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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.
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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
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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
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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).
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PROPERTIES
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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)
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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.
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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
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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.
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PROCESS DESCRIPTION
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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.
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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.
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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
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MASS BALANCE
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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
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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
Solvent from Scrubber
1038 kg of acetic acid
Reactor
90kg of water
Reflux Stream
17610 kg of acetic acid
3150 kg of water Vapour stream
19220 kg of acetic acid
3043.4 kg of water
705.6 kg of oxygen
5861.8 kg of nitrogen
240 kg of Carbon dioxide
Inlet = 37459 Kg=Outlet
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Error! 7.3 MASS BALANCE OF FIRST CRYSTALLIZER
Air65.6kg of Oxygen
215.9kg of Nitrogen
Liquid Stream Liquid Stream
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
1135kg of Acetic acid
311.84kg of water
215.9kg of Nitrogen
1.6kg of Oxygen
21.2 kg of paraxylene
Inlet = 8662.5 Kg = Outlet
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7.4 MASS BALANCE OF SECOND CRYSTALLIZER
Dilution Solvent
780kg of Acetic acid
43.9kg of water
Liquid Stream Liquid Stream
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
660kg of Acetic acid
64.78kg of water
Inlet = 7801 Kg = Outlet
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7.5 MASS BALANCE OF THIRD CRYSTALLIZER
Solvent
1476kg of Acetic acid 208.8kg of water
Liquid Stream Liquid Stream
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
1482kg of Acetic acid
208.8kg of water
.336kg of Oxygen
1.106kg of Nitrogen
Inlet = 8762.4 Kg = Outlet
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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
Inlet = 7651 Kg = Outlet
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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
Inlet = 5142 Kg = Outlet
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ENERGY BALANCE
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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
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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
Solvent from Scrubber
(a) Acetic Acid
Q = 17.3 x 103
= 16192.8
(b) Water
Q = 5 x 103 = 2700 Kcal
∆QReactant = 288057.6 Kcal
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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
Solvent from Scrubber
(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
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8.2 REACTOR
Reference Temperature = 335K
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
Solvent from Scrubber
(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
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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
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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
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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
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(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
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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
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(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
eat given to the system = 45801.46 Kcal
= 3.6 x 103 x 850.25 = 3060.9 Kcal
Q
H
∆Qreq = ∆QPdts – ∆Q
∆Q = 355219.34 - 401020.8
= -45801.46 Kcal
H
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8.5 THIRD CRYSTALLIZER:
Stream
tic acid
0
Reference Temperature = 273K
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
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(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
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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
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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
(c) erepthalic acidT
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
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(c) erepthalic acidT
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 given to the system = 12761.64 Kcal
EAT CHANGEH
∆Qreq = ∆QPdts – ∆Q
∆Q = 35169.2 – 47930.84= –12761.64 Kcal
H
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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
(c) erepthalic acidT
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
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(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
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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
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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
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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
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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 =
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COST ESTIMATION
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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
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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
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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
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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
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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
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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.
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PLANT LAYOUT
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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.
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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
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.
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INSTRUMENTATION ANDCONTROL
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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.
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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.
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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.
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SAFETY AND HAZARD
ANALYSIS
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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
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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.
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CONCLUTION
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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.
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BIBLIOGRAPHY
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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