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NEW MEXICO STATE UNIVERSITY
Department of Chemical Engineering
CHE 441 – Chemical Kinetics and Reactor Engineering
Spring 2008, Exam 1 (Closed Book/Notes)
Full Name (please print) SOLUTION
Social Security Number
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provide, or supplicate assistance from anyone during the course of this examination. Further, I understand that if
found guilty of ethical violations regarding my solution to this exam, I will be punished to the maximum extent
allowable described under "Academic Misconduct" in the NMSU Student Handbook. I understand the "Student
Code of Conduct" provides policies and procedures that will be followed by the faculty member administering this
examination should an accusation be made by the exam proctor or a fellow student. Finally, I accept and agree
to fulfill my responsibility to report (in writing) to the supervising professor should I observe any solicitation for
assistance or action that can be regarded as cheating within 24 hours of the completion of the exam.
Signature Date
Examination Rules
Closed Book & Notes
Clearly state ALL assumptions AND indicate where they apply in the solution
All pages must be stapled together in proper order at the conclusion of the exam. Any
pages not stapled to the cover page will not be graded.
Ch E 441 Spring 2008 Exam 1
Problem 1 (15 points)
Begin with the general mole balance and develop the design equation for a continuous stirred tank
reactor. Clearly state each assumption, and show how and where it applies to the solution. Simplify
for space time assuming a liquid-phase, elementary isomerization reaction of reactant A.
SOLUTION
general mole balance:
dt
dN
F rdV F
j
j
V
jo j
assume steady state operation F rdV F 0
j
V
jo j
assume perfect mixing F rV F 0
jo j j
solve for volume
j
jo j
r
F F
V
rate law for elementary isomerization
A A
r kC
liquid phase suggest constant
volumetric flowrate
A o A
F C
substitute
A
o Ao A
A
o Ao o A
kC
C C
kC
C C
V
solve for space time
C
C
k
V 1
A
Ao
o
Ch E 441 Spring 2008 Exam 1
Problem 3 (20 points)
a. Develop a relationship between the space time and the liquid phase effluent concentration from
a series of N equally sized (same volume ) CSTRs in which a first-order reaction is occuring.
b. Reduce the equation in the limit as N .
c. Identify the resulting expression relative to your knowledge of kinetics.
SOLUTION
(a) Material balance on component A about vessel i:
Ai
o i i 1
o
i
o
oi
i
r
V C X X
F
CV
Substitute rate law containing stoichiometry of = 0:
i
i 1 i
i
o i 0 i 1 0
i
kC
C C
kC
C 1 C C 1 C C
Written as
i
i
i 1
1 k
C
C
Since all reactors are the same size, the space time for each is the same, therefore:
N
N
N 1
3
2
2
1
1
0
N
0
1 k
C
C
C
C
C
C
C
C
C
C
Solving for the space time
C
C
k
N
N
1 N
N
0
N i
(b) In the limit, N ,
C
C
k
N
1 N
N
0
N
N
lim
1 N
N
0
N
0
1 N
N
0
1 N
N
0
C
C
C
C
ln
Nk
C
C
k
C
C
k
N
dN
d
(^)
X
ln
kC
C
C
ln
k
C
C
C
C
ln
Nk
C
C
k
N 0
0
1 N
N
0
N
0
1 N
N
0
N
N
lim
(c) This is the result obtained for a PFR
A Ao A
X
0 Ao A Ao
Ao
X
ln
kC
1 x
dx
kC
F
VC A
Ch E 441 Spring 2008 Exam 1
Problem 4 (25 points)
The gas-phase reaction A + 2B 2D is to be carried out in an isothermal plug flow reactor at 5.0 atm. The mole
fractions of the feed stream are A = 0.20, B = 0.50, and inerts = 0.30.
(a) What is the steady-state volumetric flow rate at any point in the reactor? (ignore pressure drop)
(b) What are the stoichiometric expressions for the concentrations of A, B, and D as a function of conversion
at any point along the reactor?
In Out C A = F A /
A F Ao -X F Ao F Ao (1-X) C Ao (1-X)/(1-0.2X)
B F Bo = B F Ao = (5/2) F Ao -2 X F Ao F Ao (2.5-2X) C Ao (2.5-2X)/(1-0.2X)
D FDo = C FAo = 0 +2 X FAo FAo (2X) CAo (2X)/(1-0.2X)
Inerts F Io = I F Ao = (3/2) F Ao 0 F Ao (1.5) C Ao (1.5)/(1-0.2X)
Total 5 FAo (5-X) FAo CAo (5-X)/(1-0.2X)
(c) What is the feed concentration (mol/ dm
3 ) of A if the feed temperature is 55°C?
(d) Determine how large the plug-flow reactor must be to achieve a conversion (based on A) of 0.70 if the
temperature in the reactor is uniform (55°C), volumetric feed rate is 50 dm
3 /min, and the rate law at 55°C
is – 𝑟 = 𝑘𝐶 𝐴
1 2 𝐶 𝐵 , where k is 2.5 L
/ mol
/min
(e) Plot concentrations, volumetric flow rate, and conversion as a function of reactor length (L = 7.6 cm).
(f) How large would a CSTR have to be to take the effluent from the PFR in part (d) and achieve a conversion
of 0.85 if the temperature of the CSTR is 55°C?
(g) How many 1-inch diameter tubes, 20 ft in length, packed with a catalyst, are necessary to achieve 95%
conversion of A using the original feed stream? Plot the pressure and conversion as a function of reactor
length. The particles are 0.5 mm in diameter and the bed porosity is 45%.
(h) Assuming the reaction is reversible, calculate the PFR volume required to achieve 70% of the equilibrium
conversion, and the CSTR size necessary to raise the conversion of the PFR effluent to 85% of the
equilibrium conversion if temperatures is constant at 100°C. The activation energy for the reaction is 30
kJ/mol, and the reaction equilibrium constant at 100°C is 10 (m
3 / kmol)
1/ .
SOLUTION
Solution must be submitted to the Ch E Main Office by the close of business on February 20, 2008.
Electronic solutions should be submitted by email attachment to drockstr@nmsu.edu.
Sign this page below, and staple to the hard copy you submit to the office.
assume properties of air
at 5 atm and 55°C
μ 197.7 10
− 7 ⋅
N s⋅
m
2
:= ⋅ MW 30
gm
mol
:= ⋅
assume a typical catalyst density (from text) (^) ρ c
120
lb
ft
3
:= ⋅ D p
:=0.5 mm⋅
tube dimensions (^) L tube
:= 20 ft⋅ D tube
:= 1 in⋅ ϕ :=0.
g c
4.17 10
8 ⋅
lb ft⋅
hr
2 ⋅lbf
:= ⋅ ρ o
P MW⋅
R T⋅
kg
m
3
:= = G
C Ao
ν o
⋅ ⋅MW
A c
kg
m
2 ⋅s
:= =
N tubes
:= 76 W cat
N tubes
L tube
⋅ π
D tube
2
2
⋅
⋅ ⋅( 1 −ϕ)
ρ c
kg
:= ⋅ =248.
β o
G
ρ o
g c
⋅ D p
⋅
1 −ϕ
ϕ
3
⋅
150 ⋅ ( 1 −ϕ)⋅μ
D p
+1.75 G⋅
⋅ 4.83 10
− 3 ×
atm
ft
:= = ⋅ α
2 β o
⋅
( 1 −ϕ) A c
⋅ ρ c
⋅ ⋅P
:=
D W Y( , ) kg
r A
Y ( 0 )
−
F Ao
ρ c
⋅
Y ( 1 )
⋅
α
2 Y 1
⋅
− 1 ε Y 0
⋅
:= ⋅ Y o
0
1
:= S rkfixed Y o
, 0 W cat
, , 1000 ,D ( ) :=
W S
0 〈 〉
:= X S
1 〈 〉
:= P S
2 〈 〉
:= X 1000
=0.
0 100 200
1
conversion
pressure
a :=− 1
P := 5 atm⋅ T := 328.15 K⋅ R 0.
L atm⋅
mol K⋅
:= ⋅ k 2.
L
mol
⋅min
:= ⋅ b :=− 2
d := 2
y
0
:= Θ B
y 1
y 0
:= Θ D
y 2
y 0
:= Θ I
y 3
y 0
:= ν o
50
L
min
:= ⋅
δ := 2 − 2 − 1 =− 1 ε y 0
:= ⋅ δ=−0.2 C Ao
y 0
⋅P
R T⋅
:= F Ao
C Ao
ν o
⋅ 1.
mol
min
:= = ⋅
C A
(X ) C Ao
1 −X
1 +ε ⋅X
:= ⋅ C B
(X ) C Ao
Θ B
b
a
− ⋅X
1 +ε ⋅X
:= ⋅
R 0.
L atm⋅
mol K⋅
:= ⋅
C D
(X ) C Ao
Θ D
d
a
− ⋅X
1 +ε ⋅X
:= ⋅ C I
(X ) C Ao
Θ I
1 +ε ⋅X
:= ⋅
K eq
10
m
3
10
3 ⋅mol
:= ⋅ k o
L
mol
⋅min
:= ⋅
E a
30
10
3 ⋅J
mol
:= ⋅ k k o
exp
E a
R
−
1
373.15 K⋅
1
328.15 K⋅
−
⋅
⋅ 9.
L
mol
⋅min
:= = ⋅
X := 0.5 Given C A
(X )
C B
⋅ (X )
C D
(X )
K eq
− = 0 X eq
:=Find X( ) =0.
r A
(X ) −k C A
(X )
C B
⋅ (X )
C D
(X )
K eq
−
:= ⋅ V PFR
F Ao
0
0.7 X eq
⋅
X
1
r A
− (X )
⌠
⎮
⎮
⌡
:= ⋅ d =8.841 L⋅
V CSTR
F Ao
⋅( 0.85 −0.7)X eq
⋅
r A
0.85 X eq
⋅ ( ) −
:= =7.287 L
