0.06
Fao = 5 mol/s
0.04
Fao = 1 mol/s
0.02
0
0
1000
2000
3000
4000
5000
W (g)
0.2
0.15
Xe
Fao = 8 mol/s
0.1
Fao = 5 mol/s
Fao = 1 mol/s
0.05
0
0
1000
2000
3000
4000
5000
W (g)
When the flowrate of A is large (high Fao), the reactants spend less time in the
reactor. That is, the residence time is short, so the exit conversion is small. At small
Fao, the residence time is long, so the exit conversion is large. Notice that at low
Fao, the conversion reaches equilibrium conversion, while at high Fao, the
conversion does not reach equilibrium conversion.
Endothermic, varying thetaI
332
330
328
326
thetaI = 4
324
322
thetaI = 1
thetaI = 0.5
320
318
316
0
1000
2000
3000
4000
5000
12-49
0.08
0.07
0.06
0.05
thetaI = 4
0.04
0.03
thetaI = 1
thetaI = 0.5
0.02
0.01
0
0
1000
2000
3000
4000
5000
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
thetaI = 4
thetaI = 1
thetaI = 0.5
0
1000
2000
3000
4000
5000
As the fraction of inerts in the entering stream increases, the concentration of A in
the entering stream decreases. Low concentration of A
slower reaction rate
lower conversion and less heat consumed by the reaction (that is, higher
temperature in the reactor).
Endothermic, varying To
355
350
345
340
335
330
325
320
315
310
305
To = 310K
To = 330K
To = 350K
0
1000
2000
3000
4000
5000
12-50
0.12
X
0.1
0.08
To = 310K
0.06
To = 330K
0.04
To = 350K
0.02
0
0
1000
2000
3000
4000
5000
Xe
W (g)
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
To = 310K
To = 330K
To = 350K
0
1000
2000
3000
4000
5000
W (g)
A higher entrance temperature causes reaction rate to be fast, thus high conversion
is observed. Conversely, a low entrance temperature causes reaction rate to be
slow, thus a lower conversion is observed.
OVERALL:
Exothermic reactions:
-equilibrium conversion varies inversely with reactor temperature
Exothermic reactions:
-equilibrium conversion varies with reactor temperature
Conversion approaches equilibrium conversion with:
-a faster reaction rate (higher T or higher amount of reactants)
-Notice in all the plots, a higher T causes X to approach Xe
-a longer residence time
P12-4 Solution is in the decoding algorithm available in the beginning of this manual
12-51
P12-5 (a) Individualized solution
P12-5 (b)
NH 4 NO3
2H 2O g
A
2W g
N 2O g
Bg
From Rate Data
k
ln 2
k1
E T2 T1
R T2T1
ln
E
R
50
E
R 970 1020
2.912
0.307
44518
E 1 1
R 970 T
k 0.307 exp
Mole Balance
FA0 X
rA
V
k
rAV
FA 0
X MB
M
V
V
FA 0
kM
FA0
Energy Balance
FA0 H A0
FA0 H A0
FA0
W
FW0 H W0 FA H A g
H W0 FA0 1 X H A g
H A g, T
H Rx
FW H W g
FA0
HA
2H W g
W
FB H B g
2FA0 X H W g
,T
H Vap
HB g
HA
0
FA0 XH B g
0
H Rx
H A0 H A g
W
CPA T 660
HA
,T
H W0 H W g
2H W g
HB g
HA
H Vap X
0
0
H A0
XE
1 X
H Vap
CP T 660
W
W
HS 500 F
H W 200 F
C PS T 500
HS 500 F
H W 200 F
C PS T 500
H Rx
12-52
H Rx X
W
0.17
0.83
FW
FA
18
80
0.9103
BTU 80lb
BTU
30.4
lb R mol
lbmol R
BTU 18lb
BTU
C PS 0.47
8.46
lb R mol
lbmol R
BTU 80lb
BTU
H Rx
336
26,880
lb
mol
lbmol
BTU
BTU
H g 200 F 168
2,916
lb
lbmol
BTU
BTU
H W 500 F 1, 202
21,636
lb
lbmol
C PA
0.38
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cpa
30.4
30.4
30.4
30.4
2 Cps
8.46
8.46
8.46
8.46
3 delH
-2.688E+04
-2.688E+04
-2.688E+04
-2.688E+04
4 Fao
257.3
257.3
257.3
257.3
5 Hg
2916.
2916.
2916.
2916.
6 Hw
2.164E+04
2.164E+04
2.164E+04
2.164E+04
7 k
0.307
0.307
0.7758943
0.7758943
8 M
500.
500.
500.
500.
9 Qg
1.604E+04
1.604E+04
4.053E+04
4.053E+04
10 Qr
2.654E+04
2.654E+04
2.73E+04
2.73E+04
11 t
0
0
20.
20.
12 T
970.
970.
990.
990.
13 thetaw
0.9103
0.9103
0.9103
0.9103
14 Xe
0.9874192
0.9874192
1.015768
1.015768
15 Xm
0.5965799
0.5965799
1.507762
1.507762
Differential equations
1 d(T)/d(t) = 1
Explicit equations
1 thetaw = 0.9103
2 Cpa = 30.4
3 Cps = 8.46
4 delH = -26880
5 k = 0.307*exp(44518*(1/970-1/T))
6 M = 500
12-53
7 Fao = 310*0.83
8 Hw = 21636
9 Hg = 2916
10 Xm = k*M/Fao
11 Xe = Cpa*(T-660)/(-delH)+thetaw*(Hw-Hg+Cps*(T-960))/(-delH)
12 Qg = -delH*Xm
13 Qr = -delH*Xe
The temperature prior to shutdown is 981.10 0R
12-54
P12-5 (c) Individualized solution
P12-5 (d) Individualized solution
P12-5 (e) Individualized solution
P12-6
A B
lb mole
hr
Tio(F)
Fio
2C
A
B
C
10
10
0.0
80
80
-
~
CPio
Btu
lb mole F
51
44
47.5
MW ,
lb
lb mol
128
94
222
63
67.2
65
20, 000
Btu
,
lb mol A
i,
lb
ft 3
HR
Energy balance with work term included is:
Q WS
X A HR
FA0
A
1, B
FB 0
FA0
i
CPi T To
10
1, X AF
10
1
Q UA(Ts T )
Substituting into energy balance,
UA(TS T ) WS FA0 H R X AF
UA(TS T ) WS
T
T0
FA0 H R
FA0 C pA C pB T T0
FA0 C pA C pB
UA(TS T ) WS FA0 H R
FA0 C pA C pB UA
Ws
63525
T
199 F
Btu
hr
12-55
UA T T0
P12-7 (a)
A B
C
Since the feed is equimolar, CA0 = CB0 = 0 .1 mol/dm3
CA = CA0(1-X)
CB = CB0(1-X)
Adiabatic:
T
X[
H R (T0 )]
X CP
i CPi
T0
CP
C pC C pB C pA
H R (T )
HC
Ci
C pA
i
30 15 15 0
HB
H A = - 41000 - (-15000 ) - (-20000) = -6000 cal/mol A
C pB
15 15 30
B
rA
6000 X
300 200 X
30
k C A2 0 (1 X ) 2 .01 k (1 X )2
VPFR
FA0
T
cal
mol K
300
dX
rA
FA0 X
rA
FA0 = CA0v0 = (.1)(2) = 0.2 mols/dm3
VCSTR
k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
See Polymath program P12-7-a.pol.
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1T
483.8314 -1.421E-13 700.
2X
0.919157 -1.516E-10 0.99
Variable Value
1 Ca0
0.1
2 Fa0
0.2
12-56
3k
5.625546
4 ra
-0.0003677
5V
500.
Nonlinear equations
1 f(T) = T- 300 - 200 * X = 0
2 f(X) = X + V*ra/Fa0 = 0
Explicit equations
1 V = 500
2 k = 0.01*exp((10000 / 2) * (1 / 300 - 1 / T))
3 Fa0 = 0.2
4 Ca0 = 0.1
5 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)
For 500 dm3 CSTR, X = 0.92
For two 250 dm3 CSTR in series, X = 0.967
P12-7 (b) Constant heat exchanger temperature Ta
When heat exchanger is added, the energy balance can be written as
H Rxn (T )
dT Ua(Ta T ) ( rA )
dV
F (
C
Cˆ )
A0
So with Cˆ P = 0,
i
C pi
i
pi
P
30 , H Rxn = -6000 cal/mol
dT
dV
Ua(Ta T ) ( rA ) 6000
FA0 (30)
Where Ua = 20 cal/m3/s/K, Ta = 450 K
See Polymath program P12-7-b.pol.
12-57
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
10
10
X
0
0
0.3634806 0.3634806
T
300
300
455.47973 450.35437
k
0.01
0.01
3.068312 2.7061663
Kc
286.49665 9.2252861 286.49665 9.9473377
Fa0
0.2
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
0.1
ra
-1.0E-04
-0.0221893 -1.0E-04 -0.0010758
Xe
0.8298116 0.3682217 0.8298116 0.3810642
DH
-6000
-6000
-6000
-6000
Ua
20
20
20
20
Ta
450
450
450
450
Fao
0.2
0.2
0.2
0.2
sumcp
30
30
30
30
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
[8] Ua = 20
[9] Ta = 450
[10] Fao = 0.2
[11] sumcp = 30
12-58
500
450
T [K]
400
350
300
250
0
1
2
3
4
5
6
7
8
9
10
6
7
8
9
10
3
V [m ]
0.4
0.35
0.3
X
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
V [m3]
0.9
0.8
0.7
0.6
Xe
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
3
V [m ]
12-59
8
9
10
P12-7 (c)
For a co-current heat exchanger,
CpC = 1cal/g/K, Ta1=450 K, m 50
g
sec
See Polymath program P12-7-c.pol.
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
10
10
X
0
0
0.3611538 0.3611538
T
300
300
442.15965 442.15965
Ta
450
434.90618 450
441.60853
k
0.01
0.01
2.1999223 2.1999223
Kc
286.49665 11.263546 286.49665 11.263546
Fa0
0.2
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
0.1
ra
-1.0E-04
-0.0160802 -1.0E-04 -0.0019246
Xe
0.8298116 0.4023362 0.8298116 0.4023362
DH
-6000
-6000
-6000
-6000
Ua
20
20
20
20
Fao
0.2
0.2
0.2
0.2
sumcp
30
30
30
30
mc
50
50
50
50
Cpc
1
1
1
1
12-60
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
[3] d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
[8] Ua = 20
[9] Fao = 0.2
[10] sumcp = 30
[11] mc = 50
[12] Cpc = 1
500
Ta
450
400
350
T
300
250
0
1
2
3
4
5
6
7
8
V [m3]
12-61
9
10
0.9
0.8
0.7
0.6
Xe
0.5
0.4
0.3
X
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
10
3
V [m ]
Next increase the coolant flow rate and run the same program to compare results.
P12-7 (d)
For counter-current flow, swap (T – Ta) with (Ta-T) in dTa/dV equation in the previous polymath
program.
See Polymath program P12-7-d.pol.
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
10
10
12-62
X
0
0
0.3647241 0.3647241
T
300
300
463.44558 450.37724
Ta
440.71
440.71
457.98124 450.00189
k
0.01
0.01
3.7132516 2.7077022
Kc
286.49665 8.2274817 286.49665 9.9439517
Fa0
0.2
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
0.1
ra
-1.0E-04
-0.0256436 -1.0E-04 -9.963E-04
Xe
0.8298116 0.3488462 0.8298116 0.381006
DH
-6000
-6000
-6000
-6000
Ua
20
20
20
20
Fao
0.2
0.2
0.2
0.2
sumcp
30
30
30
30
mc
50
50
50
50
Cpc
1
1
1
1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
[3] d(Ta)/d(V) = Ua*(Ta-T)/mc/Cpc
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
[8] Ua = 20
[9] Fao = 0.2
[10] sumcp = 30
[11] mc = 50
[12] Cpc = 1
12-63
500
Ta
450
T
400
350
300
250
0
1
2
3
4
5
6
7
8
9
10
6
7
8
9
10
V [m3]
0.9
0.8
0.7
Xe
0.6
0.5
0.4
0.3
0.2
0.1
X
0
0
1
2
3
4
5
V [m3]
P12-7 (e) Adiabatic
See Polymath program P12-7-e.pol.
12-64
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca0
0.1
0.1
0.1
0.1
2 Cpc
1.
1.
1.
1.
3 DH
-6000.
-6000.
-6000.
-6000.
4 Fa0
0.2
0.2
0.2
0.2
5 k
0.01
0.01
0.010587
0.010587
6 Kc
286.4967
276.8576
286.4967
276.8576
7 mc
50.
50.
50.
50.
8 Qg
0.6
0.6
0.6286162
0.6286162
9 Qr
0
0
0
0
10 ra
-0.0001
-0.0001048
-0.0001
-0.0001048
11 sumCp
30.
30.
30.
30.
12 T
300.
300.
301.0235
301.0235
13 Ta
450.
450.
450.
450.
14 Ua
0
0
0
0
15 V
0
0
10.
10.
16 X
0
0
0.0051176
0.0051176
17 Xe
0.8298116 0.827152
0.8298116
0.827152
Differential equations
1 d(X)/d(V) = -ra/Fa0
2 d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fa0*sumCp)
3 d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
Explicit equations
1 Kc = 10*exp(-6000/1.987*(1/450-1/T))
2 k = 0.01*exp((10000/1.987)*(1/300-1/T))
3 Fa0 = 0.2
4 Ca0 = 0.1
5 ra = -k*(Ca0^2)*((1-X)^2-X/Ca0/Kc)
6 Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
7 DH = -6000
8 Ua = 20*0
9 sumCp = 30
12-65
10 mc = 50
11 Cpc = 1
12 Qg = ra*DH
13 Qr = Ua*(T-Ta)
12-66
P12-7 (f)
We see that it is better to use a counter-current coolant flow as in this case we achieve the
maximum equilibrium conversion using a lesser volume of the PFR.
12-67
P12-8
Refer to solution P11-6
rA
dT
dV
Heat Exchange
FiCPi
(16B)
FA0
rA
dT
dV
H Rx Ua T Ta
FiCPi
iCPi
H Rx
FA0
CP X , if CP
Ua T Ta
iC Pi
A. Constant Ta (17B) Ta = 300
Additional Parameters (18B – (20B): Ta,
i C Pi , Ua,
B. Variable Ta Co-Current
(17C)
dTa
dV
Ua T Ta
, V 0
ÝCPcool
m
Ta
Tao
Ta
?
C. Variable Ta Counter Current
(18C)
dTa
dV
Ua Ta T
ÝCPcool
m
V 0
Guess Ta at V = 0 to match Ta = Tao at exit, i.e., V = Vf
(a) Variable Ta Co-Current
12-68
0 then
(b) Gas Phase Counter Current Heat Exchange Vf = 20 dm3
Matches
12-69
(c)
Constant Ta
12-70
----------------------------------------------------------------------------------------------------------------------------(d) refer to problem P11-6
(e)
Endothermic
PFR A
B
dX
dV
k1
1
1
X
KC
0
, Xe
KC
1 KC
Xe
X
XEB
X EB
i C Pi
T T0
H Rx
T0
12-71
C PA
I C PI
H Rx
T T0
H Rx X
CPA
I C PI
T T0
PFR Adiabatic
FA0
FT
1. Irreversible A
A.
B Liquid Phase, Keep FA0 Constant
First order
dX
dV
rA
FA0
F 1 X
k A0
FA0
kCA
FA0
k 1 X kCA0 1 X
FA0
Constant density liquid
volumetric flow rate without inert
0
0
FA0 FI
FA0
0 1
I
k1 X
0 1
I
dX
dV
First Order Irreversible
I
ISO
I
1
ISO
k
T
1
V
V
12-72
I
I
k 1 yI
X
k
1
I
I OP T
I
Endothermic Adiabatic
12-73
I OP T
I
12-74
12-75
-----------------------------------------------------------------------------------------------------------------------------
P12-9
Refer to solution P11-7
For Heat Exchanger
A
Energy Balance
terms
(20),
CP
15
Same as adiabatic
0
Ua
Energy Balance
1
dT
b
dW
FA0
Ta
T
rA H Rx
rA H Rx
Ua
T Ta
b
i C Pi
FA0
CP X
Ua
320
b
1400
12-76
0.23
i C Pi
(21)
dT
0.23 Ta
T
dW
500
Ta
A Constant Ta (22A)
300 K
Ua
B Co-Current Exchange (22B)
C Counter Current (22C)
T Ta
dTa
b
dW
Ýc C PCool
m
Ua
dTa
dW
rA H Rx
Ta
T
b
mcCPCool
(a) Co-Current Heat Exchange
Gas Phase Co-current Variable Ta
12-77
, mc = 018, C PCool
18
(b) Counter Current Ta
Matches
12-78
Gas Phase Counter Current Variable Ta
12-79
Gas Phase Counter Current Variable Ta
Gas Phase Counter Current Variable Ta
12-80
(c) Constant Ta
Constant Ta
12-81
Gas Phase Constant Ta
12-82
Gas Phase Constant Ta
(d) Refer to solution P11-7 for comparison
P12-10 (a)
For reversible reaction, the rate law becomes
rA
T
300 200 X
k
k (300) exp
KC
k C AC B
E 1
1
R 300 T
K C (450) exp
H Rxn 1
1
R
450 T
Stoichiometry:
CC
C A0 X
CA
C A0 (1 X )
CB
C A0 (1 X )
12-83
CC
KC
See Polymath program P12-10-a.pol.
POLYMATH Results
No Title 03-21-2006, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
10
10
X
0
0
0.0051176 0.0051176
T
300
300
301.02352 301.02352
k
0.01
0.01
0.010587 0.010587
Fa0
0.2
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
0.1
Kc
286.49665 276.85758 286.49665 276.85758
ra
-1.0E-04
-1.048E-04 -1.0E-04 -1.048E-04
Xe
0.8298116 0.827152
0.8298116 0.827152
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
Explicit equations as entered by the user
[1] T = 300+200*X
[2] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] Kc = 10 * exp(-6000 /1.987 * (1 / 450 - 1 / T))
[6] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[7] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
12-84
301.2
0.006
301
0.005
300.8
X
T [K]
0.004
300.6
0.003
300.4
0.002
300.2
0.001
300
299.8
0
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
3
4
5
V [m ]
0.83
0.8295
Xe
0.829
0.8285
0.828
0.8275
0.827
0
1
2
3
4
5
6
7
8
9
10
V [m3]
P12-10 (b)
When heat exchanger is added, the energy balance can be written as
H Rxn (T )
dT Ua (Ta T ) ( rA )
dV
FA0 (
CöP )
i C pi
C
30 , H = -6000 cal/mol
So with Cö = 0,
P
i
pi
6
3
V [m ]
Rxn
dT
dV
Ua(Ta T ) ( rA ) 6000
FA0 (30)
Where Ua = 20 cal/m3/s/K, Ta = 450 K
See Polymath program P12-10-b.pol.
12-85
7
8
9
10
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
10
10
X
0
0
0.3634806 0.3634806
T
300
300
455.47973 450.35437
k
0.01
0.01
3.068312 2.7061663
Kc
286.49665 9.2252861 286.49665 9.9473377
Fa0
0.2
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
0.1
ra
-1.0E-04
-0.0221893 -1.0E-04 -0.0010758
Xe
0.8298116 0.3682217 0.8298116 0.3810642
DH
-6000
-6000
-6000
-6000
Ua
20
20
20
20
Ta
450
450
450
450
Fao
0.2
0.2
0.2
0.2
sumcp
30
30
30
30
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
[8] Ua = 20
[9] Ta = 450
[10] Fao = 0.2
[11] sumcp = 30
12-86
500
0.4
0.35
450
0.3
0.25
X
T [K]
400
350
0.2
0.15
0.1
300
0.05
250
0
0
1
2
3
4
5
6
7
8
9
10
0
3
0.9
0.8
0.7
0.6
Xe
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
1
2
3
4
5
V [m3]
V [m ]
6
7
8
9
10
V [m3]
P12-10 (c)
For a co-current heat exchanger,
g
sec
See Polymath program P12-10-c.pol.
CpC = 1cal/g/K, Ta1=450 K, m 50
12-87
6
7
8
9
10
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
10
10
X
0
0
0.3611538 0.3611538
T
300
300
442.15965 442.15965
Ta
450
434.90618 450
441.60853
k
0.01
0.01
2.1999223 2.1999223
Kc
286.49665 11.263546 286.49665 11.263546
Fa0
0.2
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
0.1
ra
-1.0E-04
-0.0160802 -1.0E-04 -0.0019246
Xe
0.8298116 0.4023362 0.8298116 0.4023362
DH
-6000
-6000
-6000
-6000
Ua
20
20
20
20
Fao
0.2
0.2
0.2
0.2
sumcp
30
30
30
30
mc
50
50
50
50
Cpc
1
1
1
1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
[3] d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
[8] Ua = 20
[9] Fao = 0.2
[10] sumcp = 30
[11] mc = 50
[12] Cpc = 1
12-88
500
0.9
0.8
Ta
450
0.7
0.6
Xe
400
0.5
0.4
350
0.3
T
X
0.2
300
0.1
250
0
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
3
5
6
7
8
3
V [m ]
V [m ]
Next increase the coolant flow rate and run the same program to compare results.
P12-10(d)
For counter-current flow,
See Polymath program P12-10-d.pol.
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
10
10
X
0
0
0.3647241 0.3647241
T
300
300
463.44558 450.37724
Ta
440.71
440.71
457.98124 450.00189
k
0.01
0.01
3.7132516 2.7077022
Kc
286.49665 8.2274817 286.49665 9.9439517
Fa0
0.2
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
0.1
ra
-1.0E-04
-0.0256436 -1.0E-04 -9.963E-04
Xe
0.8298116 0.3488462 0.8298116 0.381006
DH
-6000
-6000
-6000
-6000
Ua
20
20
20
20
Fao
0.2
0.2
0.2
0.2
sumcp
30
30
30
30
mc
50
50
50
50
Cpc
1
1
1
1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
[3] d(Ta)/d(V) = Ua*(Ta-T)/mc/Cpc
12-89
9
10
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
[8] Ua = 20
[9] Fao = 0.2
[10] sumcp = 30
[11] mc = 50
[12] Cpc = 1
500
0.9
0.8
Ta
450
0.7
Xe
T
0.6
400
0.5
0.4
350
0.3
0.2
300
0.1
250
X
0
0
1
2
3
4
5
6
7
8
9
10
0
V [m3]
1
2
3
4
5
6
7
8
9
10
V [m3]
P12-10 (e)
We see that it is better to use a counter-current coolant flow as in this case we achieve the
maximum equilibrium conversion using a lesser volume of the PFR.
P12-10 (f)
If the reaction is irreversible but endothermic, we have
rA k C A2 0 (1 X )2 .01 k (1 X )2 as obtained in the earlier problem.
H Rxn 6000cal / mol
See Polymath program P12-10-f-co.pol.
we use 8- 7f cocurrent.pol
12-90
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
10
10
X
0
0
0.4016888 0.4016888
T
300
300
428.84625 424.16715
Ta
450
425.45941 450
425.45941
k
0.01
0.01
1.4951869 1.314808
Ca0
0.1
0.1
0.1
0.1
Fa0
0.2
0.2
0.2
0.2
ra
-1.0E-04
-0.0132694 -1.0E-04 -0.0047067
DH
6000
6000
6000
6000
Ua
20
20
20
20
Fao
0.2
0.2
0.2
0.2
sumcp
30
30
30
30
mc
50
50
50
50
Cpc
1
1
1
1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
[3] d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 2) * (1 / 300 - 1 / T))
[2] Ca0 = 0.1
[3] Fa0 = 0.2
[4] ra = -k * (Ca0 ^ 2) *(1 - X) ^ 2
[5] DH = 6000
[6] Ua = 20
[7] Fao = 0.2
[8] sumcp = 30
[9] mc = 50
[10] Cpc = 1
12-91
For counter-current flow,
See Polymath program P12-10-f-counter.pol.
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
10
10
X
0
0
0.3458817 0.3458817
T
300
300
449.27319 449.27319
Ta
423.8
423.8
450.01394 450.01394
k
0.01
0.01
2.5406259 2.5406259
Kc
0.3567399 0.3567399 9.8927301 9.8927301
Fa0
0.2
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
0.1
ra
-1.0E-04
-0.0141209 -1.0E-04 -0.0019877
Xe
0.0333352 0.0333352 0.3801242 0.3801242
DH
6000
6000
6000
6000
12-92
Ua
Fao
sumcp
mc
Cpc
20
0.2
30
50
1
20
0.2
30
50
1
20
0.2
30
50
1
20
0.2
30
50
1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
[3] d(Ta)/d(V) = Ua*(Ta-T)/mc/Cpc
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 2) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(6000 / 2 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = 6000
[8] Ua = 20
[9] Fao = 0.2
[10] sumcp = 30
[11] mc = 50
[12] Cpc = 1
12-93
P12-11 (a)
Constant Ta, PBR
Part (1)
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W
0
0
50
50
X
0
0
0.8987768 0.8987768
T
450
450
893.55511 893.55511
To
450
450
450
450
vo
20
20
20
20
Cao
0.2706726 0.2706726 0.2706726 0.2706726
Ca
0.2706726 0.0072668 0.2706726 0.0072668
k
0.133
0.133
8.5739542 8.5739542
ra
-0.0359995 -0.1539472 -0.0359995 -0.062305
Ta
323
323
323
323
Cpa
40
40
40
40
delH
-2.0E+04
-2.0E+04
-2.0E+04 -2.0E+04
Uarho
0.08
0.08
0.08
0.08
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -ra/(Cao*vo)
[2] d(T)/d(W) = (Uarho*(Ta-T)+ra*delH)/(Cao*vo*Cpa)
Explicit equations as entered by the user
12-94
[1] To = 450
[2] vo = 20
[3] Cao = 10/0.0821/450
[4] Ca = Cao*(1-X)/(1+X)*To/T
[5] k = 0.133*exp(31400/8.314*(1/450-1/T))
[6] ra = -k*Ca
[7] Ta = 323
[8] Cpa = 40
[9] delH = -20000
[10] Uarho = 0.08
Ua
rho
0.08
J
s kgcat K
Part (2)
Ua
rho
240
J
s kgcat K
12-95
In part (1), Ua/rho is so small that the reactor behaves as if there is no heat exchange occurring.
In part (2), Ua/rho is 3000 times larger, causing the heat exchange term Ua/rho*(T-Ta) to be
much larger. Therefore, in part (2), we can see the effect of heat exchange in the temperature
drop, which causes the reaction to be slower, hence the lower conversion X.
Part (3)
Keep
Ua
rho
240
J
s kgcat K
Add differential equation
dy
dW
2y
* (1
X )*
T
, with
To
And include pressure drop in expression for Ca: C a
C ao
0.019 kg -1
(1 X ) To
y
(1 X ) T
If there is pressure drop in the PBR, we observe a slight decrease in the conversion X. The
pressure drop causes Ca to decrease (y 1 always), which decreases reaction rate slightly,
which decreases conversion slightly.
P12-11 (b) Co-current and Counter-current heat exchange, PBR
For co-current heat exchange, add equation:
dTa
dW
For counter-current heat exchange, add equation:
with mc
0.2
kg
and C p ,coolant
s
5000
J
kg K
Part (1)
12-96
Ua
rho
(T Ta )
mcoolantC p , coolant
dTa
dW
Ua
rho
(Ta T )
,
mc C p , coolant
Ua
rho
0.08
J
s kgcat K
Co-current
Counter-current
Again, Ua/rho is so small that the reactor behaves as if there is no heat exchange.
Therefore, it doesn’t matter whether we use a counter-current or co-current heat exchanger;
the temperature and conversion profiles will be similar.
Part (2)
Ua
rho
240
J
s kgcat K
Co-current
12-97
Counter-current
The heat exchange is significant in this part, so we will see a difference in the temperature and
conversion profiles depending on whether a co-current or counter-current heat exchange is
used. The counter-current heat exchanger is generally better at heat transfer than the cocurrent heat exchanger. Therefore, in the counter-current case, the temperature of the reactor
is almost constant at 323K. Notice that the temperature and conversion profiles are similar to
the constant Ta case (P8-9a, part 2).
The co-current heat exchanger, on the other hand, is not able to keep the reactor temperature
as low as the counter-current. Therefore, the higher reactor temperature results in a higher
conversion in the co-current case.
Part (3)
Keep
Ua
rho
240
J
s kgcat K
Add differential equation
dy
dW
2y
* (1
X )*
T
, with
To
12-98
0.019 kg -1
And include pressure drop in expression for Ca: C a
C ao
(1 X ) To
y
(1 X ) T
Co-current
Counter-current
The pressure drop causes Ca to decrease (y 1), which decreases reaction rate slightly, which
decreases conversion slightly.
P12-11 (c)
Step 1: Find XMB = f(T):
FA0 X
W
rA '
rA '
kC A
CA
C A0
(1 X ) T0
(1 X ) T
Combining:
FA0 X
W
(1 X ) T0
kC A0
(1 X ) T
12-99
Note
V
W
0
b 0
k b (1
T
T0
X2
X (1
X ) TT0
k b )X
k b
X)
( TT0
T
X MB
Where k
T2
T0 k b
k1 exp
E 1
R T1
0
6TT0 k b
2T
T0
2 2
k2 b
2
1
T
Step 2: Find XEB = f(T):
When C P 0 and pure A enters the reactor,
X EB
UA(T Ta )
C PA (T T0 )
FA0
H RXN
0.16
0.14
0.12
X
0.1
Xmb
0.08
Xeb
0.06
0.04
0.02
0
365
370
375
380
385
T(K)
The intersection of the mass balance and energy balance lines correspond to the
conditions in the reactor. Therefore, the reactor operates at T=376.7K and
X=0.101.
Parameters:
12-100
T0
450K
4000s
0.001
b
kg
dm 3
k
0.0133* exp
Ta
323K
UA 500
1
dm 3
R 450
T
kg s
J
s K
FA0
CA 0 0
CPA
40
H RXN
1
E
10
mol
0.0821* 450 dm
3
* 20
dm 3
s
05.41
mol
s
J
mol K
20000
J
mol
P12-11 (d)
For a reversible reaction, we have all the previous equations, but the rate law is modified as:
rA k f C A kr CB CC
X T0
1 X T
Plugging the equation for kr, and solving using POLYMATH program, we get the plots.
Only the co-current program and plots are shown.
CB
CC
C A0
See Polymath program P12-11-d.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 W
0
0
80.
80.
2 X
0
0
0.057593
0.057593
3 T
450.
420.7523
450.
420.7523
4 Ta
323.
323.
426.1627
420.7565
5 T0
450.
450.
450.
450.
6 v0
20.
20.
20.
20.
12-101
7 k
0.133
0.0742131
0.133
0.0742131
8 Uarho
240.
240.
240.
240.
9 P0
1.013E+06 1.013E+06
1.013E+06
1.013E+06
10 CA0
270.8283
270.8283
270.8283
270.8283
11 CA
270.8283
258.1071
270.8283
258.1071
12 kr
0.2
0.076962
0.2
0.076962
13 CC
0
0
15.77362
15.77362
14 CB
0
0
15.77362
15.77362
15 rA
-36.02017
-36.02017
-0.0062421
-0.0062421
Differential equations
1 d(X)/d(W) = -rA / v0 / CA0
2 d(T)/d(W) = (Uarho * (Ta - T) + rA * 20000) / v0 / CA0 / 40
3 d(Ta)/d(W) = Uarho * (T - Ta) / .2 / 5000
+ sign = cocurrent, -ve sign = countercurrent in RHS of eqn.
Explicit equations
1 T0 = 450
2 v0 = 20
3 k = 0.133 * exp(31400 / 8.314 * (1 / T0 - 1 / T))
4 Uarho = .08 * 3000
5 P0 = 1013250
6 CA0 = P0 / 8.314 / T0
7 CA = CA0 * (1 - X) / (1 + X) * T0 / T
8 kr = 0.2 * exp(51400 / 8.314 * (1 / T0 - 1 / T))
9 CC = CA0 * X / (1 + X) * T0 / T
10 CB = CA0 * X / (1 + X) * T0 / T
11 rA = -(k * CA - kr * CB * CC)
12-102
P12-11 (e) Individualized solution
P12-12 (a)
Start with the complete energy balance:
dEˆ
dt
Q WS
Fi H i in
Fi H i out
Fi RV
i
out
The following simplifications can be made:
It is steady state.
In part (a), there is no heat taken away or added
There is no shaft work
That leaves us with
Fi H i in Fi H i out
Evaluating energy terms:
Fi RV
i
out
H A0 FA0 H B 0 FB 0 H C 0 FC 0
Out: H A FA H B FB H C ( FC RCV )
In:
Now we evaluate Fi
FA FA0 FA0 X
FB FB 0 FA0 X
FC FC 0 FA0 X
FB0 = FC0 = 0
12-103
Inserting these into our equation gives:
H A FA0
H A FA0 X
H B FA0 X
H C FA0 X
H C RCV
H A0 FA0
0
Combining and substituting terms gives:
FA0 H A H A0
FA0 X H RX
H C RC
0
FA0CPA T T0
FA0 X H RX
H C (T0 ) CPC T T0
RC
0
RC
0
Differentiating with respect to V with ΔCP = 0
FA0C p
dT
dV
dX
dV
FA0
(rA )
dT
dV
H Rx T
H Rx T
H C (T0 ) CPC T T0
H C (T0 ) CPC T T0
FA0
i
C pi
Combine that with the mole balance and rate law:
rA
dFA
dV
CA
rB
rC
k CA
CB CC
KC
dF
dFB
rB ; C rC RC
dV
dV
F T
F T
CT 0 A 0 CB CT 0 B 0 CC
FT T
FT T
rA ;
CT 0
FC T0
FT T
See Polymath program P12-11-a.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
0.2710027 0.1839444
0.2710027
0.1839444
2 Cb
0
0
0.0098751
0.0098751
3 Cc
0
0
0.0065147
0.0029402
4 Cpa
40.
40.
40.
40.
5 Cpb
25.
25.
25.
25.
6 Cpc
15.
15.
15.
15.
7 Ct0
0.2710027 0.2710027
0.2710027
0.2710027
8 dHrx
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
12-104
RC
9 Fa
5.42
5.143851
5.42
5.143851
10 Fa0
5.420054
5.420054
5.420054
5.420054
11 Fb
0
0
0.2761489
0.2761489
12 Fc
0
0
0.1641532
0.0822207
13 Ft
5.42
5.42
5.584153
5.502221
14 Hc
-3.735E+04 -3.735E+04
-3.48E+04
-3.48E+04
15 k
0.133
0.133
1.325591
1.325591
16 kc
1.5
1.5
1.5
1.5
17 Kc
0.0006905 0.0001596
0.0006905
0.0001596
18 ra
-0.0360434 -0.0589786
-0.0027523
-0.0027523
19 Rc
0
0
0.009772
0.0044103
20 T
450.
450.
619.7977
619.7977
21 To
450.
450.
450.
450.
22 V
0
0
30.
30.
23 Xe
0.0504144 0.0284733
0.0504144
0.0284733
Differential equations
1 d(Fc)/d(V) = -ra-Rc
2 d(Fb)/d(V) = -ra
3 d(Fa)/d(V) = ra
4 d(T)/d(V) = (Fa0*ra*dHrx - Rc*Hc)/(Fa0*Cpa)
Explicit equations
1 k = .133*exp((31400/8.314)*(1/450-1/T))
2 To = 450
3 Ct0 = 10/0.082/450
4 Ft = Fa + Fb + Fc
5 Cc = Ct0*Fc/Ft*To/T
6 Fa0 = Ct0*20
7 Ca = Ct0*Fa/Ft*To/T
8 dHrx = -20000
9 Cb = Ct0*Fb/Ft*To/T
10 Kc = 0.01*exp((dHrx/8.314)*(1/300-1/T))
11 Cpc = 15
12 ra = -k*(Ca-Cb*Cc/Kc)
12-105
13 kc = 1.5
14 Rc = kc*Cc
15 Xe = ((Kc*T/(Ct0*To))/(1+(Kc*T/(Ct0*To))))^0.5
16 Cpa = 40
17 Cpb = 25
18 Hc = -40000+Cpc*(T-273)
P12-12 (b)
Now, the heat balance equation needs to be modified.
dT
dV
Ua(Ta T ) (rA )
H Rx T
FA0
H C (T0 ) CPC T T0
i
C pi
Temperature reaches a peak value of 628 K before again dropping down.
12-106
RC
P12-13 (a)
Substrate  More cells + Product
SC+P
G(T) = X*-∆HRX
To solve for G(T) we need X as a function of temperature, which we get by solving the mass
balance equation.
rg
FA0 X FS 0 X
V
and since rS
then,
rA
rS
Yc / s
V
Yc / s FS 0 X
rg
rg
CC and
(T )
CS
KS
CS
6700
T
48000
1 exp 153
T
0.0038* T exp 21.6
where
(T )
1max
if we combine these equations we get:
Yc / s FS 0 X
V
(T )CC CS
K S CS
12-107
V
CS
V
Yc / s FS 0 X K S
CS
(T )CC CS
CS 0 1 X and CC
Yc / s FS 0 X K S
CS 0Yc / s X
CS 0 1 X
(T )CS 0Yc / s XCS 0 1 X
Canceling and combining gives:
V
FS 0 K S
(T )C
CS 0 1 X
2
S0
1 X
Now solve this expression for X:
FS 0 K S
X 1
(T )VCS20 FS 0CS 0
It can be observed from graph that if the reactor operates at a steady state temperature of
294.4<T(K)>316.79, Conversion achieved is 0.
Now that we have X as a function of T, we can plot G(T).
To get R(T) we must calculate the heat removed which is the sum of the heat absorbed by
reactants to get to the reaction temperature and the heat removed from any heat exchangers.
The heat gained by the reactants = CP 0 T T0
The heat removed by the heat exchanger = UA(T-Ta)/FS0
UA
R(T ) CPS T T0
T Ta
FS 0
12-108
Now enter the equations into polymath and specify all other constants. The adiabatic case is
shown below. The non-adiabatic case would be with explicit equation [12] as A = 1.1.
See Polymath program P12-13-a.pol.
Differential equations as entered by the user
[1] d(T)/d(t) = 1
Explicit equations as entered by the user
[1] mumax = .5
[2] Ycs = .8
[3] vo = 1
[4] Ta = 290
[5] mu = mumax*(.0038*T*exp(21.6-6700/T))/(1+exp(153-48000/T)))
[6] Ks = 5
[7] V = 6
[8] Cso = 100
[9] Fso = vo*Cso
[10] Cps = 5
[11] dH = -20000
[12] UA = 0*300
[13] kappa = UA/(Cps*Fso)
[14] To = 280
[15] X = 1-((Fso*Ks)/((mu*V*(Cso^2))-Fso*Cso))
[16] Gt = if T<294.4 or T>316.7 then 0 else X*(-dH)*Ycs
[17] Rt = Cps*(T-To)+UA*(T-Ta)/Fso
Independent variable
variable name : t
initial value : 0
T(0): 280
final value : 45
Adiabatic case
12-109
Non adiabatic case
12-110
P12-13 (b)
To maximize the exiting cell concentration, we want to maximize the conversion of substrate. If
we look at G(T) from part A, we see that it is at a maximum at about 310 K. This corresponds to
the highest conversion that can be achieved. By changing the values of UA and m c we can
change the slope of the R(T). What we are looking to do is get R(T) to intersect with G(T) at 310
K.
Since we now have a limited coolant flow rate we will use a different value for Q.
Q
mC C PC T
Ta 1 exp
UA
mC C PC
and so,
R(T )
mC C PC T
Ta 1 exp
UA
mC C PC
C PS T
T0
Now we set R(T) equal to the maximum value of G(T) which is 15600 J/h
G (T ) 15600 FS 0
mC C PC T
Ta 1 exp
UA
mC C PC
FS 0 C PS T
T0
And now plug in the known values. Assume the maximum coolant flow rate and that will give
the minimum heat exchange area.
12-111
15600
g
J
100
g
hr
60000
g
hr
1560000
UA
4.2
J
hr
J
310 K
gK
290 K 1 exp
5040000
J
hr
1 exp
UA
g
60000
hr
J
4.2
gK
UA
252000
J
hr K
J
hr K
A=1.85 m2
Plot of G(T) ,R(T) Vs T
12-112
100
11000
g
h
J
hr
5
J
310 K
gK
288 K
P12-13 (c)
From the plot of G(T) vs R(T) for
(1) Aiabatic case : There are two steady states possible- 281K (stable and X=0), 294.2K
(unstable) and 316.76K (stable and X=0.175)
(2) Non adiabatic case with large coolant flow rate : There are three steady states –289K (stable
and X=0), 294.2K (unstable) and 315.7K (stable and X=0.89)
(3) Non adiabatic case with limited coolant flow rate : There are three steady states –289K
(Stable and X=0), 294.3K (unstable) and 309K (stable and X=0.975)
P12-13 (d)
(1) Adiabatic case : Varying To
From the graph, it can be observed that increasing To will increase the steady state
temperature.
12-113
(2) Non adiabatic case with large coolant flow rate :
Varying Ta
Increasing Ta will increase steady state temperature.
(3) Non adiabatic case with limited coolant flow rate
Varying coolant flow rate
Increasing coolant flow rate will decrease steady state temperature achieved.
12-114
P12-14
Assume Adiabatic
(a) -12,000 cal/mol
(b) Ignition = 260°C
Extinction = 190°C
(c) Ignition, T = 300°C, 475°C
Extinction, T = 215°C, 375°C
(d) X (Ignition) = 0.281
X (Extinction) = 0.86
P12-15 (a)
G(T)
X
HR X
k
1
k
,k 6.6 10 3 exp
R(T) C p 0 1
T Tc
C p0
50
Tc
i C pi
UA
C p0 FA 0
8000
50 80
Ta T0
1
350K
E 1
R 350
1
T
2
To find the steady state T, we must set G(T) = R(T). This can be done either graphically or solving
the equations. We find that for T0 = 450 K, steady state temperature 399.94 K.
12-115
POLYMATH Results
02-22-2006, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t
0
0
1000
1000
T
350
350
450
450
RT
0
0
1.5E+04
1.5E+04
EoR
2.013E+04 2.013E+04 2.013E+04 2.013E+04
k
0.0066
0.0066
2346.7972 2346.7972
tau
100
100
100
100
X
0.3975904 0.3975904 0.9999957 0.9999957
GT
2981.9277 2981.9277 7499.968 7499.968
at
2981.9277 -7500.032
4336.6841 -7500.032
12-116
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = 0.1
Explicit equations as entered by the user
[1] RT = 150*(T-350)
[2] EoR = 40000/1.987
[3] k = 6.6*0.001*exp(EoR*(1/350-1/T))
[4] tau = 100
[5] X = tau*k/(1+tau*k)
[6] GT = 7500*X
[7] at = GT-RT
P12-15 (b)
First, we must plot G(T) and R(T) for many different T0’s on the same plot. From this we must
generate data that we use to plot Ts vs To.
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1T
399.9425 1.337E-10 450. ( 300. < T < 600. )
Variable Value
1 EoR
2.013E+04
2 GT
7491.375
3k
8.685615
4 RT
7491.375
5 tau
100.
6X
0.99885
Nonlinear equations
1 f(T) = RT - GT = 0
Explicit equations
1 RT = 150*(T-350)
2 EoR = 40000/1.987
3 k = 6.6*0.001*exp(EoR*(1/350-1/T))
4 tau = 100
12-117
5 X = tau*k/(1+tau*k)
6 GT = 7500*X
400
390
380
370
T
360
350
340
330
320
310
300
310
330
350
370
390
410
430
To
P12-15 (c)
For high conversion, the feed stream must be pre-heated to at least 404 K. At this temperature,
X = .991 and T = 384.2 K in the CSTR. Any feed temperature above this point will provide for
higher conversions.
P12-15 (d)
For a temperature of 369.2 K, the conversion is 0.935
P12-15 (e)
The extinction temperature is 363.3 K (90oC).
P12-16 (a)
Mol Balance :
FA0 X
v0C A0 X
V
rA
k C A CB / K e
V
v0
X
k 1 X
X / Ke
12-118
X 1
X
k (1 1/ K e )
k
k
1
k (1 1/ K e )
G (T )
H Rx X
80000 X
k 1min 1
10 min
K e 100
10
X
.901
1 10(1.01)
G (400) 72080 cal / mol
P12-16 (b)
UA
FA0C pA
R(T ) C pA (1
3600
10* 40
9
)(T TC )
400(T TC )
T0
Ta
310
1
R(T ) 400(T 310)
The following plot gives us the steady state temperatures of 310, 377.5 and 418.5 K
TC
See Polymath program P12-16-b.pol.
P12-16 (c)
12-119
P12-16 (d)
P12-16 (e)
The plot below shows Ta varied.
The next plot shows how to find the ignition and extinction temperatures. The ignition
temperature is 358 K and the extinction temperature is 208 K.
12-120
P12-16 (f)
P12-16 (g)
At the maximum conversion G(t) will also be at its maximal value. This occurs at approximately
T = 404 K. G(404 K) = 73520 cal. For there top be a steady state at this temperature, R(T) = G(T).
See Polymath program P12-16-g.pol.
UA
FA0C pA
R(T ) C pA (1
)(T TC ) 73520
T0
Ta
1
If we plug in the values and solve for UA, we get:
UA = 7421 cal/min/K
where TC
P12-16 (h) Individualized solution
P12-16 (i)
The adiabatic blowout flow rate occurs at
V
0.0041s
v0
v0 V 0.0041*10
v0
0.041
0.0041s
dm3
min
12-121
See Polymath program P12-16-i.pol.
P12-16 (j)
Lowing T0 or Ta or increasing UA will help keep the reaction running at the lower steady state.
P12-17
Given the first order, irreversible, liquid phase reaction:
A B
Ta 100 C
UA 1.0 cal / min/ C
Pure A Feed
0.5g mol / min
C pA
Cp
C pB
HR
C pA
Design Eqn : V
Rate Law :
0
FA0 X
rA
rA kC A
Stoichiometry : C A
C pB
2cal / g mol / C
100cal / g
v0
k (1 X )
C A0 (1 X )
Energy Balance : UA(T TA ) FA0 X H R
FA0
i
C pi (T T0 )
Simplifying,
k
1
X
1 k 1 1
k
HR
1 1
k
The equation for heat removal curve is:
UA
R(T ) C pA 1
T Tc ,
, Tc
C pA FA0
G (T )
HR X
Ta To
1
Plot this along with the heat generation curve for various T0.
12-122
mol / C
P12-17 (a)
(a)
Tc=156oC
R T
CP 0 1
C PS
C PA
T TC
ICP I
UA
FA0C PA
R T
Tc
C PA
2
1cal min K
mol 2cal
0.5
min mol K
2 1 1 T TC
1 Ta T0
1 1
Ta
1
4 T Tc
T0
2
100 T0
2
Draw R(T) with slope 4such that it is tangent to G(T) curve. The point of the tangency is at
G(T) = R(T) = 27.5 and T 163°C. We could solve for Tc or read it off the graph by extrapolating to R(T) =
0 to find Tc = 156.
R T
Ta
4 T
T0
4T 200 2T0
2
27.5cal mol A
From the above Figure at the point of tangency R(T) = 27.5 and T = 163oC solving for T0,
T 0 212 C
or
Tc
Ta
1
T0
156
100 T0
, T0
2
212
R(T) slope is 4. Move
27.5 4 163 Tc
Tc 156
100 T0
156
2
T0
312 100 212oC
12-123
P12-17 (b)
(b)
Ts=180oC
Ts=163oC
We see there are two intersections at which the reactor can operate. At upper intersection T = 180. T =
163oC and 180oC from Figure P12-17.
P12-17 (c)
Slope of R(T) curve is 4 if we move to left we see that there are 3 intersections. Because we heated the
RcTr above the extraction temperature it will operate at the upper steady state at the upper steady
state G(T) = R(T) = 90 cal/mol and T = 180°C. Extrapolating G(T) to large T,
X = 1 so G = (– HRx) we find – HRx = 100 cal/mol).
GT
X H Rx
X 100
X 0.9
90
4 180 Tc
100 T0
2
Solving for T
Tc 157
or get Tc at R T
T0
214
12-124
0, Tc 157
(c)
P12-17 (d)
(d)
Tc=148oC
At the point of tangency R(T) = G(T) = 80 and T = 167
R T
80
Tc 147
Tc
T0
Ta
1
194
4 167 Tc
100 T0
2
T0
*Note all the temperatures in parts (a) – (d) are 2°C due to accuracy of reading the graph
12-125
P12-17 (e) Individualized solution
P12-18
TC = Ta = T0 = 330 K
τ = V/υ0 = 1.2 h
V = FA0X/-rA = CA0υ0X/ -rA
-rA = k(CA – CB/KC) = kCA0(1-X – X/KC)
k = 0.001exp(30000/1.987(1/300-1/T)
X = τk/(1+τK+τK/KC)
G(T) = (-∆HRx)X
G(T) is plotted as a function of T.
See Polymath program P12-18.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cpo
250.
250.
250.
250.
2 DHrx
-4.2E+04
-4.2E+04
-4.2E+04
-4.2E+04
3 E
3.0E+04
3.0E+04
3.0E+04
3.0E+04
4 GT
50.33959
0.0001053
3.61E+04
0.0001053
5 k
0.001
0.001
8.48E+07
8.48E+07
6 Kc
5.0E+06
2.507E-09
5.0E+06
2.507E-09
7 t
0
0
300.
300.
8 T
300.
300.
600.
600.
9 tau
1.2
1.2
1.2
1.2
0.8596127
2.507E-09
10 X
0.0011986 2.507E-09
Differential equations
1 d(T)/d(t) = 1
Explicit equations
1 E = 30000
12-126
2 k = 0.001*exp((E/1.987)*(1/300-1/T))
3 DHrx = -42000
4 Kc = 5000000*exp((DHrx/1.987)*(1/300-1/T))
5 Cpo = 250
6 tau = 1.2
7 X = tau*k/(1+tau*k + tau*k/Kc)
8 GT = X*(-DHrx)
From the plot, the maximum conversion achieved, Xmax = 0.86
At Xmax, T = 367.6 K and G(T) = 36100 cal/mol
R(T) = G(T)
12-127
κ = 2.84
Since
Therefore, UA = (2.84)(10 mol/h)(250 cal/mol/K) = 7,100 cal/h/K
P12-19
Adiabatic
POLYMATH Report
Ordinary Differential Equations
No Title
08-May-2009
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alpha
0.019
0.019
0.019
0.019
2 Ca
1.9
0.7510709
1.902453
0.7510709
3 Ca0
1.9
1.9
1.9
1.9
4 Cc
0
0
0.1974831
0.1113793
5 Cp0
40.
40.
40.
40.
6 Cpc
4200.
4200.
4200.
4200.
7 deltaCp -30.
-30.
-30.
-30.
8 DH
-2.0E+04
-2.0E+04
-2.0E+04
9 epsilon -1.
-1.
-1.
-1.
10 Fa0
5.
5.
5.
5.
11 k
0.01
0.01
0.0261675
0.0261675
12 Kc
10000.
902.809
10000.
902.809
13 m
0.05
0.05
0.05
0.05
14 ra
-0.0361
-0.0627617
-0.014758
-0.014758
15 T
450.
450.
817.9727
817.9727
16 T0
450.
450.
450.
450.
17 Ta
500.
500.
500.
500.
18 Ua
0
0
0
0
-2.0E+04
12-128
19 W
0
0
50.
50.
20 X
0
0
0.4949381
0.4949381
21 y
1.
0.2174708
1.
0.2174708
Differential equations
1 d(X)/d(W) = -ra/Fa0
2 d(y)/d(W) = -alpha/2/y*(1+epsilon*X)*T/T0
3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0
4 d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc
Explicit equations
1 Cp0 = 40
2 DH = -20000
3 epsilon = -1
4 T0 = 450
5 Ca0 = 1.9
6 Fa0 = 5
7 alpha = 0.019
8 Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*y
9 Ua = 0.8*0
10 deltaCp = 1/2*20-40
11 Kc = 10000*exp(DH/8.314*(1/450-1/T))
12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*y
13 m = 0.05
14 Cpc = 4200
15 k = 0.01*exp(8000/8.314*(1/450-1/T))
16 ra = -k*(Ca^2-Cc/Kc)
General
Total number of equations
20
Number of differential equations 4
Number of explicit equations
16
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
12-129
Now, constant temperature Ta = 300K
POLYMATH Report
Ordinary Differential Equations
No Title
08-May-2009
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alpha
0.019
0.019
0.019
0.019
2 Ca
1.9
0.7615715
1.906858
0.7615715
3 Ca0
1.9
1.9
1.9
1.9
4 Cc
0
0
0.1991675
0.1180575
5 Cp0
40.
40.
40.
40.
6 Cpc
4200.
4200.
4200.
4200.
7 deltaCp -30.
-30.
-30.
-30.
8 DH
-2.0E+04
-2.0E+04
-2.0E+04
9 epsilon -1.
-1.
-1.
-1.
10 Fa0
5.
5.
5.
5.
11 k
0.01
0.01
0.0248669
0.0248669
12 Kc
10000.
1025.524
10000.
1025.524
13 m
0.05
0.05
0.05
0.05
-2.0E+04
12-130
14 ra
-0.0361
-0.0610755
-0.0144197
-0.0144197
15 T
450.
450.
783.9972
783.9972
16 T0
450.
450.
450.
450.
17 Ta
500.
500.
500.
500.
18 Ua
0.8
0.8
0.8
0.8
19 W
0
0
50.
50.
20 X
0
0
0.4848172
0.4848172
21 y
1.
0.2300674
1.
0.2300674
Differential equations
1 d(X)/d(W) = -ra/Fa0
2 d(y)/d(W) = -alpha/2/y*(1+epsilon*X)*T/T0
3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0
4 d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc *0
Explicit equations
1 Cp0 = 40
2 DH = -20000
3 epsilon = -1
4 T0 = 450
5 Ca0 = 1.9
6 Fa0 = 5
7 alpha = 0.019
8 Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*y
9 Ua = 0.8
10 deltaCp = 1/2*20-40
11 Kc = 10000*exp(DH/8.314*(1/450-1/T))
12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*y
13 m = 0.05
14 Cpc = 4200
15 k = 0.01*exp(8000/8.314*(1/450-1/T))
16 ra = -k*(Ca^2-Cc/Kc)
General
Total number of equations
20
12-131
Number of differential equations 4
Number of explicit equations
16
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
Now, Co-current heat exchanger
POLYMATH Report
Ordinary Differential Equations
No Title
08-May-2009
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alpha
0.019
0.019
0.019
0.019
2 Ca
1.9
0.7614153
1.906781
0.7614153
3 Ca0
1.9
1.9
1.9
1.9
4 Cc
0
0
0.1990505
0.1177523
5 Cp0
40.
40.
40.
40.
6 Cpc
4200.
4200.
4200.
4200.
-30.
-30.
-30.
7 deltaCp -30.
12-132
8 DH
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
9 epsilon -1.
-1.
-1.
-1.
10 Fa0
5.
5.
5.
5.
11 k
0.01
0.01
0.0249108
0.0249108
12 Kc
10000.
1021.01
10000.
1021.01
13 m
0.05
0.05
0.05
0.05
14 ra
-0.0361
-0.0610733
-0.0144393
-0.0144393
15 T
450.
450.
785.126
785.126
16 T0
450.
450.
450.
450.
17 Ta
500.
499.0741
521.1534
521.1534
18 Ua
0.8
0.8
0.8
0.8
19 W
0
0
50.
50.
20 X
0
0
0.4849408
0.4849408
21 y
1.
0.2296895
1.
0.2296895
Differential equations
1 d(X)/d(W) = -ra/Fa0
2 d(y)/d(W) = -alpha/2/y*(1+epsilon*X)*T/T0
3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0
4 d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc
Explicit equations
1 Cp0 = 40
2 DH = -20000
3 epsilon = -1
4 T0 = 450
5 Ca0 = 1.9
6 Fa0 = 5
7 alpha = 0.019
8 Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*y
9 Ua = 0.8
10 deltaCp = 1/2*20-40
11 Kc = 10000*exp(DH/8.314*(1/450-1/T))
12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*y
13 m = 0.05
12-133
14 Cpc = 4200
15 k = 0.01*exp(8000/8.314*(1/450-1/T))
16 ra = -k*(Ca^2-Cc/Kc)
General
Total number of equations
20
Number of differential equations 4
Number of explicit equations
16
Elapsed time
1.157 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
12-134
Now, counter – current heat exchanger
POLYMATH Report
Ordinary Differential Equations
No Title
08-May-2009
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alpha
0.019
0.019
0.019
0.019
2 Ca
1.9
0.7671508
1.910238
0.7671508
3 Ca0
1.9
1.9
1.9
1.9
4 Cc
0
0
0.2023329
0.1203647
5 Cp0
40.
40.
40.
40.
6 Cpc
4200.
4200.
4200.
4200.
7 deltaCp -30.
-30.
-30.
-30.
8 DH
-2.0E+04
-2.0E+04
-2.0E+04
9 epsilon -1.
-1.
-1.
-1.
10 Fa0
5.
5.
5.
5.
11 k
0.01
0.01
0.025358
0.025358
12 Kc
10000.
976.5939
10000.
976.5939
13 m
0.05
0.05
0.05
0.05
-2.0E+04
12-135
14 ra
-0.0361
-0.062846
-0.0149206
-0.0149206
15 T
450.
450.
796.6909
796.6909
16 T0
450.
450.
450.
450.
17 Ta
520.
500.0379
521.8302
500.0379
18 Ua
0.8
0.8
0.8
0.8
19 W
0
0
50.
50.
20 X
0
0
0.4958573
0.4958573
21 y
1.
0.2280604
1.
0.2280604
Differential equations
1 d(X)/d(W) = -ra/Fa0
2 d(y)/d(W) = -alpha/2/y*(1+epsilon*X)*T/T0
3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0
4 d(Ta)/d(W) = - Ua*(T-Ta)/m/Cpc
Explicit equations
1 Cp0 = 40
2 DH = -20000
3 epsilon = -1
4 T0 = 450
5 Ca0 = 1.9
6 Fa0 = 5
7 alpha = 0.019
8 Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*y
9 Ua = 0.8
10 deltaCp = 1/2*20-40
11 Kc = 10000*exp(DH/8.314*(1/450-1/T))
12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*y
13 m = 0.05
14 Cpc = 4200
15 k = 0.01*exp(8000/8.314*(1/450-1/T))
16 ra = -k*(Ca^2-Cc/Kc)
General
Total number of equations
20
12-136
Number of differential equations 4
Number of explicit equations
16
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
12-137
P12-20 (a)
Liquid Phase : A B
dX
dV
rA
FA0
rA
kC ACB
k
E 1
R T
.01*exp
C
1
300
C A C A0 (1 X ) CB
The energy Balance:
dT Ua(Ta T ) ( rA )(
dV
FA0 C pA C pB
4
D
Assume D
HR )
a
U
4
5J
1cal
1m 2
m 2 Ks 4.184 J 10dm 2
.0120
cal
dm 2 Ks
See Polymath program P12-20-a.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 V
0
0
1000.
1000.
2 T
300.
300.
434.7779
348.2031
3 X
0
0
0.9620102
0.9620102
4 Ta
300.
300.
300.
300.
5 R
1.988
1.988
1.988
1.988
6 E
10000.
10000.
10000.
10000.
7 cao
0.1
0.1
0.1
0.1
8 ca
0.1
0.003799
0.1
0.003799
9 cb
0.1
0.003799
0.1
0.003799
10 k
0.01
0.01
1.808628
0.1018746
11 ra
-0.0001
-0.0010427
-1.47E-06
-1.47E-06
12 cpb
15.
15.
15.
15.
13 cpa
15.
15.
15.
15.
14 fao
0.2
0.2
0.2
0.2
12-138
15 Dhr1
-6000.
-6000.
-6000.
-6000.
16 a
1.
1.
1.
1.
17 U
0.012
0.012
0.012
0.012
Differential equations
1 d(T)/d(V) = (U * a * (Ta - T) + (-ra) * (-Dhr1)) / (fao * (cpa + cpb))
2 d(X)/d(V) = -ra / fao
Explicit equations
1 Ta = 300
2 R = 1.988
3 E = 10000
4 cao = .1
5 ca = cao * (1 - X)
6 cb = cao * (1 - X)
7 k = .01 * exp(-E / R * (1 / T - 1 / 300))
8 ra = -k * ca * cb
9 cpb = 15
10 cpa = 15
11 fao = .2
12 Dhr1 = -6000
13 a = 1
14 U = .0120
12-139
P12-20 (b)
Gas Phase: A
dX
rA
dW FA0
B C
rA
k f CA
k r CB CC
CB
CC
CA0
X T0
1 X T
k
0.133 exp
E 1
R T
kr
0.2exp
Er 1
R 450
1
450
1
, Er
T
The energy Balance:
dT Ua(Ta T ) ( rA )(
dW
FA0C pA
HR
U
51.4 kJ mol
HR )
40 50 70=-20 kJ/mol
5
See Polymath program P12-20-b.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 W
0
0
50.
50.
2 X
0
0
0.0560855
0.0560855
3 T
400.
371.902
400.
371.902
4 T0
400.
400.
400.
400.
5 k
0.133
0.0651696
0.133
0.0651696
6 v0
20.
20.
20.
20.
7 kr
0.2
0.0622149
0.2
0.0622149
8 Uarho
5.
5.
5.
5.
9 Ta
323.
323.
323.
323.
10 P0
1.013E+06 1.013E+06
1.013E+06
1.013E+06
11 CA0
304.6819
304.6819
304.6819
304.6819
12-140
12 CA
304.6819
292.8948
304.6819
292.8948
13 CC
0
0
17.40323
17.40323
14 CB
0
0
17.40323
17.40323
15 rA
-40.52269
-40.52269
-0.2446624
-0.2446624
Differential equations
1 d(X)/d(W) = -rA / v0 / CA0
2 d(T)/d(W) = (Uarho * (Ta - T) + rA * 20000) / v0 / CA0 / 40
Explicit equations
1 T0 = 400
2 k = 0.133 * exp(31400 / 8.314 * (1 / T0 - 1 / T))
3 v0 = 20
4 kr = 0.2 * exp(51400 / 8.314 * (1 / T0 - 1 / T))
5 Uarho = 5
6 Ta = 323
7 P0 = 1013250
8 CA0 = P0 / 8.314 / T0
9 CA = CA0 * (1 - X) / (1 + X) * T0 / T
10 CC = CA0 * X / (1 + X) * T0 / T
11 CB = CA0 * X / (1 + X) * T0 / T
12 rA = -(k * CA - kr * CB * CC)
12-141
P12-21
First note that ΔCP = 0 for both reactions. This means that ΔHRx(T) = ΔHRx° for both reactions.
Now start with the differential energy balance for a PFR:
rij ( H Rxij ) Ua (Ta T ) r1 A ( H 1 A ) r2 B ( H Rx 2 B )
dT Ua (Ta T )
dV
F j C Pj
F j C Pj
If we evaluate this differential equation at its maximum we get
dT
0 and therefore, Ua (Ta T ) r1 A ( H Rx1C ) r2 B ( H Rx 2 B )
dV
We can then solve for r1A from this information.
Ua(Ta T ) r2 B ( H Rx 2 B )
r1 A
( H Rx1 A )
Ua(Ta T ) 2k 2 D C B CC ( H Rx 2 B )
r1 A
( H Rx1 A )
10(325 500) 2(0.4)(0.2)(0.5)(5000)
r1A
0.043
50000
1
1
r1 A
0.043
2 k1C C A C B
2 k1C (0.1)( 0.2)
k1C
4 .3
k1C (500 )
4.3
E 1
R 400
1
500
E
1
1.987 400
1
500
k1C ( 400 ) exp
0.043 exp
cal
mol
Alternate Solution:
E 18300
12-142
0
E = 76575 J /mol
P12-22
T(K)
CC
800
0.92
700
1.06
600
1.0755
500
0.78
650
1.1025
625
1.099
675
1.088
Ans: In the range between 300K to 600K, answer is
600K
600K
12-143
P12-23
Mole balance:
dFA
rA
dW
Rate Laws:
rA
r2 B
dFB
dW
rB
dFC
dW
rC
r1 A r3 A
rB
r1 A r2 B
rC
r3 A
r1 A
k1C A
r2 B
k B CB
r3 A k3CC
Stoichiometry:
FT
C A CT A 0
FT T
FBT0
FT T
Energy balance:
dT Ua(Ta T ) ( r1 A )( H R1 A ) ( rR 2 B ) ( r3 A )(
dW
FAC pA FBC pB FC C pC
CB
dT
dW
CT
H R3 A )
16(500 T ) ( r1 A )1800 ( rR 2 B )1800 ( r3 A )1100
100( FA FB FC )
k1
0.5exp 2(1 320 / T )
k2
k1
KC
12-144
k3
0.005exp 4.6(1 460 / T )
KC
10 exp 4.8(430 / T 1.5)
See Polymath program P12-23.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 w
0
0
100.
100.
2 fb
1.
0.9261241
1.368476
0.9261241
3 fa
1.
0.6296429
1.
0.8694625
4 fc
0
0
0.2044134
0.2044134
5 T
330.
330.
416.3069
416.3069
6 Ua
16.
16.
16.
16.
7 Ta
500.
500.
500.
500.
8 Dhr1a
-1800.
-1800.
-1800.
-1800.
9 Dhr3a
-1100.
-1100.
-1100.
-1100.
10 cpa
100.
100.
100.
100.
11 cpb
100.
100.
100.
100.
12 cpc
100.
100.
100.
100.
13 k1
0.5312401 0.5312401
0.7941566
0.7941566
14 k3
0.0008165 0.0008165
0.0030853
0.0030853
15 ct
2.
2.
2.
2.
16 ft
2.
2.
2.
2.
17 To
330.
330.
330.
330.
18 Kc
3.885029
1.062332
3.885029
1.062332
19 k2
0.1367403 0.1367403
0.74756
0.74756
20 ca
1.
0.5682599
1.
0.6892094
21 cb
1.
0.7341242
1.253213
0.7341242
22 r1a
-0.5312401 -0.5748799
-0.362406
-0.5473402
23 r3a
-0.0008165 -0.002138
-0.0007594
-0.0021264
24 rc
0.0008165 0.0007594
0.002138
0.0021264
25 r2b
-0.1367403 -0.5770243
-0.1367403
-0.5488019
12-145
26 rb
0.3944998 -0.0522707
0.3944998
-0.0014617
27 ra
-0.3953164 -0.3953164
0.0510521
-0.0006647
Differential equations
1 d(fb)/d(w) = rb
2 d(fa)/d(w) = ra
3 d(fc)/d(w) = rc
4
d(T)/d(w) = (Ua * (Ta - T) + (-r1a) * (-Dhr1a) + (-r2b) * (Dhr1a) * (-r3a) * (-Dhr3a)) / (fa * cpa +
fb * cpb + fc * cpc)
Explicit equations
1 Ua = 16
2 Ta = 500
3 Dhr1a = -1800
4 Dhr3a = -1100
5 cpa = 100
6 cpb = 100
7 cpc = 100
8 k1 = .5 * exp(2 * (1 - 320 / T))
9 k3 = .005 * exp(4.6 * (1 - (460 / T)))
10 ct = 2
11 ft = 2
12 To = 330
13 Kc = 10 * exp(4.8 * (430 / T - 1.5))
14 k2 = k1 / Kc
15 ca = ct * fa / ft * To / T
16 cb = ct * fb / ft * To / T
17 r1a = -k1 * ca
18 r3a = -k3 * ca
19 rc = -r3a
20 r2b = -k2 * cb
12-146
21 rb = -r1a + r2b
22 ra = -r2b + r1a + r3a
P12-23 (a) As seen in the above table, the lowest concentration of o-xylene (A) = .568 mol/dm3
P12-23 (b) The maximum concentration of m-xylene (B) = 1.253 mol/dm3
P12-23 (c) The maximum concentration of o-xylene = 1 mol/dm3
P12-23 (d) The same equations are used except that FB0 = 0.
The lowest concentration of o-xylene = 0.638 mol/dm3. The highest concentration of
m-xylene = 1.09 mol/dm3. The maximum concentration of o-xylene = 2 mol/dm3.
P12-23 (e)
Decreasing the heat of reaction of reaction 1 slightly decreases the amount of E formed.
Decreasing the heat of reaction of reaction 3 causes more of C to be formed. Increasing the
feed temperature causes less of A to react and increases formation of C. Increasing the ambient
temperature causes a lot of C to be formed.
P12-23 (f) Individualized solution
P12-24 (a)
A
B C
A D E
A C
F G
We want the exiting flow rates B, D and F
Start with the mole balance in PFR:
dFC
dFA
dFB
rA
rB
dV
dV
dV
dFE
dFF
rE
rF
dV
dV
Rate Laws:
rA
r1s
rB
r1s
rC
r1s r3T
rD
r2 B
rE
r2 B
rF
r3T
r2 B
r3T
12-147
rC
dFD
dV
dFG
dV
rG
rD
r1s
(1
r2 B
(1
r3T
(1
10925
PA
T
25000
) exp 13.2392
PA
T
11000
) exp 0.2961
PA PC
T
) exp
0.08359
FB
FD
PB PC
K
Stoichiometry:
PA
FA
PT 0
FT
PB
FB
PT 0
FT
PC
FC
PT 0
FT
FT
FA
FI
steamratio .0034
FC
FE
FF
FG
FI
Energy Balance:
dT
dV
r1s H R1 A r2 B H R 2 A r3T H R 3 A
FA *299 FB *283 FC *30 FD *201 FE *90 FF *249 FG *68 FI *40
K p1
exp b1
b2
T
b3 ln(T )
(b4T b5 )T b6 T
See Polymath program P12-24.pol.
For T0 = 800K
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 v
0
0
10.
10.
2 fa
0.00344
0.002496
0.00344
0.002496
3 fb
0
0
0.0008974
0.0008974
4 fc
0
0
0.0008615
0.0008615
5 fd
0
0
1.078E-05
1.078E-05
6 fe
0
0
1.078E-05
1.078E-05
7 ff
0
0
3.588E-05
3.588E-05
12-148
8 fg
0
0
3.588E-05
3.588E-05
9 T
800.
765.237
800.
765.237
10 Hla
1.18E+05
1.18E+05
1.18E+05
1.18E+05
11 H2a
1.052E+05 1.052E+05
1.052E+05
1.052E+05
12 H3a
-5.39E+04
-5.39E+04
-5.39E+04
-5.39E+04
13 p
2137.
2137.
2137.
2137.
14 phi
0.4
0.4
0.4
0.4
15 Kl
0.0459123 0.0196554
0.0459123
0.0196554
16 sr
14.5
14.5
14.5
14.5
17 fi
0.04988
0.04988
0.04988
0.04988
18 ft
0.05332
0.05332
0.0542282
0.0542282
19 Pa
0.1548387 0.1104652
0.1548387
0.1104652
20 Pb
0
0
0.0397155
0.0397155
21 Pc
0
0
0.0381277
0.0381277
22 r2b
2.991E-06
5.16E-07
2.991E-06
5.16E-07
23 rd
2.991E-06
5.16E-07
2.991E-06
5.16E-07
24 re
2.991E-06
5.16E-07
2.991E-06
5.16E-07
25 r3t
0
0
4.196E-06
4.151E-06
26 rf
0
0
4.196E-06
4.151E-06
27 rg
0
0
4.196E-06
4.151E-06
28 rls
0.0002138 2.481E-05
0.0002138
2.481E-05
29 rb
0.0002138 2.481E-05
0.0002138
2.481E-05
30 rc
0.0002138 2.066E-05
0.0002138
2.066E-05
31 ra
-0.0002167 -0.0002167
-2.948E-05
-2.948E-05
Differential equations
1 d(fa)/d(v) = ra
2 d(fb)/d(v) = rb
3 d(fc)/d(v) = rc
4 d(fd)/d(v) = rd
5 d(fe)/d(v) = re
12-149
6 d(ff)/d(v) = rf
7 d(fg)/d(v) = rg
8
d(T)/d(v) = -(rls * Hla + r2b * H2a + r3t * H3a) / (fa * 299 + fb * 273 + fc * 30 + fd * 201 + fe *
90 + ff * 68 + fi * 40)
Explicit equations
1 Hla = 118000
2 H2a = 105200
3 H3a = -53900
4 p = 2137
5 phi = .4
6
Kl = exp(-17.34 - 1.302e4 / T + 5.051 * ln(T) + ((-2.314e-10 * T + 1.302e-6) * T + -0.004931) *
T)
7 sr = 14.5
8 fi = sr * .00344
9 ft = fa + fb + fc + fd + fe + ff + fg + fi
10 Pa = fa / ft * 2.4
11 Pb = fb / ft * 2.4
12 Pc = fc / ft * 2.4
13 r2b = p * (1 - phi) * exp(13.2392 - 25000 / T) * Pa
14 rd = r2b
15 re = r2b
16 r3t = p * (1 - phi) * exp(.2961 - 11000 / T) * Pa * Pc
17 rf = r3t
18 rg = r3t
19 rls = p * (1 - phi) * exp(-0.08539 - 10925 / T) * (Pa - Pb * Pc / Kl)
20 rb = rls
21 rc = rls - r3t
22 ra = -rls - r2b - r3t
Fstyrene = 0.0008974
Fbenzene = 1.078E-05
12-150
Ftoluene = 3.588E-05
SS/BT = 19.2
P12-24 (b)
T0 = 930K
Fstyrene = 0.0019349
Fbenzene = 0.0002164
Ftoluene = 0.0002034
SS/BT = 4.6
P12-24 (c)
T0 = 1100 K
Fstyrene = 0.0016543
Fbenzene = 0.0016067
Ftoluene = 0.0001275
SS/BT = 0.95
P12-24 (d)
Plotting the production of styrene as a function of To gives the following graph. The
temperature that is ideal is 995K
P12-24 (e)
Plotting the production of styrene as a function of the steam gives the following graph and the
ratio that is the ideal is 25:1
12-151
P12-24 (f)
When we add a heat exchanger to the reactor, the energy balance becomes:
Ua Ta T
r1s H R1 A r2 B H R 2 A r3T H R 3 A
dT
dV FA *299 FB *283 FC *30 FD *201 FE *90 FF *249 FG *68 FI *40
With Ta = 1000 K
Ua = 100 kJ/min/K = 1.67 kJ/s/K
The recommended entering temperature would be T0 = 440 K. This gives the highest outlet flow
rate of styrene.
P12-24 (g) Individualized solution
P12-24 (h) Individualized solution
P12-25
Let
a=A
b = A2
c = A4
PART 1
POLYMATH Report
Ordinary Differential Equations
21-Jun-2010
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
12-152
1 Ca
2.
0.155599
2.
0.155599
2 Cb
0
0
0.4963834
0.1995088
3 Cpa
25.
25.
25.
25.
4 Cpb
50.
50.
50.
50.
5 Cpc
100.
100.
100.
100.
6 DH1a
-3.25E+04
-3.25E+04
-3.25E+04
-3.25E+04
7 DH2b
-2.75E+04
-2.75E+04
-2.75E+04
-2.75E+04
8 E1
4000.
4000.
4000.
4000.
9 E2
5000.
5000.
5000.
5000.
10 Fa
100.
7.779951
100.
7.779951
11 Fb
0
0
24.81917
9.975439
12 Fc
0
0
18.06729
18.06729
13 k1
0.6
0.6
0.6
0.6
14 k1a
0.6
0.6
76.97387
6.133163
15 k2
0.35
0.35
0.35
0.35
16 k2b
0.2072016 0.2072016
89.46089
3.787125
17 Qg
7.8E+04
8971.335
1.569E+06
8971.335
18 Qr
-1.5E+04
-1.5E+04
7.696E+05
1.44E+05
19 r1a
-2.4
-42.01417
-0.1484903
-0.1484903
20 r1b
1.2
0.0742452
21.00708
0.0742452
21 r2b
0
-18.03538
0
-0.1507418
22 ra
-2.4
-42.01417
-0.1484903
-0.1484903
23 rb
1.2
-8.261933
17.21274
-0.0764966
24 rc
0
0
9.017691
0.0753709
25 sumCp
2500.
2500.
2500.
2500.
26 T
300.
300.
1084.641
459.0062
27 T1
300.
300.
300.
300.
28 T2
320.
320.
320.
320.
29 Ta
315.
315.
315.
315.
30 Ua
1000.
1000.
1000.
1000.
31 V
0
0
10.
10.
32 vo
50.
50.
50.
50.
Differential equations
12-153
1 d(T)/d(V) = (Qg-Qr)/sumCp
2 d(Fc)/d(V) = rc
3 d(Fb)/d(V) = rb
4 d(Fa)/d(V) = ra
Explicit equations
1 Cpc = 100
2 Cpb = 50
3 Cpa = 25
4 k2 = .35
5 Ta = 315
6 Ua = 1000
7 DH2b = -27500
8 DH1a = - 32500
9 T1 = 300
10 E1 = 4000
11 k1 = 0.6
12 T2 = 320
13 E2 = 5000
14 k2b = k2*exp((E2/1.987)*(1/T2-1/T))
15 vo = 50
16 Cb = Fb/vo
17 Ca = Fa/vo
18 sumCp = (Fa*Cpa+Fb*Cpb+Fc*Cpc)
19 k1a = k1*exp((E1/1.987)*(1/T1-1/T))
20 r2b = -k2b*Cb^2
21 rc = -r2b/2
22 r1a = -k1a*Ca^2
23 r1b = -r1a/2
24 rb = r1b + r2b
25 ra = r1a
26 Qg = r1a*DH1a+r2b*DH2b
27 Qr = Ua*(T-Ta)
12-154
(b) The required reactor volume = 3.5 dm3
(c)
12-155
Part 2
12-156
12-157
P12-26
Let A = a
A3 = b,
A6 = c
k1 = k1A and k2 = k2A3
Thus the values of quantities exiting the reactor are as follows:
FA = 23.15 mol/sec
FA3 = 11.906 mol/sec
FA6 = 6.85 mol/sec
T = 589.73 K
POLYMATH Report
Nonlinear Equations
21-Jun-2010
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
12-158
1 Fa
23.15856 -4.263E-14 30.
2 Fb
11.90629 1.243E-14 30.
3 Fc
6.85376 8.882E-16 2.
4T
589.7334 -2.328E-10 1000.
Variable Value
1 Ca
0.5620827
2 Cb
0.2889783
3 Cpa
25.
4 Cpb
75.
5 Cpc
150.
6 Cto
2.
7 DH1a
-8.0E+04
8 DH2b
-1.0E+05
9 E1
4000.
10 E2
5000.
11 Fao
100.
12 Ft
41.91861
13 k1
0.9
14 k1a
24.32177
15 k2
0.45
12-159
16 k2b
16.41453
17 Qg
7.518E+05
18 Qr
7.518E+05
19 r1a
-7.684144
20 r1b
2.561381
21 r2b
-1.370752
22 ra
-7.684144
23 rb
1.190629
24 rc
0.685376
25 T1
300.
26 T2
320.
27 Ta
315.
28 To
300.
29 UA
100.
30 V
10.
Nonlinear equations
1 f(T) = (Qr - Qg) = 0
2 f(Fc) = 0 -Fc+rc*V = 0
3 f(Fb) = 0 - Fb + rb*V = 0
4 f(Fa) = Fao-Fa+ra*V = 0
12-160
Explicit equations
1 Cpc = 150
2 Cpb = 75
3 Cpa = 25
4 k2 = 0.45
5 Ta = 315
6 UA = 100
7 DH1a = -80000
8 DH2b = -100000
9 T1 = 300
10 E1 = 4000
11 k1 = 0.9
12 T2 = 320
13 E2 = 5000
14 k2b = k2*exp((E2/1.987)*(1/T2-1/T))
15 Cto = 2
16 Ft = Fa + Fb + Fc
17 To = 300
18 Cb = Fb/Ft*Cto*To/T
19 Ca = Fa/Ft*Cto*To/T
12-161
20 k1a = k1*exp((E1/1.987)*(1/T1-1/T))
21 r2b = - k2b*Cb^2
22 rc = -r2b/2
23 r1a = -k1a*Ca^2
24 r1b = -r1a/3
25 rb = r1b + r2b
26 ra = r1a
27 Qg = r1a*DH1a+r2b*DH2b
28 Fao = 100
29 Qr = UA*(T-Ta)+Fao*Cpa*(T-To)
30 V = 10
12-162
CD12GA1
12-163
12-164
12-165
12-166
See Polymath program CD12GA-1.pol.
12-167
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V
0
0
1.0E+08
1.0E+08
X
0
0
0.6495463 0.6495463
Fao
1200
1200
1200
1200
T
550
225.22683 550
225.22683
Cao
0.064
0.064
0.064
0.064
To
550
550
550
550
k
1.949E+05 5.047E-05 1.949E+05 5.047E-05
Kc
1.01E+06
1841.4832 1.01E+06 1841.4832
f
1
1
2.4419826 2.4419826
Ca
0.064
0.0547713 0.064
0.0547713
Cb
0.064
0.0547713 0.064
0.0547713
Cc
0
0
0.2030311 0.2030311
ra
-798.17344 -798.17344 -1.503E-07 -1.503E-07
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] Fao = 1200
[2] T = -500*X+550
[3] Cao = .064
[4] To = 550
[5] k = .035*exp(8419.5*(1/273-1/T))
[6] Kc = 25000*exp(2405.6*(1/298-1/T))
[7] f = To/T
[8] Ca = Cao*(1-X)*f
[9] Cb = Cao*(1-X)*f
[10] Cc = 2*Cao*X*f
[11] ra = -k*(Ca*Cb-Cc^2/Kc)
12-168
CD12GA2
12-169
See Polymath program CD12GA-2.pol.
POLYMATH Results
NLES Solution
Variable Value
f(x)
Ini Guess
Ca
0.0016782 -1.472E-16 0.0017
Cb
0.0016782 -1.472E-16 0.0017
Cd
0.0071436 -1.154E-16 0.0072
Cu
0.0011782 -2.017E-17 0.0012
V
3794.94 1.018E-09 3794
Cao
0.01
Cbo
0.01
vo
6000
k1
6.73
k2
1.11
tau
0.63249
rd
0.0112944
ru
0.0018628
ra
-0.0131572
rb
-0.0131572
NLES Report (safenewt)
Nonlinear equations
[1] f(Ca) = tau*ra+Cao-Ca = 0
[2] f(Cb) = tau*rb+Cbo-Cb = 0
[3] f(Cd) = Cd-tau*rd = 0
[4] f(Cu) = Cu-tau*ru = 0
[5] f(V) = 171000/(20190*Ca+6660*Cb)-V = 0
Explicit equations
[1] Cao = .01
[2] Cbo = .01
[3] vo = 6000
[4] k1 = 6.73
12-170
[5] k2 = 1.11
[6] tau = V/vo
[7] rd = k1*Ca
[8] ru = k2*Cb
[9] ra = -k1*Ca-k2*Cb
[10] rb = ra
CD12GA 2 (a)
CD12GA 2 (b)
CD12GA 2 (c) Individualized solution
______________________________________________________________________________
12-171
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13-1
Solutions for Chapter 13 – Unsteady State Non-isothermal
Reactor Design
P13-1 Individualized solution
P13-2 (a) Example 13-1
(1)
If the heat of mixing had been neglected, the shape of the graphs would have been as follows:
(2)
The new T0 of 20 ˚F (497 ˚R) gives a new
HRn and T. With T=497+89.8X the polymath program of
example 9-1 gives t= 8920 s for 90 % conversion.
13-2
(3)
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 t
0
0
9000.
9000.
2 X
0
0
1.
1.
3 T
480.
480.
894.193
894.193
4 k
8.307E-06
8.307E-06
56.65781
56.65781
5 Ta
896.4
896.4
896.4
896.4
6 Ua
0.22
0.22
0.22
0.22
7 Qr
91.608
0.4855311
91.608
0.4855311
8 V
1.2
1.2
1.2
1.2
9 Nao
1.
1.
1.
1.
10 NiCpi
403.
403.
403.
403.
11 Ca
0.8333333 -5.468E-07
0.8333333
-1.392E-09
12 ra
-6.923E-06 -0.0038571
8.639E-06
7.887E-08
13 Qg
-0.3016323 -168.055
0.3763917
0.0034364
Differential equations
1 d(X)/d(t) = k * (1 - X)
2 d(T)/d(t) = (Qr - (-36309) * (-ra) * V) / NiCpi
Explicit equations
1 k = 0.000273 * exp(16306 * ((1 / 535) - (1 / T)))
2 Ta = 896.4
3 Ua = .22
4 Qr = Ua * (Ta - T)
5 V = 1.2
6 Nao = 1
7 NiCpi = 403
8 Ca = (Nao/V) * (1 - X)
9 ra = -k * Ca
10 Qg = (-36309) * (-ra) * V
13-3
P13-2 (b) Example 13-2
To show that no explosion occurred without cooling failure.
13-4
P13-2 (c) Example 13-3
Decreasing the coolant rate to 10 kg/s gives a weak cooling effect and the maximum temperature in the
reactor becomes 315 K. An increase of the coolant rate to 1000 kg/s gives a Tmax of 312 K. A big change
to the coolant rate has, in this case, only a small effect on the temperature, and because the
temperature does not change significantly the conversion will be kept about the same.
P13-2 (d) Example 13-4
13-5
13-6
P13-2 (e) Example 13-5
Using the code from Example 9-5, we could produce the following graphs either by changing TO and
finding the steady state conversion or changing the coolant flow rate and finding the steady state
conversion and temperature. These are the graphs of those:
13-7
P13-2 (f) Example 13-6
(1) Individualised solution
(2) The transition to runaway occurs over a very small range of Ua . If U a value is changed
such that Ua is taken from 2.1 ×104 to 1.9 ×104 J/hr/K the runaway occurs. Thus it occurred
over a very narrow range of Ua values .
(3) The modelling is done and the changes are incorporated in the polymath code for Example
13.6
A new switch type variable is introduced .
13-8
Sw2 such that Sw2 = 1
if 300<T<422
else
0
and a new term is added in the energy equation such that the equation becomes :
d(T)/d(t) = SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)-SW1*UA*(T –
373.15))/SumNCp) + SW1*Sw2*4/60
Case 1: When Ua = 2.77 × 106 J/hr/K
But the cooling starts only for T>455K
Thus Ua = 2.77 × 106 J/hr/K , T> 455K
= 0 for T< 455K
See polymath Program for part 3
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
A1A
4.0E+14
4.0E+14
4.0E+14
4.0E+14
2
A2S
1.0E+84
1.0E+84
1.0E+84
1.0E+84
3
CA
4.3
0.0457129
4.3
0.0457129
4
CB
5.1
0.8457129
5.1
0.8457129
5
CS
3.
2.999997
3.
2.999997
6
Cv1
3360.
3360.
3360.
3360.
7
Cv2
5.36E+04
5.36E+04
5.36E+04
5.36E+04
8
DHRx1A -4.54E+04
-4.54E+04
-4.54E+04
-4.54E+04
9
DHRx2S
-3.2E+05
-3.2E+05
-3.2E+05
-3.2E+05
10 E1A
1.28E+05
1.28E+05
1.28E+05
1.28E+05
11 E2S
8.0E+05
8.0E+05
8.0E+05
8.0E+05
12 FD
2467.445
61.27745
8874.034
61.27745
13 Fvent
2467.445
61.27745
8874.034
61.27745
14 k1A
0.0562573
0.0562573
1.82787
0.7925113
15 k2S
8.428E-16
8.428E-16
2.367E-06
1.276E-08
16 P
4.4
4.4
4.4
4.4
17 r1A
-1.233723
-4.437017
-0.0306385
-0.0306385
18 r2S
-2.529E-15
-7.102E-06
-2.529E-15
-3.829E-08
19 SumNCp 1.26E+07
1.26E+07
1.26E+07
1.26E+07
20 SW1
1.
1.
1.
1.
21 Sw2
0
0
0
0
13-9
22 T
422.
422.
466.4882
454.9731
23 t
0
0
4.
4.
24 UA
0
0
2.77E+06
0
25 V0
4000.
4000.
4000.
4000.
26 VH
5000.
5000.
5000.
5000.
Differential equations
1 d(CA)/d(t) = SW1*r1A
mol/dm3/hr
2 d(CB)/d(t) = SW1*r1A
change in concentration of cyclomethylpentadiene
3 d(CS)/d(t) = SW1*r2S
change in concentration of diglyme
4 d(P)/d(t) = SW1*((FD-Fvent)*0.082*T/VH)
5 d(T)/d(t) = SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)-SW1*UA*(T-373.15))/SumNCp) + SW1*Sw2*4/60
Explicit equations
1 V0 = 4000
dm3
2 VH = 5000
dm3
3 DHRx1A = -45400
J/mol Na
4 DHRx2S = -3.2E5
J/mol of Diglyme
5 SumNCp = 1.26E7
J/K
6 A1A = 4E14
per hour
7 E1A = 128000
J/kmol/K
8 k1A = A1A*exp(-E1A/(8.31*T))
rate constant reaction 1
9 A2S = 1E84
per hour
10 E2S = 800000
13-10
J/kmol/K
11 k2S = A2S*exp(-E2S/(8.31*T))
rate constant reaction 2
12 SW1 = if (T>600 or P>45) then (0) else (1)
13 r1A = -k1A*CA*CB
mol/dm3/hour (first order in sodium and cyclomethylpentadiene)
14 r2S = -k2S*CS
mol/dm3/hour (first order in diglyme)
15 FD = (-0.5*r1A-3*r2S)*V0
16 Cv2 = 53600
17 Cv1 = 3360
18 Fvent = if (FD<11400) then (FD) else(if (P<28.2) then ((P-1)*Cv1) else ( (P-1)*(Cv1 +Cv2)))
19 UA = if T> 455 then (2.77e6) else (0)
no cooling
20 Sw2 = if ((T>300) and (T<422)) then (1) else (0)
13-11
(4) Solving the equations with Ua = 0 for the entirety of the process, we get that the temperature
reaches at 0.94 hours while the system reaches runaway at 3.649 hours. This implies the
maximum time in minutes that the cooling can be lost is 163 minutes.
13-12
P13-2 (g) Example RE13-1
P13-2 (h) Example RE13-2
13-13
P13-2 (i) No solution will be given
P13-3
After the feed was shut off,
See Polymath program P13-3.pol
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cpa
0.38
0.38
0.38
0.38
2 dH
-336.
-336.
-336.
-336.
3 k
0.0081711
0.0081711
4.285E+17
4.285E+17
4 T
980.
980.
3.302E+20
3.302E+20
5 t
0
0
4.
4.
Differential equations
1 d(T)/d(t) = -dH * k / Cpa
Explicit equations
1 dH = -336
2 Cpa = .38
3 k = (0.307/60)* exp(44498 * (1 / 970 - 1 / T))
13-14
As can be seen from the above plots, temperature shoots at time t = 3.16 mins
Since NA doesn’t come in the temperature equation, there will not be any effect of feed present in the
reactor.
0
If T0 = 100 F, Temperature will remain constant for next 4 minutes, which means no explosion.
0
If T0 = 500 F, there will not be any explosion and the temperature profile will look like
13-15
P13-4 (a)
See Polymath program P13-4-a.pol
13-16
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Cbo
10.
10.
10.
10.
2
Fbo
10.
0
10.
0
3
k
0.0089352
0.0089352
10.08511
10.08511
4
Na
500.
0.1396707
500.
0.1396707
5
Nao
500.
500.
500.
500.
6
Nb
0
0
42.43357
0.1397745
7
Nc
0
0
499.8603
499.8603
8
Qg
0
0
1.24E+05
11.81314
9
Qr
0
0
0
0
10 ra
0
-0.3460808
0
-1.969E-05
11 T
298.
298.
510.4441
510.4441
12 t
0
0
120.
120.
13 V
50.
50.
100.
100.
14 vo
1.
0
1.
0
15 X
0
0
0.9997207
0.9997207
Differential equations
1 d(Na)/d(t) = ra*V
2 d(Nb)/d(t) = ra*V+Fbo
3 d(Nc)/d(t) = -ra*V
4 d(T)/d(t) = ((6000*(-ra*V))-(Fbo*15*(T-323)))/(15*Na+15*Nb+30*Nc)
5 d(X)/d(t) = -ra*V/Nao
Explicit equations
1 Fbo = if(t<50)then(10)else(0)
2 Nao = 500
3 Cbo = 10
4 k = .01*exp((10000/1.987)*(1/300-1/T))
5 vo = Fbo/Cbo
6 V = if(t<50)then(50+(vo*t))else(100)
7 ra = -k*Na*Nb/(V^2)
8 Qg = -6000*ra*V
9 Qr = 0
13-17
P13-4 (b)
This is the same as part (a) except the energy balance.
Energy balance:
13-18
See Polymath program P13-4-b.pol
P13-4 (c)
This is the same as part (b) except the reaction is now reversible.
13-19
See Polymath program P13-4-c.pol
13-20
P13-5 (a)
P13-5 (b)
13-21
P13-6
13-22
See Polymath program P13-6.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
X
T
k1
Ca
Cb
k2
Cc
ra
initial value
0
0
373
0.002
0.1
0.125
3.0E-05
0
-2.236E-04
minimal value
0
0
373
0.002
0.0749517
0.0999517
3.0E-05
0
-0.1344598
maximal value
10
0.2504829
562.91803
106.13627
0.1
0.125
366.75159
0.0250483
8.644E-06
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(t) = -ra/.1
[2] d(T)/d(t) = ((40000+(10*(T-298)))*(-ra)*(1/.1))/(56.25-(10*X))
Explicit equations as entered by the user
[1] k1 = .002*exp((100000/8.314)*(1/373-1/T))
[2] Ca = .1*(1-X)
[3] Cb = .1*(1.25-X)
[4] k2 = .00003*exp((150000/8.314)*(1/373-1/T))
[5] Cc = .1*X
[6] ra = -((k1*(Ca^.5)*(Cb^.5))-(k2*Cc))
13-23
final value
10
0.2504829
562.91802
106.13612
0.0749517
0.0999517
366.75083
0.0250483
1.3E-07
P13-7 (a)
Use Polymath to solve the differential equations developed from the unsteady state heat and mass
balances. Refer to P13-7-a.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Cc
0.1
0.1
14.25628
14.25628
2
Cpc
4.2
4.2
4.2
4.2
3
Cps
5.
5.
5.
5.
4
Cs
300.
282.3047
300.
282.3047
5
Hrxn
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
6
Iprime
0.0866522
0.0004366
0.3898631
0.0004366
7
Km
5.
5.
5.
5.
8
mu
0.0426159
0.0002145
0.1916769
0.0002145
9
mu1max 0.5
0.5
0.5
0.5
10 neg_rs
0.005327
0.0038221
1.083901
0.0038221
11 Q
0
0
0
0
13-24
12 rg
0.0042616
0.0030577
0.8671209
0.0030577
13 rho
1000.
1000.
1000.
1000.
14 rs
-0.005327
-1.083901
-0.0038221
-0.0038221
15 t
0
0
300.
300.
16 T
278.
278.
334.6251
334.6251
17 V
25.
25.
25.
25.
18 Ycs
0.8
0.8
0.8
0.8
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg
[2] d(Cs)/d(t) = -rg/Ycs
[3] d(T)/d(t) = (Q+(-Hrxn)*(rg))/(rho*Cps)
Explicit equations as entered by the user
[1] Iprime = (0.0038*T*exp(21.6-6700/T))/(1+exp(153-48000/T))
[2] Km = 5.0
[3] mu1max = 0.5
[4] Ycs = 0.8
[5] mu = mu1max*Iprime*(Cs/(Km+Cs))
[6] Q = 0
[7] Cps = 5
[8] Hrxn = -20000
[9] V = 25
[10] rho = 1000
[11] rg = mu*Cc
[12] Cpc = 4.2
[13] rs = -rg/Ycs
13-25
P13-7 (b)
When we change the initial temperature we find that the outlet concentration of species C has a
maximum at T0 = 300 K. Refer to P13-7-b.pol
P13-7 (c)
13-26
Cc can be maximized with respect to T0 (inlet temp), Ta (coolant/heating temperature), and heat
exchanger area. Therefore, if we are to find the optimal heat exchanger area the inlet temperature and
coolant/heating temperature needs to be specified. If we take T0 = 310 and Ta = 290 we find that the
optimal heat exchanger area is 0.46 m2.The concentration of cells at the end of 24 hrs under these
conditions is 21.55 g/dm3 . Refer to P13-7-c.pol.
The plot of Cc Vs A:
P13-8 (a)
13-27
See Polymath program P13-8-a.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Na
Nb
Nc
X
T
vb
k
V
Fbo
Ca
Cb
Cc
ra
initial value
0
50
0
0
0
300
1.5
5.0E-04
10
6
5
0
0
0
minimal value
0
0.3324469
0
0
0
283.60209
1.5
2.3E-04
10
6
2.202E-04
0
0
-0.0028611
maximal value
1000
50
5900.6649
49.667553
0.9933511
300
1.5
5.0E-04
1510
6
5
3.9077251
0.0718765
0
final value
1000
0.3324469
5900.6649
49.667553
0.9933511
298.76383
1.5
4.73E-04
1510
6
2.202E-04
3.9077251
0.0328924
-1.59E-06
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Na)/d(t) = ra*V
[2] d(Nb)/d(t) = 2*ra*V+Fbo
[3] d(Nc)/d(t) = -ra*V
[4] d(X)/d(t) = -ra*V/50
[5] d(T)/d(t) = ((250*(290-T))-(80*vb*(T-325))+(-55000*(-ra*V)))/(35*Na+20*Nb+75*Nc)
13-28
Explicit equations as entered by the user
[1] vb = 1.5
[2] k = .0005*exp((8000/1.987)*(1/300-1/T))
[3] V = 10+(vb*t)
[4] Fbo = 4*vb
[5] Ca = Na/V
[6] Cb = Nb/V
[7] Cc = Nc/V
[8] ra = -k*Ca*Cb^2
13-29
P13-8 (b)
P13-9
13-30
P13-9 (a)
See Polymath program P13-9-a.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cc
Cb
T
UA
V
k1
k2
ra1
rb2
rb1
initial value
0
0.3
0
0
283
0
10
1.1172964
4.081E-09
-0.3351889
0
0.3351889
minimal value
0
1.34E-65
0
-2.02E-44
283
0
10
1.1172964
4.081E-09
-35.016552
-27.963241
-6.345E-60
maximal value
0.2
0.3
0.3
0.2895784
915.5
0
10
2.141E+05
1.041E+06
6.345E-60
2.103E-38
35.016552
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra1
[2] d(Cc)/d(t) = -rb2
13-31
final value
0.2
1.34E-65
0.3
-1.864E-65
915.5
0
10
2.141E+05
1.041E+06
6.345E-60
-3.974E-59
-6.345E-60
[3] d(Cb)/d(t) = rb1+rb2
[4] d(T)/d(t) = ((UA*(330-T))+(55000*(-ra1*V))+(71500*(-rb2*V)))/(200*V*(Ca+Cb+Cc))
Explicit equations as entered by the user
[1] UA = 0
[2] V = 10
[3] k1 = 3.03*exp((9900/1.987)*(1/300-1/T))
[4] k2 = 4.58*exp((27000/1.987)*(1/500-1/T))
[5] ra1 = -k1*Ca
[6] rb2 = -k2*Cb
[7] rb1 = -ra1
P13-9 (b)
13-32
P13-9 (c)
13-33
P13-10
Semi batch with parallel reactions
dNa
dt
dNb
dt
dNd
dt
dNu
dt
r1 r2 V
Fbo
r1 r2 V
r1 V
r2 V
Rate laws:
r1 k1 Ca
r2 k2.Cb
k1 10 exp( 2000 / T)
k2 20 exp( 3000 / T)
Stoichiometry:
Na
V
Nd
Cd
V
V V0 v b t
Ca
Cb
Cu
Fbo
Cbo
Cao 5mol / dm 3
Nb
V
Nu
V
vb
Cbo 1mol / dm 3
13-34
dT
dt
3000( r1 V) [5000 10(T 300)]( r2 V) 30 Fbo (T Tbo)
20Na 30Nb 50Nd 40Nu
See polymath program P13-10.pol
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
5.
0.0402932
5.
0.0402932
2
Cao
5.
5.
5.
5.
3
Cb
1.
0.1879238
1.
0.3260075
4
Cbo
1.
1.
1.
1.
5
Cd
0
0
0.7888994
0.6235848
6
Cu
0
0
0.0504077
0.0504077
7
Fbo
10.
10.
10.
10.
8
k1
0.0204583
0.0204583
0.0464803
0.0460806
9
k2
0.0018507
0.0018507
0.0063377
0.0062561
10 Na
500.
28.20523
500.
28.20523
11 Nb
100.
68.89056
228.2052
228.2052
12 Nd
0
0
436.5094
436.5094
13 Nu
0
0
35.28541
35.28541
14 r1
-0.1022916
-0.1022916
-0.0018567
-0.0018567
15 r2
-0.0018507
-0.0020395
-0.001066
-0.0020395
16 T
323.
323.
372.3484
371.7507
17 t
0
0
60.
60.
18 Tao
373.
373.
373.
373.
19 Tbo
323.
323.
323.
323.
20 V
100.
100.
700.
700.
21 vb
10.
10.
10.
10.
Differential equations
1 d(Na)/d(t) = (r1+r2)*V
2 d(Nb)/d(t) = Fbo + (r1+r2)*V
3 d(Nd)/d(t) = -r1*V
13-35
4 d(Nu)/d(t) = -r2*V
5
d(T)/d(t) = (-30*Fbo*(T-Tbo) + 3000*(-r1*V) + (5000 + 10*(T-300))*(r2*V))/(20*Na+30*Nb+50*Nd+40*Nu)
Explicit equations
1
Cbo = 1
2
Fbo = 10
3
k1 = 10*exp(-2000/T)
4
k2 = 20*exp(-3000/T)
5
vb = Fbo/Cbo
6
V = 100 + vb*t
7
Ca = Na/V
8
Cb = Nb/V
9
r1 = -k1*Ca
10 r2 = -k2*Cb
11 Cao = 5
12 Cd = Nd/V
13 Tao = 373
14 Tbo = 323
15 Cu = Nu/V
13-36
13-37