Tutorial 2 Reaction Engineering
Question Solutions 1 – 3
1.
(
29
A
batch
X
k
t


=
1
ln
1
=
(
29
min
9
.
160
8
.
0
1
ln
01
.
0
1
=


t
cycle
= t
batch
+ t
fill
+ t
empty
+ t
clean
=
hr
3
.
4
min
4
.
258
60
10
250
20
250
9
.
160
=
=
+
+
+
2.
Starting from the CSTR design equation:
(
29
(
29
A
A
A
A
A
A
A
A
r
X
kC
X
kC
X
r
X
F
V

=
=

=
1
0
0
Here we make use of our old friend:
C
A
= C
A0
(1X
A
)
Now we have to rearrange to make X
A
the subject of the equation:
(
29
A
A
A
A
r
X
X
F
kC
V

=
1
0
0
0
0
0
0
1
A
A
r
A
A
r
A
F
kC
V
F
kC
V
X
+
=
Now we note that F
A0
=
v
0
C
A0
,
that is molar flow = volumetric flow x molar concentration
and therefore
0
0
1
v
k
V
v
k
V
X
r
r
A
+
=
We substitute the known data in to the equation:
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.
0
04
.
0
)
02
.
0
(
8
.
1
1
04
.
0
)
02
.
0
(
8
.
1
=
+
=
A
X
47.3 % of the inlet reactant disappears to form product.
The outlet concentration is calculated from C
A
= C
A0
(1X
A
) = 15(10.473) = 7.905
Kmol m
3
.
3.
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 Spring '10
 reddy
 Reaction

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