Manufacture of Terepthalic Acid

8/13/2019 Manufacture of Terepthalic Acid

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ACKNOWLEDGEMENT

We take this opportunity to thank the Associate Director Dr. C..Muthamizhchelvan

for providing us with an excellent infrastructure and conducive atmosphere for developing our

project.

We would also like to thank the Head of Department of Chemical engineering

Dr.R.Karthikeyan, for encouraging us to do our project.

We sincerely thank our project guide Ms. S. Vishali for her valuable guidance, support and

encouragement in all aspects of this project and for its completion

We would also like to thank our faculty members and technicians of our Chemical Department

who helped in the successful completion of our project.

2



CONTENTS Page no

1. INTRODUCTION ……………………………………………………………........5

1.1 INTRODUCTION

1.2 HISTORY

2. PROPERTIES………………………………………………………………………7

2.1 PHYSICAL AND CHEMICAL PROPERTIES

3. APPLICATION…………………………………………………………………….8

3.1 APPLICATION OF TEREPTHALIC ACID

4. VARIOUS METHODS OF PRODUCTION………………………………………9

4.1 METHODS OF SELECTION

5. REASONS FOR SELECTION…………………………………………………….10

5.1 REASONS FOR PARTICULAR PROCESS

5.2 ADVANTAGES OF USING THIS PROCESS

5.3 DISADVANTAGES OF OTHER PROCESSES

6. PROCESS DESCRIPTION………………………………………………………..12

7. MASS BALANCE…………………………………………………………………16

7.1 MASS BALANCE FOR MIXED TANK

7.2 MASS BALANCE FOR REACTOR

7.3 MASS BALANCE FOR FIRST CRYSTALLIZER

7.4 MASS BALANCE FOR SECOND CRYSTALLIZER

7.5 MASS BALANCE FOR THIRD CRYSTALLIZER

7.6 MASS BALANCE FOR FILTER

7.7 MASS BALANCE FOR DRIER

3



8. ENERGY BALANCE…………………………………………………………….24

8.1 ENERGY BALANCE FOR MIXED TANK

8.2 ENERGY BALANCE FOR REACTOR

8.3 ENERGY BALANCE FOR FIRST CRYSTALLIZER

8.4 ENERGY BALANCE FOR SECOND CRYSTALLIZER

8.5 ENERGY BALANCE FOR THIRD CRYSTALLIZER

8.6 ENERGY BALANCE FOR FILTER

8.7 ENERGY BALANCE FOR DRIER

9. PROCESS DESIGN……………………………………………………………….42

9.1 DESIGN OF REACTOR

9.2 DESIGN OF DRIER

9.3 DESIGN OF MIXED TANK

10. COST ESTIMATION……………………………………………………………..47

11. PLANT LAYOUT………………………………………………………………….54

12. INSTRUMENTATION AND CONTROL………………………………………...58

12.1 INSTRUMENTS

12.2 OBJECTIVES

12.3 TYPICAL CONTROL SYSTEMS

13. SAFETY AND HAZARD ANALYSIS……………………………………………62

14. CONCLUSION…………………………………………………………………….65

15. BIBLIOGRAPHY………………………………………………………………….67

4





1.1 INTRODUCTION

Terephthalic acid is one isomer of the three phthalic acids. It finds important use as a

commodity chemical .Principally as a starting compound for the manufacture of polyester

(specifically PET), used in clothing and to make plastic bottles. It is also known as 1,4-

benzenedicarboxylic acid, and it has the chemical formula C6H4(COOH)2.. The acids are

produced by oxidation of the methyl group on the corresponding p- xylene[106-42-3].

Terephthalic acid are used to make saturated polyesters with aliphatic diols as the

comonomer. Terephthalic acid is commercially available as polymer grade (greater than

99.9 weight % pure, exclusive of some residual water) and technical grade (typically

greater than 97-98% pure). Impurities include p-toluic acid, 4-formylbenzoic acid, residual

water, trace metals and ash (trace metal oxides).( It has recently become an important

component in the development of hybrid framework materials.

1.2 HISTORY

Phthalic acid (the ortho isomer of terephthalic acid) was obtained by French chemist

Auguste Laurent in 1836 by oxidizing naphthalene tetrachloride. Believing the resulting

substance to be a naphthalene derivative, he named it naphthalenic acid. Swiss chemist

Jean Charles Galissard de Marignac determined its formula and showed Laurent’s

supposition to be incorrect, upon which Laurent gave it its present name, with teres

meaning well-turned, refined, elegant in latin [3] (symmetry increased over ortho and meta

isomers).

6



PROPERTIES

7



2.1 PHYSICAL AND CHEMICAL PROPERTIES

Molecular Weight : 166.14

Physical State : Solid

Melting Point : Sublimes at 402 deg C (755.6 deg F) at atmospheric

pressure, without melting.(14,15,22) Reported to melt at 427 deg C (800.6 deg F) in a sealed

tube.

Boiling Point : Sublimes

Relative Density (Specific Gravity) : 1.522 at 25 deg C (water = 1)

Solubility in Water : Practically insoluble (1.7 mg/100 g) at 25 deg C.

Solubility in Other Liquids : Soluble in dimethyl sulfoxide, dimethylformamide and

alkalies, such as potassium and sodium hydroxide; slightly soluble in cold ethanol, methanol,

formic acid and sulfuric acid; very slightly soluble in chloroform, diethyl ether and glacial

acetic acid

Coefficient of Oil/Water Distribution

(Partition Coefficient) : Log P(oct) = 1.25; 1.96; 2.0 (measured)

pH Value : .16 (saturated solution (0.002% in water))

Vapour Density : Less than 0.0013 kPa (0.01 mm Hg) at 20 deg C (13);

0.067 kPa (0.5 mm Hg) at 120 deg C

Saturation Vapour Concentration : Very low at normal temperatures

Evaporation Rate: Probably very low at normal temperatures.

Other Physical Properties : ACIDITY: Weak acid; pKa1 = 3.54 (Ka1 = 2.9 X 10(-

4)); pKa2 = 4.46 (Ka2 = 3.5 X 10(-5)) at 25 deg C.(14,15)

NOTE: Very small amounts of terephthalic acid in water are reported to substantially lower the

pH of the solution and form a fairly strong acid.(13)TRIPLE POINT: 427 deg C (800.6 deg F)

8



3.1 APPLICATIONS :

• Nearly all purified terephthalic acid (PTA) is consumed in polyester productionincluding polyester fibre, polyethylene terephthalate (PET) bottle resin and polyester

film.

• For most grades of polyester used in textiles and food and beverage containers, it is

more economical to use PTA than the alternative dimethyl terephthalate (DMT)

intermediate.

• The remaining PTA is used in making cyclohexanedimethanol, terephthaloyl chloride,

copolyester-ether elastomers, plasticisers and liquid crystal polymers.

• PTA applications include coatings and composite materials, based on unsaturated

polyester resins, and hot-melt adhesives.

• Terephthalic acid is also widely used to make dyes, medicine, and synthetic perfumes,

pesticides, and other chemical compounds.

9



4.1 VARIOUS METHODS OF PRODUCTION

The various methods of production of Terephthalic Acid are as follows:

1. Oxidation of p-xylene by oxygen from air.

2. Re arrangement of phthalic acid to terephthalic acid via the corresponding potassium

salts.

3. Oxidizing para-dederivatives of benzene

4. Oxidizing caraway oil, a mixture of cymene and cuminol with chromic acid

10



5.1 REASONS FOR PARTICULAR PROCESS :

Catalytic, liquid phase-air oxidation of p-xylene: This method is most widely used all over the world to produce technical grade

Terephthalic acid. This method was developed by Mid –Century Corp. The process generally

uses acetic acid as the solvent and a catalyst to oxidize p-Xylene in liquid phase by air

oxidation. The process is also called as the Amoco process. This uses a catalyst usually a heavy

metal eg. Cobalt. The process may use typically multivalent metals like manganese as catalyst

for oxidation and bromine serves as the renewable source of free radicals.

5.2 ADVANTAGES OF USING THIS PROCESS:

1. The reaction is very simple with a single step.

2. The raw materials used in this process are easily available since it is a byproduct of a

petroleum industry.

3. The Terephthalic acid produced in this process has a yield of almost 100% with the

presence of 4-formylbenzoic acid in trace amount.

4. The oxidation process is highly efficient when compared to the other methods it

brings about a conversion of about 95 wt%.

5. The product purity is very high 99%.

6. The process has very few pollution problems.

7. The solvent and the catalyst can be recovered and reused. The recovery of solvent is

possible till 90%.

5.3 DISADVANTAGES OF OTHER PROCESSES:

1. Henkel process is now obsolete as the Terephthalic acid produced by that method cannot

be used in the manufacture of polymers.

2. Oxidizing benzene and caraway oil are not economically viable on Industrial Scale.

11



PROCESS DESCRIPTION

12



6.1 PROCESS DESCRIPTION:

The process can be divided into different units:

• Reactor Unit

• Crystallization Unit

• Separation and Drying Unit

Reactor Unit:

The reactor unit comprises of a mixing tank and a reactor. The raw materials p-Xylene,

air, acetic acid (solvent) and the catalyst (cobalt) are fed continuously into the feed mixing

tank. The residence time is 5-10 min.The mixed stream pumps the reactor , and the air is flown

into the reactor through four inlets. The reactor is maintained at a temperature of 150ºC and a

pressure of 1500 kPa – 3000kPa. The air is added in greater stoichiometric ratio to minimize

the formation of byproducts. The heat of reaction is removed by condensing and refluxing

acetic acid The residence time of this reaction varies from 30 minutes to 3 hours. More than

95% of p-Xylene is converted to product.. The outlet from the reactor is a slurry, since it is

soluble to a limited extent in the solvent.used.

13



Crystallization Unit:

The oxidation reaction is conducted in two stages, first stage being the agitated

oxidation reactor, while the second stage is the agitated first crystallizer. Exothermic heat of

reaction is removed by condensing the boiling reaction solvent. A portion of this condensate is

withdrawn to control the water concentration in the reactor. Reactor effluent is depressurized

and cooled to filtering conditions in a series of three crystallizing vessels ( first crystallizer,

second crystallizer and third crystallizer ) for the secondary reaction and crystallization step.

Air is fed to the first crystallizer for additional reaction, which used to do polishing oxidation

of unreacted paraxylene from the reactor.

Separation And Drying Unit:

The separation and drying unit consists of a rotary filter and a rotary drier. In the filter

most of the water content is removed from the product (Terephthalic acid). There are two

streams leaving the filter. One stream is sent to the recovery unit. And the other is sent to the

rotary drier. The solid stream is sent to the drier. Preheated air is sent to the through the drier to

remove the moisture present in the final product. The product from the drier is 99% pure

Terephthalic acid.

14



MANUFACTURE OF TERETHALIC ACID

1. Paraxylene 2. Water 3. Acetic acid

4. Air

5. Mixed Tank

6. CSTR Reactor7. First Crystallizer

8. Second Crystallizer

9. Third Crystallizer

10. Rotary Filter

11. Rotary Drier

12. Crude TPA

15



MASS BALANCE

16



MASS BALANCE

BASIS: 15 Kmole of Paraxylene/ Hour ~ 14.8 Kmole of Terepthalic Acid/ Hour

7.1 MASS BALANCE OF MIXED TANK

Solvent from Scrubber

1038kg of Acetic acid

90kg of water

Raw Material

1590 kg of Paraxylene feed Reaction Mixture

Solvent 7038 kg of Acetic Acid

6000 kg of Acetic acid 428.4 kg of water

338.4 kg of water 1590 kg of ParaxyleneMixed tank

Feed

Inlet = 9056.4 Kg= Outlet

17



7.2MASS BALANCE OF REACTOR

Air

1780.8kg of oxygen

5861.8kg of nitrogen

Liquid Stream Liquid Stream

1590 kg of Paraxylene 127.2 kg of Paraxylene

6000 kg of acetic acid 5428 kg of Acetic Acid

338.4 kg of water

535 kg of Water

2290.8 kg of TPA


1038 kg of acetic acid

Reactor

90kg of water

Reflux Stream


3150 kg of water Vapour stream


3043.4 kg of water

705.6 kg of oxygen

5861.8 kg of nitrogen

240 kg of Carbon dioxide

Inlet = 37459 Kg=Outlet

18



Error! 7.3 MASS BALANCE OF FIRST CRYSTALLIZER

Air65.6kg of Oxygen

215.9kg of Nitrogen


5428kg of acetic acid 4293kg of acetic acid

2290.8kg of TPA

227.16kg of water

535 kg of water 2456.8 of TPA

127.2 kg of paraxylene

Vapour Stream

First Crystalizer


311.84kg of water

215.9kg of Nitrogen

1.6kg of Oxygen

21.2 kg of paraxylene

Inlet = 8662.5 Kg = Outlet

19



7.4 MASS BALANCE OF SECOND CRYSTALLIZER

Dilution Solvent


43.9kg of water


4293kg of acetic acid 4416kg of Acetic acid

227.16kg of water 203.4kg of water

2456.8 kg of TPA 2456.8kg of TPA

Second Crystallizer

Vapour Stream


64.78kg of water

Inlet = 7801 Kg = Outlet

20



7.5 MASS BALANCE OF THIRD CRYSTALLIZER

Solvent

1476kg of Acetic acid 208.8kg of water


4416kg of Acetic acid 4410kg of Acetic acid

203.4kg of water 203.4kg of water

2456.8kg of TPA 2456.8kg of TPA

Air

.336kg of Oxygen

1.106kg of Nitrogen

Third Crystallizer

Vapour Stream


208.8kg of water

.336kg of Oxygen

1.106kg of Nitrogen

Inlet = 8762.4 Kg = Outlet

21



7.6 MASS BALANCE OF ROTARY FILTER

Solvent552kg of Acetic acid

28.8kg of water

Liquid Stream Filter Cake

4410 kg of Acetic acid 496.2kg of acetic acid

203.4 kg of water 23.76kg of water

2456.8kg of PTA 2456.8kg of PTA

Rotary Filter

Filterate4465.8kg of acetic acid

208.44kg of water


22



7.7 MASS BALANCE OF DRIER

Air

6.72 kg O2

196 kg N2 27 kg H2O

2456.8 of TPA 2456.8kg of Crude TPA

Filter cake Dried Scrubber Bottoms

496.2 kg of acetic acid 2320.2kg of acetic acid

23.76 kg of water 158.58kg of water

Solvent

1824kg of acetic acid

Drier

111.6 kg of water

Air

6.72kg O2

196kg N2

3.78 kg Water


23



ENERGY BALANCE

24



ENERGY BALANCE

Specific heat constant

Paraxylene = 44.8 cal/Mol K

Acetic Acid (l) = 31.2 cal/Mol K

Acetic Acid (g) = 24 cal/Mol K

Terepthalic Acid = 47.6 cal/Mol K

Water (l) = 18 cal/Mol K

CO2 (g) = 10.34+0.00274T–19550 cal/Mol K

H2 O(g) = 8.22+0.00015T+0.00000134T2 cal/Mol K

N2 (g) = 6.5+0.00100T cal/Mol K

O2 (g) = 8.27+0.0002587T – 187700/T2 cal/Mol K

Latent Heat of Vaporization

Paraxylene = 81 cal/g

Acetic Acid = 96.48 cal/g

Water = 544.8 cal/g

Latent Heat of Formation

Terepthalic Acid = -731.9 KJ/Mol

H2O = 285.8 KJ/Mol

Paraxylene = -24.4 KJ/Mol

25



8.1 MIXED TANK

Reference Temperature = 283K

Reactants:Paraxylene

Q = 15 x 103

= 20160 Kcal

Solvent

(a) Acetic Acid

Q = 100 x 103

= 224640 Kcal

(b) Water

Q = 18.8 x 103

24364.8 Kcal


(a) Acetic Acid

Q = 17.3 x 103

= 16192.8

(b) Water

Q = 5 x 103 = 2700 Kcal

∆QReactant = 288057.6 Kcal

26



Products:

Paraxylene

Q = 15 x 103

= 41664 Kcal

Solvent

(a) Acetic Acid

Q = 100 x 10 = 193440 Kcal

(b) Water

Q = 18.8 x 10

3

= 20980.8 Kcal


(a) Acetic Acid

Q = 17.3 x 103

= 33465.12 Kcal

(b) Water

Q = 5 x 103

= 5580 Kcal

∆Q product = 295129.92 Kcal

HEAT CHANGE

∆Qreq = ∆QPdts – ∆Qreact + ∆HoR

= 295129.92- 288057.6 + 0

= 7072.32 Kcal

Heat to be supplied by the system = 7072.32 Kcal

27



8.2 REACTOR


Reactant:

Paraxylene

Q = 15 x 103

= 6720 Kcal

Solvent

(a) Acetic Acid

Q = 100 x 103

= 31200 Kcal

(b) Water

Q = 18.8 x 103

= 3384 Kcal


(a) Acetic Acid

Q= 17.3 x 103

= 5397.6 Kcal

(b)

WaterQ = 5 x 10

3 = 900 Kcal

Air

(a) Oxygen

Q = 55.65 x 103

+ 0.0002587T‐18770/T2 dT

= 55.65 x 103x 1044.34 = 58117.52 Kcal

(b) Nitrogen

Q = 209.35 x 103

= 209.35 x 103 x 709.31 = 14849.40 Kcal

28



Reflux

(a) Acetic Acid

Q = 293.5 x 103

= 1007292 Kcal

(b) Water

Q = 175 x 103

= 346500 Kcal

∆Q Reactants = 1474360.52 Kcal

Product:

Liquid stream

(a) Acetic acid

Q = 90.5 x 103

= 426363.6 Kcal

(b) Terepthalic acid

Q = 13.8 x 103

= 99188.88 Kcal

(c) Water

Q = 29.7 x 103

= 80724.6 Kcal

(d) Paraxylene

Q = 1.2 x 103

= 8117.6 Kcal

Vapour stream

(a) Acetic Acid

Q = 320.3 x 103

= 1160767.2 Kcal

29



Ql = mλ = 96.48 x 320.3 x 103

x 60 = 1854152.64 Kcal

(b) Wat r

169 x 10

e

Q =3

+ 0.00015T + 0.00000134T 2 dT

Ql = mλ = 169 x 10

(c) Oxygen

2.05 x 10

= 169 x 103 x 1285 = 217165 Kcal

3 x 18 x 544.8 = 1657281.6 Kcal

Q = 23

+ 0.0002587T‐18770/T2 dT

(d) Nitrogen

9.35 x 10

= 22.05 x 103 x 1454.04 = 32061.58 Kcal

Q = 20 3

=209.35 x 103 x 1043.49 = 218454.6 Kcal

(e) Carbon dioxide

Q = 6 x 103

0.00274T – 19550/T 2 dT

Q Reactant = 5248276.26 Kcal

Reaction:

2 C6H4(COOH)2+2H2O

R L 6H4(COOH)2+ 2∆HF(H2O)]–[ F 2 F 8H10]

dts – ∆Qreact + ∆HoR

10

3

× -305.8)

eat given to the system = 146125.74 Kcal

= 6 x 103 x 1860.66 = 11163.96Kcal

∆

C8H10 +2O

Ho

= [∆H (C ∆H 2O + ∆H C∆= –731.08+2(-285.8)+24.4

8 Kcal / mol= –1278.28 KJ/ mol = -305.

EAT CHANGE H

∆Qreq = ∆QP

= 5248276.26 – 1174362+(13.8 ×= – 146,125.74Kcal

H

30



8.3 FIRST CRYSTALLIZER

eference Temperature = 363K

eactants:

(a) Acetic Acid

5 x 10

R

R

3Q =90. = 347302.8 Kcal

(b) Terepthalic Acid

Q =13.8 x 103

= 80796.24 Kcal

)WaterQ =29.7 x 10

(c3

= 65755.8 Kcal

Air

.05x 10

(a) Oxygen

3Q = 2 + 0.0002587T‐18770/T2

dT

(b)Nitroge

= 2.05 x 103 x 910.54 = 1866.61 Kcal

n

Q = 7.7 x 103

= 7.7 x 103 x 517.54 = 3985.06 Kcal

Q Reactants = 499106.51 Kcal

roducts:

(a) Acetic acid

10

∆

P

1. Liquid Stream

Q = 71.55 x 3 = 247791.96 Kcal

(b) ater

62 x 10

W

Q = 12. 3 = 25214.76 Kcal

31



(c) Terepthalic Acid

Q = 14.8 x 103 = 78197.28 Kcal

2. Vapour Stream

cid(a) Acetic A

Q = 18.9 x 103 = 50439.6 Kcal

Ql = mλ = 96.4

32 x 10

8 x 18.9 x 103

x 60 = 109408.32 Kcal

(b) Water

Q = 17. 3 + 0.00015T + 0.00000134T 2 dT

Q = mλ = 17.3

= 17.32 x 10

3

x 1274.39 = 22072.43 Kcal

2 x 18 x 103 x 544.8 = 169846.84 Kcal

(c) Nitrogen

Q = 7.7 x 103

= 7.7 x 103 x 7

67.95 = 5913.22 Kcal

x 10

(d) Oxygen

Q = 0.05 3 + 0.0002587T‐18770/T2 dT

146.82 = 5

(e) = 0.05 x 103 x 1 7.34 Kcal Paraxylene

Q = 0.2 x 103 = 994.56 Kcal

l3 x 81 = 1717.2 Kcal

∆QPdts = 7116

Qreq = ∆QPdts – ∆Qreact + ∆Ho

6.51 +711653.51+(–305.8 × 14.8 × 103) = –4313293 Kcal

313293 Kcal

Q = mλ = 0.2 x 106 x 10

53.51 Kcal

∆HoR = -305.8 Kcal/mol

HEAT CHANGE

∆

∆Q Required = –49910

Heat given to the system = 4

32



8.4 SECOND CRYSTALLIZER

Acetic acid

10

Reference temperature = 363k

Reactant:

(a) Q = 71.55 x

3= 247791.96

(b) er

62 x 10

Wat

Q = 12.3

= 25214.76 Kcal

(c) Terepthalic acid

Q = 14.8 x 103

= 78197.28 Kcal

Dilution Solvent

10

(a) Acetic acid

Q = 13 x3

= 45021.6 Kcal

(b) Water

2.4 x 10Q =3

= 4795.2 Kcal

∆Q eactants = 401020.8 Kcal

Product:

Stream

cid

x 10

R

Liquid

(a) Acetic a

Q = 73.6 3 = 167631.36 Kcal

(b) Wat

11.3 x 10

er

Q =3

=14848.2 Kcal

33



(c) Terepthalic acid

Q = 14.8 x 103

= 51427.04 Kcal

Vapour Stream:

(a) Acetic acid

10

Q = 11 x3

= 19272 Kcal

Ql = mλ = 11 x 103 x 60 x 96.48 =63676.8 Kcal

e

x 10

(b) Wat r

Q = 3.63

+ 0.00015T + 0.00000134T 2 dT

l = mλ = 3.6 x 18 x 103x 544.80 = 35303.04 Kcal

∆QPdts = 355219.34 Kcal

EAT CHANGE

react + ∆HoR

req


= 3.6 x 103 x 850.25 = 3060.9 Kcal

Q

H

∆Qreq = ∆QPdts – ∆Q

∆Q = 355219.34 - 401020.8

= -45801.46 Kcal

H

34



8.5 THIRD CRYSTALLIZER:

Stream

tic acid

0


Reactant:

Liquid

(a) Ace

Q = 73.6 x 13

= 374300.16 Kcal

(b) 3 x 10

Water

Q = 11.3

= 33154.2 Kcal

(c) erepthalic acidT

Q = 14.8 x 103

= 114830.24 Kcal

Air

(a) xygen

05 x 10

O

Q = 0.013

+ 0.0002587T‐18770/T2 dT

(b) 0395 x 10

= 0.0105 x 103 x 970.146 = 10.20 Kcal

Nitrogen

Q = 0.3

= 0.0395 x 103 x 169.64 = 6.70 Kcal

Solvent

(a) cetic acid

0

A

Q = 24.6 x 1 3 = 38376 Kcal

35



(b) ater

= 11.6 x 10

W

Q 3 = 10440 Kcal

∆Qreact = 57111.75 Kcal

stream

(a) Acetic acid

73.5 x 10

Product:

Liquid

3

Q = = 213267.6 Kcal

(b) ater

= 11.3 x 10

W

Q 3 = 18916.2 Kcal

(c) erepthalic acid

= 14.8 x 10

T

Q 3 = 65516.64 Kcal

Vapour Stream

(a) Acetic Acid

= 24.7 x 10

3

Q = 55130.4 Kcal

Ql = mλ = 24.7 x 6

Q = 11.6 x 103

0 x 103

x 96.48 = 142983.36

(b) Water

+ 0.00015T + 0.00000134T 2 dT

1.6 x 103 x 781.73 = 9068.07 Kcal

Ql = mλ = 11.6 x 13

= 1

0 x 544.8 = 113754.24 Kcal

(c) Oxygen

36



Q = 0.0105 x 103

+ 0.0002587T‐18770/T2 dT

(d) Nitrogen

0395 x 10

= 0.0105 x 103 x 1018.61 = 10.70

Q = 0.3

= 0.0395 x 103 x 634.21 = 25.05 Kcal

∆QPdts = 618672.26 Kcal

EAT CHANGE

dts – ∆Qreact + ∆HoR

∆QReq = 618672.26 – 57111.75

eat to be supplied by the system = 561560.51 Kcal

H

∆Qreq = ∆QP

= 561560.51 Kcal

H

37



8.6 ROTARY FILTER

eference temperature = 353k

eactants:

Liquid Stream

tic acid

0

R

R

(a) Ace

Q = 73.5 x 1 3 = 29811.6 Kcal

(b) ater

3 x 10

W

Q = 11. 3 = 2644.2 Kcal


Q = 14.8 x 103 = 9158.24 Kcal

Solvent

Acetic Acid(a) Q = 9.2 x 103

= 5740.8 Kcal

(b) ater

x 10

W

Q = 1.6 3 = 576 Kcal

∆Qreact = 47930.84 Kcal

roduct:

Filter cake

cid

0

P

(a) Acetic a

Q = 8.27 x 1 3 = 2580.24 Kcal

(b) ater

x 10

W

Q =13.2 3 = 237.6 Kcal

38




Q = 14.8 x 103 = 7044.8 Kcal

Filterate

Acetic acid

10

(a) Q = 74.43 x 3

= 23222.16 Kcal

(b) ater

58 x 10

W

Q = 11. 3 = 2084.4 Kcal

∆QPdts = 35169.2 Kcal

react + ∆HoR

Req


EAT CHANGEH

∆Qreq = ∆QPdts – ∆Q

∆Q = 35169.2 – 47930.84= –12761.64 Kcal

H

39



8.7 DRIER

eference temperature = 303K

eactant:

Filter cake

ic acid

0

R

R

(a) Acet

Q = 8.27 x 1 3 = 15481.44 Kcal

(b) ater

2 x 10

W

Q = 1.3 3 = 1425.6 Kcal


Q = 14.8 x 103 = 42268.8 Kcal

Air

Oxygen

x 10

(a) Q = 0.21 3 + 0.0002587T‐18770/T2

dT

(b) itrogen

x 10

= 0.21 x 103 x 814.20 = 170.98 Kcal

N

Q = 7 3

= 7 x 103 x 409.98 = 2869.86 Kcal

(c) Water

x 10Q = 1.5 3 = 1620 Kcal

Solven

cetic acid

0

t

(a) A

Q = 30.4 x 1 3 = 9484.6 Kcal

40



(b) ater

x 10

W

Q = 6.2 3 = 1116 Kcal

∆QReact = 74437.28 Kcal

roduct:

ude TA

10

P

Cr

3

Q = 14.8 x = 73970.4 Kcal

rier Scrubber Bottoms

10

D

(a) Acetic acid

Q = 38.67 x 3 = 126486.36 Kcal

(b) ater

1 x 10

W

Q = 8.8 3 = 16650.9 Kcal

Dried Air

gen

x 10

(a) Oxy

Q = 0.21 3 + 0.0002587T‐18770/T2 dT

(b) itrogen

x 10

= 0.21 x 103 x 1456.5 = 305.87 Kcal

N

Q = 7 3

= 7 x 103 x 95.34 = 667.38 Kcal

(c) Wat

0.21 x 10

er

Q = 3 = 52.92 Kcal

Q Pdts = 218133.83 Kcal

Required Pdts React + ∆H0R

8133 4 0

Heat to be supplied by the system = 143696.55 Kcal

∆

Q = ∆Q – ∆Q∆

= 21 .83 – 74 37.28 +

= 143696.55 Kcal

41





PROCESS DESIGN

OR

• We have a CSTR Reactor

• Calculation of volume, length and diameter of the reactor

t are:

= 1049 kg/m3

Volumetric feed rate in m /min = Specific volume × mass of the reactants entering

V = m /ρ = (1590)/(860x60)

/min

• = (7038) / (1049×60)

/min

• r

= (428.4) / (1000×60)

14m3 /min

• gen

= (55.65 × 32)/(60×20×1.429)3 /min

• = (209.35 × 28)/( 60×20×1.251)

3 /min

Total v e = 5.093

Taking safety factor as 10%

= V / Vo

e timee × Vo

9.1 DESIGN OF REACT

• Density values of the reactan

(a) Paraxylene = 860 kg/m3

(b) Acetic acid

(c) Water = 1000 kg/m3

(d) Oxygen = 1.429 kg/m3

(e) Nitrogen = 1.251 kg/m3

Volume occupied by each reactant is found out by:-3

• Paraxylene

o o

= 0.0308 m

3

Acetic acid

Vo

= 0.1118 m3

Wate

Vo

= 0.007

Oxy

Vo

= 1.038 m

Nitrogen

Vo

= 3.905 m

olumetric feed rat

Vo = 5.6023

Space time (CSTR)

Spac = 30 minVolume = Space tim

= 30 × 56.0233

= 168.69 m

43



2πV = × D × L

4

Assumption L /P = 1.5

2πV = × D × 1 .5 D

3

4

1 6 8 0 . 7 × 4D =π×1. 5

= 1.126m

= 1.689m

gitator Design

gitator design involves the calculation of diameter and height of the impeller from the

1. Diameter of the impeller = D = D/3 = 0.375m

D

L

A A

bottom, length and width of the blade.

A

2. Height of the impeller from the bottom =D/3 = 0.375m

3. Length of the blade = LB =DA /4 = 0.0937m

4. Width of the blade = WB =DA /5 = 0.075m

44



9.2 DESIGN OF MIXING TANK

asic:

ng mixing tank

e length and the design pressure

= 860 Kg/m3

3

Volume occupied by the each reactant

= m0 / ρ

= (15 × 106)/ (860 × 60)

= (117.3 × 60)/(1049 × 60)

11 m3/min

3. = (23.8 × 18) / (1000 × 60)

4 × 10-6

m3/min

= 0.179m3

Space time = 5min

0.149 45m3

owance

= 0.815msume L / D ratio is 0.5

= π/4 × D2

× 0.5D

g = mass/Volume 0.8195

g × L5

esign pressure = 0.692 × 1.1

B

Designi

• To find the diameter , th

• To find the tank volumeDensity of inlet streams

• Paraxylene

• Acetic acid = 1049 Kg/m

• Water = 1000 Kg/m3

1. Paraxylene

V0

= 0.0308m3/min

2. Acetic acid

V0

= 0.1

Water

V0

= 7.1

V0

V/Vo = 5V

= 5 × = 0.7

Assume 10% all3

VAs

Calculation:

Volume0.81952 = = π/4 × 0.5D

3

D3 = 2.806

D = 1.277m

λ = 0.6385m

Pav = 9057.828/= 11052.9kg/m

3

Pressure(p) =p1 ×= 11052.87 × 9.8 × 0.638

= 0.692atm

Hence the d

= 0.7612atm

45



8.3 DESSIGN OF DRIER

ir flow rate = 202.72Kg dry air/hr

midity Y1

= mass flow rate of wet air (kg / hr)

× 0.02)

6.77kg/hr= 500kg/m hr

rea = ms / mv = 206.77/ 500 = 0.414

D2 )

eat duty:

t = Gs × Cs (tg1 – tg2 )e dry solid

3.55 J/kg0C

t = 14117018 KJ/hr

= 500kg /m hr

= 40 C (By psychometric chart)

A

Leaving air hu = 0.020

Area = ms /mv ms

mv = maximum air velocity (kg/ m2hr)

ms = Gs + ( Gs × Y1)

ms = 202.72 + (202.72

= 202

mv

A

Area = (π/4) ×D

2 = (4/π)(0.414

D = 0.726m

H

QCs = heat capacity of th

=124

Q

2G

0T b

0(90-40)-(44-40)ΔT = = 18.21 C

90-40ln

44-40

⎡ ⎤⎢ ⎥⎣ ⎦

T

0.67

q

0.125π DG ΔT Length of the dryer =

D = 0.726m = 2.36feet

Btu/hrq = 13380.44

T0F = 59.32 0 F

Tq

0.125π×2.38 (102.42) 0.67×59.32L = 10.86ft= 3.31mL =

46



COST ESTIMATION

47



COST ESTIMATION

umber of working days per year = 350

.2 Kg/Day

e

URN OVER RATIO:

tio of total income to fixed capital investment.

urn over ratio = Total Income

ment

or chemical industries the turn over ratio is one.

ales = Rs 144,45,98,400

IRECT COST:

f the fixed capital investment= 0.7 x 1444598400 = Rs 1011218880

inting cost

cess and auxiliary cost

N

Cost of 1kg of Terepthalic Acid =Rs 70

Production of Terepthalic Acid = 58963

Gross sales for 1 yr or total incom =58963.2 x 350 x 70

= Rs 144,45,98,400

T

It can be defined as the ra

T

Fixed capital Invest

F

Thus, Fixed capital investment = Gross Annual S

But, Fixed capital investment = Direct cost + Indirect cost

D

It is taken as 70% o

The costs involved in the direct cost are,

i. Equipment cost

ii. Installation & Pa

iii. Instrumentation Cost

iv. Electrical cost

v. Piping Cost

vi. Building, pro

vii. Service facilities & yard improvement cost

viii. Land cost

48



Equipment cost

as 24% of fixed capital investment = 0.24 x 1444598400

ainting and installation cost

equipment cost = 0.4 x 346703616

strumentation cost

s 10% of equipment cost = 0.1 x 346703616

iping cost

5% of the equipment cost = 0.25 x 346703616

lectrical cost

aken as 25% of equipment cost = 0.25 x 346703616

uilding, process and auxiliary cost

st = 0.391677 x 346703616

ervice facilities & yard improvement cost

st = 0.1 x 346703616

It is taken

= Rs 346703616

P

It is taken as 40% of the

= Rs 138681446.4

In

It can be taken a

= Rs 34670361.6

P

It is 2

= Rs 86675904

E

I t can be t

= Rs 86675904

B

It is 39.1677% of equipment co

= Rs 135795832.2

S

It can be taken as 10% of equipment co

= Rs 34670361.6

49



Land cost

usually taken as 1% of fixed capital cost

= 0.01 x 1444598400

DIRECT COST:

apital Investment – Direct Cost

ion cost

al

ision cost

cost = 0.1 x 346703616 = Rs 34670361.6

ontingency

as 3.7% of fixed capital = 0.037 x 1444598400 = Rs 53450140.8

orking capital

apital investment

tal + working capital

stimation of total product cost

ome = 0.1 x 1444598400 = Rs 144459840

It is

= Rs 14445984

IN

Indirect cost = Fixed C

= 1444598400 – 1011218880= Rs 433379520

It consists of the following items

i. Engineering and supervis

ii. Contingency

iii. Working capit

Engineering and superv

It can be taken as 10% of equipment

C

It can be taken

W

It is 20% of total c

Total capital investment = fixed capi

= 1444598400 + (0.2 x 1444598400) = Rs 1733518080

Working capital = 0.2 x 1733518080 = Rs 346703616

E

Annual income = Rs 1444598400

Gross earning is 10% of annual inc

Product cost = Total annual income – Gross earnings

= 1444598400 – 144459840 = Rs 1300138560

50



Direct production cost

60% of the total product cost = 0.6 x 1300138560

aw materials cost

total product cost = 0.02 x 1300138560

s 26002771.2

perating labor cost

s 15% of total product cost = 0.15 x 1300138560

84

irect supervisory & clinical labor cost

.2 x 195020784

tilities

aken as 15% of total product cost = 0.15 x 1300138560

95020784

aintenance & repair cost

ital investment cost = 0.036 x 1444598400

2.4

aboratory charges

67% of operating labor cost = 0.0667 x 195020784

.29

oyalties

taken as 1.45% of the fixed capital cost = 0.0145 x 1444598400

It can be taken as

= Rs 780083136

R

It is 2% of the

= R

O

It can be taken a

= Rs 1950207

D

It is 20% of operating labor cost = 0

= Rs 39004156.8

U

It can be t

= Rs 1

M

It is 3.6% of fixed cap

= Rs 5200554

L

It is taken as 6.

= Rs 13007886

R

It is

= Rs 20946676.8

51



ixed charges

taken as 20% of product cost = 0.2 x 1300138560

027712

lant overheads

es cost for general upkeep and overhead packaging, medical services, safety

and pro

epreciation

r machinery is 10% of fixed capital cost = 0.1 x 1444598400

52

= Rs 144026460.5

surance

he fixed capital cost = 0.01 x 1444598400 = Rs 14445984

ent value

of the total product cost = 0.03033 x 1300138560

= Rs 39433202.52

eneral expenses

ive cost includes cost for officer, legal fees, office supplier and

commu

al; product cost = 0.05 x 1300138560 = Rs 65006928

F

It can be

= Rs 260

P

This includ

tection, recreation, sewage, laboratories, and storage facilities.

It is 5% of the total product cost = 0.05 x 1300138560 = Rs 65006928

D

Depreciation fo

= Rs 144459840

Depreciation of building is 3% of the land cost = 0.03 x 14445984

= Rs 433379.52

Total depreciation value = 144459840 – 433379.

In

It is 1% of t

R

It is 3.033%

G

Administrat

nication.

It is 5% of the tot

52



Distribution and selling cost

total product cost = 0.07 x 1300138560

91009699.2

esearch and development cost

cost = 0.01x 1300138560 = Rs 13001385.6

inancing

2% of the total product cost = 0.02 x 1300138560 = Rs 26002771.2

et profit

d after deduction of taxes from the Gross Earnings.

s 577839360

etermination of Pay-Back period (without interest charges)

)

I t accounts for 7% of the

= Rs

R

It is 1% of the total product

F

It is

N

It is obtaine

Net profit is 40% of the Gross Earnings = 0.4 x 1444598400 = R

D

= Depreciable fixed capital investment

(Average profit + average depreciation)/yr

= 1444598400

(577839360 +144459840

= 2 yrs.

53



PLANT LAYOUT

54



PLANT LAYOUT

LANT LOCATION :

he major factors that are involved in the selection plant size and plant location are

1. Availability of raw materials

ation facilities

bility

vailability of raw materials:

he source of raw materials is one of the important factor in the selection of plant site.

arket:

he locations of markets or intermediates distribution centers affect the cost of the product

ransport Facilities:

Transportation depends on the type of raw materials and products. Water, railroads andhig

P

T

2. Markets

3. Transport

4. Water supply

5. Energy Availa

6. Climate

7. Labour

A

T

M

T

distributed.

T

hways are the three types of transportation. Plant should have at least two of these.

55



Water supply:

We use water for cooling. So plant should be located where adequate water supply

sho

nergy Availabilty:

Power and fuel are to be considered as major factors in the choice of plant site.

uld be available.

E

56



.

57



INSTRUMENTATION ANDCONTROL

58



INSTRUMENTATION AND CONTROL

The process flow sheet shows the arrangement of the major equipments and thus, the

control of such major equipments is necessary.

13.1 INSTRUMENTS

Instruments are provided to monitor the key process variable during plant operation.

They may be incorporated in automatic control loops, or used for the manual monitoring of the

process operation. It is desirable that the process variable to be monitored be measured

directly: Often, however, this is impractical and some dependent variable, that is easier to

measure, is monitored in its place.

13.2 OBJECTIVES

SAFE PLANT OPERATION

1) To keep the process variables within known safe limits.

2) To detect dangerous situations as they develop and to provide alarms and automatic shutdown systems.

3) To provide inter locks and alarms to prevent dangerous operating procedures.

PRODUCTION RATE

To achieve the design product output.

PRODUCT QUALITY

To maintain the product composition within the specified quality standards.

COST

To operate the lowest production cost, commensurate with the other objectives.

59



13.3 TYPICAL CONTROL SYSTEMS

LEVEL CONTROL

In any equipment where an interface exists between two phases (liquid-vapor), some means ofmaintaining the interface at the required level must be provided. The control valve should be

placed on the discharge line from the pump. In this process, a level control valve is used in the

reactor, neutralization tanks and Decolorization tanks.

PRESSURE CONTROL

Pressure control will be necessary for most systemshandling vapor or gas. The method of

control will depend on the nature of the process.

FLOW CONTROL

Flow control is usually associated with inventory control in a storage tank or other equipment.

There must be a reservoir to take up the changes in flow rate. To provide flow control on a

compressor or pump running at a fixed speed and supplying a next constant volume output, a

by-pass would be used. Flow control valves are used in the reactors, neutralization tanks and

Decolorization tanks to control the flow of input and output streams.

CASCADE CONTROL

With this arrangement, output of one controller is used to adjust the set point of another.

Cascade control can give smoother control in situations where direct control of the variable

would lead to unstable operation. In reactor and neutralization tank 1 temperature control is

cascaded with flow controls.

60



REACTOR CONTROL

The schemes used for reactor control depend on the process and the type of the reactor.

If a reliable on-line analyzer is available, and the reactor dynamics are suitable, the product

composition can be monitored continuously and the reactor conditions and feed flows

controlled automatically to maintain the desired product composition and yield. More often,

the operator is the final link in the control loop, adjusting the controller set points to maintain

the product within specification, based on periodic laboratory analyses.Reactor temperature

will normally be controlled by regulating the flow of the heating or cooling medium. Pressure

is held constant. Material balance control will be necessary to maintain the correct flow of

reactants, products and unreacted materials.

61



SAFETY AND HAZARD

ANALYSIS

62



SAFETY AND HAZARD ANALYSIS:

The term “engineered safety ” covers the provision in the design of control systems,

alarm traps, trips , pressure- relief devices, automatic shutdown systems, duplication of

key equipment services, fire-fighting equipment, sprinkler systems and blast walls to

contain any fire or explosion.

Any organization has a legal and moral obligation to safeguard the health and welfare

of its employees and the general public. The term “loss prevention” is an insurance term,

the loss being the financial loss caused by an accident. These losses will not only be thecost of replacing damaged plant and third party claims, but also the loss of earnings from

lost production and lost sales opportunities.

All manufacturing processes are to some extent hazardous, but in chemical processes

there are additional, special, hazards associated with the chemicals used and the process

conditions. The designer must be aware of these hazards, and ensure, through the

application of a sound engineering practice that the risks are reduced to acceptable levels.

Safety is the loss prevention in process design can be considered under the following

broad headings:

1.Identification and assessment of potential hazards.

2.Control of the hazards: Eg. Explosion & Corrosion Fine dust dispersed in air in sufficient

concentrations, and in the presence of an ignition is a potential dust explosion hazard.

3. Control of the process

4. Limitation of the losses: The damage and injury caused if an incident occurs can be

prevented by providing pressure relief, safe plant layout, fire fighting equipments etc.

Process can be divided into those that of intrinsically safe, and those for which the

safety has to be engineered in. An intrinsically safe process in one in which safe operation

63



is inherent in the nature of the process i.e a process which causes no danger or negligible

danger under all foreseeable circumstances.

The safe operation of such processes depends on the design and provision of engineered

safety devices, and on good operating practices, to prevent dangerous situations

developing, and to minimize the consequences of any accident that arises from the failure

of these safeguards.

The process designer will be concerned more with the preventive aspects of the use of

hazardous substances.

1.Substitution : Substitution of the processing route with the less hazardous material

2.Containment: Sound design of equipment and piping to avoid leaking.

3.Ventilation : Use open structures or provide adequate ventilation systems

4.Disposal: Provision of effective vent stacks to dispose material

5. Emergent Equipment : Escape routes, rescue equipments, safety devices

It is the duty of the designer to select a process that is inherently safe whenever it is

practical, and economical, to do so. However, most chemical manufacturing processes are,

to a greater or lesser extent, inherently safe, and dangerous situations can develop if the

process deviates from the design values.

Workers attitude to safety is very important and it depends on whole array of factors,

ranging from social and religious background to his own circumstances and character. It

can be said that usually workers themselves are not the driving force in accident prevention

activities. Even in technically advanced countries where workers are relatively well off,

tremendous efforts are required to make them safety minded. This seems to show that

workers are seldom spontaneously interested in safety even though their lives are at stake.

This situation can be improved by organizing safety committee which takes care of job

safety.

64



CONCLUTION

65



CONCLUSION

Terepthalic Acid is an important chemical of considerable value. As a component of a lot of

polymers, it has grown to be a large volume synthetic organic chemical.

Our project analyses the material abd energy balances involved in the process and also deal

with the design and cost estimation aspects.

66



BIBLIOGRAPHY

67



BIBLIOGRAPHY

(1) M. Gopala Rao and Marshall Sittig, “Dryden’s Outlines of Chemical Technology”,

2nd Ed., East-West press, Page No: 431-433.

(2) Kirk-Othmer, “Encyclopaedia of Chemical Technology”, 5th Ed, Volume-1, JohnWiley & Sons Inc., Page No: 286-290.

(3) S.D Shukla and Pandey “A text book of chemical Technology”Vol-1, Inorganic

(4) I.Mukhlyonov & I.Furmer, “The most important industrial chemical process” part-2

MIR publishers.

(5) R. H. Perry and Don W. Green, “Perry’s Chemical Engineers’ Hand Book”, 6th and

7th Ed. Mc-Graw Hill International edition,

(6) H.Sawistowski &W.smith, “Mass Transfer Process calculations”, Interscience

publishers, Page No:54-99

(7) R. K. Sinnott, “Coulson and Richardson’s Chemical Engineering Series, volume-6,

Chemical Equipment Design” 3rd Ed., Butter Worth-Heinemann, Page No: 828-855, 891-895

(8) Joshi M. V., “Process Equipment Design”, 2nd Ed., Mc-Millan India Ltd,

(9) Max S. Peters and Klaus Timmerhaus, “Process Plant Design and Economics For

Chemical Engineers”, 3rd Ed., Mc-Graw Hill Book Company, Page No: 207-208,

484-485.

(10) Indian standard “Specification for Shell and Tube Heat Exchangers”,

IS 4503-1967, Page No: 5-66

Documents

Manufacture of Terepthalic Acid