Plant Design of MIBK

Table of Contents 1.INTRODUCTION ............................................................................................................................1 1.1PHYSICAL PROPERTIES ........................................................................................................1 1.2USES...........................................................................................................................................2 2. MANUFACTURING PROCESS ....................................................................................................4 2.1 PROCESS SELECTION ...........................................................................................................4 2.2 OVERALL PROCESS DESCRIPTION ...................................................................................6 3. MATERIAL BALANCE .................................................................................................................9 4. ENERGY BALANCE ...................................................................................................................16 5. EQUIPMENT DEIGN ...................................................................................................................31 5.1 LIST OF ITEMS TO BE REPORTED ....................................................................................31 5.2 AVAILABLE INFORMATION .............................................................................................31 5.3 ADDITIONAL INFORMATION ...........................................................................................31 5.4 DESIGN METHODOLOGY...................................................................................................35 5.5 DESIGN COMPUTATION (PROCESS)................................................................................41 5.6 DESIGN METHEDOLOGY (MECHANICAL) .....................................................................46 5.7 DESIGN COMPUTATIONAL (MECHANICAL) .................................................................50 6. DESIGN DATA SHEET ...............................................................................................................55 7. NOMENCLEATURE ....................................................................................................................56 8. REFERENCES ..............................................................................................................................58

0

1.INTRODUCTION Methyl isobutyl ketone (MIBK) (CAS no. 108-10-1), also known as 4-methyl-2-pentanone, (chemical formula C6H12O) is a chemically stable compound possessing a bland, ketonic odor and used generally as a solvent. The structure of MIBK is as follows.

It is by far the most important of all the acetone derivative solvents. In terms of production volume MIBK lies behind acetone and 2-butanone. The total produced amount of MIBK in world in 1987 is 1,80,000 ton and in 1996 3,80,000 ton [1]. MIBK is a medium evaporating solvent. This higher ketone homolog of acetone methyl ethyl ketone is miscible with most organic solvents. MIBK also exhibits strong solvency for many natural and synthetic resins, gums, waxes, oils, cellulose esters and other coating systems. Water solubility of MIBK is not good compare to other ketone solvents like acetone and MEK (methyl ethyl ketone). This property makes MIBK an useful liquid-liquid extraction solvent. 1.1PHYSICAL PROPERTIES MIBK is a colorless liquid with a less pronounced ketone like odor. Some physical properties are given below.

Melting point

-80 0C

Boiling point

116 0C

Density₄20

0.8004

nD20

1.3959

Enthalpy of evaporation

36.15 kJ/mol

Specific heat (20 0C)

1922 kJ/g

Enthalpy of combustion

3740 kJ/mol

Flash point (DIN 51 755)

14 0C

Ignition temperature

475 0C

Ignition class (VED)

G1

Explosion class (VED)

1

Solubility at 25 0C MIBK in water Water in MIBK

1.7 wt.% 1.9 wt.%

MIBK is poorly soluble in water but is miscible with common organic solvents. It forms an azeotrope with water and a large number of solvents. For example, the azeotrope with water (bp 87.9 0C) contains 75.7 wt.% and that with n-butanol (bp 114.4 0C) 70 wt.% MIBK.

1

MIBK undergoes auto-oxidation to form peroxide. After evaporation of MIBK-water mixture in presence of air, a dangerous, increasing concentration of peroxide in the aqueous phase has been reported [1]. In condensation reactions with carbonyl compounds, the α-methyl group usually reacts. Only formaldehyde is able to undergo condensation reactions with the α-methylene group. The industrial reactions such as hydrogenation and reductive amination may be carried out with the keto group. 1.2USES More than 60% of the MIBK produced in the United States is used as a protective coating solvent. Significant volumes are also used in the manufacture of rubber anti-ozonates, in rare metal extraction, as a solvent for pesticides, in adhesives and pharmaceuticals and as a denaturant for ethyl alcohol. A profile of MIBK used in the United States in 1980 is given in Table 1 [4]. Table 1. MIBK uses in United States in 1980

Use

Percentage

Surface coatings (industrial, lacquers)

68

Rubber anti-ozonates

8

Rare metal extraction

8

miscellaneous

6

Export

10

Total

100

Surface Coating: In surface coatings, MIBK is often used as the medium-boiling component of nitrocellulose lacquer solvent blends. It is frequently blended with cheaper aromatic solvents to achieve blended active solvent characteristics in high solid lacquers. Due to its strong solvency and low density, MIBK finds application in the higher-solid coating systems. MIBK has demonstrated that it can be successfully used in high solids acrylic, acrylic/urethane polyesters, and oil-modified polyesters formulations. These high-solids coatings are required in many applications where EPA air quality regulations limit the maximum weight of organic solvent per unit volume of coating. Because of its strong solvency, MIBK can be used to increase solids content of the coating solution while maintaining desirable viscosity characteristics. A further use is as a solvent for dyes in printing industry. Rare Metal Extraction: A rapidly growing field of application for MIBK is in the recovery, separation and purification of rare transition elements, usually as their complexes. Typical applications include zirconium from hafnium for nuclear reactors, tantalum from niobium for electronics and nuclear applications, tungsten from molybdenum, and uranium from thorium. 2

Others: A derivative of MIBK, n-(1,3-dimethylbutyl)-n-phenyl-p-phenylenediamine is used as an antioxidant for rubber. Miscellaneous end uses for MIBK include pharmaceuticals, where it is used by major antibiotic producers in the extraction and purification of penicillin from the fermentation broth, in adhesive formulations based on nitrile rubber and acrylics, and in pesticides, in the removal of paraffins from mineral oil, for the production of lubricating oils where it is used as a solvent for active ingredients. The importance of MIBK as an intermediate for synthesis is relatively low. The most important product is 4-methyl-2-pentanone obtained by hydrogenation of ketone. It has some importance as a polymerization initiator for ethylene and for handling of unsaturated polyester resins [1]. 3. STORAGE AND TRANSPORTATION MIBK is available with a purity of >99%. Typical specification is given below

Content

>99 wt.%

Water content (ASTM D 1364)

0.1 % max.

Hazen colour no. (APHA, DIN 53 409)

10 max.

Nonvolatile residue (ASTM D 1353)

0.002 wt.% max.

Boiling range (DIN 51 751)

114-117 0C

Acid content (as acetic acid)

0.002 wt.% max.

Conditions that could lead to auto-oxidation (presence of air, light, heat or heavy metals) or condensation (strongly acidic or basic media) should be avoided during storage. Containers of V2A steel are recommended for both storage and transportation. Hazard codes for transportation are as follows: IMDG code: class 3.2 RID/ADR: class 3, number 3 b CFR 49: 172.101 flammable liquid UN number: 1245

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2. MANUFACTURING PROCESS There are several commercial processes discussed in the literature for the manufacture of MIBK from acetone. These include: i. Low temperature condensation of acetone in the presence of a base catalyst in the liquid phase to yield di-acetone alcohol (DAA) followed by dehydration of the DAA to mesityle oxide (MO) in the presence of acid catalyst. MO is then hydrogenated to MIBK in the presence of a nickel catalyst [2]. ii. Higher temperature condensation of acetone in the liquid phase in the presence of either a basic or acid catalyst directly to MO followed by hydrogenation to MIBK. Isophorone is a by-product of this reaction [3,4]. This process is of commercial significance only if Isophorone is recovered as a co-product since MO yields are lower than low temperature condensation process. Catalyst such as copper chr..omite[1]or zirconium phosphate [1]is used for condensation, and palladium on aluminum oxide for the hydrogenation. iii.Vapour phase condensation of acetone and hydrogen in the presence of a catalyst to directly form either MO or MIBK [5]. iv. Direct liquid phase condensation-dehydration-hydrogenation of acetone in the presence of hydrogen using a bi-functional catalyst to form MIBK. For this process a combination of a cation exchanger with palladium is generally used as a catalyst [1]. The acetone is fed with hydrogen over the palladium charged catalyst at 1300C and 0.5-5.0 MPa. A selectivity of >95% at a conversion of < 50% is thereby attained. As a result of the increasing production of acetone by dehydrogenation of isopropyl alcohol, the production of 4-methyl-2-pentanone as a byproduct in this process is becoming more important [1]. 2.1 PROCESS SELECTION The low temperature condensation of acetone route and the direct liquid phase condensationdehydration-hydrogenation processes are of commercial significance. Several advantages are clamed for the one step MIBK synthesis including: 1. Proton catalysis and metal catalysis with one specific resin. 2. The thr..ee steps of condensation, dehydration and hydrogenation all takes place at the same time. 3. High conversion rate of acetone because MO is removed from the equilibrium. 4. No need for additional separate hydrogenation unit. But this one step synthesis of MIBK is relatively latest technology and most of the advancement of the process is copyright protected. The process description of this process is not usually available in literature. On the other hand the low temperature condensation route is the oldest process for 4

manufacturing MIBK. Most of the plants in world producing MIBK follow this route. That’s why the low temperature condensation of acetone route is selected for the plant design purpose and reaction chemistry is discussed in further details. MIBK Via Low Temperature Condensation Of Acetone As said earlier, thr..ee distinguished steps are involved in this route of MIBK manufacture, viz. condensation, dehydration and hydrogenation. These steps are discussed below in detail.

Condensation of Acetone to DAA: The production of diacetone alcohol (DAA) by acetone condensation at low temperature is expressed by the following chemical equation. This is a base catalyzed aldol condensation reaction. ⇔

(

)

(

)

DAA

Acetone

The industrial production of DAA takes place in the liquid phase using an alkaline catalyst. Water inhibits the condensation reaction, so it is advantageous to use acetone containing not more than 0.5 wt.% water [6]. According to a Shell patent [7], solid catalysts, e.g., Ba(OH)2 or Ca(OH)2 , impregnated on a suitable carrier, are used in a fixed bed reactors the condensing agent. Dilute NaOH can also be used as catalyst in a well-mixed reactor. When dilute catalyst are employed, they must be removed from the reactor product or neutralized prior to recovery of the DAA. The reaction of acetone to DAA is equilibrium controlled. Increasing the temperature decreases the equilibrium concentration of DAA [2]. Temperatures of 10-30 0C are used commercially and at 20 0C

the yield of DAA is about 6 mol/1000 mol catalyst/h using the solid catalyst system. When

using a soluble catalyst in the liquid phase, the reactor residence time is reported to be 5-10 min [2]. The reaction is exothermic and the heat of reaction is 4.9 kcal/g-mol[2]. About 8-10 % conversion (80-90 % of equilibrium conversion) is obtained per pass thr..ough the reactor. Selectivity to DAA is reported to be 90-95 % [2]. By products are mainly MO and higher molecular weight condensation products according to the following reaction. (

acetone

)

MO

Dehydration of DAA to MO: DAA is converted to MO in the presence of a mineral acid at 100-120 0C. The catalyst normally employed is 0.05% phosphoric acid even though sulfuric acid and other dehydrating agents can be used. The reaction is equilibrium limited. However, since it can be carried out in the base of a distillation tower, the MO and water are continuously removed from the reaction zone and the reaction can be essentially carried to completion [2]. The reaction is highly selective (90-95%) to 5

MO [2].Mesitylene and phorone(EQ. 4 and 5) are main by-products of the reaction. The reaction scheme is shown below.

(

)

(

)

(

→

DAA

) MO

( Acetone

(

)

Mesitylene

)

Acetone

(

)

Phorone

Hydrogenation of MO to MIBK: MIBK is produced by hydrogenation of MO either in the liquid or vapour phase. Catalyst employed for either vapour or liquid phase hydrogenation are nickel [2,8], copper-chr..omium [9], or palladium on carbon or alumina [10]. The vapour phase hydrogenation is usually carried out at 150170 0C and atmospheric pressure using a space velocity of 0.5 to 1.5 volumes/h/volume of catalyst, based on the liquid MO feed. The liquid phase reaction is carried out at 60-130 0C and pressure ranging from 3 to 30 atm. [2]. Liquid hourly space velocities are about 10h-1 when feeding pure MO to 1h-1 when feeding lower concentrations of MO. Over-hydrogenation to MIBC can occur if the MIBK concentration in the reactor outlet increases above 90 %. Selectivity to MIBK is 90-98 % if the conversion of MO is less than 95%. By-products of this reaction include acetone and isopropanol. (

)

(

MO

MIBK

( )

( MIBK

)

isipropanol

Acetone

(

)

)

(

)

MIBC

2.2 OVERALL PROCESS DESCRIPTION An overall flow plan for the manufacture of MIBK from acetone is presented in the following flow sheet. Acetone is fed directly into reactor packed with Ca(OH)2. The reactor can be staged with inter-cooling between each stage to take advantage of favorable equilibrium of DAA at low temperature. The reactor may be jacketed with the arrangement of cooling water supply for the same purpose. For economic reasons, the reactor outlet temperature is maintained at 0-20 0C to achieve a conversion per pass of about 9%. After the reaction step is completed, the product can be 6

stabilized with phosphoric acid and stripped free of acetone, producing a 90 % DAA mixture. Recovery of acetone from DAA should be carried out at low pressure to avoid reversion of the DAA back to acetone at temperature greater 120 . The crude DAA is mixed with 0.05 wt.% phosphoric acid and fed to a dehydrating column where acetone and MO-water azeotrope mixture are distilled overhead. Heavy by-products are purged from the bottom of the column. Acetone is then recovered for recycle from MO in the MO recovery column. The can further purified from water prior to hydrogenation if required. The hydrogenation may be carried out in the liquid phase or the gaseous phase. Hydrogenation in the liquid phase is a simpler process, since inter alia it is possible to operate at lower temperatures, the mesityl oxide does not need to be previously evaporated, and there is no need to recondense the reaction product after completion of the reaction. However, that hydrogenation in the liquid phase has the drawback that the life of the catalyst is very short. It has now been found that this drawback can be obviated by treating the crude mesityl oxide with hydrogen peroxide and inorganic base solution prior to the hydrogenation [11]. Any suitable inorganic base can be used in the treatment, but relatively strong inorganic bases such as aqueous caustic alkali solutions are particularly advantageous. Sodium or potassium hydroxides are examples of suitable caustic alkalis which can be used successfully. The treatment with hydrogen peroxide and inorganic base solution can be carried out simply by adding an aqueous solution of hydrogen peroxide and base solution to the mesityl oxide. A good contact between the reagents can be ensured by stirring or other suitable measures. The contact time does not need to be long; in most cases 10 seconds is quite sufficient. In practice, however, it is more convenient to work with longer contact times and there is no objection to this. There is generally no advantage in extending the time of treatment beyond about 30 minutes, however. The hydrogen peroxide and the inorganic base solution can be added to the mesityl oxide in any given sequence, or simultaneously. In view of the comparatively low degree of stability of hydrogen peroxide in an alkaline medium, but generally it is advised first to admix the hydrogen peroxide well with the mesityl oxide and only then to add the inorganic base solution, in order to prevent unnecessary consumption of hydrogen peroxide. The reaction may be carried out at room temperature, although there is no abjection to using higher or lower temperatures. The operating temperature normally lies between about 5° and about 50° C. The quantities of the reagents required are not large. They can, for example, be used in amounts, per liter of crude mesityl oxide, ranging from about 0.3 to about 1.0 ml. of hydrogen peroxide solution of 8% concentration and from about 0.3 to about 1.0 ml. of caustic alkali solution of 10% concentration, or in equivalent amounts of solutions of other concentrations. The concentrations of the solutions are not critical, in practice they are so chosen as to make it easy to proportion with the desired accuracy the small amounts of the reagents required. Hydrogen peroxide solutions of about 5 to 7

about 15% w. concentration in water and aqueous inorganic base solutions of about 5 to about 15 wt.% concentration are conveniently used. After the treatment with hydrogen peroxide and inorganic base according to the invention, it is sufficient to separate the aqueous phase from the mesityl oxide in a simple manner, as, for instance, by decantation. Washing with water is possible but is not generally necessary. MIBK is produced from MO by hydrogenation in a reactive distillation tower or in a fixed bed reactor over Ni catalyst in the presence of hydrogen [11]. Raney nickel is preferably employed as nickel catalyst. Pure MIBK is finally recovered from uncovered MO in a two-step distillation scheme where lights are removed in the first tower and pure MIBK is recovered as an overhead product in the second tower.

8

3. MATERIAL BALANCE Data required for calculation: Compound

Chemical Formula

Molecular weight (kg/kmol)

Sp. Heat (kJ/kg 0C)

Boiling point @ 1atm. (0C)

Heat of vaporization (kJ/kg)

Acetone

C3H6O

58

0.084

56.5

523.00

DAA

C6H12O2

116

3.49

166

MO

C6H10O

98

3.84

129

2001.93

-253.54

MIBK

C6H120

100

1.25

117

462.94

-288.54

Mesitylene

C9H12

120

1.16

164.7

Phorone

C9H14O

138

1.50

198.5

60

1.26

108

Isopropanol C3H8O Hydrogen

H2

2

14.304

Water

H2O

18

4.188

Standard heat of formation (kJ/mol) -551.69

880.12 0

100

2245.00

-241.83 (as vapour)

Production rate of MIBK: Production rate of MIBK: Basis: 20 TPD Plant operating hour: 24 hr. /day Production rate: 833.34 kg/hr. It is assumed that 1% MIBK will be lost in the downstream. So, MIBK to be produced in the reactor: 833.34 × 1.01 kg/hr. = 841.67 kg/hr. = 8.4167 kmol/hr. 1. HYDROGENATION REACTOR (MIBK REACTOR): Assumption: 95% conversion of the reaction (6).



100% conversion of the reaction (7)  Reaction (8) is neglected since there is no possibility of over-hydrogenation. 5% excess MO is taken.  No excess hydrogen is used.  

9wt% i.e. 84.63 kg/hr. MO is present in the reactor outlet. 1.8 wt. % i.e. 18.00kg/hr.acetone is present in the reactor outlet Calculation: MIBK to be produced in the reactor: 8.4167 kmol/hr. 9

MO required: (8.4167/0.95) kmol/hr. = 8.8596 kmol/hr. = 868.24 kg/hr. Excess MO in output: 868.24 × 0.05 = 43.412kg/hr. H2 required: (8.4167/0.95)kmol/hr. = 8.8596 kmol/hr. = 17.19kg/hr. Excess H2 in outlet: 8.8595 ×0.05 kmol/hr. = 0.4429 kmol/hr. = .8859 kg/hr. MO in inlet: 868.24 + (84.68 – 43.413) kg/hr. = 910 kg/hr. H2 used for reaction (7): .4429 kmol/hr. Isopropanol formed: 0.4429 kmol/hr. = 26.57kg/hr. Acetone required: 0.4429 kmol/hr. = 25.68kg/hr. Acetone in inlet: (25.68 + 18.00) kg/hr. = 43.68 kg/hr. Component

Inlet (kg/hr.)

Outlet (kg/hr.)

MO

910.00

84.63

Acetone

43.68

18.00

Hydrogen

17.19

MIBK

841.67

Isopropanol

25.79

Total

970.87

9970.87

2. DEHYDRATING COLUMN (MO REACTOR): Assumption: 100% conversion for reaction (3).



80% conversion for reaction (4). 90% conversion for reaction (5). 1 wt. % i.e. 10.65 kg/hr.mesitylene is present in the reactor outlet.  9 wt. % i.e. 10.66 kg/hr.phorone is present in the reactor outlet.  4.34wt% i.e. 50.101 kg/hr.acetone is present in the reactor outlet. Calculation: MO to be produced: 910 kg/hr. = 9.28 kmol/hr. DAA required: 9.28 kmol/hr. = 1073 kg/hr. Water produced: 9.28 kmol/hr. = 170.79 kg/hr. Mesitylene to be produced: 10.65 kg/hr. = 0.088 kmol/hr. Acetone required: (3 × 0.088)/0.80 = .33 kmol/hr. = 19.30 kg/hr. Excess acetone in outlet: 0.33 × 0.2 = 0.66 kmol/hr. = 38.28 kg/hr. Water produced: (3 × 0.088) kmol/hr. = 0.264kmol/hr. = 4.75 kg/hr. 10

Phorone to be produced: 10.66 kg/hr. = 0.077 kmol/hr. Acetone required: (3 × 0.077)/0.90 = .2566 kmol/hr. = 14.88 kg/hr. Excess acetone in outlet: .2566 × 0.2 = 0.0513 kmol/hr. =2.59 kg/hr. Water produced: (2 × 0.077) kmol/hr. = 0.154 kmol/hr. = 2.772 kg/hr. Acetone in inlet: (50.101+ 19.30 + 14.88 – 38.28– 2.59) kg/hr. = 79.20 kg/hr. Component

Inlet (kg/hr.)

DAA

1073

Acetone

79.20

Outlet (kg/hr.)

50.101

MO

910.00

water

170.79

Mesitylene

10.65

Phorone

10.66

Total

1152.20

1152.20

3. DAA REACTOR: Assumption: Conversion and selectivity of reaction (1) is 9% and 59% respectively. 



96.6% conversion for reaction (2).  The acetone used as raw material contains 99.8 wt. % pure acetone and rest 0.2 wt. % water.  Calculation: DAA to be produced: 1073 kg/hr. = 9.28 kmol/hr. Acetone required: (2 × 9.28)/0.09 kmol/hr. = 206.22 kmol/hr. = 11993.37 kg/hr. Excess acetone in outlet: 206.22× 0.91 = 187.66kmol/hr. = 10884.29 kg/hr. We know, selectivity = (moles of desired product/moles of undesired product) Thus moles of undesired product (MO) = (9.28/59) kmol/hr. = 0.1572 kmol/hr. = 15.43 kg/hr. Water formed: 0.788 kmol/hr. = 18.18 kg/hr. Acetone required: (2 × 0.788)/0.966 kmol/hr. = 1.632 kmol/hr. = 94.68 kg/hr. This amount of acetone is used from the unreacted acetone of the reaction (1). Unutilized acetone in the reactor outlet: (10884– 94.68)kg/hr. = 10895.01 Total acetone in inlet: (100/99.8) × 11993.37× 0.998 kg/hr. = 11993.37kg/hr. Water in inlet: (100/99.8) × 11993.37 × 0.002 kg/hr. = 24.13 kg/hr.

11

Component

Inlet (kg/hr.) [2]

Outlet (kg/hr.) [3]

Acetone

11993.37

10895.01

Water

24.13

34.06

DAA

1073

MO

15.43

Total

12017.50

12017.50

4. ACETONE RECOVERY COLUMN: Assumption: Distillate composition is similar to raw material composition i.e. 99.8 wt.% acetone and rest 0.2  wt.% water. Calculation: Acetone at top: (10895.01 – 79.01) = 10816 Total distillate flow rate: (100/99.8) × 10816 kg/hr. = 10837.67 kg/hr. Water at top: (10837.67 - 10816) kg/hr. = 21.67 kg/hr. Component

Inlet (kg/hr.) [4]

Outlet (kg/hr.) Top [4A]

Bottom [5]

Acetone

10895.01

10816.00

79.01

Water

27.67

21.67

5.998

DAA

1079.40

1079.40

MO

15.43

15.43 10837.67

Total

12017.50

1179.85

12017.50

Since in the DAA reactor, some amount of MO and water is produced as the side product, they will enter into the dehydrating column (MO reactor). So the material balance of the MO reactor and thereafter that of the MIBK reactor will be changed accordingly. The revised material balances of these reactors are shown below sequentially. 5. DEHYDRATING COLUMN (MO REACTOR): The change in material balance is not complex. There is some amount of MO and water in the inlet stream and they will carry forward. Since the MO reactor is a reactive distillation (RD) tower, the material balance table is arranged like a distillation tower.

12

Component

Inlet (kg/hr.) [6]

Outlet (kg/hr.) Top [7]

DAA

1079.40

Acetone

79.01

50.10

MO

15.43

927.34

Water

5.998

181.06

Bottom [6A]

Mesitylene

10.65

Phorone

10.66 1158.51

Total

1179.84

21.32

1179.84

6. MO RECOVERY COLUMN: Assumption: Distillate composition is similar to raw material composition i.e. 99.8 wt.% acetone and rest 0.2  wt.% water. Calculation: Acetone at top: (250.53 – 225.79) kg/hr. = 24.74 kg/hr. Total distillate flow rate: (100/99.8) × 24.74 kg/hr. = 24.78 kg/hr. Water at top: (24.78 – 24.74) kg/hr. = 0.04 kg/hr. Component

Inlet (kg/hr.) [7]

Outlet (kg/hr.) Top [7A]

Bottom [8]

Acetone

50.10

4.94

45.15

Water

181.066

0.008

181.05

MO

927.35

927.35 4.95

Total

1158.52

1153.56

1158.52

7. DECANTER: Assumption: Volume of mixing is neglected.



1ml 8 wt.% H2O2 solution and 1ml 10 wt.% NaOH solution is required per liter of crude MO solution [11]. Calculation: Calculation of the volume of the inlet stream: 13

3

Component

Mass flow rate (kg/hr.) Density (kg/m )

Volume (L/hr.)

Acetone

45.15

792.5

56.97

Water

181.05

1000

181.05

MO

927.35

858.0

1080.82

Total

1318.84

Density of H2O2 solution: 1440 kg/m

3

Density of 10 wt.% NaOH solution : 1111.1kg/m

3

8 wt.% H2O2 solution = 5.7 vol% solution. 10 wt.% NaOH solution = 8.9 vol% solution. H2O2 solution required: 6.59 L It contains 0.375 L i.e. 0.54 kg H2O2 and 6.215 L i.e. 6.215 kg water. Thus total mass of the H2O2 solution added: (0.54 + 6.215) kg/hr. = 6.755kg/hr. NaOH solution required: 6.59 L It contains 0.586 L i.e. 0.651 kg NaOH and 6.004 L i.e. 6.004 kg water. Thus total mass of the H2O2 solution added: (0.651 + 6.004) kg/hr. = 6.655 kg/hr. Thus total mass of the aqueous phase: (181.05 + 6.755 + 6.655) kg/hr. = 194.46 kg/hr. Component

Inlet (kg/hr.)

Outlet (kg/hr.)

Feed stream

Settling

Aqueous

Organic

[9]

agent [9A]

phase [9B]

phase [10]

Acetone

45.15

Water

181.05

MO

927.35

45.15 181.05 927.352

H2O2 solution

6.755

6.755

NaOH solution

6.655

6.655

13.41

194.46

1153.55 Total

1166.96

972.502

1166.96

8. HYDROGENATION REACTOR (MIBK REACTOR): Assumption : Same as earlier. Calculation: 14

MIBK to be produced in the reactor: 8.4167 kmol/hr. =841.67kg/hr. MO required: (8.4167/0.95) kmol/hr. = 8.8596 kmol/hr. = 868.24 kg/hr. Excess MO in output: 868.24 × 0.05 = 43.412kg/hr. H2 required: (8.4167/0.95) kmol/hr. = 8.8596 kmol/hr. = 17.72kg/hr. Excess H2 in outlet: 8.8595 ×0.05 kmol/hr. = 0.4429 kmol/hr. = .8859 kg/hr. MO in inlet: 927.25 kg/hr. H2 used for reaction (7): .4429 kmol/hr. Isopropanol formed: 0.4429 kmol/hr. = 26.56kg/hr. Acetone required: 0.4429 kmol/hr. = 25.68kg/hr. Acetone in inlet: (25.68 + 19.47) kg/hr.= 45.15 kg/hr. Acetone in outlet: (45.15-25.68)=19.47kg/hr. MO in outlet=102.52kg/hr. Component

Inlet (kg/hr.) [12 & 13]

Outlet (kg/hr.) [14]

MO

927.35

102.52

Acetone

45.15

19.47

Hydrogen

17.72

MIBK

841.67

Isopropanol

26.56

Total

990.22

990.22

9. 1ST PURIFICATION COLUMN: Assumption: Isopropanol and acetone is separated completely as the distillate.



No intermixing of the top and bottom product occurs. Calculation: Component

Inlet (kg/hr.) [14]

Outlet (kg/hr.) Top [14A] Bottom [15]

Acetone

19.47

19.47

Isopropanol

26.56

26.56

MO

102.52

102.52

MIBK

841.67

841.67 46.03

Total

990.18

944.08 990.18 15

10. FINAL PURIFICATION COLUMN: Assumption: 1 wt.% of the total MIBK is lost with the residue.



The composition of the distillate is 99.8 wt.% MIBK and rest 0.2 wt.% water.  Calculation: MIBK at top: 4166.67 kg/hr. Total distillate flow rate: (100/99.8) × 4166.67 kg/hr. = 4175.02 kg/hr. Water at top: (4175.02 – 4166.67) kg/hr. = 8.35 kg/hr. Component

Inlet (kg/hr.) [16]

Outlet (kg/hr.) Top [17]

Bottom [16A]

MO

102.41

1.67

100.74

MIBK

841.67

833.33

8.34

835

109.08

Total

944.08

944.08

11. ACETONE RECYCLED STREAM: Assumption: Make up acetone and recycled acetone ratio = 1: 9.23 Calculation: Amount of acetone recycled (stream no. 20): (10837.67 + 4.96) kg/hr. = 10842.63 kg/hr. Amount of acetone fed to the DAA reactor (stream no. 2): 12017.41 kg/hr. Amount of make-up acetone (stream no. 1): (12017.41 – 10842.63) = 1174.78 kg/hr.

4. ENERGY BALANCE 1. ACETONE RECYCLED STREAM Assumption: Temperature of the make up acetone: 35 0C.  Temperature of the recycled stream: 50 0C  Reference temperature: 25 0C  Calculation: Heat content of the make up stream [1]: (1172.43 × 0.084 × (35-25)) + (2.34 × 4.188 × (35-25)) kJ/hr.= 1082.84 kJ/hr..

16

Heat content of the recycled stream [20]: (10820.944× 0.084 × (50-25)) + (21.68 × 4.188 × (5025)) kJ/hr.. = 24993.87 kJ/hr. Heat content of the mixed stream [2]: (11993.37 × 0.084 × (T-25)) + (24.03 × 4.188 × (T-25)) kJ/hr. Now, Heat content of the mixed stream = Heat content of the make up stream + Heat content of the recycled stream Or, (11993.37 × 0.084 × (T-25)) + (24.03 × 4.188 × (T-25)) = 1082.84 + 24993.87 Solving we get 0

T = 48.53 C 0

Thus temperature of the mixed stream [2]: 48.53 C 2. DAA REACTOR: Assumption: Only the desired reaction is taken into account.



Specific heat, Cp is calculated at reference temperature.  Cp is assumed constant over the temperature range.  0

0

Let chilling water is available at 5 C and 5 C increase in the outlet temperature is possible.



0

Reference temperature: 25 C Calculation: 0

Reactor inlet temperature: 48.53 C 0

Reactor outlet temperature: 15 C Compound

mi (kg/hr.)

niHi (kJ/s)

mo (kg/hr.)

noH0 (kJ/s)

Acetone

11993.37

?

10895.01

?

Water

24.13

?

34.06

?

DAA

1073

?

MO

15.43

?

nH values are calculated by the expression, n× H = m× Cp× ΔT Heat content of the inlet stream: (11993.37× 0.084 × (48.53-25)) + (24.13× 4.188 × (48.53 -25)) kJ/hr.= 26082.99 kJ/hr..

17

Heat content of the outlet stream: (10895.01× 0.084 × (15-25)) + (34.06× 4.188 × (15-25)) + (1073× 3.49 × (15-25)) + (15.43× 3.84 × (48.53-25)) kJ/hr. = - 49420.12 kJ/hr.. Extent of reaction, ξ: (no-ni)/γ = (10895.01-11993.37)/(-2 × 58) = 9.4686 kmol/hr.. = 9468.62 mol/hr. The standard heat of reaction, ΔHr..o[4]: - 4.9 kcal/mol = - 20.58 kJ/mol Then the enthalpy change of the reaction, ΔH = ξ×ΔHr..0 + ∑noHo - ∑niHi Putting all the values we get, ΔH = -270367.30 kJ/hr.. Energy balance:

Q – Ws = ΔH + ΔEk + ΔEp Q = ΔH = -270367.30 kJ/hr.

Where, Ws = shaft wok = 0 (no moving parts). ΔEk = Kinetic energy change = 0 (neglected). ΔEp = Potential energy change = 0 (neglected). So, 270367.30 kJ/hr..heat is to be removed from the reactor to maintain the desired conversion. Flow rate of chilling water: 270367.30 /(4.188 × 5) = 12911.52 kg/hr. Description

Heat input (kJ/hr..)

Inlet stream [2]

26082.99

Outlet stream [3]

Heat output (kJ/hr..)

49420.12

Reaction

293704.43

Chilling water

270367.30

Total

319787.42

319787.42

3. ACETONE RECOVERY COLUMN: Assumption: The solutions are ideal in nature. 



Saturated liquid feed is introduced in the column.  

Vapour leaving from top of the tower is saturated vapour.  Total condenser is used.  Bottom product exits as saturated liquid.  Heat loss from the column is negligible.  Reflux ratio = 0.10  Reference temperature: 25 0C



18

Calculation: Component

Feed [4] Massflow rate (kg/hr..) 10822.87 27.64 1079.16 15.62

Acetone Water DAA MO

Wt % 90.66 0.23 8.98 0.13

Top [4A] Massflow rate Wt % (kg/hr..) 10816.00 99.80 21.66 0.20

Bottom [5] Massflow rate Wt % (kg/hr..) 79.05 6.70 6.01 0.51 1079.44 91.49 15.33 1.30

Saturation temperature of mixture is determined using the formula Tsat = ∑ Tisat × xi Boiling point of feed: 66.53 0C Boiling point of distillate: 56.59 0C Boiling point of residue: 157.85 0C We know, R = L0/D or, L0 = R × D = 1083.760 kg/hr.. G1 = L0 + D = 11921.4 kj/hr.. HG = (11897.6 × 0.084 × (56.6-25)) + (23.844 × 4.188 × (56.6-25)) + (449 × 119.22) + (11897.6× 523) kJ/hr.. = 6310711.103 kJ/hr.. HL = (1081.6 × 0.084 × (56.6-25)) + (2.16 × 4.188 × (56.6-25)) kJ/hr.. = 5737.5 kJ/hr.. HD = (10816.00× 0.084 × (56.6-25)) + (21.66× 4.188 × (56.6-25)) kJ/hr.. = 31576.49 kJ/hr.. HW = (79.05× 0.084 × (157.9-25)) + (6.01× 4.188 × (157.9-25)) + (1079.16× 3.49 × (157.9-25)) + (15.62× 3.84 × (157.9-25)) kJ/hr.. = 512736.08 kJ/hr.. HF = (10822.87× 0.084 × (66.5-25)) + (138.35 × 4.188 × (66.5-25)) + (1079.16× 3.49 × (66.5-25)) + (15.62× 3.84 × (66.5-25)) kJ/hr.. = 220563.37 kJ/hr.. QC = HG - HL - HD = 6273397.11 kJ/hr.. QB = HD + HW + QC + QL – HF = 6687311.91 kJ/hr. Description

Heat input (kJ/hr..)

Inlet stream [4]

220563.37

Heat output (kJ/hr..)

Top Outlet stream [4A]

31576.49

Bottom Outlet stream [5]

512736.08

Reboiler

6687311.91

Condenser Total

6273397.11 6907875.28

6907875.28

4. DEHYDRATING COLUMN (MO REACTOR): Assumption: Only the desired reaction is taken into account. The solutions are ideal in nature. Vapour leaving from top of the tower is saturated vapour. Total condenser is used. 19

Bottom product exits as saturated liquid. Heat loss from the column is negligible. Reflux ratio = 0.10 The reaction is completed at the feed tray . Reactor inlet temperature [2]: 120 0C Reference temperature: 25 0C Calculation: Component

Inlet (kg/hr..) Mass flowrate (kg/hr..)

Outlet (kg/hr..) `Top

Wt %

Massflowrate (kg/hr..)

Bottom Wt %

DAA

1079.404

91.49

Acetone

70.01

6.70

50.106

4.33

MO

15.436

1.31

927.34

80.04

Water

5.998

0.50

181.066

15.63

Massflowrate (kg/hr..)

Wt %

Mesitylene

10.65

10.06

Phorone

10.67

9.99

Saturation temperature of mixture is determined using the formula Tsat = ∑ Tisat × xi Boiling point of feed: 172.9 0C Boiling point of distillate: 121.3 0C Boiling point of residue: 181.6 0C We know, R = L0/D or, L0 = R × D =115.85kg/hr.. G1 = L0 + D = 6371.82 kg/hr.. HG = (275.9 × 0.084 × (121.5-25)) + (995.92 × 4.188 × (121.5-25)) + (5100 × 3.84 × (121.5-25)) + (2245 × 195.92) + (275.9 × 523) + (5100.00 × 2001.93) kJ/hr.. = 14884564.65 kJ/hr.. HL = (25.08 × 0.084 × (121.5-25)) + (90.54 × 4.188 × (121.5-25)) + (463.64 × 3.84 × (121.5-25)) kJ/hr.. = 208607.58 kJ/hr.. HD = (50.106× 0.084 × (121.5-25)) + (927.34× 3.84 × (121.5-25)) + (181.066× 4.188 × (121.5-25)) kJ/hr.. = 417217.65 kJ/hr.. HW = (10.65× 1.16 × (181.6-25)) + (10.67× 1.50 × (181.6-25)) kJ/hr.. = 4441.02 kJ/hr.. HF = (1079.16× 3.49 × (120-25)) + (70.01× 0.084 × (120-25)) + (15.436× 3.84 × (120-25)) + (5.998× 4.188 × (120-25)) kJ/hr.. = 366371.59 kJ/hr.. QC = HG - HL - HD = 14258739.42 kJ/hr.. Extent of reaction, ξ: (no-ni)/γ = (927.34-77.18)/(1 × 98) = 46.526 kmol/hr. 20

The standard heat of reaction, ΔHr..o is calculated from standard heat of formation using the formula: ΔHr..o = ∑γi × Hfproduct - ∑γi × Hfreactant The standard heat of reaction, ΔHr..o: - 56320 kJ/kmol QB = HD + HW + QC + QL – HF - ξ ΔHr..o = 2049042.42 kJ/hr.. Description

Heat input (kJ/hr.)

Inlet stream [6]

366371.59

Heat output (kJ/hr.)

Top Outlet stream [7]

41721.52

Bottom Outlet stream [6A]

4441.02 17024536.38

Reboiler

14258739.42

Condenser Reaction

524068.86

Total

2939635.63

2939635.63

5. MO RECOVERY COLUMN: Assumption: The solutions are ideal in nature.



Saturated liquid feed is introduced in the column.  Vapour leaving from top of the tower is saturated vapour.  Total condenser is used.  Bottom product exits as saturated liquid.  Heat loss from the column is negligible.  Reflux ratio = 0.20  Reference temperature: 25 0C



Calculation: Component

Feed Massflow rate (kg/hr.)

Top Wt %

Massflow rate (kg/hr.)

Bottom Wt %

Massflow rate (kg/hr.)

Wt %

Acetone

50.106

4.33

4.94

99.80

45.158

3.91

Water

181.06

15.63

0.008

0.20

181.05

15.70

MO

927.35

80.04

927.35

80.39

Saturation temperature of mixture is determined using the formula Tsat = ∑ Tisat × xi Boiling point of feed: 121.3 0C Boiling point of distillate: 56.6 0C Boiling point of residue: 121.6 0C 21

We know, R = L0/D or, L0 = R × D = 4.96 kg/hr. G1 = L0 + D = 29.74 kg/hr. HG = (5.99 × 0.084 × (56.6-25)) + (0.06 × 4.188 × (56.6-25)) + (2245 × 0.06) + (5.99 × 523) kJ/hr. = 3291.31 kJ/hr. HL = (4.95 × 0.084 × (56.6-25)) + (0.01 × 4.188 × (56.6-25)) kJ/hr. = 14.46 kJ/hr. HD = (24.74 × 0.084 × (56.6-25)) + (0.04 × 4.188 × (56.6-25)) kJ/hr. = 70.96 kJ/hr. HW = (45.158 × 0.084 × (121.6-25)) + (181.06 × 4.188 × (121.6-25)) + (927.35 × 3.84 × (121.625)) kJ/hr. = 417611.13 kJ/hr. HF = (50.10 × 0.084 × (121.5-25)) + (927.35 × 3.84 × (121.5-25)) + (181.06 × 4.188 × (121.5-25)) kJ/hr. = 417218.88 kJ/hr. QC = HG - HL - HD = 3205.89 kJ/hr. QB = HD + HW + QC + QL – HF = 93834.66 kJ/hr. Description

Heat input (kJ/hr.)

Inlet stream [7]

417218.88

Heat output (kJ/hr.)

Top Outlet stream [7A]

14.92

Bottom Outlet stream [8]

417611.13

Reboiler

3291.31

Condenser Total

3205.89 420756.98

420756.98

6. DECANTER: Assumption: Since the concentration of H2O2 and NaOH solution is very low, their specific heat is approximated as of pure water. Temperature of H2O2 and NaOH solution is 30 0C Temperature of the feed stream is 40 0C Temperature of the organic phase outlet is 40 0C Calculation: Heat content of the inlets: (45.158 × 0.084 × (40-25)) + (181.05 × 4.188 × (40-25)) + (927.35×3.84 × (40-25)) + (13.41 × 4.188 × (40-25)) kJ/hr. = 65692.02 kJ/hr. Let the temperature of the aqueous phase outlet is T Heat content of the outlets: (45.158 × 0.084 × (40-25)) + (927.35 × 3.84 × (40-25)) + (183,76×4.188 × (T-25)) kJ/hr. = 267361.87 + 3847.51 × (T-25) We know, Heat content of the inlets = Heat content of the outlets Or, 324512.99 = 267361.87 + 3847.51 × (T-25) 22

Solving this equation, T = 39.8 0C Description

Heat Input (kJ/hr.)

Feed stream [9]

65692.02

Settling agent [9A]

56.16

Heat Output (kJ/hr.)

Aqueous phase outlet [9B]

11430.22

Organic phase outlet [10]

53472.37

Total

64902.59

64902.59

7. HYDROGENATION REACTOR (MIBK REACTOR): Assumption: Only the desired reaction is taken into account. Specific heat, Cp is calculated at reference temperature. Cp is assumed constant over the temperature range. Reference temperature: 25 0C Effect of pressure on Cp is neglected. Temperature of the hydrogen stream: 115 0C Pressure in the reactor: 20 atm. Let chilling water is available at 5 0C and 5 0C increase in the outlet temperature is possible. Calculation: Reactor inlet temperature: 115 0C Reactor outlet temperature: 116.80C Compound

mi (kg/hr..)

niHi (kJ/s)

mo (kg/hr..)

noH0 (kJ/s)

MO

927.35

?

102.41

?

Acetone

45.158

?

19.47

?

Hydrogen

17.72

?

?

MIBK

841.668

?

Isopropanol

26.58

?

nH values are calculated by the expression, n× H = m× Cp× ΔT Heat content of the inlet stream: (45.158× 0.084 × (115-25)) + (927.35× 3.84 × (115 -25)) + (17.72× 14.304 × (115-25)) kJ/hr.. = 343632.7 kJ/hr. Heat content of the outlet stream: (19.47× 0.084 × (116.8-25)) + (102.41× 3.84 × (116.8-25)) + (841.668 × 1.25 × (116.8-25)) + (26.58× 1.26 × (116.8-25)) kJ/hr.. = 135906.75 kJ/hr. Extent of reaction, ξ: (no-ni)/γ = (841.668 -0)/(1 × 100) = 42.0834 kmol/hr. = 42083.4 mol/hr.. 23

The standard heat of reaction, ΔHr.o is calculated from standard heat of formation using the formula: ΔHr.o = ∑γi × Hfproduct - ∑γi × Hfreactant The standard heat of reaction, ΔHr.o: -35.00 kJ/kmol Then the enthalpy change of the reaction, ΔH = ξ×ΔHr.0 + ∑noHo - ∑niHi Putting all the values we get, ΔH = -2511362.29 kJ/hr. Energy balance:

Q – Ws = ΔH + ΔEk + ΔEp Q = ΔH = -2511362.29 kJ/hr..

Where, Ws = shaft wok = 0 (no moving parts). ΔEk = Kinetic energy change = 0 (neglected). ΔEp = Potential energy change = 0 (neglected). So, 2511362.29 kJ/hr.heat is to be removed from the reactor to maintain the desired conversion. Flow rate of cooling water: 2511362.29/(4.188 × 10) = 59965.67 kg/hr. Description

Heat input (kJ/hr.)

Inlet stream [12 & 13]

3436932.7

Outlet stream [14] Reaction

135906.75 294583.8

Cooling water Total

Heat output (kJ/hr.)

502272.45 638178.57

638178.57

8. 1ST PURIFICATION COLUMN: Assumption: The solutions are ideal in nature. Saturated liquid feed is introduced in the column. Vapour leaving from top of the tower is saturated vapour. Total condenser is used. Bottom product exits as saturated liquid. Heat loss from the column is negligible. Reflux ratio = 0.10 Reference temperature: 25 0C

24

Calculation: Component Feed Massflow rate

Top

Bottom

Massflow rate

Massflow rate

(kg/hr.)

Wt %

(kg/hr.)

Wt %

19.47

1.97

19.47

42.30

Isopropanol 26.56

2.68

26.56

57.70

MO

102.41

MIBK

841.66

Acetone

(kg/hr.)

Wt %

10.35

102.41

10.85

85.00

841.66

89.15

Saturation temperature of mixture is determined using the formula Tsat = ∑ Tisat × xi Boiling point of feed: 116.8 0C Boiling point of distillate: 86.3 0C Boiling point of residue: 118.3 0C We know, R = L0/D or, L0 = R × D = 4.60 kg/hr. G1 = L0 + D = 253.24 kg/hr. HG = (21.42 × 0.084 × (86.3-25)) + (29.22 × 1.26 × (86.3-25)) + (21.42 × 523) + (29.22 × 880.12) kJ/hr. = 39286.96 kJ/hr. HL = (1.94 × 0.084 × (86.3-25)) + (2.65 × 1.26 × (86.3-25)) kJ/hr. = 214.67 kJ/hr. HD = (19.47× 0.084 × (86.3-25)) + (26.56× 1.26 × (86.3-25)) kJ/hr. = 2151.69 kJ/hr. HW = (102.41× 3.84 × (118.30-25)) + (841.66× 1.25 × (118.30-25)) kJ/hr. = 134849.23 kJ/hr. HF = (19.47 × 0.084 × (116.8-25)) + (102.41× 3.84 × (116.8-25)) + (841.66× 1.25 × (116.8-25)) + (26.56× 1.26 × (116.8-25)) kJ/hr.. = 132831.34 kJ/hr. QC = HG - HL - HD = 36920.6 kJ/hr. QB = HD + HW + QC + QL – HF = 137255.74 kJ/hr.

Description Inlet stream [14]

Heat input (kJ/hr.) 134849.23

Heat output (kJ/hr.)

Top Outlet stream [14A]

2151.69

Bottom Outlet stream [15]

132831.34

Reboiler

36920.6 184626.89

Condenser Total

173928.6

173928.6

9. FINAL PURIFICATION COLUMN: Assumption: The solutions are ideal in nature. 25

Saturated liquid feed is introduced in the column. Vapour leaving from top of the tower is saturated vapour Total condenser is used. Bottom product exits as saturated liquid. Heat loss from the column is negligible. Reflux ratio = 2.33 Reference temperature: 25 0C The tower operates at 3 kg pressure Calculation: Component

Feed Massflow rate (kg/hr.)

Top

Wt %

Massflow rate (kg/hr.)

Bottom

Wt %

Massflow rate (kg/hr.)

Wt %

MO

102.41

10.85

1.67

0.20

100.74

92.36

MIBK

841.66

89.15

833.33

99.80

8.33

7.64

Saturation temperature of mixture is determined using the formula Tsat = ∑ Tisat × xi Boiling point of feed: 161.9 0C Boiling point of distillate: 159.7 0C Boiling point of residue: 179.77 0C We know, R = L0/D or, L0 = R × D = 1945.56 kg/hr. G1 = L0 + D = 13902.82 kg/hr. HG = (2775.002 × 1.25 × (159.7-25)) + (5.56 × 3.84 × (159.7-25)) + (400.38 × 27.81) + (2775.002 × 462.94) kJ/hr. = 1765910.85 kJ/hr.. HL = (1941.66 × 1.25 × (159.7-25)) + (3.89 × 3.84 × (159.7-25)) kJ/hr. = 328939.09 kJ/hr. HD = (833.33× 1.25 × (159.7-25)) + (1.67 × 3.84 × (159.7-25)) kJ/hr. = 141175.74 kJ/hr. HW = (8.33 × 1.25 × (179.8-25)) + (100.74× 3.84 × (179.8-25)) kJ/hr. = 61494.93 kJ/hr. HF = (841.66× 1.25 × (161.9-25)) + (102.41× 3.84 × (161.9-25)) kJ/hr.. = 197865.59 kJ/hr. QC = HG - HL - HD = 1295796.02 kJ/hr. QB = HD + HW + QC + QL – HF = 292836.23 kJ/hr.

26

Description

Heat input (kJ/hr.)

Inlet stream [16]

197865.59

Heat output (kJ/hr.)

Top Outlet stream [17]

141175.74

Bottom Outlet stream [16A]

61494.93

Reboiler

292836.23

Condenser Total

1295796.02 1498467.47

1498467.47

10. PRE-HEATER 1: Assumption: 

High pressure (12.5 bar) saturated steam of temperature 190 0C is available for heating.

Calculation: Heat load: (1007922.47 – (- 242871.32)) kJ/hr. = 68732.72 kJ/hr. Latent heat of vaporization of saturated steam [15]: 1977.4 kJ/kg Steam required: (68732.72 /1977.4) kg/hr. = 34.758 kg/hr. 11. REBOILER 1: Calculation: Heat load: 6619328.626 kJ/hr. Latent heat of vaporization of saturated steam: 1977.4 kJ/kg Steam required: (6619328.626 /1977.4) kg/hr. = 3347.49 kg/hr. 12. CONDENSER 1: Calculation: Heat load: 6275974.58 kJ/hr. Cooling water required: (6275974.58 /4.188 × 10) kg/hr. = 149856.12 kg/hr. 13. COOLER 1: Calculation: Heat load: (366524.344-513359.93) kJ/hr. = -146835.586 kJ/hr. Cooling water required: (146835.586 /4.188 × 10) kg/hr. = 3506.102 kg/hr. 14. REBOILER 2: Calculation: Heat load: 2049042.538 kJ/hr. Latent heat of vaporization of saturated steam: 1977.4 kJ/kg Steam required: (2049042.538 /1977.4) kg/hr. = 1036.23 kg/hr. 15. CONDENSER 2: 27

Calculation: Heat load: 2517973.77 kJ/hr. Cooling water required: (2517973.77 /4.188 × 10) kg/hr. = 60123.53 kg/hr. Calculation: Heat content of the outlet stream: (53.38 × 1.16 × (30-25)) + (53.32× 1.50 × (30-25)) kJ/hr.. = 709.50 kJ/hr. Heat load: (709.50 -22221.66) kJ/hr. = -21512.16 kJ/hr. Cooling water required: (21512.16 /4.188 × 10) kg/hr. = 513.66 kg/hr. 16. REBOILER 3: Calculation: Heat load: 15658.64 kJ/hr. Latent heat of vaporization of saturated steam: 1977.4 kJ/kg Steam required: (15658.64 /1977.4) kg/hr. = 7.92 kg/hr. 17. CONDENSER 3: Calculation: Heat load: 3131.72 kJ/hr. Cooling water required: (3131.72 /4.188 × 10) kg/hr. = 74.778 kg/hr. 18. COOLER 4: Calculation: Heat content of the outlet stream: (225.79 × 0.084 × (40-25)) + (905.29 × 4.188 × (40-25)) + (4636.76 × 3.84 × (40-25)) kJ/hr.. = 64846.43 kJ/hr. Heat content of the inlet stream: 417611.06 kJ/hr. Heat load: (64846.43 - 417611.06) kJ/hr. = -352764.622 kJ/hr. Cooling water required: (1763823.11/4.188 × 10) kg/hr. = 8423.22 kg/hr. 19. PRE-HEATER 2: Calculation: Heat content of the outlet stream: (45.158 × 0.084 × (115-25)) + (927.35 × 3.84 × (115-25)) kJ/hr.. = 320834.22 kJ/hr. Heat content of the inlet stream: (45.158 × 0.084 × (40-25)) + (927.35 × 3.84 × (40-25)) kJ/hr.. = 53472.37 kJ/hr. Heat load: (320834.22 – 53472.37) kJ/hr. = 267361.87 kJ/hr. Latent heat of vaporization of saturated steam: 1977.4 kJ/kg Steam required: (267361.87 /1977.4) kg/hr. = 135.20 kg/hr. 20. REBOILER 4: Calculation: Heat load: 38022.49 kJ/hr. 28

Latent heat of vaporization of saturated steam: 1977.4 kJ/kg Steam required: (38022.49 /1977.4) kg/hr. = 19.22 kg/hr. 21. CONDENSER 4: Calculation: Heat load: 36925.27 kJ/hr. Cooling water required: (36925.27 /4.188 × 10) kg/hr. = 881.69 kg/hr. 22. PRE-HEATER 4: Calculation: Heat content of the outlet stream: 134850.88 kJ/hr. Heat content of the inlet stream: 197868.01 kJ/hr. Heat load: (197868.01 – 134850.88) kJ/hr. = 63017.13 kJ/hr. Latent heat of vaporization of saturated steam: 1977.4 kJ/kg Steam required: (63017.13 /1977.4) kg/hr. = 31.86 kg/hr. 23. REBOILER 5: Calculation: Heat load: 1300599 kJ/hr. Latent heat of vaporization of saturated steam: 1977.4 kJ/kg Steam required: (1300599/1977.4) kg/hr. = 657.732 kg/hr. 24. CONDENSER 5: Calculation: Heat load: 1295794.16 kJ/hr. Cooling water required: (1295794.16 /4.188 × 10) kg/hr. = 30940.64 kg/hr. 25. COOLER 5: Calculation: Heat content of the outlet stream: (19.47× 0.084 × (30-25)) + (26.56 × 1.26 × (30-25)) kJ/hr. = 175.55 kJ/hr. Heat content of the inlet stream: 2152.34 kJ/hr. Heat load: (175.55 - 2152.34) kJ/hr. = -1976.78 kJ/hr. Cooling water required: (1976.78 /4.188 × 10) kg/hr. = 47.2 kg/hr. 26. COOLER 6: Calculation: Heat content of outlet stream: (833.34 × 1.25 × (30-25)) + (1.67 × 3.84 × (30-25)) kJ/hr. = 5240.4 kJ/hr. Heat content of the inlet stream: 141176.41 kJ/hr. Heat load: (5240.4 – 141176.41) kJ/hr. = -135936.01 kJ/hr. 29

Cooling water required: (135936.01 /4.188 × 10) kg/hr. = 3245.84 kg/hr. 27. COOLER 7: Calculation: Heat content of the outlet stream: (8.33 × 1.25 × (30-25)) + (100.74 × 3.84 × (30-25)) kJ/hr. = 1986.33 kJ/hr. Heat content of the inlet stream: 61496.89 kJ/hr. Heat load: (1986.33 – 61496.89 ) kJ/hr. = -59510.56 kJ/hr. Cooling water required: (59510.56 /4.188 × 10) kg/hr. = 1420.97 kg/hr. 28. COOLER 8: Calculation: Heat content of the outlet stream: (10837.67 × 0.084 × (50-25)) + (4.95 × 4.188 × (50-25)) kJ/hr. = 23278.01 kJ/hr. Heat content of the inlet stream: (10837.67 × 0.084 × (56.6-25)) + (4.95 × 4.188 × (56.6-25)) kJ/hr. = 29423.40 kJ/hr. Heat load: (23278.01 – 29423.40 ) kJ/hr. = -6145.39 kJ/hr. Cooling water required: (6145.39 /4.188 × 10) kg/hr. = 146.73 kg/hr. 29. COMPRESSOR 1: Assumption: Hydrogen is available at 3 atm. pressure and 30 0C temperature. The hydrogen stream is to be pressurized upto 20 atm. Adiabatic compression is done in the compression. Heat capacity ratio, γ of hydrogen is 1.405 Calculation: Let outlet temperature is T20C The pressure-temperature relation for adiabatic compression is P(1-γ/γ)T = constant Thus, T2 = T1 × (P1/P2) (1-γ/γ) = 30 × (3/20) (1-1.405/1.405) = 52 0C 30. PREHEATER 3: Calculation: Heat content of the outlet stream: (17.72 × 14.304 × (115-25)) kJ/hr. = 22812.01 kJ/h Heat content of the inlet stream: (17.72 × 14.304 × (52-25)) kJ/hr. = 6843.60 kJ/hr. Heat load: (22812.01 – 6843.60) kJ/hr. = 15968.41 kJ/hr. Latent heat of vaporization of saturated steam: 1977.4 kJ/kg Steam required: (15968.41 /1977.4) kg/hr. = 8.07 kg/hr.

30

5. EQUIPMENT DEIGN The mass transfer equipment which is selected for detailed design is the final purification column. It is a distillation column which separates MIBK from MO and gives a product purity of 99.8 wt% MIBK solution. The column handles 4720.4 kg/hr.. mixture of MIBK and MO having composition 89.15 wt% MIBK and 10.85 wt% MO. A sieve tray column operating at 3 kg pressure is to be designed for this job. 5.1 LIST OF ITEMS TO BE REPORTED 1. Column sizing (diameter and height) 2. Tray hydraulic design 3. Tray support specification 4. Material of construction 5. Pressure drop 6. Roof and bottom specification 7. Shell, nozzle, bolt, and flange specification. 8. Other accessories and fittings. 9. Condenser –Reboiler setup. 10. Column support. 11. Drawing of the column.

5.2 AVAILABLE INFORMATION 1. Flow rate and composition of feed. 2. Top and bottom product specification. 3. Operating pressure of the column.

5.3 ADDITIONAL INFORMATION 1. Saturation temperature of MIBK and MO 2. Vapour-liquid equilibrium data at operating temperature and pressure 3. Relative volatility 4. Top and bottom temperature 5. Feed temperature at 3kg pressure 6. Complete material balance 7. Physical properties (density and surface tension) of distillate, residue and vapour

31

1. Saturation temperature of MO is determined by the help of Wagner equation [13]which is given below. *( )

(

)+

And the form of this equation which is used to determine the saturation temperature of MIBK is as follows.

Minimum for these equations is triple point and the maximum is critical temperature (Tc). The pressure is in bar and the temperature is in kelvin [13].The values of the constants: a, b, c, d are given below [13] Component

Tc (K)

A

B

C

D

Pc (bar)

Tmax (K)

MIBK

605

-8.68118

3.99203

-4.81662

-1.73164

34.1465

605

MO

571.40

3.8222

1190.6904

195.45

571.40

Putting P= 3kg = 2.94 bar, saturation temperature and the saturation pressure is calculated. From these values vapour-liquid equilibrium (VLE) data is calculated using Raoult’s law because we assumed the mixture and the vapour phase ideal. (Tsat)MIBK = 159.7

(Psat)MIBK = 2.94 bar

(Tsat)MO = 179.8

(Psat)MO = 2.94 bar

From the saturation temperature values it is clear that MIBK is more volatile than MO. So MIBK is designated as component 1 and MO is designated as component 2. 2. The VLE data is calculated using – ( )

( ) ( ) ( )

The calculated VLE data are given below. T (o C)

P1 sat (bar)

P2 sat (bar)

X1

Y1

159.7 160 161 162 163 164 165 166 167

2.940000 2.957839 3.022597 3.088397 3.155251 3.223168 3.292157 3.362229 3.433394

1.870336 1.883699 1.928778 1.974686 2.021431 2.069026 2.117479 2.166802 2.217005

1 0.983392128 0.924487848 0.866754293 0.810153982 0.754650685 0.70020938 0.646796194 0.594378366

1 0.989359 0.95046 0.910504 0.869469 0.827335 0.784081 0.739686 0.694128 32

168 169 170 171 172 173 174 175 176 177 178 179 179.77

3.50566 3.579038 3.653537 3.729168 3.80594 3.883862 3.962945 4.043198 4.12463 4.207252 4.291073 4.376103 4.442405

2.268097 2.320091 2.372995 2.426821 2.481579 2.537281 2.593936 2.651556 2.710151 2.769733 2.830312 2.891899 2.94

0.542924197 0.492403007 0.442785099 0.394041713 0.346144996 0.299067958 0.252784443 0.207269096 0.162497327 0.118445284 0.075089821 0.032408473 0

0.647384 0.599432 0.550249 0.499812 0.448098 0.395081 0.340738 0.285044 0.227973 0.1695 0.109597 0.048239 0

3.The relative volatility at top and bottom is determined by using the formula [14]

And

= 1.572

(

)

(

)

= 1.511.

The geometric mean relative volatility, α = 1.541 4.The top and bottom temperature is nothing but the saturation temperature of the top and bottom

product respectively. Top temperature = 159.7 Bottom temperature = 179.8 5.Feed temperature = (159.7 × 0.8915)+(179.8 × 0.1085) = 161.90 The overall material balance equation isF=D+W Where, F= Feed rate, kmol/hr. D= Overhead product rate, kmol/hr. W= bottom product rate, kmol/hr. Doing component balance for MIBK FXf= DXd+ WXw Where, Xf= mole fraction of MIBK in feed Xd= mole fraction of MIBK in distillate Xw= mole fraction of MIBK in residue Given, F = 944.08 kg/hr.. = 9.47kmol/hr.. Xf= 0.8891 Xd= 0.9980 Xw= 0.0750 33

So the equations becomesD+W = 9.47 9.47× 0.8891 = D × 0.998 + W × .0750 Solving these two equations we get, W= 1.116kmol/hr.and D= 8.354kmol/hr. 6. The liquid density is calculated by modified form of Rackett equation [15](

Density (g/ml) =

)

Where, A,B and n are regression coefficient. T and Tc are in Kelvin. The values of the coefficient for MIBK and MO are given bellow [15]: component

A

B

n

Tc

MIBK

0.26654

0.25887

0.28571

571.40

MO

0.27648

0.25438

0.28484

600.00

ρ(MIBK) at top temperature (159.7 )= 0.6567 g/ml ρ(MIBK) at bottom temperature (179.8 )= 0.631 g/ml ρ(MO) at top temperature (159.7 )= 0.7159 g/ml ρ(MO) at bottom temperature (179.8 )= 0.6919 g/ml So the distillate density becomes, ρ(top) = 0.0019×0.7159 + 0.9981×0.6567 = 0.65689 g/ml = 656.89 kg/m3 Similarly the residue density becomes, ρ(bottom) = 0.68739 g/ml = 687.39 kg/m3 7. The vapour density is calculated by the following formula. ρ= (PM/RT) where, P=total pressure M= molecular weight R=universal gas constant T= temperature By calculating we get, Vapour density at top, ρv(top) = 8.173 kg/m3 Vapour density at bottom, ρv(bottom) = 7.666 kg/m3 8. The liquid surface tension is determined by modified Othmer relation [15] Surface tension (dyne/cm) = A×(1-T/Tc)n The values of the constants are given below [15] component

A

n

Tc

34

MIBK

57.130

1.2040

571.4

MO

66.120

1.2560

600.0

The surface tension of distillate and residue is calculated applying same procedure for density which gives the following value. σ(top) = 10.394 dyne/cm σ(bottom) = 11.115 dyne/cm. 5.4 DESIGN METHODOLOGY STEP 1: TO FIND NUMBER OF THEORETICAL STAGES To determine the number of theoretical stages, here we used the method of McCABE and THIELE [16]. In this method what we need first is the equilibrium curve (x Vs y). The VLE data obtained by Raoult’s lay is plotted. The operating line equations for both top and bottom section are determined by appropriate mass balance. The equations of enriching section and stripping section operating lines are respectivelyYn+1

= (L/G) Xn+ (D/G) Xd……………………………………………………………(5.1) =(R/R+1) Xn+ Xd/(R+1)

Ym+1

= (Lb/Gb).Xm– (W/Gb).Xw……………………………………………………….. (5.2) = (Lb/Lb-W)Xm– (W/Lb-W)XW

Where, R = L/G = Reflux ratio. L, G = liquid & vapour flow rate in top section respectively

Lb, Gb= liquid & vapour flow rate in bottom section respectively The q-line equation is-

y = (q/q-1).x – (Zf/q-1)……………………………………………………………………….(5.3) where, q = (HG-HF)/(HG-HL) HG = enthalpy of vapour from any tray. HF = enthalpy of feed HL = enthalpy of liquid Therefore if the feed is saturated liquid then q = 1 and q/q-1 = ∞. So q-line would be vertical. The general procedure of finding number of equilibrium stage is as follows. 1. The distillate composition is marked on x-axis and travelled vertically upward upto the diagonal line. 2. Travel vertically upward from the point of feed composition on the x-axis upto the diagonal. Extend the line according to the slope of q-line and mark the point of intersection with the equilibrium line.

35

3. Draw a line joining these two points and extend up to y-axis. The intercept on y-axis is Xd/(Rm+1), from which we can find Rm. 4. Assume an operating reflux ratio 1.2-1.5 times of Rm and find the intercept (Xd/R+1) on y-axis. 5. Join the intercept and the corresponding point of the distillate composition on diagonal. This is the enrichment section operating line. 6. By staircase construction we will get the number of theoretical stage in enrichment section. 7. Similarly we can find the number of theoretical stage in the striping section. Mark the position of bottom product on x-axis and find the corresponding on the diagonal. Join this point with the q-line intersection point on the equilibrium curve. This is the striping section operating line. 8. By staircase construction we will get the number of equilibrium stage in the striping section. 9. The feed point should be located closest to the intersection of two operating line. 10. The equilibrium curve is given below.

1.0 0.9 0.8 0.7

Y1

0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

X1

In this design problem the driving force for separation is very small. The equilibrium curve is very close to the diagonal. Since the top and bottom product composition is very close to unity, it is very difficult to mark the points on the graph. So determination of theoretical stage by graphical method is difficult in this case. Based on the basic logic of the McCABE and THIELE method, a computer program is written in Turbo-C language. The logic is given bellow.

36

The equation of the equilibrium line is determined by polynomial curve fitting. The equation and the R2 value is given above. The position of the points required for drawing the graph were determined by solving equilibrium line equation and q-line equation. The equation of the connecting lines between the two points (mentioned above in 1. & 2.) was determined. The intercept of the line with y-axis was determined and from there the minimum reflux ratio was calculated. The actual reflux ratio was determined and the operating line equation was generated. Then the values of L, G, q, L, G were calculated using the following equations and the striping section operating line equation was stablished. L=R.D ………………………………………………………………………………….(5.4) G=D(R+1) ……………………………………………………………………………….(5.5) L=qF+1 ………………………………………………………………………………….(5.6) G=F(q-1)+G ……………………………………………………………………………..(5.7) The mathematical logic behind staircase construction is employed in the program to calculate number of theoretical stage separately in top and bottom section to get an idea about the feed tray location. The program section for enrichment section will terminate when calculated x value crosses the feed composition. Similarly program section for striping section will terminate when calculated x value crosses the residue composition. The program is shown below.

37

#include #include #include main() { intn,b; float m1,m2,c1,c2,a,x,y; printf("\n enter the slope of the operating line of the enriching section:"); scanf("%f",&m1); printf("\n enter the intersect of the operating line of the enriching section:"); scanf("%f",&c1); printf("\n enter the slope of the operating line of the stripping section:"); scanf("%f",&m2); printf("\n enter the intersect of the operating line of the stripping section:"); scanf("%f",&c2); n=1; y=0.9981; x=0.9981; while(x>0.8895) { x=(1.415-(sqrt((1.415*1.415)-(4*0.428*y))))/(2*0.428); a=y; y=(m1*x)+c1; n=n+1; } y=a; printf("\n the no of top tray is= %d",n); b=n; while(x>0.0755) { x=(1.415-sqrt(1.415*1.415-(4*0.428*y)))/(2*0.428); y=(m2*x)+c2; n=n+1; } printf("\n the calculated theoritical stages are= %d",n); n=n-b; printf("\n the no of bottom tray is= %d",n); getch(); }

38

STEP 2: DETERMINATION OF NUMBER OF ACTUAL TRAY We have to determine or assume the overall column efficiency to calculate the number of actual stage. If the efficiency is η, then number of actual tray is given by, Actual tray = (theoretical tray / η) ……………………………………………………………..(5.8) STEP 3: COLUMN SIZING [17] Determination of column diameter: (All the calculations are done on the basis of operating conditions ) 1. Maximum liquid rate (Lw) in Kg/sec. Lw = (liq flow rate in Kgmol/hr.. *Molecular Weight) / 3600 …………………….......(5.9a) 2. Maximum vapour rate (Vw) in Kg/sec. Vw = (Vapour flow rate in Kgmol/hr.. *Molecular Weight )/3600……………………………….(5.9b) 3. Assume tray spacing in meter 4. Flow Factor (Flv). ……………….(5.10)

√ 5. Assume: a. Hole diameter (dh) b. Hole Area= (3.14*dh2)/4. c. Plate thickness. d. Weir height (hw). e. Downcomer area (Ad). 6.

Find

Fair’s

co-efficient

(K),

From

graph

find

k1,then

K

=

k1

Ϭ

(

/.02)0.2……………………..(5.11) 7.Find flooding velocity Flooding velocity (uf) = K [(ρv-ρl)/ρv]0.5……………………………………………………..(5.12) 8. Assume % flooding. 9. Find actual velocity assuming % flooding. a. Actual velocity (ut) = uf* % flooding b. Volumetric flow rate = Vw/ρv.........................(5.13) c. Net flow area = (Flow rate/actual velocity) ……………………………………………………(5.14) d. Column area = (Net area * 100 ) / ( 100%- allowance) …………………………………(5.15) e. Diameter = [(column area *4)/3.14]0.5 ……………………………………………………(5.16) 10. The column diameter is rationalized with respect to that available in the market 11. Recalculate: 39

a. Column area (Ac) = (3.14*d2)/4 …………………………………………………….(5.17) b. Net area (AO) = Ac* % allowance ……………………………………………………(5.18) c. Down comer area (Ad) =Ac *% down comer area ……………………………………(5.19) d. Active area (Aa)= Ac – 2Ad e. Hole area (Ah) = Ac* % hole area. f. Superficial velocity (uo) = (volumetric flow rate / net area) ……………..……………(5.20) g. Find Lw/Dc from the graph Weir length, lw =value from graph *Dc …………………………………………………(5.21) Step 3: Calculation of %flooding % flooding = (uo /ut)*100 STEP 4: ENTRAINMENT CHECKING Read the value of from the graph . For sieve tray, it must be less than 1. If entrainment is large, increase the tray spacing & lower the vapour velocity & recalculate. If % flooding > 95% or < 70% , change the tray spacing ,active area and recalculate. STEP 5: WEEPING CHECKING a) Find weir liquid crest (how): how = 750 [Lw/(ϼl*lw)]2/3 ………………………………….(5.22)

b) Assume hw. c) Find K from the graph. d) Find minimum vapour velocity (Uh):Uh = [K- 0.9 (25.4-dh)] / (ϼv)0.5………………(5.23) e) Check whether uh > uh (min). If uh < uh (min),decrease the hole area & recalculate.

STEP 6: CALCULATION OF TRAY PRESSURE DROP a) Find Co from fig. b) Find dry tray pressure drop, hd. hd =51(uh/Co)2*(pv/pl) ….............................(5.24) c) Find residual head, hr.. = (12.5* 103)/pl……………………………………….(5.25) d)

Find

total

pressure

drop,

ht

=

hd

+

(

hw

+

how)

+

hr..………………………………………..(5.26) e) Check whether it is acceptable or not. If the pressure drop is beyond the tolerance level, increase the tray spacing and recalculate. STEP 7: CHECK DOWN COMER BACK UP a) Assume apron height, hap = (hw-10) b)Calculate the clearance area (Aap) under the down comer. Aap=hap×lw ……………………………………………………………(5.27) c) Find head loss for flow thr..ough this clearance ,hdc. hdc = 166 [Lwd/(ϼl* Aap)]2…………………………………………..(5.28) d) Find down comer back up, hb 40

hb= (hw+how)+ht+hdc………………………………………………..(5.29) STEP 8: CALCULATION OF FROTH HEIGHT To be on the safer side, it is taken as twice the down comer back up. Thus, Fb= 2*hb…(5.30) STEP 9: CALCULATION OF DOWNCOMER RESIDENCE TIME Down comer residence time, tr= (Ad*hb*ϼl)/Lwd…………………………………….(5.31) To get most efficient design, all the calculations are to be repeated to get minimum column diameter and minimum tray spacing. 5.5 DESIGN COMPUTATION (PROCESS) S/L No

Variables

Top portion

Bottom portion

1

Temperature ( )

159.7

179.8

2

Pressure (kg)

3

3

3

Vapour flow rate (Kmol/hr.)

27.81

27.81

4

Liquid flow rate (Kmol/hr.)

19.46

28.93

5

Vapour Mass Flow rate, kg/s

0.7724

0.7724

6

Liquid mass Flow rate, kg/s

0.5404

0.7884

7

Vapour density (Kg/m3)

8.173

7.666

8

Liquid density (Kg/m3)

656.89

687.39

9

Surface tension (dyne/cm)

10.394

11.115

10

Surface tension (N/m)

0.010394

0.011115

STEP 1: CALCULATION OF THEORETICAL STAGE The numbers of theoretical stages are calculated by computer programming which are given below. No. of tray (top): 28 No. of tray (bottom): 16 Feed enters at 29th tray from top. Total no. of tray: 45 The computer output is shown below.

41

STEP 2: DETERMINATION OF NUMBER OF ACTUAL TRAY Overall column efficiency is assumed 85% [18] From eqn. (5.8) Actual tray (top): 33 Actual tray (bottom): 19 Feed enters at 34th tray from tray. Total no. of tray: 53 STEP 3: COLUMN SIZING Determination of column diameter

1. From eqn. (5.9a) Maximum liquid rate (Lw) in Kg/s = 0.07724. 2. From eqn. (5.9b) Maximum vapour rate (Vw) in Kg/s = 0.7884. 3. Tray spacing is assumed 0.5 m. 4. From eqn. (5.10) Flow Factor (Flv), Flv (top) = 0.07804 Flv(bottom) = 0.1077 5. Assumed: a. Hole diameter (dh) = 5 mm b. Fractional Hole Area= 0.1 c. Plate thickness = 5 mm d. Weir height (hw) = 50 mm e. Fractional Downcomer area (Ad) = 0.12 6. From graph k1 = 0.08 (top) and 0.079 (bottom) Then from eqn. (5.11), K = 0.071 for both top and bottom.

42

7.From eqn. (5.12),

Flooding velocity (uf) (top) = 0.635 m/s, Flooding velocity (uf) (bottom) = 0.667 m/s 8. Assume velocity as 75% of flooding velocity. 9. Actual velocity: a. Actual velocity (ut) (top) = 0.476 m/s, Actual velocity (ut) (bottom) = 0.50 m/s b. Volumetric flow rate (top) = 0.0945m3/S, Volumetric flow rate (bottom) = 0.100m3/S c. From eqn. (5.14) Net flow area (top) = 0.198m2 Net flow area (bottom) = 0.201m2 d. From eqn. (5.15) Column area (top) = 0.2256m2 Column area (bottom) = 0.2290m2 e. From eqn. (5.16) Diameter (top) = 0.535 m Diameter (bottom) = 0.54 m

13. The column diameter is rationalized with respect to that available in the market. And the diameter is selected 0.55 m for both top and bottom. Recalculation:

a. From eqn. (5.16) Column area (Ac) (top) = 0.2290m2 Column area (Ac) (bottom) = 0.2290m2 b. From eqn. (5.18) Net area (Ao) (top)=0.2290 m2 Net area (Ao) (bottom)=0.2290 m2 c. From eqn. (5.19) Down comer area (Ad) (top) = 0.0285m2 Down comer area (Ad) (bottom) = 0.0285m2 d. Active area (Aa) (top) = Ac – 2Ad = 0.172m2 Active area (Aa) (top) = Ac – 2Ad = 0.172m2 e. Hole area (Ah) (top) = 0.0180m2 Hole area (Ah) (bottom) = 0.0180 m2 f. Single hole area (top and bottom) = 19.635 mm2 Number of holes (top and bottom) = 920 g. From eqn. (5.19) Superficial velocity (uo) (top) = 0.452 m/s Superficial velocity (uo) (top) = 0.482 m/s g. Weir length, lw (top) = 0.42 m Weir length, lw (bottom) = 0.42 m STEP 3: CALCULATION OF %FLOODING

43

Actual % flooding (top) = 71.2 % Actual % flooding (top) = 72.3 % The % flooding is in between 95% and 70%, so the result is satisfactory STEP 4: ENTRAINMENT CHECKING From eqn. (5.10) Flow Factor (Flv), Flv(top) = 0.078 Flv(bottom) = 0.107 The value of the fractional entrainment from the graph corresponding to flood% is determined. Entrainment factor (top): 0.029 Entrainment factor (bottom): 0.018 So the entrainment is less than 0.1 STEP 5: WEEPING CHECKING a) From eqn. (5.22) Weir liquid crest (how) (top) = 11.73 mm liq. Weir liquid crest (how) (bottom) = 14.64 mm liq. b) Weir height hw is assumed 50 mm for both top and bottom. c) From the graph, K (top) = 30.6 K (bottom) = 30.7 d) From eqn. (5.22) Minimum vapour velocity (uh) (top) = 5.23 m/s Minimum vapour velocity (uh) (bottom) = 5.58 m/s e) Actual hole vel - weep velocity (top) = 4.20 m/s Actual hole vel - weep velocity (top) = 4.36 m/s So uh is greater than uh (min) and therefore the result is satisfactory. STEP 6: CALCULATION OF TRAY PRESSURE DROP a) Co from fig. = 0.79 (for both top and bottom) b) From eqn. (5.24) Dry tray pressure drop, hd(top) = 27.59 mm liq. Dry tray pressure drop, hd(bottom) = 28.11 mm liq. c) From eqn. (5.25) Residual head, hr..(top) = 19.09 mm liq. Residual head, hr..(bottom) = 18.11 mm liq. d) From eqn. (5.26) Total pressure drop, ht(top) = 108.35 mm liq. Total pressure drop, ht(bottom) = 110.94 mm liq. e) The plate pressure drop is within the acceptable limit. So, further modification is required. STEP 7: CHECK DOWN COMER BACK UP 44

a) Apron height, hap is assumed 45 mm for both top and bottom. b) From eqn. (5.27) The clearance area (Aap) under the down comer (top & bottom) = 0.0189 m2. c) From eqn. (5.28) Head loss for flow thr..ough this clearance, hdc(top) = 0.314 mm liq. Head loss for flow thr..ough this clearance, hdc(bottom) = 0.610 mm liq. d) From eqn. (5.28) Down comer back up, hb(top) = 170.4 mm liq. Down comer back up, hb(bottom) = 176.2 mm liq. STEP 8: CALCULATION OF FROTH HEIGHT From eqn. (5.28) Froth height (top) = 340.8 mm liq. Froth height (bottom) = 352.4 mm liq. Tray spacing – Downcomer backup (top) = 159.2 mm Tray spacing – Downcomer backup (bottom) = 147.6 mm So the design is safe to avoid flooding. STEP 8: CALCULATION OF DOWNCOMER RESIDENCE TIME From eqn. (5.28) Down comer residence time, tr(top) = 6.33 sec. Down comer residence time, tr(bottom) = 4.4 sec. This is greater than 3 sec. So the result is satisfactory

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5.6 DESIGN METHEDOLOGY (MECHANICAL) STEP 1: CALCULATION OF DESIGN PRESSURE AND DESIGN TEMPERATURE [19] Design pressure= operating pressure + 20% allowance Design temperature = operating temperature + 10% allowance STEP 2: SELECTION OF MATERIAL OF CONSTRUCTION STEP 3: CALCULATION OF SHELL THICKNESS [19] Shell thickness,

(

)

mm

Where, P' = Design pressure D=Shell inner diameter S= Allowable stress J= Weld joint efficiency Set Corrosion allowance and determine final thickness, t'=t+(corrosion allowance/1000) m Rationalize the thickness obtained according to availability, t" STEP 4: CHECK WHETHER THE THICKNESS OBTAINED CAN WITHSTAND THE LOAD OF WHOLE TOWER [19] Select insulation material & insulation thickness,i Dead weight calculation Outer diameter of the shell=D=d+(2*t") m Outer radius of the shell=R=(D/2) Inner radius of the shell= r =(d/2) Dia including the insulation= D’= D+(2*i) m Radius including the insulation=R’=(D’/2) Density of the shell material=ρ Density of insulating material=ρ’ Tray thickness, c = 0.005 m Tray spacing, s = 0.5 m Weight of torispherical top, w1 = ((π*D2/4)*t"*ϼ*1.7) kg Weight of insulation, w2 = ((π*D’^2/4)*i*ρ’*1.7) kg Dead weight, W = (w1 + w2) kg Calculation of weight per unit length Weight of shell/unit length= (π*(D^2-d^2)/4)*ρ 46

Weight of insulation/unit length=(π*(D’^2-D^2)/4)*ρ’ Weight of liquid/unit length= ((π*d^2)/4)*density (use the larger density between water and the fluid handled) Assume weight of external affixes/unit length Weight of tray/unit length= ((π*d^2)/4)*c*ρ)/s Actual tray weight/unit length= (Weight of tray/unit length)*1.3 Total weight /unit length, w= a) + b) + c) + d) +f) Calculate total compressive force at a distance of x from top. Compressive stress=(

)

(

)

Calculation of axial stress Axial stress= (r*(Pin-Pout))/(2*t) Calculation of net tensile stress Net tensile stress = axial stress – compressive stress Calculation of wind load Velocity of air = V= 300km/hr. = 83.33 m/s (assumed) Density of air = ρ(air) = 1.126 kg/m3 (at 30oC & 1 atm) Cd=0.44 in turbulent region Effective projected area, A= 1.2*(D’*x) (allowing 20% excess for additional features) Wind force, Fw= (Cd*A*ρ(air)*V^2)/2 Bending moment at a height x from top, BM=(Fw/2)*x^2 Moment of inertia, MI= π*r’*t Where r’ = average column radius Wind stress= (BM*R/MI) Total compressive stress Total compressive stress for windward side – wind stress – net tensile stress = allowable stress Total compressive stress for leeward side— wind stress + net tensile stress = allowable stress Find the distance x from top up to which the calculated thickness can be used by solving these two equations. If the distance is less than the total column height, then the calculated is to be repeated substituting x by (x-distance obtained from above solution)

STEP 4: CALCULATION OF SEISMIC LOAD [20] Total height of the column, h = (no. of tray*s) + (1+1.5)m Modulus of elasticity, E = 27990000 psi

47

Acceleration due to gravity, g = 32.2 ft/s2 Moment of inertia, m = π*((((D+d)/2)/2)^3)*t Period of vibration, n = (2*π*((w8*h4))/(E*32.2*(m/(0.30484))))^0.5)/3.53 sec.

Seismic co-efficient, e = 0.08 Maximum bending stress, b = (8*e*w8*h2)/(π*R2*t) psi Weight of column, w9 = w8*h kg Dead weight stress, Ws = w9*9.81/(π*ds) Seismic stress, S = (2*e*w9)/(3*π*R) psi STEP 5: CALCULATION OF TORISPHERICAL HEAD[19] Thickness of head= Where C=corrosion allowance =0.118, S=max. allowable stress=17114.51psi, E=weld joint efficiency=0.85, P= design pressure=42.67psi, rc= crown radius=71.5

STEP 6: NOZZLE DIA CALCULATION [19] Nozzle dia, Dopt (mm) = Where, G= Mass flow rate = volumetric flow rate × density d= density, kg/m3 STEP 7: GASKET DESIGN Design equation:

√

(

)

Where, d0 = outer diameter of the gasket di= inner diameter of the gasket = 10 mm greater than the outside dia of shell Y = minimum yield stress m = gasket factor P = design internal pressure Actual gasket width N= (d0- di)/2 Basic seating width=N/2 G= diameter at location of gasket load reaction G= di+N 48

STEP 8: BOLT DESIGN Determination bolt load due to design pressure: G2 P)/4

Design equation: H = (3.142 Where,

H = total hydrostatic end force, P = design pressure Determination of load to keep joints tight under operating condition: Design equation: Hp = (3.14 G 2b

m

P)/4

Where, HP = load to achieve adequate compression of gasket under operating condition G = diameter of location of gasket load 2b= effective gasket pressure width Total operating load: Design equation: Wo = H + Hp Determination of bolt load under bolting up condition Design equatio :Wg = (3.14

G

b

:

Y)

Where, b = effective gasket seating width = 0.24 ft Y = minimum gasket seating stress Determination of minimum bolting area: If A0 is the bolt area required under operating condition and Ag is the area required under bolting up condition then, Design equation: A0 = W0/S0 & Ag = Wg/Sg Where, S0 = allowable stress for bolting material at design temperature Sg = allowable stress for bolting material at atm temperature Among A0 and Ag , greater is selected. STEP 9: FLANGE CALCULATION Flange outside diameter (A) = [C + d +25] mm Where, d = bolt diameter C = bolt circle diameter Flange thickness can be calculated by, 49

Bs=2d+ 6t/ (m+0.5) Where, Bs= bolt spacing d = bolt diameter t= flange thickness m=gasket factor 5.7 DESIGN COMPUTATIONAL (MECHANICAL) STEP 1: CALCULATION OF DESIGN PRESSURE AND DESIGN TEMPERATURE Design pressure= 3.48 atm Design temperature = 1980C STEP 2: SELECTION OF MATERIAL OF CONSTRUCTION Material of construction: Grade 2B of IS 2002-1962 [21] Specification of alloy steel: C% (max): 0.22 Si% : 0.10-0.35 S% (max): 0.050 P% (max): 0.050 Designation: Grade 2B STEP 3: CALCULATION OF SHELL THICKNESS Allowable stress, S = 118000000 N/m2 = 1.18 × 1014N/mm2 Weld joint efficiency=0.85 (for double welded butt joint with full penetration) Shell thickness, t = 1mm Corrosion allowance = 3mm Final thickness, t' = 4 mm Rationalized thickness, t"= 6 mm

STEP 4: CHECK WHETHER THE THICKNESS OBTAINED CAN WITHSTAND THE LOAD OF WHOLE TOWER Insulation material & insulation thickness, i Felt wool is selected as insulating material and insulation thickness is 5 cm. Dead weight calculation Outer diameter of the shell=D=0.562 m Outer radius of the shell=R= 0.281 m Inner radius of the shell= r = 0.275m

50

Dia including the insulation= D’= 0.656 m Radius including the insulation=R’= 0.0.328 m Density of the shell material=7850 kg/m3 Density of insulating material=332.31kg/m3 Tray thickness, c = 0.005 m Tray spacing, s = 0.5 m Weight of torispherical top, w1 = 19.44 kg Weight of insulation, w2 = 9.5 kg Dead weight, W = 28.94 kg Calculation of weight per unit length Weight of shell/unit length= 40.91 kg/m Weight of insulation/unit length= 31.63 kg/m Weight of liquid/unit length= 237.6 kg/m (Density of water is used since it is more than that of process fluid handled) Weight of external affixes/unit length= 50 kg/m is assumed Weight of tray/unit length= 88.74 kg/m Actual tray weight/unit length= 18.65 kg/m Total weight /unit length, w= 467.53 kg/m

Calculation of total compressive force at a distance of x from top. CS area without corrosion allowance= 0.008 m2 Compressive stress, kg/m2= 3617.5 + 58411.25x Calculation of axial stress Axial stress, kg/m2= 359223.57 Calculation of net tensile stress Net tensile stress, N/m2 = 355606.07 – 58411.25x

Calculation of wind load Velocity of air = V= 300km/hr.. = 83.33 m/s (assumed) Density of air = ρ(air) = 1.126 kg/m3 (at 30oC & 1 atm) Cd=0.44 in turbulent region Effective projected area,m2/m, A= 0.285 (allowing 20% excess for additional features) Wind force/unit length,N/m, Fw= 1961 Bending moment at a height x from top, Nm, BM= 980.5x2 Moment of inertia, m4, MI= 0.0008 Wind stress, N/m2= 340723x2 Total compressive stress Total compressive stress for windward side 340723x2+58411.25x -355606.07 = 118000000 51

Total compressive stress for leeward side – 340723x2-58411.25x +355606.07 = 118000000 Solving these two equations we get the lowest possible real solution of x x= 18.5 m. Since the value obtained is less than total column height, we have to repeat the procedure. The recalculation shows for rest of the tower we can use next higher shell thickness of 8mm. So the final inference is we will use 6mm thick shell from top to 19m. After that up to the bottom 8mm shell thickness is to be used.

STEP 5: CALCULATION OF SEISMIC LOAD Total height of the column, h = 27.5 m Modulus of elasticity, E = 27990000 psi Acceleration due to gravity, g = 32.2 ft/s2 Moment of inertia, m = 8.6×10-5 Period of vibration, n = 35 sec. Seismic co-efficient, e = 0.08 Maximum bending stress, b = 34979.37 psi Weight of column, W9= 41676.83 kg Dead weight stress, WS= 325.41 N/mm2 Seismic stress, S = 3.83psi = 26406.9 N/m2 So the seismic stress is less than the allowable stress. STEP 6: CALCULATION OF TORISPHERICAL HEAD Head Type: Torispherical Thickness (top) =12mm Thickness (bottom) =16mm

STEP 7: NOZZLE DIA CALCULATION Top nozzle dia. for vapour outlet = 120 mm, Thickness = 8 mm Top nozzle dia. for reflux inlet = 26.64mm, Thickness = 6.88 mm Bottom nozzle dia. for reboiler inlet =120 mm, Thickness = 8 mm Bottom nozzle dia. for residue outlet = 26.64 mm, Thickness = 6.88 mm Nozzle dia for feed inlet= 15.8 mm, Thickness = 5.54 mm 52

The written values are rationalized [19] STEP 8: GASKET DESIGN Material of construction: compressed asbestos sheet and various metallic reinforced sheets and clothes (IS-2825-1969) Y=25x106N/m2 m=2.75 P= 2.941x105N/m2 d0/di=1.0614 di= 0.566 m d0= 0.570 m Actual gasket width N= 2 mm Basic seating width= 1 mm Diameter at location of gasket load reaction= 0.568 m STEP 9: BOLTING CALCULATION Material of construction: Hot rolled carbon steel Determination of bolt load due to design pressure: Total hydrostatic end force, H= 3.58×105 N Determination of load to keep joints tight under operating condition: Load to achieve adequate compression of gasket under operating condition, Hp=6923.45N. Total operating load: Minimum bolt load under bolting up condition, W0=3.64×105 N Determination of bolt load under bolting up condition: Wg= 89176 N. Minimum theoretically bolt area: A0 =0.006 m2 Selection of bolts: Bolt size =M 20×2 Root area=0.0002 m2 Flange thickness =20mm Hub thickness =0.5x20mm=10mm Bolt spacing=50 mm No. of bolts=32 nos. (To achieve uniform tightening, the no of bolts should be in multiple of 4). Bolt circle diameter, C = 0.509 m 53

Optimum bolt diameter = 20 mm STEP 10: CALCULATION OF FLANGE Material of construction: carbon steel Specification: IS-2856-1964 Flange thickness, t= 5.41 mm Rationalized Thickness of flange =20mm [19] STEP 10: SUPPORT DESIGN Type of support: Skirt support

ASPECTS NOT ADDRESSED 1. The detailed process and mechanical design of the condenser and reboiler is not included in this design report. 2. Control and instruments for the equipment is not discussed here. 3. Safety and environmental aspects is not discussed. 4. Operational features are not discussed. 5. Detail support and foundation design is not included.

54

6. DESIGN DATA SHEET Column Tray Data Sheet Operating Data

TOP

Equipment no.(tag) Description(func.) Sheet no. BOTTOM OR TOP AND BOTTOM

TOWER INSIDE DIAMETER(mm) TRAY SPACE(mm) TOTAL TRAYS IN SECTION

550 500 33

550 500 19

550 500

Internal Conditions at Tray Number VAPOR TO TRAY RATE(kmol/hr.) DENSITY(kg/m3) PRESSURE(atm.) TEMPERATURE (K) LIQUID FROM TRAY RATE(kmol/hr.) DENSITY(kg/m3) TEMPERATURE VISCOSITY NUMBER OF LIQ FLOW PATHS

27.81 8.173

27.81 7.666 3.48

19.46 656.89

28.93 687.39

Technical / Mechanical Data TRAY MATERIAL TRAY THICKNESS (mm) CAP MATERIAL HOLDDOWN MATERIAL NUTS AND BOLTS MATERIAL SUPPORT RING MATERIAL SUPPORT RING SIZE DOWNCOMER BOLT BAR THICKNESS CORROSION ALLOWANCE TRAYS TOWER ATTACHMENTS TRAYS NUMBERED FROM TOP TO BOTTOM TRAY MANWAY REMOVAL FROM DATE OF ENQUIRY ORDER NO. MANUFACTURER

Carbon steel 5 Carbon steel Carbon steel Hot rolled Carbon steel Carbon steel 60 nil

DATE OF ORDER DRG NO. `

NOTES (1) INTERNAL VAPOUR AND LIQUID LOADINGS AT THE LIMITING SECTION ARE REQUIREDTO ENSURE PROPER TRAY DESIGNS. DENSITIES ARE REQUIREDAT ACTUAL INSIDE TOWER CONDITIONS OF TEMPERATURE AND PRESSURE. VISCOSIY IS NOT REQUIRED UNLESS GREATER THAN O.7 Cp. (2) CROSS OUT DIMENSION UNITS WHICH DO NOT APPLY TRAY SUPPLIES TO ADVISE

REMARKS

55

7. NOMENCLEATURE A= area Aa= Active area Ac = column area Ad = down comer area Aeff= effective area Ah = tray deak full open hole area An = net area Cp=specific heat d = diameter

D, ρ = density Fdl= Stress due to dead load Flv = flow factor g = gravitational constant G, L = gas and liquid flow rate Hi = enthalpy of ith stream. hw= weir height Lwi= weir length n=number of moles Pc, Tc = critical pressure and temperature Psat, Tsat = saturation pressure and saturation temperature Pvp = vapour pressure QB= reboiler load Qc = condenser load r = radius R= optimum reflux ratio. R= reflux ratio Rm= minimum reflux ratio. t = actual thickness of the plate th= skirt thickness W= residue flow rate Wex= weight of external connections Wh= weight of head Wh= weight of head Wint= weight of column internal Wl= weight of liquid inside the column 56

Ws= weight of shell Wtot = total weight to be supported X1 = liquid mole fraction Y1= equilibrium vapour mole fraction α = relative volatility η = overall column efficiency ν = stoichiometric coefficient σ = surface tension\

57

8. REFERENCES 1.Ullmann’s ―Encyclopedia of Industrial Chemistry‖, Vol. 18, p. 735, 6th edition (2003), WileyVCH 2. J.J. McKetta, ―Encyclopedia of Chemical Processing and Design‖, Vol. 30, p. 50, 1st edition (1989), Marcel Dekker Inc. 3. K. Schmitt, ―New acetone chemistry in Germany‖, Chem. Ind. Int., p. 204 (September 1966) 4. S.I.D.A. U.S. Patent 3,802,999 (October 31, 1961); Shell Development Co., U.S. Patent 2,344,226 (March 14, 1944) 5. Esso Research and Engineering Co., U.S. Patent 3,361,828 (January 2 , 1968) 6. Nippon Sekiyu, British Patent 968,151 (July 2, 1964) 7. McAllister et al., Shell Development Co., U.S. Patent 2,130,592 (September 20, 1938) 8. Esso Research and Engineering Co., U.S. Patent 3,153,068 (September 13, 1964) 9.Kgowa Hakko Kogyo, U.S. Patent 3,449,435 (June 10, 1969) 10.Scholven-Chemie, U.S. Patent 3,361,822 (January 2, 1968) 11. Shell Oil, U.S. Patent 3,374,272 (March 19, 1968) 12.R.M. Felder, R.W. Rousseau, ―Elementary Principles of Chemical Process‖, 3rd edition (2000), John Wiley & Sons. Inc. 13. B.E. Pouling, J.M. Pransnitz, J.P. O’Connell ―Properties of Gases and Liquids‖ , 5th edition (2000), McGraw-Hill. 14.J.M.Smith, H.C.Van Ness, M.M. Abbott, ―Chemical Engineering Thermodynamics‖, 6th Edition, Tata-McGraw-Hill Edition. 15.C.L.Yaws, ―Chemical Properties Handbook‖, 5th edition, 1999 16. Robert .E. Treybal , ― Mass Transfer Operation‖,3rd edition. 17.Sinnot, R. K., ―Coulson Richardson’s Chemical Engineering‖, Vol-6,2nd Edition, Asian books Private Ltd. 18. Perry R.H. Chemical Engineers Handbook – McGraw Hill Ltd. 5th Edition. 19. B.C Bhattacharya, ―Introduction to Chemical Equipment Design‖, 1st edition, Reprint, CBS Publishers. 20. L.E. Brownell and E.H. Young, ―Equipment Design‖, Wieley Eastern Limited, New Delhi. 21. IS 2002-1962 22. Kirk Othmer, ―Encyclopedia of Chemical Technology‖, Vol. 1, p. 190, 3rd edition (1978), John Wiley & Sons. Inc.

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