1-2
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 CD-
ROM 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 Montgomerys
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.
1-3
Summary
Assigned Alternates Difficulty Time (min)
P1-1 AA SF 60
P1-2 I SF 30
P1-3 O SF 30
P1-4 O SF 30
P1-5 AA SF 30
P1-6 AA 1-13 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
Assigned
= Always assigned, AA = Always assign one from the group of alternates,
O = Often, I = Infrequently, S = Seldom, G = Graduate level
1-4
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
1-5
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:
A
AA
r
FFV 0
Here rA is the rate of a first order reaction given by:
rA = - kCA
Given : CA = 0.1CA0 , k = 0.23 min-1, v0 = 10dm
3 min-1
Substituting in the above equation we get:
3
A0 0 0 0 0
1
0
(1 0.1) (10 / min)(0.9)
0.1 (0.23min )(0.1)
A A
A A
C v C v C v dmV
kC kC
V = 391.304 m3
(d) k = 0.23 min-1
From mole balance: dNA
dtrA V
Rate law: rA k CA
rA kNA
V
Combine:
dNA
dtk NA
1-6
at 0t , NAO = 100 mol and t t , NA = (0.01)NAO
t = A
A
N
N
k
0ln1
min100ln23.0
1
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.
1-7
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.
1-8
'
Ar 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:
dVrFFdt
dNjjj
j
0
No accumulation and no spatial variation implies
dVrFF jjj00
Also rj = b rj` and W = Vb where b is the bulk density of the bed.
=> '00 ( ) ( )j j j bF F r dV
Hence the above equation becomes
0
'
j j
j
F FW
r
We can also just apply the general mole balance as
'
0( ) ( )j
j j j
dNF F r dW
dt
Assuming no accumulation and no spatial variation in rate, we get the same form as above:
0
'
j j
j
F FW
r
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.
Lets do part c first.
1-9
[Flowrate In (moles/time)] penicillin + [generation rate (moles/time)]penicillin [ Flowrate Out(moles/time)]
penicillin = [rate of accumulation (moles/time)]penicillin
Fp,in + Gp Fp,out = dNp
dt
Fp,in = 0 (because no penicillin inflow)
Gp = .V
pr dV
Therefore,
.
V
pr dV - Fp,out = dNp
dt
Assuming steady state for the rate of production of penicillin in the cells stationary state,
dNp
dt= 0
And no variations
, ,p in p out
p
F FV
r
Or,
,p out
p
FV
r
Similarly, for Corn Steep Liquor with FC = 0
C
C
C
CC
r
F
r
FFV 00
Assume RNA concentration does not change in the stationary state and no RNA is generated or
destroyed.
P1-10
Given
21010*2 ftA RTSTP 69.491 ftH 2000
31310*4 ftV T = 534.7 R PO = 1atm
Rlbmol
ftatmR
3
7302.0 yA = 0.02
3
1010*04.2ft
lbmolCS
1-10
C = 4*105 cars
FS = CO in Santa Ana winds FA = CO emission from autos
hr
ftvA
3
3000 per car at STP
P1-10 (a)
Total number of lb moles gas in the system:
0PV
NRT
N = 13 3
3
1 (4 10 )
.0.73 534.69
.
atm ft
atm ftR
lbmol R
= 1.025 x 1011 lb mol
P1-10 (b)
Molar flowrate of CO into L.A. Basin by cars.
STP
AATAA
TR
PCvyFyF 0
carsft
lbmol
carhr
ftFT 400000
359
130003
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 102.04 10 lbmol/ft3
= 3.412 x 103lbmol/hr
P1-10 (e)
1-11
Rate of emission of CO by cars + Rate of CO by Wind - Rate of removal of CO = COdN
dt
dt
dCVCvFF cocooSA (V=constant, VCN coco )
P1-10 (f)
t = 0 , coOco CC
0
co
coO
Ct
co
A S o coC
dCdt V
F F v C
cooSA
coOoSA
o CvFF
CvFF
v
Vt ln
P1-10 (g)
Time for concentration to reach 8 ppm.
8
0 32.04 10CO
lbmolC
ft, 8
3
2.0410
4CO
lbmolC
ft
From (f),
0
34 3 13 8
3 3
3 313 4 3 13 8
3
.ln
.
6.7 10 3.4 10 1.673 10 2.04 104
ln
1.673 10 6.7 10 3.4 10 1.673 10 0.51 10
A S O CO
o A S O CO
F F v CVt
v F F v C
lbmol lbmol ft lbmol
ft hr hr hr ft
ft lbmol lbmol ft lbmol
hr hr hr hr ft
t = 6.92 hr
P1-10 (h)
(1) to = 0 tf = 72 hrs
coC = 2.00E-10 lbmol/ft3 a = 3.50E+04 lbmol/hr
ov = 1.67E+12 ft3 /hr b = 3.00E+04 lbmol/hr
Fs = 341.23 lbmol/hr V = 4.0E+13 ft3
dt
dCVCvF
tba cocoos
6sin
Now solving this equation using POLYMATH we get plot between Cco vs t
1-12
See Polymath program P1-10-h-1.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 72 72
C 2.0E-10 2.0E-10 2.134E-08 1.877E-08
v0 1.67E+12 1.67E+12 1.67E+12 1.67E+12
a 3.5E+04 3.5E+04 3.5E+04 3.5E+04
b 3.0E+04 3.0E+04 3.0E+04 3.0E+04
F 341.23 341.23 341.23 341.23
V 4.0E+13 4.0E+13 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
1-13
(2) tf = 48 hrs sF = 0 dt
dCVCv
tba cocoo
6sin
Now solving this equation using POLYMATH we get plot between Cco vs t
See Polymath program P1-10-h-2.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 48 48
C 2.0E-10 2.0E-10 1.904E-08 1.693E-08
v0 1.67E+12 1.67E+12 1.67E+12 1.67E+12
a 3.5E+04 3.5E+04 3.5E+04 3.5E+04
b 3.0E+04 3.0E+04 3.0E+04 3.0E+04
V 4.0E+13 4.0E+13 4.0E+13 4.0E+13
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
1-14
(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
A
AA
r
FFV 0
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,
05.0
10)5.0(01.010)5.0(00A0
k
vCvCV A
1-15
V = 99 dm3
FR: The general equation is
krdV
dFA
A , Now FA = CAv0 and FA0 = CA0v0 => kdV
vdCA 0
Integrating the above equation we get
VC
CA dVdC
k
v A
A 0
0
0
=> )( 00
AA CCk
vV
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
A
A
r
vCvCV 00A0
A
A
kC
Cv )01.01(00
k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1
)/5.0*01.0)(36.0(
)99.0)(/5.0)(/10(31
33
dmmolhr
dmmolhrdmV => V = 2750 dm3
PFR:
From above we already know that for a PFR
AAA kCr
dV
vdC 0
Integrating
VC
C A
A dVC
dC
k
v A
A 0
0
0
VC
C
k
v
A
A00 ln
Again k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1
1-16
Substituing the values in above equation we get V = 127.9 dm3
P1-11 (c)
- rA = kCA2 with k = 3 dm3/mol.hr
CSTR:
VCA0v0 CAv0
rA
v0CA0(1 0.01)
kCA2
Substituting all the values we get
V(10dm3 /hr)(0.5mol /dm3)(0.99)
(3dm3 /hr)(0.01*0.5mol /dm3)2 => V = 66000 dm3
PFR:
dCAv0
dVrA kCA
2
Integrating
v0
k
dCA
CA2
CA0
CA
dV
0
V
=>v0
k(1
CA
1
CA0) V
=> V10dm3 /hr
3dm3 /mol.hr(
1
0.01CA0
1
CA0) = 660 dm3
P1-11 (d)
CA = .001CA0
tdN
rAVNA
NA 0
Constant Volume V=V0
tdCA
rACA
CA 0
Zero order:
t1
kCA0 0.001CA0
.999CAo
0.059.99h
First order:
1-17
t1
klnCA0
CA
1
0.001ln
1
.0016908 s
Second order:
t1
k
1
CA
1
CA0
1
3
1
0.0005
1
0.5666h
P1-12 (a)
Initial number of rabbits, x(0) = 500
Initial number of foxes, y(0) = 200
Number of days = 500
1 2
dxk x k xy
dt .(1)
3 4
dyk xy k y
dt ..(2)
Given,
1
1
2
3
1
4
0.02
0.00004 /( )
0.0004 /( )
0.04
k day
k day foxes
k day rabbits
k day
See Polymath program P1-12-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 500 500
x 500 2.9626929 519.40024 4.2199691
y 200 1.1285722 4099.517 117.62928
k1 0.02 0.02 0.02 0.02
k2 4.0E-05 4.0E-05 4.0E-05 4.0E-05
1-18
k3 4.0E-04 4.0E-04 4.0E-04 4.0E-04
k4 0.04 0.04 0.04 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 3 0.00004 /( )k day rabbits
1-19
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:
A0 A
A
F FV =
r
1-20
Rate Law :
2
A Ar kC
Combine:
A0 A
2
F FV =
AkC
3
0 0 3
2 63 .A Adm molA molA
F v Cs dm s
3
0 3
0.1 0.33 .A Adm molA molA
F v Cs dm s
3
32
3
(6 0.3)
V = 19,000
(0.03 )(0.1 ).
mol
s dmdm mol
mol s dm
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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
Assigned
Alternates
Difficulty
Time (min)
P2-1 O 15 P2-2 A 30 P2-3 A 30
P2-4 O 75 P2-5 O M 75 P2-6 AA 7,8,11 FSF 45 P2-7 S FSF 45 P2-8 AA 6,8,11 SF 45 P2-9 S SF 15 P2-10 AA SF 1 P2-11 AA 6,7,8 SF 60 P2-12 S M 60 CDP2-A O 8,B,C,D FSF 5 CDP2-B O 8,B,C,D FSF 30 CDP2-C O 8,B,C,D FSF 30 CDP2-D O 8,B,C,D FSF 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,
2.34.62
1
2
1 011
A
A
A r
XF
r
XFV
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
Now, FAO = 0.4/2 = 0.2 mol/s,
Table: Divide each term A
A
r
F 0 in Table 2-3 by 2.
X 0 0.1 0.2 0.4 0.6 0.7 0.8
[FAO/-rA](m3) 0.445 0.545 0.665 1.025 1.77 2.53 4
Reactor 1 Reactor 2
V1 = 0.82m3 V2 = 3.2 m
3
V = (FAO/-rA)X
182.0
1
0 Xr
F
XA
A 22.3
2
0 Xr
F
XA
A
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 ;
Levenspiel Plot
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 0.2 0.4 0.6 0.8 1
Conversion
Fa
o/-
ra
0A
A
F
r1.28m3
at X=0.2 ; 0A
A
F
r.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
2
0.2
30.247
AO
A
FV dX
r
m
For next CSTR,
X3 = 0.65, 32AO
Fm
rA, V3 =
33 2( ) .3AOF X X
mrA
(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:
2
0.088
X
AO
A
FV dX
r
,
By estimation using the levenspiel plot
X2 = .183
For CSTR,
VCSTR2 =3 2 30.1
AOF X Xm
rA
=> X3 = .316
(3) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the smaller
CSTR.
Conversion Original Reactor Volumes Worst Arrangement
X1 = 0.20 V1 = 0.188 (CSTR) V1 = 0.23 (PFR) X2 = 0.60 V2 = 0.38 (PFR) V2 = 0.53 (CSTR) X3 = 0.65 V3 = 0.10 (CSTR) V3 = 0.10 (CSTR)
For PFR,
X1 = 0.2
1
1
0
X
AO
A
FV dX
r
Using trapezoidal rule,
XO = 0.1, X1 = 0.1
1
1 1
3
3
0.21.28 0.98
2
0.23
O
O
X XV f X f X
rA
m
m
For CSTR,
For X2 = 0.6, 31.32AO
Fm
rA, V2 = 2 1
AO
A
FX X
r= 1.32(0.6 0.2) = 0.53 m3
For 2nd CSTR,
For X3 = 0.65, 32AO
Fm
rA, V3 = 0.1 m
3
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
FAO/-rA (m3) 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
1 2
10
X X
Ao Ao
A AX
F FV dX dX
r r
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))
PFR
CSTR
X
X A
AO dXr
FV
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 r(mol/dm3.min) 1/-r(dm3.min/mol)
0 1 1
0.20 1.67 0.6
0.40 5 0.2
0.45 5 0.2
0.50 5 0.2
0.60 5 0.2
0.80 1.25 0.8
0.90 0.91 1.1
P2-6 (a)
To solve this problem, first plot 1/-rA vs. X from the chart above. Second, use mole balance as given
below.
CSTR:
Mole balance: min./5
4.0min/300F V
3
A0
dmmol
mol
r
X
A
CSTR =>
=>VCSTR = 24 dm3
PFR:
Mole balance:
X
A
APFRr
dXFV
0
0
= 300(area under the curve)
VPFR = 72 dm3
P2-6 (b)
For a feed stream that enters the reaction with a previous conversion of 0.40 and leaves at any
conversion up to 0.60, the volumes of the PFR and CSTR will be identical because of the rate is constant
over this conversion range.
6.
4.
6.
4.
0
6.
4.
00 Xr
FdX
r
FdX
r
FV
A
A
A
A
A
A
PFR
P2-6 (c)
VCSTR = 105 dm3
Mole balance: A
CSTRr
XA0F V
moldmmol
dm
r
X
A
min/35.0min/300
105 3
3
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.
Mole balance:
2
120
XA
A
r
XXFV
Rearranging, we get
008.040.0
0
2
2AXA
F
V
r
X
At X2 = 0.64, 008.040.0
2
2
XAr
X
Conversion = 0.64
P2-6 (e)
From part (a), we know that X1 = 0.40. Use trial and error to find X2.
Mole balance: 22
40.040.0
0 30072
X
A
X
A
APFRr
dX
r
dXFV
At X2 = 0.908, V = 300 x (area under the curve)
=> V = 300(0.24) = 72dm3
Conversion = 0.908.
P2-6 (f)
See Polymath program P2-6-f.pol.
P2-7 (a)
S
S
r
XFV 0
FS0 = 1000 g/hr
At a conversion of 40% g
hrdm
rS
3
15.01
Therefore 360)40.0)(1000(15.0 dmV
P2-7 (b)
At a conversion of 80%, g
hrdm
rS
3
8.01
FS0 = 1000 g/hr
Therefore 3640)80.0)(1000(8.0 dmV
P2-7 (c) X
S
SPFRr
dXFV
0
0
From the plot of 1/-rS Calculate the area under the curve such that the area is equal to V/FS0 = 80 / 1000
= 0.08
X = 12%
For the 80 dm3 CSTR, S
S
r
XFdmV 0380
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 CSTRs 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)
SM
CS
sCK
CkCr and 001.01.0 0 SSC CCC
SM
SSSs
CK
CCkCr
001.01.0 0
001.01.0
1
0 SSS
SM
s CCkC
CK
r
Let us first consider when CS is small.
CS0 is a constant and if we group together the constants and simplify then CskCk
CK
rS
SM
s 2
2
1
1
since CS < KM
CskCk
K
rS
M
s 2
2
1
1 which is consistent with the shape of the graph when X is large (if CS is small X is
large and as CS grows X decreases).
Now consider when CS is large (X is small)
As CS gets larger CC approaches 0:
001.01.0 0 SSC CCC and 0SS CC
If SM
CS
sCK
CkCr then
CS
SM
s CkC
CK
r
1
As CS grows larger, CS >> KM
And CCS
S
s kCCkC
C
r
11
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
2
1)(
0
X
X A
Ar
dXFV
Volume = Geometric area under the curve of (FA0/-rA) vs X)
5.01000005.04000002
1V
V = 150000 m3
.
P2-8 (b) CSTR Volume to achieve 50% conversion Mole Balance
)(
0
A
A
r
XFV
1000005.0V V = 50000m3
P2-8 (c) Volume of second CSTR added in series to achieve 80% conversion
)(
)( 1202
A
A
r
XXFV
)5.08.0(5000002V
V2 = 150000m3
P2-8 (d) Volume of PFR added in series to first CSTR to achieve 80% conversion
)3.0100000()3.04000002
1(PFRV
VPFR = 90000m3
P2-8 (e) For CSTR, V = 60000 m3 (CSTR) Mole Balance
XX
r
XFV
A
A
)500000800000(60000
)(
0
X = 0.463 For PFR, V = 60000 m3 (PFR) Mole balance
X
A
Ar
dXFV
0
0)(
dXXX
)100000800000(600000
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 = r2h = (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 = r2h = (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) = Fao (mother). The Levenspiel Plot is shown:
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0 0.2 0.4 0.6 0.8
ma
o/-
ra
M1
Conversion
Autocatalytic Reaction
Mother
Baby
31.36 *0.34 0.232
baby
A
FaoXV m
r
Since the volume of the stomach is proportional to the age of the baby hippo, and the volume of the
babys stomach is half of an adult, then the baby hippo is half the age of a full grown hippo.
4.52.25
2
yearsAge years
2) If Vmax and mao are both one half of the mothers then
0
2 2
1
2 motherA
Ao
AM AM
mm
r r
and since
max2
AAM
M A
v Cr
K C then
max
2 2
112
2baby mother
A
AM AM
M A
v C
r rK C
2 22
1
21
2
AoAo Ao
AM AM motherbaby AMmother
mm m
r rr
2
Ao
AM
m
r will be identical for both the baby and mother.
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:
0
1
A
AM
m
r= 127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499
And since Vstomach = 0
1
A
AM
m
rX,
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 = X0
|f
A
A X X
F
r
So the area under the 0A
A
F
r versus X curve for a CSTR is a rectangle but the height of rectangle
corresponds to the value of 0A
A
F
r at X= Xf
But in this case the value of 0A
A
F
r is taken at X= Xi and the area is calculated.
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:
A
A
r
XFV 0 PFR:
X
Ar
dXV
0
Until the conversion (X) reaches 0.5, the reaction rate is independent of conversion and the reactor volumes will be identical.
i.e. CSTRA
A
A
A
A
PFR Vr
XFdX
r
F
r
dXV 0
5.0
0
05.0
0
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:
A
A
A
A
r
XCv
r
XFV 000
13
38
3
3
3
00
103.
1032005
09.0
mol
sm
m
mol
s
m
m
r
Cv
VX
A
A
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).
311000
1 105.1 mr
XCv
r
XFV
A
A
A
A
The PFR volume required to go from X=0.5 to X=0.7 (reaction rate depends on conversion).
Finally, we add V2 to V1 and get:
Vtot = V1 + V2 = 2.3 x1011 m3
CDP2-A (d)
What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a CSTR to
raise the conversion to 90 %
We notice that the new inverse of the reaction rate (1/-rA) is 7*108. We insert this new value into our CSTR mole balance equation:
311000 104.1 mr
XCv
r
XFV
A
A
A
ACSTR
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: dt
dNVr AA
Stoichiometry: )1(0 XNN AA
Combine: dt
dXNVr AA 0
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):
dt
dXX
N
Vr
A
A )1(0
0
X
A
A
t
Xr
dXCdt
00
0 )1(
After integration, we have:
)1ln(1
0 XCr
t AA
Inserting the values for our variables: t = 2.02 x 1010 s That is 640 years.
CDP2-A (f)
Plot the rate of reaction and conversion as a function of PFR volume.
The following graph plots the reaction rate (-rA) versus the PFR volume:
Below is a plot of conversion versus the PFR volume. Notice how the relation is linear until the conversion exceeds 50%.
The volume required for 99% conversion exceeds 4*1011 m3.
CDP2-A (g)
Critique the answers to this problem.
The rate of reaction for this problem is extremely small, and the flow rate is quite large. To obtain the
desired conversion, it would require a reactor of geological proportions (a CSTR or PFR approximately
the size of the Los Angeles Basin), or as we saw in the case of the batch reactor, a very long time.
CDP2-B Individualized solution
CDP2-C (a)
For an intermediate conversion of 0.3, Figure above shows that a PFR yields the smallest volume, since
for the PFR we use the area under the curve. A minimum volume is also achieved by following the PFR
with a CSTR. In this case the area considered would be the rectangle bounded by X =0.3 and X = 0.7 with
a height equal to the CA0/-rA value at X = 0.7, which is less than the area under the curve.
CDP2-C (b)
CDP2-C (c)
CDP2-C (d)
For the PFR,
CDP2-C (e)
CDP2-D
CDP2-D (a)
CDP2-D (b)
CDP2-D (c)
CDP2-D (d)
CDP2-D (e)
CDP2-D (f)
CDP2-D (g)
CDP2-D (h)
CDP2-E
CDP2-F (a)
Find the conversion for the CSTR and PFR connected in series.
X -rA 1/(-rA)
0 0.2 5
0.1 0.0167 59.9
0.4 0.00488 204.9
0.7 0.00286 349.65
0.9 0.00204 490.19
CDP2-F (b)
CDP2-F (c)
CDP2-F (d)
CDP2-F (e)
CDP2-F (f)
CDP2-F (g) Individualized solution
3-1
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3-2
Solutions for Chapter 3 Rate Laws
P3-1 Individualized solution.
P3-2 (a) Example 3-1
For, E = 60kJ/mol For, E = 240kJ/mol
16 600001.32 10 expJ
kRT
16 2400001.32 10 expJ
kRT
T (K) k (1/sec) 1/T ln(k)
310 1023100 0.003226 13.83918
315 1480488 0.003175 14.2087
320 2117757 0.003125 14.56667
325 2996152 0.003077 14.91363
330 4194548 0.00303 15.25008
335 5813595 0.002985 15.57648
T (K) k (1/sec) 1/T ln(k)
310 4.78E-25 0.003226 -56.0003
315 2.1E-24 0.003175 -54.5222
320 8.77E-24 0.003125 -53.0903
325 3.51E-23 0.003077 -51.7025
330 1.35E-22 0.00303 -50.3567
335 4.98E-22 0.002985 -49.0511
3-3
P3-2 (b) No solution will be given
P3-2 (c)
A + 2
1B
2
1C
Rate law: rA kACA2CB and kA 25
1
s
dm3
mol
2
rA
1
rB
1/2
rC
1/2 25CA
2CB
kBCA2CB
1/2
kCCA2CB
1/2
kC kB 12.51
s
dm3
mol
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.
3-4
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)
1/T Flashes/min
ln(flashes/min)
294 0.003401 9 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) 1/T x10
3
chirps/min
ln(chirps/min)
287.2 3.482 80 4.382
293.3 3.409 126 4.836
300 3.333 200 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) 1/T x10
3
V(cm/s) ln(V)
298 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.
3-5
ln (V) = 44.6 1.33E4/T
At T = 40oC (313K) V = 6.4cm/s
At T = -5oC (268K) 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 x103 V(cm/s) ln(V)
283 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 Activation Energy
Cricket 52150
Firefly 54800
Ant 95570
Honeybee 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.
3-6
P3-6 Antidote did not dissolve from glass at low temperatures.
P3-7 (a) If a reaction rate doubles for an increase in 10C, at T = T1 let k = k1 and at T = T2 = T1+10, let k = k2 = 2k1.
Then with k = Ae-E/RT in general, 1/1E RTk Ae and 2/2
E RTk Ae , or
2 1
1 1
2
1
E
R T Tke
k or
2 2
1 1
1 2
1 22 1
ln ln
1 1
k k
k kE
T TR
TTT T
Therefore:
21 1
1 11
2 1
ln 10ln 2 10
10
kT T
T TkE R R
T T
1 1
1010
ln 2
ET T
R
which can be approximated by 0.510
ln 2
ET
R
P3-7 (b)
Equation 3-18 is E
RTk Ae
From the data, at T1 = 0C, 1/
1
E RTk Ae , and at T2 = 100C, 2/
2
E RTk Ae
Dividing gives 2 11 1
2
1
E
R T Tke
k, or
2
1 1 2 2
1 2 1
2 1
ln
ln1 1
kR
k RTT kE
T T k
T T
1.99 273 373.050
ln 7960100 .001
calK K
mol K calE
K mol
3-7
1 3 1 1
1
7960
10 min exp 2100min
1.99 273
E
RT
cal
molA k ecal
Kmol K
P3-7 (c) Individualized solution
P3-8
From the given data -rA(dm
3/mol.s) T(K)
k= -rA/(4*1.5)
1/T ln (k)
0.002 300 0.00033333
0.003333 -8.00637
0.046 320 0.00766667
0.003125 -4.87087
0.72 340 0.12
0.002941 -2.12026
8.33 360 1.38833333
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
3-8
(a) Activation energy (E), E = 14999*8.314 =124700 J/mol = 124.7 kJ/mol
(b) Frequency Factor (A), ln A = 41.99 A = 1.72X1018
(c) k = 1.72X1018 exp(- (1)
Given T0 = 300K
Therefore, putting T = 300K in (1), we get k(T0) = 3.33X10-4
Hence,
k(T) = k(T0)exp[
P3-9 (a)
From the web module we know that (1 )dX
k xdt
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 100C.
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 = 62HC
kC
2) C2H4 + 1/2O2 C2H4O Rate law: -rA = 2/1
0242CkC HC
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[104HnC
C 104HiC
C /Kc]
5) CH3COOC2H5 + C4H9OH CH3COOC4H9 + C2H5OH
A + B C + D
Rate law: -rA = k[CACB CCCD/KC]
3-9
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 Rate law: -rHBr =
2
22
2
2/1
1
Br
HBr
BrH
C
Ck
CCk
(2) H2 + I2 2HI Rate law: -rA = 2 21 H I
k C C
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:
3-10
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:
3-11
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 T(K)
1/T ln k
6.5 300
0.003333 1.871802
13 310
0.003226 2.564949
18 313
0.003195 2.890372
3-12
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
2( )DimerA AnthraceneC
Cr k C
K where, Kc = k+/k-
Similarly for the second reaction:
3-13
Norbornadiene Quadricyclane
( )Quadricyclane
Norbornadiene Norbornadiene
C
Cr k C
K 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 (-HR)
We have to calculate E for the reaction
CH3 + RBr CH3Br + R
Given: HR = - 6 kcal/mol
From the given data table, we get
6.8 = C (17.5)
And, 6.0 = C (20)
=> C = 12.4 KJ/mol and = 0.32
Using these values, and HR = - 6 kcal/mol, we get E = 10.48 KJ/mol
4-1
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4-2
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 (B XA/3) = 10 ( - ) = 2.33 mol/dm3
For 90 % Conversion
CD = same as example & CB = CAo (B XA/3) = 10 ( ) = 0 mol/dm3
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: 21 0.5(1 0.14 )
(1 )(0.54 0.5 )A
X
r X X
1/(-ra) vs X
0
20
40
60
80
100
120
140
160
180
0 0.2 0.4 0.6 0.8 1 1.2
X
1/(
-ra
)
The rate of reaction decreases drastically with increase in conversion at higher conversions.
4-3
P4-2 (d) Example 4-4
POLYMATH Report No Title Ordinary Differential Equations 05-May-2009
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
4-4
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
4-5
Therefore, catalyst weight required for 30% conversion is 1.611x10
5 g = 161.1 kg
4-6
Therefore, catalyst weight required for 60% conversion is 2.945x105 g = 294.5 kg
Therefore, catalyst weight required for 99% conversion is 4.041x10
5 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.
N2O4 2NO2
A 2B
Equilibrium rate constant is given by: KCCB,e
2
CA,e
Stoichiometry: yA0 yA0(2 1) yA0
Constant volume Batch:
CANA0(1 X)
V0CA0(1 X) and CB
2NA0X
V02CA0X
Plug flow reactor:
CAFA0(1 X)
v0(1 X)
CA0(1 X)
(1 X) and CB
2FA0X
v0(1 X)
2CA0X
(1 X)
4-7
CAOyAOPO
RTOyAO 0.07176 mol /dm
3
Combining: For constant volume batch:
KCCB,e
2
CA,e
4CAo2 X 2
CAO (1 X) Xe
KC (1 Xe)
4CA0
For flow reactor:
)1)(1(
4 22
,
2
,
XXC
XC
C
CK
AO
Ao
eA
eB
C 04
)1)(1(
A
eeC
eC
XXKX
See Polymath program P4-2-e.pol.
POLYMATH Results
NLES Report (safenewt)
Nonlinear equations [1] f(Xeb) = Xeb - (kc*(1-Xeb)/(4*Cao))^0.5 = 0
[2] f(Xef) = Xef - (kc*(1-Xef)*(1+eps*Xef)/(4*Cao))^0.5 = 0
Explicit equations [1] yao = 1
[2] kc = 0.1
[3] Cao = 0.07174*yao
[4] eps = yao
4-8
Yinert Yao Xeb Xef
0 1 0.44 0.508
0.1 0.9 0.458 0.5217
0.2 0.8 0.4777 0.537
0.3 0.7 0.5 0.5547
0.4 0.6 0.525 0.576
0.5 0.5 0.556 0.601
0.6 0.4 0.5944 0.633
0.7 0.3 0.6435 0.6743
0.8 0.2 0.71 0.732
0.9 0.1 0.8112 0.8212
0.95 0.05 0.887 0.89
0.956 0.044 0.893 0.896
P4-3 Solution is in the decoding algorithm available separately from the author.
P4-4
A1
2B
1
2C
yA01
2
1
2
1
21
1
2
(a) CB0 CA0 0.1mol
dm3
(b) CA CA01 X
1 X0.1
1 0.25
11
2X
0.1 0.75
0.8750.086
mol
dm3
0 3
0.25
0.1 0.1252 20.1 0.0143
11 0.8751
2
C A
X
molC C
X dmX
(c) CB CA0
11
2X
1 X
CA0 11
2X
11
2X
CA0 0.1mol
dm3
(d) CB 0.1mol
dm 3
4-9
(e) rA
4
rC
2, rA 2rC 4 mol dm
3 min
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/dm
3
Stoichiometric Table:
Species Symbol Initial Change Remaining
Ethylene
oxide
A CAO=16.13 mol/dm3 - CAOX CA= CAO(1-X)
= (1-X) lbmol/ft3
Water B CBO= 55.5 mol/dm3,
B =3.47
-CAOX CB = CAO( B -X)
=(3.47-X) lbmol/ft3
Glycol C 0 CAOX CC = CAOX
= X lbmol/ft3
Rate law: -rA = kCACB
Therefore, -rA = k 2
AOC (1-X) ( B -X) = k (16.13)2(1-X) (3.47-X)
At 300K E = 12500 cal/mol, X = 0.9,
k = 0.1dm3/mol.s
CSTR = AO
A
C X
r =
2
16.13 0.92.186sec
0.1 16.13 1 0.9 3.47 0.9 and, V = 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 = AO
A
C X
r =
2
16.13 0.90.109sec
19.99 16.13 1 0.9 3.47 0.9,
and, V = X FA0 = 0.109 X 200 liters = 21.8 liters
4-10
P4-5 (b) Isothermal, isobaric gas-phase pyrolysis,
C2H6 C2H4 + H2
A B + C
Stoichiometric table:
Species symbol Entering Change Leaving
C2H6 A FAO -FAOX FA=FAO(1-X)
C2H4 B 0 +FAOX FB=FAOX
H2 C 0 +FAOX FC=FAOX
FTO=FAO FT=FAO(1+X)
= yao = 1(1+1-1) = 1
v = vo(1+ X) => v = vo(1+X)
CAO = yAO CTO = yAO P
RT
= 3
1 6
0.082 1100.
atm
m atmK
K kmol
= 0.067 kmol/m3 = 0.067 mol/dm3
CA = 1(1 )
(1 ) 1
AOAAO
O
XF XFC
v v X X mol/dm3
CB = ( )
(1 ) 1
AOBAO
O
F XF XC
v v X X mol/dm3
CC = ( )
(1 ) 1
C AOAO
O
F F X XC
v v X X mol/dm3
Rate law:
-rA = kCA= kCAO1
1
X
X =0.067 k
1
1
X
X
If the reaction is carried out in a constant volume batch reactor, =>( = 0)
CA = CAO(1-X) mol/dm3 CB = CAO X mol/dm
3 CC = CAO X mol/dm3
P4-5 (c) Isothermal, isobaric, catalytic gas phase oxidation,
C2H4 + 1
2O2 C2H4O
A + 1
2B C
4-11
Stoichiometric table:
Species Symbol Entering Change Leaving
C2H4 A FAO -FAOX FA=FAO(1-X)
O2 B FBO -
1
2FAOX
FB=FAO( B -X/2)
C2H4O C 0 +FAOX FC=FAOX
B =
112
2
AOBO
AO AO
FF
F F
2
3
AO AOAO
TO AO BO
F Fy
F F F
2 11 1 0.33
3 2AOy
33
620.092
3 .0.082 533
.
AO AO TO AO
atmP molC y C y
RT dmatm dmK
mol K
1 1 0.092 1
1 1 0.33 1 0.33
AO AOAA
O
F X C X XFC
v v X X X
0.046 12
1 1 0.33
AO B
BB
O
XF
XFC
v v X X
0.092
1 1 0.33
C AOC
O
XF F XC
v v X X
If the reaction follow elementary rate law
Rate law: 0.5A A Br kC C
0.5
0.092 1 0.046 1
1 0.33 1 0.33A
X Xr k
X X
P4-5 (d) Isothermal, isobaric, catalytic gas-phase reaction in a PBR
C6H6 + 2H2 C6H10
A + 2B C
Given:
min/50 30 dm
Stoichiometric Table:
Species Symbol Entering Change Leaving
C6H6 A FA0 -FA0X FA=FA0(1-X)
H2 B FB0=2FA0 -2FA0X FB=FA0(B-2X)
C6H10 C 0 FA0X Fc=FA0X
4-12
22
0
0
0
0
A
A
A
B
BF
F
F
F
3
1
00
0
0
0
0
BA
A
T
A
AFF
F
F
Fy
3
2)121(
3
10Ay
3000055.0
3
1
)443(*0821.0
6
3
13
dm
mol
K
atm
RT
PyCC
KmolatmdmATA
)1(
)1(
)1(
)1(
320
0
0
X
XC
X
XFFC A
AAA
)1(
)1(*2
)1(
)22(
)1(
)2(
320
320
0
0
X
XC
X
XC
X
XFFC AA
BABB
)1()1(320
0
0
X
XC
X
XFFC A
AC
C
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. 2
' BAA PkPr
2
3**
minminatmatm
atmkgcat
mol
kgcat
mol
Notice that if you use concentrations in this rate law, the units will not work out.
RTCRTCyPyP AAAA *)( 00
3
3
32
33
0
322 )()1(
)1(4)( RT
X
XkCRTCkCPkPr ABABAA
Design Equation for a fluidized CSTR:
'
0
A
A
r
XFW
333
0
3
32
0
)()1(4
)1(
RTXkC
XXFW
A
A
332
0
3
32
0
)()1(4
)1(
RTXkC
XXW
A
Evaluating the constants:
Katatmkgcat
molk 300
min53
3
At 170C (443K),
4-13
atmkgcat
mol
KK
Kmol
Jmol
J
TTR
Ekk A
min1663000
443
1
300
1
314.8
80000
5311
443300
300443
Plugging in all the constants into the design equation:
X = 0.8
kgcatK
WKmolatmdm
dm
mol
atmkgat
mol
dm7
332
min
3
32
min 1025.5)4430821.0()8.01()055.0(16630004
)8.01(8.0503
33
3
At 270C (543K),
atmkgcat
mol
KK
Kmol
Jmol
J
TTR
Ekk A
min9079000
543
1
300
1
314.8
80000
5311
543300
300543 Plu
gging in all the constants into the design equation:
X = 0.8
kgcatK
WKmolatmdm
dm
mol
atmkgat
mol
dm8
332
min
3
32
min 1022.5)5430821.0()8.01()055.0(90790004
)8.01(8.0503
33
3
P4-6 (a) Let A = ONCB C = Nibroanaline
B = NH3 D = Ammonium Chloride
A + 2B C + D
- A A Br kC C
P4-6 (b) Species Entering Change Leaving
A FA0 - FA0X FA0(1-X)
B FB0 = BFA0 =6.6/1.8 FA0
-2 FA0X FB=
FA0(B 2X)
C 0 FA0X FC= FA0X
D 0 FA0X FD=FA0X
P4-6 (c) For batch system,
CA=NA/V -rA = kNANB/V2
P4-6 (d)
4-14
- A A Br kC C
00 0
0 0 0
1 1 , 1AA A A AA A A ANN N F F
F X C X C C XV V V v v
00 0
0 0 0
2 2 , 2AB B BB B A B B A BNN N F
F X C X C C XV V V v
2
0 1 2A A Br kC X X
0
0
6.63.67
1.8
BB
A
C
C
0 31.8A
kmolC
m
21.8 1 3.67 2Ar k X X
P4-6 (e) 1) At X = 0 and T = 188C = 461 K
min0202.067.38.1
min0017.0
3
2
3
32
00m
kmol
m
kmol
kmol
mkCr BAA
0 3
kmol0.0202
m minAr
2) At X = 0 and T = 25C = 298K
TTR
Ekk
O
O
11exp
min.
31003.2
298
1
461
1
.987.1
11273
expmin.
0017.0
6
3
kmol
m
Kmol
calmol
cal
kmol
mk
-rAO = kCAOCBO = 2.41 X 10-5 kmol/m3min
3)
0
0
1 1exp
Ek k
R T T
4-15
3 11273m 1 10.0017 exp
kmol min 461 5611.987
cal
molkcal K K
mol K
3m0.0152
kmolmink
0 0 0A A Br kC C
3
3 3
m0.0152 1.8 6.6
kmolminA
kmol kmolr
m m
3
kmol0.1806
m minAr
P4-6 (f) rA = kCAO
2(1-X)(B-2X)
At X = 0.90 and T = 188C = 461K
1) at T = 188 C = 461 K
min00103.0
9.0267.39.018.1min.
0017.0
3
2
3
3
m
kmol
m
kmol
kmol
mrA
2) at X = 0.90 and T = 25C = 298K
min1023.1
9.0267.39.018.1min.
1003.2
3
6
2
3
36
m
kmol
m
kmol
kmol
mrA
3) at X = 0.90 and T = 288C = 561K
min0092.0
9.0267.39.018.1min.
0152.0
3
2
3
3
m
kmol
m
kmol
kmol
mrA
P4-6 (g) FAO = 2 mol/min
1) For CSTR at 25oC -rA min
1023.13
6
m
kmol
4-16
3
3
3
9.0
60.162
min1023.1
1.0min/2
1
m
m
mol
mol
r
XFV
XA
AO
2)At 288oC, -rA min
0092.03m
kmol
3
3
9.0
739.21
min0092.0
1.0min/2
1
m
m
mol
mol
r
XFV
XA
AO
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 O 6 + 2a = 1.2c + d + 2e
For N b = 0.86c For H 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
4-17
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.
2 2 3
1 3
2 2N H NH
Making H2 as the basis of calculation:
2 2 3
1 2
3 3H N NH
1 2
3 3A B C
Stoichiometric table:
Species Symbol Initial change Leaving
H2 A FAO -FAOX FA=FAO(1-X)
N2 B FBO= B FAO -FAOX/3 FB=FAO( B -X/3)
NH3 C 0 +2FAOX/3 FC=(2/3)FAOX
4-18
P4-8 (b)
2 1 21
3 3 3
2 10.5
3 3AOy
3
16.40.5
.0.082 500
.
AO
atmC
atm dmK
mol K
= 0.2 mol/dm3
2
3
3
3
1 0.2 10.1 /
11
3
0.22 20.1 /
3 1 31
3
AO
H A
AO
NH C
C X XC C mol dm
XX
C X XC C mol dm
XX
P4-8 (c)
kN2 = 40 dm3/mol.s
(1) For Flow system:
2 2 2 2
312 2
312 2
2
113
40
1 13 3
N N N H
AO
r k C C
X
XC
X X
rN2 1.61 X
1X
3
32
(2) For batch system, constant volume.
4-19
312 2
2 2 2 2
2
2
12
32
2
12
32
0
0 0
0
0
0
2
0
1
1
3
13
40 1 13
1.6 1 13
N N N H
AA AA
A A
A B
BB
H O
N A
r k C C
N XN NC
V V V
C C X
XN
NC
V V
XC
Xr C X
XX
P4-9 (a) Liquid phase reaction assume constant volume
Rate Law (reversible reaction):
CA A B
C
Cr k C C
K
Stoichiometry:
0 1A AC C X , 0 1B AC C X , 0C AC C X
To find the equilibrium conversion, set -rA = 0, combine stoichiometry and the rate law, and solve for Xe.
A B C CC C K C
22
0 01A e C A eC X K C X
2
0
12 1 0e e
A C
X XC K
0.80eX
To find the equilibrium concentrations, substitute the equilibrium conversion into the stiochiometric
relations.
0 3 31 2 1 0.80 0.4A A
mol molC C X
dm dm
0 3 31 2 1 0.80 0.4B A
mol molC C X
dm dm
4-20
0 3 32 *0.80 1.6A Amol mol
C C Xdm dm
P4-9 (b) Stoichiometry:
0 1 3 1 2Ay and 0C
0
0
0
1 1
1 1 2
AAA A
N X XNC C
V V X X
00
0
3 3
1 1 2
C AC A
N N X XC C
V V X X
Combine and solve for Xe.
3
0 0
1 3
1 2 1 2
e eC A A
e e
X XK C C
X X
2 2 3
01 1 2 27C e e A eK X X C X
230274 3 1 0A e e
C
CX X
K
0.58eX
Equilibrium concentrations:
00 33
0
100.305
400 0.082
A
P atm molC
RT dmdm atmK
mol K
3
1 0.580.305 0.059
1 2 0.58A
molC
dm
3
3 0.58 0.3050.246
1 2 0.58C
molC
dm
P4-9 (c) Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction.
Stoichiometry:
0
0
0
11
AAA A
N XNC C X
V V
4-21
00
0
33C AC A
N N XC C X
V V
Combine and solve for Xe
3
0 01 3C A e A eK C X C X
0.39eX
Equilibrium concentrations
30.305 1 0.39 0.19A
molC
dm
30.305 0.39 0.36C
molC
dm
P4-9 (d) Gas phase reaction in a constant pressure, batch reactor
Rate law (reversible reaction):
3
CA A
C
Cr k C
K
Stoichiometry:
0 1 3 1 2Ay and 0C
0
0
0
1 1
1 1 2
AAA A
N X XNC C
V V X X
00
0
33
1 1 2
C AC A
N N X XC C
V V X X
Combine and solve for Xe:
3
0 01 3
1 2 1 2
C A e A e
e e
K C X C X
X X
0.58eX
Equilibrium concentrations:
3
0.305 1 0.580.059
1 2 0.58A
molC
dm
4-22
3
3 0.305 0.580.246
1 2 0.58C
molC
dm
P4-10 Given: Gas phase reaction A + B 8C in a batch reactor fitted with a piston such that
V = 0.1P0 2
3
21.0
sec
ftk
lbmol
2
A A Br kC C
NA0 = NB0 at t = 0
V0 = 0.15 ft3
T = 140C = 600R = Constant
P4-10 (a)
00
0 0
0.5AAA B
Ny
N N
8 1 1 6
0 3Ay
Now 0 0
0
1V P
V XT
PT
and 0
1T
T, 0 010P V , and 10P V
Therefore 2
010 110
VV X
V or
2 2
0 1V V X
0 1A AN N X 0B A BN N X 0
0
1BBA
N
N
332
02
333
20
1
1
AA BA A B
kN XkN Nr kC C
VV X
0 00 0
AA
y PN V
RT
Therefore
4-23
3
0 0
3
2
1
1
AA
Xy Pr k
RTX
3
9
3 3
2
15.03*10
sec1 3
A
X lbmolr
ftX
P4-10 (b)
2 2
0 1V V X
2 20.2 0.15 1 X
0.259X
10
38.63*10
secA
lbmolr
ft
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.
= yA0 = 0.6 (3+5-2-3) = 1.8 is the correct value.
(3) The expression for CA and CB will therefore be,
0 1
1
A
A
C XC
X
0
21
3
1
A
B
C X
CX
(4) According to the system of units being used in the calculations, R = 0.0821 atm. Liter/
mol & Temperature = 700 K should be used.
5-1
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these materials .
5-2
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)
5-3
FB = 200X106 / (365 X 24 X 3600 X 28) lbmol/sec = 0.226 lbmol/sec
FAO = 0.026
0.452 / sec0.5
BF lbmolX
Also, 30.00415 /AOC lbmol ft
Now, we have from the example
0
0
11 ln
1
A
A
FV X
kC X
1 3
0.452 / sec 11 1 ln 1 0.5
3.07sec 0.00415 / 1 0.5
lbmolX
X lbmol ft
= 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
Dp2
Dp10.0775
1
0.750.1033
1
21
12
2
2 0.103 602
1 1 0.2410o
atmX X ft
L fty
P atm
Now, 02
1o
o
P
L, 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 (
198073.0
60
O
OAO
AORT
PyC )
5-4
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.
V1
P
P5-2 (f) Example 5-6 For turbulent flow
1
Dp and
1
P0
0112 1 1 1
2 02
1 1
1 5
5
P
P
PD
D P
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.
5-5
P5-4 We have to find the time required to cook spaghetti in Cuzco, Peru.
Location Elevation (km) Pressure (mm Hg) Boiling Point (C) Time (min)
Ann Arbor 0.21 739 99.2 15
Boulder 1.63 625 94.6 17
Cuzco 3.416 493 88.3 ?
Assume reaction is zero order with respect to spaghetti conversion:
E
ARTA
dCr k Ae
dt
so that
0
A
A
EC
RT
CC Ae t
For complete conversion (i.e.: well cooked) CA = 0 at time t.
Therefore
0
0
E
RTA
E
A RT
C tAe
Cte
A
0 1ln ln lnA
b
C Ek t
A R T
Now, plot the natural log of the cooking time versus 1/Tb and get a linear relationship. Extrapolation to
Tb = 88.3C = 361.45 K yields t = 21 minutes.
5-6
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.
5-7
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.
A B
k = 5 min-1 vo = 5 dm3/min.
Xexpected = 0.632 Xactual = 0.618
0
1ln
1
X
AO o
A Expected
F dX vV
r k X=
5 1ln 1.0
5 1 .632
Now,
For PFR, 1
1ln
1PV
X.(1)
For CSTR, 11( )
1
actualAO actualC
A actual
X XF X
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