HW # 2
1.
General equation
Fin – F out + Generation = accumulation
Generation = (-rA) Δ V (1-ε)
FA, V – FA, V+ΔV –rA Δ V (1-ε) = 0
=>
2.
dFA
= −rA (1 − ε )
dV
Fi − Fout + G =
dN A
dt
Fi = 0
Gj = 2.5 × 10-7 mole/m2 s
C0, stink = 5 × 10-3 mole/m3
ν0 = 0.1 m3/s
Fout = ν0 Cs, out
V = 5 × 8 × 2.5 = 100 m3
A = 5 × 8 m2
(a) steady-state
dN A
dt
-ν0 Cs, out + Gj A = 0
-0.1×Cs, out + 2.5 × 10-7 × 40 = 0
Cs, out = 10-4 mole/m3
Fi − Fout + G =
(b) not steady state
Fi − Fout + G =
− Fout + G =
dN A
dt
dC S V
dt
−ν 0 CS + G j A
dC S
V
dt
ν
dC S
− 0 dt =
Gj A
V
(C S −
)
ν0
∫
3600
0
=
(−10 − 4 )dt = ∫
CS
CS , 0
dC S
(C S − 10 −7 )
CS = 2.34 × 10 -4 mole/m3
3.
AÆ B
X = 0.99
FA0 = 5 mol / h
ν0 = 10 dm3 / h
CA0 = 0.5 mole / dm3
For CSTR
V =
FA 0 X
− rA
dFA
=r A
dV
FA = FA0(1-X) Î dFA = -FA0dX
For PFR
FA0
dX
= −r A
dV
(a) -rA = k
k =0.05
0
For CSTR
V =
For PFR
5∫
0.99
0
5 × 0.99
= 99dm 3
0.05
dX
= dV
0.05 ∫
V = 99 dm3
(b) -rA = kCA
k = 0.0001s-1
ν 0 C A0 X
ν0X
F A 0 X FA 0 X
=
=
=
= 2750dm 3
r
kC
kC
X
k
X
− A
)
(1 − )
A
A0 (1 −
For CSTR
V=
For PFR
dFA
dX
= − FA 0
= rA = − kC A
dV
dV
dX
= kC A0 (1 − X )
ν 0 C A0
dV
ν 0 0.99 dX
= dV
k ∫0 (1 − X ) ∫
V= 127.92 dm 3
(c) –rA=kCA2
k= 3 dm3/mol h
For CSTR V =
FA0 X
ν 0 C A0 X
ν0X
=
=
= 66000dm 3
2
2
2
− rA
kC
(
1
−
X
)
kC A0 (1 − X )
A0
dFA
dX
2
2
= − FA 0
= rA = −kC A = −kC A0 (1 − X ) 2
dV
dV
ν0
dX
= dV
kC A0 (1 − X ) 2
For PFR
V = 660dm 3
4. ( for 3rd edition)
X
0
0.2
0.4
0.6
0.8
0.9
rA
10
16.67
50
50
12.5
90.9
1/rA
0.1
0.06
0.02
0.02
0.8
0.11
0.12
0.1
0.08
-1/rA
y = -0.2x + 0.1
y = 0.3x - 0.16
0.06
0.04
0.02
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
X
(a)
VCSTR =
FA0 X 300 × 0.4
=
= 2.4 dm3
− rA
50
X
VPFR = FA0
dX
∫−r
0
A
= 300 ×
(0.1 + 0.02)
× 0.4 = 7.2 dm3
2
(b) Between 0.4 ~ 0.6, because the rate is constant over this conversion range.
(c) VCSTR = 10.5 dm3
FA0 X
X
= 300 ×
− rA
= − rA
1
=>
X
=0.035
− rA
You can trial and error until
X
=0.035, or just calculate area from the plot.
− rA
From X=0.4
V1= 2.4
X=0.6
V2 =3.6
10.5 = 300 X (0.3X-0.16)
Maximum conversion X = 0.7
(d)
From part (a) we know X1 = 0.4
VCSTR = 2.4 = 300(X2-0.4)(0.3X2-0.16)
X= 0.64
(e)
From part (a) we know X1 = 0.4
VPFR = 7.2 = FA0
X2
X2
X
dX
dX
∫0 − rA = 300 0∫.4 − rA = 1.2 + 300 0∫.6(0.3 X − 0.16)dX
X2 = 0.905
(f)
55
50
45
40
Rate of Reaction
35
30
25
20
15
10
5
0
0
1
2
3
4
5
6
7
8
9
10
PFR Volume (dm3)
1
0.9
0.8
0.7
conversion
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
PFR Volume (dm3)
6
7
8
9
10
4. for 4th edition
X
0
0.2
0.4
0.6
0.8
0.9
rA
1
1.67
5.0
5.0
1.25
0.91
1/rA
1
0.6
0.2
0.2
0.8
1.1
1.2
1
0.8
-1/rA
y = -2x + 1
y = 3x - 1.6
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
X
(a)
VCSTR =
FA0 X 300 × 0.4
=
= 24 dm3
− rA
5
X
VPFR = FA0
dX
∫−r
0
= 300 ×
A
(1 + 0.2)
× 0.4 = 72 dm3
2
(b) Between 0.4 ~ 0.6, because the rate is constant over this conversion range.
(c) VCSTR = 10.5 dm3
FA 0 X
X
= 300 ×
− rA
= − rA
=>
X
=0.035
− rA
You can trial and error until
X
=0.035, or just calculate area from the plot.
− rA
From X=0.4
V1= 24
10.5 = 300 X (-2X+1)
Maximum conversion X = 0.0378
1
(d)
From part (a) we know X1 = 0.4
VCSTR = 24 = 300(X2-0.4)(3X2-1.6)
X= 0.64
(e)
From part (a) we know X1 = 0.4
X
VPFR = 72 = FA0
X2 = 0.905
X2
X2
dX
dX
∫0 − rA = 300 0∫.4 − rA = 12 + 300 0∫.6(3 X − 1.6)dX
(f)
6
Rate of Reaction
4
2
0
0
10
20
30
40
50
60
70
80
90
100
PFR Volume (dm3)
1
0.9
0.8
0.7
conversion
0.6
0.5
0.4
0.3
0.2
0.1
0
0
10
20
30
40
50
PFR Volume (dm3)
60
70
80
90
100
5.
(a) CSTR is used before PFR.
(b) One can calculate the amount of catalyst needed to carry out the same reaction to
80 % conversion using a single CSTR by determining the area of the rectangle.
90
80
70
60
-FA0/rA'
(kg catalyst)
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.6
0.7
0.8
0.9
1
conversion X
approximately 22.4 kg catalyst.
(c) for 40 % conversion
90
80
70
60
-FA0/rA'
(kg catalyst)
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
conversion X
Î 8 kg catalyst
(d)
90
80
70
60
-FA0/rA'
(kg catalyst)
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.9
0.8
1
conversion X
Calculate the area under the curve between 0 and 0.8. Approximately 30 kg
(e)
90
80
70
60
-FA0/rA'
(kg catalyst)
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
conversion X
Calculate the area under the curve between 0 and 0.4 Approximately 15 kg
0.9
1
(f)
(g) For different (-rA) vs. (X) curves, reactors should be arranged so that the smallest
amount of catalyst is needed to give the maximum conversion. This can be done
by minimizing the area that is occupied be a given reactor. One useful heuristic is
that for curves with a negative slope, it is generally better to use a CSTR.
Similarly, when the curve has a positive slope, it is generally better to use a PFR.