T.R. EGE UNIVERSITY Chemical Engineering Department
CHEMICAL ENGINEERING DESIGN PROJECT REPORT I
Submitted to: Prof.Dr.Ferhan ATALAY
Prepared by;
Res.Assist. Nilay GİZLİ Sezai ERDEM Tuğba GÜRMEN
05078901 Ürün ARDA 05068091 M.Serkan ACARSER 05068076 Müge METİN 05078875 Sıla Ezgi GÜNGÖR 05068052 Ali KÜÇÜK March 2009 Bornova-İZMİR
Date: 30.03.2009
SUMMARY The process purpose is to produce acetone from isopropyl alcohol (IPA) at the given conditions. This report is formed, some properties, manufacturing process of acetone. In manufacturing process, feed drum, vaporizer, heater, reactor, furnace, cooler, condenser, flash unit, scrubber, acetone and IPA columns are used.
i
INTRODUCTION Acetone (dimethyl ketone, 2-propane, CH3COCH3 ), formulation weight 58,079, is the simplest and the most important of the ketones. It is a colourless, mobile, flammable liquid with a mildly pungent and somewhat aromatic odour. It is miscible in all proportions with water and with organic solvents such as ether, methanol, ethyl alcohol, and esters. Acetone is used as a solvent for cellulose acetate and nitrocellulose, as a carrier for acetylene and as a raw material for the chemical synthesis of a wide range of products such as ketene, methyl methacrylate, bisphenol A, diacetone alcohol mesityl oxide, methyl isobutyl ketone, hexylene glycol ( 2-methyl-2, 4-pentanediol ), and isophorone.
Acetone is produced in various ways;
The Cumene Hydroperoxide Process for Phenol and Acetone
Isopropyl Alcohol Dehydrogenation
Direct Oxidation of Hydrocarbons to a Number of Oxygeanted Products
Including Acetone
Catalytic Oxidation of Isopropyl Alcohol
Acetone as a By-Product of the Propylene Oxide Process Used by Oxirane
The p-Cymene Hydroperoxide Process for p Cresol and Acetone
The Diisopropylbenzene Process for Hydroquinone (or Resorcinol ) and
Acetone
In this report isopropyl alcohol dehydrogenation was investigated.
ii
TABLE OF CONTENTS: Summary
i
Introduction
ii
1.0 Describing of Process
1
2.0 Results
2
3.0 Discussion
4
4.0 Nomenclature
7
5.0 Appendix
8
5.1 Flowchart
8
5.2 Mass Balance
9
5.2.1 Reactor
9
5.2.2 Flash Unit
10
5.2.3 Scrubber
11
5.2.4 Acetone Column
14
5.2.5 IPA Column
15
5.2.6 Feed Drum
16
5.3 Energy Balances
References
17
5.3.1 Feed Drum
17
5.3.2 Vaporizer
18
5.3.3 Pre - Heater
19
5.3.4 Reactor
20
5.3.5 Cooler
22
5.3.6 Condenser
23
5.3.7 Scrubber
26
5.3.8 Acetone Column
27
5.3.9 IPA Column
30 32
1.0 DESCRIPTION OF THE PROCESS
At the beginning of the process, feed including i-propyl alcohol and water, and recycle
stream are mixed in feed drum. From here, this mixture is send to vaporizer to change stream’s phase as vapour. After vaporizer, mixture is heated to reaction temperature in the heater. Reactor used is a tubular flow reactor. Acetone, hydrogen gas (H2) are produced and water and i-propyl-alcohol are discharged. The mixture with acetone, hydrogen, water, ipropyl-alcohol are sent to cooler and then to condenser. After condenser the mixture is sent to flash unit. Hydrogen, acetone, i-propyl-alcohol and water are obtained as top product. This top product is sent to scrubber to remove hydrogen. The bottom product of flash unit which is formed by acetone, water, i-propyl-alcohol are mixed with the bottom product of scrubber before acetone column. In acetone column, acetone is obtained from top product with 99 wt%. İ-propyl alcohol and water and also 0,1% of acetone is sent to i-propyl-alcohol column from bottom product. The top product of this column is sent to feed drum and bottom product is thrown away as waste water.
‐ 1 ‐
2.0 RESULTS
Table1: Properties of Substances Property
H2O
Acetone
IPA
H2
Molecular Weight(kg/kmol)
18,015
58,08
60,096
2,01
Freezing Point(°C)
0
-95
-88,5
-259,2
Boling Point(°C)
100
56,2
82,2
-252,8
Critical Temperature (°C)
647,3
508,1
508,3
33,2
Critical Pressure (bar)
220,5
47
47,6
13
Critical Volume (m3/min)
0,056
0,209
0,220
0,065
Liquid Density(kg/m3)
998
790
786
71
Heat of Vaporization(J/mol)
40683
29140
39858
904
658,25
273,84
1139,70
13,82
283,16
131,63
323,44
5,39
-242,0
20,43
-272,60
0
-228,77
62,76
-173,5
0
32,243
3,710
32,427
27,143
1,923x10-3
2,345x10-1
1,886x10-1
2,73x10-3
1,055x10-5
-1,160x10-4
6,405x10-5
-1,380x10-5
-3,596x10-8
2,204x10-8
-9,261x10-8
7,645x10-9
11
-113
0
-259
168
-33
111
-248
Constants in the liquid viscosity equation (A) Constants in the liquid viscosity equation (B) Standard Enthalpy of Formation at 298K(kJ/kmol) Standard Gibbs Energy of Formation at 298K (kJ/kmol) Constant in The Ideal Gas Heat Capacities Equation(A) Constant in The Ideal Gas Heat Capacities Equation(B) Constant in The Ideal Gas Heat Capacities Equation(C) Constant in The Ideal Gas Heat Capacities Equation(D) Minimum Temperature For Antoine Constant (°C) Maximum Temperature For Antoine Constant (°C)
-2-
Table 2: Calculated mol and mass values of substances
Acetone 1
Basis:100kmol/h
ipropylalcohol
multiplied scale factor
kmol/h ton/year kmol/h
kg/h
Basis:100kmol/h
multiplied scale factor
ton/year kmol/h ton/year kmol/h
‐
‐
‐
‐
‐
2
‐
‐
‐
‐
‐
100
52644,1
3
‐
‐
‐
‐
‐
100
4
‐
‐
‐
‐
‐
5
90
6 7
Water
kg/h
multiplied scale factor
ton/year kmol/h ton/year kmol/h
kg/h
Basis:100kmol/h
multiplied scale factor
ton/year kmol/h ton/year kmol/h
90,937 47872,96 228,797 13749,785 120448,4 44,786 7067,741 112,682 2029,966 17782,44
kg/h
ton/year
‐
‐
‐
‐
‐
251,6 15120,154 132452,6 49,25 7772,211 123,913 2232,293 19554,88
‐
‐
‐
‐
‐
52644,1
251,6 15120,154 132452,6 49,25 7772,211 123,913 2232,293 19554,88
‐
‐
‐
‐
‐
100
52644,1
251,6 15120,154 132452,6 49,25 7772,211 123,913 2232,293 19554,88
‐
‐
‐
‐
‐
45790,27 226,44 13151,635 115208,3
10
5264,41
25,16 1512,015 13245,26 49,25 7772,211 123,913 2232,293 19554,88
90
1584,68 226,44 455,144 3987,07
90
45790,27 226,44 13151,635 115208,3
10
5264,41
25,16 1512,015 13245,26 49,25 7772,211 123,913 2232,293 19554,88
90
1584,68 226,44 455,144 3987,07
90
45790,27 226,44 13151,635 115208,3
10
5264,41
25,16 1512,015 13245,26 49,25 7772,211 123,913 2232,293 19554,88
90
1584,68 226,44 455,144 3987,07
1,952
90
1584,68 226,44 455,144 3987,07
8
24,148 12286,04 60,756 3528,708 30911,67 0,776 408,518
9
65,789 33472,18 165,525 9613,692 84216,01 9,194 4840,098 23,132 1390,141 12177,69 45,491 7178,998 114,455 2061,907 18062,36
10 24,124 12273,83 60,696 3525,224 30880,95 0,776 408,518
1,952
117,307 1027,832 3,731 588,794
9,387
169,107 1481,407
117,307 1027,832 607,772 95913,35 1529,15 27547,709 241318
11
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
12
0,024
12,211
0,06
3,485
30,722
‐
‐
‐
‐
‐
604,041 95324,56 1519,77 27378,603 239836,6 ‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
90
1584,68 226,44 455,144 3987,07
13 89,913 45746,01 226,221 13138,916 115097
9,97 5248,616 25,085 1507,508 13205,52 653,263 103092,4 1643,61 29609,634 259380,4
‐
‐
‐
‐
‐
14 89,824 45700,73 225,997 13125,906 114983
0,907 477,482
‐
‐
‐
‐
‐
15
0,089
45,281
0,224
13,010
113,928
9,063 4771,134 22,803 1370,369 12004,17 653,263 103092,4 1643,61 29609,634 259380,4
‐
‐
‐
‐
‐
16
‐
‐
‐
‐
‐
‐
‐
‐
‐
‐
17
0,089
45,281
0,224
13,010
113,928
‐
‐
‐
‐
‐
‐
‐
2,282
‐
137,139 1201,345
‐
‐
‐
‐
‐
‐
‐
648,799 102387,9 1632,38 29407,290 257607,9
9,063 4771,134 22,803 1370,369 12004,17 4,464
‐ 3 ‐
Basis:100kmol/h
Hydrogen
704,47 11,231
202,326 1772,447
3.0 DISCUSSION
Feed drum is a kind of tank used for the mixing of the recycle stream and feed stream. Recycle stream concentration was assumed to be same with the feed stream. The temperature of the feed stream is assumed to be 250 C at 2 bar pressure, which is assumed to be constant. The temperature of recycle stream was calculated as 111,50 C. The temperature of the leaving stream was calculated as 32,890 C, by the energy balance around feed drum.
In the vaporizer molten salt was used for heating. The temperature at the entrance of the unit is the temperature of the mixture leaving the feed drum, which is 32,890 C. And the leaving temperature is the bubble point temperature of the mixture, which is 109,50 C. The pressure is 2 bars, and assumed to be constant.
Since the temperature leaving the vaporizer is not enough for the reaction a pre-heater was used. The unit is working at 2 bars, and assumed to be constant. The entrance and leaving temperatures are 109,50 C and 3250 C.
The reactor was the starting point for the calculations. The temperature values for the entering and leaving streams were found from literature, which are 3250 C and 3500 C, respectively. The reaction taken place inside is endothermic, for this reason the reactor has to be heated. For heating, molten salt was used. The pressure is 1,8 bar, and assumed to be constant.
The entrance temperature of the cooler is 3500 C and leaving is 94,70 C. For cooling, water was used. Instead of water a refrigerant may be used. Better results may get. But since it costs too much, it wasn’t chosen as the cooling material. From the temperature values it’s easily seen that the load is on the cooler not on the condenser, for this process. But in reality the unit cannot cool that much, and the load is mostly on the condenser. In this process, the mixture cooled down to its dew point. The pressure is 1,5 bar, and assumed to be constant.
-4-
The temperature of the entering stream is the dew point and the leaving temperature is the bubble point of the mixture. In the condenser water was used as cooling material. In the calculation of the dew and bubble points Antoine Equation was used. Trial and error was used with the help of Excel. The mixture includes acetone, i-propyl-alcohol, water and hydrogen. But hydrogen was not taken into consideration in the calculations. Since the condensation temperature of hydrogen is very low, it is not condense in the condenser. It stays in the for this reasons it has no affect on bubble and dew point calculations. Also since it does not affect the temperature calculations it’s not taken into consideration on mole and mass fraction calculations. The leaving and entering temperatures are 94,70 C and 810 C, respectively. The pressure is 1,5 bar, and assumed to be constant.
Flash unit was assumed to be isothermal, for this reason temperature was not changed. It is 810 C in the entrance and exit. The pressure is 1,5 bar, and assumed to be constant. By trial and error method, (V / F) value was found to be 0,2. The entrance temperature of the unit is the bubble point of the mixture, but if it was its dew point the (V/F) value would be much higher.
Scrubber was assumed to be adiabatic. The temperature of water entering the unit was assumed to be 250 C. The temperature of the off gas, including hydrogen and a very little amount of acetone, was assumed to 700 C. But this assumption is too high, a lower temperature should have been assumed, since a lot of water is used in the unit. It should have been around 400 C - 500 C. The temperature of the leaving stream was found to be 28.10 C. The pressure of the unit is 1,5 bar, and assumed to be constant.
The streams leaving the scrubber and flash unit are mixed together before entering the acetone column. The temperature leaving the flash unit and scrubber are 810 C and 28.10 C, respectively. The temperature of the mixture was found to be 450 C. This result was getting by using energy balance around the mixing point.
-5-
The acetone column is used to separate the acetone from the mixture. The entrance temperature is 450 C. The leaving temperatures for the top and bottom product are 102,3 and 105, respectively, which are the bubble and dew points. Top product of the unit includes acetone i-propyl-alcohol and 99wt% of the product is acetone. This amount is assumed to be the desired acetone production rate, which is 115000 ton/year. From the bottom i-propylalcohol, water and a very little amount of, 0,1 %, acetone is discharged. The pressure is 1,1 bar, and assumed to be constant.
In the distillation column, i-propyl-alcohol and water are separated. The entrance temperature is 1050 C. The leaving temperatures of the top and bottom products are both 111,50 C. The top product is recycled to the feed drum. For this reason it’s assumed to have the same concentration with the feed stream. But in reality a very little amount of acetone exists in the stream. It’s calculated but neglected on the recycle stream calculations. The bottom product is assumed to be pure water and it’s thrown away. Since its temperature is very high it cannot be recycled to the scrubber. But if a cooler is used, a recycle can be used. The pressure is 1,1 bar, and assumed to be constant.
In the calculations one year is assumed to be 360 working day and 8600 hours. If it was 300 working day and 7200 hours, the results may be higher.
Since approximated values are used in the calculations, some errors may occur. The values were taken in three decimal digits. If four or more decimal digits were taken, more accurate results would get. Also during the calculations of the specific heats, approximated values used.
-6-
4.0 NOMENCLATURE MW = Molecular Weight [kg/kmol] n = mole[mol/h] y = mol or mass fraction of gas stream x = mol or mass fraction of liquid stream PT = Total Pressure [bar] Pi* = Vapour Pressure of Component [bar] Pv* = Vapour Pressure [bar] F = Feed Flow Rate [kmol/h] V = Flow Rate of Vapour [kmol/h] L = Flow Rate of Liquid [kmol/h] T = Temperature [°C] ∆Hvap = Latent Heat of Vaporisation [kJ/kg] TC = Critical Temperature [°C] PC = Critical Pressure [bar] Tb = Normal Boiling Point [°C] Q = Heat [kJ] m = Mass Flow Rate [kg/h]
-7-
12
STACK GAS
OFF GAS H2
WATER FURNACE
11
SCRUBBER
NATURAL GAS
AIR
14
ACETONE
17
8
1
MOLTEN SALT
4
13
IPA COLUMN
CONDERSER
ACETONE COLUMN
3
7
FLASH
VAPORIZER
6
REACTOR
FEED DRUM
2
10
COOLER
HEATER 17
9
WASTE WATER
15 16
5
RECYCLE IPA -8-
17
5.0 APPENDIX 5.1 MASS BALANCES Production Rate : 115000 ton/year
5.1.1 REACTOR
4
I-propylalcohol=100 kmol/h H2O = 49.25 kmol/h
R E A C T O R
5
acetone H2 H2 O i-propyl-alcohol
conversion = 90 %
nacetone5 = 100*0.9 = 90 kmol / h nH 5 = 100*0.9 = 90 kmol / h 2
nH O 5 = 49.25 kmol / h 2
ni − propylalcohol 5 = 100*0.1 = 10 kmol / h nTotal 5 = nacetone5 + nH 5 + nH O 5 + ni − propylalcohol 5 = 239.25 kmol / h 2
2
90 = 0.376 239.25 90 yH 5 = = 0.376 2 239.25 49.25 yH O 5 = = 0.206 2 239.25 10 yi − propylalcohol 5 = = 0.042 239.25 yacetone5 =
-9-
5.1.2 FLASH UNIT
8
acetone H2 H 2O i-propyl-alcohol
acetone = 90 kmol/h H2 = 90 kmol/h
F L A S H
7
H2O = 49.25 kmol/h i-propyl-alcohol = 10 kmol/h
9
acetone H 2O i-propyl-alcohol
• It is assumed that there is no change at temperature and pressure. P* y K i= i = i PT xi At buble point (T = 81°C)
For acetone 1161 224 + 81 = 1651.6 mmHg
* = 7.02447 − log Pacetone
* Pacetone
K acetone =
1651.6 = 1.467 ((1.5 /1.013) *760)
For i-propyl-alcohol * = 8.37895 − log PIPA
1788.02 227.438 + 81
* PIPA = 381.89 mmHg
K IPA =
381.89 = 0.339 1125.092
For water 1668.21 228 + 81 = 369.89 mmHg
log PH* O = 7.96681 − 2
PH* O 2
K H 2O =
369.89 = 0.328 1125.092
- 10 -
From trial-error; (V/F) = 0.2 F = n acetone7 + n H O 7 + n IPA 7 = 149.25 kmol/h 2
F=V+L 0.2 =
V F
V = 29.85 kmol/h L = 119.4 kmol/h
yv = K × xL F × zF = V × yv + L × xL
For acetone yv = 1.467× xL 90 = 29.85 × yv + 119.4 × xL xL = 0.551 yv = 0.809
For i-propyl-alcohol yv = 0,339 × xL 10 = 29.85 × yv + 119.4 × xL xL = 0.077 yv = 0.026
For water yv = 0.328 × xL 49.25 = 29.85 × yv + 119.4 × xL xL = 0.381 yv = 0.125
ªAt stream 8; V = 29.85 kmol/h yacetone = 0.809⇒
nacetone 8 = (0.809) ×(29.85) = 24.148 kmol/h
yi-propyl-alcohol = 0.026 ⇒ ywater = 0.125
⇒
ni-propyl-alcohol 8 = (0.026) ×(29.85) = 0.766 kmol/h nwater 8 = (0.125) ×(29.85) = 3.731kmol/h - 11 -
ªAt stream 9; L = 119.4 kmol/h ⇒
xacetone = 0.551
xi-propyl-alcohol = 0.077 ⇒ ⇒
xwater = 0.381
nacetone 9 = (0.551) ×(119.4) = 65.789 kmol/h ni-propyl-alcohol 9 = (0.077) ×(119.4) = 9.194 kmol/h nwater 9 = (0.381) ×(119.4) = 45.491 kmol/h
12
5.1.3 SCRUBBER
OFF-GAS H2=90 kmol/h Acetone
11 H2 O
H2 = 90 kmol/h H2O = 3.731 kmol/h 8 Acetone = 24.148 kmol/h i-propyl-alcohol = 0.776 kmol/h
10
T = 810C (354.15 K); P = 1.5 bar (1.48 atm) Assume 1/1000 of inlet acetone is in off-gas.
∴ n acetone12 = 0.024148 kmol / h n acetone10 = 24.148 − 0.024148 = 24.124 kmol / h n Total8 = n acetone8 + n H 8 + n H O8 + n IPA8 2
2
n Total8 = 24.148 + 90 + 3.731 + 0.776 = 118.655 kmol / h n Total12 = n acetone12 + n H 12 2
n Total12 = 0.024148 + 90 = 90.024 kmol / h yacetone12 = 0.024148 / 90.024 = 2.68*10−4 yacetone8 = 24.148 /118.655 = 0.203 - 12 -
Acetone H2 O i-propyl-alcohol = 0.776 kmol/h
y acetone12 1 − A L ; A = 11 = 6 y acetone8 1 − A mV8 m=
e
3598 ⎞ ⎛ ⎜10.92 − ⎟ T ⎠ ⎝
P
⇒
m=
e
3598 ⎞ ⎛ ⎜10.92 − ⎟ 354.15 ⎠ ⎝
1.48
= 1.445
y acetone12 2.68*10−4 1− A = = 1.320*10−3 = y acetone8 0.203 1 − A6
From trial-error A is found as 3.523
L11 = mAV8 = 1.445*3.523*118.655 L11 = 604.041 kmol / h n H O10 = n H O8 + n H O11 2
2
2
n H O10 = 3.731 + 604.041 = 607.772 kmol / h 2
n Total10 = n acetone10 + n H O10 + n IPA10 2
n Total10 = 24.124 + 607.772 + 0.776 = 632.672 kmol / h
- 13 -
5.1.4 ACETONE COLUMN
14
acetone i-propyl-alcohol
A C E T O N E
acetone = 89.913 kmol/h i-propyl-alcohol = 9.97 kmol/h
13
water = 653.263 kmol/h
C O L U M N
acetone 15
i-propyl-alcohol water
nacetone 13 = nacetone 9 + nacetone 10 = 65.789 + 24.124 = 89.913 kmol/h nı-propyl-alcohol 13 = ni-propyl-alcohol 9 + nı-propyl-alcohol 10 = 9.194 + 0.776 =9.97 kmol/h nwater 13 = nwater 9 + nwater 10 = 45.491 +607.772 = 653.263 kmol/h nT 13 = nacetone 13+ nwater 13 + nı-propyl-alcohol 13 nT 13 = 89.913 + 653.263 + 9.97 = 753.146 kmol/h
Assume that 1/1000 of acetone is in bottom product
∴ nacetone 15 =
89.913 = 0.089 kmol/h 1000
nacetone 14 =89.913-0.089 = 89.824 kmol/h
Since acetone purity is 99%
nı-propyl-alcohol 14 = 89.824 ×
0.01 = 0.907 kmol/h 0.99
nı-propyl-alcohol 15 = nı-propyl-alcohol 13 - nı-propyl-alcohol 14 = 9.97-0.907 =9.063 kmol/h nwater 15 = nwater 13 = 653.263 kmol/h - 14 -
5.1.5 IPA COLUMN
17
acetone = 0.089 kmol/h i-propyl-alcohol = 9.063 kmol/h water = 653.263 kmol/h
15
acetone i-propyl-alcohol water
C I O P L A U M N
16
water
since all the i-propyl-alcohol is at the top product
nı-propyl-alcohol 17 = nı-propyl-alcohol 15 = 9.063 kmol/h nacetone 17 = nacetone 15 = 0.089 kmol/h
Assume the composition of the recycle stream is as feed stream so that; ywater=0.33 ; yIPA=0.67
nwater 17 = 9.063 ×
0.33 = 4.464 kmol/h (neglecting acetone composition) 0.67
nwater 16 = nwater 15 - nwater 17 = 653.263 – 4.464 = 648.799 kmol/h
- 15 -
5.1.6 FEED DRUM
i-propyl-alcohol water 1
FEED DRUM
i-propyl-alcohol=9.063 kmol/h water=4.464 kmol/h
2
i-propyl-alcohol=100 kmol/h water=49.25 kmol/h
17
Input = Output nı-propyl-alcohol 2 = ni-propyl-alcohol 1 + nı-propyl-alcohol 17 nı-propyl-alcohol 1 = 100 – 9.063 = 90.937 kmol/h nwater 2 = nwater 1 + nwater 17 nwater 1 = 49.25 – 4.464 = 44.786 kmol/h ¾ since 115000 tons/year acetone is wanted to produce, all of these calculations should be
correlated as this amount. These new values are shown in Table 1.
amount = 89.824 kmol/h *
58.08kg 1ton 8760h × × = 45700.726ton / year 1kmol 1000kg 1year
Scale Factor: 115000 ton year = 2.516 45700.726 ton year
- 16 -
5.2 ENERGY BALANCES 5.2.1 FEED DRUM
T=25o C mi-propyl-alcohol = 13749.785 kg/h mwater = 2029.966 kg/h
1 2
FEED DRUM
T=111.5o C mi-propyl-alcohol = 1370.369 kg/h mwater = 202.326 kg/h
T=32.89o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h
17
Tref = 25oC ; Cp,I-propyl-alcohol = 2,497 kJ/kg ; Cp,water = 4,178 kJ/kg. For stream 1,2 and 17 calculate Cp,mix; Cp,mix = 2,497×0,87+4,178×0,13 Cp,mix =2,715 kJ/kgK mTotal,1=13749.785 + 2029.966 = 15779.75 kg/h mTotal,2=15120.154 + 2232.293 = 17352.447 kg/h mTotal,17=1370.369 + 202.326 = 1572.695 kg/h QIN = QOUT 15779.75*2,715*(25-25) + 1572.695*2,715*(111,5-25) = 17352.447*2,715×(T-25) T = 32,830C
- 17 -
5.2.2 VAPORIZER T=32.83 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h
2
VAPORIZER
3
T=109.5 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h
At 32.83 oC Cp i-propyl-alcohol = 145 kJ/kmol.K = 2.413 kJ /kg.K Cp water = 4.179 kJ /kg.K
For Water:
TC = 508.3 K Tb = 394.399 K ∆Hf = 39838 kJ/kmol
∆H vap,H O 2
⎡ T −T ⎤ = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦
0,38
⎡ 508.3K − 382.5 K ⎤ ∆H vap,H O = 39838 ⎢ 2 ⎣ 508.3 K − 394.399 K ⎥⎦
For IPA :
0.38
= 41370.970kj / kmol = 2296.473 kJ / kg
TC = 647.3 K Tb = 375 K ∆Hf = 40683 kJ/kmol
∆H vap,IPA
⎡ 647,3K − 382,5 K ⎤ = 40683 ⎢ ⎥ ⎣ 647,3 K − 375 K ⎦
0,38
- 18 -
= 40253,505kj / kmol = 669,82 kj / kg
Q = mi-propyl-alcohol×Cp i-propyl-alcohol×∆T+mwater×Cp,water ×∆T + mwater×∆Hvap,water+mIPA×∆Hvap,IPA
Q = 15120.154*2.413* (109.5 − 32.83) + 2232.293*4.179* (109.5 − 32.83) +2232.293*2296.473 + 15120.154*669.82 Q = 9.652 ×106 kJ We assume ∆T = 20
Molten Salt :
Q = m × Cp,molten salt × ∆T 9.652 × 106 kJ= 1,56 kJ /kg × m × (20)
m= 309.358 tons
5.2.3 PRE-HEATER T=109.5 o C mwater = 2232.293 kg/h mi-propyl-alcohol = 15120.154 kg/h
4
3
T=325 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h
HEATER
Tref = 109,5 oC ; Cp i-propyl-alcohol = 2.468 kJ /kg.K ; Cp water = 2.019 kJ /kg.K Q = mwater × Cp,water × ∆T + mi-propyl-alcohol ×Cp -propyl-alcohol × ∆T Q = (2232.293 ×2,468 ×(325-109.5)) + (15120.154 ×2,019 ×215,5) Q = 1.845 ×106 kJ
Molten Salt :
We assume ∆T = 150
Q = m × Cp,molten salt × ∆T 1.845 × 106 kJ= 1,56 kJ /kg × m × (150) m= 7.885 ton
- 19 -
5.2.4 REACTOR
(CH 3 )2 CHOH → (CH 3 )2 CO + H 2 Table 3: mole and Hf values of acetone, i-propyl-alcohol and H2 nin kmol/h
Hf kJ/kmol
nout kmol/h
(CH3)2CHOH
251.6
-272.290
25.16
(CH3)2CO
0
-216.685
226.44
H2
0
0
226.44
4 T=325 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h
325
∆Hin ,
IPA = − 272,29 +
R E A C T O R
∫ ( 32,427 +1,886×10
−1
5
T=350 o C mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h mH2 = 455.144 kg/h
T + 6,405×10−5 T2 − 9,261×10−8 T3 )dT
25
∆Hin,IPA = -272,29 + 20,104 = -252,186 kJ/kmol 350
∆Hout, IPA = −272,29 +
∫ ( 32,427 +1,886×10
−1
T + 6,405×10−5 T2 − 9,261×10−8 T3 )dT
25
∆Hout, IPA = -249,691 kJ/kmol 350
∆Hout,acetone = − 216,685 + ∫ ( 71,96 + 20,1×10−2 T +12,78×10−5 T2 + 34,76×10−8 T3 ) dT 25
∆Hout,acetone= -182,745 kJ/kmol
- 20 -
350
∆Hout,H = 2
∫ ( 28.84×10
−3
+ 0.00765×10−5 T + 0.3288×10−8 T2 − 0.8698×10−12 T3 ) dT
25
∆Hout,H = 9.466 kj/kmol 2
∆Hr0=(-216,685kJ/kmol) – (-272,29 kJ/kmol) ∆Hr0= 55.605 kJ/kmol ∆H r =
226.44 kmol × 55.605 kJ kmol = 12591.196 kJ 1
Q = ∑ out n i H i − ∑ in n i H i + ∆H r
Q=[25.16 (-249.691)+ 226.44(-182.745)+226.44(9.466)]-[251.6(-252.186)] + 12591.196 Q=30521.67 kJ
Molten Salt : Cp (molten salt between 360°C – 410°C) = 1,56 kJ/kg Q = m × Cp,molten salt × ∆T 30521.67 kJ= 1,56 kJ /kg × m × (50) m= 391.300 kg/h
- 21 -
5.2.5 COOLER T=350 o C mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h m H = 455.144 kg/h
6
5 COOLER
2
T=94.7o C mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h m H = 455.144 kg/h 2
Tref=94.7 oC
Cp,H = 12.608 kJ / kg.K 2
Cp,water = 2.035 kJ / kg.K Cp,IPA = 2.536 kJ / kg.K Cp,acetone = 1.896 kJ / kg.K Q = [(m H *C p,H ) + (m water *C p,water ) + (m IPA *C p,IPA ) + (m acetone *Cp,acetone )]* ∆T 2
2
Q = [(455.144*12.608) + (2232.293*2.035) + (1512.015*2.536) + (13151.635*1.896)]* (94.7 − 350)
Q= - 10.123 ×106 kJ
Water :
∆ T for the Water = (35-15)=20 Cpwater = 4.179 kJ/kg Q = m × Cp,water × ∆T 10.123 × 106 kJ= 4.179 kJ /kg × m × (20) m= 121.117 ton/h
- 22 -
5.2.6 CONDENSER T=94.7o C (Tdp) mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h m H = 455.144 kg/h
6
7
CONDENSER
2
log P* = A −
B C + Tdp
T=81o C (Tbp) mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h m H = 455.144 kg/h 2
; P* mm Hg
Assumption = PT = 1.5 bar = 1125 mmHg
0
y i − propyl − alcohol × PT y acetone × PT y water × PT y × PT + + + H* 2 =1 * * * Pacetone × Tdp Pwater × Tdp Pi − propyl − alcohol × Tdp PH 2 × Tdp
From literature; For acetone:
A=7.02447 B=1161 C=224
For water:
A=7.96681
Using;
B = 1668.21
yacetone =0.6 ywater=0.33 yi-propyl-alcohol= 0.07
C=228
by trial and error Tdp= 94.7 °C found
For i-propyl-alcohol: A= 8.37895
B=1788.02 C=227.438
- 23 -
P = x A PA* (Tbp) + x B PB* (Tbp) + ...
Using;
m ×Cp ×∆T + m∆Hf = Qtot
yacetone = 0.6 ywater= 0.33 yi-propyl-alcohol= 0.07 by trial and error Tbp = 81°C
For Acetone:
At 94.7 oC and 1.5 bar Cp,Acetone = 1.297 kJ/kg.K Qacetone = m×Cp ×∆T Qacetone = 13151.635 kg × (1.297 kJ/ kg.K) ×[(81+273.15) –( 94.7+273.15)] = -233.690*103 kJ ∆H vap
⎡ T −T ⎤ = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦
0,38
∆Hf,acetone = 29140 kJ/kmol Tc = 508.1 K Tb = 341.5 K ∆H vap
⎡ 508.1K − 354 K ⎤ = 29140* ⎢ ⎣ 508.1 K − 341.5 K ⎥⎦
0,38
= 28289.029kJ / kmol = 487.07kJ / kg
For IPA:
At 94.7 oC and 1.5 bar Cp,i-propyl-alcohol = 1.761 kJ/kg.K Q,i-propyl-alcohol = 1512.015 kg × (1.761 kJ/ kg.K) ×(354.15-367.85) = -36.478*103 kJ ∆H vap
⎡ T −T ⎤ = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦
0,38
∆Hf,i-propyl-alcohol = 39858 kJ/kmol Tc = 508.3 K Tb = 366.6 K 0,38
∆H vap
⎡ 508,3K − 354 K ⎤ = 39858* ⎢ ⎥ = 41169,35kJ / kmol = 685,128kJ / kg ⎣ 508,3 K − 366,6 K ⎦ - 24 -
For Water :
At 94.7 oC and 1.5 bar Cp,water = 1.898 kJ/kg K Q water =2232.293 kg (1,898 kJ/ kg.K) ×(354.15-367.85) = -58.045*103 kJ ⎡ T −T ⎤ ∆H vap = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦
0,38
∆Hf,water = 40683 kJ/kmol Tc = 647.3 K Tb = 385.186 K ∆H vap
⎡ 647,3K − 354 K ⎤ = 40683* ⎢ ⎣ 647.3 K − 385.186 K ⎥⎦
0,38
= 42442.0561kJ / kmol = 2356.845 kJ / kg
For Hydrogen :
At 94.7 oC and 1.5 bar C p,H = 13.225 kJ/kg K 2
Q
= 455.144 kg (13, 225 kJ / kg.K ) * ( 354.15 − 367.85 ) = -82.464 *103 kJ
H2
∑m C i
p,i
∆T = −410.677 *103 kJ ;
i
∑ m ∆H i
vap,i
i
QTotal = ∑ mi C p,i ∆T + ∑ mi ∆H vap,i = 12.3*106 kJ i
i
Water :
∆ T for the Water = (35-15)=20 Cpwater = 4.182 kJ/kg Q = m × Cp,water × ∆T 682691.799 kJ= 4.182 kJ /kg × m × (20) m= 147.058 ton/h
- 25 -
= 12.702*106 kJ
5.2.7 SCRUBBER
12 mwater = 27378.603 kg/h
T=70o C macetone = 3.485 kg/h m H = 455.144 kg/h 2
11
T=81o C mi-propyl-alcohol = 117.307 kg/h mwater = 169.107 kg/h macetone = 3528.708 kg/h m H = 455.144 kg/h
8
2
10
T=28.1o C mi-propyl-alcohol = 117.307 kg/h mwater = 27547.709 kg/h macetone = 3525.224 kg/h
Qin = Qout TRef = 25 oC ;
455.144x14.419 x (81-25)+ 3528.708x1.259x(81-25) + 169.107x4.193x(81-25) + 117.307x1.716x (81-25) = 455.144 x14,401x(70-25) + 3.485x1,229x(70-25) +3525.224x1,249x(T-25) +27547.709x4,183x(T-25) + 117.307x1,710x(T-25) 42228,319 = 18777,661 + (T – 25) x 7551,149
T = 28.1 oC
- 26 -
5.2.8 ACETONE COLUMN T=102.3o C mi-propyl-alcohol = 137.139 kg/h macetone = 13125.906 kg/h
14
T=45o C mi-propyl-alcohol = 1507.508 kg/h mwater = 29609.634 kg/h macetone = 13138.916 kg/h
A C E T O N E
13
C O L U M N
T=105o C mi-propyl-alcohol = 1370.369 kg/h mwater = 29609.634 kg/h macetone = 13.010 kg/h
15
∆H vap
⎡ T −T ⎤ = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦
0,38
Before the application the formula boiling temperatures ( Tb ) for each of the component must be find at 1,1 bar pressure.
For the boiling point calculation; ln P sat = A −
B T
will be used.
¾ CONDENSER: For acetone: Pc = 47 bar
Tc = 508.1 K
P = 1.0133 bar
T = 329.2 K ( normal boiling point )
ln 1.0133 = A −
B 329.2
ln 47 = A −
B 508.1
Then; A = 10.91 and B = 3587.3 At 1.1 bar pressure, boiling point is;
- 27 -
ln 1.1 = 10.91 −
3587.3 Tb
Tb = 331.706 K
For i-propyl-alcohol Pc = 47.6 bar
Tc = 508.3K
P = 1.0133 bar
T = 355.35 K ( normal boiling point )
ln 1.0133 = A −
B 355.35
ln 47.6 = A −
B 508.3
Then; A = 12.807 and B = 4546.375
At 1.1 bar pressure, boiling point is; ln 1.1 = 12.807 −
4546.375 Tb
Tb = 357.653 K
Substituting the results to the first equation;
∆ H acetone
⎡ 508.1 − 375.3 ⎤ = 29140 × ⎢ ⎣ 508.1 − 331.706 ⎥⎦
0,38
∆Hacetone = 26160,195 kJ/kmol ( 450,417 kJ/kg ) at 102,30C
⎡ 508.3 − 375.3 ⎤ ∆H IPA = 39858 × ⎢ ⎣ 508.3 − 357.653 ⎥⎦
0,38
∆Hi-propyl-alcohol = 38014 kJ/kmol (632,618 kJ/kg ) at 102,3 0C
For the mixture; ∆Hmixture = 450.417×0.99+632.618×0.01 ∆Hmixture =452.24 kJ/kg mT =13263.045 kg
For the energy balance for the mixture; Q = mT ×∆Hmixture =6 × 106 kJ
- 28 -
For water: Pc = 220.5 bar
Tc = 647.3 K
P = 1.0133 bar
T = 373.15 K ( normal boiling point )
ln 1.0133 = A −
B 373.15
ln 220.5 = A −
B 647.3
Then; A = 12.72 and B = 4743.39
At 1.1 bar pressure, boiling point is; ln 1.1 = 12.72 −
4743.39 Tb
Tb = 375.723 K
¾ REBOILER:
⎡ 508.1 − 378 ⎤ ∆H vap,acetone = 29140 × ⎢ ⎣ 508.1 − 331.706 ⎥⎦
0,38
∆H vap,acetone = 25956.795kJ / kmol = 446.915kJ / kg
For Water: ∆H vap,water
⎡ 647.3 − 378 ⎤ = 40683 × ⎢ ⎣ 647.1 − 375.723 ⎥⎦
0,38
∆H vap,water = 40553, 043kJ / kmol = 674,872kJ / kg
⎡ 508.3 − 378 ⎤ ∆H vap,i −propyl−alcohol = 39858 × ⎢ ⎣ 508.3 − 357.653 ⎥⎦
0,38
∆H vap,i − propyl−alcohol = 37719.801kJ / kmol = 627.722kJ / kmol
yacetone = 4.364*10-4 ; ywater = 0.955 ; yIPA = 0.045 - 29872 - × 0, 955 + 627, 722 × 0, 045 = 672, 945 kJ kg ∆ H vap,mixture = 446, 915 × 4, 364 × 10 −4 + 674,
Balance;
Q=mT∆Hvap,mixture=30993.013×672,945=20,86×106 kJ
5.2.9 IPA COLUMN 17
T=105o C 15 mwater = 29609.634 kg/h macetone = 13.010 kg/h mi-propyl-alcohol = 1370.369 kg/h
C I O P L A U M N
16
Same procedure is followed as in acetone column. Tb,i-propyl-alcohol = 84.653 0 C Tb,water = 102.723 0 C ∆Hf,water = 40683 kJ/kmol ∆Hf,i-propyl-alcohol = 39858 kJ/kmol ∆Hf,acetone = 29140 kJ/kmol ∆Hvap,water = 40294.194 kJ/kmol = 2236.081 kJ/kg ∆Hvap,i-propyl-alcohol = 38014 kJ/kmol = 632.618 kJ/kg ∆Hvap,acetone = 26160.195 kJ/kmol
Since acetone is neglected; ywater=0.13 ; yIPA=0.87 ∆Hvap,mixture = 2236.081×0,13+632,618×0.87 = 841.068 kJ/kg
For the energy balance for the mixture; Q = mT ×∆Hmixture = 1941.326 kg × 841.068 kJ/kg - 30 -
T=111.5o C mi-propyl-alcohol = 1370.369 kg/h mwater = 202.326 kg/h macetone = 13.010 kg/h
mwater = 29407.290 kg/h
Q = 1.633*106 kJ
Reboiler:
∆H vap,WATER
⎡ 647,3 − 384,5 ⎤ = 40683 × ⎢ ⎥ ⎣ 647,1 − 375, 723 ⎦
0,38
= 40179,523kJ / kmol = 2230,892 kJ / kg Q=mT∆Hvap,water=2230,892 ×29407.290=65,604×106 kJ
- 31 -
REFERENCES ¾
Treybol, R.E, Mass-Transfer Operations, 3rd Edition, McGraw-Hill Book Company, 1980
¾
Coulson, J.M., Richardson,J.F, Chemical Engineering Volume6, Great Britain Pergamon Press, 1977
¾
Yaws, C., Physical Properties, McGraw-Hill Book Company, USA, 1977
¾
Othmer-K, Encyclopaedia Of Chemical Technology Volume-1, John Willey and Sons, 1978
¾
Foust, A.S., Wenzel, L.A., Clump, C.W., Meus, L., Anderson, L.B., Principles of Unit Operations, John Willey and Sons Inc, USA, 1960
¾
Perry, R.H., Green, D., Perry’s Chemical Engineers’ Handbook, 5th Edition, McGrawHill International Ed., 1984
¾
McCabe, W.L., Smith, J.C., Horriott, P., Units Operations of Chemical Engineering, McGraw-Hill International Edition, USA, 1993
¾
Felder, R.M., Rousseau, R.W., Elementary Principles of Chemical Process, 2nd Edition, John Willey and Sons Inc, USA, 1986
‐ 32 ‐
