acetaldehyde.doc

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1. INTRODUCTION Acetaldehyde CH3CHO (ethanal) first used extensively during World War І as a starting material for making Acetone from Acetic Acid. It is an important intermediate in industrial organic synthesis. Acetic acid, acetic anhydride n-butanol and 2-ethyl hexanol are the major products derived from acetaldehyde. Smaller amounts of acetaldehyde are also consumed in the manufacture of pentaerythritol, trimethylolpropane, pyridines, peracetic acid, crontoaldehyde, chloral, 1, 3-butylene glycol and lactic acid.

HISTORY Acetaldehyde was first prepared by Scheele in 1774 by the action of manganese dioxide and sulfuric acid on ethanol. The structure of acetaldehyde was established in 1835 by Liebig from a pure sample prepared by oxidizing ethyl alcohol with chromic acid. Liebig named the compound aldehyde from the Latin word translated as al (cohol) dehyd(rogenated). The formation of acetaldehyde by the addition of water to acetylene was observed by Kutscherow in 1881. The first commercial application was the production of acetone via acetic acid between 1914 and 1918 in Germany (Wacker-Chemie and Hoechst) and in Canada(Shawinigan).

Production of Acetaldehyde

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OCCURRENCE Acetaldehyde is a product of most hydrocarbon oxidations. It is an intermediate product in the respiration of higher plants. It occurs in traces in all ripe fruits that have a tart taste before ripening. Acetaldehyde is an intermediate product of alcoholic fermentation but it is reduced almost immediately to ethanol. It may form in wine and other alcoholic beverages after exposure to air, and imparts an unpleasant taste; the aldehyde reacts to form diethyl acetal and ethyl acetate. Acetaldehyde is an intermediate product in the decomposition of sugars in the body and, hence, occurs in traces in blood. Acetaldehyde also has been detected in plant juices and essential oils, roasted coffee and tobacco, smoke. Commercial production processes include ∗

Dehydrogenation or Oxidation of Ethanol.

∗

Addition of Water to Acetylene.

∗

Partial Oxidation of Hydrocarbon (Propane etc).

∗

Direct oxidation of Ethylene. The ethylene route has become the most important.


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PHYSICAL PROPERTIES Acetaldehyde is a colorless, mobile liquid having a pungent, suffocating odor that is somewhat fruity and quite pleasant in dilute concentrations. Acetaldehyde is completely miscible with water and most organic solvents. It forms no azeotrope with water, methanol, ethanol, acetone, acetic acid or benzene. Binary azeotropes are formed with butane (BP - 70 C, 84 wt% of butane) and diethyl ether (BP – 18.90 C, 23.5 wt % of ether). General physical properties of acetaldehyde are presented in Table 1 through 3. Table 1: Density of Acetaldehyde Temperature 0 C 0

Density g/cm3 0.805

3

0.800

6

0.797

9

0.792

Temperature 0 C 12

Density g/cm3 0.789

16

0.784

18

0.781

20

0.779


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Sr. No. 1.

Properties

Values

Formula weight

44.052

2.

Boiling point at 101.3 KPa (1 atm) 0 C

20.16

3.

Melting point, 0 C

-123.5

4.

Coefficient of thermal expansion / 0 C (0-300 C)

0.00169

5.

Vapor density (air =1)

1.52

6.

Surface tension at 20 0 C (dyn/cm)

21.2

7.

Heat capacity liquid (J/g 0 C, 00 C)

2.1856

8.

Heat capacity vapor (J/g 0 C,250 C)

1.4068

9.

γ = Cp/Cv at 30 0 C and 101.3 KPa

1.145

10.

Latent heat of fusion KJ/mol (cal /g)

3.24 (17.6)

11.

Latent heat of vaporization KJ/mol (cal/g)

25.71 (139.5)

12.

Heat of solution in water KJ/mol At 00 C At 250 C Heat of combustion of liquid at constant pressure KJ/mol At 250 C

-8.20 -6.82 12867.9

13.

1169.01 14.

15.

Heat of formation KJ/mol At 00 C At 250 C Critical temperature 0 C

-165.48 -192.18 181.5

16.

Critical pressure MPa

6.4

17.

Ignition temperature in air 0 C

165

18.

19.

Free energy of formation KJ/mol At 00 C At 250 C Explosive limits of mixtures with air vol % CH3CHO

-136.40 3064.298 4.5 - 60.5

20.

Relative density (t 0 C)

21.

Relative index (t 0 C)

22.

Molar volume of gas at 101.3 KPa L/mol At 20.16 0 C

dt4=0.80450.001325t ntD=1.342400.0005635t 23.40 Production of Acetaldehyde

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23.

At 25.00 0 C Thermal conductivity of liquid at 20 0 C Jm-1s-1K-1

23.84 0.174

24.

Thermal conductivity of vapor at 25 0 C Jm-1s-1K-1

1.09x10-2

25. 26. 27. 28. 29.

Entropy for gaseous acetaldehyde at 25 0 C J/mol K Entropy for liquid acetaldehyde at 20.16 0 C J/mol K Entropy of vaporization at 20.16 0 C J/mol K Dissociation constant at 0 0 C mol/lit Ionization potential eV

265.9 172.9 91.75 0.7x10-14 10.5

Table 3: Vapor pressure of acetaldehyde Temperature 0 C

Vapor Pressure KPa

20

100.6

20.16 20.18

Temperature 0 C -50

Vapor Pressure KPa 2.5

101.3 101.33

-20

16.4

0

44.0

30

145.2

5

54.8

44.5

202.65

6.7

59.07

50

279.4

9.3

65.87

58.3

303.98

10

67.7

68.0

405.3

11.6

70.8

70

492.6

13.3

76.93

75.7

506.63

15

82.9

100

1014

17.6

90.93


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CHEMICAL PROPERTIES Acetaldehyde is a highly reactive compound which is commonly consumed at the site of manufacture. Oxidation, hydrogenation and the numerous condensation, addition and polymerization reactions are example of its reactivity. ∗

Decomposition: - Acetaldehyde decomposes at temperature above 400

0

C

forming principally methane and carbon monoxide. The activation energy of the pyrolysis reaction is 97.7 KJ/mol. ∗

Oxidation: - The liquid phase oxidation of acetaldehyde with air (oxygen) is of great industrial importance. Most acetic acid is produced by this route. Acetaldehyde is readily oxidized with oxygen or air to acetic acid, acetic anhydride and peracetic acid. The principal product depends on the reaction conditions. Nitric acid oxidation of acetaldehyde yields glyoxal. When oxidized with oxygen, mono peracetate is formed as an intermediate and decomposes into peracetic acid and acetaldehyde at elevated temperature. In the presence of Mn +2 salts acetic acid is obtained from acetaldehyde monoperacetate and in presence of Co+2 and Cu+2 salts acetic anhydride can be formed.

∗

Reduction: - Acetaldehyde is readily reduced to ethanol. A wide variety of catalyst may be used. Raney nickel, palladium, supported nickel and copper oxide etc. Prior to 1939, i.e. before petro chemically produced ethylene became available in Europe; the reaction to produce ethanol from acetaldehyde was used industrially.

∗

Polymerization: - Paraldehyde 2, 4, 6-trimethyl-1, 3-5-trioxane a cyclic trimer of acetaldehyde is formed from mineral acid such as H2SO4, HCl, and Phosphoric. Distillation of paraldehyde in the presence of small amounts of Production of Acetaldehyde

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H2SO4 regenerates acetaldehyde. Being readily depolymerised paraldehyde is used to replace acetaldehyde in reactions where acetaldehyde volatility is problem. Metaldehyde a cyclic tetramer of acetaldehyde is formed at temperature below 0 0 C in presence of dry HCl. Polyacetaldehyde a rubbery polymer is unstable and depolymerises in few days to acetaldehyde. ∗

Reaction with Nitrogen Compounds: - With primary amines, Schiff bases CH3CH=NR are formed.

∗

Reaction with Halogens and Halogen Compounds: - Halogen readily replaces the hydrogen atoms of the acetaldehyde’s methyl group. Phosgene is produced by the reaction of the carbon tetrachloride with acetaldehyde in the presence of anhydrous aluminum chloride.

∗

Miscellaneous reactions: - Hydrocyanic acid adds to acetaldehyde in the presence of an alkali catalyst to form cyanohydrins. Grignard reagents add readily to acetaldehyde the final product being a secondary alcohol.


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HANDLING In handling acetaldehyde one has to remember that it is an extremely reactive compound that can be easily oxidized, reduced or polymerized and is highly reactive with oxygen. It has to be treated as a volatile, flammable and toxic material. The following is a list of precautions recommended when handling acetaldehyde. 1. Nitrogen or other inert gases should be used as a blanketing material whenever exposure to air is a possibility. 2. Safety goggles should be used. 3. Transfer should be made in open air structures or using suitable gas mask or self contained breathing equipment if necessary. 4. Drums should be stored out of doors avoiding direct exposure to sunlight. 5. Acetaldehyde should be chilled before transferring and a nitrogen blanket should be used.


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SHIPPING AND STORAGE Acetaldehyde is shipped in 5-10 or 55-gal drums insulated tank trucks and insulated tank cars. Acetaldehyde in the liquid state is non corrosive to most metals, but it can be easily oxidized to acetic acid, especially in vapor stage. Suitable materials of constructioN are stainless steel and aluminum. Drums coatad with phenolic res)ns Have9aLso `een u3eD. If a darker cmlor9and soe iron contamInation are Jot objectionable, carbon steel may be used. Because acetaldehyde is classed as flammable liquid, It requires a ‘red DOT’ (Department of Transportation) shipping label. Bulk storage held at low temperature and pressure is recommended over storage in a pressure vessel. According to regulation in Japan, outer storage tanks must not be made of copper, magnesium, silver, mercury or alloys of these metals. They must be equipped with cooling facilities in order to keep the temperature below 15

0

C and

also with an inert gas sealing system.


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CAPACITY AND MARKET SURVEY Today the most important production process worldwide is the direct oxidation of ethylene. In Western Europe there is also some capacity for the production of acetaldehyde by oxidation of ethanol (<15%) and hydration of acetylene (2%). In Eastern Europe the hydration is even more important. It is estimated that in Eastern about 235000 T (44% of the annual capacity) of acetaldehyde can be produced by this route. The worldwide production of acetaldehyde has been nearly constant since the early 1980s. Table 1 gives production data for acetaldehyde. Table 4: - Production of Acetaldehyde (103 T)

Year 1923 1987 1990 1993

USA 259 270 274 147

W. Europe 554 564 596 606

Mexico 212 193 190 249

Japan 256 285 284 349

Total 1281 1312 1444 1351

No accurate data are available for world production. Production data are difficult to obtain because most of the acetaldehyde produced is consumed internally by the manufacturer. Therefore estimated world production is only crude and should be taken with caution. Table 5 gives important producers and their production capacities (103 T) are listed. Table 6 gives the major producers and estimated capacities in the United States in 1970.

Table 5:- Worldwide Production & their capacities (103 T) Production of Acetaldehyde

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Sr. No.

Producers

Capacity

1.

Azot PO, Ukraine

100

2.

BP Chemicals Ltd, Italy

140

3.

Doljchim, Romania

4.5

4.

Eastman Chemical Company, US

250

5.

Erkimia SA, Spain

75

6.

Hoechst AG, Germany

290

7.

Huls AG, Germany

80

8.

Japan Aldehyde Company Ltd, Japan

69

9.

Kyowa Yuka Company Ltd, Japan

62

10.

Mitsui Petrochemical Industries Ltd, Japan

53

11.

Morelos, Mexico

150

12.

Neftechim, Bulgaria

90

13.

Omskiy Zavod Synnth. Caoutch, Russia

100

14.

Pajaritos, Mexico

44

15.

Petroleos Mexicanos (PEMEX), Mexico

100

16.

Salavatnephteorgsyntez, Russia

100

17.

Showa Denko KK, Japan

140

18.

Societe Francaise Hoechst SA, France

96

19.

Tokuyama Petrochemical Company Ltd, Japan

100

20.

Wacker Cheme GmbH, Germany

65

21.

Zacklady Chemiezne, Poland

90


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Table 6:- Acetaldehyde United States Capacity 1970 Sr. No.

Producer and Location

Capacity (million lb/year)

1.

Celanese, Bay City, Texas

250

2.

Celanese, Bishop, Texas

240

3.

Celanese, Clear Lake City, Texas

500

4.

Celanese, Pampa, Texas

10

5.

Eastman, Longview Texas

500

6.

Publicker, Philadelphia, Pennysylvania

70

7.

Union Carbide Institute, West Virginia, Texas 650 City, Texas

8.

Others

20

Total

2240


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MARKET SURVEY IN INDIA Acetaldehyde is a very common organic compound, large amounts of which are being used in the manufacture of various useful products. The consumer industries are showing rapid growth in our country. With the increasing population and industrialization these industries are likely to grow more rapidly in the coming years. This ensures a very good future demand for acetaldehyde in our country.

1.9 MANUFACTURERS OF ACETALDEHYDE IN INDIA 1. Somaiya Organo Chemicals, Mumbai. 2. Laxmi Organic Industries Ltd, Mumbai. 3. Shree Ganesh Chemicals, Mumbai. 4. Apra Enterprise, Mumbai. 5. Ashok Chem-Pharma International, Mumbai. 6. Indian Organics Limited, Pune.


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1.10 ECONOMIC ASPECTS AND FUTURE TRENDS The economics of acetaldehyde processes are highly sensitive to product yield and raw material prices. This is apparent from the raw material and energy requirements and the relative simplicity of the processes. Therefore the process with a significantly lower net raw material cost will have an advantage. This is the reason ethylene oxidation processes have dominated the acetaldehyde for the last few years and they will probably continue to do so at least until the mid 1980s. The production pattern for acetaldehyde has undergone significant changes since the principal industrial routes to acetaldehyde were hydration of acetylene and oxidation of ethanol. First increasing acetylene costs made this feedstock economically unattractive. Then the two aldehyde WackerHoechst GmbH processes for the liquid phase oxidation of ethylene to acetaldehyde began commercial operation. Western Europe and China are the largest consumers of acetaldehyde. Import and export of acetaldehyde are negligible as it is mainly consumed at the process site.


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1.11 HEALTH AND SAFETY FACTORS The main toxic effect of acetaldehyde is irritation of the eyes and membranes. Anesthesia is the dominant effect of acetaldehyde. Acetaldehyde appears to paralyze respiratory muscles, causing panic. It has a general narcotic action which accelerates heart action. During industrial exposure acetaldehyde vapors produce local irritation to the eyes and to a lesser extent to the throat and nose. Coughing, burning sensation, headaches and stupefaction are symptoms of gross exposure. Prolonged exposure causes a decrease of both red and white blood cells; there is also a sustained rise in blood pressure. In normal industrial operations there is no health hazard in handling acetaldehyde provided normal precaution are taken. Mixtures of acetaldehyde vapor and air are flammable; they are explosive if the concentrations of aldehyde and oxygen rise above 4% and 9% respectively. The maximum allowable concentration in air has been set to 200 ppm. In water concentration are generally less than 0.1 µg/L, therefore contribution from drinking water are considered negligible. Acetaldehyde and its vapors have a highly irritative effect on the eyes, nose and respiratory passages and a slightly irritative effect on the skin. Irreversible injury is possible. Those working with acetaldehyde should wear protective clothing, protective gloves and tightly fitting goggles. Any splashes on the skin or in the eyes should be rinsed away with plenty of water (a doctor should be called if necessary). Contaminated clothing should be changed at once. If acetaldehyde is swallowed, a doctor should be called immediately In the case of fire, water, carbon dioxide, and dry extinguishing agents should be used for fire fighting. Foam is less suitable because acetaldehyde vapors break through the foam blanket owning to their high pressure at normal temperature.


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2. PRODUCTION Raw materials that have been used for the production of acetaldehyde are: 1. Ethanol from fermentation of carbohydrates or from hydration of ethylene. 2. Acetylene 3. Ethylene 4. Lower hydrocarbon e.g. butane 5. Carbon monoxide and hydrogen (Synthesis gas) 6. Methanol

1. i) From Ethanol and Air : Acetaldehyde is produced commercially by the catalytic oxidation of ethyl alcohol. Passing alcohol vapors and preheated air over a silver catalyst at 4800 C carries out the oxidation. CH3CH2OH + ½ O2 → CH3CHO + H2O, Ethanol

∆H = 242 KJ/mol (57.84 kcal / mol)

Oxygen With a multitubular reactor, conversions of 74-82% per pass can

be obtained while generating steam to be used elsewhere in the process.

1. ii) From Dehydrogenation of Ethyl alcohol : Acetaldehyde

also,

produced

commercially

by

the

dehydrogenation of ethyl alcohol. Reaction: C2H5OH Ethanol

→

CH3CHO Acetaldehyde

+

H2


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2. Oxidation of Ethylene: Wacker – Chemie and Farbwerke Hoechst developed the direct liquid phase oxidation of ethylene in 1957 – 1959.

The catalyst is an aqueous

solution of PdCl2 and CuCl2. In 1894, F.C. Phillips observed the reaction of ethylene with an aqueous palladium chloride solution to form acetaldehyde. C2H4+PdCl2 + H2O → CH3CHO +Pd +2HCl The metallic palladium is reoxidized to PdCl2 with CuCl2 and the cuprous chloride formed is reoxidized with oxygen or air. Pd + 2CuCl2→ PdCl2 +2CuCl 2CuCl+1/2 O2 + 2HCl → 2CuCl2 + H2O The net result is a process in which ethylene is oxidized continuously through a series of oxidation – reduction reactions. C2H4 + ½ O2→ CH3CHO

∆H = -244 kJ(102.1 kcal)

Studies of the reaction mechanism of the catalytic oxidation have suggested that a cis – hydroxyethylene – palladium π complex is formed initially, followed by an intramolecular exchange of hydrogen and palladium to give a gem – hydroxyethyl palladium species which leads to acetaldehyde and metallic palladium. There are two variations for the production of acetaldehyde by the oxidation of ethylene; the two – stage process developed by Wacker – Chemie and the one – stage process developed by Farbwerke Hoechst. In the two – stage process ethylene and oxygen (air) react in the liquid phase in two stages. In the first stage ethylene is almost completely converted to acetaldehyde in one pass in a tubular plug-flow reactor made of titanium. The reaction is conducted at 125-1300C Production of Acetaldehyde

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and 3 atm palladium and cupric chloride catalysts. Acetaldehyde produced in the first reactor is removed from the reaction loop by adiabatic flashing in a tower. The flash step also removes the heat of reaction. The catalyst solution is recycled from the flash – tower base to the second stage (or oxidation) reactor where the cuprous salt is oxidized to the cupric state with air. The high pressure off – gas from the oxidation reactor, mostly nitrogen, is separated from the liquid – catalyst solution and scrubbed to remove acetaldehyde before venting. A small portion of the catalyst stream is heated in the catalyst regenerator to destroy undesirable copper oxalate. The flasher overhead is fed to a distillation system where water is removed for recycle to the reactor system and organic impurities, including chlorinated aldehydes, are separated from the purified acetaldehyde product. In the one-stage process ethylene, oxygen, and recycle gas are directed to a vertical reactor for contact with the catalyst solution under 3 atm pressure and 1300 C  . The water evaporated during the reaction absorbs the heat evolved, and make – up water is fed as necessary to maintain the catalytic solution concentration.

The gases are water – scrubbed and the resulting acetaldehyde

solution is fed to a distillation column. The tail – gas from the scrubber is recycled to the reactor. Inerts are eliminated from the recycle gas in a bled – stream which flows to an auxiliary reactor for additional ethylene conversion. This oxidation process for olefins has been exploited commercially principally for the production of acetaldehyde, but the reaction can also be applied to the production of acetone from propylene and methyl ethyl ketone from butanes. Careful control of the potential of the catalyst with the oxygen stream induced commercially by a variation of this reaction.


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3. From Acetylene Although acetaldehyde has been produced commercially by the hydration of acetylene since 1916, this procedure has been replaced almost completely by the direct oxidation of ethylene. HC≡ CH

+

Acetylene

H2O

→

Water

CH3CHO Acetaldehyde

4. From Synthesis Gas A rhodium catalyzed process capable of converting synthesis gas directly into acetaldehyde in a single step has been reported. CO

+

H2

→

CH3CHO

+

other products


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3. SELECTION OF PROCESS The economy of the commercial processes depends essentially upon prices and the availability of raw materials. In highly industrializedCountries maintaining high prices for ethanol by Fiscal measures or where petrochemical ethanol was not available, as in Germany or Japan before 1939, Acetylene was the favored starting material for acetaldehyde. The acetylene process is still operated in some Eastern European Countries and also by companies where cheap acetylene is available. Petro chemically produced ethanol, however, was the favored raw material in the other countries whereas ethanol made by fermentation was and still is used on a small scale in countries with less chemical industry. In Western countries including Japan, all these processes have now been almost completely replaced by the direct oxidation process developed in the late 1950s by Wacker-Chemie and Hoechst. This is because ethylene is available at lower price then acetylene. Generally all processes based on acetylene, ethylene and ethanol are more selective than the oxidation of saturated hydrocarbons. This is because in the later case, other oxidation products are formed in addition to acetaldehyde. Because of the great expense of separating the product mixture, such processes are economically only in large units and when all main and secondary products obtained in the process are utilized. The ethylene route has become the most important. Since 1960, the liquid phase oxidation of ethylene has been the process of choice for the manufacture of acetaldehyde. In the 1970s the world capacity of this process increases to over 2x106 T/annum.


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4. MANUFACTURE OF ACETALDEHYDE FROM ETHYLENE Process Description: A flow sheet for the single stage reactor process using oxygen is shown in Figure. Fresh oxygen and ethylene are fed independently to a vertical ceramic- lined reactor vessel which contains a water solution of catalyst. During the process development serious problems have been caused by the extremely corrosive aqueous CuCl2 – PdCl2 solution. In one stage process the reactor is lined with acid proof ceramic material, the tubing is made of titanium, and certain others parts are of titanium .The oxygen content of the recycle gas is limited to a maximum of 9 mol % to stay safely below the flammable range. The reactor is operated to 120 – 130

0

C and about 3 atm. The heat of reaction is removed by evaporation of

acetaldehyde and water from the catalyst solution. The ethylene conversion per pass is about 75 %. The vapor from the reactor is cooled by 20

0

C and scrubbed with

water to condense and remove the acetaldehyde. The scrubbing operation, which is done at low pressure, requires a high scrub water volume and is responsible for the low concentration of acetaldehyde (∼ 9 wt %) in the crude product. Most of the vent gas is recycled to recover ethylene, but a gas purge is necessary to remove inerts from the system. For this reason, high purity ethylene and oxygen are required to minimize ethylene losses in the purge. The residue from the scrubber contains 8 to 10 wt % acetaldehyde and is fed to distillation column. There are two way of manufacture:


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1. One Stage Process 2. Two Stage Process One Stage versus Two Stage Process →

Advantages of the One Stage Over the Two Stage Process are : ∗

Less by product formation.

∗

No liquid recycle, the catalyst always stays in one vessel, minimizing possibilities of catalyst loses and avoiding expensive liquid recirculation equipment.

∗

Lower operating pressure which results in lower compression costs for oxygen vs. air.

∗

Lower volume of vent gases.

∗

Effluent may be biologically treated because of the higher dilution of the chlorinated acetaldehydes.

∗

Lower in unit capital costs due to use of ceramic equipment and less titanium because the diluted streams are less corrosive.

→

Advantages of the Two Stage over the One Stage Process are : ∗

Safety (Ethylene and the air are reacted in different vessels).

∗

No ethylene or oxygen recycle (almost complete conversion).

∗

Less liquid effluent.

Temp 0 C

P

One

phase Liquid

Stage Two

Liquid

Process

Reactor

Catalyst

Conversio

Yield %

100-130

(atm) 3

Palladium

n% 75

93-95

125-130

10

Palladium

99.5

93-95

Stage Production of Acetaldehyde

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As it can be seen that in the above comparison both variations of the process offer advantages and disadvantages. Also since the process are very raw material sensitive and the ethylene yields are the same then the economics are probably very close. All of this reflected in the fact that there are about the same number of one and two stage units in operation.


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5. Thermodynamics:The word ‘thermodynamics’ implies flow of heat. It deals with energy changes accompanying all types of physical and chemical processes. Thermodynamics also helps to lay down the criteria for predicting feasibility or spontaneity of a process, including a chemical reaction, under a given set of conditions. In other words, it helps to predict whether a given process or a chemical reaction is feasible under the given conditions of temperature, pressure and concentration. It also helps to determine the extent to which a process, including a chemical reaction, can proceed before attainment of equilibrium. Therefore to determine the feasibility of a chemical reaction Gibb’s free energy change of chemical reaction is determine and the conditions for the feasibility of a chemical reaction are mentioned below: ΔG < 0, reaction can take place spontaneously, i.e. it is feasible 0 < ΔG <40,000 KJ/kmol, the reaction may or may not be possible and needs further study. ΔG > 40,000 KJ/kmol, the reaction is very unfavorable. Free

energy

of

formation

is

considered

as

a

function

of

temperature, while doing calculations.


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Thermodynamic properties of raw materials & product are as given below:Property as a function of temperature can be written as Property = A + BT + CT2 +DT3 + ET4 (i)

Ethylene :-

Property

A

B

C

D

E

Specific Heat(Cp), KJ/kmol K ∆Hf (KJ/mol) ∆G (KJ/mol)

32.003

-1.4831 x 10-2

2.4774 × 10-4

-2.3766× 10-7

6.8274 × 10-11

63.053

-4.1076 × 10-2

1.6598×10-5

-

-

51.752

4.9338 × 10-2

1.7284 × 10-5

-

-

(ii)

Acetaldehyde :-

Property

A

B

C

D

E

Specific Heat(Cp), KJ/kmol K ∆Hf (KJ/mol)

34.140

4.002 × 10-2

2.5034 × 10-4

-1.6445 × 10-7

4.7248×10-11

-154.122

-4.7166 ×10-2

2.0279 × 10-5

-

-

-167.052

1.0714 × 10-1

1.8665 × 10-5

-

-

∆G (KJ/mol)


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Reaction :C2H4 Ethylene

+

0.5O2 Oxygen

→

CH3CHO Acetaldehyde

2C2H4 Ethylene

+

O2 Oxygen

→

2CH3CHO Acetaldehyde

a)

Heat of Reaction :-

Ethylene :∆H0 = A + BT + CT2 (KJ/mol) ∆H0 = 63.053 – 4.1076×10-2T + 1.6598×10-5×T2

Therefore. at 298K ∆H0298 = 63.053 – 4.1076×10-2 x 298 + 1.6598×10-5×2982 = 52.286 KJ/mol

And, at 130 ºC that is at 403 K ∆H0403 = 63.053 – 4.1076×10-2 x 403 + 1.6598×10-5×4032 = 49.195 KJ/mol

Acetaldehyde:∆H0 = A + BT + CT2 (KJ/mol) ∆H0 = -154.122 – 4.7166×10-2×T + 2.0279×10-5×T2


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Therefore. at 298K ∆H0298 = -154.122 – 4.7166×10-2×298 + 2.0279×10-5×2982 = -166.377 KJ/mol

And at 403K ∆H0403 = -154.122 – 4.7166×10-2×403 + 2.0279×10-5×4032 = -169.836 KJ/mol Therefore, At 298K ∆H0reaction = ∑ H products – ∑ H reactants = 2 [(∆H0acetaldehyde ) – (∆H0ethylene )] ] = 2 [(-166.377) – (52.286)] = - 437.336 KJ/mol

And at 403K ∆H0reaction = ∑ H products – ∑ H reactants = 2[ (∆H0acetaldehyde) – (∆H0ethylene)] = 2[(- 169.836) – (49.195)] = - 438.062 KJ/mol

(b) Feasibility of Reaction :-


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Ethylene :-

∆G0 = A + BT + CT2 (KJ/mol) ∆G0 = 51.752 + 4.9338×10-2×T + 1.7284 ×10-5×T2 Therefore, At 298K ∆G0298 = 51.752 + 4.9338×10-2×298 + 1.7284 ×10-5×2982 = 67.9896 KJ/mol

Acetaldehyde :∆G0 = A + BT + CT2 (KJ/mol) ∆G0 = -167.052 + 1.0714×10-1×T + 1.8665×10-5×T2 Therefore, At 298K ∆G0298 = -167.052 + 1.0714×10-1×298 + 1.8665×10-5×2982 = -135.1243 KJ/mol Thus, ∆G0reaction = 2[∑ ∆G products – ∑ ∆G reactants]

= 2[(∆G0acetaldehyde ) – ( ∆G0ethylene)] Therefore, at 298K ∆G0298

= 2(-135.1243) – (-67.9896) = - 406.2278 KJ/mol

Now, ∆G0298 = -RT ln K ln K1 = (406.2278×1000)/(8.314 × 298) = 163.962 Production of Acetaldehyde

L.I.T. Nagpur

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Therefore, K298 = 1.61×1071

Now, dlnK ΔH o = dT RT

Therefore, ln(K2/K1) = (-∆H0298 /R) × [(1/T2)-(1/T1)] ln(K2/K1) = (-437.336×1000/ 8.314) × [(1/403)-(1/298)] = 45.99

Therefore, (K2/K1) = 9.412 × 1019 K2 = 1.711 x 1051 ∆G0 403 = - RT (ln K2) = -8.314 × 403 × ln (1.711 x 1051) = -395.258 KJ/mol Since, ∆G0 403 is negative, the reaction is feasible at 403K


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6. Material Balance Plant capacity = 4500 Ton/Annum = 4500/300 =15 T/day = (15 x 1000)/3600 = 625 Kg/hr = 14.21 Kmol/hr Therefore Acetaldehyde to be produced = 625 Kg/hr The main reaction taking place in the reactor is C2H4

+

0.5O2

Ethylene

→

Oxygen

CH3CHO Acetaldehyde

From available data, For 1000 Kg of Acetaldehyde to be produced, Ethylene required is 670 Kg and Oxygen required is 410 Kg The above reaction can again be written as C2H4 Ethylene

+

0.5O2

→

CH3CHO

Oxygen

Acetaldehyde

Quantity

670

410

1000

(Kg) Molecular

28

32

44

Weight Moles Stiochiometric

23.93 1

12.81 0.5

22.73 1

amount For finding out the Limiting and excess reactant the above equation can be written as A

+

0.5B →

C Production of Acetaldehyde

L.I.T. Nagpur

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Where, A = Ethylene B = Oxygen C= Acetaldehyde The stiochiometric ratio of the reactant can be calculated as,

The feed ratio is calculated as,

As Stiochiometric ratio is more than the feed ratio, ‘A’ i.e. C2H4 is Limiting and ‘B’ i.e. O2 is in Excess. Basis : For 1 hr of operation, % conversion based on C2H4 fed = C2H4 reacted/C2H4 available 0.75 ∴

=

x /y

x = 0.75 y

% yield based on C2H4 reacted = CH3CHO produced/ C2H4 reacted 0.95

=

CH3CHO produced/ x

As acetaldehyde formed is 99.5 % pure ∴ Acetaldehyde to be manufactured = 14.21/0.995 = 14.28 Kmol Assume 6 % overall loss, ∴ Acetaldehyde produced in the first column i.e. reactor = 14.28 (1+0.06) = 15.14 Kmol


L.I.T. Nagpur

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∴

0.95

=

15.14/ x

⇒

x

=

15.94 Kmol

∴

y

=

x /0.75

⇒

y

=

21.25 Kmol

∴ Ethylene fed

=

21.25 Kmol

Ethylene reacted

=

15.94 kmol

Unreacted Ethylene

=

21.25 – 15.94

=

5.31Kmol

Calculation of O2 requirement: From the main reaction, 1 Kmol of C2H4

≡

0.5 Kmol of O2

15.94 Kmol of C2H4

≡

7.97 Kmol of O2

O2 reacted = 7.97 Kmol = O2 consumed Assume 42 % excess air to be supplied, O2 in air supplied = 1.42 x 7.97 = 11.32 Kmol N2 in air supplied = (1/99) x 11.32 = 0.114 Kmol ∴ O2 in air = 11.32 Kmol N2 in air = 0.114 Kmol This N2 will remain inert over the complete process, ∴ N2 in product from reactor = 0.114 Kmol Oxygen fed

=

11.32 Kmol

Oxygen reacted

=

7.97 Kmol


L.I.T. Nagpur

33

Oxygen unreacted = =

11.32 – 7.97 3.347 Kmol

Over the Reactor and Scrubber : Mixed Components Fed are Ethylene C2H4

=

21.25 Kmol

Oxygen O2

=

11.32 Kmol

Nitrogen

=

0.114 Kmol

Quantity reacted are Ethylene C2H4

=

15.94 Kmol

Oxygen O2

=

7.97 Kmol

Quantity unreacted are Ethylene C2H4

=

5.31 Kmol

Oxygen O2

=

3.347 Kmol

Nitrogen

=

0.114 Kmol

Components which are unreacted passes through the scrubber ∴ Gas leaving the scrubber from top

= 5.31 + 3.347 +0.114 = 8.771 Kmol

mol % of N2 in gas leaving the scrubber from top = (0.114/8.771) = 0.01299 Concentration of N2 in gas leaving the scrubber from the top, in purge and in recycle will be same. Production of Acetaldehyde

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Let, P be the Kmol of Purge and R be the Kmol of Recycle feed For steady state operation to avoid the built of N2 in the system we have, N2 in fresh feed = N2 in purge = 0.01299 P Let, m be the kmol of O2 in purge n be the kmol of O2 in fresh feed

O2 balance after the scrubber will be, O2 in gas leaving the scrubber from top = O2 in purge + O2 in recycle ∴ O2 in recycle =3.347 – m

O2 balance before reactor will be, O2 in fresh feed + O2 in recycle = O2 in mixed feed ∴ n + 3.347 – m = 11.32 n = 7.973 + m Air contains 99 mol % of O2 and 1 mol % N2 ∴ N2 in fresh feed = (1/99) x (7.973 + m) = 0.01 (7.793 + m)

Gas leaving the scrubber = 8.771 kmol ∴ mole fraction of O2 in gas leaving the scrubber from the top = 3.347/8.771 = 0.3816 mole fraction of O2 in purge = 0.3816


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As concentration of O2 in gas leaving scrubber from the top, purge, and recycle is same. N2 in fresh feed = 0.01 (7.973 + m) = 0.01299P ⇒

P = 0.769 (7.973 + m) (O2 in purge/Total purge) = (m/P)

∴ (m/P) = (m)/ [0.769(7.973 + m)] = 0.3816 ⇒

m = 3.31 kmol

O2 in purge = 3.31 kmol

N2 in fresh feed = 0.01 (7.973 + m) = 0.1128 kmol N2 in purge = N2 in fresh feed = 0.1128 kmol

P = 0.769 (7.973 + m) ⇒

P = 8.677 kmol

C2H4 in purge = P - O2 in purge - N2 in purge = 8.6771 – 3.31 – 0.1128 C2H4 in purge = 5.2542 kmol

We have, R + P = 8.771 R = 8.771 – 8.6771


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R = 0.094 kmol

N2 balance after scrubber: N2 in gas leaving the scrubber from top = N2 in recycle + N2 in purge N2 in recycle = 0.114 – 0.1128 N2 in recycle = 0.0012 kmol

O2 balance after scrubber: O2 in recycle = O2 in gas leaving the scrubber from top - O2 in purge = 3.347 – 3.31 O2 in recycle= 0.037 kmol

C2H4 balance after scrubber: C2H4 in recycle = C2H4 in gas leaving the scrubber from top - C2H4 in purge = 5.31 – 5.2542 C2H4 in recycle = 0.0558 kmol

O2 balance before reactor: O2 in fresh feed = O2 in mixed feed - O2 in recycle = 11.32 – 0.037 O2 in fresh feed = 11.283 kmol

C2H4 balance before reactor: C2H4 in fresh feed = C2H4 in mixed feed - C2H4 in recycle = 21.25 – 0.0558 C2H4 in fresh feed = 21.1942


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Fresh feed to the process = F F = 21.1942 + 11.283 + 0.1128 = 32.59 kmol

Recycle ratio = (R/F) = (0.094/32.59) = 0.288 x 10-2

Common feed ratio = (M F/F F) = (21.25 + 11.32 + 0.114)/32.59 = 1.003

C2H4 into the process = C2H4 in fresh feed C2H4 out = C2H4 in purge

Overall conversion = [C2H4 in - C2H4 out]/C2H4 in = (21.1942 – 5.2542)/21.1942 = 75.21 % After reactor: C2H4 unreacted = 5.31 kmol = 148.68 kg O2 unreacted = 3.347 kmol = 104.104 kg N2 unreacted = 0.114 kmol = 3.192 kg Total kmoles = 8.771 Total mass coming out = mass of unreacted C2H4, O2, N2 + mass of CH3CHO Outlet of reactor consist 15.14 kmol of acetaldehyde


L.I.T. Nagpur

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∴ kg of acetaldehyde will be = 15.14 x 44 = 666.16 kg Total output coming from reactor in kg = 666.16 + 148.68 +107.104 +3.192 = 925.136 kg

Taking balance over Srubber : Taking overall balance on scrubber: Input mass = Output mass K+L=M+N Where, K = total output in kg from reactor L = total water supplied M = Reused N = Scrubbed

10 % of acetaldehyde is present in scrubbed ∴ 925.136 + L =M + N

Taking acetaldehyde balance on weight basis, 666.16 + L x 0 = M x 0 + N x 0.1 As water used is pure and no acetaldehyde is reused. ⇒

N = 6661.6 kg

∴ 925.136 + L = (148.68 + 107.104 + 3.192) + 6661.6 ⇒

L = 5995.44 kg

Water to be fed = 5995.44 kg ≈ 6 m3 Production of Acetaldehyde

L.I.T. Nagpur

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Taking balance over Distillation Column: Total feed to the distillation column = 6661.6 kg Feed contains 10 % acetaldehyde and remaining 90 % water

Xf1 (CH3CHO) = 0.1 Xf2 (H2O) = 0.9 Top Product (D):Mole fraction of components in the top product:Xd1 (CH3CHO) = 0.96 Xd2 (H2O) = 0.04

Bottom Product (W): Mole fraction of components in the bottom product:Xw1 (CH3CHO) = 0.01 Xw2 (H2O) = 0.99 Taking overall material balance:F= D + W Putting the values in above equation :6661.6 = D + W --------------------------------------------- (1)

Now, taking component balance:F Xf= D Xd +W Xw Production of Acetaldehyde

L.I.T. Nagpur

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For Acetaldehyde:6661.6 × 0.1 = D × 0.96 + W × 0.01 666.16 = 0.96 D + 0.01 W -------------------------------- (2)

Solving equations 1 & 2, we get:Top product (D) = 631.1 Kg Bottom product (W) = 6030.5 Kg

Hence, acetaldehyde in the top product = 0.96 × 631.1 = 605.94 Kg And, Water in the top product = 0.04 × 631.1 = 25.26 Kg

Acetaldehyde in the bottoms product = 0.01 × 6030.5 = 60.46 Kg Water in the bottoms product = 0.99 × 6030.5 = 5969.94 Kg

Reflux Ratio = 0.69 (calculated below) Hence, Ln = 0.69 × D = 0.69 × 631.1 = 435.46 Kg Vn = Ln + D = 435.46 + 631.1 =1066.56 Kg Lm = Ln+ F = 435.46 + 6661.6 = 7097.06 Kg Vm = Lm – W = 7097.06-6030.5 = 1066.56 Kg

Degasser: Feed contains 96 % of acetaldehyde and Output from bottom contains 95.5 % acetaldehyde Taking overall balance on Degasser,

I=G+O Production of Acetaldehyde

L.I.T. Nagpur

41

Where, I = input to degasser G = Gas leaving from top O = final output 631.1 = G + O

Taking acetaldehyde balance, 631.1 x 0.96 = 0 x G + 0.995 x O Solving equations, we get O = 608.9 kg G = 22.2 kg


L.I.T. Nagpur

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7. Energy Balance Properties required for energy balance calculation 1. Specific Heat as a function of temperature

Component Ethylene Oxygen Nitrogen Acetaldehyd e Water

A 32.003 29.526 29.342 34.140 92.053

B C -2 -1.4831× 10 2.4774× 10-4 -8.8999× 10-3 3.8083× 10-5 -3.5395× 10-3 1.0076× 10-5 4.0020× 10-2

2.5034× 10-4

D -2.3766× 10-

E 6.8274× 10-

7

11

-3.2629× 10-

8.8607× 10-

8

12

-4.3116× 10-

2.5935× 10-

9

13

-1.6445× 10-

4.7248× 10-

7

11

-3.9953× 10-2 -2.1103× 10-4 5.3469× 10-7

-

2. Latent Heat of Vaporization KJ/kg

Component Acetaldehyde Water

i.

Temperature K 294 294

λ 1131.24 2451.32

Reactor

Reaction temperature = 403 K Reference temperature = 298 K Outlet temperature = 383 K

Heat in Feed:

Q1 + Q2 + Q3 Production of Acetaldehyde

L.I.T. Nagpur

43

Q1 = Heat energy associated with Ethylene Q2 = Heat energy associated with Oxygen Q3 = Heat energy associated with Nitrogen a. Ethylene Moles of Ethylene entering = 21.25 kmol Tenter

Q1

=

∫ CpdT

n×

Tref

403

= 21.25 ×

(32.003 − 1.4831× 10−2 T + 2.4774× 10−4 T 2 −

∫ 2.3766× 10

298

−7

T 3 + 6.8274× 10−11T 4 )dT

= 21.25 (5048.5) = 107.28 × 103 KJ

b. Oxygen Moles of Oxygen entering = 11.32 kmol Tenter

Q2

=

n×

∫ CpdT

Tref

403

= 11.32 ×

(29.526 − 8.8999× 10−3 T + 3.8083× 10−5 T 2 −

∫ 3.2629× 10

298

−8

T 3 + 8.8607× 10−12 T 4 )dT

= 11.32 (3130.74) = 35.44 × 103 KJ

c. Nitrogen


L.I.T. Nagpur

44

Moles of Nitrogen entering = 0.114 kmol Tenter

Q3

=

n×

∫ CpdT

Tref

403

= 0.114×

(29.342 − 3.5395× 10−3 T + 1.0076× 10−5 T 2 −

∫ 4.3116× 10

298

−9

T 3 + 2.5935× 10−13 T 4 )dT

= 0.114 (3062.02) 3

= 0.3491 × 10 KJ

Heat in input = Q1 + Q2 + Q3 = (107.28 + 35.44 + 0.34907) × 103 = 143.07× 103 KJ

Heat associated with output = H1 +H2 + H3 + H4 H1 = Heat energy associated with Ethylene H2 = Heat energy associated with Oxygen H3 = Heat energy associated with Nitrogen H4 = Heat energy associated with Acetaldehyde


L.I.T. Nagpur

45

a) Ethylene Moles of Ethylene leaving = 5.31 kmol Texit

H1 =

∫ CpdT

n×

Tref

383

(32.003 − 1.4831× 10−2 T + 2.4774× 10−4 T 2 −

∫ 2.3766× 10

= 5.31×

298

−7

T 3 + 6.8274× 10−11T 4 )dT

3

= 25.419 × 10 KJ

b) Oxygen

Moles of Oxygen leaving = 3.35

Texit

H2 =

n×

∫ CpdT

Tref

383

= 3.35 ×

(29.526 − 8.8999× 10−3 T + 3.8083× 10−5 T 2 −

∫ 3.2629× 10

298

−8

T 3 + 8.8607× 10−12 T 4 )dT

3

= 9.985 × 10 KJ

c) Nitrogen Production of Acetaldehyde

L.I.T. Nagpur

46

Moles of Nitrogen leaving = 0.114 kmol Texit

H3 =

∫ CpdT

n×

Tref

383

= 0.114 ×

(29.342 − 3.5395× 10 −3 T + 1.0076× 10 −5 T 2 −

∫ 4.3116× 10

298

−9

T 3 + 2.5935× 10 −13 T 4 )dT

3

= 0.332× 10 KJ

d) Acetaldehyde

Moles of Acetaldehyde leaving =15.14 kmol

Texit

H4 =

n×

∫ CpdT

Tref

383

=

15.14 ×

(34.140 + 4.002 × 10 −2 T + 1.5634× 10−4 T 2 −

∫ 1.6445× 10

298

−7

T 3 + 4.7248× 10 −11 T 4 )dT

3

= 109.38 × 10 KJ

Heat associated with output

= H1 +H2 + H3 + H4 = (25.419 + 9.985 + 0.332 + 109.38) × 103 = 145.116 × 103 KJ

Heat generated during the reaction = - 244 KJ/kmol Production of Acetaldehyde

L.I.T. Nagpur

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∴ For 15.14 kmol of acetaldehyde heat generated will be = - 244 × 15.14

∴ Heat released during the reaction

= Input – Output + Generation = (143.07 – 145.116 – 3.695) × 103 = - 5.741 × 103 KJ

Negative sign indicates that reaction is exothermic

ii.

Around Cooler

Heat lost or removed = Heat gained by the cooling medium (water) Inlet temperature of water = 25ºC Outlet temperature of water = 40ºC Cooling water flowrate =? Heat gained = mwater × Cpwater × ∆T m water = 5.741 × 103 / [4.184 × (313-298)] = 91.44 kg ≈ 100 kg

iii.

Balance around Scrubber Production of Acetaldehyde

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Taking Energy Balance around the scrubber, we get

I–O+G-C=0 As there is no generation of heat term G will be = 0

I–O-C=0 Where I = Input O = Output C = Consumption or Removed

Input Heat Calculation: I1 + I2 + I3 + I4 + I5 Where I1 = Heat associated with Ethylene I2 = Heat associated with Oxygen I3= Heat associated with Nitrogen I4 = Heat associated with Acetaldehyde I5 = Heat associated with water entering Inlet temperature = 383 K Outlet temperature = 313 K Water Inlet temperature = 298 K A. Ethylene Moles of ethylene entering = 5.31 kmol Texit

I1 =

n×

∫ CpdT

Tenter


L.I.T. Nagpur

49

(32.003 − 1.4831× 10 −2 T + 2.4774× 10 −4 T 2 −

313

= 5.31×

∫ 2.3766× 10

383

−7

T 3 + 6.8274× 10 −11 T 4 )dT

= - 13.455 × 103 KJ B. Oxygen Moles of oxygen entering = 3.35 kmol Texit

I2 =

∫ CpdT

n×

Tenter

313

= 3.35 ×

(29.526 − 8.8999× 10 −3 T + 3.8083× 10−5 T 2 −

∫ 3.2629× 10

383

−8

T 3 + 8.8607× 10−12 T 4 )dT

= - 6.988 × 103 KJ

C. Nitrogen Moles of nitrogen entering = 0.114 kmol Texit

I3 =

n×

∫ CpdT

Tenter

313

= 0.114 ×

(29.342 − 3.5395× 10−3 T + 1.0076× 10−5 T 2 −

∫ 4.3116× 10

383

−9

T 3 + 2.5935× 10−13 T 4 )dT

= - 0.2356 × 103 KJ

D. Acetaldehyde Moles of acetaldehyde entering = 15.14 kmol


L.I.T. Nagpur

50 Texit

∫ CpdT

n×

I4 =

Tenter

313

=

15.14 ×

(34.140 + 4.002 × 10 −2 T + 1.5634× 10 −4 T 2 −

∫ 1.6445× 10

383

−7

T 3 + 4.7248× 10−11T 4 )dT

= - 64.404 × 103 KJ

E. Water

5

water

I

= (m

Cp∆T)

= 5995.44 × 4.186 × (40 - 25) = 376.45 × 103 KJ

Total heat in input

= (-13.455 – 6.988 – 0.2356 – 64.404 + 376.45) × 103 3

= 291.37× 10 KJ

1

2

3

4

5

Outlet heat calculation: V + V + V + V + V

Where,


L.I.T. Nagpur

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1

V = Heat associated with ethylene from top

2

V = Heat associated with oxygen from top

3

V = Heat associated with nitrogen from top

4

V = Heat associated with acetaldehyde from bottom

5

V = Heat associated with water from bottom

Temperature remains constant at output i.e. temperature in scrubber = outlet

temperature = 313 K 1)

Ethylene from top -2

1

-4

2

-7

3

-11

4

V = 5.31 (32.003 – 1.4831× 10 T+2.4774× 10 T -2.3766× 10 T +6.8274× 10 T )

= 238.94 KJ

2) Oxygen from top V2 = 3.35 (29.526 – 8.8999× 10-3 T+3.8083× 10-5 T2-3.2629× 10-8 T3+8.8607× 10-12T4) = 99.012 KJ Production of Acetaldehyde

L.I.T. Nagpur

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3) Nitrogen from top V3 = 0.114 (29.342 – 3.5395× 10-3 T+1.0076× 10-5 T2-4.3116× 10-9 T3+2.5935× 1013

T4) = 3.316 KJ

4) Acetaldehyde from bottom V4 =15.14 (34.14+4.002× 10-2 T+2.5034× 10-4 T2-1.6445× 10-7 T3+4.7248× 10-11T4) = 1.0084 × 103 KJ

5) Water from bottom V5 = 333.08 (92.053 –3.9953× 10-2 T-2.1103× 10-4 T2+5.3469× 10-7 T3) = 25.071× 103 KJ

Total heat in output = 238.94 + 99.012 + 3.316 + (1.0084 + 25.071) × 103 KJ = 26.42 × 103 KJ

Heat removed

= Total Input heat – Total Output heat = (291.37 – 26.42) × 103 KJ = 264.95 × 103 KJ

iv.

Distillation Column :-

Temperature of feed = 313K


L.I.T. Nagpur

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Temperature of distillate = 294K Temperature of the bottoms product = 374K Reference Temperature = 298K

Heat in feed (F×HF) = Heat in acetaldehyde + Heat in water

Texit

=

∫ CpdT

nacetaldehyde ×

Texit

nwater ×

+

Tenter

313

=

15.14 ×

∫ CpdT

Tenter

(34.140 + 4.002× 10−2 T + 1.5634× 10−4 T 2 −

∫ 1.6445× 10

298

−7

T 3 + 4.7248× 10−11T 4 )dT +

313

333.08 ×

(92.053 − 3.9953× × 10−2 T −

∫ 2.1103× 10

298

−4

T 2 + 5.3469× 10− 7 T 3 )dT

=

13.088 × 103 + 1.131× 103

=

14.219 × 103 KJ/hr

Heat in Distillate (D×Hd) = Heat in acetaldehyde + Heat in water Texit

=

nacetaldehyde ×

∫ CpdT

Tenter

Texit

+

nwater ×

∫ CpdT

Tenter


L.I.T. Nagpur

54 294

=

13.77 ×

(34.140 + 4.002× 10−2 T + 1.5634× 10−4 T 2 −

∫ 1.6445× 10

298

−7

T 3 + 4.7248× 10−11T 4 )dT +

294

1.403×

(92.053 − 3.9953× × 10−2 T −

∫ 2.1103× 10

298

(D×Hd)

=

-3.073× 103 – 0.424 × 103

=

-3.497 × 103 KJ/hr

−4

T 2 + 5.3469× 10− 7 T 3 )dT

Heat in bottoms (WHW) = Heat in acetaldehyde + Heat in water

Texit

=

nacetaldehyde ×

∫ CpdT

Texit

+

nwater ×

Tenter

374

=

1.374 ×

∫ CpdT

Tenter

(34.140 + 4.002 × 10 −2 T + 1.5634× 10 −4 T 2 −

∫ 1.6445× 10

298

−7

T 3 + 4.7248× 10 −11T 4 )dT +

374

331.66 ×

(92.053 − 3.9953× × 10−2 T −

∫ 2.1103× 10

298

= =

−4

T 2 + 5.3469× 10 − 7 T 3 )dT

6.2249 × 103 + 1896.63 × 103

1920.86 × 103 KJ/hr

Average Latent heat ( Lavg ) = Lacetaldehyde × Xacetaldehyde + Lwater × Xwater = 0.96 × 1131.24 + 2451.32 × 0.04 Production of Acetaldehyde

L.I.T. Nagpur

55

= 1184.04 KJ/kg

Condenser duty ( Qc ) = Vn × Lavg = 1066.56 × 1184.04 × = 1262.85 × 103 KJ/hr

Taking overall Heat balance on distillation column:-

Heat in Feed (FHF) + Reboiler duty (Qr) = Heat in distillate (DHd) +Heat in bottoms WHW) + Condenser duty (Qc) Hence, (Qr) = (DHd) + (WHW) + (Qc) - (FHF) = (-3.497 + 1920.86 + 1262.85 – 14.219) × 103 = 3165.994 × 103 KJ/hr

Therefore, quantity of steam required = (Qr) / Lsteam = 3165.994 × 103 / 2200 = 1439.089 Kg/hr = 0.39975 Kg/sec


L.I.T. Nagpur

v.

56

Taking Energy Balance around degasser :

Inlet temperature of feed = 294 K Outlet temperature = 291 K Taking energy balance over degasser we get

I–O=0 There is one input and two outputs, one for final product and other for gases Heat associated with the input: Input contains 14.34 kmol of Acetaldehyde Texit

Heat in input =

nacetaldehyde ×

∫ CpdT

Tenter

291

= 14.34 ×

(34.140 + 4.002× 10−2 T + 1.5634× 10−4 T 2 −

∫ 1.6445× 10

294

−7

T 3 + 4.7248× 10−11T 4 )dT

= -2.386 × 103 KJ/hr

Texit

Heat in output

=

nacetaldehyde ×

∫ CpdT

Tenter

with acetaldehyde

As temperature remains constant at output i.e. 291 K = 13..84(34.140 + 4.002 × 10− 2 T + 1.5634 × 10− 4 T 2 − 1.6445 × 10− 7 T 3 + 4.7248 × 10− 11T 4 ) Production of Acetaldehyde

L.I.T. Nagpur

∴ Heat associated with acetaldehyde output

57

=

0.765 KJ/hr

Input – Output from top – Output from Bottom = 0 Output from top = Input – Output from Bottom ∴ Heat associated with gases at output from top will be = input – output from bottom = (- 2.386 - 0.765) × 103 = -3.151 × 103 KJ/hr


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8. Equipment Design:Distillation column :Input Feed (F): 1) Acetaldehyde = 666.16 Kg/hr 2) Water = 5995.44 Kg/hr Total feed to the distillation column

= 5995.44+666.16 = 6661.6 Kg/hr

Mole fraction of components in the feed:Xf1 (CH3CHO) = 666.16/6661.6 = 0.1 Xf2 (H2O) = 5995.44/6661.6 = 0.9 Top Product (D) :Mole fraction of components in the top product:Xd1 (CH3CHO) = 0.96 Xd2 (H2O) = 0.04 Bottom Product (W): Mole fraction of components in the bottom product:Xw1 (CH3CHO) = 0.01 Xw2 (H2O) = 0.99 Taking overall material balance:F= D + W Putting the values in above equation:6661.6 = D + W --------------------------------------------- (1) Now, taking component balance:F Xf= D Xd +W Xw Production of Acetaldehyde

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For Acetaldehyde:6661.6 x 0.1 = D × 0.96 + W × 0.01 666.16 = 0.96 D + 0.01 W -------------------------------- (2) Solving equations 1 & 2, we get :Top product (D) = 631.05 Kg/hr Bottom product (W) = 6030.5 Kg/hr Hence, acetaldehyde in the top product = 0.96 × 631.05 = 605.94 Kg/hr Water in the top product =0.04 × 631.05 = 25.26 Kg/hr Acetaldehyde in the bottoms product = 0.01 × 6030.5 = 60.46 Kg/hr Water in the bottoms product = 0.99 × 6030.5 = 5969.94 Kg/hr Boiling point of acetaldehyde = 20 ºC Boiling point of water = 100 ºC Hence, acetaldehyde is more volatile component (MVC)

Sr. No

Componen t

In

Out

F Kmol/hr

D Kg/hr

Kmol/hr

W

Kg/hr

Kmol/hr

Kg/hr

1.

CH3CHO

15.14

666.16

13.77

605.94

1.37

60.46

2.

H2O

333.08

5995.44

1.41

25.26

331.66

5969.94

Mass in

6661.6

Mass out

6661.6

x – y data can be calculated with the help of avg. value of relative volatility. 0

α = PA 0 PB Where A is MVC and B is LVC Numerator and Denominator are the respective Vapor Pressure. Vapor pressure can be calculated using Equation. Production of Acetaldehyde

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0

log10 PA (mm Hg) = A + B/T +C log10 T + D T + E T2

Sr.

Component

A

B

C

D

E

1.

CH3CHO

87.37

-3682.2

-31.548

0.020114

5.534E-13

2.

H2O

29.8605

-3.1522E3

-7.3037

2.4274E-9

1.8090E-6

No.

Using the above equation, the vapor pressure can be calculated at the top and bottom.

Sr.

Condition

No.

0

0

Temperature

PA

PB

°K

(CH3CHO)

(H2O)

0

0

α=PA /PB

1.

Top

294

801

20

40.05

2.

Bottom

374

7727

812

9.52

αavg= (α1 x α2)0.5 = (40.05 x 9.52)0.5 = 19.53 Y = αavg X / (1+ (aavg-1) X Using the above equations the X-Y data is generated as follows X

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

Y

0.51

0.69

0.78

0.83

0.87

0.89

0.91

0.93

0.94

0.95

X

0.55

0.60

0.65

0.70

0.75

0.80

0.85

0.90

0.95

1.0

Y

0.96

0.97

0.97

0.98

0.98

0.99

0.99

0.99

0.99

1.0

From Graph [Xd/(R+1)] min = 0.66 [0.96/(R+1)] min = 0.66 Therefore, Rmin = 0.46 But, Ropt = 1.5 Rmin Production of Acetaldehyde

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Therefore, Ropt = 1.5 x 0.46 = 0.69 R = Ln/D Hence, Ln = 0.69 x 631.1 = 435.46 Kg/hr As D = 631.1 Kg/hr Vn = Ln + D = 435.46 + 631.1 =1066.56 Kg/hr Lm = Ln+ F = 435.46 + 6661.6 = 7097.06 Kg/ hr Vm = Lm – W = 7097.06-6030.5 = 1066.56 Kg/hr Therefore equation of upper of upper operating line : yn =

x R x n +1 + d R +1 R +1

yn =

0.96 0.69 x n +1 + 1.69 1.69

Therefore, yn = 0.408xn+1 + 0.568 Hence y intercept = 0.568 and corresponding slope = 0.408 Equation of Lower operating line (LOL): y m +1 = ∴ y m +1 =

Wx w LM xM − VM VM

6030.5 × 0.01 7097.06 xm − 1066.56 1066.56

∴ y m +1 = 6.654 x m − 0.0565 Therefore from graph Number of theoretical plates = 5-1 = 4 One plate is accounted for rebolier Efficiency of the column = 60 % Therefore actual number of plates required = 4/0.6 = 7 Feed plate = 3

Let tray spacing = 500 mm Height of distillation column = (N+1) x 0.5 = 8 x 0.5


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=4m

Take 100 mm of water as the pressure drop per plate in the column. Therefore, column pressure drop = 100 × 10−3 × 1000 × 9.81× 7 = 6867 Pa = 6.867 KPa. Therefore, Top pressure = 101.325 KPa. Bottom pressure = 101.325 + 6.867= 108.192KPa. At the bottom of the distillation column vapor density can be calculated as follows; ( ρ v ) bottom =

P × M avg RT

∴ ( ρ v ) bottom =

108.192 × (44 × 0.01 + 18 × 0.99) × 273 101.325 × 22.414 × 374

∴ ( ρ v ) bottom = 0.6349 Kg / m 3 . The density of liquid at the bottom of the column can be calculated by multiplying the density of each component with the mole fraction. ( ρ l ) bottom = 784 × 0.01 + 1000 × 0.99 ∴ ( ρ l ) bottom = 997.84 Kg / m 3 . At the top of the distillation column vapor density can be calculated as follows; ( ρ v ) top =

P × M avg

∴ ( ρ v ) top =

RT 101.325 × (44 × 0.96 + 18 × 0.04) × 273 101.325 × 22.414 × 294

∴ ( ρ v ) top = 1.7798 Kg / m 3 .

The density of liquid at the bottom of the column can be calculated by multiplying the density of each component with the mole fraction. Production of Acetaldehyde

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( ρ l ) top = 784 × 0.96 + 1000 × 0.04 ∴ ( ρ l ) top = 792.64 Kg / m 3 .

Calculation of Parachor value (P):Ptop = 118 × 0.96 + 48.4 × 0.04 = 115.216 Pbottom = 118 × 0.01 + 48.4 × 0.99 = 49.096

Surface tension:4

σ top

 P × ( ρ L − ρV )  −12 =  × 10 M   4

∴ σ top

115.216 × (792.64 − 1.727 )  −12 =  × 10  (44 × 0.96 + 18 × 0.04) 

∴ σ top = 20.25dyne / cm = 0.02025 N / m

Similarly:4

σ bottom

 P × ( ρ L − ρV )  −12 =  × 10 M  

∴ σ top = 0.05165 N / m

Liquid-vapor flowrate factor FLV is given by :-

∴ FLV ( top ) =

=

Ln Vn

ρV ρL

435.46 1.7798 = 0.01934. 1066.56 792.64


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FLV ( bottom )

64

ρV ρL

L = m Vm

= 0.1678

Tray spacing = 0.5 m. From graph in RC-6 page-568 we get; Bottom k1 = 0.08, Top k1 = 0.09 For liquid surface tension (0.02 N/m) take K1 as it is otherwise: σ  K 1 ( bottom ) = 0.08  0.02 

0.2

= 0.096 K 1 (top ) = 0.09

Maximum velocity at top and bottom can be calculated as follows; Bottom: u f = k1

Top: u f = k1

ρ L − ρV 997.84 − 0.6349 = 0.096 = 3.81m / s. ρV 0.6349

ρ L − ρV 792.64 − 1.7798 = 0.09 = 1.897 m / s. ρV 1.7798

Now we will design the column for 70% flooding, therefore the velocities will be given by; Bottom: uf = 3.81 × 0.7 = 2.667 m/s. Top: uf = 1.897 × 0.7 = 1.3279 m/s.

Maximum volumetric flow rate of vapor can be calculated as follows:


L.I.T. Nagpur

Vm 1066.56 = = 0.4667 m 3 / s. ρV × 3600 0.6349 × 3600

Bottom Q =

Top Q =

65

1066.56 = 0.1665m 3 / s. 1.7798 × 3600

Now area at the bottom and at the top of the tower can be calculated as follows: Net area (An) bottom = Net area (An) top =

Q 0.4667 = = 0.1749m 2 . uv 2.667

Q 0.1665 = = 0.1254m 2 . u v 1.3229

Now let us take 20% of the area as the downcomer area, Ad = 0.2AT But, AT = An + Ad Therefore, An/0.8 = AT Therefore,

Bottom area =

Top area =

0.1749 = 0.21863m 2 . 0.8

0.1254 = 0.15675m 2 . 0 .8

Now, area =

π 2 × di . 4

Column diameter can be calculated as follows; Bottom =

0.21863 × 4 = 0.5276m π

. Top =

0.15675 × 4 = 0.4467 m. π

Take 0.5276 m as inside diameter of the column.


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LIQUID FLOW PATTERN: Now, maximum volumetric flow rate at the bottom =

7097.06 = 1.9756 × 10 −3 m 3 / sec . 997.84 × 3600

PROVISIONAL PLATE DESIGN: Column diameter Dc = 0.5276 m. Area of column, Ac =

π × 0.5276 2 = 0.2186m 2 . 4

Now area of downcomer is 20% of the area of the column.

Ad = 0.2 × Ac = 0.04372m 2 . 2 Also Net area is given by, An = Ac − Ad = 0.2186 − 0.04372 = 0.1749m . 2 Active area, Aa = Ac − 2 Ad = 0.2186 − 2 × 0.04372 = 0.13116m .

Take hole area as 10% of active area. Therefore

Ah = 0.1×0.13116 = 0.013116 m2.

For 20% downcomer area, from graph of (Ad/Ac) ×100 Vs For, (Ad/Ac) = 0.2,

Iw Dc

Iw = 0.86 Dc

Therefore Length of the weir, Iw = 0.86 × 0.5276 = 0.4537 m. Take weir height as 50 mm. Plate thickness = 5mm Hole diameter = 5mm CHECK WEEPING: Maximum liquid rate, LW =

7097.06 = 1.9714kg / sec . 3600

At 70% turndown the liquid rate is, = 0.7 × 1.9714 = 1.3799 kg/sec.


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The height of the liquid crest over the weir can be estimated using the Francis weir formula. For a segmental down comer this can be written as:

how = 750( Where,

Lw 2 / 3 ) ρ L lw

lw = weir length, m,

how = weir crest, mm liquid, Lw = liquid flow-rate, kg/s. At the maximum flow rate of liquid the liquid crest over the weir can be calculated using the maximum liquid flow rate calculated above, while for the conditions at minimum flowrate are assumed to be the 70% turndown conditions. Maximum how = 750(

Minimum how = 750(

1.9714 ) 2 / 3 = 19.99 mm. 997.84 × 0.4537

1.3799 ) 2 / 3 = 15.76 mm. 997.84 × 0.4537

At minimum rate, hw + how = 50 + 15.76 = 65.76 mm. From graph of K2 Vs (hw + how), K2 = 30.4 Where, K2 is a constant dependent on the depth of the clear liquid on the plate.

The minimum design vapour velocity is given by ,

uh = Where,

[ K 2 − 0.9(25.4 − d h )] ρV

uh = minimum vapour velocity through the holes(based on the

hole

area), m/s, dh = hole diameter, mm, K2 = a constant, dependent on the depth of clear liquid on the plate

uh =

30.4 − 0.9(25.4 − 5) 0.6349

= 15.11m / sec .


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Actual minimum vapor velocity is given by, Actual uh = Minimum vapor rate/Ah

uh =

0.7 × 0.4667 = 24.91m / sec . 0.013116

This is well above the weeping velocity (minimum vapor velocity), therefore the design is acceptable. PLATE PRESSURE DROP:Maximum vapor velocity through the holes is given by, u h (max) =

Vmax 0.4667 = = 35.58m / sec . hole area 0.013116

For (plate thickness)/(hole dia) = 1, from graph of (Ah/Aa)×100 Vs Co For, (Ah/Aa)= 0.1; Co = 0.84 The pressure drop through the dry plate can be estimated using expressions derived for flow through orifices.

hD = 51[

= 51[

u h 2 ρV ] Co ρ L 35.58 2 0.6349 ] × = 58.22mm. ……..(bottom) 0.84 997.84

RESIDUAL HEAD:

hr =

12.5 ×10 3 ρL

= 12.53 mm.


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69

TOTAL HEAD:

ht = hD + (hw + how ) + hr = 58.22 + (50 + 15.76) + 12.53 = 136.51mm. of H 2 O Comment:This is higher than the assumed per plate pressure drop of 100 mm, however it won’t cause a large change in the physical properties, even if everything were recalculated, thus it is acceptable.

DOWNCOMER LIQUID BACK UP: The height above the bottom edge of the apron is calculated as follows: hap = hw − 10 = 50 − 10 = 40mm. Thus the clearance area under the apron is given by, Aap = haplw = 0.04×0.4531 = 0.0182 m2. This is less than the area of the downcomer Ad. Thus use Aap in the equation given below :Head loss in the downcomer can be estimated by the following equation:

hdc = 166[

Lwd 2 ] ρ L Am

Where, Am = Ad or Aap whichever is smaller. Lwd = liquid flowrate in the downcomer.

∴ hdc = 166[

1.9714 ]2 = 1.956mm 997.84 × 0.0182


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70

In terms of clear liquid, the downcomer backup is given by;

hb = ( hw + how ) + ht + hdc = 50 + 15.76 + 136.51+ 1.956 = 204.23 mm. = 0.204 m. Now tray spacing + weir height = 0.4537 + 0.05 = 0.504 m.

1 hb 〈 × 0.504 2 Therefore he tray spacing is within acceptable limits. DOWNCOMER RESIDENCE TIME: The residence time of the liquid over the downcomer is given by;

tr =

Ad hbc ρ L 0.04372 × 0.204 × 997.84 = = 4.52 sec . Lwd 1.9714

This is greater than 3 seconds therefore it is acceptable. ENTRAINMENT: The entrainment can be estimated by the following relationships which give entrainment as a function of percentage flooding.

uv =

max imun bottom vapor. flowrate 0.4667 = = 2.1m / sec . An 0.1749

% flooding =

uv u f ( bottom )

=

2 .1 × 100 = 75.5% 2.78

Therefore from graph of FLV Vs % flooding. (fractional entrainment), ψ = 0.013 , this is less than 0.1, therefore it is acceptable.


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71

TRIAL LAYOUT:

0.4537m

0.5276m

NUMBER OF HOLES: Area of one hole,

A1 =

π × d 2 = 1.964 × 10 −5 m 2 / hole 4

Number of holes, N=

Ah 0.013116 = = 668.33 ≈ 669 holes. A1 1.964 × 10 −5

MECHANICAL DESIGN

Design of shell:Taking material of construction as Stainless Steel Maximum allowable stress f = 1420 kg/cm2 Operating pressure = 101.325 KN/m2 = 101325/(9.81 x 10000) = 1.033 Kg/cm2 Design pressure is 10% excess of operating pressure = 1.033 x 1.1 = 1.1363 kg/cm2 Thickness of shell (ts) = (P x Di)/ (2fJ –P) = (1.1363 x 52.76)/(2 x 1420 x 0.85 – 1.1363) = 0.248 mm Taking allowance = 3 mm Shell thickness = 3.248 Use thickness of 3.5 mm Production of Acetaldehyde

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Therefore, Outer diameter of the column (Do) = Di + 2t = 0.5346m

Design of heads:Elliptical heads are used th

= (P DiV)/(2fJ)

Where, Di = internal diameter of the column P = design pressure = 1.1363 kg/cm2 th = thickness of head J = welded joint efficiency = 0.85 V = stress intensity factor And, V = (2 + k2)/6 k = ratio of major axis to minor axis = 2:1 =2 Therefore, V = 1 th = (1.1363 x 52.76 x 1) / (2x1420 x0.85) =0.02484 cm = 0.2484 mm Taking allowance = 3 mm Thickness of head = 3.2484 mm Therefore, use thickness of 3.5mm

Design of gasket and bolt size:-

Gaskets are used for making leak proof joint between two surfaces Gasket: Asbestos with suitable binder (3mm thick) Gasket factor m = 2.0 Production of Acetaldehyde

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Minimum design sitting stress for asbestos with suitable binder (3 mm thick) is Ya = 112 kg / cm2

Go/Gi = [(Ya – Pi x m)/(Ya – Pi (m + 1))]0.5 = [(112 – 1.1363 x 2)/(112 – 1.1363(2+1)] 0.5 = 1.010 Gi = 52.76 + 2 x 0.0816 = 52.9232 cm Go = 1.010 x 52.9232 = 53.45cm Mean gasket diameter G = (Go + Gi)/2 = 53.185 cm Basic gasket sitting width bo = ( Go - Gi )/4 = 0.1325 cm = 1.4 mm Taking it as 1 mm. Effective gasket sitting width as bo is less than 6.3 mm b = bo b = 1.4 mm

Force acting on bolt under atmospheric condition Wm1 = 3.14 x b x G x Ya = 3.14 x 0.14 x 53.185 x 112 = 2619.9 kg Force acting on the bolt under operating condition Wm2 = 3.14 x 2b x G x m x Pi + 3.14 x G2 x Pi / 4 Production of Acetaldehyde

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= 3.14 x 2 x0.12 x 53.185 x2 x1.1363 + 3.14 x 53.1852 x 1.1363 / 4 = 2630.74 kg Maximum bolting area:Bolting material is rolled carbon steel. Am1 = Wm1 / fa Am2 = Wm2 / fb Where, fa = allowable stress for bolt material under atmospheric Conditions = 545 kg / cm2 fb = allowable stress for bolt material under operating condition = 545 kg / cm2 Am1

= 2619.9/545 = 4.801 cm2

Am2

= 2630.74/545 = 4.827 cm2

Therefore, minimum bolting area is taken as 4.827 cm2 No. of bolts = mean diameter of gasket/2.5 = 53.185/2.5 = 21.274 Since the total no. of bolts must be a multiple of 4 No. of bolts = 24 If Am is the area of one bolt then, Am x 24 = 4.827 Am = 0.201125 cm2 Therefore, 3.142 x db2 /4 = 0.201125 db = 0.5060 cm Diameter of bolt = 0.506 cm Pitch circle diameter = outside diameter of gasket + 2xdiameter of bolt + 1.2 Production of Acetaldehyde

L.I.T. Nagpur

75

= Go + 2 x db + 1.2 = 53.45+ 2 x 0.506 + 1.2 = 55.662 cm

Bolt spacing = (3.14 x pitch circle diameter)/no. of bolts = 3.14 x 55.662 / 24 = 7.283 cm

Flange design:Outside diameter of the flange = Pitch circle diameter + 2 x diameter of bolt = 55.662 + 2 x 0.5060 = 56.674 cm

Thickness of the flange tf = G x (Pi/kf) 0.5 Where, k = 1/ [0.3 + (1.5 x Wm x Hg) /(HxG)] G = diameter of the gasket load reaction or mean diameter of gasket Wm = maximum bolt load = 53678.106 kg Hg = radial distance from gasket load reaction to the bolt circle = (pitch circle diameter – G)/2 = (55.662 – 53.185)/2 = 1.2385 cm H = hydrostatic end force = (3.14 x G2 x Pi)/4 = (3.14 x 53.1852x 1.1363)/4 = 2523.14 Kg k = 1/[0.3 +(1.5 x 53678.106 x 1.2045)/(1956.6x 46.835)] Production of Acetaldehyde

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76

= 0.9587 tf = 53.185 x (1.1363/0.9587x1420)0.5 = 1.536 cm

Design of skirt support:The stresses due to vessel dead weights, wind load and seismic load are taken into account while the column is designed to withstand maximum values of tensile or compressive stresses. The stresses are:

1. Due to dead weight fd = W/(3.14 x Dsk x tsk) Dsk = outside diameter of skirt support tsk = thickness of skirt support W = total weight of the vessel including attachments

Weight of shell can be calculated as:lnW = 0.694 + 0.882 lnB B

= {[(L/D) + 1.82] P D3}/(25600 + 1.2P) + 20L

Where W = weight of column without internals (kg) D = diameter of column (inches) D = 0.5276 m = 20.78 inches P = design pressure (Psig) = 1.1363 kg / cm2 = (14.7 x 1.1363)/1.1 = 15.185 Psig Production of Acetaldehyde

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77

L = column height of cylindrical shell and heads (inches) L = 4 m = 157.48 inches Therefore, B = 3159.56 kg W = 2443.83 kg

Weight of the contents = volume of contents x Avg density = [(3.14 x 0.52762 x 4) x (0.1 x 784 + 0.9 x 1000)]/4 = 1025.465 kg

Assume, Weight of accessories = 200 kg ∑W = 5552.86 + 1025.465 + 200 = 3669.295 kg Dsk = Gi = 52.93 cm fd

= ∑W/(3.14 x Dsk x tsk) = 3669.295/(3.14 x 52.93 x tsk) = 22.078 / tsk kg/cm2

2. Due to wind load PLW = K1 x K2 X P x H x Do Where K1 = coeff depending on shape factor = 0.7 K2 = 1

P = wind pressure = 128 kg/m2 (assuming) Production of Acetaldehyde

L.I.T. Nagpur

78

Do = 0.5346 m H=4+2=6m PLW = K1 x K2 x P x H x Do = 0.7 x 1 x 128 x 0.5346 x 6 = 287.4 kg

The bending moment due to wind at the base of skirt is Mw = PLW x H/2 = 287.4 x 6/2 = 862.2 kg-m

f w = Mw / Z = 862.2/[(3.14 D2sk tsk)/4] = 3920.45/tsk kg / m2 = 0.392 / tsk kg /cm2

1. Due to seismic load fs = 4 Mw / (3.14 D2o tsk) but, Mw = (2 x C x H x W)/3 therefore, fs = (2 x 4 x C x H x W)/(3 x 3.14 D2o tsk) Where, C = seismic coefficient = 0.8 H=4+2=6m W = 3669.295 kg; Do = 0.5346 m


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79

fs = (2 x4 x 0.8 x 6 x 3669.295)/(3 x 3.14 x 0.5346 2 x tsk) = 5.234/tsk kg/cm2

Maximum tensile stress at the bottom of the skirt = fd - (fw or fs) = (21.863-5.234)/tsk = 16.629/tsk

kg/ cm2

Permissible tensile stress = 1420 kg/ cm2 tsk = 16.629 / 1420 = 0.01171 cm = 0.117 mm fc (max) = (21.863/tsk) + (5.234/tsk) = 27.097/tsk kg/ cm2 fc (permissible ) = yield point stress / 3 = 2000 / 3 = 666 kg/ cm2 tsk

= 27.097/ 666 = 0.0407 cm = 0.41 mm

Therefore, thickness of 1 mm is used.


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80

Material of Construction: Many factors have to be considered when selecting engineering materials, but for chemical process plant the overriding consideration is usually the ability to resist corrosion. The process designer will be responsible for recommending materials that will be suitable for the process conditions. He must also consider the requirements of the mechanical design engineer; the material selected must have sufficient strength and be easily worked. The most economical material that satisfies both process and mechanical requirements should be selected; this will be the material that gives the lowest cost over the working life of the plant, allowing for maintenance and replacement. One of the major problems with direct ethylene oxidation processes is the extremely corrosive properties of the catalyst solution. Titanium or brick lined equipment must be used.For the distillation column mild steel can be used as there is no problem of corrosion. Pressure maintained in the column is 1 atmosphere.


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9. COST ESTIMATION All prices are in Lakh

Chemical Engineering Plant Cost Index: Cost index in 1999 = 365 Cost index in 2006 = 606.5 Estimation of Capital Investment Cost: I. Direct Costs: material and labor involved in actual installation of complete facility a) Equipment + installation + instrumentation + piping + electrical + insulation + Painting 1.

Purchased equipment cost (PEC): i.e., PEC = Rs. 1450

Present cost of equipment will be

= 1450 × 606.5/365 = 2409.383562

2.

Installation, including

insulation and painting:

(25-55% of purchased equipment cost.) Consider the Installation cost = 40% of Purchased equipment cost = 40% of 2409.383562 = Rs. 963.7534247 3. Instrumentation and controls, installed: (6-30% of Purchased equipment cost.) Consider the installation cost equipment cost

= 15% of Purchased = 15% of 2409.383562 = Rs. 361.4075343


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82

4. Piping installed: (10-80% of Purchased equipment cost) Consider the piping cost

= 40% Purchased equipment cost = 40% of 2409.383562 = Rs. 963.7534247

5. Electrical, installed: (10-40% of Purchased equipment cost) Consider Electrical cost = 25% of Purchased equipment cost = 25% of 2409.383562 = Rs. 602.3458905 Hence total cost of (1+2+3+4+5) =Rs. 5300.643836 B. Buildings, process and Auxiliary: (10-70% of Purchased equipment cost) Consider Buildings, process and auxiliary cost = 40% of PEC = 40% of 2409.383562 =Rs. 963.7534247 C. Service facilities and yard improvement: (40-100% of Purchased equipment cost) Consider the cost of service facilities and yard improvement = 62% of PEC = 62% of 2409.383562 = Rs. 1493.817808 D. Land: (1-2% of fixed capital investment or 4-8% of Purchased equipment cost) Consider the cost of land = 5% of PEC = 5% of 2409.383562 = Rs. 120.4691781 Thus, Direct cost = Rs. 7878.684247 II. Indirect costs: expenses which are not directly involved with material and labour of actual installation of complete facility


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A. Engineering and Supervision: (5-30% of direct costs) Consider the cost of engineering and supervision = 15% of Direct costs i.e., cost of engineering and supervision = 15% 7878.684247 = Rs. 1181.802637 B. Construction Expense and Contractor’s fee: (6-30% of direct costs) Consider the construction expense and contractor’s fee = 10% of Direct costs i.e., construction expense and contractor’s fee = 10% of 7878.684247 = Rs. 787.8684247 C. Contingency: (5-15% of PEC) Consider the contingency cost = 10% of 2409.383562 i.e., Contingency cost = Rs. 240.9383562 Thus, Indirect Costs = Rs. 2210.609418 III. Fixed Capital Investment: Fixed capital investment = Direct costs + Indirect costs = 7878.684247 + 2210.609418 i.e., Fixed capital investment = Rs. 10089.29367 IV. Working Capital: (10-20% of Fixed-capital investment) Consider the Working Capital = 15% of Fixed-capital investment i.e., Working capital = 15% of 10089.29367 = Rs. 1513.39405


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V. Total Capital Investment (TCI): Total capital investment = Fixed capital investment + Working capital = 10089.29367 + 1513.39405 i.e., Total capital investment = Rs. 11602.68772 Estimation of Total Product cost: I. Manufacturing Cost = Direct production cost + Fixed charges + Plant overhead cost. A. Fixed Charges: (10-20% total product cost) i. Depreciation: (depends on life period, salvage value and method of calculation-about 10% of FCI for machinery and equipment, and 2-3% for Building Value for Buildings) Consider depreciation = 10% of FCI for machinery and equipment, and 3% for Building Value for Buildings) i.e., Depreciation = (0.10× 10089.29367) + (0.03× 963.7534247) = Rs. 1037.84197 ii. Local Taxes: (1-4% of fixed capital investment) Consider the local taxes = 4% of fixed capital investment i.e. Local Taxes = 0.04× 10089.29367= Rs. 403.5717468 iii. Insurances: (0.4-1% of fixed capital investment) Consider the Insurance = 0.6% of fixed capital investment i.e. Insurance = 0.006× 10089.29367= Rs. 60.53576202


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iv. Rent: (8-12% fixed capital investment) Consider rent = 10% of fixed capital investment = 10% of 10089.29367 Rent = Rs. 1008.929367 Thus, Fixed Charges = Rs. 2510.878846 B. Direct Production Cost: Now we have Fixed charges = 10-20% of total product charges – (given) Consider the Fixed charges = 15% of total product cost Total product cost = fixed charges/15% Total product cost =2510.878846/0.15 Total product cost (TPC) = Rs.16739.19231 i. Raw Materials: (10-50% of total product cost) Consider the cost of raw materials= 30% of TPC Raw material cost = 30% of 16739.19231 Raw material cost = Rs. 5021.757692 ii. Operating Labor (OL): (10-20% of total product cost) Consider the cost of operating labor = 15% 1673.19231 Operating labor cost = 15% of 16739.19231 Operating labor cost = Rs. 2510.878847 iii. Direct Supervisory and Clerical Labor (DS & CL):(10-25% of OL) Consider the cost for Direct supervisory and clerical labor = 12% of OL Direct supervisory and clerical labor cost = 12% of 2510.878847 Direct supervisory and clerical labor cost = Rs. 301.3054616


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iv. Utilities: (10-20% of total product cost) Consider the cost of Utilities = 15% of total product cost Utilities cost= 15% of 16739.19231 Utilities cost = Rs. 2510.878847 v. Maintenance and repairs (M & R): (2-10% of fixed capital investment) Consider the maintenance and repair cost = 5% of fixed capital investment i.e. Maintenance and repair cost = 0.05× 10089.29367 = Rs. 504.4646835 vi. Operating Supplies: (10-20% of M & R or 0.5-1% of FCI) Consider the cost of Operating supplies = 15% of M & R Operating supplies cost = 15% of 504.4646835 Operating supplies cost = Rs. 75.66970253 vii. Laboratory Charges: (10-20% of OL) Consider the Laboratory charges = 15% of OL Laboratory charges = 15% of 2510.878847 Laboratory charges = Rs. 376.6318271 viii. Patent and Royalties: (2-6% of total product cost) Consider the cost of Patent and royalties = 5% of total product cost Patent and Royalties = 5% of 16739.19231 Patent and Royalties cost = Rs. 836.9596155 Thus, Direct Production Cost = 5021.757692+2510.878847+301.3054616 +2510.878847+504.4646835+75.66970253+376.6318271+836.9596155 Direct Production Cost = Rs. 12138.54668


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C. Plant overhead Costs (50-70% of Operating labor, supervision, and maintenance or 5-15% of total product cost); includes for the following: general plant upkeep and overhead, payroll overhead, packaging, medical services, safety and protection, restaurants, recreation, salvage, laboratories, and storage facilities. Consider the plant overhead cost = 60% of OL, DS & CL, and M & R Plant overhead cost = 60% of (2510.878847+301.3054616+504.4646835) Plant overhead cost = Rs. 1995.389395 Manufacture cost = Direct production cost + Fixed charges + Plant overhead costs. Manufacture cost = 12138.54668+2510.878846+1995.389395 Manufacture cost = Rs. 16644.81492 II. General Expenses = Administrative costs + distribution and selling costs + research

and development costs

A. Administrative costs:(40-60% of operating labor) Consider the Administrative costs = 50% of operating labor Administrative costs = 50% of 2510.878847 Administrative costs = Rs. 1255.439424 B. Distribution and Selling costs: (2-20% of total product cost); Includes costs for sales offices, salesmen, shipping, and advertising. Consider the Distribution and selling costs = 10% of total product cost Distribution and selling costs = 10% of 16739.19231 Distribution and Selling costs = Rs. 1673.919231


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C. Research and Development costs: (about 3% of total product cost) Consider the Research and development costs = 3% of total product cost Research and Development costs = 3% of 16739.19231 Research and Development costs = Rs. 502.1757693 Thus, General Expenses = Rs.1255.439424+1673.919231+502.1757693 = Rs. 3431.534424 III. Total Production cost

= Manufacture cost + General Expenses = 16644.81492+ 3431.534424

Total production cost = Rs. 20076.34934 IV. Gross Earnings/Income: Selling Price of acetaldehyde per kg = Rs. 570 × 10-5 Total Income

=

Selling

×

price

Quantity

of

product

manufactured = 570× (15× 103 /day) × (300 days/year) × 10-5 Total Income

= Rs. 25650

Gross income

= Total Income – Total Production Cost = (25650) – (20076.34934)

Gross Income

= Rs. 5573.65066

Let the Tax rate be 45% (common) Taxes

= 45% of Gross income = 45% of 5573.65066

Taxes

= Rs. 2508.142797

Net Profit

= Gross income - Taxes =5573.65066 - 2508.142797

Net profit

= Rs. 3065.507863


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Rate of Return: Rate of return

= Net profit× 100/Total Capital Investment

Rate of Return

= 3065.507863/ (11602.68772)

Rate of Return

= 0.26421

Rate of Return

= 26.421%

Pay Back Period Pay back period

= 1/rate of return = 3.785 years

Pay Back Period

= 45.42 months


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10. Process control, Instrumentation & Automation Process control and automation, together with their associated instruments can be regarded as the mechanical brain and nerves of modern chemical processing. Thus a properly instrumented and controlled plant ensures: 1. Increased productivity and low cost. 2. Better and more uniform quality of the product. 3. Greater safety for operating personnel. Instrumentation usually will incure a proportionately large increment of engineering design cost because of the extensive analysis and relatively large amount of detailed associated with the intelligent specification of even the lowest cost instrument item. This coupled with the high cost of installation emphasizes the need for close scrutiny into the essentiality of every element in the instrumentation system.

Direct

Digital

Feedforward-Feedback

Control

of

Distillation

Column:The direct digital feedforward- feedback control of distillation column is shown in Figure. The two principal disturbances inlet feed flowrate and the inlet feed composition are measured and the signals are sampled through the sampler switches. The discrete time signals are converted into digital signals by the analog to digital converters. The electronic comparators are used to generate the errors which are the difference between the set points and the measured values. The generated errors are minimized by the electronic feedforward controller monitored by the computer. The output command signals from the computer which are in the digital forms are converted into the analog forms by the digital to analog converters. The discrete time signals are converted into continuous signals by the hold elements. Production of Acetaldehyde

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The electro-pneumatic transducers are used to convert electric signals into the pneumatic signals. The pneumatic signals of the electronic pneumatic transducer in the transmission line of feed composition act on the diaphragm actuator of the control valve which in turns regulates the steam pressure in the reboiler. The composition of the distillate is measured and the signals are sampled through the sampler switches. The discrete time signal is converted into the digital signals by the analog to digital converter. The electronic comparator is used to generate the error which is the difference between the set point and the measured value. The generated error is minimized by the electronic controller which is monitored by the computer. The output command signal from the computer which is in the digital form is converted into analog form by the digital to analog converter. The hold element is used to convert the discrete time signal into the continuous signal. The electro-pneumatic transducer is used to convert the electric signal into the pneumatic signal. The pneumatic signal of the electro-pneumatic transducer in the transmission line of the inlet feed composition measurement in the feedforward control and the pneumatic signal of the electro-pneumatic transducer in the transmission line of the distillate composition measurement act on the diaphragm actuator of the control valve which in turn regulates the reflux ratio.

SCADA CONTROL The supervisory control and data acquisition SCADA refers to a type of industrial control system which is used to collect data from the various sensors of the plant that are either in close proximity or off site sometimes miles away from the process plants and then sends this data to the SCADA master computer control station from a remote location. The SCADA system is used for controlling and monitoring chemical or transport processes and for the distribution and transmission of power. The SCADA is used to control the complete site spread out over a very long distance from few kilometers to several miles for efficient control.


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SCADA Configuration:The SCADA system configuration consists of following control units. 1) Supervisory control station 2) Programmable Logic Controller & Remote Terminal Unit

3) Computer network segment. The process information such as temperature, flowrate, composition etc. are communicated between the process plant and PLC’s of control system.

Sr. No. 1 2 3 4 5

Control Equipment Reactor Cooler Scrubber

Sensor Thermocouple Thermocouple Thermocouple,

Distillation Column

Mass spectrometer Venturimeter,

Degasser

Mass spectrometer Mass spectrometer


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11. Plant Location & Layout:The location of the plant can have a crucial effect on the profitability of a project, and the scope for future expansion the plant should be located where the minimum cost of production and distribution can be obtained, but other factors, such as room for expansion and safe living conditions for plant operation as well as the surrounding community, are also important.. Many factors must be considered when selecting a suitable site, the principal factors to consider are: 1. Location, with respect to the marketing area. 2. Raw material supply. 3. Transport facilities. 4. Availability of labour. 5. Availability of utilities: water, fuel, power. 6. Availability of suitable land. 7. Flood and Fire Protection. 8. Environmental impact, and effluent disposal. 9. Local community considerations. 10. Climate. 11. Political and strategic considerations.

Marketing Area:Acetaldehyde is widely used as an intermediate. Hence, it is necessary to locate the plant near to factories where it is major raw material. From this prospective Mumbai is the ideal place for acetaldehyde plant since many factories are located close to Mumbai. That’s why most of acetaldehyde manufacturing unit are located in Mumbai.


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Raw Materials:The raw material for the plant is ethylene. As Most of the ethylene is produced from petroleum, hence it can be obtained very easily from the refineries. Hence it is advised that the site of the plant should be nearer to the refineries, so that the transportation cost is reduced. From this prospective Mumbai or any place nearer to Mumbai is the most suitable region for the plant.

Transport:The transport of materials and products to and from the plant will be an overriding consideration in site selection. If practicable, a site should be selected that is close to at least two major forms of transport: road, rail, waterway (canal or river), or a sea port. Road transport is being increasingly used, and is suitable for local distribution from a central warehouse. Rail transport will be cheaper for the long-distance transport of bulk chemicals. Air transport is convenient and efficient for the movement of personnel and essential equipment and supplies, and the proximity of the site to a major airport should be considered. All these facilities of transport are very easily available in a place like Mumbai; hence from the prospective transportation facilities Mumbai is the ideal place for the project.

Availability of Labour:Labour will be needed for construction of the plant and its operation. Skilled construction workers will usually be brought in from outside the site area, but there should be an adequate pool of unskilled labour available locally; and labour suitable for training to operate the plant. Skilled tradesmen will be needed for plant maintenance. Local trade union customs and restrictive practices will have to be considered when assessing the availability and suitability of the local labour for recruitment and training. Production of Acetaldehyde

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Utilities (services):Chemical processes invariably require large quantities of water for cooling and general process use, and the plant must be located near a source of water of suitable quality. Process water may be drawn from a river, from wells, or purchased from a local authority. At some sites, the cooling water required can be taken from a river or lake, or from the sea; at other locations cooling towers will be needed. Electrical power will be needed at all sites. A competitively priced fuel must be available on site for steam and power generation.

Environmental impact and Effluent disposal:All industrial processes produce waste products, and full consideration must be given to the difficulties and cost of their disposal. The disposal of toxic and harmful effluents will be covered by local regulations, and the appropriate authorities must be consulted during the initial site survey to determine the standards that must be met. An environmental impact assessment should be made for each new project, or major modification or addition to an existing process.

Local Community Considerations:The proposed plant must fit in with and be acceptable to the local community. Full consideration must be given to the safe location of the plant so that it does not impose a significant additional risk to the community. On a new site, the local community must be able to provide adequate facilities for the plant personnel: schools, banks, housing, and recreational and cultural facilities


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Flood and fire protection:Many industrial plants are located along rivers ornear large bodies of water, and there are risks of flood or hurricane damage. Before selecting a plant site, the regional history of natural events of this type should be examined and the consequences of such occurrences considered. Protection from losses by fire is another important factor in selecting a plant location. In case of a major fire, assistance from outside fire departments should be available. Fire hazards in the immediate area surrounding the plant site must not be overlooked.

Land (site considerations):Sufficient suitable land must be available for the proposed plant and for future expansion. The land should ideally be flat, well drained and have suitable load-bearing characteristics. A full site evaluation should be made to determine the need for piling or other special foundations.

Climate:A

dverse climatic condition at a site will increase costs.

Abnormally low temperatures will require the provision of additional insulation and special heating for equipment and pipe runs. Stronger structures will be needed at locations subject to high winds (cyclone/hurricane areas) or earthquakes.


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Political and strategic Considerations:Capital grants, tax concessions, and other inducements are often given by governments to direct new investment to preferred locations; such as areas of high unemployment. The availability of such grants can be the overriding consideration in site selection.

PLANT LAYOUT After the process flow diagrams are completed and before detailed piping, structural, and electrical design can begin, the layout of process units in a plant and the equipment within these process units must be planned. This layout can play an important part in determining construction and manufacturing costs, and thus must be planned carefully with attention being given to future problems that may arise. Since each plant differs in many ways and no two plant sites are exactly alike, there is no one ideal plant layout. However, proper layout in each case will include arrangement of processing areas, storage areas, and handling areas in efficient coordination and with regard to such factors as: 1) New site development or addition to previously developed site 2) Economic distribution of utilities and services 3) Health and safety considerations 4) Waste-disposal requirements 5) Auxiliary equipment 6) Space available and space required 7) Roads and railroads 8) Possible future expansion


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acetaldehyde.doc

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