Index

Sr. No. Topics Page No.

1 Introduction 1

2 Physiochemical Properties/ Uses 2

3 List of all Manufacturing processes 6

4 Detail flow Sheet / Manufacturing Process 7

5 Material Balance 9

6 Energy Balance 21

7 Design of Equipment 33

8 Instrumentation & Process control 50

9 Plant Location & Plant layout 51

10 Safety & Environment 53

11 Material Data Safety Sheet 56

12 Cost Estimation 58

13 Manufacturing Industries 63

14 Conclusion 64

15 References 65

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Chapter 1: Introduction

Phenol, also known as carbolic acid, is a toxic, white crystalline solid with a sweet tarry odor, commonly referred to as a "hospital smell". Its chemical formula is C6H5O H and its structure is that of a hydroxyl group (-OH) bonded to a phenyl ring; it is thus an aromatic compound.

Phenols are similar to alcohols but form stronger hydrogen bonds. Thus, they are more soluble in water than are alcohols and have higher boiling points. Phenols occur either as colorless liquids or white solids at room temperature and may be highly toxic and caustic.

Phenols are widely used in household products and as intermediates for industrial synthesis. For example, phenol itself is used (in low concentrations) as a disinfectant in household cleaners and in mouthwash. Phenol may have been the first surgical antiseptic. In 1865 the British surgeon Joseph Lister used phenol as an antiseptic to sterilize his operating field. With phenol used in this manner, the mortality rate from surgical amputations fell from 45 to 15 percent in Lister’s ward. Phenol is quite toxic, however, and concentrated solutions cause severe but painless burns of the skin and mucous membranes. Less-toxic phenols, such as n-hexylresorcinol, have supplanted phenol itself in cough drops and other antiseptic applications. Butylated hydroxytoluene (BHT) has a much lower toxicity and is a common antioxidant in foods.

In industry, phenol is used as a starting material to make plastics, explosives such as picric acid, and drugs such as aspirin. The common phenol hydroquinone is the component of photographic developer that reduces exposed silver bromide crystals to black metallic silver. Other substituted phenols are used in the dye industry to make intensely colored azo dyes. Mixtures of phenols (especially the cresols) are used as components in wood preservatives such as creosote.

Natural sources of phenols

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Phenols are common in nature; examples include tyrosine, one of the standard amino acids found in most proteins; epinephrine (adrenaline), a stimulant hormone produced by the adrenal medulla; serotonin, a neurotransmitter in the brain; and urushiol, an irritant secreted by poison ivy to prevent animals from eating its leaves. Many of the more complex phenols used as flavorings and aromas are obtained from essential oils of plants. For example, vanillin, the principal flavoring in vanilla, is isolated from vanilla beans, and methyl salicylate, which has a characteristic minty taste and odour, is isolated from wintergreen. Other phenols obtained from plants include thymol, isolated from thyme, and eugenol, isolated from cloves.

Chapter 2: Physiochemical Properties

1. Molecular weight: 94.11

2. Boiling point (at 760 mm Hg): 181.7 degrees C (359.1 degrees F)

3. Specific gravity (water = 1): 1.07 at 20 degrees C (68 degrees F)

4. Vapor density: 3.24

5. Melting point: 43 degrees C (109.4 degrees F)

6. Vapor pressure at 35 degrees C (77 degrees F): 0.35 mm Hg

7. Solubility: Soluble in water (8.3 g/100 ml) and benzene; very soluble in alcohol, chloroform, ether, glycerol, carbon disulfide, petrolatum, volatile and fixed oils, and aqueous alkali hydroxides; almost insoluble in petroleum ether.

Flammability

The National Fire Protection Association has assigned a flammability rating of 2 (moderate fire hazard) to phenol.

1. Flash point: 79 degrees C (175 degrees F) (closed cup)

2. Auto ignition temperature: 715 degrees C (1319 degrees F)

3. Flammable limits in air (percent by volume): Lower, 1,7; upper, 8.6

4. Extinguisher: For small fires use dry chemical, water spray, or regular foam. Use water spray, fog, or regular foam to fight large fires involving phenol.

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Chemical Reactions of phenols:-

Acidity of phenol:

Although phenols are often considered simply as aromatic alcohols, they do have somewhat different properties. The most obvious difference is the enhanced acidity of phenols. Phenols are not as acidic as carboxylic acids, but they are much more acidic than aliphatic alcohols, and they are more acidic than water. Unlike simple alcohols, most phenols are completely deprotonated by sodium hydroxide (NaOH).

Oxidation:

Like other alcohols, phenols undergo oxidation, but they give different types of products from those seen with aliphatic alcohols. For example, chromic acid oxidizes most phenols to conjugated 1,4-diketones called quinones. In the presence of oxygen in the air, many phenols slowly oxidize to give dark mixtures containing quinones.

Hydroquinone (1, 4-benzenediol) is a particularly easy compound to oxidize, because it has two hydroxyl groups in the proper relationship to give up hydrogen atoms to form a quinone. Hydroquinone is used in developing photographic film by reducing activated (exposed to light) silver bromide (AgBr) to black metallic silver (Ag↓). Unexposed grains of silver bromide react more slowly than the exposed grains.

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Electrophilic aromatic substitution:

Phenols are highly reactive toward electrophilic aromatic substitution, because the nonbonding electrons on oxygen stabilize the intermediate cation. This stabilization is most effective for attack at the ortho or para position of the ring; therefore, the hydroxyl group of a phenol is considered to be activating (i.e., its presence causes the aromatic ring to be more reactive than benzene) and ortho- or para-directing.

Picric acid (2, 4, 6-trinitrophenol) is an important explosive that was used in World War I. An effective explosive needs a high proportion of oxidizing groups such as nitro groups. Nitro groups are strongly deactivating (i.e., make the aromatic ring less reactive), however, and it is often difficult to add a second or third nitro group to an aromatic compound. Three nitro groups are more easily substituted onto phenol, because the strong activation of the hydroxyl group helps to counteract the deactivation of the first and second nitro groups.

Phenoxide ions, generated by treating a phenol with sodium hydroxide, are so strongly activated that they undergo electrophilic aromatic substitution even with very weak electrophiles such as carbon dioxide (CO2). This reaction is used commercially to make salicylic acid for conversion to aspirin and methyl salicylate.

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Phenolic resins account for a large portion of phenol production. Under the trade name Bakelite, a phenol-formaldehyde resin was one of the earliest plastics, invented by American industrial chemist Leo Baekeland and patented in 1909. Phenol-formaldehyde resins are inexpensive, heat-resistant, and waterproof, though somewhat brittle. The polymerization of phenol with formaldehyde involves electrophilic aromatic substitution at the ortho and para positions of phenol (probably somewhat randomly), followed by cross-linking of the polymeric chains.

Uses

- Phenol finds extensive use in the production of phenol formaldehyde resin which is a thermoset

- It is used in industrial and decorative laminates, molding molds, textile auxiliaries and varnishes.

- Both acetone and phenol are condensed in the presence of a catalyst to produce bisphenol-A whish is used in the manufacture of epoxy resins, polycarbonate resins, polyester resins rubber chemicals and fungicides.

- Phenol has a sweet odour that is detectable at 0.06 PPM, which enables it to be used in an air freshener.

- Phenol is also a powerful disinfectant and bacteria killer.

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- In industry, phenol is used as a starting material to make plastics, explosives such as picric acid, and drugs such as aspirin.

- The common phenol hydroquinone is the component of photographic developer that reduces exposed silver bromide crystals to black metallic silver.

- Other substituted phenols are used in the dye industry to make intensely colored azo dyes

Chapter 3: List of all Manufacturing Process

Small quantities of phenol are isolated from tars and coking plant water produced in the coking of hard coal and the low temperature carbonization of brown coal as well as from the wastewater from cracking plants.

By far the greatest proportion is obtained by oxidation of benzene or toluene. Although direct oxidation of benzene is in possible, the phenol formed is immediately oxidized further.

Therefore, alternative routes must be chosen, e.g. via halogen compounds which are subsequently hydrolyzed or via cumene hydroperoxide which is then cleaved catalytically.

The following process was developed as industrial synthesis for production of phenol.

1. Alkylation of benzene with propene to isopropyl benzene (cumene), oxidation of cumene to the corresponding tert-hydroperoxide, and cleavage to phenol and acetone (Hock process).

2. Toluene Oxidation to benzoic acid and subsequent oxidizing decarboxylation to phenol ( Dow process )

3. Sulfonation of Benzene and production of phenol by heating the benzene sulfonate in moltan alkali hydroxide.

4. Chlorination of benzene and alkali hydrolysis of the chlorobenzene.

5. Chlorination of Benzene and steam hydrolysis of the chlorobenzene ( Rasching process, Rasching – Hocoker, Gulf oxychlorination)

6. Dehydrogenation of Cyclohexanol – cyclohexanone Mixtures ( Scientific Design)

Justification for the selected process:-

Of the processes named, only the Hock process (cumene oxidation) and the toluene oxidation is important industrially. The other processes were given up for economic reasons. In the Hock process acetone is formed as byproduct.

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This has not, however, hindered the expansion of this process; because there is a market for acetone .New plants are now run predominantly on the cumene process.

Chapter 4: Description of selected process

Cumene Process:-

The Cumene – phenol process is based on the discovery of cumene hydeoperoxide and its cleavage to phenol and acetone published in 1944 by H. hock and S.Lang. This reaction was developed into an industrial process shortly after World War II by Distilliers Co. in the United Kingdom and the Hercules Powder Co. in the United States.

Process:-The raw material for the product phenol i.e. cumene is obtained by alkylation of Benzene with propene using phosphoric acid or aluminium chloride as catalyst.

Two reaction steps form the basis of the production of phenol from cumene.1. Oxidation of cumene with oxygen to cumene hydroperoxide:

2. Cleavage of cumene hydroperoxide in an acidic medium to Phenol and Acetone.

Oxidation:-The oxidation of cumene with air or oxygen – enriched air is carried out in S.S. (304) reactors R1 & R2 which work on bubble column principal. The reactors are connected in cascade series to achieve an optimal residence time distribution. The oxidation is

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performed at 90-130 ºC and 0.5 – 0.7 Mpa. It is carried out in an aqueous emulsion stabilized by an alkali such as sodium carbonate in the 8.5 -10.5 pH range. The waste gas from the reactor is purified by two stage condensation of organic impurities, which consist predominanatly of cumene. Water is used as coolant in the first stage and refrigerant in the second stage.

The oxidation is autocatalytic, i.e. the reaction rate increases with increasing hydroperoxide concentration. The reaction is exothermic; ca. 800KJ are released per kilogram of cumene hydroperoxide. The heat of reaction is removed by cooling.

Cleavage:-

The acid-catalyzed cleavage of cumene hydroperoxide to phenol and acetone follows an ionic mechanism. Sulfuric acid is used exclusively as the catalyst in industry. The acid-catalyzed cleavage is carried out in homogeneous phase in which an excess of acetone is charged to the cleavage reactor and 10-12% sulfuric acid is added. The reaction temperature is the boiling points of the cumene hydroperoxide – acetone mixture i.e. 50-60 ºC. The heat removal from the strongly exothermic reaction (ca. 1680 KJ per kilogram of cumene hydroperoxide) is achieved by means of evaporation of acetone from the reaction system.

Separation:-

The first distillation column is a multicomponent one in which acetone is obtained as a top product and the bottom products are sulfuric acid, phenol, water. The next distillation column is operated under vacuum in which the top product is the waste water and bottom product is mainly phenol associated with sulfuric acid. In the next section which is a simple separator the phenol is obtained as top product and sulfuric acid as bottom product.

Phenol is obtained with a purity of > 99.9% by subsequent distillation.

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Chapter 5: Material BalanceBasis: - 1 t/h of cumene in feed. 24 t/d of cumene.

Let cumene in air-vapor mixture- 8%

Air/cumene= 92/8= 11.5 Let volume ratio= Mole ratio (assuming ideal gas mixture)Mole ratio=11.5

As per stoichiometric rxn:

Cumene + oxygen Cumene hydroperoxide

Therefore, O2/Cumene = 1 and Air/Cumene = 1/0.21 = 4.762

M.W. of air = 28.84 kg/kmol M.W. of cumene = 120 kg/kmol; So, % excess of air/O2 used = (11.50 – 4.762) / 4.762 = 141.5%

Now, Stream 1:- 1t/p of cumene = 1000/120 = 8.33 kmol/h

Stream 2:- Molar flow rate of air = 11.5 * 8.33 = 95.8 kmol/h

Oxygen in stream 2 = 95.8* .21= 20.118 kmol/h = 20.118 * 32 kg/h = 643.74kg/h

Nitrogen in stream 2 = 95.8 * .79 = 75.682 kmol/h = 75.682 * 28 kg/h = 2119.096 kg/h

Let the conversion be 25% in the reactor; the conversion in 1st reactor 17% & in 2nd is 8%.

Air in stream 3 = (95.8 * 17) / 25 = 65.144 kmol/h = 65.144 * 28.84 kg/h = 1878.75 kg/h

O2 in 3 = (20.118 * 17) / 25 = 13.68 kmol/h = 13.68 * 32 kg/h = 437.74 kg/h N2 in 3 = (75.682 * 17) / 25 = 51.46 kmol/h = 51.46 * 28 kg/h = 1440.98 kg/h

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Air in stream 4 = (95.8 * 8) / 25 = 30.656 kmol/h = (30.656 * 28.84) / 25 = 884.12 kg/h

O2 in 4 = (20.118 * 8) / 25 = 6.437 kmol/h = 6.437 * 32 kg/h = 205.99 kg/h

N2 in 4 = (75.682 * 8) / 25 = 24.218 kmol/h = 24.218 * 28 = 678.11 kg/h

Now,Stream 6 will contain O2, N2, vapour of cumene & cumene hydroperoxide.

O2 in stream 6 = O2 in stream 3 – O2 consumed in 1st converter.

In 1st converter conversion based on cumene is 17%; Therefore, cumene consumed in 1st reactor = 8.33 * 0.17 = 1.4161 kmol/hO2 consumed in 1st reacter = 1.4161 kmol/h (as per stoichiometry)

Therefore, O2 unconsumed = 13.68 – 1.4161 = 12.2639Cumene unconsumed = cumene feed – cumene consumed in 1st reactor = 8.33 – 1.1461 = 6.9138 kmol/h

Now, Pt = 6 atm = Po2 + Pn2 + ∑ Pvi * XiComposition of liquid phase of cumene: X c = F.R.of unconverted cumene/(F.R. of unconverted cumene + F.R. of cumene hydroperoxide = F.R. of feed–consumed / (F.R. of unconverted cumene + F.R. of cumene hydroperoxide) = (8.33 – 1.4161)/ (6.9139 + 1.4161) = 0.83

Therefore, X cup = 1- Xc = 1- 0.83 = 0.17

At 100º C Pvc = 180 mm Hg & Pvcup = 5 mm HgPt = Po2 + Pn2 + ∑ Pvi Xi

760 * 6 = Po2 + Pn2 + (180 * 0.83 + 0.17 * 5)Po2 + Pn2 = 4409.75 mm Hg

Also, Po2 + Pn2 / Pt = No2 + Nn2 / Nt

Nn2 = same as the feed = 51.46 kmol / h

No2 = unreacted o2 = 13.68 – 1.4161 = 12.2639 kmol/h

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Po2 + Pn2 / Pt = No2 + Nn2 / Nt 4409.75 / 4560 = (51.46 + 12.2639)/ NtNt = 65.8951 kmol/h

Nc = Nt * Yc = Nt * (Pvc * Xc / Pt) = 65.8951 * (180 * 0.83 / 760 * 6) = 2.1589 kmol / hNcup = Nt * Ycup = 65.8951 * (0.17 * 5 / 760 * 6) = 0.01228 kmol /h

In stream 8, assuming total condensation in C.W. condenser (for 1st trial calculation)Xc = Yc / Yc +Ycup & Xcup = 1- Xc

Yc = Nc / Nt = 2.1589 / 65.8951 = 0.0327Ycup = Ncup / Nt = 0.01228 / 65.8951 = 0.0001864

Xc = 0.032 / (0.0327+0.00018) = 0.973 Xcup = 1-Xc = 0.0269

Now, considering the condensation temp at 40º C , we getPvc = 10 mm Hg & Pvcup = negligible

Pt = Po2 + Pn2 + ∑ Pvi Xi6 * 760 = Po2+Pn2+ 10 * 0.973+0 * 0.0269Po2+Pn2 = 4550.27

Now, Po2+Pn2 /Pt = 4550.27 / 4560 = No2+Nn2 /Nt & No2+Nn2 = 51.46+12.2639Therefore, Nt = (4560 * 63.7239) / 4550.27

Nc = Nt * (Pvc * Xc / Pt) = 63.860 * (10 * 0.973 / 6 * 760) = 0.136 kmol /h

X= liquid phase mole fraction Y= mole fraction in gas phase

Yc= 0.00212 Xc= Yc/Yc = 1

Hence major amount of vapors are considered in C.W. condenser, so that revision in values of Xi (for calculation) is not required.

In stream 8a, let the brine be the cooling medium in the next condenser to recover remaining cumene vapor, so let temp = 5º C, Pvc = 1 mm Hg

Pt = Po2 + Pn2 + ∑ PviXi6 * 760 = Po2 + Pn2 + 1 * 1 + 0Po2 + Pn2 = 4559 mm Hg

Po2+Pn2 /Pt = 4559 / 4560 = 0.9997 = No2+Nn2 /Nt & No2+Nn2 = 51.46+12.2639 = 63.7239Nt = 63.7239 / 0.9997 = 63.7378

Nc = Nt * (Pvc * Xc / Pt)

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= 63.860 * (1 * 1 / 6 * 760) = 0.01397 kmol / h

Yc = 0.01397 / 63.7378 = 0.000219

Wc = (0.000219 * 120)/ (0.000219 * 120 + 51.46 * 28 + 12.2639 * 32) = 0.00001433 = 14.33 ppm of cumene is lost in vent.

Now, cumene & cumene hydroperoxide in stream 5 is ,

Ncup = 1.416 (cumene consumed in 1st reactor – cumene lost in vent)

Nc = Cumene in feed – cumene lost = 8.33 – 1.416 – 0.01397 = 2.14493 kmol /h For stream 9 & 10;

Cumene consumed in 2nd reactor assuming 25% conversion (i.e. 17 +8)

Hence cumene consumed in 2nd reactor = 8.33 * 0.25 – 8.33 * 0.17 = 0.6664 kmol/h which implies,

O2 consumed = 0.6664 kmol/h & cumene hydroperoxide formed = 0.6664 kmol /h

Therefore cumene in Stream 7 = stream 6 – stream8 + stream 8a = 2.14493 kmol/h

Cumene hydroperoxide = complete recycleThe amount of various components after the reaction in 2nd reactor,

Unconverted cumene = unreacted cumene from 1st reactor – cumene consumed in 2nd reactor = 6.9139- 0.6664 = 6.2475 kmol/h

Cumene hydroperoxide = cumene hydroperoxide formed in (1st + 2nd) reactor = 1.1461+0.6664 = 2.0825 kmol /h

O2 in stream 9 = O2 in 4 – O2 consumed No2 = 6.437 – 0.6664 = 5.4706 kmol /h

N2 in stream 9 = N2 in 4 Nn2 = 24.219 kmol /h

Mole fraction of cumene in liquid phase of 2nd rector Xc = 6.2475 / (6.2475 + 2.0825) = 0.75

Xcup = 1 – Xc =0.25

At 100º C, Pvc = 180 mm Hg Pvcup = 5 mm Hg

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Pt = Po2 + Pn2 + ∑ PviXi 760 * 6 = Po2 + Pn2 + 180 * 0.75 + 5 * 0.25Po2 + Pn2 = 4423.75 mm Hg

Po2+Pn2 /Pt = 4423.75 / 4560 = No2+Nn2 /Nt & No2+Nn2 = 5.7706+ 24.214 = 29.9896

Nt = (29.9896 * 4560) / 4423.75 = 30.9132

Nc = Nt * Yc = Nt * (Pvc * Xc / Pt) = 30.9132 * (5 * 0.25 / 760 * 6 ) = 0.008259 kmol /h

In stream 11, assuming total condensation in C.W. condenser (1st trial calculation) Xc = Yc/ (Yc + Ycup) Xcup = 1- Xc

Yc = Nc / Nt = 0.9152 / 30.9132 = 0.0296Ycup = Ncup / Nt = 0.008259 / 30.9132 = 0.000267Xc = Yc / Yc + Ycup = 0.0296 / 0.0296 + 0.000267 = 0.991

Xcup = 1-Xcu = 0.0089

Now, considering the condensation temperature 40º C, we get Pvc = 100 mm Hg & Pvcup = negligible

Pt = Po2 + Pn2 + ∑ PviXi6 * 760 = Po2 + Pn2 + 10 * 0.991 + 0.0089 * 0 Po2 + Pn2 = 4550.09

Po2 + Pn2 / Pt = 4550.09 / 4560 = No2 + Nn2 / NtTherefore, Nt = (29.9896 * 4560) / 4550.09 = 30.055

Nc = Nt * (Pvc * Xc / Pt) = 30.055 * (10 * 0.991 / 760 * 6) = 0.06532

Ncup = 0.008259 (considering complete conversion)

Cumene in stream 11 = cumene in 9 – Nc = 0.9152 – 0.06532 = 0.84988

Yc = Nc / Nt =0.06532 / 30.055 = 0.00217 Xc = 1

Hence major amount of vapour are condensed in C.W. condenser. So, the revision in values of stream Xi is not required.

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Im stream 12, let the brine be the cooling medium in the next condenser to recover remaining cumene vapour, temp = 5º C Pvc = 1 mm HgPt = Po2 + Pn2 + ∑ PviXi6 * 760 = Po2 + Pn2 + 1 * 1 + 1 * 0

Po2 + Pn2 = 4559 mm Hg

Po2 + Pn2 / Pt = 4559 / 4560 = 0.997 = No2 + Nn2 / NtTherefore, Nt = 29.9896 / 0.997 = 30.0798

Now, cumene in stream 12 = cumene in 9 – cumene in 11 = 0.9152 – 0.84988 = 0.06532

Nc = Nt * (Pvc * Xc / Pt) = 30.0798 (1 * 1 / 760 * 6) = 0.006596 kmol / h

Yc = Nc / Nt = 0.006596 / 30.0798 = 0.000219

Wc = 0.000219 * 120 / (0.000219 * 120 + 24.219 * 28 + 5.7706 * 32) = 0.000916Therefore, 916.5 ppm of cumene is lost in vent Now, cumene & cumene hydroperoxide in stream 12b, Nc = Cumene in 12 – Cumene in 12a = 0.06532 – 0.006596 = 0.05872

Dissolved gases will be released in flash vessel. Since the boiling point of cumene & cumene hydroperoxide are higher their loss is negligible.

Therefore, the content of stream 10 ≈ stream 13,

Now, in distillation column, Xd = 0.999 Xw = (50 * 0.00001 * 152) / 120 = 0.0000633 Zf= 0.75

F = D + W & F * Zf = D * Xd + W * Xw

Cuemen in stream 10 = unconverted cumene in 2nd reactor = 6.9 – 0.6664 – 0.006596 = 6.227 kmol /h

Cumene hydroperoxide in stream 10 = cumene hydroperoxide formed in 2nd reactor = 2.0825

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F = D + W = 6.2275 + 2.0825 = 8.3095 kmol /h

F * Zf = D * Xd + W * Xw8.3095 * 0.75 = D * 0.999 + W * 0.0000633

Therefore, W = 2.0713 kmol /h (nearly pure cumene) D = 6.238 kmol/h (cumene)

The heat removal from the strongly exothermic cleavage reaction (ca. 1680 KJ / Kg ) of cumene hydroperoxide is achieved by means of evaporation of acetone from the reaction system.[1,2]

W = 2.076 kmol/h * 152 kg/kmol = 315.552 kg/h

Therefore, heat released in cumene hydroperoxide = 1680 * 315.552 = 530127.26 KJ/hλ acetone = 145 cal / gm = 606.39 kj /kgTc acetone = 235 T = 50 º C Tc- T = 185 º C Evaporation rate of acetone in reactor = 530127.26 / 606.39 = 874.2349 kg/h

Cumene hydroperoxide a high boiling point component (n.b.p. is 246.6 º C)

Vapour pressure of water at 50 º C is 92.51 mm Hg.Partial pressure of water vapour in vapour mixture = v.p. * liquid mole fraction = 92.51 * Xw

Op. pressure, Pt = 760 mm Hg = Pac + PwPac / Pw = Nac / Nw

Cumene hydroperoxide = 315.552 kg /h

Therefore, acetone = 315.552 kg/h

H2SO4 (catalyst) = 10% of 315.552 = 31.55 kg /hAs per the stoichiometry of the reaction the no. of moles of cumene hydroperoxide is same as that of phenol and acetone formed.

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Phenol = 2.076 kmol/h Acetone = 2.076 kmol/h = 182.688 kg/h = 120.408 kg /h

Water added along with H2SO4 = (31.55 * 0.9) / 0.1 = 283.95 kg /h

Water coming from scrubber along with the condensed acetone vapour considering 20% acetone soln with water = (874.2349 * 0.8) / 0.2 = 3496.9396 kg/h

In stream 18,Phenol = 2.076 kmol/h = 182.688 kg /h

Acetone = 2.076 kmol/h + 15.073 kmol /h = 17.179 kmol/h = 994.6729 kg/h

H2SO4 = 31.55kg/h = 0.3219 kmol/h

Water = 283.95 + 3496.9396 = 3780.8896 kg/h = 210.048 kmol/h

Total moles = 260.824 kmol/h

Therefore, mole fraction of water Xw= (210.049 / 260.824) = 0.8053

Partial pressure of water vapour in vapour mixture = v.p. * liq. Mole fraction = 92.51 * Xw = 92.51 * 0.8053 Pw = 74.4983 mm HgOperating pressure = 760 = Pt = Pac + PwPac = 760 – 74.7983 = 685.5016

Pac / Pw = Nac / Nw (Nac= mole of acetone evaporated)Nw = (Nac * Pw) / Pac = (15.073 * 74.4983) / 985.5016 = 1.638 kmol / h

In stream 16, the water and acetone vapour evaporated is,Nw = 1.638 kmol /h = 29.484 kg /h

Na = 15.073 kmol/h = 874.2349 kg/h

In stream 17,Acetone = 15.073 kmol/h (same the previous)

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= 874.249 kg/h

Water = 194.2744 kmol/h = 3496.9396 kg/h

Therefore, water to be added in the scrubber = 194.2744 – 1.638 = 192.6364 kmol/h = 3467.4556 kg/hFor stream 19 & 20,Xd = 0.999 Xw = 0.00001551

F = D + WF * Zf = D * Xd + W * Xw

260.824 = D + W (F is taken as the total moles in stream 18 i.e. feed for the dist. Col.)

Solving the above equation, we get D= 17.1495 & W = 243.674 kmol/h

Xd = 0.999 Xw = (0.00005 * 18) / 94 = 0.000095745

F = D + WF * Zf = D * Xd + W * Xw Zf = 243.674 / (243.674 + 17.1495) = 0.9342Solving we get, W = 15.805 kmol/h D = (243.674 – 15.805) = 227.869 kmol /h

Therefore, the phenol + H2SO4 content is 15.805 kmol/h & assuming 100% separation in decanter we get,Pure phenol = 15.805 – 0.3219 =15.4831kmol/h = 1455.41 kg/h1 ton = 1000 kg 1.45 t/h = 1.45 * 24 = 34.9298 t/d ≈ 35 t/d Now for fresh feed = Feed – cumene recycled from stream 15 = 8.33 – 6.238 =2.092kmol/h

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Material Balance Stream wise:-

Stream Component kg/h kmol/h wt% mol%       

1 Cumene 999.6 8.33 100 100       

2 O2 643.74 20.118 23.3 21N2 2119.096 75.682 76.7 79

Total 2762.87 95.8 100 100       

3 O2 437.74 13.68 23.3 21N2 1440.98 51.46 76.7 79

Total 1878.75 65.144 100 100       

4 O2 205.98 6.437 23.3 21N2 678.11 24.218 76.7 79

Total 884.12 30.656 100 100       

5 Cumene 828 6.9 79.4 83Cumene

hydroperoxide 215.232 1.416 20.6 17Total 1043.232 8.316 100 100

       6 O2 392.44 12.2639 19.27 18.61

N2 1440.88 51.46 70.69 78.09Cumene 202.9366 2.1589 9.95 3.28Cumene

hydroperoxide 1.8666 0.01228 0.09 0.02Total 2038.1232 65.8951 100 100

       7 Cumene 257.388 2.1449 99.3 99.43

Cumene hydroperoxide 1.8666 0.01228 0.7 0.0051

Total 259.2546 2.15718 100 100       

8 Cumene 16.32 0.136 0.88 0.21O2 392.44 12.2639 21.12 19.61N2 1440.88 57.46 78 80.58

Total 1849.64 63.8599 100 100       

8a Cumene 14.6436 0.12203 100 100       

18

8b Cumene 1.6764 0.01397 0.09 0.022O2 392.44 12.2639 21.4 19.24N2 1440.88 51.46 78.51 80.74

Total 1834.996 63.73787 100 100       

9 O2 184.6592 5.7706 18.94192 18.6672N2 678.132 24.219 69.56126 78.3455

Cumene 109.824 0.9152 11.26549 2.9606Cumene

hydroperoxide 1.255 0.008259 0.12873 0.0267Total 974.8702 30.91306 100 100

       10 Cumene 747.24 6.227 70.2438 74.94

Cumene hydroperoxide 316.54 2.0825 29.7562 25.06

Total 1063.78 8.3095 100 100       

11 Cumene 101.9856 0.84988 98.7844 99.04Cumene

hydroperoxide 1.255 0.008259 1.2156 0.96Total 103.2406 0.858239 100 100

       12 Cumene 7.8384 0.06532 0.9003 0.2173

O2 184.6592 5.7706 21.2098 19.2002N2 678.132 24.219 77.8898 80.5825

Total 870.6296 30.05492 100 100       

12a Cumene 0.79152 0.006596 0.09 0.021989O2 184.6592 5.7706 21.39 19.24N2 678.132 24.219 78.52 80.74

Total 863.5827 29.9962 100 100       

12b Cumene 7.04688 0.05878 100 100

13 Cumene 747.24 6.227 70.24 74.94Cumene

hydroperoxide 316.54 2.0825 29.86 25.06Total 1063.78 8.3095 100 100

14Cumene

hydroperoxide 315.552 2.076 100 100

15 Cumene 748.5787 6.238 100 100

16 Water 29.484 1.638 3.26 9.8Acetone 874.2349 15.073 96.74 90.2

Total 903.7189 16.711 100 100

19

17 Water 3496.9396 194.2744 80 92.8Acetone 874.2349 15.073 20 7.2

Total 4371.1745 209.3474 100 100

18 Phenol 182.688 2.076 3.6612 0.9041Acetone 994.6429 17.149 19.93 7.4692H2SO4 31.55 0.3219 0.63 0.1402Water 3780.8896 210.049 75.773 91.486Total 4989.7705 229.5959 100 100

19 Acetone 994.671 17.1495 100 100

20 Water 4386.132 243.674 8.57 33.33Phenol 22905.356 243.674 44.76 33.33H2SO4 23880.052 243.674 46.67 33.34Total 51171.54 731.022 100 100

21 Waste water 4101.642 227.869 100 100

22 Phenol 1482.67 15.805 48.96 50H2SO4 1548.89 15.805 51.04 50Total 3034.56 31.61 100 100

23 Phenol 1455.414 15.4831 100 100

24 H2SO4 31.5462 0.3219 100 100

20

Chapter 6: Energy Balance

The energy balance of a particular system can be achieved from the first law of thermodynamics which states that total energy of an isolated system remain constant. The first law of thermodynamics relates to the conservation of energy of energy.

Now let’s take equipment one by one and make energy balance each around.Energy balance around 1st rector :-

The reaction is exothermic; ca 800 KJ are released per kilogram of cumene hydroperoxide.

Equation:-

Rate of energy in with feed stream (1 & 3) + Heat generated by reaction = Heat removed in overhead condenser + rate of energy out with product stream + heat removed in external heat exchanger.

Let the temp. of feed streams (1 & 3) and let the reference temp. For this energy balance is 25 ºC.

Enthalpy of feed stream = 0.

Heat generation by rxn = ∆ Hr in KJ/Kg * Kg/h of product formed = 800 * (1.416 * 152) = 172185.6 KJ/h

Heat duty of cooling water overhead condenser = qci = ∑ miλi mc = 2.14493 kmol/h = 257.3916 kg/h mcup = 0.01228 kmol/h = 1.86656 kg/h

Latent heat of cumene = 330 KJ/kg

Since cumene hydroperoxide is a similar compound its latent heat = 330 KJ/kg

qc1 = ∑ miλi

21

= mcλc + mcupλcup = 257.3916 * 330 + 1.8666 * 330 = 85555.01 KJ/h

mw cw ∆t = qc1

mw = 85555.01 / (4.186 * 40) (∆t = 40 ºC assumed) = 510.9592 kg/h

Cumene Properties

Freezing point = 96.03 ºCBoiling point = 152.39 ºCDensity, g/cm30 ºC 0.878620 ºC 0.8619 40 ºC 0.8450

Heat of vaporization at b.p. = 312J/gHeat of vaporization at 25 ºC = 367J/gHeat of formation at 25 ºC = -41300 J/molHeat of combustion at constant pressure & 25 ºC Gross = 43400 J/g Net = 41200 J/g

Almost an organic peroxides are photo and thermally sensitive because of the facile cleavage of the weak oxygen – oxygen bond; ∆H = -125.6 to -184.2 kJ / mol (i.e. -30 to -44 Kcal / mol)

Critical temperature = 351.4 ºCCritical pressure = 3220 KPa = 31.8 atm

Density g/cm3 = 0.280Heat capacity of ideal vapour at 25 ºC = 153 J/molHeat capacity at 25 ºC = 197 J / (mol k)Surface tension of 20 ºC = 28.8 mN/m

Vapour pressure KPa t ºC 1 33.22 20 99.08 40 119.790 80 143.50 200 180.68 Now, we take Ethanol – water soln as Brine soln.Cp brine = 0.727 kcal/kg ºC * 4.186 = 3.0432 KJ / kg ºCmc=0.01397 kmol/h = 1.6764 kg/h λc = 350 KJ/kgqc2 = miλi = 1.6764 * 350 = 586.74 KJ/h

22

mb cb ∆t = 586.74 KJ/hmb = 586.74 / (0.727 * 4.186 * 2) = 96.40kg/hTherefore, required in condensed qc = qc1 + qc2 = 86141.75 KJ/h

278Rate of energy out with product stream Hg = ∫ Cp avg dT 298 Cpo2 = 8.27 + 0.000258 T -187700 / T2 cal / mol ºC Cpn2 = 6.50 + 0.00100 T cal/mole ºC

Multiplying the above two equations with 4.186 to convert to KJ/mol ºC, we get

Cpo2 = 34.6182 + 0.00107999 T – 78571122 T2 KJ/mol ºC Cpn2 = 27.029 + 0.004186 T KJ/mol ºC

At avg temp. = 288 ºC

Cpo2 = 34.9292 – 9.4728 = 25.4564 KJ/mol ºCCpn2 = 28.4146 KJ/mol ºC

Cp avg = ∑ Yi Cpi = Yo2 Cpo2 + Yn2 Cpn2 = 25.4564 * 0.1924 + 28.4146 * 0.8074 = 26.0569 KJ/mol ºC

278 Hg = ∫ Cp avg dT = 26.0569 * (278 - 298) 298 = -521.138 KJ/kmol * 63.73787 kmol /h = -33216.2261 KJ/h

For stream 5, liquid stream

Yc= 0.8297 Ycup = 0.1703

Cp c = 205 KJ/kmolºCCp cup = 205 KJ/kmol ºC

Cp avg = ∑ Yi Cpi = 0.8297 * 205 + 0.1703 * 205 = 205 KJ/mol ºC

373 Hg = ∫ Cp avg dT = 205 * (373 - 298) = 15375 KJ/kmol * 8.316 kmol / h 298 = 127858.5 KJ/kg

23

Rate of energy in with feed stream (1 & 3) + Heat generated by rection = Heat removed in overhead condenser + rate of energy out with product stream + heat removed in external heta exchanger.

0 + 172185.6 = 86141.75 + 127858.5 – 33216.226 + Hext = 180784.024 + H extHext = -8598.64 KJ/h

Circulation rate required of reaction mass in reactor1 , for rise in reaction temp 1 ºCTherefore, mc clc ∆t = 8598.64 KJ/hmc = 8598.64 / (1.642 * 1) = 5236.6869

ms λs = 8598.64 ms = 8598.64 / 2201.6 = 3.905 kg /h

Rate of energy in with feed stream (5) + Heat generated by reactor = Heat removed in overhead condenser + rate of energy out with product stream + heat removed in external heta exchanger.

Rate of energy in with stream (5) = Hr(5) = 127858.5 KJ/kg

Heat generated in reation = ∆ Hr in KJ/kg * kg/h of product formed = 800 * 101.2928 = 81034.24

Heat removed in overhead condenser:-mc = 0.84988 kmol/h = 101.956 kg/hmcup = 0.008259 kmol/h = 1.2554 kg/h

Latent heat of cumene = 330 KJ/kgLatent heat of cumene hydroperoxide = 330 KJ/kg

qc1 = ∑ miλi = mcλc + mcupλcup = 101.9856 * 330 + 1.2554 * 330 = 34069.53

mw cw ∆t = qc1mw = 34069.53 / (4.186 * 40) = 203.4731 kg/h

Distillation column is operated in vacuum.

mc=0.006596 kmol/h = 0.76152 kg/h λc = 350 KJ/kg

qc2 = miλi = 0.79152 * 350 = 277.032 KJ/hmb cb ∆t = 277.032 KJ/hmb = 277.032 / (0.727 * 4.186 * 2) = 45.5162 kg/h

24

Therefore, required in condensed qc = qc1 + qc2 = 34346.562 KJ/h

278Rate of energy out with product stream Hg = ∫ Cp avg dT 298

Cpo2 = 25.4564 KJ/mol ºC (same as previous)Cpn2 = 28.4146 KJ/mol ºC (same as previous)

Cp avg = ∑ Yi Cpi = Yo2 Cpo2 + Yn2 Cpn2 = 25.4564 * 0.1924 + 28.4146 * 0.8074 = 26.0569 KJ/mol ºC

278 Hg = ∫ Cp avg dT = 26.0569 * (278 - 298) 298 = -521.138 KJ/kmol * 63.73787 kmol /h = -33216.2261 KJ/h

For stream 10,

Yc= 0.7494 Ycup = 0.2506

Cp c = 205 KJ/kmolºCCp cup = 205 KJ/kmol ºC

Cp avg = ∑ Yi Cpi = 0.7494 * 205 + 0.2506 * 205 = 205 KJ/mol ºC

373 Hg = ∫ Cp avg dT = 205 * (373 - 298) = 15375 KJ/kmol * 8.316 kmol / h 298 = 127858.5 KJ/kg

From the energy balance equation:-8595.6 + 81034.24 = 34346.562 + 127858.5 – 33216.2261 + ∆ Hext ∆ Hext = -39555.956 KJ/kg

Circulation rate required of reaction mass in reactor1 = mc clc ∆t = 39555.96mc = 39555.956 / (1.642 * 1) = 24090.12 Kg/h

Q = Hg – Hf / Hg - Hl = (Hg – Hl ) + ( Hl – Hf ) / Hg – Hl = λf + Cpf ∆t / λf

λf = λc Xf + (1- Xf) λcup

25

Cpf = Cpc Xf + (1- Xf) Cpcup

λc = 330 KJ/kg = 39600 KJ/kmolλcup = 330 KJ/kg = 50160 KJ/kmol

λf = 42240 KJ/kmolCpf = 205 KJ/mol ºC (as Cpc = Cpcup)

Q = 42240 + 205 (50) / 42240 = 1.3882 α = v.p. of cumene / v.p. of cumene hydroperoxide at tb = 110 ºC = 30 / 5 (acc. to v.p. data ) = 6

Rm = 1/ α - 1 [xd / xf – α (1- xd) / (1 - xf)] (for q =1) = 1 / 6 -1 [0.999 / 0.75 – 6 (1- 0.999) / (1 – 0.75)] = 1 / 5 (1.332 – 0.024) = 0.2616R = 2 Rm = 0.5232

Qc = (R+1) * D * λc

D = 6.238 kmol/hλc = 300 * 120 KJ/kmol = 36000 KJ/kmol

Therefore, Qc = 1.5232 * 6.328 * 36000 = 346997.1456 KJ/h

mw clw ∆t = 346997.1456 mw = 346997.1456 / (4.186 * 40.32) = 10361.8354 kg /h

Reboiler heat duty for ref. temp.90 ºC, Qb = qc + Hd*D + Hw*W – Hf*F + ql

Hf*F = m Cpmix ∆t Cpmix = Cpc * Xc + Cpcup (1 - Xc) = 205 KJ/kmolHf*F = 8.3095 * 205 * 10 = 17034.475 KJ/h

Hd*D = m Cpmix ∆t = 6.238 * 205 * (99.08 -90) = 11611.4132 KJ/h

Let the temp. at the bottom of distillate 150 ºC,Hw*W = 2.076 * 205 * (150 – 90.08) = 21670.5336 KJ/h

26

Considering 10% heat loss we will multiply it by 1.1 ql = 1.1

Qb = (346997.1456 + 11611.4132 + 21670.5339 – 17034.475) * 1.1 = 399569.0791 KJ /h = ms * λs

At t = 182.02 λs = 2005.9ms = 399569.0791 / 2005.9 = 199.195 kg/h

The heat removal from the strongly exothermic cleavage reaction (ca. 1680 KJ/kg i.e. 0.011053 KJ/kmol of cumene hydroperoxide) is achieved by means of evaporation of acetone from the reaction system.

Enthalpy of feed stream + heat of dilution of sulfuric acid + enthalpy of liquid stream + energy evolved due to rxn = enthalpy of liquid product stream + gas stream leaving from the top + heat removed in C.W. circulated.

Heat of dilution of sulfuric acid = - 4 KJ/ mol H2SO4 = 4000 * 0.3219 = 1287.6 KJ/h

Heat duty of feed stream = m Cp ∆t = 2.076 * 205 * (150 - 25) = 53197.5 KJ/h

Enthalpy evolved due to the reaction ∆ Hr = 1680 KJ/kg * 315.552 Kg /h = 530127.36 KJ/h

Cumene hydroperoxide = 2.076 * 152 = 315.552 kg/h

Stream 18, at 60 ºC Enthalpy of liquid product stream = ∑ m Cp ∆t = (m Cp ∆t) acetone + (m Cp ∆t) phenol + (m Cp ∆t) aq. S.A. Cp acetone = 0.514 cal/g ºC = 0.514 * 4.186 * 58 = 124.79 KJ/kmol ºC = 2.1516 KJ/kgºCCp phenol = 1.105 KJ/kg ºC = 1.105 * 94 = 103.87 KJ/kmol ºC Cp aq.S.A. = 0.9177 cal / g ºC = 0.9177 * 4.186 * (0.02 * 98 + 0.98 * 18) = 74.8009 KJ /kmol ºC = 3.8415 KJ/kg ºCConsidering 10% aq. H2SO4 soln = (10/98) / (10/98 + 90/18) = 0.102 / 5.102 = 0.02

Enthalpy of liquid product stream = (2.076 * 103.871*(60-25) + 17.149 * 124 *(60 – 25) + 210.3709 * 74.809 * (60 -25)) = 633205.665 KJ/h

Stream 16:- Gas stream in the scrubber

27

Cp water = 0.44 kcal / kg ºC * 4.186 * 18 = 33.1531 KJ / kmol ºC

Cp acetone = 300 KJ/kmol ºC = 15.073 * 300 * (60 - 25) + 1.638 * 33.153 * (60-25) = 160167.1672 KJ/h

Stream 17 :- Liquid stream∆ H = ∑ m Cp ∆t = (194.2744 * 4.186 * (40 - 25)) water + (15.073 * 124.79 * (40 -25))acetone = 12198.4896 + 28214.3651 = 40412.8547 KJ/h

According to energy balance eqn :-38302.2 + 14895.3 + 530127.36 + 12198.4896 + 28214.3651 = 633205.66 + 160167.1672 + Heat removed in external circulated water.

∆ Hr = - 169635.1125 KJ/h = ms λs

At 110ºC λs of steam is 2226.2 KJ/kg , ms = 169635.1125 / 2226.2 = 76.1994 kg/h

In this case rxn is not highly exothermic & it is necessary to remove all the acetone formed during the rexn by vaporization to maintain the rxn temp.

Across Scrubber:-

Heat of the feed + Heat released in acetone formed = Heat required in condenser + Heat of product stream. 160167.1672 + 35.082 = Qc + 40412.8547Qc = 119789.3949 KJ/h

Heat released in abs. of acetone soln = 2500 * 1.8 B.t.u. / lb mole = 2500* 1.8 * 0.252 * 4.1868 * 2.205 KJ/kmol = 2.3275 KJ/kmol * 15.073 kmol /h = 35.082 KJ/h

Qc= mw Clw ∆t = 119789.3949 mw = 119789.3949 / 4.187 * (40 - 32) = 3574.5224 kg/h

Now, using Underwood’s eqn:-

∑ (αi xif) / (αi - υ) = 1 – q & ∑ (αi xid) / (αi - υ) = Rm + 1

At avg temp = 104ºC = 377 K , determine αTopmost temp = 58ºC & Bottommost temp = 150ºC

ln P = A – B / (T+C)

Component   A B C

28

         Acetone Light Key 14.7171 2975.95 -34.5228         Phenol   15.2767 4027.98 -76.7014         Water Heavy Key 16.5362 3985.44 -38.9974

Since the conc. of H2SO4 is negligible so it will not create any influence on the vapour pressure of water.ln Pacetone = 14.7171 – 2975.95 / (377 - 345228) Pacetone = 414.7185 KPa.

ln Pphenol = 15.2764 – (4027.98) / (377 – 76.7014) Pphenol = 6.4456 KPa.

ln Pwater = 16.5362 – (3985.44) / (377 – 38.9974) Pwater = 115.0078 KPa.

Component α xif xid       Acetone Pphenol/Pwater = 0.0560 0.9041 -       Phenol Pacetone/Pwater = 3.606 7.4692 0.999       Water Pwater/Pwater = 1 91.486 0.001

Assuming negligible V.P. of H2SO4, α H2SO4 = 0.

Assuming q =1, we get∑ (αi xif) / (αi - υ) = 1 – q = 0

Let υ = 3.0 L.H.S = -1.31466 ≠ 0 = R.H.S.

Let υ = 3.015 L.H.S = 0.1539 ≈ 0 = R.H.S.

So taking υ = 3.015 in eqn:∑ (αi xid) / (αi - υ) = Rm + 1

Rm +1 = 6.09492 = 5.09492

R = 2 * Rm = 10.18984

Energy balance around distillation column;

Qb = qc + Hd*D + Hw*W – Hf*F + ql

Qc = (R+1) * D * λc

29

D = 17.1495 kmol/hλc = 17.1495 KJ/kmol at 56.4 ºC Therefore, Qc = 11.18984 * 17.1495 * 29121 = 5588324.58 KJ/h

At ref. temp. 58 ºC Hd*D = 0

Hf*F = ∑m Cp ∆t = (2.076 * 103.87 * (104 - 58))phenol + (0.3219 * 74.8009 * (104 - 58))H2SO4 + (210.049 * 74.8009 * (104 - 58))water + (17.149 * 124.79 * (104 - 58))acetone= 1754024.097 KJ/h

Let the temp. at the bottom of distillate 150 ºC,Hw*W = 243.674 * (103.87 + 74.8009) * (150 – 58.5) = 4005445.666 KJ/h

Considering 10% heat loss we will multiply it by 1.1 ql = 1.1

Qb = (5588324.58 – 1754024.097 + 4005445.66) * 1.1 = 8623720.757 KJ /h = ms * λs

At t = 182.02 λs = 2005.9ms = 8623720.757 / 2005.9 = 4298.963 kg/h

Now, using Underwood’s eqn:-

∑ (αi xif) / (αi - υ) = 1 – q & ∑ (αi xid) / (αi - υ) = Rm + 1

Topmost temp = 58ºC & Bottommost temp = 150ºC

This column is perated under vacuum to avoid thermal cracking of phenol & to facilitate the use of sat. steam as a heating medium in reboiler.

ln P = A – B / (T+C), using this eqn as in earlier case,Pphenol = 38.2895 KPa

Let pressure drop in distillation column 20KPa, so op. press. at top = (18.289 * 760) / 101.325 = 137.1825 KPa.At avg. temp. = (150 + 58.5) / 2 = 104 ºC

Pwater = 115.0078 KPa.Pphenol = 6.4456 KPa.PH2SO4 = 0

α = 115.0078 / 6.4456 = 17.8428

30

Rm = 1/ α - 1 [xd / xf – α (1- xd) / (1 - xf)] (for q =1)

= 1 / (17.8428 -1) [0.999 / 0.9342 – 17.8428 (1- 0.999) / (1 – 0.9342)] = 0.0474As Rm is very small we take R=1;

Qc = (R+1) * D * λc

D = 227.869 kmol/hλc = 2361.1 KJ/kmol Therefore, Qc = 2 * 227.869 * 2361.1 = 8642152.639 KJ/h

At ref. temp. 58.5 ºC, Hd*D = 0

Qb = qc + Hd*D + Hw*W – Hf*F + ql

Hf*F : Residue of previous distillation column is used to preheat its feed stream.

Heat duty of preheater = [2.076 * 103.87 * (104 - 60)) phenol + (210.049 * 74.8009 * (104 - 60)) water + (17.149 * 124.79 * (104 - 60)) acetone] = (243.674 * (103.87 + 74.8009)) * (150 - t)

150 – t = 18.2837 t = 150 – 18.2837 = 131.716 ºC

Hf*F = (243.674 * (103.87 + 74.8009)) * (131.716 – 58.5) = 3860425.23 kg/h

Let the temp. at the bottom of distillate 150 ºC,Hw*W = 15.805 * (103.87 + 74.8009) * (150 – 58.5) = 258386.2621 KJ/h

Considering 10% heat loss we will multiply it by 1.1 ql = 1.1

Qb = (8643152.639 + 0 + 258386.2621 – 3860425 .23) * 1.1 = 5545225.042 KJ /h = ms * λs

At t = 182.02 λs = 2005.9ms = 5545225.042 / 2005.9 = 2764.31956 kg/h

Utility Equipments:-

Heat duty of brine = 586.74 + 277.032 = 863.772 KJ/h

31

Heat duty of C.W. = 85555.01 + 34069.53 + 346887.1456 + 119789.3949 + 5588324.58 + 8643152.639 = 14817887.9 KJ/h

Water added in process = 3574.5224 scrubber + 3780.8896 reactor = 7355.412 Kg/h * 1.25 = 9.19375 tph

ms = 2764.31956 + 4298.963 + 199.795 + 3.905 = 7266.38256 kg/h

Capacity of boiler req. = 7.266 – 1.25 = 9.0825 tph ≈ 10 tph

Pressure at sat, steamreq. = 12 kg/cm2 gauge

I. Type of boiler = Fire tube boiler

Heat duty of brine plant req. = 863.772 * 1.25 = 1076.715 KJ/h

1 TR = 12000 Btu/h = 12660.67 KJ/hTherefore, 1076.715 / 12660.67 = 0.08504 TR

II. Type of brine = Ethanol – water soln. Inlet temp. = 0ºC & outlet temp. = -2ºC

III. Heat duty of cooling tower = (14817887.9 + 863.772) * 1.25 = 18523439.59 kg/h 18523439.59 / 12660.67 = 1463.069 TR ≈ 1500 TR

IV. DM water plant capacity = water req. for generation of steam + process = 10 + 10 = 20 tph

32

Chapter 7: Design of Equipment

Process Design Type OF the Reactor = Continuous Stirred tank Reactor Design of Continuous Stirred Tank Reactor for 1 ton/hour of Cumene.

Technical Data: =

1. Reaction :-

2. Catalyst: - The 1st reaction is autocatalytic while in the 2nd reaction sulfuric acid is used as catalyst i.e. 10% in Acetone soln.

3. Reaction is only slightly exothermic so to maintain the reaction temp. 60 ºC Saturated steam is used in jacket.

Solution:-

ρlav = 1 / ∑ ( Wi / ρli )

Density (acetone) = 729 kg /m3

Density (cumene hydroperoxide) = 800 kg /m3

Density (H2SO4) = 1834 kg /m3

Density (water) = 1000 kg /m3

Ρlav = 1 / [(0.6159 / 800) + ( 0.1942 / 729 ) + ( 0.0062 / 1834) + ( 0.7381 / 1000) ] = 921.7775 kg /m3

Vo = m / ρlav = 5122.6345 / 921.7775 = 5.5573 m3 / h = 0.09262 m3 / min

33

Γ = V / Vo V = 0.09262 * 15 = 1.3893 m3

Let, H/D = 1 where H= Depth of liq. In reactor shell in meter & D = Inside diameter of vessel

Let the type of bottom head = Torisherical

Inside volume of torisperical head, V= 0.084672 Di 3 + π / 4 Di 2 Sf

Where Di = Inside diameter of reactor,mV = Inside volume, m3

Sf = straight flange, m

Vworking = π / 4 Di 2 h + 0.084672 Di 3 + π / 4 Di 2 Sf

Let Sf= 1.5 in. = 0.0381 m & H/D = 1

1.3893 = π /4 Di3+ 0.084672 Di 3 + π / 4 Di 2 Sf

On solving the above eq. we get Di = 1.5624 m

H = D = 1.5624 m

Consider provision of 20% extra space for vapour – liquid disengagement then actual height of shell of reactor is H = 1.87488m

Energy balance across reactor:-

The heat removal from the strongly exothermic cleavage reaction (ca. 1680 KJ/kg i.e. 0.011053 KJ/kmol of cumene hydroperoxide) is achieved by means of evaporation of acetone from the reaction system.

Enthalpy of feed stream + heat of dilution of sulfuric acid + enthalpy of liquid stream + energy evolved due to rxn = enthalpy of liquid product stream + gas stream leaving from the top + heat removed in C.W. circulated.

Heat of dilution of sulfuric acid = - 4 KJ/ mol H2SO4 = 4000 * 0.3219 = 1287.6 KJ/h

Heat duty of feed stream = m Cp ∆t = 2.076 * 205 * (150 - 25) = 53197.5 KJ/h

Enthalpy evolved due to the reaction ∆ Hr = 1680 KJ/kg * 315.552 Kg /h = 530127.36 KJ/h

Cumene hydroperoxide = 2.076 * 152 = 315.552 kg/h

Stream 18, at 60 ºC

34

Enthalpy of liq. product str. = ∑ m Cp ∆t = (m Cp ∆t) acetone + (m Cp ∆t) phenol + (m Cp ∆t) aq. S.A. Cp acetone = 0.514 cal/g ºC = 0.514 * 4.186 * 58 = 124.79 KJ/kmolºC = 2.1516 KJ/kg ºCCp phenol = 1.105 KJ/kg ºC = 1.105 * 94 = 103.87 KJ/kmol ºC Cp aq.S.A. = 0.9177 cal / g ºC = 0.9177 * 4.186 * (0.02 * 98 + 0.98 * 18) = 74.8009 KJ /kmol ºC = 3.8415 KJ/kg ºCConsidering 10% aq. H2SO4 soln = (10/98) / (10/98 + 90/18) = 0.102 / 5.102 = 0.02

Enthalpy of liquid product stream = (2.076 * 103.871*(60-25) + 17.149 * 124 *(60 – 25) + 210.3709 * 74.809 * (60 -25)) = 633205.665 KJ/h

Stream 16:- Gas stream in the scrubber

Cp water = 0.44 kcal / kg ºC * 4.186 * 18 = 33.1531 KJ / kmol ºC

Cp acetone = 300 KJ/kmol ºC = 15.073 * 300 * (60 - 25) + 1.638 * 33.153 * (60-25) = 160167.1672 KJ/h

Stream 17 :- Liquid stream∆ H = ∑ m Cp ∆t = (194.2744 * 4.186 * (40 - 25)) water + (15.073 * 124.79 * (40 -25))acetone = 12198.4896 + 28214.3651 = 40412.8547 KJ/h

According to energy balance eqn :-38302.2 + 14895.3 + 530127.36 + 12198.4896 + 28214.3651 = 633205.66 + 160167.1672 + Heat removed in external circulated water.

∆ Hr = - 169635.1125 KJ/h = ms λs

At 110ºC λs of steam is 2226.2 KJ/kg , ms = 169635.1125 / 2226.2 = 76.1994 kg/h

In this case rxn is not highly exothermic & it is necessary to remove all the acetone formed during the rexn by vaporization to maintain the rxn temp.Now, μacetone at 60 ºC = 0.6 mPa.sμcum. hyp. at 60 ºC = 2.8 mPa.s μs.a. at 60 ºC = 3.1mPa.sμwater at 60 ºC = 0.5 mPa.s

1 / μmix = ( W/ μ ) acetone + ( W/ μ ) cup + ( W/ μ ) water + ( W/ μ ) s.a. = 1.86273

Therefore, μmix = 0.5368 mPa.s

35

It is difficult to find the suitable correlation for the reaction mass side heat transfer coefficient as the reaction is taking place. hi ca be governed by boiling coefficient or by convective film coefficient. If flat blade 45º C turbine agitator is used to improve the reaction rate and to improve the convective film heat transfer coefficient, thenhi = hnb or hic whichever is less.hic = convective film coefficient, W/(m2 * ºC)hnb = nucleate boiling coefficient, W/(m2 * ºC)

To calculate the convective film coefficient hic, let the tip velocity of turbine agitator V = 200 m / min = π * Da * n ,

where Da = Diameter of agitator n = Rotational spped of agitator in revolutions per min.

For 45 ºC flat blade turbine,Da = shell I.D. / 3 = 1.5624 / 3 = 0.5208 mn = 200 / (π * 0.5208) = 122.23≈ 123 rpm.

For 45 ºC flat blade turbine, baffled vessel for Re > 400,

(hi * Da ) / k = 0.74 * Re 0.67 * Pr 0.33 * (μ / μw ) 0.14

where, Re = Reynolds’s number =(n * Da 2 * ρmix ) / μmix = 123160 * 0.52082 * 921.77575 / (0.5369 * 10 -3) = 954793.33995

Pr = Cpmix * μmix / Kmix

Cpmix = ∑ Cpi * Xi = ( Cp * X )acetone + ( Cp * X )cup + ( Cp * X )aq. s.a. = (2.1516 * 0.1993 ) + (1.105 * 0.0366) + ( 3.8415 * 0.76403 ) = 3.4042 KJ / kg ºC

Thermal conductivity of liquids in B.t.u. / ( hr * sq. ft. * ºF / ft. ) ; Kacetone = 0.102K sul. acid = 0.30K water = 0.363K cup = 0.1502

To convert B.t.u. / ( hr * sq. ft. * ºF / ft. ) into W/ m ºC multiply by 1.73

K acetone = 0.17646 W / m ºCK sul. acid = 0.5190 W / m ºCK water = 0.62799 W / m ºCK cup = 0.2598 W / m ºC

Kmix = ∑ Ki Xi

36

= (0.71646*0.07469) acetone + (0.5190*0.00140)s.a. + (0.91486*0.62799)water + ( 0.2598 * 0.0090 ) cup = 0.590768 W / m ºC

Pr = Cpmix * μmix / Kmix = ( 3.4042 * 0.5368 * 10 -3 * 10 3 ) / 0.59077 = 3.0932

hi = 0.74 * (K / Da ) * Re 0.67 * Pr 0.33 * (μ / μw ) 0.14

= 0.74 * ( 0.59077 / 0.5208 ) * ( 954793.39) 0.67 * ( 3.0932) 0.33 * (1) 0.14

= 12369.52773 W / m ºC

Nucleate boiling coefficient for the boiling of azeotropic mixture can be determined by mostinski’s equation.

hnb = 0.104 * Pc 0.6p (φ / A ) 0.7* [1.8 (P/Pc) 0.17 + 4 * (P/Pc) 1.2+ 10 * (P/Pc) 10]φ = 169635.1125 KJ /h = 47.1209 * 1000 J/sec or W

A = π * D * h = ( π * 1.5625 * ( 0.75 * 1.5624 ) = 5.75169 m2

P = 1 bar & Pc = 47.1 bar Pr = P/ Pc = 1 / 47.1 = 0.02123

Substituting all the above values in the eqn. of hnb, we get

hnb = 11845.94072 W / m ºC

Since convective film coefficient hi = 12369.5277 W / m ºC > hnb, hnb = nucleate boiling coefficient is taken for heat transfer area calculations.

hi = 11845.94072 W / m ºC

Jacket side or saturated side heat transfer coefficient ho’ = 6000 W / m ºC

Thermal conductivity of reactor shell material (S.S. 304) = 16 W / m ºC

Let thickness of reactor shell = 6.35 cm. (Assumed but it should be decided based on mechanical design)

Overall heat transfer coefficient:

1/Uo = (1/ho) + (1/hod ) + (Do* ln (Do/Di) / 2*Kw ) + ( Do/Di) * ( 1/hi ) + ( Do/Di ) * (1/ hid )

Here Di = 1.5624m Do = Di + 2 * ts = 1.5721 m

Putting the above values we get, 1 / Uo = 0.00085188 Uo = 1173.8698 W / m ºC

Heat duty φt = 47.1209 * 1000 J/sec or W ΔTm = 35 ºC

Heat transfer area required A = φt / (Uo * ΔTm ) = 47120.9 / ( 1173.8698 * 35 )

37

= 1.1469 m 2

Outside heat transfer area provided with 75% height covered jacket Ao av. = 5.75169 m 2

% Excess heat transfer area ≈ 401.577 %

Heat transfer area provided minimum is Ao = 1.1469 * 1.5 = 1.72035 cm 2

= π * do * L’L’ = 1.172035 / (π * 1.5624 ) = 0.23878 m

where L’ = length of cylinder that must be covered by jacket = 0.23780 m.

However, entire height of liquid pool ( hl= 1.5624) must be covered by plain jacket as shown in fig to keep uniform temp. of entire liq. pool . Reaction temp. can be controlled by controlling the flow rate of cooling water as shown. Mechanical Design

Resulting data from the process design:-

Top head = TorisphericalOperating Pressure = 1 atmosphereDensity of Liquid = 921.7775 kg / m3 (mix)Viscosity of Liquid (mix) = 0.5368 CPType = 45 pitched blade turbineTip velocity = 200 m/ minuteDa / Dt = 113 = 0.333H / Dt = 1.0N = 123 rpm (as previous) No. of agitator required = 1 (as H/D=1)

Da = 0.5208 mF = 1317.9 KgHm2 (Stress)P = 1 atmosphere

Np = Pgc /9n3Da5

5.0 = P X 1 / (921.78 * (128 / 60)3 * (0.5208)5

P = 1521.29 W= 2.0669 hp

Output Power of Electric Motor required = 2.0669 X 1.1 X 1.2= 2.72831 hp

≈ 3 HP motor

Shaft material (55 – 304)

UTS = 75000 Psi = (75000 * 1.033) / 14.696= 5271.8426 Kg / cm2

Shear stress = 18750 * 0.8 Psi = 15000 Psi

38

= 1054.3685 Kg / cm2

Ultimate yield stress yield = 0.6 * 75000 = (45000 * 1.033) / 14.696= 3163.1056 kg / cm2

Average rated torque shaft:-Te = (Power out * 75 * 60) / 2N = (3 * 75 * 60) / 2 * 123

= 17.4682 Kgf.m

TM = 1.5 Te = 25.2023 Kgf.m

Maximum shear stress created in shaft :-

Fs = Tm / Zp = 25.2023 / (d3 / 16)d3 = (25.2023 X 16) / ( X 1054.3685 X 104)d = 0.02304 m d = 23.04 mm

Bm = Fm * l

Fm = Zm / (0.75 Rb) = 25.2023 / (0.75 X (0.5208)) = 129.044 Kgf

l = Total height of vessel – (Da) + 0.5= 2.3228 m

Bm = Fm X l = 299.7434 Kgf.m

Me = 1 / 2 [Bm + (Bm2 + Tm2)]= 1 / 2 [299.7434 + (299.7434)2 + (5.2023)2]= 300.2722 Kgf.m

Stress created by equivalent Bm

f = MelZ3163.105 X 104 = 300.2722 / (d3 / 32 )d3 = (300.2722 X 32) / X 3163.105 X 104

d = 0.04589 m = 45.89 mm

Maximum deflection in shaft :-

δ = (Wl3 /3EI) = (Fml3/ 3EI)

I = d4 δ = 129.044 X 2.32283 X 106 X 6464 3 X 2 X 106 X X (4.589)4

= 12.3816 cm

Nc = Critical Speed = 60 * 4.987 / (12.3816) ½

39

= 85.035 rpm

N = Actual Speed = 123 rpm

N / Nc = 1.4465

Taking d = 6 cm = 4.2368 Nc = 145.3689 N / Nc = 123 /145.3689 = 0.8461 = 84 %

Let diameter be 7.0 cm.

= 2.28696Nc = Critical Speed = 60 X 4.987 / (2.28696) ½

= 197.8612 rpm

N = Actual Speed = 123 rpm

N / Nc = 123 / 197.8612 = 0.6216

62 % of critical speed which is within the range & maximum stress induced in shaft

Blade material:

= Zm / Z & Z = Bt * Bw2 / 6

Bw = Da / 8 = 0.5208 / 8 = 0.0651 = 65.1 mm

= Zm / Z 0.5 * 5271.8426 * 104 = 25.2023 * 6 / (Bt * Bw)2

Bt = 25.2023 * 6 * 1000 / 0.06512 * 0.5 * 5271.8426 * 104

Bt = 1.35 mm

Actual length of shaft = 2.3228 m

Wt. of 6 blades = 6 * length * width * thickness * size= 6 * (0.5208 / 2) * 0.0651 * 0.00135 * 800 = 1.09849 Kg

Wt. of shaft = / 4 d2* l * ρ= / 4 * (0.072 * 2.3228 * 8000) = 71.513 kg

Designer of Top Head (Torispherical):-

40

Internal design pressure = (max. op. pr. within head – min. op. pr. outside the head) * 1.1 = (1.033 – 0) X 1.1 = 1.1363 Kgf / cm2

th’ = (PRcW ) / (2 FJ – 0.2 P) + CA

Rc = I D of reactor R1 = 10 % of Rc= 1.5624 m = 156.24 mm= 1562.4 mm

W = 1 / 4 [3 + (1562.4 / 156.24) ]= 1.54

f = 1317.9 Kgf / m2 J = 1 (Seamless Torispherical)

th’ = (PRcW ) / (2 FJ – 0.2 P) + C = (1.1363 * 1562.4 * 1.54 / (2 * 1317.9 * 1 – 0.2 * 1.1363) + 3

= 4.0377 mm

th = 1.06 th1

= 4.28

Take 5 mm thick plate of SS (304) for the fabrication of head.

OD = I.D. + 2 tn= 1562.4 + 2 X 5= 1572.4 mm

Blank dia. = OD + OD / 24 + 2 * SF + 2 /3 * iCr

SF = 3 tn = 3 X 5= 15 mm < 1.5 inch

SF = 1.5 inch = 38.1 mm

iCr = inside corner radius= 156.24 mm

B.D. = 1562.4 + 1562.4 + 2 X 38.1 + 2 X 156.24 24 3

= 1807.86 mm

Wt. of head = / 4 (1.80786)2 * 0.004 * 800 = 82.1427 Kg

Height or length of head :-

OA = th + B + SF

B = Rc – [(Rc – R1)2 – (ID / 2 – R1)2] 0.5

= 1562.4 - [(1562.4 – 156.24)2 – (1562.4 / 2– 156.24)2]

41

= 302.7528 mm

OA = 5 + 302.7528 + 38.1= 345.8528 mm

Vol. within head is V1 + / 4 * Di2 * SF

V1 = 0.000049 Di3

Di = 1562.4 mm = 61.5118 in

V1 = 11.4043 f3 = 11.4043 * (0.3048)3 + ( / 4 * 1.5622 * 0.0381)= 0.3959 m3

Pint = 1.1363 Kgf / cmDesigner of bottom Head (Torispherical):-

Considering steam at 180 C temp.P = 1.05 MPa = 10.5 * 100 KPa = (10.5 * 100 * 1.033) / 101.325

= 10.705 Kgf / cm2

Pe = (10.705 – 0) X 1.1 = 11.775 Kgf / cm2

External design pressure = 10.705 Kgf / cm2

Analytical Method:-

th1 = PR1W + CA2FJ – 0.2P

P = 1.67 Pe = 19.6644 Kgf / cm2

th1 = 19.6644 X 156.24 X 1.54 + 32 X 1317.9 X 1 – 0.2 X 19.6644

= 4.79775 mm

O.D. = 1 D + 2th = 1544.2 mm

th = 5.085615 < 25.4 mm

th’ = 4.4 r (p / 2E) [3(1-μ2)]½ + CA= 22.4880

tn = 21.87 < 25.4 mm

Blank dia. = OD + OD / 24+ 2 * SF + 2 / 3 * iCr + th

O.D. of head = 1526.4 + 2 * 25.4 = 1577.2 mm Blank dia. = 1577.2 + 1577.2 / 24 + 2 * 38.1 + 2 / 3 * 156.24 + 25.4

= 1848.6766

Wt. of bottom head = / 4 * (1.8486)2 * 0.0254 * 8000= 545.425 Kg

Height or length of head:-

42

B = 302.4528 mm

OA = tn + b + SF = 25.4 + 302.7528 + 38.1= 366.2528 mm

V = V1 + / 4 Di2 *SF = 11.4043 ft3 = 11.4043 * (0.3098)3 + ( / 4 * (1.562)2 * 0.0762 ) = 2.3153 m3

Design of Shell:-

Pin = 1.1363 Kgf / cm2

t = Pri / FE – 0.6 P + CA= (1.1363 * 1562.412) / [(1317.9 * 2 * 106 ) – (0.6 * 1.1363) ] + 3= 3.004

≈ 4 mm

Pext. = (10.705 – 0) X 1.1= 11.7755 Kg / cm2

Select 12.7 mm thick plate of S.S. (304)

t = Actual thickness – CA = 12.7 – 1.5= 11.2 mm

Do / t = (1562.4 + 25.4) / 11.2 = 141.7678

L / Do = 1874.88 /1587.8 = 1.1808

At t = 140 F (i.e. 60 C) ; A = 0.0007 B = 7500 PSi

Maximum allowable external pressureP = B / (Ddt) = 7500/ 141.7678 = 52.9034 psi

Pex > P

P = (52.9037 * 1.033) / 14.696 = 3.718 Kgf / cm2

Taking t = 25.4

Wt. of the shell = / 4 * (Do2 – Di2) * l *ρ = / 4(1.61322 – 1.56242) X 1.84488 X 8000 =1900.390 Kg

Jacket Shell:-

Pin. = (10.705 – 1.033) * 1.1 = 10.6392

43

di = (O.D. of Shell) + 100 = 1713.2

th = Pri / (FE – 0.6 P) + CA= (10.6392 * 856.6) / (1317.9 * 2 * 106 – 0.6 * 10.6392)= 3.0000034 4 mm

Pext = 1 atm

ta = 8 mmt = Actual thickness – CA = 8 – 1.5

= 6.5 mm

Do = D.I. + 2 (E) = 1713.2 + 2 X 8= 1729.2 mm

Do / t = 1729.2 / 6.5 = 266.031

L / Do = (0.75 * 1874.88) / 1729.2 = 0.81318

A = 0.00045, At t = 180 + 50 = 230 C = 4460 FB = 5500 psi

Maximum allowable external pressureP = B / (Do / t ) = 5500 / 266.031 = 20.6742

P = 20.6742 *1.033 / 14.696 = 1.453 Kgf / cm2

Pext > Pd

Now assuming tn = 25.4 mm

t = 25.4 – 1.5 = 23.9 mm

Do = 1764 = 73.8075t 23.9L = 0.75 *1874.88 = 0.79714Do 1764

A = 0.0035At = 446 F B = 9500 psi

P = B = 9500 = 128.7132 psi(Do/t) 73.8075

44

P = 128.7132 *1.033 = 9.0474 Kgf / cm2

14.696Pext < Pd

Thickness required for external pressure > thickness required for internal pressure. Therefore stiffing ring is used to reduce cost of equipment stiffing ring is attached on the outer side.

Dimensions of stiffing ring :- W = 50 mm thick = 8 mm

As = W * T = 400 mm2

L1 = 200 mm

B = PDo = 10.6392 * Dot + AsL t + 400/200

Select ta = 8 mm t = 6.5 mm

Do = 1713.2 + 2 * 8 =1729.2B = 10.6392 X 1729.2 = 2164.38

6.5 + 2At t = 446 F A = 0.00015 in / in.

Moment of inertia req for stiffing ring

I req = Do2L1 (t + As/L1)A / 14= (1729.2)2 * 200 (6.5 + 400/200) * 0.00015 / 14= 54463.13 mm4 = 5.4463 cm4

I pr. = Thk * Width3 / 12 = 8 X (50)3 / 12 = 83333.33 mm4

= 8.3333 cm4

I pro = 8.337 cm4 I req = 5.44 cm4

Wt. of Jacket Shell = (Do2 – Di2) * H * ρ / 4 (1.7292 – 1.71322) * 0.75 * 1.87488 * 8000 / 4

=480.5159 Kg

Design of shell using stiffening ring. Stiffening ring is attached on the inside of shell.Dimensions of stiffing ring: - W = 50 mm thick = 8 mmAs = W * T = 400 mm2

L1 = 200 mmB = PDo = 11.7755 * Do

t + AS/ L1 t + 400/200

Select ta = 8 mm t = 6.5 mmDo = 1562.4 + 2 * 8 = 1578.4B = 11.7755 X 1578.4

6.5 + 2= 2186.64

45

At t = 140 F A = 0.00015 in/in

I req = Do2L1 (t + As/L1)A / 14= (1578.4)2 * 200 (6.5 + 400/200) * 0.00015 / 14 = 15378.098 mm4

= 4.5378 cm4

I provided = Thk X Width3 / 12 = 8 X (50)3 / 12 = 83333.33 mm4

= 8.3333 cm4

Ir < I pro t = 8 mm thicknessWt. of Shell = (Do2 – Di2) * H * ρ / 4

= (1.57842 – 1.56242) * 1.87448 * 8000 / 4= 591.8626 Kg

Jacket Bottom Head:-

Pint = 10.6392 SAME ASPent = 1 atm JACKET SHELL

tn1 = PRcW + CA2FJ – 0.2P

Rc = I.D. of Jacket Ri = 171.32 mm= 1713.2 mm

W = 1 (3 + 1713.2)4 171.32

= 1.54

f = 1317.9 Kgf / m2 J = 1

CA = 1.5 X 2= 3 mm

tn1 = 10.6392 X 1713.2 X 1.54 + 32 X 1317.9 X 1 – 0.2 X 10.6392

= 13.658 mm

Take 14 mm thickness

For external pressure:-

Pent = 1 atmtn1 = PRcW + CA

2FJ – 0.2P

P = 1.67 Pe = 1.67 atm.

tn1 = 1.67 X 1562.4 X 1.54 + 32 X 1317.4 X 1 – 0.2 X 1.67

46

= 4.5252 mm

SF = 3 * thk =15 mm or 38.1 mm

Thk 5 mmOD = 1 D + 2 X 5 = 1572.4 mm

B.D. = OD + OD / 24 + 2 SF + 2 /3 iCr= 1572.4 + 1572.4 / 24 + 2 X 38.1 + 2 / 3 X 156.24= 1818.276

Wt. of Jacket Bottom Head= * (1.8183)2 * 0.005 * 8000 / 4 = 103.8679 kg.

Height or Length of head:-

Volume contained within head =V1 + Di2SF / 4

V1 = 0.000049 Di3

Di = 1562.4 mm = 61.5118 in

V1 = 11.4043 ft3

V = 11.4043 * (0.3048)3 + * 1 * 0.038 / 4= 0.3529 m3

Wt of empty vessel = wt of top head + wt of bottom head + wt of Jacket bottom head + wt of shell + jacket shell

= 81.1424 + 545.425 + 103.8678 + 591.8626 + 480.5159 =1803.814 Kg X 1.2= 2164.5768 Kg

Cost of the vessel = 2164.5768 * 400 = 865827.2

Inside vol. of vessel = Di2h / 4= (1.5629)2*(1.87488)/4 = 3.5946 m3 + 2* 2.3153= 8.2252 m3

Wt of vessel = 8.2252 X 921.775= 7581.7837 Kg

Weight of vessel + Liquid = 2164.5768 +7581.7837+ Shaft + 6 Blades = 9746.3605 Kg + 1.09849 + 71.513

= 9818.97199 Kg

Design of Bracket Support:-

P = 4Pw (H – F) + EW H = 1.87488 to 75 nDb n F = 0.75

PW = KP1h1Do = 0.7 X 100 X 2 X 1.7292= 242.088 Kgf

P = 4Pw (H – F) / nDb + EW / n

47

Db = Jacket shell OD + 150 mm= 17292 + 150 = 17442

P = 4 X 242.088 X (1.87488 + 0.75 – 0.75) / (4 X 1.744) + 9818.97 /4= 2714.998 Kgf

Bracket Design :- Given size of base plate = 150 mm X 150 mm A= 150 mm B = 150 mm a = 140 mm (assumed) distance between two gusset

plateAverage pressure = P /B = 2714.998 / 14 X 15 = 12.9285 Kgf / cm2

Maximum bending stress created in the base plate:-

Fmax = 0.7 Pav B2 (a4)T2 B4+a4

T12 = 0.557789 cm2 = 7.468 mm

T1 = 0.7468 cmUse 8 mm thick base plate Gusset plate thicknessMax bending stress acting in a distance parallel to the edge of gusset platefmax = 3 PC X 1

T2h2 CoSP = 2714.998 KgfC = Bolt circle diameter – OD of reactor / 2

= 1.744 – (1.5724) / 2= 0.0858 m

h = 145 mm (assumed)tan = 150 / 145-10 = 48fmax = Max allowable bending stress

= 1575 Kgf / cm2

1575 = 3 X 2714.998 X 8.58 / T2 X 14.52 X COS48T2 = 3.15 mm

Take T2 = 4 mm

48

Chapter 8: Instrumentation and Process Control

Instrumentation and process control use for Phenol continuous stirred tank reactor is shown in the figure.

Reaction taking place in this reactor is exothermic but heat of reaction is very low 0.0110526 KJ/mol & feed streams are introduced at room temperature. Hence to control the reaction temperature 60 ºC saturated steam is supplied through jacket.

Following instruments and control methodology are used controlling process variable in the reactor.

1) Temperature Controller 2) Pressure Controller 3) Liquid Level Controller

1) Temperature Controller

Temperature of Phenol reactor is control by controlling the flow rate of saturated steam to the temp of Phenol reactor depend on Acetone to oxygen ratio in the reactor. Increase in Acetone air ratio favors the low exothermic reaction.

49

Example: Increase in the temp of the reactor above the control valve will give the signal to the control valve placed in the inlet line of the saturated steam to the carburetor and it will increase flow rate of saturated steam. Increase flow rate of saturated steam will increase the vaporization rate of Acetone and hence increase the Acetone to air ratio.

2) Pressure Controller

Reactor is operated at pressure close to atmospheric pressure. Hence pressure control (on reaction side) is not required. Hence pressure indicator is provided on reaction side. However acetone of rxn mass is continuously removed in a vapor form. Hence Pressure Indicator & Pressure safety valves are provided.

3) Liquid level controller: - It is controlled by controlling the flow rate of product.

Chapter 9: Plant Layout & Location

Plant layout:-

Plant layout can play an important part in determining construction and manufacturing cost and thus must be planned carefully with attention being given to future problems that may arise. Since each plant differs in many ways and no two plants are exactly alike. There is no one ideal plant layout. However proper plant layout in each case will include arrangement of processing areas, storage areas, and handling areas in efficient coordination and with regard to such factors as:

Factors in planning layout:-

(1) New site development or addition to previously developed Site.(2) Type and quantity of products to be produced.(3) Type of process and product control.(4) Operational convenience and accessibility.(5) Economic distribution of utilities and services.(6) Type of building and building code requirement.(7) Health and safety consideration.(8) Waste disposal problems.

50

(9) Auxiliary equipment.(10)Space available and space required.(11)Roads and railroads.(12)Possible future expansion.

Principal of planning layout:-

(1) Storage area should not be provided nearby the process if explosive material is used, here Phenol & Cumene as stored at the topmost area quite away from process area with a space for future expansion.

(2) Control room may be provided in the process area at sufficient height with glass windows, so that plants become visible to the engineer sitting inside.

(3) Boiler and other utility should not be provided in the process area.Fuel for the boiler should be near to the boiler.

(4) Three cooling towers are located in one line at the RH.S. along with the water storage facilities with E.T.P. plant following it.

(5) Administrative office, marketing office, canteen should be provided on the other side of the process area.

(6) Figure shows the suggested layout based on unit concept.(7) Also, all the basic arrangements for road, security etc. are been incorporated in the

layout.

Plant Location :-

The location of plant is one of the prime factors in deplaning of new establishment. The geographical location of the final plant can have strong influence on the success of an industrial venture.

Following factor should consider in choosing the plant location:-Availability of Raw Material:-When cumene is used as a raw material it should be located near a petrochemical refinery where benzene can be obtained easily which is a raw material for cumene production.

Location of market:Product is used in laminates, varnishes, resins, air freshener, pharmaceutical, plastics, dye Industries.Market is easily available if plant situated near to this companies.

Transportation and communication:People and goods must be conveniently transported to and from the plant satisfactorily. Plant must have both road and rail facility.

Labor and skilled workers:A large portion of product cost of any manufactured item is represented by labor costs. So location should be such that labor and skilled workers are easily available.

51

Utilities:Location must have adequate water supply and ample supply of electricity should be available at or new location economic, availability of fuel, coal or gas is also prime consideration.

Suitable land:Sufficient land should be available for proposed plant and for future expansion. The land should be flat and should have suitable load bearing characteristics.

Effluent Disposal:Full consideration must be given to the difficulties and cost of disposal of waste.The plant must have separate disposal unit. Near plant location a centralized system for drainage may be available.

Climate:It is desirable to locate a plant where climate conditions are most favorable for ease of operation for plant and personnel.

Local community consideration:The plant should be closed to local community and should not imposing any big risk to people and environment.

Chapter 10: Safety & Environment

List of Safety Equipments at Industry

a) They provide HELMET, EAR PLUG, GOGGLES, FACE SHIELD, HAND GLOVES, APRONS, SAFETY SHOES personal protective equipments.

b) Warning Instruments:- Oxygen, Carbon dioxide, Cumene, cumene hydroperoxide indicator with respectable sensor.

c) Fire Extinguisher is also provided on each an every point and a First aid Boxes at all floor where employees are working. Type of Fire Extinguisher:-

1. Soda Acid Fire Extinguisher2. Foam Type Fire Extinguisher3. Carbon Dioxide Fire Extinguisher4. Dry Chemical Fire Extinguisher5.

Product Identification:-

Synonyms: Carbolic acid; Phenic acid; Phenylic acid; Hydroxybenzene; Phenol, fused; Monohydroxybenzene; Phenol, solid

Hazards Identification:-

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Emergency Overview: POISON! DANGER! MAY BE FATAL IF SWALLOWED, INHALED OR ABSORBED THROUGH SKIN. RAPIDLY ABSORBED THROUGH SKIN. CORROSIVE. CAUSES SEVERE BURNS TO EVERY AREA OF CONTACT. AFFECTS CENTRAL NERVOUS SYSTEM, LIVER AND KIDNEYS. COMBUSTIBLE.

Lab Protective Equip: GOGGLES & SHIELD; LAB COAT & APRON; VENT HOOD; PROPER GLOVES; CLASS B EXTINGUISHER Storage Color Code: White Stripe (Store Separately)

Inhalation: Breathing vapor, dust or mist results in digestive disturbances (vomiting, difficulty in swallowing, diarrhea, loss of appetite). Will irritate, possibly burn respiratory tract. Other symptoms listed under ingestion may also occur.

Ingestion:-Poison. Symptoms may include burning pain in mouth and throat, abdominal pain, nausea, vomiting, headache, dizziness, muscular weakness, central nervous system effects, increase in heart rate, irregular breathing, coma, and possibly death. Acute exposure is also associated with kidney and liver damage. Ingestion of 1 gram has been lethal to humans.

Skin Contact:-Corrosive. Rapidly absorbed through the skin with systemic poisoning Effects to follow. Discoloration and severe burns may occur, but may be disguised by a loss In pain sensation.

Eye Contact:-Corrosive. Eye burns with redness, pain, blurred vision may occur. May cause severe damage and blindness.

First Aid Measures:-

Inhalation:-Remove to fresh air. If not breathing, give artificial respiration. If breathing is difficult, give oxygen. Get medical attention immediately.

Ingestion:-If swallowed, immediately administer castor oil or other vegetable oil. Never give anything by mouth to an unconscious person. Be ready to induce vomiting at the advice of physician or poison control center. Castor oil (or vegetable oil) dosage should be between 15 and 30 cc. Get medical attention immediately.

Skin Contact:-In case of skin contact, immediately flush skin with large amounts of water while removing contaminated clothing and shoes. As soon as possible, repeatedly apply polyethylene glycol to affected area. Destroy contaminated clothing and shoes. Flush skin with water for at least 30 minutes. It is very important to avoid rubbing or wiping affected parts which would aggravate irritation and cause product dispersion. Continue treatment until the burned area changes color from white to pink. Expect that this can take a long period of time (20 minutes or more). The polyethylene glycol application should be done during transportation to the hospital. If polyethylene glycol is not available, flush with water for at least 30 minutes prior to going to hospital. Get medical attention immediately.

Eye Contact:-Immediately flush eyes with plenty of water for at least 15 minutes, lifting lower and upper eyelids occasionally. Get medical attention immediately.

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Fire Fighting Measures:-

Explosion:-Above flash point, vapor-air mixtures are explosive within flammable limits noted above. Sealed containers may rupture when heated.

Fire Extinguishing Media:-Water spray, dry chemical, alcohol foam, or carbon dioxide. Water spray may be used to keep fire exposed containers cool.

Special Information:-In the event of a fire, wear full protective clothing and NIOSH-approved self-contained breathing apparatus with full face piece operated in the pressure demand or other positive pressure mode. Structural firefighter's protective clothing is ineffective for fires involving this material. Stay away from sealed containers.

Handling and Storage:-

Keep in a tightly closed container. Store in a cool, dry, ventilated area away from sources of heat or ignition. Protect against physical damage. Store separately from reactive or combustible materials, and out of direct sunlight.

Avoid dust formation and control ignition sources. Employ grounding, venting and explosion relief provisions in accord with accepted engineering practices in any process capable of generating dust and/or static electricity. Empty only into inert or non-flammable atmosphere.

All phenol operations should be enclosed to eliminate any potential exposure routes. Containers of this material may be hazardous when empty since they retain product residues (dust, solids); observe all warnings and precautions listed for the product.

Ecological Information:-

Environmental Fate:-When released into the soil, this material is expected to readily biodegrade. When released into the soil, this material is not expected to leach into groundwater. When released into the soil, this material may evaporate to a moderate extent. When released into the soil, this material is expected to have a half-life between 1 and 10 days.

When released into water, this material is expected to readily biodegrade. When released into water, this material is not expected to evaporate significantly. When released into water, this material is expected to have a half-life between 10 and 30 days. This material has an estimated bioconcentration factor (BCF) of less than 100. This material is not expected to significantly bioaccumulate. When released into the air, this material is expected to be readily degraded by reaction with photochemically produced hydroxyl radicals.

When released into the air, this material may be moderately degraded by photolysis. When released into the air, this material is expected to have a half-life of less than 1 day.

Environmental Toxicity:-This material is expected to be toxic to aquatic life. The LC50/96-hour values for fish are between 10 and 100 mg/l.

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Toxicological Information:-

Oral rat LD50: 317 mg/Kg; skin rabbit LD50:630 mg/kg; inhalation rat LC50: 316 mg/m3; irritation data: skin rabbit, standard Draize, 500 mg/24H severe; eye rabbit, standard Draize 5 mg/30S rinse, mild. Investigated as a tumorigen, mutagen, and reproductive effector.

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Chapter 12: Material Safety Data Sheet

General

Synonyms: benzenol, carbolic acid, hydroxybenzene, monohydroxybenzene, monophenol, oxybenzene, phenic acid, phenylic acid, phenyl alcohol, phenyl hydrate, phenyl hydroxide, phenylic alcohol Molecular formula: C6H5OH CAS No: 108-95-2 EC No: 203-632-7 Annex I Index No: 604-001-00-2

Physical data

Appearance: colourless crystals with a characteristic odour Melting point: 40 - 42 C Boiling point: 182 C Specific gravity: 1.07 Vapour pressure: 0.35 mm Hg at 20 C Flash point: 79 C Explosion limits: 1.5 % - 8.6 % Autoignition temperature: 715 C

Stability

Stable. Substances to be avoided include strong oxidizing agents, strong bases, strong acids, alkalies, calcium hypochlorite. Flammable. May discolour in light.

Toxicology

This material is a systemic poison and constitutes a serious health hazard. The risks of using it in the laboratory must be fully assessed before work begins. Vesicant. Typical MEL 2 ppm; typical OEL 1 ppm. Acute poisoning by ingestion, inhalation or skin contact may lead to death. Phenol is readily absorbed through the skin. Highly toxic by inhalation. Corrosive - causes burns. Severe irritant.

Toxicity data ORL-HMN LDLO 140 mg kg-1 ORL-RAT LD50 317 mg kg-1 IPR-RAT LD50 127 mg kg-1 ORL-RBT LDLO 420 mg kg-1 SKN-RAT LD50 669 mg kg-1

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Transport information

(The meaning of any UN hazard codes which appear in this section is given here.) UN No 1671. Hazard class 6.1. Packing group II.

Personal protection

Safety glasses, gloves, good ventilation.

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Chapter 12: Cost estimation

Plant capacity: 11550 TPA

Cost of major equipments:SR.NO. Equipment Nos. Total price in lac

Rs.

1. Air Filter 1 0.502. Silica gel tower 1 0.753. Reactor 1 1 154. Reactor 2 1 155. Reactor 3 1 96. Dist. column 1 1 207. Dist. column 2 1 208. Dist. column 3 1 209. Flash vessel 1 110. Scrubber 1 1.511. Cooling water

condenser9 36

12. Brine condenser 2 1013. Cumene storage tank 1 3014. Cooling tower 5 7515. Brine plant 1 1516. Boiler 1 15017. D.M. water plant 1 3018. Phenol storage tank 1 3019. Heat exchanger

across reactor 1 & 21 12

20. Total cost 1 490.75

Fixed Capital Investment :1. Total purchased equipment cost = 1.25 x cost of major equipments

= 1.25 x 490.75 = 61343750 Rs.

2. Cost of installation =40% of (1)

= 2600000 Rs.

3. Insulation cost = 2% of (1) = 1300000 Rs.

4. Cost of instruments and control = 30% of (1) = 19500000 Rs.

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5. Cost of piping = 30% of (1) = 19500000 Rs.

6. Cost of electrical installation = 20% of (1) = 13000000 Rs.

7. Building including service & land = 20000000 Rs.

8. (i) Roads & walls = 8000000 Rs. (ii) Fences = (30% of (7)) = 902500 Rs. (iii) Parking area = 500000 Rs. 9. Account, construction & engineering = 20%of (1) = 13000000 Rs. 10. Fixed capital = Sum of (1) to (10) = 186702500 Rs.

11. Contingency = 10% of fixed cost = 186702500 Rs.

Total fixed cost = (10) + (11) = 205372750 Rs.

Manufacturing Cost: (A) Direct manufacturing cost:

(a) Raw materials i. Propylene 25 Rs./Kg ii. Air 00 Rs. /Kg

Capacity of plant: 7920 TPA

Total cost: Cumene 198000000 Rs.

(b) Operating Direct Supervision Electrical Labor

Sr. No. Designation Nos. Pay/month(Rs.)

1. Head of department 1 1000002 Production manager 3 750003. Asst.manager 6 400004. Shift engineer 4 250005. Shift operator 6 150006. Chemist 3 100007. Unskilled operator 8 80008. Maintenance

engineer3 15000

9. Boiler attendants 3 500010. Maintenance staff 3 5000

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11. Clerical staff 6 5000

Total Salary = 954000 Rs. /month = 11448000 Rs. /annum.

(1) Utilities 10% of raw materials = 19800000 Rs.(2) Maintenance & repairs 6% of fixed capital investment = 12300000

Rs.(3) Operating supplies 20% of maintenance cost = 2460000 Rs.(4) Laboratory charges 20% of operating supplies = 492000 Rs

Direct Manufacturing Cost = 244500000 Rs.

(B) Fixed Charges: 1. Depreciation:

a. 10% of fixed capital for machinery & equipments = 20540000 Rs.b. 3% of initial cost of building = 600000 Rs. Total sum of a & b = 21140000 Rs.

2. Local Taxes = 2% of fixed capital = 4107000 Rs.

3. Insurances = 1% of fixed capital = 2054000 Rs.

Fixed Manufacturing cost = sum of (1) to (3) = 27300000 Rs.

(C)Plant overhead : = 60% of operating labor, supervision & electrical labor = 6868800 Rs.

Total Manufacturing cost = (A)+(B)+(C) = 279000000 Rs.

General Expenses:

(a) Administration cost = 30% of labor= 5000000 Rs.

(b) Distribution cost = 5% of mfg. cost= 1365000 Rs.

(c) Research & development cost = 3% of mfg. cost

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= 819000 Rs.

Total general expenses = 7184000 Rs.

Total production cost = mfg cost + general expenses= 286184000 Rs.

Working capital investment: 25% of fixed capital investment = 46676000 Rs.

Total Capital investment = total fixed capital + working capital= 233378500 Rs.

Profit:

Product Quantity Quantity Price Total costPhenol

Acetone

11550 TPA

6924

11550000 Kg/A

6924000 Kg/A

42 Rs./kg

28 Rs./kg

485100000 Rs.

193900000 Rs.

Gross Profit = sales – Total production cost= (485100000 + 193900000) – 286184000= 392816000 Rs.

1. Income tax Calculation :-

Sales tax deduction = 2% of sales = 13580000 Rs.

Taxable income = Gross profit – Sales tax = 392800000 – 13580000 = 379220000 Rs.

Tax is considered to be 40%.

Income tax = 40% of (Gross profit – Sales tax) = 151688000 Rs.

Net Profit = 392800000 – 151688000 = 241100000 Rs.

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Profitability Analysis:

Annual investment after income tax = (net profit/ total capital investment) x 100 = (241100000/233378500) x100 = 103.308 %

Pay out period without interest = (depreciable fixed capital investment/(avg. profit + avg. depreciation/yr.)) = 186702500 / (241100000 + 21140000) = 8 months.

Break even analysis:

Total production cost = 286184000 Rs.Total Production = 11550 TPA

At Break even point, Total Sales = Total Production Cost X x 11550 x 70 x1000 = 286184000 X = 0.22 yr. = 3 months.(approx.)

In terms of tons, Phenol = 75600 & Acetone = 1887.84

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Chapter 13: Manufacturers of Phenol

Industries in India:-

- Herdillia Chemical Ltd. Mumbai, Maharashtra- Kinjal chemicals A’ bad Gujarat- Hindustan organic chemical ltd., Mumbai, Maharashtra- Nile chemicals, Mumbai, Maharashtra- Ultimate chemicals Pvt. Ltd. Mumbai, Maharashtra- Bombay lubricants oil company, Mumbai, Maharashtra- Lawrice labs, Mumbai, Maharashtra- Rocky chemicals, Ankleshwar- BL chemicals, Mumbai, Maharashtra

Industries in abroad:-

- Mitsui chemicals, Japan (World’s 1st manufacturer)- Kumho P & B chemicals, Korea (World’s 2nd manufacturer)- Arian chemie co. , Iran- A.K. Aromatics, Saudi Arabia- Christopher KIM, Canada- J.N. Global, South Korea- Beijing Jiyi chemical and industry co. ltd., China

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Chapter 13: Conclusion

From the above process I can conclude that the process which I have selected is the most appropriate and the most of the industry in India and abroad use the same for the manufacturing of phenol as Acetone which has got a great market demand is also obtained as co-product.

As phenol is also absorbed by skin on long exposure wear impervious protective clothing, including boots, gloves, lab coat, apron or coveralls, as appropriate, to prevent skin contact. Butyl rubber and neoprene are suitable materials for personal protective equipment. Use chemical safety goggles and/or full face shield where dusting or splashing of solutions is possible. Maintain eye wash fountain and quick-drench facilities in work area.

On the basis of material balance one can conclude that 1tph of cumene gives approx. 1.45 tph of phenol and 0.87 tph of acetone as product and co-product respectively.

Also, here the optimization is done at various places especially at stream 18 where the residue of the previous distillation column is used to heat the feed to the reactor-3 and inturn the residue is being cooled.

From the design point of view the thing which is clear is M.O.C. of th reactor which I have designed should be S.S. 304 as phenol is corrosive and also sulfuric acid is being added as a catalyst in the process. Also as there is a great density difference in that of sulfuric acid and phenol an agitator is very much compulsory.

Lastly, the cost estimation shows that the process is quite a profitable process with Payout period = 8 months Break Even point = 3 monthNet Profit = 24.11 crore.

Thus, on the basis of above mentioned points if one is interested in manufacturing the Phenol as a product then it’s a profitable venture.

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Chapter 14: References

HAND BOOK AND ENCYCLOPEDIA

1. Kirk-Othmer, Concise “Encyclopedia of Chemical Technology”, vol- A 19 Page – 299-308

2. Ullmans “ Encyclopedia of industrial chemistry”, vol-A11 Page – 385-3923. Perry J.H. “Chemical Engineering Hand book”, 6th Edition. Page-48-98

BOOKS

4. Dryden “Outlines of Chemical Technology”, 3rd edition, East West Press. Page. no – 504-520

5. B.I Bhatt and S.M. Vora “ Stoichiometry” 4th edition ,Tata McGraw hill publisher Page no-129-133

6. Coulsen Richard & R. K. Sinnot “Process Engineering and Design” Vol-67. S.B.Thakore & B.I. Bhatt “Process Engineering and Design, Tata McGraw hill

publisher Page. no- 720-725,219, 8. Edwin H Young & Cloyd E Brownell “Process Equipment Design”.Page.no- 246-

247.9. S.B.Thakore & D.A.Shah “Illustrated Process Equipment Design”,1st edition, Atul

Prakashan.Page.no-79-82,232-240, 70-71.10. Klaus Timmerhaus “Process Plant design & economics for chemical engineer”,3rd

edition , Tata McGraw Hill Publisher.Page-No-95-102,150-20911. C S Rao “Environmental pollution control engg.”, New Age International (P) Ltd.

OTHER

12. http://www.indianindustry.com/solvents/8311.html

13. http://www.researchandmarkets.com/reportinfo.asp?report_id=344110

14. http://www.jtbaker.com/msds/englishhtml/p1949.htm

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