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
Difficulty
Time (min)
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
P1-1
Alternates
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
rA
rA V
dt
k
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
penicillin
+ [generation rate (moles/time)]penicillin – [ Flowrate Out(moles/time)]
= [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 Cco
t
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
vo Fs
a
a = 3.50E+04 lbmol/hr
= 1.67E+12 ft3 /hr
b = 3.00E+04 lbmol/hr
= 341.23 lbmol/hr
V = 4.0E+13 ft3
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
C
2.0E-10
v0
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
V
4.0E+13
2.0E-10 1.67E+12
341.23 4.0E+13
72 2.134E-08
1.877E-08
1.67E+12
341.23
1.67E+12
341.23
4.0E+13
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
C
2.0E-10
v0
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
2.0E-10 1.67E+12
48 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) (0.36hr 1 )(0.01 * 0.5mol / dm 3 )
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
=> V = 2750 dm3
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
dN rAV
NA0 NA
Constant Volume V=V0
t
CA0 CA
dC A 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, 1
k1
0.02day
k2
0.00004 /(day
k3
0.0004 /(day rabbits )
k4
0.04day
foxes )
1
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
x
500
2.9626929
519.40024
4.2199691
y
200
1.1285722
4099.517
117.62928
k1
0.02
k2
4.0E-05
0.02 4.0E-05
500
0.02
0.02
4.0E-05
4.0E-05
1-17
k3
4.0E-04
k4
0.04
4.0E-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
FA0 FA kC A2
v0C A v0C A
dm3 2molA 3 . s dm3 3
dm3 0.1molA . s dm3
mol s V= dm3 mol (0.03 )(0.1 3 ) 2 mol.s dm
6molA s 0.3molA s
(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 X2 rA
X 1 = 1.32(0.6 – 0.2) = 0.53 m3
For 2nd CSTR, For X3 = 0.65,
FAO rA
2m3 ,
V3 = 0.1 m3
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 X CSTR
FAO dX 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 0
Mole balance:
dX rA
= 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 .4
FA 0 dX rA
FA0 .6 dX rA .4
P2-6 (c) VCSTR = 105 dm3 Mole balance: VCSTR
FA0 X rA
FA 0 X rA
.6 .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 rA
Mole balance: V
X1 X2
Rearranging, we get
X2
0.40 rA
X2
At X2 = 0.64,
V FA 0
X2
0.008
0.40 rA
0.008
X2
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 0.40
dX rA
X2
300
At X2 = 0.908, V = 300 x (area under the curve) => V = 300(0.24) = 72dm3 Conversion = 0.908.
0.40
dX rA
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 0
dX rS
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 V2
FA0 ( X 2 X 1 ) ( rA ) 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 KM CA
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
X
PFR: V 0
dX rA
Until the conversion (X) reaches 0.5, the reaction rate is independent of conversion and the reactor volumes will be identical. 0.5
i.e. VPFR 0
dX 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
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).
13
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:
Combine: rAV
NA
N A0 (1 X )
N A0
dX dt
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
t
X
dt 0
C A0 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 0 0.2 0.1 0.0167 0.4 0.00488 0.7 0.00286 0.9 0.00204
1/(-rA) 5 59.9 204.9 349.65 490.19
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
ln
E R
or
Ae
E / RT1
k2 k1
1 T2
and k2
Ae
E / RT2
, or
k2 k1 T1 T2 T1T2
ln
1 T1
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 (c) k = 1.72X1018 exp(-
A = 1.72X1018
(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
1/ 2 Rate law: -rA = kCC2 H 4 C 02
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) -rA = kCACB2 (2) -rA = kCB (3) -rA = k (4) -rA = kCACB-1
P3-11 (b) (1) H2 + Br2 → 2HBr
(2) H2 + I2 → 2HI
Rate law: -rHBr
1/ 2 k1C H 2 C Br 2 = C HBr k2 C Br2
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] [2] [3] [4]
yao = 1 kc = 0.1 Cao = 0.07174*yao 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 (b) C A
CC
C A0
C A0 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
Rate law:
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
-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 1 X 1 X
CA =
FA v
FAO (1 X ) vO (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
C AO
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( FAOX 2
FAO FAO FBO
B
-X/2)
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
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
Change -FA0X -2FA0X FA0X
4-11
Leaving FA=FA0(1-X) FB=FA0(θB-2X) Fc=FA0X
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 FA0 FB 0
1 3
6atm 1 0.0821 * (443K ) 3
FA0 (1 X ) X) 0 (1
FB
FA0 ( B 2 X ) X) 0 (1
C A0
FA0 X X) 0 (1
1 (1 2 1) 3
2 3
mol dm3
(1 X ) (1 23 X )
C A0
C A0
y A0
0.055
dm3 atm mol K
FA
FC
FA 0 FT 0
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
rA
( y A0 C ) * RT
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
kC A2 0 1 X
rA B
C A0
rA
CB 0 C A0
6.6 1.8
1.8
kmol m3 2
k 1.8
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
rA0 rA0
kC
2 A0
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
rA0
1 561 K
m3 kmol min
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 A
1 N2 3 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) 2 1 1 3 3 y AO C AO
2 3 2 3
0.5
16.4atm
0.5 0.082
CH 2
CNH3
CA
CC
1 3
atm.dm mol.K
C AO 1 X 1 X
= 0.2 mol/dm3
3
500 K 0.2 1 X X 1 3
2 C AO X 3 1 X
0.1mol / dm3
0.2 X X 1 3
2 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
X e2
2
Xe
0.80
2
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
N A0 1 X
CA
NA V
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
C A0
1 2Xe
KC 1 X e 1 2 X e
4 Xe
2
3X e 1 2Xe 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
N A0 1 X
CA
NA V
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
X
T T0
1 , P0
or
V2
NB
N A0
kN A3 0 1 X V03 1
X
B
10V0 , and P 10V
V02 1
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
1
X
1 X
9
1 3X
3 3 2
3 3 2
lb mol ft 3 sec
P4-10 (b) V2
V02 1
0.22
X rA
X
0.152 1
X
0.259 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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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
2
1
o
L
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
0.24
1 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
1 and Dp
2
1
DP1 DP 2
1 P0
P01 P02
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
C A0 C A0 A
ln
tAe te
C A0 A
E RT
E RT
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 0
FAO dX rA
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
X
For PFR,
V
FA0 0
=> X PFR
0.4 0.6
0.67
dX , k kC A0 1 X
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
X
=> kC AO
1 X
1 kC AO
For PFR,
=> X
1
1
X
0
0.4 0.62
2
dX 1 X 1 kC AO
2
1.111
1 X kC AO 1 X
2
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
0
dX (1 X )2
(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
X
0
1 X 1
k
dX
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 X
kC A0
1
1 2
1
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
1
1
E
2.8
E
1.21
R 300
E R
300 320 20
ln
320 2.8 1.21
E 8010 5-15
20
R 320 300 , 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
V
kC A0
1
X 2 X1 1 1 X2
k
1 0.5 1 0.5
X 22
k 0.886
X2 1
X 2 X1 1 X2 1 X2
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
:
For Fluidized CSTR
:
Here,
-rBuʹ = K*PBu
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
2902.2dm3 , Z
KC
k kO
2
V 1 X
Z
z
C BO X
X2 KC
2
FAO kOC AO 2
Z z
CB
X
Z
V
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= ?
V
(1)Mole Balance:
FA0 X rA
rA
(2)Rate Law: (3)Stoichiometry:
k CACB
Liquid
0
Stoichiometric Table
A B C
FA0 FB0
FA0 X
BFA0
FA
FA0 X
0
FB
2FA0 X
FA0 1 X FA0
FC
Concentration
CA CB
FA
FA0 1 X
0
FB
FA0
0
51.2M B
1M
C A0 1 X
0 B 0
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
k k C B0 1.95 10 FA0 X
V
rA
dm3
4
mol s
0C A0 X
0
kC A0 1 X
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
1 dm3
V
3.3 10 3 dm3 s
0
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
dV
FA0
kC A0 1 X 0C A0
Separate variables and Integrate with limits V = 0 then X = 0 and V = V then X = X X 0
ln
dX 1 X
0
1
kV
1 X
k
V
dV
0
k
0
X 1 e
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 0
dX rA
1 1 0
rA
yA
1.0
kC A0 1 X 0.9
V
yA
FA0 0
C A0 0.9
dX rA
FA0 0
PA RT
FA0 1 ln kC A0 1 0.9
dX kC A0 1 X
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
V
1
3219hr
FA0 RT ln10 kP
V
0.72 ft 3
V
nt Dt2 L 4
1 T2
0.0015exp
25000 1 1.104 540
1 720
ft 3 psia lb mol R
720 R
53.6 min
1
19.1
lb mol hr
10.73
3219 hr
1
146.7 psia
5-34
ln10
1
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
1
X
2
0
1 X
2
dX
2 1
ln 1 X
2
2
X
1 X 1 X
(for the integration, refer to Appendix A) from the Ideal Gas assumption.
C AO
y AO CTO
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.
Since yA = 1.0,
rA
kC A0
B
= 0. For a liquid phase reaction CA
X K eq
1 X
For the first reactor, X1
V1
FA0 0
dX rA
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
kC A0VF FA0 2
2
kC A0V1 FA0
1.766
dX 1
1
kC A0V1 FA0
2 0.883
1 1 1 5.8
ln 1
1
1 X K eq 1 1
1 K eq
1 K eq
1
ln 1
1
ln 1
1
1 X K eq 0
1 X2 K eq
1.766
1
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
V2
FA0,2 0
kC A0V2 FA0 1 X 1
FA0 1 X
dX rA
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 1 K eq
0.356
1 0.45
0.766
1.174
Overall conversion for this scheme:
X
X
FA0
FA0,2 1 X 2
FA0
FA0 1 X1 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 0
X
dX rM X
CM 0 v0 dX rM
V 0
v0 0
v0 ln 1 X k
dX k 1 X
CM0, k, and V should be the same for Case 1 and Case 2. Therefore,
kV kV
Case1
v0
Case1
ln 1 X Case1
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
kV
v0 Case1
1 X Case1
v0 Case 2
Case 2
gal 0.37 hr 1 0.37
1468
gal 0.50 hr 1 0.50
1667
2500
X Case1 Case1
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
FA0
FA0 0
rA
1
FA0 0
40 6.14 ft 3
6.86
kC A0 1 X
X
dX rA
X
FA0 0
dX kC A0 1 X
y A0 P RT FA0 RT 1 ln or k ky A0 P 1 X
with C A0
lb mol hr
dX rA
kC A0 1 X X
V
lb hr lb 73 lb mol
2
500
mA MWA X
V
V
50
nt D 2 L 4
V
0.75 12 4
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
0.740 19,500 R 0.00152
E RT1
A k1 exp so k
100
ln
E 1 R T
k1 exp
1 T1
From above we have
FA0 RT 1 ln Vy A0 P 1 X
k 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 T
19500
1
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
1
DP1 DP 2
Ac1 Ac 2
3
0.018kg
2 1
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 X2
1 0.5 1 X2
10 3 2 20atm atm kg 3 0.00316kg
1
1 1 0.00316kg
0.81
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
1
50kg
3/ 2
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 X P0 0 1
FA
CA
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 dX dW
kC A0 1 X 1
kC A0 1 X 1 FA0
Separate Variables dX 1 X
k
1
W
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
1
2
W
2
kC A0 1 X 10 k 300 K
0.1
k 300 K
2000 101
k 400 K
k 300 K exp
y2
W dW
2
W2
99 2
100
2
2
19.8
101 2
dm 6 mol 2 min
E 1 R 300
1 400
19.8 exp
10,000 1 1.987 300
1311.179
dm 6 mol 2 min
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
kC A0
dX A 1 XA 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 1 XA
kC A0
kC A0
y A0
1(1 1 1) 1
XA 1 XA 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 FA0 X 2 kC A0 1 X
1
2
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 2
13.47 0.244 10
2
4.99 10 3 kg
4.99 10 3 100 100 2
0.0802 100 24.95
X
1
2
6.02
0.86
P5-30 Individualized solution
5-60
1
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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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
t
dX kC A0 1 X
C A0 0
1 ln 1 X 52.2 weeks k
1 ln 1 X 0.0022 weeks 1
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
0.0048 weeks
dX kC A0 1 X
26weeks
1
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
0
rA
CO 2
CA
V
0C B0
0
V
CO 2
V 0
CO 2
CB
rA
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
1
C B0
0.8 1.0
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
FT
CTO
FA 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 AC 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
C AO
t
vo t VO
v 1 X C BO O t C AO X VO
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 AC 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
C AO
t
vo t VO
v 1 X C BO O t C AO X VO
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] [2] [3] [4] [5] [6]
k = 0.00009 ra = -k*Ca*Cb vo = 0.05 Vo = 200 V = Vo + vo*t X = 1 - Ca/7.72
CDGA 6-1 (e) CDGA 6-1 (f)
Individualized solution 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 2
C A3
rA3
2
rA3V
dN A3 dt
dC A3 dt
Mole balance for reactor 3 species B:
CB 2 v2 CB 3v2
CB 2 2
CB 3 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 0.05 0.038 0.0306 0. 256 0.0222 0.0195 0.0174
X 0 0.024 0.0388 0.0488 0.0556 0.061 0.0652
CB 0.5 0.4988 0.49806 0.49756 0.49722 0.49695 0.49674
error(%) 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 45.7 106 moles / s 3.96 103 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)(1m ) C A(1m ) 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)( 1n ) C A( 1n ) 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 dP k 1 P0 P or 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 Oz + alkene products
rO z
Rate law:
dCO z dt
k1 k2
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 dt 100 AC S
Vw MWF
kMWS d w where dt 100 AC S
V MWF
d ln ln ln w dt d ln dt ln w
-16.629 2.079
-15.425 2.996
-14.32 3.497
-13.816 3.689
7-16
-13.479 3.871
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
MWF
kMWS 1001.775 AC S
V
AC 10*106 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*109 6 g 10 3 9 m 1.2986*10 20 g gmol
1.775
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:
dN dN F 6 S dt dt
dw w 6k V 100MWF dt 100 MWF
V
10.7
20
weight fraction = 20%
dw 6k 1.775 w 100MWF 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*104 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:-
AB+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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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
C AO exp
'
k1
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
k1
'
1 k1 and
X opt
1 exp
k1
'
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] [2] [3] [4] [5]
tau = 0.5 K1 = 55.2 K2 = 30.2 CHo = 0.016 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
~ S X /T CH CM CX
YMX
8.3mole xylene produced mole toluene produced CD-ROM example 0.0089 0.0029 0.0033 0.5 0.41
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] [2] [3] [4] [5] [6] [7]
Ca0 = 1 k1 = 1 k2 = 100 K1a = 10 K2a = 1.5 t = 100 X = 1-Ca/Ca0
P8-6 (a)
A
X
rX
1/ 2 k1C A
A
B
rB
k 2C A
A
Y
rY
k 3C A
2
1/ 2
k1
mol 0.004 dm 3
k2
0.3 min
k3
dm 3 0.25 mol. min
Sketch SBX, SBY and SB/XY as a function of CA
8-13
1
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
=> V
rA
rX
rB
rY
v0 (C A0 C A* ) rA
(k1C A1/ 2
k 3C A 2 )
k 2C A
v0 (C A0 * 1/ 2
(k1 (C A )
C A* )
k 2C A
*
* 2
k 3 (C A ) )
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 *
CB* 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
0.11 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
A B
U
SDU =
rD rU
and
r1 A 100 exp( 1000 K / T )C AC B
and
r2 A 106 exp( 8000K / T )C ACB
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
B D
and
r2 A
U
10 exp( 8000 K / T )C ACB 109 exp( 10, 000 K / T )CB CD
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 D U1 A
and
r1 A
4280 exp( 12000 K / T )C A
and
r2 D
10,100 exp( 15000 K / T )C D
U 2 and
r3 A
rD
S D / U 1U 2
rU 1
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
rU 2
At T = 300K k1 = 1.18 x 10-14 &
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 1.94 10
At T = 1000K k1 = 0.026 &
14
C A 1.94 10 18 C D 18 C D 6.03 10 15 C A
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
(1)
S D / U1
800 exp 10 exp
8000 / T C A0.5CB
80 exp
300 / T C ACB
exp
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
0
0
100.
100.
10 t
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
Similarly,
k2'
3.3 10 3 s
And, k3'
1
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
1
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
k3'
0.78 10 3 dm3 / mol.s
1 663K
0.42 10 3 s
1
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
CA
0.0069 M C B
k 3 E CC C D
4 k 3 E CC C D 3
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
Ini Guess 3 0 0 0 0
NLES Report (safenewt)
Nonlinear equations [1] [2] [3] [4] [5]
f(Ca) = Cao-Ca+ra*tau = 0 f(Cb) = Cb - rb*tau = 0 f(Cc) = Cc-rc*tau = 0 f(Cd) = Cd-rd*tau = 0 f(Ce) = Ce - re*tau = 0
Explicit equations [1] [2] [3] [4] [5] [6] [7]
0.51M C D
kd = 3 ka = 7 rb = ka*Ca/3 ra = -(ka*Ca+kd/3*Ca*Cc^2) ke = 2 rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc 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: Mole balance:
dFA dV
rA
Rate law:
rA
k1 A C A
rB
1 k1 A C A 3
rC
1 k1 A C A 3
rD
k 2 D C A CC2
rE
k 3E CC C D
Stoichiometry: C A
FT
dy dV
dFC dV
dFB rB dV 1 k 2 D C A CC2 3
CTO
FA y FT
FA
FB
2 k 2 D C A CC2 3
rC
dFD dV
rD
CD
CTO
dFE dV
K 3 E CC C D
4 k 3 E CC C D 3
CC FC
FD
CTO
FC y FT
FD y FT
FE
FT 2 y FTO
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
See Polymath program P8-12-j.pol.
P8-13(a)
8-40
1
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
0.01 1 5 10 100 1000
5 5 5 5 5 5
0.002 0.2 1 2 20 200
YC
SC/E
SD/G
SC/D
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
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
50
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
RE
0.06 0.0093 W
From Polymath program:
and FE o
0
0.000558 mol 2 kg.s
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
AD
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 CS CP
YS / P
1
YP / S
YS / C
238.7 245 5.03 2.14 1 2.18
YS / P
1 P
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 k2
CH 3
CH 3CHO
CHO
CH 3CHO
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
k3
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
k1C AC
2k4CCH3
2
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
kC AC
3
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
rCOCl2 k4
0.5
k1 k5 k3 CO Cl2 k2 k4 k5 Cl2
Cl2
k5 k3
k1 k2 k4
1/ 2
CO Cl2
3 2
k5 Cl2
k5 Cl2
rCOCl2
k11/ 2 k3 k5 CO Cl2 k21/ 2 k4
3 2
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
k P 2 RO 2 k P1 O2
0
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 E.S E.S P E
k1 k2 k3 k4
E.S E S P E 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
rE
k1 E S
S
k4 E P
k2 E S
E S
k5 E P
k3 E S
since E is not consumed: ET
E
k4
S
k3 E S k 4 k5
E P
rE
E
E
0 k3 E S
S
k2
E S
ET
k3
E S
k4
k5
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
E S1
k2
k3
E S1 S2
E S1S2
k4
k5
E S1S2 (1) rE
S1
(2) rE
S1S2
P E
0 k1 E S1
k2 E S
0 k3 E S S2
k3 E S S2
k4 E S1S2
k4 E S1S2
k5 E S1S2
If we add these two rates we get: (3) rE
rE
S1
0 k1 E S
S1S2
k2 E S
k5 E S1S2
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)
k4
0 k1 E S
S
k1 S
rP
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 k1 S k2
k5 k4 P
k3
k5
k1k3 S ET
rP
k1 S k2 k3
k2 k3 1
rP
k5 E P
k4 E P
ET
rP
k3 E S
k3
0 k4 E P
E P
rP
k2 E S
E
k2
P
rP
P
E P
k5
E S rE
E S
k2
E P
rE
k3
k1
E S
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
K M EH
kK M EH
EH 2 EH
E
ET
ET
rP
S
S
K2 H K1 H
EH E
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 =
-
rDPP-4 =
dCa = k1*Ca+ k2*Ca*Cb dt
Sub to: - (Assumed values) at t=0, Ca0 = Cb0 =1 gmol/liter
dCb = k2*Ca*Cb dt
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
V v
=
τ
=
XA =
k1CAo 2 1
XA
M1
XA
k2 CAo 2 1
XA
M2
XA k1CAo 1
XA
M1
XA
k2 C Ao 1
X A M2
XA
k2 C Ao 1
X A M2
XA
XA k1CAo 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
rS
,
rg
CC
CCV
Divide by CCV,
D
CS
max
KmD / (
CS
max
KM
Cs Cs
D)
9-44
Cs Km Cs max
,
Now
rS
DC C
CC
YS / C rg
DYC / S C SO
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 / (
D)
max 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 CC
rS , C CO
rS YC / S C SO
rg
YS / C rC , CS
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
1hr 1 1 e
20 g / dm3 / 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
CS
0.8hr
1
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
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
rg
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
0.8hr
1
0.57hr
1
4 g / dm 3 4 g / dm 3 10 g / dm 3
1
0.37 hr
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
1
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 KS max
CS KS
CS 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 =
} } *0.0375*
}
1.332
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 KB
Cv KT PT Cv K B PB 1.39
KT PT 0 (1 X ) K B PT 0 X 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
rS
k S C E SPH
CT
K E PE C V
CV
rA
CE
k SK E C T
S
PE PH 1 K E PE
P10-5 (b) Individualized solution
P10-6 O2
2S
2O S
C3 H 6 O S
rS
C3 H 5OH
k A PA2 CV2
rAD kA
0
CA S rB
rC
k3 PBCV K A PA
k D CC
S
rC S kD
0
CC
K C PC CV
CT
rB
S
C A2 S KA
CV K A PA rS
S
C S
S
k3 PB C A S
rAD
CV
rS
CA
S
PC CV KD
CC
S
k D CC
CV 1
S
K C PC CV
K A PA
K C PC
k3CT PB K A PA 1
K A PA
2S
2A S
B A S
C3 H 5OH S
C3 H 6OH S
rB
A2
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'
1 O2 2
S ' H2
1 O2 2
S
Rate of adsorption: rAD
Rate of surface reaction: rs Rate of desorption: rD
CS '
k AD (Cv ' PH 2O k s CS '
k D (CS '
K H 2O
H 2O
H2
H 2O
)
---------- (1)
K H 2 PH 2 Cv ' )
Assuming, surface reaction to be rate limiting, we get rAD rD 0 k AD k D CS '
H 2O
K H 2OCv ' PH 2O and CS '
H2
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 '
k D (CS
H 2O
CS
H2
) ---------- (1)
K H 2 PH 2 Cv )
H2
Assuming, surface reaction to be rate limiting, we get rAD rD 0 k AD k D CS '
K H 2OCv ' PH 2O and CS
H 2O
H2
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
H2
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
ln
11.2 0.4
E R
E 1 R T1 E R
1 T2
E T2 T1 R T1T2
473 393 473 393
7743
E 15330.65
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
C A2 .S C A2 .S
A2 .S S K S C A2.s / CV 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
Ct
CV (1 K A PA
K S C A2.S CV
K A PACV
K A2 PB K S
K S K A2 PA 2
C A2 . S 2
CV
K A2 K S K A2 PB PA2 )
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
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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 ) i
T
Ci
HC
C pA
HB B
C pB
30 15 15 0
H A = -41000-(-15000-(-20000) = -6000 cal/mol A cal 15 15 30 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) VPFR
FA0
VCSTR
dX rA
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
FT
C A CI FA
C A01
FI C A 0 CI 0
FA0 FA0 FI 0
C A 0 CI 0 1 I
C A01
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
i
i
X H RX
CPA
CPA
i
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) Mole Balance
Adiabatic (1)
dX dW W dX dV
Rate Law
Stoichiometry
rA FA0 bV
rA B FA0 k CA
rA FA0 CB KC
(2)
rA
(3)
k
(4)
KC
KC2 exp
(5)
CA
CA0 1 X y T0 T
(6)
CB
CA0Xy T0 T
k 1 exp
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 , ,
Energy Balance
CP
Adiabatic and
(16A)
0
Additional Parameters (17A) & (17B) Heat Exchange
(16B)
H Rx X iC Pi
T T0
T0 ,
dT dV
rA
FiCPi
FA0
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
b
(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)
Rate Law
(9)
T2
Rate Law
(10)
K C2 1,000 KC 1000exp
H Rx
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
2CA 0X
dy dW
11-23
T0 y T
T0 y T
T 2y T0
= 0.00015
(b)
T0 T
0
rA
k C A0 1 X y
rA
kC 2A0
T0 T
C A0 2 X y
T
4X 2
1 X 2 X
2C A0 Xy
T0
KC
T0 T
KC
y
T0
2
T
(c)
Equilibrium Conversion
KC
4C 2A0 X 2e 1 Xe 2 Xe
C 2A0
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
FA02 CPi X
C pA
i
.12 T T0 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 X
C pA
i
T T0 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
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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
Ѳ=
FX
0
FX
FX
0
1
=
(1.5C A0 2C A0 ) (1.5C A0 C A0 ) (1.5C A0 C A0 )
1 2.5
0.4
(iii) Ѳ I = 3 ∑Cp = ѲI*CpI + CpA = 3×50 + 1×163 = 313J/molA/K
FX Ѳ=
0
FX
FX
0
1
=
(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
3.
244
275, 563, 644
4.
261
292, 409, 571, 661
265, 625
12-31
5.
283
310, 363, 449, 558, 677
6.
292
373, 463, 562, 674
7.
304
475, 551, 686
8.
338
9.
350
518, 713 720
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
ln
k2 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 HS 500 F
H W 200 F H W 200 F
H Rx 12-52
C PS T 500 C PS T 500
H Rx X
W
0.17 0.83
FW FA
18 80
0.9103
80lb BTU 30.4 mol lbmol R 18lb BTU C PS 8.46 mol lbmol R 80lb BTU H Rx 26,880 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
BTU lb R BTU 0.47 lb R BTU 336 lb 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
2C
A
B
C
10
10
0.0
80
80
-
Btu lb mole F
51
44
47.5
lb lb mol
128
94
222
63
67.2
65
lb mole hr Tio(F) Fio
~
CPio MW ,
i,
lb ft 3
HR
20, 000
Btu , lb mol A
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 ) i
T
Ci
HC
C pA
HB B
C pB
30 15 15 0
H A = - 41000 - (-15000 ) - (-20000) = -6000 cal/mol A
15 15 30
cal mol K
6000 X 300 200 X 30 k C A2 0 (1 X ) 2 .01 k (1 X )2
300 rA
VPFR
FA0
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
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
Ta
?
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
T T0
H Rx X I C PI
CPA
PFR Adiabatic FA0 FT
1. Irreversible A A.
B Liquid Phase, Keep FA0 Constant
First order
dX dV
rA FA0
kCA FA0
k
FA0 1 X 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
dTa
B Co-Current Exchange (22B)
Ýc C PCool m
Ua
C Counter Current (22C)
T Ta
b
dW
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
V
Note
W
0
k T T0
b
(1
X2
b 0
X)
X (1
( TT0
k
T
T0 k
X MB Where k
k1 exp
b
)X
X ) TT0 k T2
b
E 1 R T1
b
0
6TT0 k 2T
b
T0
2 2
k2
2 b
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
k
kg dm 3 E
0.0133* exp
Ta
1
R 450
1
dm 3
T
kg s
323K
UA 500
J s K
FA0
CA 0
CPA
40
H RXN
10 0
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
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
out
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
rB
dFA dV CA
rC
k CA
CB CC KC
dF dFB rB ; C rC RC dV dV FA T0 F T CB CT 0 B 0 CC FT T FT T
rA ; CT 0
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 SC+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
rg
Yc / s FS 0 X 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 294.4316.79, 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
5040000
290 K 1 exp
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
i C pi
UA C p0 FA 0 Tc
Ta T0 1
8000 50 80
E 1 R 350
1 T
2
350K
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 Cp
0.5g mol / min
C pA
Design Eqn : V Rate Law :
C pB
Stoichiometry : C A
C pB
HR
0
FA0 X rA rA kC A
C pA
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 C PS
CP 0 1
T TC
C PA
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 k
C
rA FA0 kC ACB E 1 R T
.01*exp
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
rA
k f CA
CB k
CC
B C
k r CB CC
CA0
X T0 1 X T
0.133 exp
kr
0.2exp
E 1 R T
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
k1C ( 400 ) exp
0.043 exp
E 1 R 400
E 1 1.987 400
1 500 1 500
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
r2 B
rB
r1s
rC
r1s r3T
rD
r2 B
rE
r2 B
rF
r3T
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
PB PC K
Stoichiometry:
PA
FA PT 0 FT
PB
FB PT 0 FT
PC
FC PT 0 FT
FT
FA
FI
steamratio .0034
FB
FC
FD
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
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