Ethylene Glycol Chemical Engineering Final Year Project

CHAPTER I INTRODUCTION 1.1 HISTORY: Ethylene Glycol (1, 2 – ethanediol), HOCH2CH2OH usually called glycol is the simplest Diol. Diethylene glycol and Triethylene glycol are Oligomers of Mono ethylene glycol. Ethylene glycol was first prepared by Wurtz in 1859; treatment of 1,2 dibromoethane with silver acetate yielding ethylene glycol diacetate via saponification with potassium hydroxide and in 1860 from the hydration of ethylene oxide. There to have been no commercial manufacture or application of ethylene glycol prior to World War-I when it was synthesized from ethylene dichloride in Germany and used as substituted for glycerol in the explosives industry and was first used industrially in place of glycerol during World War I as an intermediate for explosives (ethylene glycol dinitrate) but has since developed into a major industrial product. The use of ethylene glycol as an antifreeze for water in automobile cooling systems was patented in the United States in 1917, but this commercial application did not start until the late 1920s. The first inhibited glycol antifreeze was put on the market in 1930 by National Carbon Co. (Now Union Carbide Corp.) under the brand name “prestone”. Carbide continued to be essentially the sole supplier until the late 1930s. In 1940 DuPont started up an ethylene glycol plant in Belle, West Virginia based on its new formaldehyde methanol process. In 1937 Carbide started up the first plant based on Lefort’s process for vapor phase oxidation of ethylene oxide. The worldwide capacity for production of Ethylene Glycol via hydrolysis of ethylene oxide is estimated to be 7×106 ton/annum [1, 2]. 1.2 CHEMISTRY: Compound contains more than one –oly group is called Polyhydric Alcohol (Dihydric alcohol) or polyols (Diols). Diols are commonly known as Glycols, since they have a sweet taste (Greek, glycys= Sweet).

1

Dihydric alcohols because compounds contain two –OH groups on one carbon are seldom encountered. This is because they are unstable and undergo spontaneous decomposition to give corresponding carbonyl compound and water.

Figure-1[10] According to IUPAC system of nomenclature, IUPAC name of glycol is obtained by adding suffix Diol to the name of parent alkanes. HO OH H--C---C--H H H

H

H

HO--C---C--OH H

H

H H H--C---C--H HO

OH

1, 2 Glycol

1, 3 Glycol

1, 4 Glycol

(α- Glycol)

(β- Glycol)

(γ- Glycol)

Glycols are Diols. Compounds containing two hydroxyl groups attached to separate carbon in an aliphatic chain. Although glycols may contain heteroatom can be represented by the general formula C2nH4nOn-1(OH) 2. [3, 4] Formula

Common name

IUPAC name

CH2OHCH2OH

Ethylene Glycol

Ethane-1, 2-Diol

1.3 USES: The following is a summary of the major uses of ethylene glycol: 1.3.1 Antifreeze  A major use of ethylene glycol is as antifreeze for internal combustion engines. Solutions containing ethylene glycol have excellent heat transfer properties and higher boiling points than pure water. Accordingly, there is an 2

increasing tendency to use glycol solutions as a year-round coolant. Ethylene glycol solutions are also used as industrial heat transfer agents.  Mixtures of ethylene glycol and propylene glycol are used for defrosting and de-icing aircraft and preventing the formation of frost and ice on wings and fuselages of aircraft while on the ground. Ethylene glycol-based formulations are also used to de-ice airport runways and taxiways as de-icing agent.  Asphalt-emulsion paints are protected by the addition of ethylene glycol against freezing, which would break the emulsion. Carbon dioxide pressurized fire extinguishers and sprinkler systems often contain ethylene glycol to prevent freezing. 1.3.2 Explosives 

Ordinary dynamite will freeze at low temperatures and cannot then be

detonated. Ethylene glycol dinitrate, which is an explosive itself, is mixed with dynamite to depress its freezing point and make it safer to handle in cold weather. 

Mixtures of glycerol and ethylene glycol are nitrated in the presence of

sulfuric acid to form solutions of nitroglycerin in ethylene glycol dinitrate, which are added to dynamite in amounts ranging from 25 to 50%. 1.3.3 Polyester Fibers  The use of ethylene glycol for fibers is becoming the most important consumer of glycol worldwide. These fibers, marketed commercially under various trade names like Dacron, Fortel, Kodel, Terylene etc are made by the polymerization of ethylene glycol with BisHydroxyEthyl Terephthalate (BHET).  These Polyester fibers are used for recyclable bottles. 1.3.4 Resins  Polyester resins made from maleic and phthalic anhydrides, ethylene glycol, and vinyl-type monomers have important applications in the low-pressure lamination of glass fibers, asbestos, cloth and paper.  Polyester-fiberglass laminates are used in the manufacture of furniture, automobile bodies, boat hulls, suitcases and aircraft parts. Alkyd-type resins are produced by the reaction of ethylene glycol with a dibasic acid such as ophthalic, maleic or fumaric acid. These resins are used to modify synthetic 3

rubbers, in adhesives, and for other applications.  Alkyds made from ethylene glycol and phthalic anhydride is used with similar resins based on other polyhydric alcohols, such as glycerol or pentaerythritol in the manufacture of surface coatings. Resin esters made with ethylene glycol are used as plasticizers in adhesives, lacquers and enamels. 1.3.5 Hydraulic Fluids  Ethylene glycol is used in hydraulic, brake and shock absorber fluids to help dissolve inhibitors, prevent swelling of rubber, and inhibit foam formation.  Hydro lubes, which are water-based mixtures of polyalkylene glycols and presses and die casting machines, and in airplane hydraulic systems because of their relatively low viscosity at high pressure. An added advantage of primary importance is that these hydro lubes are inflammable. 1.3.6 Capacitors  Ethylene glycol is used as a solvent and suspending medium for ammonium perborate, which is the conductor in almost all electrolytic capacitors.  Ethylene glycol, which is of high purity (iron and chloride free), is used because it has a low vapor pressure, is non-corrosive to aluminum and has excellent electrical properties. 1.3.7 Other uses  Ethylene glycol is used to stabilize water dispersions of urea-formaldehyde and melamine-formaldehyde from gel formation and viscosity changes. It is used as humectants (moisture retaining agent) for textile fibers, paper, leather and adhesives and helps make the products softer, more pliable and durable.  An important use for ethylene glycol is as the intermediate for the manufacture of Glyoxal, the corresponding dialdehyde. Glyoxal is used to treat polyester fabrics to make them “permanent press.”  Ethylene glycol derivatives mainly ether and ester are used as absorption fluids, Diethylene Glycol is used as a softener (Cork, adhesives, and paper) dye additive (Printing and stamping), deicing agent for runway & air craft, drying agent for gases (natural gas).  Triethylene glycol is used for same purpose as Diethylene glycol. 4

 Poly (ethylene glycol) with varying molecular masses and numerous uses in Pharmaceutical industry (Ointments, Liquids and tabletting) and cosmetic industry (cream lotion, pastes, cosmetic sticks, soaps). They are also used in textile industry (Cleaning and dyeing agents), in Rubber industry (lubricating & Mold parting agents), in ceramics (bonding agents and plasticizers).[3,4]

CHAPTER II PROPERTIES 2.1 PHYSICAL PROPERTIES:  Monoethylene glycol and its lower polyglycols are clear, odorless, colorless, syrupy liquid with a sweet taste.  It is a hygroscopic liquid completely miscible with many polar solvents, such as water, alcohols, glycol ethers, and acetone.  Its solubility is low however in non polar solvents, such as benzene, toluene, dichloroethane, and chloroform. It is miscible in ethanol in all proportion but insoluble in ether, completely miscible with many polar solvents, water, alcohols, glycol ethers and acetone. Its solubility is low, however in nonpolar solvents, such as benzene, toluene, dichloromethane and chloroform.  It is a toxic as methyl alcohol when taken orally. Ethylene glycol is difficult to crystallize, when cooled; it forms a highly



viscous, super-cooled mass that finally solidifies to produce a glasslike substance. The widespread use of ethylene glycol as an antifreeze is based on its ability



to lower freezing point when mixed with water. [3, 4] Table 2.1 Physical Properties. [1, 2] Sr.

Physical Properties

no. 1.

Molecular formula

C2H6O2

2.

Molecular weight

62

3.

Specific gravity at 20/20oC

1.1135

4.

Boiling point oC at 101.3 KPa

197.60

5

5.

Freezing point oC

-13

6.

Heat of vaporization at 101.3 KPa; KJ/mol

52.24

7.

Heat of combustion (25oC) MJ/mol

19.07

8.

Critical Temp. oC

372

9.

Critical pressure, KPa

6513.73

10.

Critical volume, L/mol

0.1861

11.

Refractive index, ŋ

1.4318

12.

Cubic expansion coefficient at 20 oC, K-1

0.62 × 10-3

13.

Viscosity at 20oC; mPa S

19.83

14.

Liquid density (20oC) gm/cm3

1.1135

15.

Flash point, oC

111

16.

Auto-ignition temp in air oC

410

17.

Flammability limits in air; vol% Upper

53

Lower

3.2

2.2 CHEMICAL PROPERTIES: Ethylene Glycol contains two primaries –OH groups. Its chemical reactions are therefore, those of primary alcohols twice over. Generally, one –OH group is attacked completely before other reacts. 2.1.1 Dehydration  With Zinc chloride, it gives Acetaldehyde HOCH2CH2OH

CH3CHO + H2O

(Ethylene Glycol)

(Acetaldehydes)

 On heating alone at 500 oC, it gives Ethylene oxide.  With H2SO4 it gives dioxane which is important industrial solvent. 2.1.2 Oxidation Ethylene glycol is easily oxidized to form a number of aldehydes and carboxylic acids by oxygen, Nitric acid and other oxidizing agents. 6

The typical products derived from alcoholic functions are Glycolaldehyde (HOCH2CHO), Glycolic acid (HOCH2COOH), Glyoxylic acid (HCO-COOH), Oxalic Acid (HOOCCOOH), formaldehyde & formic acid.  With HNO3 oxidation it yields nos. of substance as one or both primary –OH groups may be oxidized to aldehydes and these carboxylic groups. HNO3 HOCH2CH2OH

[O] HOCH2CHO

(Ethylene Glycol)

[O] HOCH2CH2COOH

(Glycol aldehydes)

CHOCOOH

(Glycolic acid)

(Glyoxylic acid) [O] HOOC-COOH (Oxylic acid) [O]

HNO3 HOCH2CH2OH

[O] HOCH2CHO

(Ethylene Glycol)

[O] CHOCHO

(Glycol aldehydes)

CHOCOOH

(Glyoxal)

(Glyoxylic acid)

2.1.3 Other reactions The hydroxyl groups on glycols undergo the usual alcohol chemistry giving a wide variety of possible derivatives. Hydroxyls can be converted to aldehydes, alkyl halides, amides, amines, azides, carboxylic acids, ethers, mercaptans, nitrate esters, nitriles, nitrite esters, organic esters, peroxides, phosphate esters, and sulfate esters.  Reaction with sodium at 50 oC to form monoalkoxide and dialkoxide when temperature is raised. Na at 50 oC

Na at 160 oC

HOCH2CH2OH

HOCH2CH2ONa

NaOCH2CH2ONa

(Ethylene Glycol)

(Mono Alkoxide)

(Di Alkoxide)

 Reaction with Phosphorus pentahalide (PCl5) it first gives Ethylene chlorohydrins and then 1, 2 dichloroethane. PBr5 reacts in same way. PCl5 HOCH2CH2OH

PCl5 HOCH2CH2Cl

(Ethylene Glycol)

(Ethylene chlorohydrins)

ClCH2CH2Cl (1, 2-Dicholorochlorohydrins)

 With Phosphorus trihalide (PBr3) to form responding dihalide PBr3

PBr3 7

HOCH2CH2OH

HOCH2CH2Br

(Ethylene Glycol)

BrCH2CH2Br

(Ethylene Bromohydrins)

(1, 2-Dibromohydrins)

 With HCl in two step reaction, form ethylene chlorohydrins at 160oC and second forms 1, 2 dichloroethane at 200oC. 160 oC HOCH2CH2OH

200 oC HOCH2CH2Cl

(Ethylene Glycol)

ClCH2CH2Cl

(Ethylene chlorohydrins)

(1, 2-Dicholorochlorohydrins)

 The largest commercial use of ethylene glycol is its reaction with dicarboxylic acids (1) to form linear polyesters. Poly (Ethylene Terephthalate) (PET) (2) is produced by esterification of teraphthalic acid to form BisHydroxyEthyl Terephthalate (BHET) (3). BHET polymerizes in a transesterification reaction catalyzed by antimony oxide to form PET. 2HOCH2CH2OH + HOOC

COOH

+

(2)

(1)

Sb2O3

H*

COOCH2CH2OH

HOCH2CH2OOC

OOC

*H

COOCH2CH2

+ HOCH2CH2OH

n (3)

Ethylene glycol esterification of BHET is driven to completion by heating and removal of the water formed. PET is also formed using the same chemistry starting with dimethyl Terephthalate and ethylene glycol to form BHET also using an antimony oxide catalyst.  Ethylene glycol also produces 1, 4-dioxane by acid-catalyzed dehydration to Diethylene glycol followed by cyclization. Cleavage of Triethylene and higher glycols with strong acids also produces 1, 4-dioxane by catalyzed ether hydrolysis with subsequent cyclization of the Diethylene of the Diethylene glycol fragment. Diethylene glycol condenses with primary amines of form

8

cyclic structures, e.g., methylamine reacts with Diethylene glycol to produce N-methylmorpholine.

HOCH2CH2OCH2CH2OH

+

CH3NH2

O

N

CH3

+ 2H2O

(6)

 Ketones and aldehydes react with ethylene glycol under acidic conditions to Form 1, 3-dioxolanes cyclic ketals and acetals.

HOCH2CH2OH

+ RCOR+

+

H

O

R'

+

R

H2O

(7)

O

 Ethylene glycol reacts with ethylene oxide to form di, tri, tetra and polyethylene glycols.  Ethylene glycols is stable compound, but special care is required when ethylene glycol is heated at a higher temperature in presence of NaOH, which is exothermic reaction at temperature above 250 oC of evolution of H2 (-90 to -160 KJ/Kg).[1,3,4]

9

CHAPTER III LITERATURE SURVEY The literature survey has been done with an aim to obtain information concerning Ethylene Glycol and its production from number of sources. Such information sources include chemical abstracts, periodicals and books on chemical technology, handbooks, encyclopedias and internet websites. The literature survey yielded a lot of information on Ethylene Glycol. A brief review of information obtained from the literature survey is presented hereafter. During the project many Journals, Manuals and Hand book have been sited The manufacturing process have been taken from “Chemical Engineering Journal 107(2005), 199-204.” The selectivity and other process parameters have been taken from “Chemical Engineering Journal 107(2005), 199-204.” The demand growths, Major producer in India & World have been taken from Internet. 3.1 DERIVATIVES OF MONO ETHYLENE GLYCOL: In addition to Oligomers ethylene glycol dervative classes include monoethers, diethers, esters, acetals, and ketals as well as numerous other organic and organometalic molecules. These derivatives can be of ethylene glycol, Diethylene glycol, or higher glycols and are commonly made with either the parent glycol or with sequential addition of ethylene oxide to a glycol alcohol, or carboxylic acid forming the required number of ethylene glycol submits. 3.1.1 Diethylene Glycol: Physical properties of Diethylene glycol are listed in Table. Diethylene glycol is similar in many respects to ethylene glycol, but contains an ether group. It was originally synthesized at about the same time by both Lourenco and Wurtz in 1859, and was first marketed, by Union Carbide in 1928. It is a co product (9 - 10%) of ethylene glycol produced by ethylene oxide hydrolysis. It can be made directly by the reaction of ethylene glycol with ethylene oxide, but this route is rarely used because more than an adequate supply is available from the hydrolysis reaction. Manufacture of unsaturated polyester resins and polyols for polyurethanes consumes 45% of the Diethylene glycol. Approximately 14% is blended into antifreeze. Triethylene glycol from the ethylene oxide hydrolysis does not meet market 10

requirements, which leads to 12% of the Diethylene glycol being converted with ethylene oxide to meet this market need. About 10% of Diethylene glycol is converted to morpholine. Another significant use is natural gas dehydration, which uses 6%. The remaining 13% is used in such applications as plasticizers for paper, fiber finishes, and compatiblizers for dye and printing ink components, latex paint, antifreeze, and lubricants in a number of applications. 3.1.2 Triethylene Glycol: Triethylene glycol is a colorless, water-soluble liquid with chemical properties essentially identical to those of Diethylene glycol. It is a co product of ethylene glycol produced via ethylene oxide hydrolysis. Significant commercial quantities are also produced directly by the reaction of ethylene oxide with the lower glycols. Triethylene glycol is an efficient hygroscopicity agent with low volatility, and about 45% is used as a liquid drying agent for natural gas. Its use in small packaged plants located at the gas wellhead eliminates the need for line heaters in field gathering systems as a solvent (11 %) Triethylene glycol is used in resin impregnants and other additives, steam-set printing inks, aromatic and paraffinic hydrocarbon separations, cleaning compounds, and cleaning poly (ethylene Terephthalate) production equipment. The freezing point depression property of Triethylene glycol is the basis for its use in heat-transfer fluids. Approximately 13% Triethylene glycol is used in some form as a vinyl plasticizer. Triethylene glycol esters are important plasticizers for poly (vinyl butyral) resins, Nitrocellulose lacquers, vinyl and poly (vinyl chloride) resins, poly (vinyl acetate) and synthetic rubber compounds and cellulose esters. The fatty acid derivatives of Triethylene glycol are used as emulsifiers, emulsifiers, and lubricants. Polyesters derived from Triethylene glycol are useful as low pressure laminates for glass fibers, asbestos, cloth, or paper. Triethylene glycol is used in the manufacture of alkyd resins used as laminating agents and adhesives. 3.1.3 Tetra ethylene Glycol: Tetra ethylene glycol has properties similar to Diethylene and Triethylene glycols and may be used preferentially in applications requiring a higher boiling point, higher molecular weight, or lower hygroscopicity.

11

Tetra ethylene glycol is miscible with water and many organic solvents. It is a humectants that, although less hygroscopic than the lower members of the glycol series, may find limited application in the dehydration of natural gases. Other possibilities are in moisturizing and plasticizing cork, adhesives, and other substances. Tetra ethylene glycol may be used directly as a plasticizer or modified by esterification with fatty acids to produce plasticizers. Tetra ethylene glycol is used directly to plasticize separation membranes, such as silicone rubber, poly (Vinyl acetate), and cellulose triacetate. Ceramic materials utilize tetra- ethylene glycol as plasticizing agents in resistant refractory plastics and molded ceramics. It is also employed to improve the physical properties of cyanoacrylate and polyacrylonitrile adhesives, and is chemically modified to form Polyisocyanate, polymethacrylate, and to contain silicone compounds used for adhesives. Tetra ethylene glycol has found application in the separation of aromatic hydrocarbons from nonromantic hydrocarbons (BTX extraction). In general, the critical solution temperature of a binary system, consisting of a given alkyl-substituted aromatic hydrocarbon and tetra ethylene glycol, is lower than the critical solution temperature of the same hydrocarbon with Triethylene glycol and is considerably lower than the critical solution temperature of the same hydrocarbon with Diethylene glycol. Hence, at a given temperature, tetra ethylene glycol tends to exact the higher alkyl benzenes at a greater capacity than a lower polyglycols. 3.2 STORAGE AND TRANSPORTATION: Pure anhydrous ethylene glycol is not aggressive toward most metals and plastics. Since ethylene glycol also has a low vapor pressure and is non caustic. It can be handled with out any problems: it is transported in railroad tank cars, tank trucks, and tank ships. Tanks are usually made of steel: high grade materials are only required for special quality requirements. Nitrogen blanketing can protect ethylene glycol against oxidation. At ambient temperature, aluminum is resistant to pure glycol. Corrosion occurs, however, above 100oC and hydrogen is evolved. Water air and acid producing impurities (aldehydes) accelerate this reaction. Great care should be taken when phenolic resins are involved, since they are not resistance to ethylene glycol.

12

3.3 SHIPPING DATA FOR ETHYLENE GLYCOL: •

Weight per Gallon at 20°C

9.29 lb

•

Coefficient of Expansion at 55°C

0.00065

•

Flash Point, Tag Closed Cup

260°F

Net Contents and Type of Container •

1–Gallon Tin Can 9.0 lb

•

5–Gallon DOT 17E, Pail 47 lb

•

55–Gallon DOT 17E, Drum 519 lb

3.4 ENVIRONMENTAL PROTECTION AND ECOLOGY: Ethylene glycol is readily biodegradable, thus disposal of waste water containing this compound can proceed without major problems. The high LC 50value of over 10000 mg/lit account for its low water toxicity. 3.5 PRODUCT SAFETY: When considering the use of ethylene glycol in any particular application, review and understand our current Material Safety Data Sheet

for

the

necessary

safety

and

environmental

health

information. Before handling any products you should obtain the available product safety information from the suppliers of those products and take the necessary steps to comply with all precautions regarding the use of ethylene glycol. No chemical should be used as or in a food, drug, medical device, or cosmetic, or in a product process in which it may come in contact with a food, drug, medical device, or cosmetic until the user has determined the suitability of the use. Because use conditions and applicable laws may differ from one location to another and may change with time, Customer is responsible for determining whether products and the information are appropriate for Customer’s use [5, 6]

13

CHAPTER IV MARKET SURVEY 4.1 ECONOMIC ASPECTS: Ethylene glycol is one of the major products of the chemical industry. Its economic importance is founded on its two major commercial uses as antifreeze and for fiber production. Since Ethylene glycol is currently produced exclusively from ethylene oxide production plant are always located close to plant that produce ethylene oxide. The proportion of ethylene oxide that is converted to Ethylene glycol depends on local condition, such as market situation and transport facilities. About 60% of total world production is converted to ethylene glycol. About 50% of the ethylene glycol that is used as antifreeze. Another 40% is used in fiber industry. Consequently the ethylene glycol demand is closely connected to the development of these two sectors In view of the increasing price of crude oil, alternative production method based on synthesis gas is likely to become more important and increasing competitive. 4.2 LEADING PRODUCERS IN WORLD:  BASF, Geismer, La. (America).  DOW, Plaquemine, La .(America)  OXYPETROCHEMICALS, Bayport, Tex .(America)  PD Glycol ,Beaumont, Tex. (America)  SHELL, Geismer,La. (America)  TEXACO ,Port Neches, Tex.(America)  UNION CARBIDE, Taft,La.(America)  BP Chemicals, Belgium, (West Europe).  IMPERIAL Chemicals Ind. United Kingdom, (West Europe)  BPC (NAPTHACHIMIE),France , (West Europe)  STATE COMPLEXES ,USSR, (West Europe)

14

 PAZINKA, Yugoslavia, (West Europe)  EASTERN PETROCHEMICAL CO. Saudi Arabia, (Middle East)  National Organic Chemical, India, (Asia).  Mitsubishis Petrochemicals, (Japan) 4.3 LEADING PRODUCER IN INDIA:  India Glycol, Uttaranchal (North India).  Reliance Industries Ltd. Hazira (Gujarat).  Indian Petrochemical Corporation Ltd, Baroda (Gujarat).  NOCIL, Thane.  SM Dye chem. Pune. 4.4 MEG PRICE TREND: Table 4.1 MEG Price Trend Sr. No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

Year 2004 2005

November

Month November December January February March April May June July August September October 1st week 2nd week

Price(US$/MT) 1095 988 1045 1095 1095 971 734 736 808 836 883 883 830 822

4.5 DEMAND SUPPLY BALANCE (IN KT): Table 4.2 Demand supply balance (In KT) MEG Capacity Production Imports Exports Demand Demand Growth %

2002 590 548 11 8 551

2003 615 647 64 29 682 24%

2004 654 691 106 104 750 10%

2005 830 833 103 133 803 7%

2006 830 830 90 60 860 7%

4.6 QUALITY SPECIFICATION: Since ethylene glycol is produce in relatively high purity difference in quality are not accepted. The directly synthesized product meets high quality demands (fiber grade). 15

The ethylene glycol produce in the wash water that is use during ethylene oxide production is normally of a somewhat inferior quality (antifreeze grade). The quality specifications for mono ethylene glycol are compiling in table-2. [5, 6] Table 4.3 Quality Specification OF Ethylene Glycol DESCRIPTION

FIBER GRADE

INDUSTRIAL GRADE

Color, Pt-Co, max

5

10

Suspended matter

Substantially free

Substantially free

Diethylene glycol, wt.% max

0.08

0.6

Acidity, as acetic acid, wt%

0.005

0.02

max Ash, wt% max

0.005

0.005

Water, wt% max

0.08

0.3

Iron, ppm wt max Chlorides, ppm wt max Distillation range, ASTM at

0.07

0.05

760mm Hg: IBP, °C min DP, °C max Odor UV transmittance, % min at: 220 nm 250 nm 275 nm 350 nm Specific gravity, 20/20°C Water solubility, 25°C

196 200 Practically none 70 90 98 1.1151-1.1156 Completely miscible

196 199

70° 90° 95° 99° 1.1151-1.1156

CHAPTER V PROCESS SELECTION AND DESCRIPTION 5.1 MANUFACTURING PROCESSES: Up to the end of 1981, only two processes for manufacturing ethylene glycol have been commercialized. The first, the hydration of ethylene oxide, is by far the most important, and from 1968 through 1981 has been the basis for all of the ethylene glycol production. 16

Manufacturing process involves laboratory methods and industrial methods. 5.1.1 Laboratory methods: [3, 4]  By passing Ethylene in to cold dilute Alkaline permanganate solution i.e. Oxidation of Ethylene to Glycol 

By hydrolysis of Ethylene Bromide by boiling under reflux with aqueous sodium carbonate solution. This reaction mixture is refluxed till an oily globule of ethylene bromide disappears. The resulting solution is evaporated on a water bath and semi solid residue is extracted with ether-alcohol mixture. Glycol is recovered from solution by distillation. The best yield of glycol (8384%) can be obtained by heating ethylene bromide with potassium acetate in Glacial acetic acid.

 Ethylene glycol can be produced by an electrohydrodimerization of formaldehyde. 

An early source of glycols was from hydrogenation of sugars obtained from

formaldehyde condensation. Selectivity to ethylene glycol was low with a number of other glycols and polyols produced. Biomass continues to be evaluated as a feedstock for glycol production. 5.1.2 Industrial methods: [1, 2, 7, 8]  The production of ethylene glycol by the hydration of ethylene oxide is simple, and can be summarized as follows: ethylene oxide reacts with water to form glycol, and then further reacts with ethylene glycol and higher homologues in a series of consecutive reactions as shown in the following equations.

17

H2C O

O O H H Ethylene Glycol

Ethylene Oxide CH2

H2C

+ H2C

O

CH2

CH2

H2C

+ H2C

O

O H

O H

CH2 O H

CH2 O H

O H H2C

CH2

Diethylene Glycol CH2

O

CH2

O

CH2

H2C

O H

O H

CH2

H2C

+ H2O

CH2

CH2

CH2

O

CH2

O

CH2

Triethylene Glycol

CH2 O H

Ethylene oxide hydrolysis proceeds with either acid or base catalysis or uncatalyzed in neutral medium. Acid-catalyzed hydrolysis activates the ethylene oxide by protonation for the reaction with water. Base-catalyzed hydrolysis results in considerably lower selectivity to ethylene glycol. The yield of higher glycol products is substantially increased since anions of the first reaction products effectively compete with hydroxide ion for ethylene oxide. Neutral hydrolysis (pH 6-10), conducted in the presence of a large excess of water at high temperatures and pressures, increases the selectivity of ethylene glycol to 89-91%. In all these ethylene oxide hydrolysis processes the principal byproduct is Diethylene glycol. The higher glycols, i.e., Triethylene and Triethylene glycols, account for the remainder. Although catalytic hydration of ethylene oxide to maximize ethylene glycol production has been studied by a number of companies with numerous materials patented as catalysts, there has been no reported industrial manufacture of ethylene glycol via catalytic ethylene oxide hydrolysis. Studied catalyst include sulfonic acids, carboxylic acids and salts, cation-exchange resins, acidic zeolites, halides, anionexchange resins, metals, metal oxide and metal salts. Carbon dioxide as a co catalyst with many of the same materials has also received extensive study. 18

Ethylene glycol was commercially produced in the United States from



ethylene chlorohydrins which was manufactured from ethylene and hypochlorous acid. Chlorohydrins can be converted directly to ethylene glycol by hydrolysis with a base, generally caustic or caustic/bicarbonate mix. An alternative production method is converting chlorohydrins to ethylene oxide with subsequent hydrolysis. CH2

CH2

+

HOCH2CH2Cl

+ HOCl

NaOH

HOCH2CH2Cl HOCH2CH2OH

(8)

+

NaCl

(9)

+

NaCl

(10)

O HOCH2CH2Cl

+

Ca(OH)2

CH2

CH2

O CH2

CH2

+ H2O

HOCH2CH2OH

(11)

Du Pont commercially produced ethylene glycol from carbon monoxide,



methanol, hydrogen, and formaldehyde until 1968 at Belle, West Virginia. The process consisted of the reaction of formaldehyde, water, and carbon monoxide with an acid catalyst to form glycolic acid. The acid is esterified with methanol to produce methyl glycolate. Subsequent reduction with hydrogen over a chromate catalyst yields ethylene glycol and methanol. Methanol and formaldehyde were manufactured on site from syngas.

CO

+

CH2O

HOOCCH2OH

+

+

CH3OOCCH2OH

H2 O

H+

CH3OH

+

H2

HOOCCH2OH CH3OOCCH2OH

Cr2O3

HOCH2CH2OH

+ + +

NaCl

(12)

H2O

(13)

CH3OH

(14)

Coal was the original feedstock for syngas at Belle; thus ethylene glycol was commercially manufactured from coal at one time. Ethylene glycol manufacture from syngas continues to be pursued by a number of researchers. 

Ethylene glycol can be produced from acetoxylation of ethylene. Acetic acid,

oxygen, and ethylene react with a catalyst to form the glycol mono and diacetate. Catalysts can be based on palladium, selenium, tellurium, or thallium. The esters are 19

hydrolyzed to ethylene glycol and acetic acid. The net reaction is ethylene plus water plus oxygen to give ethylene glycol. This technology has several issues which have limited its commercial use. CH3COOH

+ CH2

CH2

+

O2

Te2Br2

CH3COOCH2CH2OH + CH3COOCH2CH2OOCCH3

CH3COOCH2OOCCH3

3 H2O

2 HOCH2CH2OH + 3 CH3COOH

(15) (16)

The catalysts and acetic are highly corrosive, requiring expensive construction materials. Trace amounts of ethylene glycol mono-and diacetates are difficult to separate from ethylene glycol limiting the glycol’s value for polyester manufacturing. This technology (Halcon license) was practiced by Oxirane in 1978 and j1979 but was discontinued due to corrosion problems. Ethylene glycol can be manufactured by the transesterification of ethylene



carbonate. A process based on the reaction of ethylene carbonate with methanol to give dimethyl carbonate and ethylene glycol is described in a Texaco patent; a general description of the chemistry has also been published. O

+

C O

2 CH3OH

Zr2Cl4

HOCH2CH2OH

+

CO(CH3O)2

(18)

O

Selectivity to ethylene glycol are excellent with little Diethylene glycol or higher glycols produced. A wide range of catalysts may be employed including ion exchange resins, zirconium and titanium compounds, tin compounds, phosphines, acids and bases. The process produces a large quantity of dimethyl carbonate which would require a commercial outlet.  Oxalic acid produced from syngas can be esterified and reduced with hydrogen to form ethylene glycol with recovery of the esterification alcohol. Hydrogenation requires a copper catalyst giving 100% conversion with selectivity to ethylene glycol of 95%. HOOCCOOH ROOCCOOR

+

2 ROH

+

4 H2

ROOCCOOR Cu

+

HOCH2CH2OH 20

+

2 H2O 2 ROH

(20) (21)



The Teijin process, which has not been commercialized to date, produces ethylene glycol by the reaction of ethylene with thallium salts in the presence of water and chloride or bromide ions. The major by-product in the reaction is acetaldehyde. A redox metal compound (such as copper) oxidizable with molecular oxygen is added to the reaction medium to permit the regeneration of the thallium salt.

 The DuPont process, based on feeds derived from synthesis gas (CO and formaldehyde), became economically obsolete because of low-priced ethylene. With the high price of oil and natural gas, there has been increasing interest in coal gasification to produce fuel and also synthesis gas for petrochemical manufacture. In 1976, Union Carbide announced that a process for the production of ethylene glycol from synthesis gas was being developed for commercialization in the early 1980.The proposed reaction was based on using a rhodium-based catalyst in tetrahydrofuran solvent at 190-230°C and high pressure (3400 atm). The equi molar mixture of CO and H2 would be converted mainly to ethylene glycol and by-product glycerol and propylene oxide. Methanol, methyl formate, and water would also be produced.[10] 5.2. PROCESS SELECTION: The process selection is based on different advantages and parameters of the industrial methods. 5.2.1 Comparison of different Processes: Hydration of ethylene oxide is an industrial approach to glycols in general, and ethylene glycol in particular. Ethylene glycol is one of the major large-scale products of industrial organic synthesis, with the world annual production of about 15.3 million t/yr in 2000. Hydration of ethylene oxide proceeds on a serial-to-parallel route with the formation of homologues of glycol:

21

Table 5.1 Comparison of different Processes SR. NO 1.

PROCESSES Hydrolysis

PARAMETER

of 1) Non- catalytic

Ethylene Oxide

Yield

:

CATALYST

DVANTAGES Use large excess water

1)Non

98% Catalytic

to increase the yield

Selectivity: 98% Temp:105oC Pressure

ADVANTAGES/DISA

which 2)

leads to high

Catalytic: energy consumption

: Sulfonic acids, 1) Use less excess water

1.5MPa

Carboxylic

2) catalytic:

acids and salts, energy consumption

Yield

:

which

95% Ion-exchange

2)

leads

High

to yield

low &

Selectivity: 90% resins, Acidic selectivity

2.

4.

zeolites,

Pressure :

halides, Metal temp & pressure

1-30 bar

oxide

Ethylene Glycol Yield :50% from

3.

Temp:200oC

3) permit use of low and 4) Acid catalyst makes

Metal salts.

the

Non Catalytic

highly corrosive.  very low yield &

Ethylene Selectivity: 75%

chlorohydrins Ethylene glycol Yield : 90-95% from

Temp: 200oC

CO,H2,CH3OH

Pressure:

&

100atm

 very costly  High pressure

Cromate Catalyst

process  Discontinued now a day

Alkali

from

or ammonium

carbonate

Temp:180oC

 Low selectivity halide  Give high yield and

salt.

Pressure:13bar 5.

Transesterificati on of ethylene

Low yield

solution

selectivity

Formaldehyde Ethylene glycol Yield :98% ethylene Selectivity: 95%

reaction

selectivity  Utility saving  Extra

Zirconium titanium 22

cost &  Produced amount

purification large of

6.

carbonate.

compound.

Esterification of Yield : 70%

Copper

 High conversion but

Oxalic acid and Selectivity: 90%

catalyst

catalyst removal is

Reduction with 7.

byproducts

very difficult.

H2 Direct one stage Selectivity: 65%

Rhodium

synthesis

catalyst

increase this process

Ethylene glycol 190-230 C

(Homogeneous

will become more

from syn gas

catalyst route.)

economical.

of Temp: o

Pressure: 3400atm

 As

crude

prices

 Use of very high pressure  Not

prove

to

be

indirect route may be viable or not.  Catalyst

is

sensitive 8.

Hydrolysis

of Yield : 90%

Pd complexes

glycol diacetate. Selectivity: 95%

very and

expensive.  Very low conversion

pdcl2+NaNO3

Temp: 160oC Pressure: 2.4MPa H2O+C2H4O

Ko

C2H4O + HOCH2CH2OH

HOCH2CH2OH

Ki

HO (CH2CH2O)2 H

----------------> (1) ----------------> (2)

Where k0,and k1 are the rate constants. Now all ethylene and propylene glycols is produced in industry by a non catalyzed reaction. Product distribution in reaction (1) is regulated by the oxide/water ratio in the initial reaction mixture. The distribution factor b = k1/k0 for a non catalyzed reaction of ethylene oxide with water is in the range of 1.9–2.8. For this reason large 23

excess of water (up to 20 molar equiv.) is applied to increase the monoglycol yield on the industrial scale. This results in a considerable power cost at the final product isolation stage from dilute aqueous solutions. i.e. energy consumption for the distillation of large amount of excess water is high. Also the selectivity of ethylene oxide hydrolysis is low i.e. 10% is converted to Diethylene glycol and tri ethylene glycol. One of the ways of increasing the monoglycol selectivity and, therefore, of decreasing water excess is the application of catalysts accelerating only the first step of the reaction (1). There are much research has been carried out to improve this process. The search for better catalyst is an objective for increase the selectivity and decrease the excess water. As evident from the kinetic data the distribution factor b = k1/k0 is reduced -0.1–0.2 at the concentration of some salts of about 0.5 mol/l. This enables to produce monoethyleneglycol with high selectivity at water–ethyleneoxide molar ratio close to 10. 5.2.2 Catalyst: A cross-linked styrene–divinylbenzene anion exchange resin (SBR) in the HCO3−/ CO3- form, activated by anion exchanging with sodium bicarbonate solution used as catalysts. (Dow Chemical produced anion-exchange resins: DOWEX SBR). The ethylene oxide hydration process in a catalytic fixed-bed tube reactor was studied .The properties of initial resins are summarized below: Functional group

:

- [PhN (CH3)3] +

Total exchange capacity (equiv./l)

:

1.4

Particle size (mm)

:

0.3-1.2

5.3 PROCESS DESCRIPTION: This process produced mono ethylene glycol by the catalytic hydrolysis of ethylene oxide in the presence of less excess of water. After the hydrolysis reaction is completed the glycol is separated from the excess water and then refined to produce mono ethylene glycol (MEG). The process is devided in to five different sections. 5.3.1

MEG reaction unit:

Ethylene oxides mixed with recycle water and pumped to glycol reactor where it is reacted with water at 1050C &1.5 MPa in the presence of catalyst. The Reactor is Catalytic Plug flow Fixed bed type. The reaction volume consists of two phase, the 24

liquid phase and ionite (catalyst) phase. The liquid streams through catalyst bed in a plug flow regime. The catalytic and non catalytic ethylene oxide hydration takes place in the ionite phase, and only non catalytic reaction takes place in the liquid phase. The distribution of the components of the reaction mixture between liquid and ionite phases is result of the rapid equilibrium. The glycol reactor operate at approximately 1.5MPa.pressure which is supplied by the reactor feed pump. The reactor effluent goes to the evaporation unit for the evaporation of excess water. 5.3.2

MEG evaporation unit:

The glycol evaporation system consists of multiple effect evaporation system(three effects). The reactor effluent flows by difference in pressure from one evaporator to the next the water content of glycol is reduced to about 15% in the evaporators. The remaining water is removed in drying column, the pressure of the system is such that the reactor effluent is maintained as a liquid and is fed as such in to the vapor portion of the first effect evaporator. Evaporation in the first effect is accomplished by 12Kg/cm2 (g) pressure steam. The overhead vapor from the first effect is used as heating media in the second effect. The steam condensate from the first effect is goes to the medium pressure condensate header. The overhead vapor from the second effect is used as heating media in the third effect. The third effect operated under vacuum. The vacuum is maintained by using steam jet ejector. The bottom of the third effect containing 15% water is fed to crude glycol tank via glycol pump, which is then fed to the drying unit. The condensate from first two effects and the vapor from third effect containing water and some amount of glycol are fed to the glycol recovery unit. 5.3.3 MEG drying unit: The concentrated glycol from the third effect is containing approximately 15% water. Essentially all the water is removed from the aqueous ethylene glycol solution in the drying column. Normally the drying column is fed from the crude glycol tank. The drying column operated under vacuum which is maintained by steam jet ejector. Drying column bottom which are free from water are transferred by column bottom pump to MEG refining column. Where the MEG is separated from the higher glycol, Water vapors leaving the top of the drying column are fed to MEG recovery unit for glycol recovery. (An inert gas line is provided at the base of the drying column for breaking the vacuum). 25

5.3.4

MEG refining unit:

Drying column bottoms essentially free of water are fed to the MEG refining column. (PACKED COLUMN). About 15% of the feed to the MEG column enters as vapor due to flashing. MEG product is withdrawn from the top of the column. Some MEG is purged in the overhead to the vacuum jets to reduce the aldehydes in the product. The MEG column bottoms primarily di-ethylene glycols are pumped from the column bottom to the storage tank. The MEG column operates at a pressure of 10mmHg (A). The vacuum is maintained by MEG column ejector system. The MEG column condenser is mounted directly on the top of the MEG column. 5.3.5

MEG recovery unit:

The MEG leaving along with water from the Top of the multiple effect evaporator & drying column are recovered in the MEG Recovery Column (PLATE COLUMN). The column is operated under Atmospheric pressure.MEG leaving from the bottom of the column and the water leaving from the top of the column are Recycle to reactor.

CHAPTER VI MATERIAL BALANCE

26

Material balances are the basis of process design. A material balance taken over complete process will determine the quantities of raw materials required and products produced. Balances over Individual process until set the process stream flows and compositions. The general conservation equation for any process can be written as Material out = material in + accumulation For a steady state process the accumulation term is zero. If a chemical reaction is taking place a particular chemical species may be formed or consumed. But if there is no chemical reaction, the steady state balance reduces to: Material out = Material in A balance equation can be written for each separately identifiable species present, elements, compounds and for total material. [10] 6.1 BASIS: Basis: 100000TPA The process is planned and developed as a continuous process. A plant is operated for 24 Hours per day and 333 per year. No of working days = 333days Capacity =

1000000 333

= 300.3 T/days = 201.47 Kmol/hr. 6.2 MOLECULAR WEIGHT (KG / KMOL): Ethylene Glycol

: 62

Water

: 18

Carbon Dioxide [CO2]

: 44.01

Water [H2O]

: 18

Nitrogen [N2]

: 28

6.3 MATERIAL BALANCE OF INDIVIDUAL EQUIPMENT: This is the amount of MEG obtained from the distillation column,

27

So assuming that 99% of MEG in the feed to the Distillation column (Refining Column) is obtained in the distillate & also 93% of MEG in feed to the Recovery Column is recovered from Recovery Column. ∴ Kmol of MEG in feed to the distillation column = 204.70 Kmol/hr. 6.3.1 Reactor:

Ethylene Oxide = 9190.54 Kg = 208.876 Kmols

Mono Ethylene Glycol = 204.7Kmols = 12691.4 Kg

REACTOR Temp. = 100 0C

Water

Water = 1881.972 Kmols = 33875.496Kg

Conversion = 100 %

= 37597.68 Kg = 2088.76 Kmol

Higher glycol = 2.088 Kmol = 221.328Kg

Pressure = 1.5-2MPa

In the reactor following reaction take place C2H4O +

H2O

HOCH2CH2OH

(Ethylene oxide) (Water)

--------- (1)

(Mono Ethylene Glycol)

C2H4O + HOCH2CH2OH

HOCH2CH2OH

(Ethylene oxide) (Mono Ethylene Glycol)

(Higher Glycol)

As selectivity = 98% Moles of undesired product formed =

204.70 98

= 2.088 Kmol Moles of MEG to be produced from reactor = 206.788kmol Moles of ethylene oxide reacted by reaction –I = 206.788 Kmol. Moles of ethylene oxide reacted by reaction –I I = 2.088 Kmol. Total Moles of ethylene oxide reacted = 206.788 + 2.088 28

--------- (2)

= 208.876 Kmol. As conversion = 100% [6] Moles of ethylene oxide charged = 208.876kmol From the literature we know that the ratio of WATER TO ETHYLENE OXIDE =10 Amount of water fed to reactor = 2088.76 Kmol. (Including excess) From the reaction moles of water reacted = 206.788 Kmol. M.B.ON WATER: Moles of water fed = Moles of water reacted + Moles of water unreacted 2088.76

=

206.788 + Moles of water unreacted

∴ Moles of water unreacted = 1881.972kmol M.B.ON MEG: Moles of MEG in the product = 206.788 – 2.088 = 204.7 Kmol Table 6.1 Material balance over reactor Component Ethylene oxide Water Mono Ethylene Glycol Higher Glycol

In, Kg 9190.54 37597.68 -

Out, kg 33875.496 12691.4 221.328

6.3.2 Triple Effect Evaporator: Consider the water content of glycol is reduced to 15% i.e. 85% of water is to be removed. Consider triple effect evaporator as single unit. Amount of water removed

= 0.85× 1881.972 = 1599.6762 Kmol. = 28794.1715 Kg

Total quantity of water at the top = 1599.6762 Kmol. = 28794.1716 kg. Remaining 15% water are still in the bottom along with the MEG and Higher glycol.

29

∴ Amount of water in the bottom

= 1881.972-1599.6762 = 282.2958 Kmol. = 5081.324 Kg

There is some quantity of glycol carry over along with water from the top of evaporator. Amount of glycol carry over along with water from 1st effect

= 165.58 kg

W1= 8285.66kg MEG = 165.58kg H2O = 8120.08kg

F = 2088.76 Kmol = (46788.224 kg) M.E.G = 204.7Kmol = 12691.4 Kg Water =1881.972 Kmol = 33875.496Kg

1st effect evaporator To 2nd effect evaporator

Pressure = 7 kg/cm2 Temp = 159 oC

Amount of glycol carry over along with water from IIst effect = 189.139kg W2= 9689.31kg MEG = 189.139kg H2O = 9500.171kg

Temp = 141 oC Pressure = 3.5 kg/cm2 To 3rd effect evaporator

2nd effect evaporator

From 2nd effect evaporator

To MEG Recovery column Y= 1610.8012kmol

Amount of glycol carry over along with water from IIst effect = 335.064 kg

30

W3= 11508.96kg MEG = 335.064kg H2O = 11173.89kg

From 3rd effect evaporator

To MEG Refining column X = 477.9588 Kmol

3rd effect evaporator Pressure = 0.25 kg/cm2 Temp = 118 oC To MEG Recovery column Y= 1610.8012kmol

(Finding using VLE calculation) Total amount of glycol carry over along with water = 689.783 Kgm. =11.125 Kmol Total quantity (water +MEG) leaving from the top of effect = 1599.6762+11.125 Y = 1610.8012 Kmol. TAKING OVERALL M.B F=Y+X 2088.76 = 1610.8012 + X X = 477.9588 Kmol. (Total quantity leaving from the bottom of last effect) Table 6.2 Material balance over Triple effect evaporator Component Water MEG HG

In, Kg 33875.496 12691.4 221.328

Liquid phase 5081.355 12001.617 221.328

Out, Kg Vapor phase 28794.141 689.783 -

6.3.3 Drying Column: Y= 289.295 Kmol = 5537.385 kg MEG = 456.061kg

F = 477.9588 kmol = 17304.2585 kg MEG = 12001.606kg H2O = 5081.324 kg. HG = 221.328kg.

Drying column Pressure = 0.21 kg/cm2 31

Temp = 87 oC

H2O = 5081.324 kg .

X = 188.306 Kmol = 11766.873 kg MEG = 186.218kmol = 115453545kg HG = 2.088 Kmol = 221.328kg

Consider all the water are removed in the drying column Amount of water removed

= 5018.324 Kgm = 282.295 Kmol.

There is some quantity of glycol carry over along with water from the top of drying column Amount of glycol carry over along with water from drying column = 456.061kg =7.3558 Kmol. (Finding using VLE calculation) Total quantity leaving from top of drying column = (Amount of water +Amount of MEG) = 282.295 +7.3558 = 289.65 Kmol. TAKING OVERALL M.B F =Y+X 477.9588 = 289.65 + X X = 188.306 Kmol. (Total quantity leaving from the bottom of drying column)  Now , Total amount of MEG leaving along with water during evaporation of water = (Amount of MEG leaving from top of TEE + Amount of MEG leaving from top of drying column) = 689.783+456.061 = 1145.844 Kgm. = 18.4813 Kmol.  Amount of feed to MEG Recovery column

32

= (Amount of MEG leaving along with water during evaporation + Amount of water removed) = 18.4813+1881.973 = 1900.451 Kmol. Table 6.3 Material balance over drying column Component Water MEG HG

In, Kg 5081.324 12001.606 221.328

Out, Kg Liquid phase Vapor phase 5081.324 11545.3545 456.061 221.328 -

6.3.4 MEG Refining Column (Packed Column):

D= 184.54 Kmol = 11448.8616 kg MEG = 184.355kmol (0.999.high purity)

F = 188.306 Kmol = 11766.873 kg

HG = 0.18454kmol

MEG refining column Pressure = 10 mmHg

MEG = 186.218kmol =11545.545kg

Temp = 93.2 oC

HG =2.088kmol = 221.328kg

W = 3.766 Kmol = 317.664 kg MEG = 1.8523kmol HG = 1.9136kmol

Assuming 99% recovery, of total MEG feed to distillation column, is obtained in the distillate. 

Kmol of MEG in Distillate

= 188.306 × 0.99 x 0.98891 = 184.355 Kmol / hr. = 11431.0818 Kg/hr.



Kmol of Distillate

(D)

= 184.355 / 0.999 = 184.54 Kmol / hr.

Avg. M.W. of distillate

= (0.999 x 62) + (0.001 x 106) = 62.044 kg / Kmol. 33

Amt. of Distillate (D)

= 184.54 x 62.04 = 11448.8618 kg / hr.



Amt. of HG in Distillate

= 184.54 x 0.001 = 0.18454 Kmol / hr. = 0.18454 x 106 = 19.561 kg / hr.



Kmol of feed (F)

= 188.306 Kmol / hr. = 11766.873 kg/hr

TAKING OVER ALL M.B. F=D+W 188.306 = 184.54 + W W = 3.766 Kmol /hr. M.B. ON MEG F x (Xf MEG) = D x (Xd MEG) + W x (Xb MEG) 188.306 x 0.9889 = 184.54 x 0.999 + 3.766 x Xb MEG Xb MEG = 0.4918 (mol.fr.of MEG in Bottoms) XbHG = (1- 0.4918) = 0.5081 (mol.fr.of HG in Bottoms) 

Kmol of MEG in Bottoms

= 0.4918 x 3.766 = 1.8521 Kmol / hr

Mol. Weight of MEG = 62 kg/Kmol = 114.831 kg/hr. 

Kmol of HG in Bottoms

= 0.5081 x 3.766 = 1.9135 Kmol / hr.

Mol. Weight of HG =106 kg/Kmol = 1.9135 x 106 = 202.83 kg/hr. Table 6.4 Material balance over Refining packed column Component MEG HG

In, Kg 11545.545 221.328

Liquid phase 114.8426 202.8416

Out, Kg Vapor phase 11430.01 19.56124 D= 1881.97kmol

34

= 11766.873 kg MEG =1.88kmol H2O =1880.08kmol

6.3.5 MEG recovery column (Plate column):

F = 1900.451kmol = 35021.339 kg

MEG recovery column

MEG = 18.481kmol =1145.844kg

Plate column

H2O =1881.97kmol = 33875.496kg.

W = 18.481kmol =1205.55 kg MEG = 17.122kmol H2O = 1.3584kmol

Assuming 99.9 % of total water in feed to distillation column is obtained in the distillate. 

Kmol of Water in Distillate

= 1881.97 x 0.999 = 1880.08 Kmol / hr



Kmol of Distillate

(D)

= 1880.08 / 0.999 = 1881.97 Kmol / hr.

Avg. M.W. of distillate

= (0.999 x 18) + (0.001 x 62) = 18.044 kg / Kmol.

Amt. of Distillate (D)

= 1881.97 x 18.044 = 33958.266 kg /hr



Amt. of MEG in Distillate

= 1881.97 x 0.001 = 1.88 Kmol / hr = 1.88 x 62 = 116.56 kg/ hr.



Amount of feed ( F )

= 1900.451 Kmol/hr = 35021.339 kg/hr.

TAKING OVERALL M.B. F = D+ W 1900.451 = 1881.47 + W 35

W = 18.481kmol / hr M.B. ON WATER F x (Xf H) = D x (Xd H) + W x (Xb H) 1900.451 x 0.99 = 1881.97 x 0.999 + 18.481 x Xb W Xb W = 0.0735 (mol.fr.of Water in Bottoms) Xb MEG = 1- 0.0735 = 0.9264 (mol.fr.of MEG in Bottoms) 

Amount of MEG in Bottoms

= 18.481 x 0.9264 = 17.122 Kmol / hr = 17.122 x 62 = 1061.56 kg/hr.



Kmol of Water in Bottoms

= 18.481 – 17.130 = 1.3584 Kmol / hr = 1.3584 x 18 = 143.99 kg/ hr.

Table 6.5 Material balance over Recovery plate column Component Water MEG

In, Kg 33875.496 1145.844

Liquid phase 24.4512 1061.546

Out, Kg Vapor phase 33841.44 116.56

Table 6.6 Overall material balances Equipment Reactor

Triple

Component

Ethylene oxide Water MEG HG effect Water

In, kg 9190.54 37597.68 33875.496

Out, Kg Liquid phase Vapor phase 33875.496 12691.4 221.328 5081.355 28794.141

evaporator MEG HG Drying column Water MEG HG MEG refining MEG

12691.4 221.328 5081.324 12001.606 221.328 11545.545 36

12001.617 221.328 11545.3545 221.328 114.8426

689.783 5081.324 456.061 11430.01

column HG MEG recovery Water

221.328 33875.496

202.8416 24.4512

19.56124 33841.44

1145.844

1061.546

116.56

column MEG

CHAPTER VII ENERGY BALANCE The first law of thermodynamics demands that energy be neither created nor destroyed. The following is a systematic energy balance performed for each unit of the process. The datum temperature for calculation is taken as 0C. The different properties like specific heat, heat of reaction, heat of vaporization, etc. are taken to be constant over the temperature range. 7.1 REACTOR: [9,11]

REACTOR Ethylene Oxide = 9190.54 Kg = 208.876 Kmols Water

= 37597.68 Kg = 2088.76 Kmol

Temp. = 100 0C Conversion = 100 % Pressure = 1.5-2MPa

37

Mono Ethylene Glycol = 204.7Kmols = 12691.4 Kg Water = 1881.972 Kmols = 33875.496Kg Higher glycol = 2.088 Kmol = 221.328Kg

In the reactor following reaction take place C2H4O + (Ethylene oxide)

H2O

HOCH2CH2OH

(Water)

(Mono Ethylene Glycol)

C2H4O + HOCH2CH2OH

HOCH2CH2OH

(Ethylene oxide) (Mono Ethylene Glycol) Table 7.1 Heat capacity and Enthalpy data COMPONENT ∆H 0 ( kj f 298

IN Ethylene oxide Water OUT MonoEthyleneGlyocol Di-EthyleneGlyocol Water

------------- (1) ------------ (2)

(Higher Glycol)

kmol

)

C p ( kj

kmol k

-77704 -285830

99.106 75.673

-454800 -285831 -562570

75.673 189.39 441.602

Assume reference temp. = 250C 7.1.1

Enthalpy of formation of reaction •

For first reaction

∆ H 0 f = ∑ ∆ H 0 fp − ∑ ∆ H 0 f R = [-454800] - [-(77704) + (-285830)] = -91266 KJ/ Kmol of EO Reacted = -91266 x 206.788 = -18.872 x 106 KJ / hr • ∆H 0

f

For second reaction

= ∑ ∆H 0 fp − ∑ ∆H 0

f R

= [-562570] – [(-77704) + (-454800)] = -30066 KJ/ Kmol of EO Reacted = -30066 x 2.088 = -62.77x103 KJ / hr Total enthalpy of formation = (-18.872 x 106 ) + (-62.77x103 ) = -18.9347 x 106 KJ / hr  Enthalpy of reactants As reactants are added at 250C, so, its Enthalpy becomes 0. 38

)

 Enthalpy of products

[

]

∆H p = ( m C p ) MEG + ( m C p )WATER + (mCp ) HG ∆T

= [ ( 204.7 x 189.39) + ( 1881.972 x 75.673 ) + (2.088 x 441.60) ] ( 105 – 25 ) = 14.5683 x 106 KJ / hr  Enthalpy of reaction ∆H R0 = ∆H p + ∆H 0

f

− ∆H R

= (14.5683 x 106) + (-18.9347 x 106) - 0 = - 4.3043 x 106 KJ / hr So, it indicates that it is an exothermic reaction. So, to control temp. Inside the reactor, cooling water is passed on shell side to remove the heat. Assuming cooling water entered at 25 o C and leave at 45 o C Q

= m x Cp x ∆T

- 4.3043 x 106 m

= m x 75.79627 x 20 = 2.8394 x 103 Kg / hr (cooling rate) [9,11]

7.2 TRIPPLE EFFECT EVAPORATOR: Water to be evaporated = 28794.716Kg/hr Total feed wF = 46788.224 Kg/hr The balances applying to this problem are: First effect:

F = 2088.76 Kmol = (46788.224 kg) M.E.G =Second 204.7Kmol effect: = 12691.4 Kg Water =1881.972 Kmol = 33875.496Kg

wSλS + wF (tF – t1) Cp = w1λ1 W1= 8285.66kg MEG = 165.58kg H2O = 8120.08kg

1st effect evaporator w1λ1 + (wF – w1) ( t1 – t2) Cp = w2λ2

To 2nd effect evaporator

Pressure = 7 kg/cm2

W2= 9689.31kg

Temp = 159 oCMEG = 189.139kg

H2O = 9500.171kg

Temp = 141 oC Pressure = 3.5 kg/cm2 2nd effect evaporator

From 2nd effect evaporator

39

To 3rd effect evaporator

To MEG Recovery column Y= 1610.8012kmol

Third effect:

w2λ2 + (wF – w1-w2) (t2 – t3) Cp = w3λ3 W3= 11508.96kg MEG = 335.064kg H2O = 11173.89kg

From 3rd effect evaporator

To MEG Refining column X = 477.9588 Kmol

rd

3 effect evaporator Pressure = 0.25 kg/cm2

To MEG Recovery column Temp = 118 oC Material balances: w + w + w = w 1 2 3 1-3 Y= 1610.8012kmol

tF = 1050C Consider steam is entered at 12 kg/cm2 so Ts = 190.8250C (After finding boiling point of solution) Also last effect operates at a vacuum of 0.25 Kg/cm2 So t3 = 106.15oC (steam temp at 0.25 kg/cm2) Consider for forward feed multiple effect evaporator pressure differences between effects will be nearly equal. So average pressure difference

= 385.056 KPa /effect

Table-7.2 Breaking up the total pressure difference: Pressur

Steam or

λ, KJ/Kg

λ, KJ/Kg

e, KPa

vapor

(Steam)

(MEG)

temp. C Steam chest, 1st

1179.69

TS= 190.82

λS = 2210.8

794.63

t1=175.17

λ1 = 2244.1

λ1 = 982.935

409.57

t2=152.585

λ2 = 2284.0

λ2 = 1001.15

24.53

t3= 106.155

λ3 = 2379.1

λ3 =1022.317

effect Steam chest, 2nd effect Steam chest, 3rd effect Vapor to condenser 7.2.1 First effect: Cp avg. = Σ xiCpi = 4.196 KJ/Kg o K 40

λ avg = 2016.38 KJ/Kg WSλS + wF (tF – t1) Cp = w1λ1 (WS x 1973.62) + (46788.224 x - 70.17 x 4.196) = w1 x 2016.38

w1 = 0.9787WS – 6830.42

----------------------------- (1)

7.2.2 Second effect: Cp avg. = Σ xiCpi = 4.105 KJ/Kg o K λ avg

= 2088.28 KJ/Kg

w1λ1 + (wF – w1) (t1 – t2) Cp = w2λ2 w1 X 2016.38 + (46788.224 -w1) (175.17-152.585) x 4.05 = w2 2088.28 Put value of w1 from equation (1) and finally

w2 = 0.9022WS – 4245.22 ---------------------------- (2) 7.2.3 Third effect: Cp avg. = Σ xiCpi = 3.873 KJ/Kg o K λ avg = 2207.35 KJ/Kg

w2λ2 + (wF – w1-w2) ( t2 – t3) Cp = w3λ3 w22088.28 + (46788.224 – w1 – w2) (152.585 – 106.155)3.873 = w3 2207.35 Put value of W2 from equation 2 and finally we get

w3 = 0.70WS – 697.42 ----------------------------------- (3) Taking overall Material balances:

w1 + w2 + w3 = w1-3 0.9787WS – 6830.42 +0.9022WS – 4245.22 + 0.70WS – 697.42 = 28794.1716 + 689.783 WS = 15.445 x 103 Kg/hr ( steam rate is required.) From above equations we calculated,

w1 = 8285.66 Kg/hr w2 = 9689.31 Kg/hr w3 = 11508.96 Kg/hr Now , Enthalpy out from the bottom of the last effect, Tbottom = 122oC

Trefrence = 25oC 41

∆T = 97oC. Enthalpy out from Bottom = (mCp∆T )MEG + ( mCp∆T )WATER + ( mCp∆T )HG = [(12001.606 x 3.077) + (5081.324 x 4.378) + (221.328 x 4.1032)] x 97 = 5.828 x 106 KJ / hr

7.3 DRYING COLUMN: Y= 289.295 Kmol = 5537.385 kg MEG = 456.061kg H2O = 5081.324 kg .

F = 477.9588 kmol = 17304.2585 kg MEG = 12001.606kg H2O = 5081.324 kg.

Drying column Pressure = 0.21 kg/cm2

HG = 221.328kg.

Temp = 87 oC X = 188.306 Kmol = 11766.873 kg MEG = 186.218kmol = 115453545kg

Toperating = 87 oC

Trefrence = 25 oC

HG = 2.088 Kmol = 221.328kg

Hence ∆T = 62oC. Poperating = 0.25 kg /cm2 Enthalpy in = 2.802 x 106 kJ / hr 7.3.1 Enthalpy out from Top = ( mλ )water + ( mλ )MEG +( mCp∆T ) = [(5081.324 x 2366.1) + (456.061 x 1109 .75)] + [289.65 x 75.2 x 64] = 12.529 x 106 kJ / hr 7.3.2 Enthalpy out from Bottom = (mCp∆T )MEG + ( mCp∆T )HG 42

= [(186.218 x 187.90) + (432.72 x 2.088)] x 62 = 2.225 x 106 kJ / hr Total Enthalpy out = Enthalpy out from (Top + Bottom) = 12.529 x106 + 2.225 x 106 = 14.75 x106 kJ / hr Q

= Total Enthalpy out - Enthalpy of feed

Enthalpy of feed = 5.828 x 106 kJ / hr Q = 14.75 x106 +5.828 x 106 = 8.926 x 106 kJ / hr Amount of steam required, Consider the steam enter at 2 kg/cm2 & 118.719oC λSteam = 2205.82 kJ / kg Q = m λsteam 8.926 x 106 = m x 2205.82 m = 4046.6 kg / hr (Rate of steam) D= 184.54 Kmol

7.4 MEG REFINING COLUMN:

= 11448.8616 kg MEG = 184.355kmol (0.999.high purity) HG = 0.18454kmol

F = 188.306 Kmol = 11766.873 kg MEG = 186.218kmol =11545.545kg

MEG refining column Pressure = 10 mmHg Temp = 93.2 oC

HG =2.088kmol = 221.328kg

W = 3.766 Kmol = 317.664 kg MEG = 1.8523kmol HG = 1.9136kmol

43

7.4.1 for top: Ttop = 91.8 oC Trefrence = 25 oC ∆T = 66.8 oC Poperating = 10 mmHg Cpmean of MEG = 189.70 kJ / kmol oK Cpmean of DEA = 441.6 kJ / kmol oK Total Enthalpy out with Distillate = (mCp∆T ) MEG + (mCp∆T )DEG = [(184.355 x 189.70) + (0.18454 x 441.6)] x 66.8 QD = 2.341 x 106 kJ / hr Reflux Ratio = 0.71 (finding using Mc Cabe & Thiel Method) i.e. L/D = 0.71 L = 0.71D Vapor formed at the top

V=L+D = 0.71D + D = 0.71 x 184.355 V = 315.247 kmol / hr

Reflux L = 0.71D = 0.71 x 184.355 L = 130.89 kmol / hr  Enthalpy out with vapor: QV = latent heat + sensible heat associated with that vapor = mλ + (mCp∆T) λMEG = 68.578 x 103 kJ / kmol λDEG = 72.067 x 103 kJ / kmol λAVEG = 68.58 x 103 kJ / kmol 44

QV = [(315.247 x 68.58 x 103) + (315.247 x 188.298 x 66.8)] = 25.58 x 106 kJ / hr  Enthalpy out with Reflux: QReflux = ( mCp∆T )Reflux = [ 130.89 x 188.551 x 66.8 ] = 1.6485 x106 kJ / hr  Condenser load, QC : = QV – ( QReflux + QD ) = [(25.58 x 106 ) – (1.6485 x106 + 2.341 x 106 )] = 21.595 x 106 kJ / hr Assuming cooling water enters the condenser at 25oC & leave at 45oC QC = (mCp∆T )cooling water 21.595 x 106 = m x 75.7962 x 20 m = 11.63 x 103 kg / hr (Rate of cooling water) 7.4.2 For bottom: Tbottom = 94.6 oC

∆T

= 69.6 oC

Cpmean of MEG = 188.531 kJ / kmol 0K Cpmean of DEG = 443.2 kJ / kmol 0K Enthalpy out with Residue, QResidue = ( mCp∆T )liq = [(1.8528 x 188.531) MEG + (1.9136 x 443.2) DEG] x 69.6 = 83.34 x 106 kJ / hr  Reboiler Load: Reboiler heat load is determined from a balance over complete system QB + QFeed = QD + QW + QC QFeed = 2.252 x 106 kJ / hr 45

QB

=

(21.595 x 106 + 83.34 x 106 - 2.252 x 106 + 2.341 x 106 )

=

21.794 x 106 kJ / hr

Amount of steam required, Consider the steam enter at 2 kg/cm2 & 118.719oC λsteam = 2205.82 kJ / kg QB = mλ steam 21.794 x 106 = m x 2205.82 m = 9.88 x 10 3 kg / hr ( Rate of steam ) 7.5 MEG RECOVERY COLUMN:

D= 184.54 Kmol = 11448.8616 kg MEG = 184.355kmol (0.999.high purity) HG = 0.18454kmol

F = 188.306 Kmol = 11766.873 kg

MEG recovery column

MEG = 186.218kmol =11545.545kg

Plate column

HG =2.088kmol = 221.328kg

W = 3.766 Kmol = 317.664 kg MEG = 1.8523kmol HG = 1.9136kmol

7.5.1 For top: Ttop = 194oC

Trefrence = 25oC

∆T = 169 oC Poperating = 760 mmHg Cpmean of MEG = 197.24 kJ / kmol oK Cpmean of H2O

= 76.55 kJ / kmol oK

Total Enthalpy out with Distillate = (mCp∆T )MEG + (mCp∆T ) WATER = [(1881.08 x 76.55) + (1.874 x 197.24)] x 169 46

QD = 24.40 x 106 kJ / hr Reflux Ratio = 0.51

(finding using Mc Cabe & Thiel Method) i.e. L/D = 0.51 L = 0.51D

Vapor formed at the top V = L + D = 0.51D + D = 0.51 x 1881.97 V

= 2841.77kmol / hr

Reflux L = 0.71D = 0.51 x 1881.97 L = 959.80 kmol / hr Enthalpy out with vapor: QV = latent heat + sensible heat associated with that vapor = mλ + (mCp∆T) λMEG = 1023.184 kJ / Kg

λH2O = 2231.8 kJ / Kg

λAVEG = 40.195 x 103 kJ / kmol QV = [(2841.77 x 40.195 x 103) + (2841.77 x 197.24x 169)] = 208.95 106 kJ / hr Enthalpy out with Reflux: QReflux = (mCp∆T) Reflux = [959.80 x 76.67 x 169] = 12.43 x106 kJ / hr Condenser load QC = QV – ( QReflux + QD ) = [(208.95 106 ) – (12.43 x106 + 24.40 x 106 )] 47

= 172.12 x 106 kJ / hr Assuming cooling water enter the condenser at 25oC & leaves at 45oC QC = (mCp∆T )cooling water 172.12 x 106 = m x 75.7962 x 20 m = 113.63 x 103 kg / hr (Rate of cooling water)

7.5.2 For bottom: Tbottom = 198 oC ∆T

Trefrence = 25oC

= 173 oC

Cpmean of MEG = 197.6285 kJ / kmol 0K Cpmean of H2O = 76.607 kJ / kmol 0K  Enthalpy out with Residue: QResidue = ( mCp∆T )liq = [(17.122 x 197.6285) MEG + (1.3584 x 76.607 )DEG ] x 173 = 603.39 x 103 kJ / hr  Reboiler Load: Reboiler heat load is determined from a balance over complete system QB + QFeed = QD + QW + QC QFeed

= ( mCp∆T )feed = [(18.481 x 187.97) MEG + (1881.97x 74.51) WATER] x 80 = 143.71x 103 kJ / hr

Reboiler load QB = [( 603.39 x 103 + 172.12 x 106 + 24.40 x 106 ) – (143.71x 103) ] = 196.97 x 106 kJ / hr Amount of steam required, 48

Consider the steam enter at 2 kg/cm2 & 118.719oC λsteam =2037.51 kJ / kg QB = mλ steam 196.97 x 106 = m x 2037.51 m = 97.67 x 10 3 kg / hr ( Rate of steam )

[9,11]

CHAPTER VIII REACTIONS KINETICS & THERMODYNAMICS

8.1 REACTOR KINETICS:

REACTOR E. O = 9190.54 Kg = 208.876 Kmols Water = 37597.68 Kg = 2088.76 Kmol

Temp. = 50 0C Conversion = 100 %

Here, FAO

=

208.876 Kmol/hr

=

9190.544 Kg/hr

∴ V0 = FAO

10.649 m3/hr

= CAO V0

208.87 = Cao x 10.649 CAO

= 19.6136 Kmol/m3

= 19.6136 mol/lit

Similarly, CBO

= 5.555 Kmol/m3

=

55.555 mol/lit

Now, Rate of reaction is given by 49

Product =2088.76 kmol = (46788.224 kg) M.E.G = 204.7Kmol = 12691.4 Kg Water =1881.972 Kmol = 33875.496Kg H.G = 2.088 Kmol = 221.328Kg

{K0([H2O] + b Σ [Gyi]) + Kct [HCO3–]} X

– d (C2H4O) =

{{[H2O] + p Σ [Gyi]}x [Oxide]}

dt Where, Gyi

= concentration of reactant (mol/lit)

[H2O] = concentration of water (mol/lit) [Oxide] = concentration of oxide (mol/lit) b = distribution factor =2.8 p = 1.88 – d (C2H4O)

=

– rA

= {K0(CB + b Σ (CA + CB)) + Kct [HCO3–] } X {(CB + p Σ (CA + CB)) x [CA]}

dt Rate constant Ko is given by, K0

=

exp [9.077 – 9355] T

where T = temperature o K K0

=

exp [9.077 – 9355] 378

= 1.5627 x 10-7

K0

L2

L2

= 0.0005625

mol2. Sec

mol2. hr

Similarly catalyst Rate constant Kct is given by, Kct

= exp [18.24 – 10574] T = 5.926 x 10-5

Kct

L2

= 0.21334

mol2. Sec

L2 mol2. hr

Now, CAO XA =

CBO XB

a 19.6136 XA 1

b =

55.555 XB 1

XB = 0.3530 XA - rA

={ K0[(CBO(1 – 0.3530 XA)) + 2.8 (CAO (1 – XA) + CBO (1 – 0.3530 XA))] + Kct [0.25]} x {[CBO (1 – 0.3530 XA) + 1.88 [CAO (1 – XA) + CBO (1 – 50

0.3530 XA)]}x CAO (1 – XA)} - rA

={ 0.0005625 [55.555 (1 – 0.3530 XA)) + 2.8 (19.6136 (1 – XA) + 55.555 (1 – 0.3530 XA)] + 0.05533} x {55.555 (1 – 0.3530 XA) + 1.88 [19.6135 (1 – XA) + 55.555 (1 – 0.3530 XA)]}x 19.6136 (1 – XA)}

Table-8.1 Different value of XA and finding corresponding rate (-rA), XA = -rA = (1/rA) (lit.hr /mol)

0.1 654.5

0.005

0.2 532.40

0.3 424.57

0.4 330.131

0.5 248.286

0.6 0.7 178.2369 119.1836

0.8 70.31

0.9 30.86

0.98 53.60

0.0018

0.0023

0.00302

0.00402

0.00561

0.01

0.03

0.186

From the above data plotting a XA Vs (

0.0084

1 ) & finding the area under he curve at − rA

Xa = 0.98

Fig-4 Reactor volume Chart AREA UNDER THE CURVE = 0.01227 Plug flow equation related to volume is given by [15] XA

51

V

=

FAO

d XA 0

- rA

XA

d XA 0

=Area under the curve

- rA

Area under the curve = 0.01227 V FAo

=

∴V

= 2562.90 lit

V

0.01227

= 2.56 m3

Now we have ∈

=

Empty volume in bed Total bed Volume

1–∈

=

1 – Empty volume in bed Total bed Volume

1–∈

=

Total Bed Vol – Empty volume in bed Total bed Volume

But Total Bed – Empty volume in Bed = Volume of Catalyst. 1–∈ =

Volume of Catalyst Total bed Volume

1–∈ =

Volume of Catalyst 2.56 m3

Consider ∈ = 0.6 ∴ Total Volume of Catalyst = 1.024 m3 8.2 THERMODYNAMICS As we know the Gibb’s free energy is given by following equation ΔG = -RT lnK Where K = Kequ = (ka * kb)/ (kc * kd) From reaction, ka and kb are rate constants for the products.

52

Similarly kc and kd are rate constants for the reactants. But assuming reaction is exothermic and irreversible; the values of kc and kd will not be in consideration to finding out equilibrium rate constant. Hence, Kequ is given by Kequ = ka kb Enthalpy, Gibbs free energy and specific heat data are below at reaction temperature 100 0C in the form of functional group. [11] -O- group: Cp = 19.28 KJ /.Kg K, G = -15.38 KJ / Kg K H = -1467.62 KJ /.Kg K -CH2- group: Cp = 20.43 KJ /.Kg K, G = -3.87 KJ / Kg K H = -1516.94 KJ /Kg K -OH- group: Cp = -1.83 KJ /.Kg K, G = -36.785 KJ / Kg K H = 96.75 KJ /.Kg K H2O: G = -8.728 KJ / Kg K

∆G total = ∆G product - ∆G reactant C2H4O

+

H2O

HOCH2CH2OH

(Ethylene oxide) (Water)

(Mono Ethylene Glycol)

∆G = [(-81.31)]-[(-23.12) + (-8.728)] ∴∆ G = - 49.53 KJ/Kmol K ∴ Ka = exp [-(-49.53) / (8314*373)] ∴ Ka =1.00 ∴ ln(K2/K1)= E/R[(1/T1)-(1/T2)]

--------- (1)

[15]

∴ ln(9.5/8)= E/8.314[(1/368)-(1/373)] T2 reaction at 100 oC= 373K T1 reaction at 95 oC=368K ∴ 0.26236 = 6.2382* E *10-6 ∴ E = 7.52*E*-7 J/mol

53

CHAPTER IX MAJOR EQUIPMENT DESIGN 9.1 DESIGN OF REACTOR AS A SHELL & TUBE HEAT EXCHANGER : Consider the reactants are flow on tube side and cooling water are on shell side Catalysts are fill inside the tube. 9.1.1 Process Design: [22] Consider length of tube Diameter of tube

= 4m = 2.5 cm

∴ Volume of one tube

π (d)2 (L) 4 π (2.5 x 10-2)2 (4) 4 1.9634 x 10-3 m3

= = =

Table-9.1 Properties at arithmetic mean temperature. Props. Cp µ K ρ

Shell Side (Water) (30oC ) 5.1865 (KJ/Kg oK) 9 x 10-4 (Kg/m.Sec) 0.62 (w/m.oK) 995.40 (Kg/m3)

No. of tube

=

= No. of tubes

=

Tube side (Ethylene oxide + H2O) (65oC) 4.840 (KJ/Kg oK) 4.187 x 10-4 (Kg/m.Sec) 0.54 (w/m.oK) 973.09 (Kg/m3)

Volume of reactor Volume of one tube 2.56 0.001963 1304 Nos.

54

 Area of tube per pass: Atp

= =

Atp

=

π (d) 2 (Ntp) 4 π (2.5 x 10-2)2 (1304) 4 0.64 m2

 Velocity: m U= ρAtp m = 12.996 Kg/Sec 12.996 ∴U = 973.09 x0.64 U = 0.02 m/sec Now,

NRe

= =

Now,

NRe

=

AO

= = =

AO

duρ υx (1 − ε ) (2.5 x 10-2) x (0.02) x (973.09) (4.187 x 10-4) x (1 -0.6) 2905.09 Nt x π x d xL (1304) x π x (2.5 x 10-2) x (4) 409.66 m2

 Shell diameter: CL  Aox( PR 2 xdo)  0.637 L  Ctp 

0.5

Ds = Consider the Triangular pitch CTP = 0.9 CL = 0.7 Take PR

= =

Pube pitch ratio 1.25

Ds

=

0.637

Ds

=

1.123 m

409.66 x (1.25)2 x (2.5 x 10-2) 4

0.7 0.9

Now, No. of tubes that can be accommodate Nt

=

__(Ds)2

0.875 CTP

(PR)2 (d)2

CL =

____(1.123)2

0.875 0.9

(1.25)2 (0.025)2

0.7 55

0.5

=

1452.8 > Total No of tube that is required.

 Shell side H.T.C : (hoxDe) k

 DexGs  = 0.36 x   µ 

0.55

 Cpxµ x  k

  

0.333

µ  x b   µw 

0.14

For triangular pitch De

=

√3 PT2 – πd2

4

2

4 πd

Pitch ratio

= PR 1.5=

De

=

PT d

PT 0.0025

PT

= 0.03125 m

=

4

√3 (0.31252 – π x (2.5 x 10-2)2 2

4 π x 0.0025

De =

As

0.018 m

Gs =

m As

=

DsxCxB PT

C

= = =

PT – do 0.03125 – 0.025 0.00625

B

= = =

0.4 Ds 0.4 x 1.125 0.45

∴ As

=

1.123 x 0.00625 x 0.45 0.03125

∴ As = 0.101 m2 m ∴ Gs = As 56

=

12.996 0.10

∴ Gs

= 128.58 Kg/m2.sec

From the above equation, (hox 0.018) 0.62 ho

 0.018 x128.58  0.36 x  9 x10 − 4   w 1825.04 20 m k

= =

0.55

 5.1865 x103 x9 x10−4   x 0.62  

 Tube side H.T.C: Nu

0.023 (NRe)0.8 (Pr)0.4

=

hixdi k

0.8 = 0.023 x(2905.09) x(

(Cpπ ) 0.4 ) k

hix 0.025 (4.84 x103 x 4.187 x10−4 ) 0.4 = 0.023 x(2905.09) 0.8 ( ) 0.54 0.254 hi

=

672.163

w m 20 k

 Over all H.T.C 1 Uo

1  1   + + 0.005   ho hi 

= =

Uo =

1  1  + + 0.005    1825.08 672.163  142.133

w m 20 k

Now, Total H.T area available = 409.66 m2 Rate of H.T. is given by ∴Q

= UA∆T

(1.1956 x 106) = A

=

117.03 x A x 20 420.59 m2 > A available

 Tube side pressure drop : Tube side pressure drop is given by 57

0.333

x(1)

0.14

∆P

gc x (∈)3 x Dp x ρ

Z

150 (1 – ∈)

= (1 – ∈) (G’)2

Nre

(∆P/Z) (1) (0.6)3 (1.2 x 10-3) x (973.09)

=

(0.4) (19.4608 )2 ∆P Z

+ 1.75

=

150 x (0.4) 2905.09

1.063 KN /m2 < 30 - 40 KN/m2

 Shell side ∆P: ∆P

f x (Gs2) x (Nb + 1) x Ds

=

2 x ρ x Dex φ f

s

=

exp [(0.57 – 0.19 ln (Res)]

=

exp [0.57 – 0.19 ln(2571.6)]

f

= 0.3977

φ s

= 1

Nb

= L/B = 4 / 0.45 = 8.88 ≅ 9

∴∆ P =

0.3977 x 128.58 x (9 + 1) x 1.123 2 x 995.40 x 0.018 x 1

∴∆ P =

15 N / m2

9.1.2 Mechanical Design:[23,24,26] Internal pressure inside the reactor = P = 1.5 MPa = 1.5 MN/m2 Design pressure = Pd = 1.05 P = 1.05 x 1.5 = 1.575 MN/m2 Vessel is IS: 2002-1962 Class 2B Vessels, So, allowable stress =98.1 M N / m2 Welding joint efficiency factor = J = 0.85  Shell Design: Thickness of Vessel based on internal pressure, t=

Pd D +C 2 f J − Pd

58

+ 1.75

=

1.575 x 1.123

+ 0.2

2 x 98.1 x 0.85 - 1.575 = 12.7 mm But standard shell thickness available in the the market = 14 mm So

ts = 14 mm

9.2 DISTILLATION COLUMN (PACKED COLUMN):

D= 184.54 kmol = 11448.8616 kg

D I S T I L L A T I O N

F = 188.306 kmol = 11766.873 kg MEG = 186.218kmol =11545.545kg HG =2.088kmol = 221.328kg.

MEG = 184.355kmol (0.999.high purity) HG = 0.18454kmol

W = 3.766 kmol = 317.664 kg MEG = 1.8523kmol HG = 1.9136kmol

Steam

Condensate

PROCESS DESIGN: Let , Mole fraction of MEG

XF= 0.9889 XD = 0.999 XW = 0.4918

9.2.1 Nos of theoretically stages Use McCabe Theile method for determining the theoretical stages. McCabe Theile Method: [17,21] 59

Assumption: Binary mixture separation between MEG & DEG. P So, at top

= total pressure of the system = 10 mmHg

Psat MEG = 1.334 KPa Psat DEG = 0.1331 KPa αtop

So,

=

Psat MEG / Psat DEG

=

1.334 / 0.1333

= 10.077 Psat MEG = 1.4757 KPa

Now at bottom

Psat DEG = 0.1673 KPa αbottom = Psat MEG / Psat DEG

So,

= 1.4757 / 0.1673 = 8.8174 Thus, αaverage

= { αtop x αbottom }0.5 = {10.007 x 8.8174}0.5

αaverage = 9.3933 Now we have the equation Y = α X / {1+ (α-1)} From the above eqn we can generate the vapor- liquid equilibrium data given as follow. Table-9.2 Vapor-Equilibrium data X 1 0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Y 0 0.51069 0.70134 0.80102 0.8623 0.9037 0.9337 0.9563 0.9740 1

Now plotting x-y diagram & Using McCabe Theile method for determining the theoretical stages. no. of theoretical stages = 18 9.2.2 Packed Column Diameter And Column Height: The necessary data required is as follows : 60

Reflux Ratio

= 0.71

Feed

= saturated liquid (assume)

Kmol of Feed F

= 188.306 kmol

Kmol of Distillate D

= 184.54 kmol

Kmol of Bottom W

= 184.54 kmol

Liquid density ρL

= 954 Kg/m3

Vapor Density ρG

= 2.099Kg/m3

Liquid Viscosity µ

= 2.2 x 10-3 Kg/m s

No. of theoretical stages = 18 ( from vapor liquid Equilibrium diagram) Owing to its low pressure drop per theoretical stage metal Pall Ring is often preffered to other packings for vacuum distillation. Here, L/D =0.71 L = 131.0234 Kmol / hr Vapor rate = V = L+D = 184.54 + 131.0234 = 375.5634 Kmol / hr.  Mass flow rate of vapor Mg =

(375.5634 x 62.044) (3600) M g = 6.4726 Kg /s

 Mass flow rate of liqid Ml =

131.0234 x62.044 3600

M l = 2.2581 Kg /s  Flow parameter X=

Ml Mg

 ρ g     ( ρ g − ρ l ) 

--------------------------- (2)

Where, X = flow parameter Ml = Mass flow rate of liquid = 2.2581 Kg/s Mg = Mass flow rate of vapor = 6.4726 Kg/s ρg = Density of vapor ρl = Density of liquid

=2.099Kg/m3 =954 Kg/m3 61

From equation (2) =

2.2581 6.4726

2.099 (954 − 2.099)

X = 0.01638 By using plot of flooding and Pressure drop.  σ g     (σ g − σ l ) 

Ml X= Mg

Y= V/S

G ′ 2 f P µ l0.1

σ

g

(σ

g−

σl )

Here column is operated under vacuum, plot value at X = 0.01638 Take, ∆P = 50 Z

N

m 2 ref. m

We get Y = 0.0125 Y=

G ′ 2 f P µ l0.1

σ

g

(σ

g−

σl )

----------------------------------------- (3)

Owing to its low-pressure drop, the metal pall ring is used as a packing for vacuum distillation, Pall Ring (Metal): Size: 38 mm (1.499”) Packing Factor fp (m-1) = 131 Deigned H.E.T.P. (m) = 0.5425 µ = Viscosity of liquids = 2.2 x 10-3 Kg/m s ρg = Density of gas = 2.099 kg / m3 ρl = Density of liquid = 954 kg / m3 G' in kg / m2. s From equation (3) Y= G'2

G ′ 2 f P µ l0.1

σ

g

(σ

g−

σl )

= [(0.0125 x 2.099 x (954 – 2.099)] / [131 x (2.2 x 10-3) 0.1 ]

G'2 = 0.3515 kg / m2 s G’ = 0.5929 kg / m2 s Taking vapor mass velocity as 60% if flooding velocity. 62

Hence, Area of tower = At =

Mg 0.6 G ′

= 6.4726 / (0.6 x 0.5929) = 18.19 m2 At =

π 2 D 4 = 18.19

Diameter of tower = D = 4.81 ≈ 5 m Total height of packed bed, Z = No.of theoretical stages x H.E.T.P. HETP = height equivalent to one theoretical plate For pall ring metal (1.5”) H.E.T.P. = 0.5425 Z = No .of theoretical stages x H.E.T.P. Z = 18 x 0.5425 Z = 9.765 m = 10 m In packed tower, height of each bed is approximately ≥ its diameter. Hence, 2 beds of 5 m heights are provided. Spacing between each bed is = 2 m Hence, total spacing = 0.5 x 3 = 1.5 m Taking both disengaging space and distribution space = 2 m Hence total height of tower

=

10 + 2 + 2

= 14 m MECHANICAL DESIGN [23, 24, 26] 9.2.3 Shell Design: Shell Thickness (based on external pressure): Diameter of tower = Di = 5m Height of column = 14 m Pressure inside column is vacuum. Outside pressure is 1 atm = 0.1 MN /m2 Hence design pressure = Pd = 0.1 x 1.05 = 0.105 MN /m2 Shell is I.S. 2825-1969

63

Allowable stress = f = 98.1 MN /m2 Welding joint efficiency factor = J = 0.85 Thickness of shell, The inside depth of the end can be calculated from the following correlation, hi = Ri -

( Ri − Di / 2) x( Ri + Di / 2 − 2ri

Ri = 5m Di = Do =5m (initial approximation) ri = 0.1 x Ro = 0.5m hi = 5 -

(5 − 5 / 2) x(5 + 5 / 2 − 2 x0.5)

= 0.9686 m Effective length of tower W/O stiffner = (Tangent to Tangent length + 2/3 x hi) = 12 + 2/3x 0.9688 = 12.6459m Do / L = 5/ 12.6459 = 0.3953 K value corresponding to this = 0.246 & m =2.43 Modulus of elasticity E = 1x 105 MN/m2 Corroded shell thickness, P = KxE x (t/Do)m 0.105 = 0.246 x 1 x 105 (t/4)2.43 t = 0.02467 m Take standard thickness = 26 m t = 26 m Checking for plastic deformation, P = 2x f x t/Do x

1 1 + [1.5U (1 − 0.2 Do / l ) / 100 x(t / Do)]

U = 1.5 % (take for new vessel) P = 2 x 98.1 x 4.82 x10-3 x

1 1 + [1.5 x1.5(1 − 0.2 x0.3953 / 100 x 0.00482)]

= 0.2446 › 0.105 MN/m2

9.2.4 Head Design: Selecting Torispherical head 64

Outside diameter of shell, Do = Di + 2 ts = 5 + 2 (0.026) = 5.052 m So, Ro = 0.9 Do = 0.9 x 505.2 = 454.68 cm Temperature = Td

= T + 50 oC = 95 + 50 = 145 oC

Table-9.3 Head thickness t(cm)

Ro / t

Ro / 100t

Factor B

(assumed) 1.2 1.3 1.4 1.5

378.9 349.75 324.77 303.2

3.78 3.49 3.24 3.03

4400 4500 4800 5000

P=

factB 14.22 xRo / t

0.7402 0.8201 0.9421 1.051› Pdesign

So, we take thickness of head t = 1.5 cm Consider 2mm corrosion allowance,

th = 1.5 +0.2 th = 17 mm Standard thickness available

th = 18 mm

9.2.5 Check For Stresses In MEG Column [Design Of Vertical Column]: DATA: Vessel Type = Class 1

Pdesign =1.05 MN/m2 Total Column height = 14 m Column Diameter = 5m Skirt height = 4 m Packing Type = 38mm Metal Pall ring Column MOC = SS304 Wind Pressure = 1000 N/m2 Weight of attachments (ladder + pipes + liquid distributor) = 1373.4N ρliquid = 1113.2 Kg/m3 =10.917 x 103 N/m3 65

Head = Torispherial Head thickness =18mm Tdesign = 145o C Permissible Stress =91.8 MN/m2 ρs = 78480 N/m3 Insulation = Asbestos ρin = 5.64*103 N/m3 tin =50 *10-3 m Shell thickness without corrosion allowance = 26 = 0.026m Do = 5.052m Vessel is Located in non seismic zone  Determination Of Longitudinal Stresses: •

Axial Stress due to pressure

ƒp = P x Do / 4 x ts = 1.05 x 5.052 / 4 x 0.026 = 51.005 MN/ m2 Axial Stress Due to dead Load up to X m (a) Stress due to Shell weight ƒs fs = Ws / Π x D x ts Ws = weight of shell = Π x D x t x ρs x X fs = Π x D x t x ρs x X / Π x D x ts =

ρs x X = 0.7848 X MN / m2

( b) Stress due to insulation ƒin ƒin = tin x ρin x X/ ts = 0.05 x ( 5.64*103 ) x X / 0.026 ƒin = 0.01084 X MN / m2 (c ) Stress due to liquid and packing load in column Volume of bed = Π x D2x H /4 = Π x 52 x 10

= 196.3495 m3 66

Consider void fraction for packing є = 0.5 Liquid hold up = Vb x є

= 196.3495 x 0.5 = 98.1747 m3 Weight of liquid = Vl x ρl = 137.44 x 954 = 93658 kg = 918.511 x 103 N Total weight of packing = packed bed volume x density of packing = Π x D2x H /4 x ρp = Πx52x 10 /4 x 430 = 83.43 x 103 kg = 824.08 x 103 N ƒl = Weight of (liquid + packing) / π x D x ts Weight of ( liquid + packing) =Weight of liquid + total weight of packing = ( 918.511 x 103 + 824.08 x 103 ) = 1.7465MN ƒl = 1.7465 x X / π x ( 5.052) x( 0.026 ) = 4.243 x X MN/m2 ( d ) Stress due to attachments Weight of head = π/4 x D2 x thead x ρs = 0.785 x ( 5.052 )2 x ( 0.018 ) x ( 784480 ) Whead =0.02831 MN Weight of ladder + liq distributor = 1373.4 x X N ƒa = Whead + (Wladder & liquid distributor) / π x Di x t = 0.02831 + 0.0013734 x X / π x ( 5 ) x (0.026) ƒa = 0.06931 + 0.003362 x X MN/m2 Total dead load stress ƒdead = ƒs + ƒin + ƒl + ƒa = 0.7848 X + 0.01084 X + 4.243 X + 0.06931 + 0.003362 X

67

ƒdead =0.06931 +4.3246 xX N/m2 

Longitudinal Stress due to dynamic loads The axial Stress due to wind load in self supporting tall vessel, The total load due to wind acting on the bottom and upper part of thr vessel are determined by Pw = K1 x K2 x Pwindpress x H x Do ( insulated ) Where, Pw = wind pressure = 1000 n/m2 K1 = coefficient depending upon the shape factor = 0.7 for cylindrical surface K2 = coefficient depending upon the period of vibration of the vessel = 1 if period of vibration is 0.5 sec or less =2 if period of vibration is exceed 0.5 Do( insulated ) = Di + 2 tin Period of Vibration is given by T = 6.35 x 10-5 x ( H / D )1.5 x ( W / t ) 0.5 H = L + Skirt height W = total weight Now, Weight of shell Ws = π x Dits x L x ρs = (3.14) x (5) x (0.026) x ( 14 ) x (78480) Ws = 0.4487 M N Weight of insulation Win = π x D x tin x L x ρin = ( 3.14 ) x ( 5 ) ( 0.05 ) x ( 14 ) x ( 5.64*103 ) Win = 0.06266 MN Weight of ( liquid + packing) = 1.7465 MN Weight of attachment = Weight of two heads + Weight of ladder, pipes, liq distributor = 2 (0.02831) + 1.3734 x 10-3 Wa = 0.058 N Total Weight = WS + Win + W (liquid + packing) + Wa = 0.4487 + 0.06266 + 1.7465 + 0.058 W = 2315.86 KN H = L + Skirt height 68

= 14 + 4 H = 18 m a) Period of Vibration T = 6.35 x 10-5 x ( H / D )1.5 x ( W / ts ) 0.5 T = 6.35 x 10-5 x ( 18 / 5.052 )1.5 x ( 2315.86 / 0.026) 0.5 T = 0.127 sec < 0.5 sec

∴ K2 =1

K1 = 0.7 for cylindrical vessel

b) Wind load Pw = K1 x K2 x Pwindpress x H x Do ( insulated ) Do( insulated ) = Di + 2 tin = 5 + 2 (0.05) Pw = ( 0.7 ) x ( 1 ) x ( 1000 ) x ( X ) x ( 5.1 ) Pw = 3.571 x X KN / m •

•

Do = 5.1 m

Bending moment i.e Wind moment at the base of the vessel due to wind load is given by Mwind = { Pwmd x H }/2 = 3.571 x X x X /2 Mwind = 1.785 x X2 KN m Resulting bending stress i.e Wind Stress in axial direction is given by ƒwind = 4 x Mwind / π x ts x D2 = 4 x ( 1.785 x X2) / 3.14 x ( 0.02 6) x ( 5.052 )2 ƒwind

= 0.00349 x X2 MN/m2

 Calculation of Resultant longitudinal stress : Tensile stress on upwind side at X m from top ƒtensile = -ƒp - ƒdead + ƒwind f (max) = f x J = 98.1 x 0.85 = 83.385 MN/m2 Substituting f max for f tensile 83.385 = 0.00349 X2 - 4.3246X - 0.0693 – 51.005

69

Solving this equation 0.00349X2 – 4.3246 X -134.46 = 0 ax2 + bx + c =0 By Solving equation X = 1496.59 >>> 18m So This design is Ok Compressive stress on downwind site at X m from top Down wind side , f compression =ƒp + ƒdead +ƒwind ƒcompressive = 0.125 x E x ( t / Do ) = 0.125 x ( 2 x 105) x ( 0.026 / 5.052 ) = 128.661 MN/m2 Equating maximum value of f comp. 128.661 = 0.00349X2 – 4.3246 X+ 0.0693 – 51.005 Solving this Equation ∴ 0.00349X2 – 4.3246 X -77.5867 = 0 ax2 + bx + c = 0 X = 18.683 >>> 18 m So, This design is ok. 9.2.6 Skirt Support Design For MEG Column: 

The minimum weight of vessel with two head and the shell is given by Wmin = Ws + 2 Whead = 0.4487 + ( 2x 0.03831) Wmin =0.50532 MN



The maximum weight of vessel with two head and the shell is given by Wmax = Wmin + Winsulation + Wattachment + W( liquid + packing ) = 0.50532 + 0.062665 + 0.058 + 1.74657 Wmax = 2.37248 MN  Period of Vibration is given by At minimum dead weight Tmin = 6.35 * 10-5 x ( H / D )1.5 x ( Wmin / ts )0.5 = 6.35 * 10-5 x ( 18 / 5.052 )1.5 x ( 505.32 / 0.026 )0.5 Tmin = 0.059 sec < 0.5 sec ∴ K2 =1 K1 = 0.7 for cylindrical vessel At maximum dead weight 70

Tmax = 6.35 x 10-5 x ( H/ D )1.5 x ( Wmax / ts )0.5 = 6.35 x 10-5 x ( 18 / 5.052 )1.5 x ( 2372.48 / 0.026 )0.5 Tmax = 0.13 sec < 0.5 sec ∴ K2 = 1 K1 = 0.7 for cylindrical vessel 

Minimum and maximum wind moments are computed by Maximum and minimum Wind load Pw = K1 x K2 x ( Pw ) x H x Do (based on insulation thickness) Do = Di + 2 x tin = 5+ 2 x ( 0.05 ) = 5.1 m For minimum Weight condition Do = 5.052 m For maximum Weight condition Do = 5.1 m ( insulated ) Pwmin = 0.7 x ( 1 ) x ( 1000 ) x ( 18 ) x ( 5.052 ) = 63.665KN Pwmax = ( 0.7 ) x ( 1 ) x ( 1000 ) x ( 18 ) x ( 5.1 ) = 64.26 K N Wind moment Mwind = Pw x H / 2 Mwindmi = 63.665 x 18 / 2 = 572.89 KN m Mwinsmax = 64.26 x 18 /2 = 578.34 KN Wind stress ƒwind ƒwind min ƒwind max ƒdead,min ƒdead max

= 4 x Mwind / π x Di2 x tsk = 4 x 572.89 / 3.14 x ( 5 )2 * tsk = 30.277 / tsk KN / m2 = 4 x 578.34 / 3.14 x ( 5 )2 tsk = 29.4821 / tsk KN / m2 = Wmin / π x Di x tsk = 0.50532 / 3.14 x 5 x tsk = 0.0328 / tsk M N / m2 = Wmax / π x Di x tsk = 2.37248 / ( 3.14 ) x ( 5) x tsk = 0.151 / tsk MN / m2

Maximum tensile stress without any eccentric load is given by ƒtensile = ƒwindmin - ƒdeadmin 71

Take J = 0.7 for double weld joint ∴ ƒtensile = ƒpermissible x J = 98.1 x 0.85 = 83.385 MN/m2 ƒtensile = ƒwindmin - ƒdeadmin 83.385 = ƒwindmin - ƒdeadmin 83.385 = 0.0328 / tsk + 0.030277 / tsk tsk = 3.11 x10 5 m Maximum tensile stress without any eccentric load is given by ƒcompressive in skirt wall = 0.125 E tsk / Do = 0.125 (2 x105) tsk / 5.052 = 4948.53 tsk ƒcompressive in skirt wall = ƒwindmax + ƒdeadmax 4948.53 tsk = 0.02948 / tsk + 0.151 / tsk tsk = 6.054 mm But minimum corroded skirt thickness = 8mm (as per IS : 2825-1969) ∴ Skirt thickness tsk = 8 Table-9.4 Specification for Distillation Column (Packed column) SR. NO. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

PARAMETER

DESCRIPTION

Tower MOC Tower ID Tower OD Shell thickness Shell head thickness Height of tower(Without support) Height of packed Bed Type of Head No. of Beds of packing ∆P/Z Selected Tower Support Skirt MOC Skirt Height Skirt thickness Type MOC Bulk Density HETP

72

SS304 5m 5.052 m 26 mm 18mm 14m 10 m Torrispherical 2 , each of height 5 m 50 N/M2/m Skirt Support SS304 4m 8 mm Pall Ring Metal (S S) 430 kg/m3 0.5425m

CHAPTER X INSTRUMENTATION AND CONTROL

10.1 WHY REQUIRED? Instruments are provided to monitor the key process variables during plant operation. Instruments monitoring critical process variables will be fitted with automatic alarms to alert the operators to critical and hazardous situations. The primary objectives of the designer when specifying instrumentation and control schemes are: 10.1.1 Safe Plant Operation: 

To keep process variables within known safe operating limits.



To detect dangerous situations as they develop and to provide alarms and



Automatic shutdown systems.



To provide interlocks and alarms to prevent dangerous operating procedures.

10.1.2 Production Rate and Quality: 

To achieve the designed product output.



To maintain the product composition within the specified quality standards.

10.1.3 Cost: 

To operate at the lowest production cost and to compensate with other

objectives. Process instrumentation is thus brain and nerves of a process plant. The instrumentation can be pneumatic, hydraulic or electric. The recent trend is to go for electronic instrumentation, but pneumatic instrumentation is still in use. The instrumentation is required to measure temperature, pressure, flowrate, level, physical properties as density, pH, humidity, chemical composition etc. 10.2 TYPICAL MONITORING SYSTEMS: 73

10.2.1 Flow Measurement: Due to nature of flow it is necessary to provide effective flow measuring devices in each supply lines. The various types of flow meters available are orifice meter, venturi meter, pitot tube etc. In spite of these the various types of area flow meters can also be used. Depending on temperature and velocity condition the suitable meter is selected for measurement of flow rates and velocity. 10.2.3 Temperature Measuring Devices: Many devices are used to measure the temperature variations in the process such as mercury in glass thermometer, bimetallic thermometer, pressure spring thermometer, thermocouples,

resistance

thermocouples,

radiation

pyrometers

and

optical

pyrometers are used. Out of all these the industrial thermocouples are competitively good as they provide large measuring range, without introducing error. Automatic control is also possible with such devices. Table-10.1 List of Thermometers with temperature Range Measuring Instrument Mercury in glass – thermometer Mercury in pressure thermometer Vapor pressure thermometer Resistance Thermocouple Thermister Pyrometer

Temp. Range ºC -27 to 400 -40 to 540 -85 to 425 -200 to 1700 -250 to 1700 Up to 300 1300 to 2500

10.2.4 Pressure Measuring Devices: Equipments, in which the important monitoring parameter is pressure, pressure measuring devices like pressure gauges are widely used. Safety of chemical plants depends up on the timely measurement of pressure and its control at a specified level. Any excess pressure development than the design pressure may damage the equipment in addition to the fire and other explosion hazard. Mainly in filter pressures where the pressure is an important criterion, this device is used. Various pressure measuring devices are: •

U – Tube Manometer 74

•

Differential Manometer

•

Inclined Manometer

•

Bourdon Tube

•

Bellows

•

Diaphragm valve

•

Mc Leod gauge

•

Pirani gauge

In addition to all measuring devices described above various measurements like density, viscosity, pH measurements etc. are installed. For measuring quality standards in laboratory various laboratory instruments are also necessary. 10.2.5. Liquid Level: Liquid level detectors measure either the position of a free liquid surface above a datum level or the hydrostatic head developed by the liquid is measured. The liquid level is measured both by direct and indirect means. Direct methods involve direct measurement of the distance from the liquid level to a datum level. Indirect method follows changing liquid surface position on bubble tube method, resistance method, radiation method, etc. 10.3 DISTRIBUTED CONTROL SYSTEM (DCS): Caprolactam production process deals with the benzene and cyclohexane which are having low boiling points. So the process is risky and also the product quality is important. Therefore for the faster control DCS can be used. It provides ease of constant monitoring the process at a distance much far away from the site and the changes can be made in the process parameters very accurately from the control room itself. 10.3.1 Merits of DCS: 1. From quality point of view: 

More accuracy and reliability.



Self tuning of any control loop is possible, so optimization of any process is

possible. 2. Management Consideration:  Less cost of cables. 75

 Less cost of installation.  Less space required.  Less hardware required.  Inventory information can be made available.  Less man power is required.  Less production cost.  Management information can be generated at regular intervals which assist management to take decisions. 3. Operational point of view:  Ease in operation.  Any combination of control group, trend group, over view path can be formed.  Because of dynamic graphic role picture of process is available.  Easy diagnostics of trip and emergency conditions.  Automatic logging of data is done by printer and hence eliminating weakness related human being.  Control is available through dynamic graphics which gives feeling to operator as if he is inside the plant and controlling the process.  Alarm systems can be regrouped to various sub groups so that operator can detect the error and causes easily. 4. Engineer’s Point of view:  Latest software is available for all types of complex function.  Required less time for designing and detail engineering.  Operators action can be logged which eliminates confusions in the event of plant trips and consequent analysis.  Flexibility is available at each level of hardware and software. 10.3.2 Demerits of DCS: In present control room lot of paramagnets are seen without any intentional, efforts hence operator feels himself existing in between the information. In case of new DCS systems, all information and data though presented in a systematic format, is hidden behind the CRT and hence to be called by operator. This requires more skill and knowledge. With acceptance of DCS, number of operators in

76

control room decreased and hence, in case of emergency decision has to be taken by almost single handed as against group decision in present situation. In single loop control system failure of one controller affects only one control loop, while case of DCS one component / card carries out lot of functions and hence failure of it causes failure of many loops. This calls for very high MTBF (Mean Time Between Failures) and high degree of redundancy making such systems costly. A limitation may be felt in operating number of control loops simultaneously in case of emergency, if adequate numbers of CRT consoles are not installed. personnel are required.

77

Skilled

CHAPTER XI PLANT UTILITY

The utilities such as water, air, steam, electricity etc. are required for most of the chemical process industries. These utilities are located at a certain distance from processing area, from processing area hazardous and storage area etc, where a separate utility department works to fulfill the utilities requirements.  Steam Generation  Cooling water  Water  Electricity  Compressed air

The utilities required for the plant are summarized as below. 11.1 STEAM GENERATION: Steam is used in plants for heating purpose, where direct contact with substance is not objectionable. The steam, for process heating, is usually generated in water tube boiler using most economical fuel available i.e. coal, fuel oil on the site. In reboiler of distillation column,drying column and evaporator steam is used at different temperature depending on requirement. 11.2 COOLING WATER: Cooling water is generally produced in plant by cooling towers. Cooling tower is used to cool the water of high temperature coming from process. Cooling tower mainly decreases temperature of water from process. There are two types of cooling tower. 11.2.1 Natural Type: In this cooling tower the water from the process is allowed to fall in a tank. From some height when falling it comes in contact with an air & gets cool. 11.2.2 Mechanical Type: 78

They are classified in three types:  Induced draft  Forced draft  Balanced draft In induced draft a fan is rotating at the bottom while in balanced draft fan is rotating at the centre. In forced draft a fan rotating at top.  Cooling by sensible heat transfer  Cooling by evaporation 11.3 WATER: A large reservoir has to be made which received water from nearby river. Storage also must provide to such extent that turbidity is settling and then sent to raw water plant for further treatment. Chlorine dose must be given to kill bacteria which prevent organic matter. Then this water is sent to further treatment. To cooling tower, DM plant, service water system, drinking water system, fire water system. Cooling water is required for heat cooler, condenser etc. for cooling effect. Here in cooling tower water is fall from high level and contacted with cross flow of air. Latent heat of water is high that even a small amount of water evaporates produce large cooling effect. The temp of CW is up to 25 to 30 ˚C. DM water is use for process . DM water is produced by removing impurities salts, pass through anion exchanger. 11.4 ELECTRICITY: It is required for motor drives , lighting and general use. It may be generated on site or purchase from GEB & G.I.P.C.L. Transformers will be to step-down the supply voltage to the voltage used on the site. A three-phase 415-volt system is used in general industrial purposes and 240-volt single phase for lighting and other low power requirements. For large motors, high voltage 600 to 1100 is used. 11.5 COMPRESSED AIR: Compressed air is used during the chocking of pipes and for cleaning purpose. Compressed air can be obtained from air compressor.

79

CHAPTER XII SAFETY, HEALTH AND POLLUTION CONTROL

12.1 SAFE OPERATIONS: The goal of chemical plant is not only to produce the chemicals, but to produce them safely. In the plant’s chain of processes and operations, loss of control anywhere can lead to accidents and losses of life and property from hazards. Attempts should be to prevent troubles from the inspection, while designing, fabricating and operating. Safety generally involves: (1)

Identification and assessments of the hazards

(2)

Control of hazards

(3)

Control of the process by provision of automatic control system, interlocks, alarm trips, etc

(4)

Limitation of the loss, by press relief, plant layout, etc.

12.2 MSDS FOR ETHYLENE OXIDE: [10] 



MATERIAL NAME:

ETHYLENE OXIDE

USES:

Chemical intermediate

SYNONYMS:

Oxirane

HAZARDS IDENTIFICATION

Appearance and Odour: Clear liquid under pressure. Sweet Ethereal Health hazards: Toxic by inhalation. Irritating to respiratory system. Causes burns. May cause cancer. Environmental Hazards: Harmful to aquatic organisms. May cause long-term adverse effects in the aquatic environment. Health Hazards Inhalation: Toxic by inhalation. Vapours may cause drowsiness and dizziness. 80

Skin Contact: Exposure to rapidly expanding gases may cause frost burns to eyes and/or skin. Liquid solutions of ethylene oxide cause serious chemical burns of the skin and eye lesions. Onset of effects may be delayed for several hours. Skin Protection: Wear protective gloves and clean body-covering clothing. Eye Contact: Couses burns Eye Protection: Use chemical safety goggles. Maintain eye wash fountain and quickdrench facilities in work area. FIRST AID MEASURES:

General Information: Do not attempt to rescue the victim unless proper repiratory protection is worn. Inhalation: Remove to fresh air. Do not attempt to rescue the victim unless proper Expiratory protection is worn Skin Contact: Remove contaminated clothing. Flush exposed area with water and follow by washing with soap if available Eye Contact: Immediately flush eyes with large amounts of water for at least 30 minutes while holding eyelids open. Transport to the nearest medical facility for additional treatment. Advice to Physician: Contact a Poison Control Center or toxicologist for guidance.

FIRE FIGHTING MEASURES:

(Clear fire area of all non-emergency personnel.) Flash point

-57oC / -71oF (PMCC / ASTM D93)

Explosion

2.6 – 99.99% (V)

Flammability limits in air Auto ignition: 428oC / 802oF Specific Hazards: 81

The vapour is heavier than air, spreads along the ground and distant ignition is possible. Sustained fire attack on vessels may result in a Boiling Liquid Expanding Vapour Explosion (BLEVE Extinguishing Media: Shut off supply. If not possible and no risk to surroundings, let the fire burn itself out. Unsuitable:

Do not use water in a jet.

Extinguishing Media Protective Equipment : Wear full protective clothing and self-contained breathing apparatus. Additional Advice: If the fire cannot be extinguished the only course of action is to evacuate immediately. Large fires should only be fought by properly trained fire fighters. Evacuate the area of all non-essential personnel. 

ACCIDENTAL RELEASE MEASURES:

Protective measures: Avoid contact with spilled or released material. Isolate hazard area and they deny entry to unnecessary or unprotected personnel. Stay unwinds and keeps out of low areas. Extinguish ant make flames. Do not smoke. Remove ignition sources. Avoid sparks. Clean Up Methods: Use water spray (fog) to reduce vapors or divert vapour cloud drift. Do not use water in ajet. Alcohol foam applied to surface of liquid pools may slow release of EO vapors into the atmosphere.

HANDLING AND STORAGE:

Handling: Ventilate workplace in such a way that the Occupational Exposure Limit (OEL) is not exceeded. The vapor is heavier than air spreads along the ground and distant ignition is possible. Electrostatic charges may be generated during pumping. Electrostatic discharge may cause fire. Storage: Ethylene oxide (EO), an extremely flammable and toxic gas, and other hazardous vapours may evolve and collect in the headspace of storage 82

tanks, transport vessels and other enclosed containers. Storage Temperature: 30oC / 86oF maximum. Product Transfer: Electrostatic charges may be generate during pumping. Electrostatic discharge may cause fire. Lines should be purged with nitrogen before and after product transfer. Refer to supplier for further product transfer instructions if required.

DISPOSAL CONSIDERATIONS:

Material Disposal:

Do not dispose into the environment in drains or in watercourses. Waste arising from a spillage or tank cleaning should be disposed of in accordance with prevailing regulation, preferably to a recognized collector or contactor. The competence of the collector or contactor should be established beforehand. Local Legislation: Disposal should be in accordance with applicable regional, national, and local laws and regulations. 12.2 MSDS FOR MONO ETHYLENE GLYCOL (PRODUCT):[10]

 PRODUCT NAME : SYNONYMS:

MONO ETHYLENE GLYCOL

1,2 - ethanediol

 HAZARDS IDENTIFICATION: Color:

Colorless

Physical State:

Liquid

Odor:

Sweet

Hazards of product: May cause eye irritation. Isolate area.  POTENTIAL HEALTH EFFECTS: Eye Contact: 83

May cause slight eye irritation. Corneal injury is unlikely. Vapor or mist may cause eye irritation. Skin Contact: Brief contact is essentially nonirritating to skin Skin Absorption: Prolonged skin contact is unlikely to result in absorption of harmful amounts. Inhalation: At room temperature, exposure to vapor is minimal due to low volatility. With good ventilation, single exposure is not expected to cause adverse effects. Effects of Repeated Exposure: Repeated excessive exposure may cause irritation of the upper respiratory tract. In humans, effects have been reported on the following organs: Central nervous system. Birth Defects Based on animal studies, ingestion of very large amounts of ethylene glycol appears to be the major and possibly only route of exposure to produce birth defects.

 FIRST-AID MEASURES: Eye Contact: Flush eyes thoroughly with water for several minutes. Remove contact lenses after the initial 1-2 minutes and continue flushing for several additional minutes. Skin Contact: Wash skin with plenty of water. Inhalation: Move person to fresh air. If not breathing, give artificial respiration; if by mouth to mouth use rescuer protection (pocket mask, etc).

 FIRE FIGHTING MEASURES: Extinguishing Media:

84

Water fog or fine spray. Dry chemical fire extinguishers. Carbon dioxide fire extinguishers. Foam. Do not use direct water stream. May spread fire.

Special Protective Equipment for Firefighters: Wear positive-pressure self-contained breathing apparatus (SCBA) and protective fire. 

HANDLING AND STORAGE HANDLING:

Handling: Do not swallow. Avoid contact with eyes. Wash thoroughly after handling. Spills of these organic materials on hot fibrous insulations may lead to lowering of the autoignition temperatures possibly resulting in spontaneous combustion. Storage: Do not store near food, foodstuffs, drugs or potable water supplies. Additional storage and handling information on this product may be obtained by calling your sales or customer service contact. Ask for a product brochure. PERSONAL PROTECTION:

Eye/Face Protection: Use safety glasses. If exposure causes eye discomfort, use a full-face respirator. Skin Protection: Use protective clothing chemically resistant to this material. Selection of specific items such as face shield, boots, apron, or full body suit will depend on the task. Remove contaminated clothing immediately. Hand protection: If hands are cut or scratched, use gloves chemically resistant to this material even for brief exposures. Use gloves with insulation for thermal protection, when needed. Examples of preferred glove barrier materials include: Butyl rubber. Natural rubber 85



STABILITY AND REACTIVITY:

Stability/Instability: Thermally stable at recommended temperatures and pressures. Thermal Decomposition Decomposition products depend upon temperature, air supply and the presence of other materials.Decomposition products can include and are not limited to: Aldehydes. Alcohols. Ethers.  TOXICOLOGICAL INFORMATION: Acute Toxicity: Ingestion For ethylene glycol: Lethal Dose, Human, adult 3 Ounces LD50, Rat 6,000 - 13,000 mg/kg Skin Absorption LD50, Rabbit > 22,270 mg/kg Inhalation LC50, 7 h, Aerosol, Rat > 3.95 mg/l 

ECOLOGICAL INFORMATION:

Chemical Oxygen Demand: 1.19 mg/mg Theoretical Oxygen Demand: 1.29 mg/mg 

DISPOSAL CONSIDERATIONS:

All disposal practices must be in compliance with all Federal, State/Provincial and local laws and regulations. Regulations may vary in different locations. Waste characterizations and compliance with applicable laws are the responsibility solely of the waste generator. 12.3 GOOD MANUFACTURE TECHNIQUES TO PREVENT ACCIDENTS Filling drum - Keep hose pipe little inside the drum rather than on the hole. Using fuming chamber - In laboratory while working with hazardous chemicals like H2S, Reduce heat of reaction -Add sulfuric acid to bucket full of water and not water to bucket full of sulfuric acid. 86

Opening flanges - While opening a flange on pipeline containing corrosive liquid, chances of liquid coming out with a spray are there. To avoid accident due to such spray or acid or alkali use plastic sheet while opening valve. So that it will not contact with body. Location of gauge glass - Gauge glass for reading level in the tank should be located away from path where many people may be working. Location of safety valve/ vent line - The vent pipe should not be located in a closed area. Location of flammable material - Storage should be away from any source of flame. Smoking - Do not smoke in unauthorized area where flammable materials are likely to be present. Purging with inert atmosphere - Before entering a reactor or a distillation column containing hazardous vapors, the equipment must be purged with air/inert gas for sufficiently long time. Machinery guards - Install guards on moving machinery parts. Incompatible chemicals - Do not mix incompatible materials together. Earthling of equipment - When two phase mixture is being separated into different tanks, the tank should be earthed to avoid spark due to accumulation of static electricity. Explosion due to dust - In the operation of fine grinding, solid temperature increases which can lead to dust explosion initiated by hot metal. It can be prevented by cooling grinder with water or inert gas purging. Drying and ignition of flammable liquids - Keep air flow rate high so that air vapor mixture is not near flammable limit. Mixing - It should be effective to take care of exothermic heat of reaction. Good house keeping - Do not store waste flammable materials near flame source. Labeling of chemicals - Label the chemicals so that they do not get mixed up with incompatible chemicals. Pipetting - Do not suck with mouth, use rubber bulb. Free excess energy exit - Do not store anything in passage way destructing free movement in emergency. 12.4 FIRE PREVENTION AND PROTECTION: 1.

Regulation for the prevention of Fire: 87

Ban on carrying of a potential source of ignition, Ban on lighting fires in battery area. Ban on smoking. Ban on carrying lamps. Use of Sparks’s arrestors. 2.

General Precautions: Maintain good housekeeping. Follow the laid down procedure strictly. Sampling and draining of hydrocarbon should be done under strict supervision. Do not operate an equipment unauthorized. Use only approved type of tools. Anticipate the hazards during vessel cleaning and take prevention steps in advance.

3.

Fire emergency mock drill:

An emergency manual can be prepared to outline procedures and drills and detail responsibilities of each individual involved.  Training  Valuable Check On The Adequacy and Condition of exits and Alarm system  Instills a Sense of Security Among The Occupiers if Careful Plans Are Made.  Exits Drills  Plant Drills (Mock drills in plant area)  Mutual Aid Drills  On-Site / Off site Drills etc. 12.5 SAFETY IN PROCESS DESIGN: Accidents are minimized by correct deign using scientific and performance data without false economy. 12.5.1 Reactor: The reactor is a heart of plant and vital for safety. Most reactions have hazard potential. Here, reaction is exothermic and at higher pressure compared to atmospheric pressure and also deals with the materials like Benzene and Cyclohexane which are highly volatile. 12.5.2 Heat Transfer: For safe operation,  Prevent mixing.

88

 Provide different surface, for cleaning, insulation, expansion.  Prevent flame travel in furnace.  Use safety over design factor of 15 – 20 %. 12.5.3 Mass Transfer: Safe guards are,  Prevent liquid injection and vigorous flashing in hot column.  Provide both pressure and vacuum relief.  Use detection and warning devices for build up of hazardous material.  Provide thermal expansion in system. 12.5.4 Pressure Vessels: It includes,  Corrosion allowance must be provided.  Take care weld joint efficiency.  Design pressure is maximum operating pressure plus static pressure plus 5 %.  Design temperature is 25-30 ºC above maximum operating temperature.  Use safety over design factor of 15 – 20 %. 12.5.5 Instrumentation and Safety devices:  Thermocouple burnout, stem or cooling water failure.  Fusible plugs to relive pressure above design value.  Combustible gas monitor with alarms for flammable.  Over temperature switch. 12.6 ENVIORNMENTAL CONSIDERATIONS The environmental considerations include: 1. Control of all emission from the plant. 2. Waste management. 3. Smells. 4. Noise pollution prevention. 5. The visual impact. 89

6. Liquid effluent specifications 7. Environmental friendliness of the products. 12.6.1WASTE MANAGEMENT:

Waste arises mainly as byproducts or unused reactants from the process, or as offspecification product produced through mis-operation. In emergency situations, material may be discharged to the atmosphere through vents normally protected by bursting discs and relief valves. 12.6.2 GASEOUS WASTES:

It is to be remembered that practice of relying on dispersion from tall stacks is seldom entirely satisfactory. The gaseous pollutants can be very easily controlled by using adsorption or absorption. Dispersed solids can be removed by scrubbing, or ESP If the gas is flammable it is to be burnt. As in the present case the gaseous waste being carbon dioxide. But the gases should not be sent to vent or to atmosphere and hence the suitable scrubber system requires to be installed down stream to minimize pollution. 12.6.3 LIQUID WASTE:

If the liquid effluent is flammable, it can be burnt in the incinerator. But as in this case if it contains salts; acids and substantial amount of alkali it is to be subjected to effluent treatment. Generally common effluent treatment plant (if the facility is situated in and Industrial area with the CETP) serves the purpose. The level of effluent treatment up to secondary treatment is sufficient for the effluent from the plant like one on the hard. 12.6.3 SOLID WASTE:

Solid wastes can be burnt in suitable incinerators or disposed by burial at licensed land fill sites. Dumping of toxic solid waste should be avoided. 12.6.4 AQUEOUS WASTE:

The principle factors which determine the nature of an aqueous industrial effluent and on which strict controls will be placed by responsible authority are:    

pH Suspended solids Toxicity Biological oxygen demand

For the present case pH of the effluent stream is expected to be alkaline and hence addition of acids is recommended to neutralize the same. The suspended solids can be 90

removed by settling, using Chemical treatment may be given to remove some of the chemicals. Oxygen concentration in waste course must be maintained at a level sufficient to support aquatic life. For this reason the biological oxygen of an effluent is of paramount importance. Standard BOD 5 tests can be applied for the determination of the same. The test measures the quantity of oxygen which a given volume of effluent will absorb in 5 days at constant temperature of 20 0C. It is a measure of the organic matter present in the effluent. Ultimate oxygen demand test can be performed if required. Waste water should be discharged into sewers with the agreement of the local water pollution control authorities or state pollution control boards. 12.6.5 NOISE:

Noise can cause serious nuisance in the neighbourhood of a process plant. Care need to be taken when selecting and specifying equipments such as compressors, air-cooler fans, induced and force draft fans for furnaces, and other noisy plant. Excessive noise can also be generated when venting through steam and other relief valves, and from flare stacks. Such equipments should be fitted with silencers.Noisy equipments should be as far away form the site boundary. 12.6.6 VISUAL IMPACT:

The appearance of the plant should be considered at the design stage. Few people object to fairyland appearance of a process plant illuminated at the night, but it is different scene at daytime. There is little that can be done to change the appearance of modern style plant, where most of the equipment and piping will be outside and in full view but some steps should be taken to minimize the visual impact. 12.6.7 ENVIRONMENTAL AUDITING:

The company should go for a systematic examination of how a business operation affects the environment. It will include all emissions to air, land and water and cover the legal constraints the effect on the community the landscape and the ecology. Following are some of the objectives of the environmental audit:  To identify environmental problems associated with the manufacturing process and the use of the products before they become liabilities.  To develop standards for good working practices.  To ensure compliance with environmental legislation.  To satisfy requirements of insurers. 91

 To be seen to be concerned with environmental questions: important for public relation  To minimize the production of waster: an economic factor

12.7 HAZOP OPERATION OF STORARGE TANK TO REACTOR SYSTEM Table 12.1 Hazardous operation of Storage tank to reactor system GUIDE WORD

NONE

DEVIATION

NO FLOW

POSSIBLE CAUSE

1) No EO/H2O available at the storage tank 2) J1/J2 pump fails (motor, fault, loss of drive, etc.) 3) Line valve close (in error or fail) 4) Line fracture

MORE OF

MORE FLOW

MORE PRES

Control Valve fail to open or LCV by pass open in error 1) V1. Valve closed in error with J1 and J2 pump running. 2) Thermal expansion in a isolated valved section due to fire. 1) High 92

CONSEQUENCES

ACTION REQUIRED

Loss of Feed to 1) Ensure good reaction section and communication with reduced output. Ethylene oxide plant. 2) Install low level alarm on storage tank. Loss of Feed to 1) Check design of reaction section and pumps and install a reduced output. stand by pump. 2) Install a kickback on pump. Loss of Feed to Check Operation of reaction section and Control valve or reduced output. Install a by pass line with other control valve.(stand by ). 1) Loss of feed to Institute regular the reaction section. patrolling and 2) Feed discharge inspection of transfer into area. line. Over filling (excess Check proper feed in reactor) operation of control valve Transfer line subjected to full pump delivery or surge pressure.

1) Install kick back on pump. 2) Install a P.G.

Line fracture or flange leak.

Install thermal expansion relief on valved section

High pressure in

1) Check whether

MORE TEMP

LESS OF

LESS FLOW

MORE THAN

HIGH E.O CONC IN STREAM.

PART OF

WATER FLOW TO REACTOR

OTHER

MAINT.

storage temperature

transfer line

2) cooling water supply to reactor cooling system stop

Affect the rate of reaction.

1) Leaking flanges of valve stab not blanked and leaking. 2) Pump J2 fails or not running properly and J1 running well. Valve V2-1, and V2-2 are not working properly

Material loss to adjacent area

Institute regular patrolling and inspection of transfer line.

Formation of more Higher glycol in the reaction which will affect the product. (excess E.O in reactor) Leads to formation of more higher glycol

Valve V2, and V2-2 are not working properly Equipment failure flange leak etc.

Leads to formation of more higher glycol

1) Install a stand by pump. 2) Check design of pump. 3) Install kickback on pump. 1) Institute regular inspection of valve. 2) Install by pass line with standby Valve. 1) Institute regular inspection of valve. 2) Install by pass line with standby valve. Install low point drain and N2 purge point and N2 vent line.

93

Line cannot be completely drainage or purge.

there is adequate. 2) Warring of high temperature at storage, if not install. 1) Check design of pumps and install a stand by pump. 2) Install a kickback on pump. 3) Check proper operation of cooling water system.

CHAPTER XIII PLANT LOCATION AND LAY OUT

13.1 PLANT LOCATION: Plant location means to discover an exact place where an industrial experience can be started more profitable & a plant is a place where men, material, money, equipment, machinery etc. are brought together for manufacturing products. Plant location involves two major activities. Plant location plays a major role in the design or production as it determines the cost of :  Getting suitable raw material.  Processing raw material to finished products.  Finished products distribution to customers. The final selection of the plant location has a strong influence on the success of any industrial venture. The following eighteen factors should be considered in choosing a plant side. 13.1.1 Raw material supply The source of raw material is one of the most important factors influencing the selecting of the plan. The raw material should be cheaply & regularly available at the plant site because this permits considerable reduction in transportation & storage charges. The major raw material used in this plant is Ethylene oxide. This can be easily available in the places nearer to Baroda (because of the huge plants at IPCL , Baroda itself) & hence any industrial area near by Baroda can be a suitable place for the plant location. Therefore, the industrial area around Baroda can be comfortably chosen as an ideal place for our plant.

94

13.1.2 Markets The location of markets or intermediates distribution centers can heavily affect the cost of product distribution. Primarily to large market can be beneficial in the following three ways:  The cost of transportation of the finished goods to the market is brought drastically down.  The delay in supplying the goods to the market can be continently reduced & avoided.  The market is studied properly & easily i.e. the future requirements can be easily & accurately predicted. 13.1.3 Energy availability Electricity power, steam supply & heating oil requirements are high in most of the chemical plants. The power & fuel can be considered as one major factor in the choice of the plant site. The local cost of power can help in determining whether power should be purchased or self generated. As far as our plant is concerned, both electricity power & fuel (gaseous, liquid or solid) as well as heating oil can be made available easily in Baroda or from nearby sources. 13.1.4 Water supply The chemical process industries use large quantities of water for cooling, heating, washing & as a raw material. Therefore the plant site should be nearer to the source of water. Baroda has plenty of such source like Ajwa lake, Mahi river & so. So, the situation favors Baroda. 13.1.5 Climate Weather can have serious effect on the economic operating of the plant. Temperature & humidity of weather should be favorable. 13.1.6 Transportation The everyday products are always needed to be transported from the plant site to the marked or other plants & the raw materials necessary from the sources to the plant.

95

Hence transportation holds a great deal in the final product cost. A plant should have easy access to transport facilities. Not only that, the transport facilities available to the plant should also be sufficient, quick & available at reasonable rates. Water, railway & national high-ways are the most common means of transportation. These facilities are very much necessary for the transfer of raw material & product transportation. Luckily Baroda has all of these facilities. 13.1.7 Labor supply Availability of skilled laborers with constant supply & reasonable pay rate should be considered in the selection of the plant site. Labor problem should be minimum. 13.1.8 Waste disposal The plant site should be such that it should have the best & adequate facilities for waste disposal. The permissible tolerance levels for various methods of waste disposal should be considered carefully & attention should be given to potential requirements for waste treatment facilities. 13.1.9 Taxes & legal phases The state & local tax rates on property (such as plant machinery, building etc.), income, and unemployment insurance & similar items have major influence on the plant site selection. 13.1.10 Site characterization The characteristics of the land at the proposed plant site should be examined carefully. The topography of the tract of the land & the soil structure must be considered, since either or both have pronounced effect on construction costs. 13.1.11 Fire & explosion protection The site should be such that it should have the best possible & quickest fire protection facility available during the emergency. If possible (means if the company can afford) the plant should be having its own fire station, fully equipped latest fire fighting equipments & skilled firemen team. So, in case of emergencies it won’t have to rely totally on the external sources. 13.1.12 Advanced library & training center 96

This is the characteristic of a good & developed organization. To develop the plant properly, trained staff is a prime requirement & for further research & development of in-house technologies, advance library facilities covering the subjects in detail is necessary. The training center should be fully equipped with skilled trainers & training facilities. 13.1.13 Community attitude Success of an industry depends very much on the attitude of the local people & whether they want work or not. 13.1.14 Presence of related industries This means the industries supplying raw materials or the power or energy requirements should be in a handy reach to avoid chaotic situations to take place.

13.1.15 Existence of hospitals, marketing centers, schools, banks, post offices, clubs An ideal industry or organization is that which takes full care of its employees & persons who are directly or indirectly involved with it. To cope up with the situation of casualties or accidents pressure of a well-equipped hospital is a must. Other than this a reputed school & the banking & postal facilities are the prime requirement of the families of the employees. 13.1.16. Housing facilities Housing facilities ( i.e., residential quarters ) for the company employees should be well maintained & provided with constant supply of water, electricity & things necessary for life. 3.1.17 Securities The security of the plant site & the housing facility from the unsocial elements is necessary & should be given equal attention. 13.1.18 Facilities for expansion

97

Considering all the major factors discussed above affecting the plant location, it is quite reasonable to select Baroda, to establish an industrial estate for the plant location. Justification for the same is discussed below.  All the transportation facilities (rail, road & water) are easily available to Baroda & are very adequate.  Waste disposal will not be a much problem as it is a separate chemical estate & no specific attention is required.  Electricity & water supply are easily in abundant in supply.  The raw materials necessary are easily available from the nearby industrial area & the industrial estate is always running with large number of chemical industries, & hence getting skilled & experienced labor at reasonable rates is not a problem. 13.2 PLANT LAYOUT: After the process flow diagrams are completed & before detailed piping, structural & electrical design can begin, the layout of process units in the plant must be planned. Plant layout means the allocation of space, arrangement of equipment & machinery in such a manner so that maximum utilization of manpower, machines & material is done & minimum material handling is required. The following factors should be considered in selecting the plant layout.  New site development or additions to previously developed site.  Type & quantity of product to be produced.  Possible future expansion.  Economic distribution of utilities & services.  Type of building & building code requirements.  Health & safety considerations.  Waste disposal problems.  Sensible use of floor & elevation space.  Operational convenience & accessibility.  Type of process & product control.  Space available & space required.

98

 Maximum advantages of gravity flow are taken to reduce the operational cost in the piping & flow design. 13.2.1 Storage Layout: Adequate storage of raw materials, intermediate products, final products, recycle materials & fuel are essential to the operating of process plants. Storage of intermediate products may be necessary during plant shutdown for emergency repairs, while storage of final products makes it possible to supply customer even during a plant difficulty of unforeseen shutdown. An additional use of adequate storage is often encountered when it is necessary to meet seasonal demands from steady production. 13.2.2 Equipment Layout: In making plant layout, a due consideration should be given to that an ample space should be assigned to each piece of equipment & their accessories. The relative levels of the several pieces of equipment & their accessories determine their placement. Gravity flow is preferable to reduce material handling cost during production, however it is not altogether necessary because liquids can be transported by pumping & solids can be moved by mechanical means. In making the equipment’s layout, the grouping should be done so that the service of equipment’s performing similar function is grouped together & so the better co-ordination of the operating is achieved.

99

CHAPTER XIV COST ESTIMATION

A plant design obviously must present a process that is capable of operating under conditions, which will yield a profit. Since net profit equals total income minus all expenses, it is essential that the chemical engineer be aware of many different types of costs involved in manufacturing processes. 14.1 ESTIMATION CAPITAL COST: 14.1.1 Purchased Equipment Cost: Table-14.1 Purchase Cost of equipments Sr. No.

1 2 3 4 5 6

7 8 9

Equipment MEG STORAGE TANK HG STORAGE TANK ETHYLENE OXIDE STORAGE TANK PLUG FLOW REACTOR TRIPPLE EFFECT EVAPORATOR VACUUM DISTILLATION COLUMN (PACKED COLUMN) DISTILLATION COLUMN (PLATE COLUMN) DRYING COLUMN CENTRIFUGAL

No. of EquipMents

Cost per Unit (Rs. Thusd)

Estimated Cost (Rs. Thusd)

2

200

400

1

200

200

2

200

400

1

500

500

1

85

85

1

70

70

1

75

75

1 6

50 10

50 60

100

10 11 12 13

PUMP HEAT EXCHANGER BOILER DM WATER PLANT COOLING TOWER

7 2

20 50

140 100

1

25

25

2

10

20

TOTAL PURCHASE EQUIPMENT COST (PEC) = 2125 Thusd Rs.

14.1.2 Direct Costs: Table-14.2 Direct Cost Sr. No. 1 2 3 4 5 6 7 8 9

Item

% of PEC

Cost (Rs. Thusd)

100

2125

40

850

15

318.75

Building cost Yard improvement cost

60 12 18 10

1275 255 382.5 212.5

Service facilities cost Land purchase cost

70 10

1487.5 212.5

Purchased Equipment Delivered cost Purchased Equipment Installation cost Instrumentation & Control cost (Installed) Piping cost (Installed) Electrical Installation cost

TOTAL DIRECT COST = 7118.75 Thusd Rs.

14.1.3 Indirect Costs: Table-14.3 Indirect Cost Sr. No.

Item

% of PEC

Cost (Rs.

1 2 3 4

Engineering & Design Cost Construction expenses Contractors Fees Contingencies

15 20 5 10

Thusd) 1067.81 1423.75 355.93 711.875

Total Indirect Cost = 3559.36 THUSD RS. Fixed Capital Investment (FCI) = Total Direct Cost + Total Indirect Cost = 7118.75 + 3559.36 101

= 10678.115 Thusd Rs. Working Capital Investment (WCI) = 20% of Fixed Capital Investment (FCI) = 0.2 x10678.115 = 2135.625 thusd Rs. TOTAL CAPITAL INVESTMENT (TCI) =

Fixed Capital Investment +Working Capital Investment = 10678.115 + 2135.625 = 12831.73 Thusd Rs.

14.2 ESTIMATION OF TOTAL PRODUCTION COST: 14.2.1 Manufacturing Cost Direct Production Cost (1)Raw Material Cost Working Days = 333 Table-14.4 Raw material cost Raw Material

Amount

Ethylene oxide Styrene Divinylbenzene

73450827.65 (Kg/Yr)

anion exchange Resin

1.024 M3

Cost

Cost ( thuds

42.14 (Rs / Kg)

Rs) 30952.1787

91491.47 (Rs/m3)

(Catalyst) TOTAL COST OF RAW MATERIAL

= 30953.115 thusd Rs.

(2) Utilities Cost = 20% of Raw Material Cost = 0.2 x 30953.115 = 2135.623 thusd Rs. (3) Maintenance and Repair Cost = 10 % of Fixed Capital Investment = 0.1 x 10678.115 = 1067.875 thusd Rs. (4) Operating Labour & Supervision Cost = 5% of Raw Material Cost = 0.05 x 30953.115 = 1547.665 thusd Rs

102

0.9368

(5) Lab & Other Service Cost

= 1% of Raw Material Cost = 0.01 x 30953.115 = 309.531 Thuds Rs

DIRECT PRODUCTION COSTS

= Raw Material Cost + Utilities Cost + Maintenance and Repair Cost + Operating Labour & Supervision Cost + Lab & Other Service Cost [19] =

30953.115 + 2135.623 + 1067.875 + 1547.665 + 309.531

= 36013.735 thusd Rs. Fixed Cost: (1) Depreciation

= 10 % of Fixed Capital Investment = 0.1 x 10678.115 = 1067.875 thusds Rs.

(2) Local Taxes = 2 % of Fixed Capital Investment = 0.02 x 10678.115 = 213.562 thusds Rs. (3) Insurance Cost = 3 % of Fixed Capital Investment = 0.03 x 10678.115 = 320.343 thusds Rs. TOTAL FIXED COST

= Depreciation + Local Taxes + Insurance Cost = 1601.7169 thusds Rs.

Plant overhead Cost: These costs are 100% of Labour cost, So, plant overhead cost is 1547.655 thusds Rs. TOTAL MANUFACTURING COST =

Direct Production Costs + Total Fixed Cost + Plant overhead cost 103

= 36013.735 + 1601.7169 + 1547.655 = 39163.1069 thusds Rs. 14.2.2 General Expenses: (1) Administrative Cost = 1% Of Manufacturing cost = 0.01 x 39163.1069 = 391.631 thusds Rs. (2) Distribution & Marketing cost = 2% Of Manufacturing cost = 0.02 x 39163.1069 = 783.262 thusds Rs. TOTAL GENERAL EXPENSES = 1174.893 thusds Rs. TOTAL PRODUCTION COST =

Total Manufacturing Cost +Total General Expenses = 39163.1069 + 1174.893 = 40338. thusds Rs.

14.3 BREAK EVEN POINT: Let N TPA be the break even production rate. Raw material Cost/(ton product) = 99829190.88/(100X365) = 2735.04 Rs /(Ton Product) Fixed Cost = 1601.7169 thusds Rs/yr At break even production, Fixed charges + Direct Production Cost = Selling Cost (1601.7169 + 36013.735) thuds Rs. = 60 X 100 X N N = 6269 TPA = 17.17 TPD Hence, the break even production rate is 17.17 TPD of the considered plant capacity. 14.4 PROFITABILITY ANALYSIS: Working days = 333 Table-14.5 Selling price Product

Amount kg/year

Selling Price Rs./kg

MEG

99829190.88

60

Total Selling Cost (TSC) = 59897.514 thusds Rs. Gross Profit = Total Selling Cost – Total Production Cost 104

= 59897.514 – 40338.00 = 19559.51 thusds Rs. Tax Paid

= 0.4 x Gross Profit = 0.4 x 19559.51 = 7823.80 thusds Rs.

Net Profit = Gross Profit – Tax Paid = 19559.51– 7823.80 = 11735.70 thusds Rs.

PAY OUT PERIOD:

It is given by P.O.P Fixed Capital Investment per year P.O.P

=

Net Profit + Depreciation 10678.115 =

11735.70 +

1067.8115

= 0.83 Years RATE OF RETURN:

It is given by R.O.R Net Profit R.O.R =

=

Total Capital Investment

11735.70 x100 12813.73

= 91.85 %

105

X 100

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