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 H H H

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

H H H 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 increasing

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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 o-phthalic,

maleic or fumaric acid. These resins are used to modify synthetic 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

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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.

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]

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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.

no.

Physical Properties

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. 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

5

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.

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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 [O] [O]

HOCH2CH2OH HOCH2CHO HOCH2CH2COOH CHOCOOH

(Ethylene Glycol) (Glycol aldehydes) (Glycolic acid) (Glyoxylic acid)

[O]

HOOC-COOH

(Oxylic acid)

[O]

HNO3 [O] [O]

HOCH2CH2OH HOCH2CHO CHOCHO CHOCOOH

(Ethylene Glycol) (Glycol aldehydes) (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 PCl5

HOCH2CH2OH HOCH2CH2Cl ClCH2CH2Cl

(Ethylene Glycol) (Ethylene chlorohydrins) (1, 2-Dicholorochlorohydrins)

With Phosphorus trihalide (PBr3) to form responding dihalide

PBr3 PBr3

HOCH2CH2OH HOCH2CH2Br BrCH2CH2Br

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(Ethylene Glycol) (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 200 oC

HOCH2CH2OH HOCH2CH2Cl ClCH2CH2Cl

(Ethylene Glycol) (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

+

+ HOCH2CH2OH

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 cyclic structures, e.g.,

methylamine reacts with Diethylene glycol to produce N-methylmorpholine.

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Ketones and aldehydes react with ethylene glycol under acidic conditions to

Form 1, 3-dioxolanes cyclic ketals and acetals.

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]

CHAPTER III

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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

requirements, which leads to 12% of the Diethylene glycol being converted with

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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.

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

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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.

3.3 SHIPPING DATA FOR ETHYLENE GLYCOL:

Weight per Gallon at 20°C 9.29 lb

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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]

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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)

PAZINKA, Yugoslavia, (West Europe)

EASTERN PETROCHEMICAL CO. Saudi Arabia, (Middle East)

National Organic Chemical, India, (Asia).

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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. Year Month Price(US$/MT)

1. 2004 November 1095

2. December 988

3. 2005 January 1045

4. February 1095

5. March 1095

6. April 971

7. May 734

8. June 736

9. July 808

10. August 836

11. September 883

12. October 883

13. November 1st week 830

14. 2nd week 822

4.5 DEMAND SUPPLY BALANCE (IN KT):Table 4.2 Demand supply balance (In KT)

MEG 2002 2003 2004 2005 2006

Capacity 590 615 654 830 830

Production 548 647 691 833 830

Imports 11 64 106 103 90

Exports 8 29 104 133 60

Demand 551 682 750 803 860

Demand Growth % 24% 10% 7% 7%

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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).

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%

max

0.005 0.02

Ash, wt% max 0.005 0.005

Water, wt% max 0.08 0.3

Iron, ppm wt max 0.07 0.05

Chlorides, ppm wt max

Distillation range, ASTM at

760mm Hg:

IBP, C min 196 196

DP, C max 200 199

Odor Practically none

UV transmittance, % min at:

220 nm 70 70

250 nm 90

275 nm 90 95

350 nm 98 99

Specific gravity, 20/20C 1.1151-1.1156 1.1151-1.1156

Water solubility, 25C Completely miscible

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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.

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 (83-84%) 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.

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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, anion-

exchange resins, metals, metal oxide and metal salts. Carbon dioxide as a co catalyst

with many of the same materials has also received extensive study.

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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.

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.

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.

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.

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%.

20

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 190-230C 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:

Table 5.1 Comparison of different Processes

SR.

NO

PROCESSES PARAMETER CATALYST ADVANTAGES/

DISADVANTAGES

21

1. Hydrolysis of

Ethylene Oxide

1) Non- catalytic

Yield : 98%

Selectivity: 98%

Temp:105oC

Pressure :

1.5MPa

2) catalytic:

Yield : 95%

Selectivity: 90%

Temp:200oC

Pressure :

1-30 bar

1)Non

Catalytic

2) Catalytic:

Sulfonic acids,

Carboxylic

acids and salts,

Ion-exchange

resins, Acidic

zeolites,

halides, Metal

oxide and

Metal salts.

Use large excess water

to increase the yield

which leads to high

energy consumption

1) Use less excess water

which leads to low

energy consumption

2) High yield &

selectivity

3) permit use of low

temp & pressure

4) Acid catalyst makes

the reaction solution

highly corrosive.

2. Ethylene Glycol

from Ethylene

chlorohydrins

Yield :50%

Selectivity: 75%

Non Catalytic very low yield &

selectivity

very costly

3. Ethylene glycol

from

CO,H2,CH3OH

&

Formaldehyde

Yield : 90-95%

Temp: 200oC

Pressure:

100atm

Cromate

Catalyst

High pressure

process

Discontinued now a

day

Low selectivity

4. Ethylene glycol

from ethylene

carbonate

Yield :98%

Selectivity: 95%

Temp:180oC

Pressure:13bar

Alkali halide

or ammonium

salt.

Give high yield and

selectivity

Utility saving

Extra purification

cost

5. Transesterificati

on of ethylene

carbonate.

Low yield Zirconium &

titanium

compound.

Produced large

amount of

byproducts

6. Esterification of

Oxalic acid and

Yield : 70%

Selectivity: 90%

Copper

catalyst

High conversion but

catalyst removal is

22

Reduction with

H2

very difficult.

7. Direct one stage

synthesis of

Ethylene glycol

from syn gas

Selectivity: 65%

Temp:

190-230oC

Pressure:

3400atm

Rhodium

catalyst

(Homogeneous

catalyst route.)

As crude prices

increase this process

will become more

economical.

Use of very high

pressure

Not prove to be

indirect route may

be viable or not.

Catalyst is very

sensitive and

expensive.

8. Hydrolysis of

glycol diacetate.

Yield : 90%

Selectivity: 95%

Temp: 160oC

Pressure:

2.4MPa

Pd complexes

pdcl2+NaNO3

Very low conversion

H2O+C2H4O Ko HOCH2CH2OH ---------------- (1)

C2H4O + HOCH2CH2OH Ki HO (CH2CH2O)2 H ---------------- (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

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

23

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

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

24

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).

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

25

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

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

26

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 =

= 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,

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:

27

In the reactor following reaction take place

C2H4O + H2O HOCH2CH2OH --------- (1)

(Ethylene oxide) (Water) (Mono Ethylene Glycol)

C2H4O + HOCH2CH2OH HOCH2CH2OH --------- (2)

(Ethylene oxide) (Mono Ethylene Glycol) (Higher Glycol)

As selectivity = 98%

Moles of undesired product formed =

= 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

= 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.

28

Ethylene Oxide = 9190.54 Kg = 208.876 Kmols

Water = 37597.68 Kg = 2088.76 Kmol

Mono Ethylene Glycol = 204.7Kmols = 12691.4 Kg

Water = 1881.972 Kmols = 33875.496Kg

Higher glycol = 2.088 Kmol = 221.328Kg

REACTOR

Temp. = 100 0C

Conversion = 100 %

Pressure = 1.5-2MPa

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 In, Kg Out, kg

Ethylene oxide 9190.54 -

Water 37597.68 33875.496

Mono Ethylene Glycol - 12691.4

Higher Glycol - 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 = 0.851881.972

= 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.

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.

29

Amount of glycol carry over along with water from 1st effect = 165.58 kg

Amount of glycol carry over along with water from IIst effect = 189.139kg

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

30

1st effect evaporator

Pressure = 7 kg/cm2

Temp = 159 oC

F = 2088.76 Kmol = (46788.224 kg)

M.E.G = 204.7Kmol = 12691.4 Kg

Water =1881.972 Kmol = 33875.496Kg

W1= 8285.66kgMEG = 165.58kgH2O = 8120.08kg

2nd

effect evaporator

Pressure = 3.5 kg/cm2

Temp = 141 oC

3rd effect evaporator


Temp = 118 oC

W2= 9689.31kgMEG = 189.139kgH2O = 9500.171kg

To MEG Recovery column

Y= 1610.8012kmol

To 3rd effect evaporator

To 2nd effect evaporator

From 3rd effect evaporator

From 2nd effect evaporator

To MEG Refining column X = 477.9588 Kmol

W3= 11508.96kgMEG = 335.064kgH2O = 11173.89kg

(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 In, Kg Out, Kg

Liquid phase Vapor phase

Water 33875.496 5081.355 28794.141

MEG 12691.4 12001.617 689.783

HG 221.328 221.328 -

6.3.3 Drying Column:

31

F = 477.9588 kmol = 17304.2585 kgMEG = 12001.606kg

H2O = 5081.324 kg.

HG = 221.328kg.

Y= 289.295 Kmol = 5537.385 kgMEG = 456.061kg

H2O = 5081.324 kg.

X = 188.306 Kmol = 11766.873 kgMEG = 186.218kmol = 115453545kg

HG = 2.088 Kmol = 221.328kg

To MEG Recovery column Y= 1610.8012kmol

Drying columnPressure = 0.21 kg/cm2

Temp = 87 oC

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

= (Amount of MEG leaving along with

water during evaporation + Amount of

water removed)

= 18.4813+1881.973

32

= 1900.451 Kmol.

Table 6.3 Material balance over drying column



Water 5081.324 - 5081.324

MEG 12001.606 11545.3545 456.061

HG 221.328 221.328 -

6.3.4 MEG Refining Column (Packed Column):

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.

Amt. of Distillate (D) = 184.54 x 62.04

= 11448.8618 kg / hr.

33

F = 188.306 Kmol = 11766.873 kg

MEG = 186.218kmol =11545.545kg

HG =2.088kmol = 221.328kg

D= 184.54 Kmol = 11448.8616 kg

MEG = 184.355kmol(0.999.high purity)

HG = 0.18454kmol

W = 3.766 Kmol = 317.664 kg

MEG = 1.8523kmol

HG = 1.9136kmol

MEG refining columnPressure = 10 mmHg

Temp = 93.2 oC

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



MEG 11545.545 114.8426 11430.01

HG 221.328 202.8416 19.56124

6.3.5 MEG recovery column (Plate column):

34

D= 1881.97kmol = 11766.873 kg

MEG =1.88kmol

H2O =1880.08kmol

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

W = 18.481kmol / hr

M.B. ON WATER

35

F = 1900.451kmol = 35021.339 kg

MEG = 18.481kmol =1145.844kg

H2O =1881.97kmol = 33875.496kg.

W = 18.481kmol =1205.55 kg

MEG = 17.122kmol

H2O = 1.3584kmol

MEG recovery column

Plate column

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



Water 33875.496 24.4512 33841.44

MEG 1145.844 1061.546 116.56

Table 6.6 Overall material balances

Equipment Component In, kg Out, Kg


Reactor Ethylene oxide 9190.54 - -

Water 37597.68 33875.496 -

MEG - 12691.4 -

HG - 221.328 -

Triple effect

evaporator

Water 33875.496 5081.355 28794.141

MEG 12691.4 12001.617 689.783

HG 221.328 221.328 -

Drying column Water 5081.324 5081.324

36

MEG 12001.606 11545.3545 456.061

HG 221.328 221.328 -

MEG refining

column

MEG 11545.545 114.8426 11430.01

HG 221.328 202.8416 19.56124

MEG recovery

column

Water 33875.496 24.4512 33841.44

MEG 1145.844 1061.546 116.56

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]

37

REACTOR

Temp. = 100 0C

Conversion = 100 %

Pressure = 1.5-2MPa

Ethylene Oxide = 9190.54 Kg = 208.876 Kmols

Water = 37597.68 Kg = 2088.76 Kmol

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 + H2O HOCH2CH2OH ------------- (1)


C2H4O + HOCH2CH2OH HOCH2CH2OH ------------ (2)

(Ethylene oxide) (Mono Ethylene Glycol) (Higher Glycol)

Table 7.1 Heat capacity and Enthalpy dataCOMPONENT

INEthylene oxide -77704 99.106Water -285830 75.673OUTMonoEthyleneGlyocol -454800 75.673Di-EthyleneGlyocol -285831 189.39Water -562570 441.602

Assume reference temp. = 250C

7.1.1 Enthalpy of formation of reaction

For first reaction

= [-454800] - [-(77704) + (-285830)]

= -91266 KJ/ Kmol of EO Reacted

= -91266 x 206.788

= -18.872 x 106 KJ / hr

For second reaction

= [-562570] – [(-77704) + (-454800)]

= -30066 KJ/ Kmol of EO Reacted

= -30066 x 2.088

= -62.77x103 KJ / hr

38

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.

Enthalpy of products

= [ ( 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

= (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 x 75.79627 x 20

m = 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: wSS + wF (tF – t1) Cp = w11

Second effect: w11 + (wF – w1) ( t1 – t2) Cp = w22

39

1st effect evaporator

Pressure = 7 kg/cm2

Temp = 159 oC

W1= 8285.66kgMEG = 165.58kgH2O = 8120.08kg

F = 2088.76 Kmol = (46788.224 kg)

M.E.G = 204.7Kmol = 12691.4 Kg

Water =1881.972 Kmol = 33875.496Kg

To 2nd effect evaporator

2nd

effect evaporator


Temp = 141 oC

W2= 9689.31kgMEG = 189.139kgH2O = 9500.171kg

Third effect: w22 + (wF – w1-w2) (t2 – t3) Cp = w33

Material balances: w1 + w2 + w3 = w1-3

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

e, KPa

Steam or

vapor

temp. C

, KJ/Kg

(Steam)

, KJ/Kg

(MEG)

Steam chest, 1st

effect

1179.69 TS= 190.82 S = 2210.8

Steam chest,

2nd effect

794.63 t1=175.17 1 = 2244.1 1 = 982.935

Steam chest, 3rd

effect

409.57 t2=152.585 2 = 2284.0 2 = 1001.15

Vapor to

condenser

24.53 t3= 106.155 3 = 2379.1 3 =1022.317

40

3rd effect evaporator


Temp = 118 oC

To MEG Recovery column

Y= 1610.8012kmol

To 3rd effect evaporator

From 2nd effect evaporator

From 3rd effect evaporator

W3= 11508.96kgMEG = 335.064kgH2O = 11173.89kg

To MEG Refining column X = 477.9588 Kmol

To MEG Recovery column Y= 1610.8012kmol

7.2.1 First effect:

Cp avg. = xiCpi

= 4.196 KJ/Kg o K

avg = 2016.38 KJ/Kg

WSS + wF (tF – t1) Cp = w11

(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. = xiCpi

= 4.105 KJ/Kg o K

avg = 2088.28 KJ/Kg

w11 + (wF – w1) (t1 – t2) Cp = w22

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. = xiCpi

= 3.873 KJ/Kg o K

avg = 2207.35 KJ/Kg

w22 + (wF – w1-w2) ( t2 – t3) Cp = w33

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

41

Now , Enthalpy out from the bottom of the last effect,

Tbottom = 122oC Trefrence = 25oC

T = 97oC.

Enthalpy out from Bottom = (mCpT )MEG + ( mCpT )WATER + ( mCpT )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:

Toperating = 87 oC Trefrence = 25 oC

Hence T = 62oC.

Poperating = 0.25 kg /cm2

Enthalpy in = 2.802 x 106 kJ / hr

7.3.1 Enthalpy out from Top

= ( m )water + ( m )MEG +( mCpT )

= [(5081.324 x 2366.1) + (456.061 x 1109 .75)] + [289.65 x 75.2 x 64]

= 12.529 x 106 kJ / hr

42

Drying columnPressure = 0.21 kg/cm2

Temp = 87 oC

Y= 289.295 Kmol = 5537.385 kgMEG = 456.061kg

H2O = 5081.324 kg.

X = 188.306 Kmol = 11766.873 kgMEG = 186.218kmol = 115453545kg

HG = 2.088 Kmol = 221.328kg

F = 477.9588 kmol = 17304.2585 kgMEG = 12001.606kg

H2O = 5081.324 kg.

HG = 221.328kg.

7.3.2 Enthalpy out from Bottom

= (mCpT )MEG + ( mCpT )HG

= [(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

Steam = 2205.82 kJ / kg

Q = m λsteam

8.926 x 106 = m x 2205.82

m = 4046.6 kg / hr (Rate of steam)

7.4 MEG REFINING COLUMN:

43

MEG refining columnPressure = 10 mmHg

Temp = 93.2 oC

W = 3.766 Kmol = 317.664 kg

MEG = 1.8523kmol

HG = 1.9136kmol

F = 188.306 Kmol = 11766.873 kg

MEG = 186.218kmol =11545.545kg

HG =2.088kmol = 221.328kg

D= 184.54 Kmol = 11448.8616 kg


HG = 0.18454kmol

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 = (mCpT ) MEG + (mCpT )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

= m + (mCpT)

MEG = 68.578 x 103 kJ / kmol

44

DEG = 72.067 x 103 kJ / kmol

AVEG = 68.58 x 103 kJ / kmol

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 = ( mCpT )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 = (mCpT )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 = ( mCpT )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

45

QB + QFeed = QD + QW + QC

QFeed = 2.252 x 106 kJ / hr

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

= 21.794 x 106 kJ / hr



steam = 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:

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

46

MEG recovery column

Plate column

D= 184.54 Kmol = 11448.8616 kg


HG = 0.18454kmol

W = 3.766 Kmol = 317.664 kg

MEG = 1.8523kmol

HG = 1.9136kmol

F = 188.306 Kmol = 11766.873 kg

MEG = 186.218kmol =11545.545kg

HG =2.088kmol = 221.328kg

Total Enthalpy out with Distillate = (mCpT )MEG + (mCpT ) WATER

= [(1881.08 x 76.55) + (1.874 x 197.24)] x 169

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

= m + (mCpT)

MEG = 1023.184 kJ / Kg H2O = 2231.8 kJ / Kg

AVEG = 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 = (mCpT) Reflux

= [959.80 x 76.67 x 169]

= 12.43 x106 kJ / hr

Condenser load

47

QC = QV – ( QReflux + QD )

= [(208.95 106 ) – (12.43 x106 + 24.40 x 106 )]

= 172.12 x 106 kJ / hr

Assuming cooling water enter the condenser at 25oC & leaves at 45oC

QC = (mCpT )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 Trefrence = 25oC

T = 173 oC

Cpmean of MEG = 197.6285 kJ / kmol 0K

Cpmean of H2O = 76.607 kJ / kmol 0K

Enthalpy out with Residue:

QResidue = ( mCpT )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 = ( mCpT )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) ]

48

= 196.97 x 106 kJ / hr



steam =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:

Here,

FAO = 208.876 Kmol/hr

= 9190.544 Kg/hr

V0 = 10.649 m3/hr

FAO = CAO V0

208.87 = Cao x 10.649

CAO = 19.6136 Kmol/m3 = 19.6136 mol/lit

Similarly,

49

REACTOR

Temp. = 50 0C

Conversion = 100

%

E. O = 9190.54 Kg= 208.876 Kmols

Water = 37597.68 Kg = 2088.76 Kmol

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

CBO = 5.555 Kmol/m3 = 55.555 mol/lit

Now, Rate of reaction is given by

– d (C2H4O) = {K0([H2O] + b [Gyi]) + Kct [HCO3–]} X

dt {{[H2O] + p [Gyi]}x [Oxide]}

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 (CA + CB)) + Kct [HCO3–] } X

dt {(CB + p (CA + CB)) x [CA]}

Rate constant Ko is given by,

K0 = exp [9.077 – 9355]

T

where T = temperature o K

K0 = exp [9.077 – 9355]

378

K0 = 1.5627 x 10-7 L 2 = 0.0005625 L 2

mol2. Sec mol2. hr

Similarly catalyst Rate constant Kct is given by,

Kct = exp [18.24 – 10574]

T

Kct = 5.926 x 10-5 L 2 = 0.21334 L 2

mol2. Sec mol2. hr

Now,

CAO XA = CBO XB

a b

19.6136 XA = 55.555 XB

1 1

XB = 0.3530 XA

- rA ={ K0[(CBO(1 – 0.3530 XA)) + 2.8 (CAO (1 – XA) + CBO (1 – 0.3530 XA))]

50

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

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 = 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.98

-rA = 654.5 532.40 424.57 330.131 248.286 178.2369 119.1836 70.31 30.86 53.60

(1/-rA) (lit.hr/mol)

0.005 0.0018 0.0023 0.00302 0.00402 0.00561 0.0084 0.01 0.03 0.186

From the above data plotting a XA Vs & finding the area under he curve at

Xa = 0.98

Fig-4 Reactor volume Chart

AREA UNDER THE CURVE = 0.01227

Plug flow equation related to volume is given by [15]

51

XA

V = d XA

FAO 0 - rA

XA

d XA =Area under the curve

0 - rA

Area under the curve = 0.01227

= 0.01227

V = 2562.90 lit

V = 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 --------- (1)


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)] [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 = 4mDiameter of tube = 2.5 cm

Volume of one tube = (d)2 (L)4

= (2.5 x 10-2)2 (4)4

= 1.9634 x 10-3 m3

Table-9.1 Properties at arithmetic mean temperature.

Props. Shell Side(Water) (30oC )

Tube side(Ethylene oxide + H2O) (65oC)

Cp 5.1865 (KJ/Kg oK) 4.840 (KJ/Kg oK) 9 x 10-4 (Kg/m.Sec) 4.187 x 10-4 (Kg/m.Sec)K 0.62 (w/m.oK) 0.54 (w/m.oK) 995.40 (Kg/m3) 973.09 (Kg/m3)

No. of tube = Volume of reactor Volume of one tube

=

No. of tubes = 1304 Nos.

Area of tube per pass:

54

Atp = (d) 2 (Ntp) 4= (2.5 x 10-2)2 (1304)

4Atp = 0.64 m2

Velocity:

U =

m = 12.996 Kg/Sec

U =

U = 0.02 m/sec

Now, NRe =

= (2.5 x 10 -2 ) x (0.02) x (973.09) (4.187 x 10-4) x (1 -0.6)

NRe = 2905.09

Now, AO = Nt x x d xL= (1304) x x (2.5 x 10-2) x (4)

AO = 409.66 m2

Shell diameter:

Ds = Consider the Triangular pitch

CTP = 0.9CL = 0.7

Take PR = Pube pitch ratio= 1.25

Ds = 0.637 0.7 409.66 x (1.25) 2 x (2.5 x 10 -2 ) 0.5

0.9 4

Ds = 1.123 m

Now, No. of tubes that can be accommodate

Nt = 0.875 CTP __(Ds) 2

CL (PR)2 (d)2

= 0.875 0.9 ____(1.123) 2

0.7 (1.25)2 (0.025)2

= 1452.8 > Total No of tube that is required.

55

Shell side H.T.C :

=

For triangular pitch

De = 4 3 PT2 – d 2

2 4

d

Pitch ratio = PR =

1.5=

PT = 0.03125 m

De = 4 3 (0.31252 – x (2.5 x 10-2)2

2 4

x 0.0025

De = 0.018 m

Gs =

As =

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 0.450.03125

As = 0.101 m2

Gs =

= 12.996 0.10

56

Gs = 128.58 Kg/m2.sec

From the above equation,

=

ho = 1825.04

Tube side H.T.C:

Nu = 0.023 (NRe)0.8 (Pr)0.4

=

=

hi = 672.163

Over all H.T.C

=

=

Uo = 142.133

Now,

Total H.T area available = 409.66 m2

Rate of H.T. is given by

Q = UAT

(1.1956 x 106) = 117.03 x A x 20

A = 420.59 m2 > A available

Tube side pressure drop :

Tube side pressure drop is given by

P gc x ()3 x Dp x

Z = 150 (1 – ) + 1.75

(1 – ) (G’)2 Nre

57

( P/Z) (1) (0.6) 3 (1.2 x 10 -3 ) x (973.09) = 150 x (0.4) + 1.75

(0.4) (19.4608 )2 2905.09

= 1.063 KN /m2 < 30 - 40 KN/m2

Shell side P:

P = f x (Gs 2 ) x (Nb + 1) x Ds

2 x x Dex s

f = exp [(0.57 – 0.19 ln (Res)]

= exp [0.57 – 0.19 ln(2571.6)]

f = 0.3977

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,

= 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

58

So ts = 14 mm

9.2 DISTILLATION COLUMN (PACKED COLUMN):

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]

Assumption:

Binary mixture separation between MEG & DEG.

P = total pressure of the system = 10 mmHg

So, at top Psat MEG = 1.334 KPa

59

DISTILLATION

F = 188.306 kmol = 11766.873 kg

MEG = 186.218kmol =11545.545kg

HG =2.088kmol = 221.328kg.

D= 184.54 kmol = 11448.8616 kg

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

Psat DEG = 0.1331 KPa

So, top = Psat MEG / Psat DEG

= 1.334 / 0.1333

= 10.077

Now at bottom Psat MEG = 1.4757 KPa

Psat DEG = 0.1673 KPa

So, bottom = Psat MEG / Psat DEG

= 1.4757 / 0.1673

= 8.8174

Thus, average = { top x bottom }0.5

= {10.007 x 8.8174}0.5

average = 9.3933

Now we have the equation

Y = X / {1+ (-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 0.5106

9

0.7013

4

0.8010

2

0.862

3

0.903

7

0.933

7

0.956

3

0.974

0

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 :

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

60

Liquid density L = 954 Kg/m3

Vapor Density G = 2.099Kg/m3

Liquid Viscosity = 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

M g =

M g = 6.4726 Kg /s

Mass flow rate of liqid

M l =

M l = 2.2581 Kg /s

Flow parameter

--------------------------- (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

g = Density of vapor =2.099Kg/m3

l = Density of liquid =954 Kg/m3

From equation (2)

=

X = 0.01638

By using plot of flooding and Pressure drop.

61

V/S

Here column is operated under vacuum, plot value at X = 0.01638

Take, ref.

We get Y = 0.0125

----------------------------------------- (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

= Viscosity of liquids = 2.2 x 10-3 Kg/m s

g = Density of gas = 2.099 kg / m3

l = Density of liquid = 954 kg / m3

G' in kg / m2. s

From equation (3)

G'2 = [(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.

Hence, Area of tower =

= 6.4726 / (0.6 x 0.5929)

= 18.19 m2

62

= 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

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 = 5m

63

Di = Do =5m (initial approximation)

ri = 0.1 x Ro

= 0.5m

hi = 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

U = 1.5 % (take for new vessel)

P = 2 x 98.1 x 4.82 x10-3 x

= 0.2446 › 0.105 MN/m2

9.2.4 Head Design:

Selecting Torispherical head

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

64

= 95 + 50

= 145 oC

Table-9.3 Head thickness

t(cm)

(assumed)

Ro / t Ro / 100t Factor B P =

1.2 378.9 3.78 4400 0.7402

1.3 349.75 3.49 4500 0.8201

1.4 324.77 3.24 4800 0.9421

1.5 303.2 3.03 5000 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

Head = Torispherial

Head thickness =18mm

Tdesign = 145o C

Permissible Stress =91.8 MN/m2

ρs = 78480 N/m3

65

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

p = 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 s

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 in

in = tin x ρin x X/ ts

= 0.05 x ( 5.64*103 ) x X / 0.026

in = 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

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

66

= 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

l = Weight of (liquid + packing) / 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

l = 1.7465 x X / x ( 5.052) x( 0.026 )

= 4.243 x X MN/m2

( d ) Stress due to attachments

Weight of head = /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

a = Whead + (Wladder & liquid distributor) / x Di x t

= 0.02831 + 0.0013734 x X / x ( 5 ) x (0.026)

a = 0.06931 + 0.003362 x X MN/m2

Total dead load stress

dead = s + in + l + a

= 0.7848 X + 0.01084 X + 4.243 X + 0.06931 + 0.003362 X

dead =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

67

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 heightW = total weight

Now,Weight of shell Ws = 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 = 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

= 14 + 4 H = 18 ma) 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

68

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 ) Do = 5.1 m Pw = 3.571 x X KN / 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

wind = 4 x Mwind / x ts x D2

= 4 x ( 1.785 x X2) / 3.14 x ( 0.02 6) x ( 5.052 )2

wind = 0.00349 x X2 MN/m2

Calculation of Resultant longitudinal stress :

Tensile stress on upwind side at X m from top

tensile = -p - dead + wind

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

Solving this equation

0.00349X2 – 4.3246 X -134.46 = 0ax2 + bx + c =0By Solving equation

X = 1496.59 >>> 18m

So This design is Ok

69

Compressive stress on downwind site at X m from top Down wind side ,

f compression =p + dead +wind

compressive = 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 = 0ax2 + bx + c = 0X = 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 byAt 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 vesselAt maximum dead weight 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 = 1K1 = 0.7 for cylindrical vessel

Minimum and maximum wind moments are computed by Maximum and minimum Wind load

70

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 stresswind = 4 x Mwind / x Di

2 x tsk

wind min = 4 x 572.89 / 3.14 x ( 5 )2 * tsk

= 30.277 / tsk KN / m2

wind max = 4 x 578.34 / 3.14 x ( 5 )2 tsk

= 29.4821 / tsk KN / m2

dead,min = Wmin / x Di x tsk

= 0.50532 / 3.14 x 5 x tsk

= 0.0328 / tsk M N / m2

dead max = Wmax / 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 tensile = windmin - deadmin

Take J = 0.7 for double weld joint

tensile = permissible x J = 98.1 x 0.85 = 83.385 MN/m2

tensile = windmin - deadmin

83.385 = windmin - deadmin

83.385 = 0.0328 / tsk + 0.030277 / tsk tsk = 3.11 x10 5 m

71

Maximum tensile stress without any eccentric load is given by compressive in skirt wall = 0.125 E tsk / Do

= 0.125 (2 x105) tsk / 5.052 = 4948.53 tsk

compressive in skirt wall = windmax + deadmax

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. PARAMETER DESCRIPTION

1 Tower MOC SS3042 Tower ID 5 m3 Tower OD 5.052 m4 Shell thickness 26 mm5 Shell head thickness 18mm6 Height of tower(Without support) 14m7 Height of packed Bed 10 m8 Type of Head Torrispherical9 No. of Beds of packing 2 , each of height 5 m10 P/Z Selected 50 N/M2/m11 Tower Support Skirt Support12 Skirt MOC SS30413 Skirt Height 4 m14 Skirt thickness 8 mm15 Type Pall Ring16 MOC Metal (S S)17 Bulk Density 430 kg/m318 HETP 0.5425m

CHAPTER XINSTRUMENTATION AND CONTROL

72

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:

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.

73

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 Temp. Range ºC

Mercury in glass – thermometer -27 to 400

Mercury in pressure thermometer -40 to 540

Vapor pressure thermometer -85 to 425

Resistance -200 to 1700

Thermocouple -250 to 1700

Thermister Up to 300

Pyrometer 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

Differential Manometer

Inclined Manometer

Bourdon Tube

74

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.

Less cost of installation.

Less space required.

Less hardware required.

Inventory information can be made available.

75

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

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.

76

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. Skilled

personnel are required.

77

CHAPTER XIPLANT 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:

They are classified in three types:

Induced draft

Forced draft

78

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 XIISAFETY, 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.

Skin Contact:

Exposure to rapidly expanding gases may cause frost burns to eyes

and/or skin. Liquid solutions of ethylene oxide cause serious chemical

80

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 quick-drench 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:

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

81

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

tanks, transport vessels and other enclosed containers. Storage

Temperature: 30oC / 86oF maximum.

82

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 : MONO ETHYLENE GLYCOL

SYNONYMS: 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:

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

83

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:

Water fog or fine spray. Dry chemical fire extinguishers. Carbon

dioxide fire extinguishers. Foam. Do not use direct water stream. May

spread fire.

84

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

STABILITY AND REACTIVITY:

Stability/Instability:

Thermally stable at recommended temperatures and pressures.

Thermal Decomposition

85

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.

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.

86

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:

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.

87

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.

Provide different surface, for cleaning, insulation, expansion.

Prevent flame travel in furnace.

Use safety over design factor of 15 – 20 %.

88

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.

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 off-

specification product produced through mis-operation. In emergency situations,

89

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

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

90

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.

To be seen to be concerned with environmental questions: important for public

relation

To minimize the production of waster: an economic factor

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12.7 HAZOP OPERATION OF STORARGE TANK TO REACTOR SYSTEM

Table 12.1 Hazardous operation of Storage tank to reactor system

GUIDE WORD

DEVIATION POSSIBLE CAUSE

CONSEQUENCES ACTION REQUIRED

NONE NO FLOW 1) No EO/H2O available atthe storage tank

Loss of Feed to reaction section and reduced output.

1) Ensure good communication with Ethylene oxide plant.2) Install low level alarm on storage tank.

2) J1/J2 pump fails (motor, fault, loss of drive, etc.)


1) Check design of pumps and install a stand by pump.2) Install a kickback on pump.

3) Line valve close (in error or fail)


Check Operation of Control valve or Install a by pass line with other control valve.(stand by ).

4) Line fracture

1) Loss of feed to the reaction section.2) Feed discharge into area.

Institute regular patrolling and inspection of transfer line.

MORE OF

MORE FLOW Control Valve fail to open or LCV by pass open in error

Over filling (excess feed in reactor)

Check proper operation of control valve

MORE PRES 1) V1. Valve closed in error with J1 and J2 pump running.

Transfer line subjected to full pump delivery or surge pressure.

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

2) Thermal expansion in a isolated valved section due to fire.

Line fracture or flange leak.

Install thermal expansion relief on valved section

MORE TEMP

1) High storage temperature

High pressure in transfer line

1) Check whether there is adequate.2) Warring of high temperature at storage, if not install.

2) cooling water supply to reactor cooling

Affect the rate of reaction.

1) Check design of pumps and install a stand by pump.2) Install a kickback

92

system stop on pump.3) Check proper operation of cooling water system.

LESS OF

LESS FLOW 1) Leaking flanges of valve stab not blanked and leaking.

Material loss to adjacent area

Institute regular patrolling and inspection of transfer line.

2) Pump J2 fails or not running properly and J1 running well.

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

1) Install a stand by pump.2) Check design of pump.3) Install kickback on pump.

MORE THAN

HIGH E.O CONC IN STREAM.

Valve V2-1, and V2-2 are not working properly

Leads to formation of more higher glycol

1) Institute regular inspection of valve.2) Install by pass line with standby Valve.

PART OF

WATER FLOW TO REACTOR

Valve V2, and V2-2 are not working properly

Leads to formation of more higher glycol

1) Institute regular inspection of valve.2) Install by pass line with standby valve.

OTHER MAINT. Equipment failure flange leak etc.

Line cannot be completely drainage or purge.

Install low point drain and N2 purge point and N2 vent line.

93

CHAPTER XIIIPLANT 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.

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:

94

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.

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.

95

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

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.

96

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

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.

97

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.

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

98

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 XIVCOST 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. Equipment No. of Equip-Ments

Cost perUnit (Rs.Thusd)

EstimatedCost

(Rs. Thusd)

1MEG STORAGE TANK

2 200 400

2HG STORAGETANK

1 200 200

3ETHYLENE OXIDESTORAGETANK

2 200 400

4PLUG FLOW REACTOR

1 500 500

5TRIPPLE EFFECT EVAPORATOR

1 85 85

6

VACUUM DISTILLATIONCOLUMN (PACKED COLUMN)

1 70 70

7DISTILLATIONCOLUMN (PLATE COLUMN)

1 75 75

8 DRYING COLUMN 1 50 50

9CENTRIFUGALPUMP

6 10 60

10 HEAT EXCHANGER 7 20 14011 BOILER 2 50 100

12DM WATERPLANT

1 25 25

13COOLINGTOWER

2 10 20

100

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

14.1.2 Direct Costs:Table-14.2 Direct Cost

Sr. No. Item % of PEC Cost (Rs. Thusd)

1Purchased Equipment

Delivered cost 100 2125

2PurchasedEquipment

Installation cost40 850

3Instrumentation & Control cost

(Installed) 15 318.75

4 Piping cost (Installed) 60 1275

5 Electrical Installation cost 12 2556 Building cost 18 382.5

7 Yard improvement cost 10 212.5

8 Service facilities cost 70 1487.5

9 Land purchase cost 10 212.5

TOTAL DIRECT COST = 7118.75 Thusd Rs.

14.1.3 Indirect Costs:Table-14.3 Indirect Cost

Sr. No. Item % of PEC Cost (Rs.

Thusd)

1 Engineering & Design Cost 15 1067.81

2 Construction expenses 20 1423.75

3 Contractors Fees 5 355.93

4 Contingencies 10 711.875

Total Indirect Cost = 3559.36 THUSD RS.

Fixed Capital Investment (FCI) = Total Direct Cost + Total Indirect Cost

= 7118.75 + 3559.36

= 10678.115 Thusd Rs.

Working Capital Investment (WCI) = 20% of Fixed Capital Investment (FCI)

= 0.2 x10678.115

= 2135.625 thusd Rs.

101

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 Cost Cost ( thuds

Rs)

Ethylene oxide 73450827.65 (Kg/Yr) 42.14 (Rs / Kg) 30952.1787

Styrene Divinylbenzene

anion exchange Resin

(Catalyst)

1.024 M391491.47

(Rs/m3)0.9368

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

(5) Lab & Other Service Cost = 1% of Raw Material Cost

= 0.01 x 30953.115

102

= 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


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

= 36013.735 + 1601.7169 + 1547.655

= 39163.1069 thusds Rs.

103

14.2.2 General Expenses:

(1) Administrative Cost = 1% Of Manufacturing cost

= 0.01 x 39163.1069


(2) Distribution & Marketing cost = 2% Of Manufacturing cost

= 0.02 x 39163.1069


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 NN = 6269 TPA = 17.17 TPDHence, the break even production rate is 17.17 TPD of the consideredplant 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

= 59897.514 – 40338.00

= 19559.51 thusds Rs.

104

Tax Paid = 0.4 x Gross Profit

= 0.4 x 19559.51


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 X 100

R.O.R = Total Capital Investment

=

= 91.85 %

105

REFERENCES

[1] Ullmann, “Ullamann’s Encyclopedia of Industrial Chemistry”, Wiley-VCH-

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Ethylene Glycol Chemical Engineering Final Year Project

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