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Material Balance
CHAPTER NO. 4
MATERIAL BALANCE
4.1 Preliminary Calculations: Basis: = 60 metric tons /day of 99.5% pure Maleic Anhydride = Oxidation yield is 92 %wt based on benzene fed = Selectivity of Maleic Anhydride is73 %wt = Recovery and purification yield is 98 %wt = 1 hr operation 60 tons of 99.5% purity Day
99.5 tons of pure MAN
1000 kg
100 tons of 99.5% pure MAN
1 metric ton
1 day 24 hr
= 2487.5 kg of Maleic Anhydride / hr Overall loses = 2% (Assuming 1% loss in scrubber and 1% loss in distillation column + producing Fumaric Acid) MAN to be produced = 2487.5 + (2487.5 x .02) = 2536.7 kg MAN / hr Conversion in the reactor is 73 %wt Consider the main reaction
C6H6 +4.5O2 (78)
C4H2O3 + 2CO2 +2H2O (98) 27
Material Balance
CHAPTER NO. 4
For 2536.7 kg/hr of MAN benzene required = (78 x 2536.7) / 98
= 2019 kg of benzene/hr
As 73 %wt benzene converts to produce MAN, Benzene actually required = 2019 / 0.73
= 2768.8 kg of benzene / hr
For given process Benzene : Air = 1: 24
By wt
Air required for oxidation = 2768.8 x 24 = 66445.5 kg of Air / hr = 2295.4 kgmol of Air / hr As Air has normally 21 %mole Oxygen and 79 %mole Nitrogen Oxygen in Air = 2295.4 x 0.21 = 482 kgmole of Oxygen / hr Nitrogen in Air = 2295.4 x 0.79 = 1813.2 kgmole of Nitrogen / hr Assuming 65 % Air humidity at 30 oC From Humidity chart = 17.5 gm moisture / kg dry air Moisture in Air = (17.5 x 66379.3) / 1000 = 1162.78 kg of water / hr
28
Material Balance
CHAPTER NO. 4
Stream 4 and 6 has Composition: Kg
%wt
kgmole
%mole
Nitrogen
50787.6
75.38
1813.2
76.84
Oxygen
15423.6
22.89
482
20.42
Water
1162.7
1.73
64
2.74
Total
67374.1
2359.8
4.2 Material Balance around Mixer MIX-100:
(3) 2768.8
(6) +
(7)
67374.1
=
70142.6
kg/hr
Stream (7) has composition:
Nitrogen Oxygen Benzene Water
kg 50787.6 15423.6 2768.5 1162.7 70142.6
%wt 72.40 21.98 3.94 1.65
kgmole 1813.2 482 35.44 64.38
%mole 75.70 20.12 1.47 2.68
2395
4.3 Material Balance around Reactor PFR-100: 29
Material Balance
CHAPTER NO. 4
In Reactor operating conditions are: Temperature 350 – 400 oC Pressure
202KPa
Oxidation yield
92 %wtMAN
selectivity
73 %wt
Of Benzene feed 73% is converted to MAN, 1.6% to Quinone and 17.4% is converted to CO2 while 8% Benzene feed leaves along with product stream as unconverted Benzene.
Chemical Reactions: C6H6 + 4.5O2
C4H2O3 + 2CO2 + 2H2O (R1)
C6H6 + 7.5O2
6CO2 + 3H2O
(R2)
C6H6 + 1.5O2
C6H4O2 + H20
(R3)
Consider Reaction (R1) Benzene converted to MAN
= 2768.5 x 0.73 = 2021 kg of Benzene/hr = 25.87 kgmole of Benzene/hr
MAN produced
= 25.87 kgmole of MAN/hr
Oxygen required
= 25.87 x 4.5 = 116.43 kgmole of O2/hr
CO2 produced
= 25.87 x 2 = 51.75 kgmole of CO2/hr
Water produced
= 25.87 x 2 = 51.75 kgmole of water/hr
30
Material Balance
CHAPTER NO. 4
Consider Reaction (R2) Benzene converted to CO2
= 2768.5 x 0.174
= 481.73 kg of Benzene/hr
= 6.17 kgmole of Benzene/hr Oxygen required
= 6.17 x 7.5 = 46.25 kgmole of Oxygen/hr
CO2 produced
= 6.17 x 6
= 37.01 kgmole of CO2/hr
Water produced
= 6.17 x 3
= 18.50 kgmole of water/hr
Consider Reaction (R3) Benzene converted to Quinon = 2768.5 x 0.016
= 44.30 kg of Benzene/hr
= 0.57 kgmole of Benzene/hr Oxygen required
= 0.57 x 1.5 = 0.85 kgmole of oxygen
Quinone produced
= 0.57 kgmole of Quinon/hr
Water produced
= 0.57 kgmole of water/hr
Total Oxygen consumed
= 163.54 kgmole of Oxygen/hr
Unreacted Oxygen
= 318.45 kgmole of Oxygen/hr
Total Water produced
= 70.82 kgmole of Water/hr
Water in product stream
=
CO2 in product stream
= 88.75 kgmole of CO2/hr
135.20 kgmole of water/hr
Composition of stream (8) kg
% wt
kgmole
% mole
Nitroge n
50787.6 4
72.41
1813.19
76.04
Oxygen
10190.3
14.52
318.44
13.35 31
Material Balance
CHAPTER NO. 4
4 Water
2433.66
3.46
134.75
5.65
CO2
3905.13
5.56
88.75
3.72
MAN
2537.25
3.61
25.87
1.08
Benzene
221.48
0.31
2.83
0.11
Quinone
61.30
0.08
0.56
0.02
69976.5 2
2379.36
4.4 Material Balance around Crude MAN Separator V-100:
Operating conditions: Operating Pressure
= 152kpa
Operating temperature
= 76.5oC
Of stream (9) 41.17 %wt crude MAN is separated as stream (10) and remaining MAN is sent along with gases in stream (11).
32
Material Balance
CHAPTER NO. 4
Stream (10) contains: Maleic Anhydride
= 4225.31 x 0.4117
= 1739.56 kg of MAN/hr
Water condensed
= 30.22 kg of water/hr
Quinone condensed
= 24.87 kg of Quinon/hr
Composition of stream (10) Kg
%wt
kgmole
%mole
MAN
1267.1
93.9
12.92
78.34
Water
60.94
4.5
3.38
20.50
Quinone
20.4
1.5
0.19
1.14
Total
1348.47
16.49
Composition of stream (11) Kg
%wt
kgmole
%mole
Nitrogen
50664.59
73.82
1808.8
76.55
Oxygen
10155.5
14.79
317.35
13.43
Water
2376.8
3.46
131.89
5.58
CO2
3902.8
5.68
88.68
3.75
MAN
1267.1
1.84
12.92
0.55
Benzene
221.4
0.32
2.83
0.12
Quinone
39.8
0.06
0.37
0.02
Total
68628.1
2362.87
33
CHAPTER NO. 4
Material Balance
34
Material Balance
CHAPTER NO. 4
4.5 Material Balance around Scrubber T-100:
Absorption is done in 40 %wt solution of Maleic Acid (MAC). Let 1% of the MAN produced in the reactor is lost in vent stream. All the Benzene and 1% MAN leaves from the top along with vent stream while Quinone leaves with scrubbed liquid. MAN lost MAN converted to MAC
= 2537.35 x0 .01 = 25.37 kg of MAN/hr = 1268.75 – 25.37 = 1243.25 kg of MAN/he = 12.67 kgmole of MAN/hr
Operating Conditions: Operating Temperature: 45oC Operating pressure:
135.2Kpa
Assuming Liquid to vapor ratio in tower as 1.27 Stream (12) is G = 68813.1 kg/hr Stream (15) will be
L = 68813.1 x 1.27 = 87392.6kg of 40 % MAC solution/hr
Reaction taking place: C4H2O3 (g) + H2O(l) (98)
(18)
C4H4O4(l) (116) 35
Material Balance
CHAPTER NO. 4
As absorption is done in 40% solution of MAC stream so (15) has composition: Kg
%wt
kgmole
MAC
34957.0
40
301.6
Water
52435.3
60
2913
87392.3
%mole 9.36 90.63
3214.6
MAC produced
= 12.67 kgmole of MAC/hr
Water consumed
= 12.67 kgmole of water/hr
MAC leaving in stream (13) = 314.2 kgmole of MAC/hr Water left in column = (133.52 + 2913) – 12.67 = 3033 kgmole of water/hr Total moles in vent stream excluding water: Kg
kgmole
Nitrogen
50787.9
1813.9
Oxygen
10190.0
318.5
CO2
3905.1
88.7
Benzene
221.5
2.83
MAN
25.37
0.25
Total
65129.4
2224.5
Assuming gases leaving in vent stream (14) are saturated with water vapors Partial pressure of water vapors at 50 oC and 136 Kpa = 12.55 Kpa As at saturation condition partial pressure is equal to vapor pressure and pressure ratio is same as molar ratio yH2O
=
PH2O/P
= 12.55/136
= 0.092
36
Material Balance
CHAPTER NO. 4
And for rest of gases excluding water y rest
(O2,
N2,
CO, Benzene, 1% MAN
CO2)
= 0.907
Total moles in vent stream (14)
= 2224.5/0.907
= 2450 kgmole/hr
Water in vent stream (14)
= 2450 – 2224.5
= 226.57 kgmole of water/hr
Water in scrubbed liquid (13) = 3033– 226.57
= 2832.62 kgmole of water/hr
Composition of vent stream (14): Kg
%wt
kgmole
%mole
Nitrogen
50787.9
73.81
1813.4
74.78
Oxygen
10190.9
14.82
318.05
13.12
CO2
3905.0
6.67
88.58
3.63
Water
3623.61
5.27
201.5
8.31
Benzene
221.6
0.33
2.83
0.12
MAN
25.37
0.04
0.25
0.01
Total
68753.0
2425.72
Composition of scrubbed liquid stream (13): Kg
%wt
kgmole
%mole
Water
50987.58
58.30
2832.43
90.01
MAC
36428.58
41.49
313.81
9.97
Quinone
36.45
0.04
0.33
0.01
Total
87452.7
3146.8
37
Material Balance
CHAPTER NO. 4
4.6 Material Balance around Mixer MIX-101:
(13)
+
(16)
=
(15)
+
(17)
As stream (15) and (17) contains 40% solution of MAC. While (16) contains pure water to ensure the composition of stream (15) and (17) at 40%wt of MAC. MAC in stream (13)
= 36428 kg of MAC/hr
40% solution will have the mass
= 36428.17/0.40 = 91071.42 kg of MAC solution/hr
So water added from stream (16)
=
91071.42 – 87452.1
= 3619.28 kg of water/hr Stream (17)
= 87452.15 + 3619.28 – 87392.1 = 3679.0 kg of MAC solution/hr
Water in stream (17)
= 3619.0 + 50987.2 – 52435 = 2170.8 kg of water/hr
MAC in stream (17)
= 36428.17 – 34597 = 1471.1 kg of MAC/hr
38
Material Balance
CHAPTER NO. 4
Composition of stream (17) Kg
%wt
kgmole
%mole
Water
2170.8
59.00
120.60
90.25
MAC
1471.1
40.00
12.67
9.480
Quinone
36.45
0.99
0.33
0.25
Total
3678.5
133.62
4.7 Material Balance around Dehydrator T-101:
In dehydrator Maleic Acid is converted back to Maleic Anhydride according to the following reaction:
C4H4O4
C4H2O3 + H20
In addition to above reaction, some Maleic Acid may irreversibly convert to Fumaric Acid according to the following reaction:
C4H2O3 + H2O
C4H4O4
C4H4O4
Maleic Acid
Fumaric Acid 39
Material Balance
CHAPTER NO. 4
Assuming 0.5% MAC converts irreversibly to Fumeric Acid and rest 99.5% converts back to Maleic Anhydride. And 95% of water is removed by azeotropic distillation from the dehydrator. Operating Conditions: Operating pressure
= 15.2 KPa
Azeotropic composition: Water = 77.5 %mole
&
MAC coming from stream (18)
O- xylene =
22.5 %mole
= 1736.78 kg of MAC/hr
0.5% is converted to Fumaric Acid Fumaric Acid produced
= 2889.17 x 0.005= 8.68 kg of Fumaric Acid/hr
MAC converted to MAN
= 1736.78 x 0.995
= 1728.095 kg of MAC/hr
= 14.89 kgmole of MAC/hr MAN produced
= 14.89 kgmole of MAN/hr
Water produced
=
Total water in Dehydrator O-xylene
14.89 kgmole of water/hr
= 14.89 + 142.71
= 157.60 kgmole of water/hr
= 157.60 x (22.5 / 77.5) = 45.75 kgmole of O-xylene/hr
As 95 % water is removed from the dehydrator, Stream (19) contains Water
= 157.60 x 0.95
= 149.72 kgmole of water/hr
O-xylene
= 149.72 x (22.7 / 77.5) = 43.47 kgmole of O-xylene/hr
40
Material Balance
CHAPTER NO. 4
Composition of stream (19) Kg
%wt
kgmole
%mole
Water
2694.89
36.87
149.72
77.5
O-xylene
4614.78
63.13
43.46
22.5
Total
7309.67
193.18
Assuming stream (22) contains pure water and pure O-xylene stream (21) is refluxed back to dehydrator. O-xylene in stream (26)
= 45.75– 43.47
= 2.29 kgmole of O-xylene/hr
So make up O-xylene required (23) = 2.29 kgmole of O-xylene/hr Water in stream (22) = 149.72 kgmole of water/hr Composition of stream (26) Kg
%wt
kgmole
%mole
Water
141.84
7.46
7.90
30.78
MAN
1460.7
76.93
14.89
58.34
Fumaric Acid
8.68
0.44
.07
0.28
O-xylene
242.88
12.74
2.289
8.93
Quinone
36.43
2.43
0.338
1.67
Total
1890.54
25.47
41
Material Balance
CHAPTER NO. 4
4.8 Material Balance around Mixer MIX-102
Mixer MIX-102 mixes crude MAN from stream (10) with crude MAN obtained from Bottom of dehydrating column stream (27)
(27) 1890.54
(10) +
(28)
1099.68
=
2990.21
Composition of stream (28) kg
%wt
Kg mole
%mole
water
172.05
5.75
9.55
25.13
MAN
2505.28
83.78
25.54
67.16
O-xylene
242.88
8.12
2.28
6.01
Quinone
61.30
2.05
0.56
1.49
Fumeric acid
8.68
0.29
0.07
0.19
Total
2990.21
38.03
42
Material Balance
CHAPTER NO. 4
4.9 Material Balance around Distillation column T-102 30
32 31
29
36
Bottom product stream (37) contains 99.5 % pure MAN. 0.5% impurity will be all the Fumeric Acid (BP = 290oC) produced in Dehydrating column and remaing is Quinone (BP = 180oC).
Operating conditions: Top plate temperature = 136oC Re-boiler temperature = 220oC Operating pressure
= 110Kpa
MAN converted to Fumaric Acid
= 8.68 x 98.06 / 116.07 = 7.333 kg of MAN/hr = 0.30 % of initial MAN produced
MAN lost in distillation column
= 2505.29 x 0.007 = 17.54 kg of MAN/hr
MAN in bottom product stream (37) = 2505.29 – 17.54= 2487.75 kg of MAN/hr 43
Material Balance
CHAPTER NO. 4
Bottom product of distillation column= 2487.75.26 / 0.995 = 2500.25 kg of product/hr Quinone in Bottom product
= 2500.25 – (2487.75 + 8.68) = 3.82 kg of Quinone/hr
Composition of bottom product stream (36) kg
%wt
Kg mole
%mole
MAN
2487.74
99.5
25.36
99.56
Fumaric acid
8.68
0.34
0.07
0.29
Quinone
3.81
0.15
0.03
0.14
Total
2500
25.47
Composition of overhead product stream (32) kg
%wt
Kg mole
%mole
Water
172.05
35.11
9.55
76.12
MAN
17.53
3.58
0.17
1.42
O-xylene
242.88
49.57
2.28
18.21
Quinone
57.48
11.73
0.53
4.23
Total
489.9637
100
12.557
100
For required purity, Reflux ratio L/D So
= 6
reflux rate as stream (31) = 6 x 489.96 = 2939.76Kg/hr Overhead product as stream (30)
= 7 x 489.96 = 3429.72 Kg/hr
44
CHAPTER NO. 4
MATERIAL BALANCE
4.1 Preliminary Calculations: Basis: = 60 metric tons /day of 99.5% pure Maleic Anhydride = Oxidation yield is 92 %wt based on benzene fed = Selectivity of Maleic Anhydride is73 %wt = Recovery and purification yield is 98 %wt = 1 hr operation 60 tons of 99.5% purity Day
99.5 tons of pure MAN
1000 kg
100 tons of 99.5% pure MAN
1 metric ton
1 day 24 hr
= 2487.5 kg of Maleic Anhydride / hr Overall loses = 2% (Assuming 1% loss in scrubber and 1% loss in distillation column + producing Fumaric Acid) MAN to be produced = 2487.5 + (2487.5 x .02) = 2536.7 kg MAN / hr Conversion in the reactor is 73 %wt Consider the main reaction
C6H6 +4.5O2 (78)
C4H2O3 + 2CO2 +2H2O (98) 27
Material Balance
CHAPTER NO. 4
For 2536.7 kg/hr of MAN benzene required = (78 x 2536.7) / 98
= 2019 kg of benzene/hr
As 73 %wt benzene converts to produce MAN, Benzene actually required = 2019 / 0.73
= 2768.8 kg of benzene / hr
For given process Benzene : Air = 1: 24
By wt
Air required for oxidation = 2768.8 x 24 = 66445.5 kg of Air / hr = 2295.4 kgmol of Air / hr As Air has normally 21 %mole Oxygen and 79 %mole Nitrogen Oxygen in Air = 2295.4 x 0.21 = 482 kgmole of Oxygen / hr Nitrogen in Air = 2295.4 x 0.79 = 1813.2 kgmole of Nitrogen / hr Assuming 65 % Air humidity at 30 oC From Humidity chart = 17.5 gm moisture / kg dry air Moisture in Air = (17.5 x 66379.3) / 1000 = 1162.78 kg of water / hr
28
Material Balance
CHAPTER NO. 4
Stream 4 and 6 has Composition: Kg
%wt
kgmole
%mole
Nitrogen
50787.6
75.38
1813.2
76.84
Oxygen
15423.6
22.89
482
20.42
Water
1162.7
1.73
64
2.74
Total
67374.1
2359.8
4.2 Material Balance around Mixer MIX-100:
(3) 2768.8
(6) +
(7)
67374.1
=
70142.6
kg/hr
Stream (7) has composition:
Nitrogen Oxygen Benzene Water
kg 50787.6 15423.6 2768.5 1162.7 70142.6
%wt 72.40 21.98 3.94 1.65
kgmole 1813.2 482 35.44 64.38
%mole 75.70 20.12 1.47 2.68
2395
4.3 Material Balance around Reactor PFR-100: 29
Material Balance
CHAPTER NO. 4
In Reactor operating conditions are: Temperature 350 – 400 oC Pressure
202KPa
Oxidation yield
92 %wtMAN
selectivity
73 %wt
Of Benzene feed 73% is converted to MAN, 1.6% to Quinone and 17.4% is converted to CO2 while 8% Benzene feed leaves along with product stream as unconverted Benzene.
Chemical Reactions: C6H6 + 4.5O2
C4H2O3 + 2CO2 + 2H2O (R1)
C6H6 + 7.5O2
6CO2 + 3H2O
(R2)
C6H6 + 1.5O2
C6H4O2 + H20
(R3)
Consider Reaction (R1) Benzene converted to MAN
= 2768.5 x 0.73 = 2021 kg of Benzene/hr = 25.87 kgmole of Benzene/hr
MAN produced
= 25.87 kgmole of MAN/hr
Oxygen required
= 25.87 x 4.5 = 116.43 kgmole of O2/hr
CO2 produced
= 25.87 x 2 = 51.75 kgmole of CO2/hr
Water produced
= 25.87 x 2 = 51.75 kgmole of water/hr
30
Material Balance
CHAPTER NO. 4
Consider Reaction (R2) Benzene converted to CO2
= 2768.5 x 0.174
= 481.73 kg of Benzene/hr
= 6.17 kgmole of Benzene/hr Oxygen required
= 6.17 x 7.5 = 46.25 kgmole of Oxygen/hr
CO2 produced
= 6.17 x 6
= 37.01 kgmole of CO2/hr
Water produced
= 6.17 x 3
= 18.50 kgmole of water/hr
Consider Reaction (R3) Benzene converted to Quinon = 2768.5 x 0.016
= 44.30 kg of Benzene/hr
= 0.57 kgmole of Benzene/hr Oxygen required
= 0.57 x 1.5 = 0.85 kgmole of oxygen
Quinone produced
= 0.57 kgmole of Quinon/hr
Water produced
= 0.57 kgmole of water/hr
Total Oxygen consumed
= 163.54 kgmole of Oxygen/hr
Unreacted Oxygen
= 318.45 kgmole of Oxygen/hr
Total Water produced
= 70.82 kgmole of Water/hr
Water in product stream
=
CO2 in product stream
= 88.75 kgmole of CO2/hr
135.20 kgmole of water/hr
Composition of stream (8) kg
% wt
kgmole
% mole
Nitroge n
50787.6 4
72.41
1813.19
76.04
Oxygen
10190.3
14.52
318.44
13.35 31
Material Balance
CHAPTER NO. 4
4 Water
2433.66
3.46
134.75
5.65
CO2
3905.13
5.56
88.75
3.72
MAN
2537.25
3.61
25.87
1.08
Benzene
221.48
0.31
2.83
0.11
Quinone
61.30
0.08
0.56
0.02
69976.5 2
2379.36
4.4 Material Balance around Crude MAN Separator V-100:
Operating conditions: Operating Pressure
= 152kpa
Operating temperature
= 76.5oC
Of stream (9) 41.17 %wt crude MAN is separated as stream (10) and remaining MAN is sent along with gases in stream (11).
32
Material Balance
CHAPTER NO. 4
Stream (10) contains: Maleic Anhydride
= 4225.31 x 0.4117
= 1739.56 kg of MAN/hr
Water condensed
= 30.22 kg of water/hr
Quinone condensed
= 24.87 kg of Quinon/hr
Composition of stream (10) Kg
%wt
kgmole
%mole
MAN
1267.1
93.9
12.92
78.34
Water
60.94
4.5
3.38
20.50
Quinone
20.4
1.5
0.19
1.14
Total
1348.47
16.49
Composition of stream (11) Kg
%wt
kgmole
%mole
Nitrogen
50664.59
73.82
1808.8
76.55
Oxygen
10155.5
14.79
317.35
13.43
Water
2376.8
3.46
131.89
5.58
CO2
3902.8
5.68
88.68
3.75
MAN
1267.1
1.84
12.92
0.55
Benzene
221.4
0.32
2.83
0.12
Quinone
39.8
0.06
0.37
0.02
Total
68628.1
2362.87
33
CHAPTER NO. 4
Material Balance
34
Material Balance
CHAPTER NO. 4
4.5 Material Balance around Scrubber T-100:
Absorption is done in 40 %wt solution of Maleic Acid (MAC). Let 1% of the MAN produced in the reactor is lost in vent stream. All the Benzene and 1% MAN leaves from the top along with vent stream while Quinone leaves with scrubbed liquid. MAN lost MAN converted to MAC
= 2537.35 x0 .01 = 25.37 kg of MAN/hr = 1268.75 – 25.37 = 1243.25 kg of MAN/he = 12.67 kgmole of MAN/hr
Operating Conditions: Operating Temperature: 45oC Operating pressure:
135.2Kpa
Assuming Liquid to vapor ratio in tower as 1.27 Stream (12) is G = 68813.1 kg/hr Stream (15) will be
L = 68813.1 x 1.27 = 87392.6kg of 40 % MAC solution/hr
Reaction taking place: C4H2O3 (g) + H2O(l) (98)
(18)
C4H4O4(l) (116) 35
Material Balance
CHAPTER NO. 4
As absorption is done in 40% solution of MAC stream so (15) has composition: Kg
%wt
kgmole
MAC
34957.0
40
301.6
Water
52435.3
60
2913
87392.3
%mole 9.36 90.63
3214.6
MAC produced
= 12.67 kgmole of MAC/hr
Water consumed
= 12.67 kgmole of water/hr
MAC leaving in stream (13) = 314.2 kgmole of MAC/hr Water left in column = (133.52 + 2913) – 12.67 = 3033 kgmole of water/hr Total moles in vent stream excluding water: Kg
kgmole
Nitrogen
50787.9
1813.9
Oxygen
10190.0
318.5
CO2
3905.1
88.7
Benzene
221.5
2.83
MAN
25.37
0.25
Total
65129.4
2224.5
Assuming gases leaving in vent stream (14) are saturated with water vapors Partial pressure of water vapors at 50 oC and 136 Kpa = 12.55 Kpa As at saturation condition partial pressure is equal to vapor pressure and pressure ratio is same as molar ratio yH2O
=
PH2O/P
= 12.55/136
= 0.092
36
Material Balance
CHAPTER NO. 4
And for rest of gases excluding water y rest
(O2,
N2,
CO, Benzene, 1% MAN
CO2)
= 0.907
Total moles in vent stream (14)
= 2224.5/0.907
= 2450 kgmole/hr
Water in vent stream (14)
= 2450 – 2224.5
= 226.57 kgmole of water/hr
Water in scrubbed liquid (13) = 3033– 226.57
= 2832.62 kgmole of water/hr
Composition of vent stream (14): Kg
%wt
kgmole
%mole
Nitrogen
50787.9
73.81
1813.4
74.78
Oxygen
10190.9
14.82
318.05
13.12
CO2
3905.0
6.67
88.58
3.63
Water
3623.61
5.27
201.5
8.31
Benzene
221.6
0.33
2.83
0.12
MAN
25.37
0.04
0.25
0.01
Total
68753.0
2425.72
Composition of scrubbed liquid stream (13): Kg
%wt
kgmole
%mole
Water
50987.58
58.30
2832.43
90.01
MAC
36428.58
41.49
313.81
9.97
Quinone
36.45
0.04
0.33
0.01
Total
87452.7
3146.8
37
Material Balance
CHAPTER NO. 4
4.6 Material Balance around Mixer MIX-101:
(13)
+
(16)
=
(15)
+
(17)
As stream (15) and (17) contains 40% solution of MAC. While (16) contains pure water to ensure the composition of stream (15) and (17) at 40%wt of MAC. MAC in stream (13)
= 36428 kg of MAC/hr
40% solution will have the mass
= 36428.17/0.40 = 91071.42 kg of MAC solution/hr
So water added from stream (16)
=
91071.42 – 87452.1
= 3619.28 kg of water/hr Stream (17)
= 87452.15 + 3619.28 – 87392.1 = 3679.0 kg of MAC solution/hr
Water in stream (17)
= 3619.0 + 50987.2 – 52435 = 2170.8 kg of water/hr
MAC in stream (17)
= 36428.17 – 34597 = 1471.1 kg of MAC/hr
38
Material Balance
CHAPTER NO. 4
Composition of stream (17) Kg
%wt
kgmole
%mole
Water
2170.8
59.00
120.60
90.25
MAC
1471.1
40.00
12.67
9.480
Quinone
36.45
0.99
0.33
0.25
Total
3678.5
133.62
4.7 Material Balance around Dehydrator T-101:
In dehydrator Maleic Acid is converted back to Maleic Anhydride according to the following reaction:
C4H4O4
C4H2O3 + H20
In addition to above reaction, some Maleic Acid may irreversibly convert to Fumaric Acid according to the following reaction:
C4H2O3 + H2O
C4H4O4
C4H4O4
Maleic Acid
Fumaric Acid 39
Material Balance
CHAPTER NO. 4
Assuming 0.5% MAC converts irreversibly to Fumeric Acid and rest 99.5% converts back to Maleic Anhydride. And 95% of water is removed by azeotropic distillation from the dehydrator. Operating Conditions: Operating pressure
= 15.2 KPa
Azeotropic composition: Water = 77.5 %mole
&
MAC coming from stream (18)
O- xylene =
22.5 %mole
= 1736.78 kg of MAC/hr
0.5% is converted to Fumaric Acid Fumaric Acid produced
= 2889.17 x 0.005= 8.68 kg of Fumaric Acid/hr
MAC converted to MAN
= 1736.78 x 0.995
= 1728.095 kg of MAC/hr
= 14.89 kgmole of MAC/hr MAN produced
= 14.89 kgmole of MAN/hr
Water produced
=
Total water in Dehydrator O-xylene
14.89 kgmole of water/hr
= 14.89 + 142.71
= 157.60 kgmole of water/hr
= 157.60 x (22.5 / 77.5) = 45.75 kgmole of O-xylene/hr
As 95 % water is removed from the dehydrator, Stream (19) contains Water
= 157.60 x 0.95
= 149.72 kgmole of water/hr
O-xylene
= 149.72 x (22.7 / 77.5) = 43.47 kgmole of O-xylene/hr
40
Material Balance
CHAPTER NO. 4
Composition of stream (19) Kg
%wt
kgmole
%mole
Water
2694.89
36.87
149.72
77.5
O-xylene
4614.78
63.13
43.46
22.5
Total
7309.67
193.18
Assuming stream (22) contains pure water and pure O-xylene stream (21) is refluxed back to dehydrator. O-xylene in stream (26)
= 45.75– 43.47
= 2.29 kgmole of O-xylene/hr
So make up O-xylene required (23) = 2.29 kgmole of O-xylene/hr Water in stream (22) = 149.72 kgmole of water/hr Composition of stream (26) Kg
%wt
kgmole
%mole
Water
141.84
7.46
7.90
30.78
MAN
1460.7
76.93
14.89
58.34
Fumaric Acid
8.68
0.44
.07
0.28
O-xylene
242.88
12.74
2.289
8.93
Quinone
36.43
2.43
0.338
1.67
Total
1890.54
25.47
41
Material Balance
CHAPTER NO. 4
4.8 Material Balance around Mixer MIX-102
Mixer MIX-102 mixes crude MAN from stream (10) with crude MAN obtained from Bottom of dehydrating column stream (27)
(27) 1890.54
(10) +
(28)
1099.68
=
2990.21
Composition of stream (28) kg
%wt
Kg mole
%mole
water
172.05
5.75
9.55
25.13
MAN
2505.28
83.78
25.54
67.16
O-xylene
242.88
8.12
2.28
6.01
Quinone
61.30
2.05
0.56
1.49
Fumeric acid
8.68
0.29
0.07
0.19
Total
2990.21
38.03
42
Material Balance
CHAPTER NO. 4
4.9 Material Balance around Distillation column T-102 30
32 31
29
36
Bottom product stream (37) contains 99.5 % pure MAN. 0.5% impurity will be all the Fumeric Acid (BP = 290oC) produced in Dehydrating column and remaing is Quinone (BP = 180oC).
Operating conditions: Top plate temperature = 136oC Re-boiler temperature = 220oC Operating pressure
= 110Kpa
MAN converted to Fumaric Acid
= 8.68 x 98.06 / 116.07 = 7.333 kg of MAN/hr = 0.30 % of initial MAN produced
MAN lost in distillation column
= 2505.29 x 0.007 = 17.54 kg of MAN/hr
MAN in bottom product stream (37) = 2505.29 – 17.54= 2487.75 kg of MAN/hr 43
Material Balance
CHAPTER NO. 4
Bottom product of distillation column= 2487.75.26 / 0.995 = 2500.25 kg of product/hr Quinone in Bottom product
= 2500.25 – (2487.75 + 8.68) = 3.82 kg of Quinone/hr
Composition of bottom product stream (36) kg
%wt
Kg mole
%mole
MAN
2487.74
99.5
25.36
99.56
Fumaric acid
8.68
0.34
0.07
0.29
Quinone
3.81
0.15
0.03
0.14
Total
2500
25.47
Composition of overhead product stream (32) kg
%wt
Kg mole
%mole
Water
172.05
35.11
9.55
76.12
MAN
17.53
3.58
0.17
1.42
O-xylene
242.88
49.57
2.28
18.21
Quinone
57.48
11.73
0.53
4.23
Total
489.9637
100
12.557
100
For required purity, Reflux ratio L/D So
= 6
reflux rate as stream (31) = 6 x 489.96 = 2939.76Kg/hr Overhead product as stream (30)
= 7 x 489.96 = 3429.72 Kg/hr
44
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