acetone production process from iso-propyl-alcohol (ipa)
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T.R. EGE UNIVERSITY
Chemical Engineering Department
CHEMICAL ENGINEERING DESIGN PROJECT REPORT I
Prepared by;
05078901 Ürün ARDA 05068091 M.Serkan ACARSER
05068076 Müge METİN 05078875 Sıla Ezgi GÜNGÖR
05068052 Ali KÜÇÜK
Submitted to: Prof.Dr.Ferhan ATALAY
Res.Assist. Nilay GİZLİ
Sezai ERDEM Tuğba GÜRMEN
March 2009 Bornova-İZMİR
Date: 30.03.2009
i
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.
ii
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.
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 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
References 32
‐ 1 ‐
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, i-
propyl-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.
- 2 -
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
Constants in the liquid viscosity equation (A) 658,25 273,84 1139,70 13,82
Constants in the liquid viscosity equation (B) 283,16 131,63 323,44 5,39
Standard Enthalpy of Formation at 298K(kJ/kmol) -242,0 20,43 -272,60 0
Standard Gibbs Energy of Formation at 298K (kJ/kmol) -228,77 62,76 -173,5 0
Constant in The Ideal Gas Heat Capacities Equation(A) 32,243 3,710 32,427 27,143
Constant in The Ideal Gas Heat Capacities Equation(B) 1,923x10-3 2,345x10-1 1,886x10-1 2,73x10-3
Constant in The Ideal Gas Heat Capacities Equation(C) 1,055x10-5 -1,160x10-4 6,405x10-5 -1,380x10-5
Constant in The Ideal Gas Heat Capacities Equation(D) -3,596x10-8 2,204x10-8 -9,261x10-8 7,645x10-9
Minimum Temperature For Antoine Constant (°C) 11 -113 0 -259
Maximum Temperature For Antoine Constant (°C) 168 -33 111 -248
‐ 3 ‐
Table 2: Calculated mol and mass values of substances
Acetone ipropylalcohol Water Hydrogen Basis:100kmol/h multiplied scale factor Basis:100kmol/h multiplied scale factor Basis:100kmol/h multiplied scale factor Basis:100kmol/h multiplied scale factor
kmol/h ton/year kmol/h kg/h ton/year kmol/h ton/year kmol/h kg/h ton/year kmol/h ton/year kmol/h kg/h ton/year kmol/h ton/year kmol/h kg/h ton/year
1 ‐ ‐ ‐ ‐ ‐ 90,937 47872,96 228,797 13749,785 120448,4 44,786 7067,741 112,682 2029,966 17782,44 ‐ ‐ ‐ ‐ ‐
2 ‐ ‐ ‐ ‐ ‐ 100 52644,1 251,6 15120,154 132452,6 49,25 7772,211 123,913 2232,293 19554,88 ‐ ‐ ‐ ‐ ‐
3 ‐ ‐ ‐ ‐ ‐ 100 52644,1 251,6 15120,154 132452,6 49,25 7772,211 123,913 2232,293 19554,88 ‐ ‐ ‐ ‐ ‐
4 ‐ ‐ ‐ ‐ ‐ 100 52644,1 251,6 15120,154 132452,6 49,25 7772,211 123,913 2232,293 19554,88 ‐ ‐ ‐ ‐ ‐
5 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
6 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
7 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
8 24,148 12286,04 60,756 3528,708 30911,67 0,776 408,518 1,952 117,307 1027,832 3,731 588,794 9,387 169,107 1481,407 90 1584,68 226,44 455,144 3987,07
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 1027,832 607,772 95913,35 1529,15 27547,709 241318 ‐ ‐ ‐ ‐ ‐
11 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ 604,041 95324,56 1519,77 27378,603 239836,6 ‐ ‐ ‐ ‐ ‐
12 0,024 12,211 0,06 3,485 30,722 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ 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 2,282 137,139 1201,345 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐
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 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ 648,799 102387,9 1632,38 29407,290 257607,9 ‐ ‐ ‐ ‐ ‐
17 0,089 45,281 0,224 13,010 113,928 9,063 4771,134 22,803 1370,369 12004,17 4,464 704,47 11,231 202,326 1772,447 ‐ ‐ ‐ ‐ ‐
- 4 -
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.
- 5 -
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.
- 6 -
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-propyl-
alcohol, 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.
- 7 -
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]
- 8 -
RE
AC
TO
R
FLA
SH
IPA C
OL
UM
N
AC
ET
ON
E C
OL
UM
N
FEED DRUM
FURNACE
HEATER COOLER
VAPORIZER CONDERSER
SCR
UB
BE
R
RECYCLE IPA
MOLTEN SALT
1
17
2 3
4
5
6 7
8
11
12
14
10
13
9 15 16
STACK GAS
NATURAL GAS
AIR
WATER
OFF GAS
H2
ACETONE
17
17
WASTE WATER
- 9 -
5.0 APPENDIX
5.1 MASS BALANCES
Production Rate : 115000 ton/year
5.1.1 REACTOR
conversion = 90 %
2
2
2 2
2
5
5
5
5
5 5 55 5
5
5
100*0.9 90 /
100*0.9 90 /
49.25 /
100*0.1 10 /
239.25 /
90 0.376239.2590 0.376
239.25
acetone
H
H O
i propylalcohol
Total acetone i propylalcoholH H O
acetone
H
n kmol h
n kmol h
n kmol h
n kmol h
n n n n n kmol h
y
y
−
−
= =
= =
=
= =
= + + + =
= =
= =
2 5
5
49.25 0.206239.25
10 0.042239.25
H O
i propylalcohol
y
y −
= =
= =
REACTOR
4I-propylalcohol=100 kmol/h H2O = 49.25 kmol/h
5acetone H2 H2O i-propyl-alcohol
- 10 -
5.1.2 FLASH UNIT
• It is assumed that there is no change at temperature and pressure.
i
i
T
ii x
yPP
K ==*
At buble point (T = 81°C)
For acetone * 1161log 7.02447
224 81acetoneP = −+
* 1651.6acetoneP mmHg= 1651.6 1.467
((1.5 /1.013)*760)acetoneK = =
For i-propyl-alcohol
* 1788.02log 8.37895227.438 81IPAP = −
+
* 381.89IPAP mmHg= 381.89 0.339
1125.092IPAK = =
For water
2
* 1668.21log 7.96681228 81H O
P = −+
2
* 369.89 mmHgH O
P =
2
369.89 0.3281125.092H OK = =
F L A S H
8
9
7acetone = 90 kmol/h
H2 = 90 kmol/h
H2O = 49.25 kmol/h
i-propyl-alcohol = 10 kmol/h
acetone H2 H2O i-propyl-alcohol
acetone H2O i-propyl-alcohol
- 11 -
From trial-error; (V/F) = 0.2
2acetone7 IPA 7H O 7
F n n n 149.25 kmol/h= + + =
F = V + L V = 29.85 kmol/h
V0.2F
= 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
At 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 ⇒ ni-propyl-alcohol 8 = (0.026) ×(29.85) = 0.766 kmol/h
ywater = 0.125 ⇒ nwater 8 = (0.125) ×(29.85) = 3.731kmol/h
- 12 -
At stream 9;
L = 119.4 kmol/h
xacetone = 0.551 ⇒ nacetone 9 = (0.551) ×(119.4) = 65.789 kmol/h
xi-propyl-alcohol = 0.077 ⇒ ni-propyl-alcohol 9 = (0.077) ×(119.4) = 9.194 kmol/h
xwater = 0.381 ⇒ nwater 9 = (0.381) ×(119.4) = 45.491 kmol/h
5.1.3 SCRUBBER
T = 810C (354.15 K); P = 1.5 bar (1.48 atm)
Assume 1/1000 of inlet acetone is in off-gas.
H2O
H2 = 90 kmol/h H2O = 3.731 kmol/h Acetone = 24.148 kmol/h i-propyl-alcohol = 0.776 kmol/h
OFF-GAS
H2=90 kmol/h Acetone
Acetone H2O i-propyl-alcohol = 0.776 kmol/h
2 2
2
acetone12
acetone10
Total8 acetone8 IPA8H 8 H O8
Total8
Total12 acetone12 H 12
Total12
n 0.024148 kmol / h
n 24.148 0.024148 24.124 kmol / h
n n n n n
n 24.148 90 3.731 0.776 118.655 kmol / h
n n n
n 0.024148 90 90.024 kmol / h
∴ =
= − =
= + + +
= + + + =
= +
= + =4
acetone12
acetone8
y 0.024148 / 90.024 2.68*10
y 24.148 /118.655 0.203
−= =
= =
8 10
11
12
- 13 -
acetone12 116
acetone8 8
3598 359810.92 10.92T 354.15
43acetone12
6acetone8
y L1 A ; Ay 1 A mV
e em m 1.445P 1.48
y 2.68*10 1 A1.320*10y 0.203 1 A
⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
−−
−= =
−
= ⇒ = =
−= = =
−
From trial-error A is found as 3.523
2 2 2
2
2
11 8
11
H O10 H O8 H O11
H O10
Total10 acetone10 IPA10H O10
Total10
L mAV 1.445*3.523*118.655
L 604.041 kmol / h
n n n
n 3.731 604.041 607.772 kmol / h
n n n n
n 24.124 607.772 0.776 632.672 kmol / h
= =
=
= +
= + =
= + +
= + + =
- 14 -
5.1.4 ACETONE COLUMN
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 0.0891000
= kmol/h
nacetone 14 =89.913-0.089 = 89.824 kmol/h
Since acetone purity is 99%
nı-propyl-alcohol 14 = 0.0189.8240.99
× = 0.907 kmol/h
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
A C C O E L T U O M N N E
14
13
15
acetone = 89.913 kmol/h
i-propyl-alcohol = 9.97 kmol/h
water = 653.263 kmol/h
acetone
i-propyl-alcohol
acetone
i-propyl-alcohol
water
- 15 -
5.1.5 IPA COLUMN
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 = 67.033.0063.9 × = 4.464 kmol/h (neglecting acetone composition)
nwater 16 = nwater 15 - nwater 17 = 653.263 – 4.464 = 648.799 kmol/h
C I O P L A U M N
17
15
16
acetone = 0.089 kmol/h i-propyl-alcohol = 9.063 kmol/h water = 653.263 kmol/h
acetone i-propyl-alcohol water
water
- 16 -
5.1.6 FEED DRUM
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
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.
58.08kg 1ton 8760hamount 89.824 kmol/h * 45700.726ton / year1kmol 1000kg 1year
= × × =
Scale Factor:
ton
year
tonyear
1150002.516
45700.726=
FEED DRUM
1
17
2
i-propyl-alcohol water
i-propyl-alcohol=100 kmol/h water=49.25 kmol/h
i-propyl-alcohol=9.063 kmol/h water=4.464 kmol/h
- 17 -
5.2 ENERGY BALANCES 5.2.1 FEED DRUM
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
FEED DRUM
1
17
2
T=111.5o C mi-propyl-alcohol = 1370.369 kg/h mwater = 202.326 kg/h
T=25o C mi-propyl-alcohol = 13749.785 kg/h mwater = 2029.966 kg/h
T=32.89o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h
- 18 -
5.2.2 VAPORIZER
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
2
0.38
vap,H O
508.3K 382.5 KH 39838 41370.970kj / kmol 2296.473 kJ / kg508.3 K 394.399 K
−⎡ ⎤∆ = = =⎢ ⎥−⎣ ⎦
For IPA : TC = 647.3 K
Tb = 375 K
∆Hf = 40683 kJ/kmol
0,38
vap,IPA647,3K 382,5 KH 40683 40253,505kj / kmol 669,82 kj / kg647,3 K 375 K
−⎡ ⎤∆ = = =⎢ ⎥−⎣ ⎦
T=32.83 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h
VAPORIZER 2 3 T=109.5 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h
2
0,38
cfvap,H O
c b
T TH HT T
⎡ ⎤−∆ = ∆ ⎢ ⎥−⎣ ⎦
- 19 -
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 Molten Salt : We assume ∆T = 20
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
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
HEATER
4 3 T=109.5 o C mwater = 2232.293 kg/h mi-propyl-alcohol = 15120.154 kg/h
T=325 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h
- 20 -
5.2.4 REACTOR
( ) ( ) 22323 HCOCHCHOHCH +→
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
∆Hin,IPA = -272,29 + 20,104 = -252,186 kJ/kmol
( )350
1 5 2 8 3out, IPA
25
H 272,29 32,427 1,886 10 T 6,405 10 T 9,261 10 T dT− − −∆ = − + + × + × − ×∫
∆Hout, IPA = -249,691 kJ/kmol
( )350
2 5 2 8 3out,acetone
25
H 216,685 + 71,96 20,1 10 T 12,78 10 T 34,76 10 T dT− − −∆ = − + × + × + ×∫
∆Hout,acetone= -182,745 kJ/kmol
R E A C T O R
4
5
T=325 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h 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
( )325
1 5 2 8 3in IPA
25
H , 272,29 32,427 1,886 10 T 6,405 10 T 9,261 10 T dT− − −∆ = − + + × + × − ×∫
- 21 -
( )2
3503 5 8 2 12 3
out,H25
H 28.84 10 0.00765 10 T 0.3288 10 T 0.8698 10 T dT− − − −∆ = × + × + × − ×∫
2out,HH 9.466 kj/kmol∆ =
∆Hr0=(-216,685kJ/kmol) – (-272,29 kJ/kmol)
∆Hr0= 55.605 kJ/kmol
kJkmol
r226.44 kmol 55.605H 12591.196 kJ
1×
∆ = =
i i i i rout inQ n H n H H= − + ∆∑ ∑
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
- 22 -
Q [(455.144*12.608) (2232.293*2.035) (1512.015*2.536) (13151.635*1.896)]*(94.7 350)= + + + −
5.2.5 COOLER
Tref=94.7 oC
2p,H
p,water
p,IPA
p,acetone
C 12.608 kJ / kg.K
C 2.035 kJ / kg.K
C 2.536 kJ / kg.K
C 1.896 kJ / kg.K
=
=
=
=
2 2water p,water IPA p,IPA acetone p,acetoneH p,H
Q [(m *C ) (m *C ) (m *C ) (m *C )]* T= + + + ∆
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
COOLER
6 5 T=350 o C mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h
2Hm = 455.144 kg/h
T=94.7o C mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h
2Hm = 455.144 kg/h
- 23 -
5.2.6 CONDENSER
; P* mm Hg
Assumption = PT = 1.5 bar = 1125 mmHg
1*2
2*** =
××
+×
×+
××
+××
−−
−−
dpH
TH
dpalcoholpropyli
Talcoholpropyli
dpwater
Twater
dpacetone
Tacetone
TPPy
TPPy
TPPy
TPPy
From literature;
For acetone: A=7.02447
B=1161
C=224
For water: A=7.96681
B = 1668.21
C=228
For i-propyl-alcohol: A= 8.37895
B=1788.02
C=227.438
CONDENSER 6 7T=94.7o C (Tdp) mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h
2Hm = 455.144 kg/h
0
T=81o C (Tbp) mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h
2Hm = 455.144 kg/h
dp
Blog P* AC T
= −+
Using;
yacetone =0.6 ywater=0.33 yi-propyl-alcohol= 0.07
by trial and error Tdp= 94.7 °C found
- 24 -
0,38
vap508,3K 354 KH 39858* 41169,35kJ / kmol 685,128kJ / kg
508,3 K 366,6 K−⎡ ⎤∆ = = =⎢ ⎥−⎣ ⎦
...)()( ** ++= TbpPxTbpPxP BBAA
m ×Cp ×∆T + m∆Hf = Qtot
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
0,38
cvap f
c b
T TH HT T
⎡ ⎤−∆ = ∆ ⎢ ⎥−⎣ ⎦
∆Hf,acetone = 29140 kJ/kmol
Tc = 508.1 K
Tb = 341.5 K
0,38
vap508.1K 354 KH 29140* 28289.029kJ / kmol 487.07kJ / kg
508.1 K 341.5 K−⎡ ⎤∆ = = =⎢ ⎥−⎣ ⎦
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
0,38
cvap f
c b
T TH HT T
⎡ ⎤−∆ = ∆ ⎢ ⎥−⎣ ⎦
∆Hf,i-propyl-alcohol = 39858 kJ/kmol
Tc = 508.3 K
Tb = 366.6 K
Using;
yacetone = 0.6 ywater= 0.33 yi-propyl-alcohol= 0.07
by trial and error Tbp = 81°C
- 25 -
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
0,38
cvap f
c b
T TH HT T
⎡ ⎤−∆ = ∆ ⎢ ⎥−⎣ ⎦
∆Hf,water = 40683 kJ/kmol
Tc = 647.3 K
Tb = 385.186 K
0,38
vap647,3K 354 KH 40683* 42442.0561kJ / kmol 2356.845 kJ / kg
647.3 K 385.186 K−⎡ ⎤∆ = = =⎢ ⎥−⎣ ⎦
For Hydrogen :
At 94.7 oC and 1.5 bar
2p,HC = 13.225 kJ/kg K
( ) ( )2
3H
Q 455.144 kg 13, 225 kJ / kg.K * 354.15 367.85 -82.464*10 kJ = − =
3 6i p,i i vap,i
i i
6Total i p,i i vap,i
i i
m C T 410.677*10 kJ ; m H 12.702*10 kJ
Q m C T m H 12.3*10 kJ
∆ = − ∆ =
= ∆ + ∆ =
∑ ∑
∑ ∑
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
- 26 -
5.2.7 SCRUBBER
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
T=81o C mi-propyl-alcohol = 117.307 kg/h mwater = 169.107 kg/h macetone = 3528.708 kg/h
2Hm = 455.144 kg/h
T=70o C macetone = 3.485 kg/h
2Hm = 455.144 kg/h
T=28.1o C mi-propyl-alcohol = 117.307 kg/h mwater = 27547.709 kg/h macetone = 3525.224 kg/h
mwater = 27378.603 kg/h
8
11
12
10
- 27 -
5.2.8 ACETONE COLUMN
0,38
cvap f
c b
T TH HT T
⎡ ⎤−∆ = ∆ ⎢ ⎥−⎣ ⎦
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;
CONDENSER:
For acetone:
Pc = 47 bar Tc = 508.1 K
P = 1.0133 bar T = 329.2 K ( normal boiling point )
2.3290133.1ln BA −=
1.50847ln BA −=
Then; A = 10.91 and B = 3587.3
At 1.1 bar pressure, boiling point is;
T=45o C mi-propyl-alcohol = 1507.508 kg/h mwater = 29609.634 kg/h macetone = 13138.916 kg/h
T=102.3o C mi-propyl-alcohol = 137.139 kg/h macetone = 13125.906 kg/h
T=105o C mi-propyl-alcohol = 1370.369 kg/h mwater = 29609.634 kg/h macetone = 13.010 kg/h
A C C O E L T U O M N N E
14
13
15
sat Bln P AT
= − will be used.
- 28 -
bT
3.358791.101.1ln −= 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 )
35.3550133.1ln BA −=
3.5086.47ln BA −=
Then; A = 12.807 and B = 4546.375
At 1.1 bar pressure, boiling point is;
bT375.4546807.121.1ln −= Tb = 357.653 K
Substituting the results to the first equation;
0,38
acetone508.1 375.3H 29140
508.1 331.706−⎡ ⎤∆ = × ⎢ ⎥−⎣ ⎦
∆Hacetone = 26160,195 kJ/kmol ( 450,417 kJ/kg ) at 102,30C
0,38
IPA508.3 375.3H 39858
508.3 357.653−⎡ ⎤∆ = × ⎢ ⎥−⎣ ⎦
∆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
- 29 -
For water:
Pc = 220.5 bar Tc = 647.3 K
P = 1.0133 bar T = 373.15 K ( normal boiling point )
15.3730133.1ln BA −=
3.6475.220ln BA −=
Then; A = 12.72 and B = 4743.39
At 1.1 bar pressure, boiling point is;
bT
39.474372.121.1ln −= Tb = 375.723 K
REBOILER:
0,38
vap,acetone508.1 378H 29140
508.1 331.706−⎡ ⎤∆ = × ⎢ ⎥−⎣ ⎦
vap,acetoneH 25956.795kJ / kmol 446.915kJ / kg∆ = =
For Water:
0,38
vap,water647.3 378H 40683
647.1 375.723−⎡ ⎤∆ = × ⎢ ⎥−⎣ ⎦
vap,waterH 40553,043kJ / kmol 674,872kJ / kg∆ = =
0,38
vap,i propyl alcohol508.3 378H 39858
508.3 357.653− −
−⎡ ⎤∆ = × ⎢ ⎥−⎣ ⎦
vap,i propyl alcoholH 37719.801kJ / kmol 627.722kJ / kmol− −∆ = =
yacetone = 4.364*10-4 ; ywater = 0.955 ; yIPA = 0.045
4 kJkgvap,mixtureH 446,915 4,364 10 674,872 0,955 627, 722 0, 045 672,945−∆ = × × + × + × =
- 30 -
Balance; Q=mT∆Hvap,mixture=30993.013×672,945=20,86×106 kJ
5.2.9 IPA COLUMN
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
C I O P L A U M N
17
15
16
T=105o C mwater = 29609.634 kg/h macetone = 13.010 kg/h mi-propyl-alcohol = 1370.369 kg/h
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
- 31 -
Q = 1.633*106 kJ
Reboiler:
0,38
vap,WATER647,3 384,5H 40683
647,1 375,723−⎡ ⎤
∆ = × ⎢ ⎥−⎣ ⎦
40179,523kJ / kmol 2230,892 kJ / kg= =
Q=mT∆Hvap,water=2230,892 ×29407.290=65,604×106 kJ
‐ 32 ‐
REFERENCES
Treybol, R.E, Mass-Transfer Operations, 3rd Edition, McGraw-Hill Book Company,
1980
Coulson, J.M., Richardson,J.F, Chemical Engineering Volume6, Great Britain
Pergamon Press, 1977
Yaws, C., Physical Properties, McGraw-Hill Book Company, USA, 1977
Othmer-K, Encyclopaedia Of Chemical Technology Volume-1, John Willey and Sons,
1978
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
Perry, R.H., Green, D., Perry’s Chemical Engineers’ Handbook, 5th Edition, McGraw-
Hill International Ed., 1984
McCabe, W.L., Smith, J.C., Horriott, P., Units Operations of Chemical Engineering,
McGraw-Hill International Edition, USA, 1993
Felder, R.M., Rousseau, R.W., Elementary Principles of Chemical Process, 2nd
Edition, John Willey and Sons Inc, USA, 1986
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