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Elements of Chemical Reaction Engineering instructor solution manual. It helps students analyzing problems and solve themselves
Typology: Exercises
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General : The goal of these problems are to reinforce the definitions and provide an
understanding of the mole balances of the different types of reactors. It lays the
foundation for step 1 of the algorithm in Chapter 4.
P1-1. This problem helps the student understand the course goals and objectives.
P1-2. Part (d) gives hints on how to solve problems when they get stuck. 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 and P1- 10. The results of these problems will appear in later chapters. Straight
forward application of chapter 1 principles.
P1-11. Straight forward modification of the mole balance. Assigned for those who
emphasize bioreaction problems.
P1-12. Can be assigned to just be read and not necessarily to be worked. It will give
students a flavor of the top selling chemicals and top chemical companies.
P1-13. 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- 17.
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
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,
Introduction to the CD-ROM 1,2,3,
Make a calculation 6
Open-ended 8,
The general equation for a CSTR is:
A
A A
r
0
Here rA is the rate of a first order reaction given by: rA = - kCA
Given : CA = 0.1CA0 , k = 0.23 min
Substituting in the above equation we get:
( 0. 23 min )( 0. 1 ( 0. 5 / ))
( 0. 5 / )( 10 /min) 0. 1 ( 0. 5 / )( 10 /min) 1 3
3 3 3 3 A0 0 0
mol dm
mol dm dm mol dm dm
kC
C v C v V
A
A !
V = 391.3 dm 3
1 k 0.23min
From mole balance:
dN A
dt
r A
Rate law: r A ! k C A
r A ! k
Combine:
dN A
dt
!k N A
0
t
1 t
d
k
N A
NA
at! = 0 , NAO = 100 mol and! =!, NA = (0.01)NAO
t = !
A
A
N
k
0 ln
= t = 20 min
Also rj = ρb rj` and W = Vρb where ρb is the bulk density of the bed.
=>
'
Hence the above equation becomes
0 '
j j
j
r
We can also just apply the general mole balance as
' ( 0 ) ( )
j j j j
dN F F r dW dt
Assuming no accumulation and no spatial variation in rate, we get the same form as
above:
0 '
j j
j
r
Mole balance on species j is:
V j j j j dt
dN F F rdV
0
0
Let Mj = molecular wt. of species j
Then F (^) j 0 Mj = wj 0 = mass flow rate of j into reactor:
N (^) j Mj = mj = mass of species j in the reactor
Multiplying the mole balance on species j by Mj
V j j j j j j j j dt
dN F M FM M rdV M
0
0
Now Mj is constant:
( ) ( )
dt
dm
dt
dM N F M FM M rdV
j
V j j
0
0
dt
dm w w M rdV
j
V
0
0
Applying mole balance to Penicillin: Penicillin is produced in the cells stationary state (See Chapter 7), so
there is no cell growth and the nutrients are used in making product.
Let’s do part c first.
[In flowrate (moles/time)] (^) penicillin + [generation rate (moles/time)]penicillin – [Out flowrate (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
Therefore,
V
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
r
Or,
p out ,
p
r
Similarly, for Corn Steep Liquor with FC = 0
C
C
C
C C
r
r
0 0
Assume RNA concentration does not change in the stationary state and no RNA is generated or destroyed.
Annual Production rate of ethylene for year 2002 is 5.21x 10 10 lb/year
Annual Production rate of benzene for year 2002 is 1.58 x 10 10 lb/year
Annual Production rate of ethylene oxide for year 2002 is 7.6 x 9 lb/year
Because the basic raw material ‘coal and petroleum’ for organic chemicals is very limited and their
production is not increasing as production of raw material for inorganic chemicals.
Type Characteristics Phases Usage Advantage Disadvantage
Batc
h
All the reactants
fed into the
reactor. During reaction nothing
is added or
removed. Easy
heating or cooling.
phase
Solid
pdn.
Conversion per
unit volume.
using for
multiple
reactions.
Operating
cost.
product
quality.
CST R
Continuous flow of reactants and
products.
Uniform composition
throughout.
liquid
liquid
Configuration possible for
different
configuration streams
2.Good
Temperature Control
reactions possible.
4.Good Control
operating cost
per unit
volume.
possible with
poor agitation.
3 High power
Input reqd.
PFR One long reactor
or number of CSTR’s in
series.
No radial variations. Conc.
changes along
the length.
Primarily gas Phase
pdn.
reactions
reactions
conversion per unit volume
maintain (No moving parts)
operating cost
operation
thermal gradient.
temperature control
and cleaning expensive.
PBR Tubular reactor
that is packed
with solid
catalyst particles.
Phase
(Solid
Catalyst) 2.Gas –
solid
reactions.
in the
heterogeneous gas
phase reaction with solid catalyst
e.g Fischer
tropsch synthesis.
conversion per
unit mass of
catalyst
operating cost
thermal
gradient.
control
Channeling
expensive.
Given
10 2 A = 2 * 10 ft TSTP = 491. 69 R H = 2000 ft
13 3 V = 4 * 10 ft T = 534.7 ϒR PO = 1atm
lbmolR
atmft R
3
= 0. 7302 yA = 0.
3
10
lbmol C (^) S
C = 4* 5 cars
FS = CO in Santa Ana wind FA = CO emission from autos
hr
ft v (^) A
3
= 3000 per car at STP
Total number of lb moles gas in the system:
N =
13 3
3
atm ft
atm ft R lbmol R
= 1.025 x 10 11 lb mol
Molar flowrate of CO into L.A. Basin by cars.
STP
A A A T A RT
v CP F y F y
0 = =!
cars ft
lbmol
hrcar
ft FT 400000 359
3
3 =!! (See appendix B)
FA = 6.685 x 10 4 lb mol/hr
N
P 0
⋅V
R⋅ T
:=
(1) to = 0 tf = 72 hrs
Cco = 2.00E-10 lbmol/ft 3 a = 3.50E+04 lbmol/hr
vo = 1.67E+12 ft 3 /hr b = 3.00E+04 lbmol/hr
Fs = 341.23 lbmol/hr V = 4.0E+13 ft 3
dt
dC F vC V
t a b
co " + s! o co =
sin (
Now solving this equation using POLYMATH we get plot between Cco vs t
See Polymath program P 1 - 14 - 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+ a 3.5E+04 3.5E+04 3.5E+04 3.5E+ b 3.0E+04 3.0E+04 3.0E+04 3.0E+ F 341.23 341.23 341.23 341. V 4.0E+13 4.0E+13 4.0E+13 4.0E+
ODE Report (RKF45)
Differential equations as entered by the user [1] d(C)/d(t) = (a+bsin(3.14t/6)+F-v0*C)/V
Explicit equations as entered by the user [1] v0 = 1.6710^ [2] a = 35000 [3] b = 30000 [4] F = 341. [5] V = 410^
(2) tf = 48 hrs Fs = 0 dt
dC vC V
t a b
co "! o co =
sin (
Now solving this equation using POLYMATH we get plot between Cco vs t
See Polymath program P 1 - 14 - 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+ a 3.5E+04 3.5E+04 3.5E+04 3.5E+ b 3.0E+04 3.0E+04 3.0E+04 3.0E+ V 4.0E+13 4.0E+13 4.0E+13 4.0E+
ODE Report (RKF45)
Differential equations as entered by the user [1] d(C)/d(t) = (a+bsin(3.14t/6)-v0*C)/V
Explicit equations as entered by the user [1] v0 = 1.6710^ [2] a = 35000 [3] b = 3000 0 [4] V = 410^
(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 v 0 As the value of v 0 is increased the graph changes to a “shifted sin-curve”. And as v 0 is decreased graph changes to a smooth increasing curve.
CSTR: The general equation is
A
A A
r
0
Here CA = 0.01CA0 , v 0 = 10 dm 3 /min, FA = 5.0 mol/hr
Also we know that FA = CAv 0 and FA0 = CA0v 0 , CA0 = FA0/ v 0 = 0.5 mol/dm 3
CSTR:
C A0 v 0 " C (^) A v 0
" rA
v 0 C (^) A 0 (1" 0.01)
kC (^) A
2
Substituting all the values we get
(10 dm
3 / hr )(0.5 mol / dm
3 )(0.99)
(3 dm
3 / hr )(0.01* 0.5 mol / dm
3 )
2 => V = 66000 dm
3
PFR:
dC (^) A v 0
dV
= rA = kC (^) A
2
Integrating
v 0
k
dC (^) A
2 C (^) A 0
C (^) A
0
V
v 0
k
=>
10 dm
3 / hr
3 dm
3 / mol. hr
) = 660 dm 3
CA = .001CA
t =
dN
N (^) A " rA V
N (^) A 0
Constant Volume V=V 0
t =
dC (^) A
C (^) A " rA
C (^) A 0
Zero order:
t =
k
[ C^ A 0 "^ 0.001 C^ A 0 ] =^
.999 C (^) Ao
= 9.99 h
First order:
t =
k
ln
ln
' = 6908 s
Second order:
t =
k
( =^666 h
Initial number of rabbits, x(0) = 500
Initial number of foxes, y(0) = 200
Number of days = 500
1 2
dx k x k xy dt
3 4
dy k xy k y dt
Given,
1 1 2 3 1 4
k day
k day foxes
k day rabbits
k day
!
!
See Polymath program P 1 - 17 - 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. y 200 1.1285722 4099.517 117. k1 0.02 0.02 0.02 0. k2 4.0E- 05 4.0E- 05 4.0E- 05 4.0E- 05 k3 4.0E- 04 4.0E- 04 4.0E- 04 4.0E- 04 k4 0.04 0.04 0.04 0.
ODE Report (RKF45)
Differential equations as entered by the user [1] d(x)/d(t) = (k1x)-(k2xy) [2] d(y)/d(t) = (k3xy)-(k4y)
Explicit equations as entered by the user [1] k1 = 0. [2] k2 = 0. [3] k3 = 0. [4] k4 = 0.
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^3y-4y^2+3x-1 = 0 [2] f(y) = 6y^2-9xy-5 = 0
No preheating of the benzene feed will diminish the rate of reaction and thus lesser conversions will be
achieved.
An interpolation can be done on the logarithmic scale to find the desired values from the given data.
Now we can interpolate to the get the cost at 6000 gallons and 15000 gallons
Cost of 6000 gal reactor = 1.905 x 10 5 $
Cost of 15000 gal reactor = 5.623 x 10 5 $
We are given CA is 0.1% of initial concentration
CA = 0.001CA
Also from Example 1.3,
0 0
A
A
Substituting v0 = 10 dm 3 /min and
k = 0.23 min
3 V = 300 dm
which is three times the volume of reactor used
in Example 1- 3.
Cost vs Volume of reactor (log - log plot)
0
1
2
3
4
5
6
0 0.5 1 1.5 2 2.5 3 3.5 4 4. log volume (gallons)
How many moles of A are in the reactor initially? What is the initial concentration of A?
If we assume ideal gas behavior, then calculating the moles of A initially present in the reactor is quite
simple. We insert our variables into the ideal gas equation:
3
3
PV atm dm kPa n moles RT (^) kPa dm atm
molK
Knowing the mole fraction of A (yAo) is 75%, we multiply the total number of moles (NTo) by the yA:
molesA = NAo = 0.75! 97.5 =73.
The initial concentration of A (CAo) is just the moles of A divided by the volume:
3 3
Ao Ao
moles N moles C moles dm volume V dm
Time (t) for a 1 st
A A
dC r dt
Our first order rate law is:! r (^) A = kCA
mole balance: A A
dC kC dt
0
A
Ao
t C A
C A
ln A
Ao
kt C
for the time of the reaction: 1
ln(0.01) 46.1min 0.1min
t k
!
Time for 2nd order reaction to consume 80% of A and final pressure (P) at T = 127 C.
rate law:
2 A A ! r = kC
mole balance:
A^2 A
dC kC dt
2 0
A
Ao
t C A
C A
dC k dt C
A Ao
kt C C
We can solve for the time in terms of our rate constant (k = 0.7) and our initial
concentration (CAo):
Ao Ao
kt C C