CE 561, Exam 2, December 11, 2009
This exam consists of three questions, each with multiple parts.
You should be careful not
to get stuck on one part.
If you do not know how to do a problem, move on and return to it
if you have time at the end.
If you cannot find the numerical answer to a problem, explain
how you would find the answer if you had more time.
You may use three pages (2sided) of
handwritten notes.
Carefully explain any assumptions you make, clearly indicate what part of what problem
you are working on, and define the symbols that you use.
The point value of each subpart
is indicated – budget your effort accordingly.
There are 100 points total.
Good luck.
1.
The reversible, exothermic reaction
A
↔
B, with rate =
k
1
C
A
–
k
2
C
B
is to be carried out in solution in a wellmixed
adiabatic
batch reactor.
At the start of each
batch, the reactor contains 100 L of a solution of A at a concentration of 5 mol per liter (
C
Ao
= 5 mol/L,
C
Bo
= 0) and a temperature of 300 K.
Emptying, cleaning, and refilling the
reactor between batches requires 2 hours.
The reactor contents have a density of 1 kg per
liter and a specific heat of 4200 J/(kg K). The enthalpy of reaction is 42 kJ/mol. The
forward and reverse rate constants are given by
k
1
= 5×10
5
exp(5000/
T
)
hr
1
k
2
= 10
12
exp(10000/
T
)
hr
1
where
T
is the temperature in Kelvins.
(a)
Write the species mole balance and energy balance equations for the reactor and derive
relationships that allow you to write the temperature and both species concentrations in
terms of a single concentration or conversion variable. (5 pts.)
The species mole balances are
1
2
1
2
A
A
B
B
A
B
dC
r
k C
k C
dt
dC
r
k C
k C
dt
=−
=−
+
=+
=
−
And the energy balance is
(
)
1
2
ˆ
ˆ
A
B
p
p
dT
H
H
r
k C
k C
dt
C
C
ρ
ρ
−∆
−∆
=
=
−
Adding the species balances, we obtain
(
)
0
A
B
d
C
C
dt
+
=
From which
5 mol/L
A
B
Ao
C
C
C
+
=
=
If we define the conversion
x
as
Ao
A
Ao
C
C
x
C
−
=
Then we have
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(
)
5 1
mol/L
C
5
mol/L
A
B
C
x
x
=
−
=
This is one way of writing all of the concentrations in terms of a single conversion variable.
Likewise, we can combine the species A mole balance with the energy balance to obtain
(
)
ˆ
ˆ
ˆ
Ao
A
p
p
p
H C
dC
dT
H
H
dx
r
dt
dt
dt
C
C
C
ρ
ρ
ρ
−∆
−∆
−∆
=
=
−
=
From which
(
)
ˆ
Ao
o
p
H C
T
T
x
C
ρ
−∆
=
+
Evaluating the second term gives
(
)
42000*5
50
K
ˆ
1*4200
Ao
p
H C
C
ρ
−∆
=
=
So,
T
= 300 + 50
x
(b)
Compute the maximum conversion of A to B that can be obtained in this adiabatic batch
reactor for the conditions given. (10 pts.)
The equilibrium conversion is attained when the net reaction rate is zero.
Thus, we want to
set
k
1
C
A
–
k
2
C
B
= 0.
Substituting into this the expressions for the rate constants in terms of
T
gives
5×10
5
exp(5000/
T
)
C
A
= 10
11
exp(10000/
T
)
C
B
Writing the temperature and all of the concentrations in terms of the fractional conversion
x
gives
(
)
(
)
5
12
7
5000
10000
5
10 exp
*5 1
10
exp
*5
300
50
300
50
5000
5
10
1
exp
300
50
x
x
x
x
x
x
x
−
−
−
×
−
=
+
+
=
×
−
+
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 Fall '09
 Enthalpy, Chemical reaction, Trigraph, Qh, residence time

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