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Material Type: Exam; Class: Chemical Kinetics and Reactor Engineering; Subject: CHEMICAL ENGINEERING; University: New Mexico State University-Main Campus; Term: Spring 2008;
Typology: Exams
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Spring 2008, Exam 1 (Closed Book/Notes)
By my signature, I attest that the work contained within this exam is a result of my own efforts. I did not receive,
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.
Ch E 441 Spring 2008 Exam 1
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Ch E 441 Spring 2008 Exam 1
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Ch E 441 Spring 2008 Exam 1
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/ .
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
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