PROJECT REPORT
on
MANUFACTURE OF L-CYSTINE
Submitted in partial fulfillment for the award of the degree of
BACHELOR OF TECHNOLOGY in
CHEMICAL ENGINEERING by
DIVYA PRABHU (10704006) DHEERAJ.T.S (10704023)
under the guidance of
Ms.K.SOFIYA, M.TECH., (Lecturer, School of Chemical Engineering)
FACULTY OF ENGINEERING AND TECHNOLOGY
SRM UNIVERSITY (under section 3 of UGC Act,1956)
SRM Nagar, Kattankulathur – 603 203 Kancheepuram Dist.
MAY 2008
BONAFIDE CERTIFICATE
Certified that this project report “MANUFACTURE OF L-CYSTINE SUGAR”
is the bonafide work of DIVYA PRABHU (10704006), DHEERAJ.T.S (10704021)
who carried out the project work under my supervision.
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HEAD OF THE DEPARTMENT INTERNAL GUIDE Date: EXTERNAL EXAMINER INTERNAL EXAMINER Date :
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ACKNOWLEDGEMENT
ACKNOWLEDGEMENT
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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.K.Sofiya 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.
TABLE OF CONTENTS
S.NO CHAPTER PAGE NO.
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1. ABSTRACT 6
2. INTRODUCTION 8
3. PROPERTIES 12
4. APPLICATIONS OF L-CYSTINE 15
5. PROCESS DESCRIPTION 17
6. MATERIAL BALANCE 22
7. ENERGY BALANCE 33
8. DESIGN 43
9. COST ESTIMATION 55
10. PLANT LAYOUT AND LOCATION 61
11. SAFETY AND LOSS PREVENTION 69
12. MATERIALS OF CONSTRUCTION 72
13. INSTRUMENTATION AND CONTROL 75
14. CONCLUSION 79
15. NOMENCLATURE 81
16. BIBLIOGRAPHY 84
LIST OF TABLES
TABLE PAGE NO.
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6.1 Glass-Lined Reactor 24
6.2 Neutralization Tank 1 25
6.3 Filter Press 1 26
6.4 Decolorization Tank 1 26
6.5 Filter Press 2 27
6.6 Neutralization Tank 2 28
6.7 Filter Press 3 29
6.8 Decolorization Tank 3 29
6.9 Filter Press 4 30
6.10 Final Neutralization Tank 31
6.11 Centrifuge 32
6.12 Tray Dryer 32
7.1 Glass Lined Reactor 35
7.2 Neutralization Tank 1 37
7.3 Neutralization Tank 2 39
7.4 Final Neutralization Tank 41
7.5 Tray Dryer 42
LIST OF FIGURES FIGURE PAGE NO.
5.1 Flow sheet for the manufacture 18
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of L-Cystine From human hair
8.1 Design of Reactor Vessel 48
8.2 Design of Filter Press 51
8.3 Design of Storage Vessel 54
10.1 Plant Layout 68
12.1 Glass Lined Equipment 73
12.2 Neutralization Tank 74
12.3 Decolorization Tank 74
13.1 Control Scheme 78
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ABSTRACT
ABSTRACT
The amino-acid L-cystine is manufactured from human hair. Human hair is rich in two
basic amino acid compounds such as L-cystine and L-Tyrosine, with L-Cystine
comprising about 12% of the hair.
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Chemically hair is a biopolymer largely of cystine cross linked proteins termed
keratins. L-cystine is a hydrolytic product of human hair, wool, horn, nail, feathers. But
human hair is considered to be the cheapest source and it has the major content of L-
Cystine than any other sources.
The human hair is hydrolyzed by hydrochloric acid; the hydrolysate consists of L-
Cystine which is separated by repeated neutralization and filtration. pH maintenance is
important in this process to obtain the crystals of L-Cystine.
L-cystine is used as a food additives, flavor enhancer, nutrient supplement and dough
strengthener.
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INTRODUCTION
INTRODUCTION
2.1 L-CYSTINE L-Cystine is a non-essential amino acid which can be produced by human beings. It is
present in many proteins and is a major constituent of keratin, the principal protein of
hair, skin and wool. Cystine is known as a disulfide amino-acid because it consists of two
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cysteine segments with their respective sulfur atoms bonded firmly together. Cystine
plays a special role as a cross-linking agent in protein structure.
2.2 NOTABLE FEATURES L-Cystine is found in the form of plates and prisms. It has a bland taste and can be taken
in powder or capsule form. The powder is water soluble. L-Cystine is the stable oxidized
form of L-Cysteine. L-refers to levorotatory, a type of optical rotation of a compound
under plane polarized light.
2.3 FUNCTIONS OF L-CYSTINE 1. It is necessary for the formation of skin and accelerates healing after injury or surgery.
2. L-Cystine strengthens the immune system, reduces damage from free radicals.
3. It promotes leucocyte formation and helps in the assimilation of vitamin B6.
4. This amino-acid is an integral part of the insulin molecule and thus important in
glucose metabolism.
5. Its sulfur rich amino group has been found useful in soothing for a variety of skin
conditions.
6. Allows skin to appear smooth and supple instead of rough and patchy.
7. It functions as an anti-oxidant and is a powerful aid to the body in protecting against
pollution and radiation.
8. It can help slow down the aging process and neutralize toxins. It serves to spare
Methanione.
2.4 HISTORY AND ISOLATION FROM NATURAL SOURCES In 1810 Watlaston described an organic compound which he had isolated from urinary
stones; it was soluble in both acids and alkalis, and it separated from alkaline solution on
acidification with acetic acid in the form of hexagonal plates. He called it an oxide
because of its apparently amphoteric nature and since it was derived from material
accumulating in the bladder, he decided on the term “cystic oxide”. The compound was
analyzed by Thaulow in Leibig’s laboratory and yielded results from which correct
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formula C6H12N2O4S2 was deduced. The name cystine instead of cystic acid was
introduced by Berrelius in 1833. In 1899 Morner isolated it from a horn hydrolysate.
Mauthner was the first to measure the optical rotation of L-Cystine and reported in 1882
an (x)D value at 20° of -205.88° in HCl solution.
2.5 SELECTION OF RAW MATERIAL The amino acid L-Cystine is produced from many sources like wool, horn, nail, feathers,
horse hair and human hair. Among these sources human hair is preferable as its
adequately available and cheap. Human hair consists of 12% of L-Cystine, 2.6% L-
Tyrosine, 82.4% of other amino acids, water, lipids, pigments and trace elements. In the
acid hydrolysis extraction process only barbed hair is suitable. Lengthy hair create
problem in the cleaning process. Dyed hair is also not preferable.
2.6 HAIR CHEMISTRY 2.6.1 HUMAN HAIR DESCRIPTION
Human hair is a keratin containing appendage that grows from large cavities or sacs
called follicles. The human hair fiber can be divided into three distinct zones along its
axis.The zone of biological synthesis and orientation resides at and around the bulb of the
hair. The next zone is the zone of keratinisation which is present in the outward direction
along the hair shaft, where stability is built into the hair structure via the formation of
cystine linkages, the third zone that eventually emerges through the skin surface is the
region of the permanent hair fibre which consists of dehydrated cornified cuticle, cortical
and sometimes medullary cells and inter-cellular cement. The diameter of human hair
fibres varies from 15 to 100µm.
2.6.2 COMPONENTS OF THE HUMAN HAIR
Hair is a complex tissue consisting of several morphological components and each
component consists of several chemical species. Depending on its moisture content (up to
32% by weight) human hair consists of approximately 65% to 95% proteins. Its
remaining constituents are water, lipids, pigment and trace elements. Proteins are
condensation polymer of amino acids, and those amino acids are Glycine, Alanine,
Valine, Isoleucine, Leucine, Phenylalanine, Tyrosine, lysine, Arginine, Histidine,
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Citrulline, Aspartic acid, Glutamic acid, Threonine, Serine, Cystine, Methionine,
Cysteine, Cysteic acid, Proline and Typtophan.
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PROPERTIES
PROPERTIES
3.1 GENERAL CHARACTERISTICS Crystal stem – Hexagonal system
Specific rotation – [α]²°D = -200 °~ -225° (1N HCl, C=2)
Melting point – 260°C
Stability – Decomposes in hot alkaline
aqueous solution
Solubility – H2O g/dl 0.0110(25°)
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0.0239(50°)
0.0523(70°)
0.1140(100°)
3.2 PHYSICAL & CHEMICAL PROPERTIES Physical state – powder
Appearance – white crystals
Specific gravity – 1.677
pH – 5.0~6.5
State of solution – ≥98%
Molecular weight – 240.31
EMPIRICAL FORMULA
C6H12N2O4S2
C= 29.99% H2= 5.03% O2= 26.63% N2= 11.66% S=26.69%
3.3 STRUCTURE –OOC–CH–CH2–S–S–CH2–CH–COO–
NH3+ NH3+
3.4 METABOLISM Glycogenic L-Cystine undergoes a reciprocal conversation with L-Cystine in the
oxidation reduction metabolism. It is metabolized to form pyruvic acid. L-Cystine is not
essential for human as it is derived from L-Methanione.
L-Cystine (1) is reduced to form L-Cysteine (2).
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3.5 ANALYTICAL REACTIONS A large number of gravimetric, colorimetric and titrimetric procedures have been offered
for the determination of cystine either as such or after its conversion to cysteine.
Such procedures depend in part upon
a) The ability of the disulfide grouping of cystine to undergo
disproportionate cleavage with heavy metals and with certain chemical reagents.
b) The kind of oxidation reduction reactions in which the sulfahydryl function of
cysteine can participate.
c) The property to form colored conjugates with various reagents and hence can
be estimated by colorimetric or spectrophotometric means; titration of the
sulfahydryl group may be achieved through the application of iodometric,
acidimetric and electrometric methods.
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APPLICATIONS
APPLICATIONS
The demand for L-cystine is 3000 tpa.
The applications of L-Cystine in different fields are:
4.1 PHARMACEUTICALS
• Expectorant for respiratory diseases such as chronic bronchitis.
• As a nutrients (transfusion solution) before and after operation on patients having
Hypoproteinemia and Hypo alimentation
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4.2 FOOD
• Components in flavors.
• Meat and meat products.
• Baby foods.
• Milk and milk products.
• Dough conditioner.
4.3 OTHERS
• Used to manufacture L-Taurine and L-glutathione .
Taurine plays a key role in central nervous system function. Glutathione is
critical for immune system function.
• Cystine strengthens the protective lining of the stomach and intestines.
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PROCESS DESCRIPTION
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5.1
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PROCESS DESCRIPTION Acid- hydrolysis extraction method is a batch process. Total batch time is 10.5 hrs. Hence
there are two batches in a day.
5.1 RAW MATERIAL PREPARATION Hair is the raw material used in the manufacture of L-cystine. Hair is first cleaned
manually, removing unwanted contents such as matches, paper, hair pins and cigarette
butts. The other principal contamination is sand which causes trouble and is removed by
beater, then hair is sent for screening and the sand particles are screened out. Then hair is
immersed in a container of hot water where it is totally rid of all impurities including oil.
5.2 GLASS-LINED REACTOR The reaction is carried out in a Glass-lined reactor which is a jacketed vessel. HCl is used
for the hydrolysis of human hair. The temperature of the hydrolysis process should be
105°C to 110°C. Acid and water are added to the reactor initially followed by preheating
which is done by indirect steam heating. Hair is charged and temp. is maintained at
110°C. Hair is hydrolyzed into hydrolysate by the breaking up of the peptide bonds in the
hair. The amino-acids present in the hair separates from the pigments, lipids which are
the constituents of hair and forms a hydrolysate. The reactor consists of an agitator,
which agitates the charged hair and accelerates the conversion of solid hair into
hydrolysate. The hydrolysate liquid is pumped to the Neutralization tank 1.
5.3 NEUTRALISATION TANK 1 In the Neutralization tank 1 the acidic condition is neutralized by adding Sodium
carbonate. The hydrolysate consists of 21 amino-acids, other impurities, acid is cooled to
50°C by water cooling. The pH is maintained at 5.2. From the hydrolysate the amino-acid
L-cystine precipitates down due to the pH maintenance. The reaction that takes place in
the tank by the addition of soda ash is as follows.
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
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CO2 formed is emitted out and the neutralized solution with salt formed is pumped to
the filter press 1 for the filtration process.
5.4 FILTER PRESS 1 The black cake and other amino-acids are separated by filtration process. The black cake
consists of L-cystine, salt formed, and certain amount of impurities. The filtration
solution is the mother liquor of other amino-acids like glycine, alanine, valine, lysine,
tyrosine, cysteine etc. The black cake consists of some traces of other amino acids which
are removed in further process. The black cake is then sent to the decolourisation process.
5.5 DECOLOURISATION TANK 1 This decolourisation tank consists of HCl and the black cake is dropped into the tank and
agitation is done to dissolve the cake. Then activated carbon is further added to the
dissolved cake for decolourisation process. The black cake is decolourised to form a
brown acidified solution of L-Cystine.
5.6 FILTER PRESS 2 Activated carbon is separated in the filtration stage. The carbon separated is considered to
be waste cake, which should be properly discarded. The filtrate solution is brown in color
which is sent to the Neutralization tank 2 to neutralize the acidified solution.
5.7 NEUTRALIZATION TANK 2 The brown color solution is pumped inside the neutralization tank 2 which consist of
Sodium hydroxide. The acidic solution is neutralized and the pH is maintained at 1.8. The
salt is formed by the following reaction
2HCl + NaOH → NaCl + H2O
The L-cystine in the solution precipitates down due to pH maintenance.
5.8 FILTER PRESS 3 The neutralized solution is filtered to remove impurities. The solution is pumped to filter
press 3 to remove the water and separate the brown colored cake. The water removed in
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this filtration stage is sent to the Effluent Treatment Plant. The brown colored L-cystine
obtained is to be further decolorized to give white colored final product.
5.9 DECOLORIZATION TANK 2 The brown colored cake is washed off from filter press 3 and dropped into the
decolorization tank 2 which consists of HCl. The cake is dissolved in the acid by
agitation, and then activated carbon is added to decolorize the brown cake. The carbon
added adsorbs impurities present in the dissolved cake.
5.10 FILTER PRESS 4 The decolorized solution is acidic in nature and it consists of carbon. The carbon content
is separated at this filtration stage. The decolorized solution is obtained as a filtrate
solution and is sent to the final neutralization tank.
5.11 FINAL NEUTRALIZATION TANK The acidic solution is neutralized by adding NaOH, the pH is maintained at 2. The L-
cystine precipitates down. The salt is formed as follows,
HCl +NaOH → NaOH + H2O
The solution is then pumped to the centrifuge.
5.12 CENTRIFUGE The neutralized solution consists of excess water which is removed by the centrifugation
process. The centrifuge is top suspended type. The salt formed in the final neutralization
is removed in this stage. The decanted water consists of iron and traces of impurities.
5.13 TRAY DRYER The product obtained consists of moisture content to about 10% which is removed by the
tray dryer. The humidity is reduced by passing hot air at 120°C. The temperature of the
product is reduced by natural cooling. The dried product obtained is further blended and
packed.
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MATERIAL BALANCE
MATERIAL BALANCE
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Molecular weight: Hydrochloric Acid (HCl) = 36.46
Sodium Hydroxide (NaOH) = 40
Sodium Carbonate (Na2CO3) = 106
Water (H2O) = 18
Sodium Chloride (NaCl) = 58.46
Carbon di-oxide (CO2) = 44
Basis: 1000 Kg of human hair is taken as feed per batch.
Pure human hair is 80% of the feed = 800 kg
Impurities is 20% of the feed = 200 kg
6.1 GLASS LINED REACTOR Acid hydrolysis: 60% HCl is required
HCl required to hydrolyze 800 kg of human hair = 800 x 0.6
= 480 kg
H2O added to hydrolyze 800 kg of human hair = 800x 0.4
= 320 kg
2% of acid added is vaporized.
2% of water added is evaporated.
Hydrolysate H2O=313.6 kg HCl=470.4 kg Total Amino Acids=1000 kg
Human hair Pure content=800 kg Impurities=200 kg H2O=320 kg HCl=480 kg
GLASS LINED REACTOR
Evaporated Solution
Water =2% of 320 =6.4 kg HCl =2% of 480
=9.6 kg
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Output (kg)
Components
Input (kg)
Evaporated
Acid
Hydrolysate
Human Hair 800
Water 320 6.4 313.6
HCl 480 9.6 470.4
Impurities 200
Total amino acids 1000.0
16.0
1800 1800
6.2 NEUTRALIZATION TANK 1 Reaction:
2HCl + NaCO3 → 2NaCl+CO2+H2O
Amount of HCl required = (470.4x2)/36.46 = 25.8036 moles
Amount of Na2CO3 required = 25.8036 x 106/2 = 1367.5908 kg
Amount of NaCl formed =25.8036 x 58.46 = 1508.4785 kg
Amount of CO2 formed =25.8036 x 44/2 =567.6792 kg
Amount of H2O formed =25.8036 x 18/ 2 =232.2324 kg
76.6587% of NaCO3 solution is required to neutralize the hydrolysate.
Water required in neutralization =1784 x (1-0.766587) = 416.4092 kg
CO2 formed is emitted out =567.6792 kg
Amino acids contain 12% cystine.
CO2=567.6792 kg
Cystine=120 kg NaCl=1508.4785kg H2O=416.4092 +232.2324+313.6 =962.2416 kg Other amino Acids=409.6007 kg
Total amino acids=1000kg H2O=313.6kg HCl=470.4 kg
NEUTRALIZATION TANK 1
Water=416.4092 kg Na2CO3=1367.5908
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Output (kg) Components Input
(kg) Evaporated
Gas
Neutralized
Solution
HCl 470.4000
H2O 730.0092 962.2416
Na2CO3 1367.5908
Total amino acids 1000.0000
Cystine 120.0000
Other amino acids 409.6007
NaCl 1508.4785
CO2 67.6792
3568.0000 3568.0000
6.3 FILTER PRESS 1 The efficiency of the filter press1 is 90%.
The cystine loss is 10%.
Black cake consists of 5% of NaCl, 10% of H2O from their weights in the feed and 2% of
other amino acids in the form of impurities.
Black cake Cystine=108 kg NaCl=75.4239kg H2O=96.2242kg Impurities=8.192kg
Cystine =120 kg FILTER PRESS 1
NaCl=1508.4785 kg Water=962.2416 kg Other amino Acids=409.6007 kg
Filtrate Cystine=10% of 120 =12 kg NaCl=5% of 1508.4785 =1433.0545 kg Water=10% of 962.2416 =866.0174 kg Other amino Acids =401.4087 kg
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Output (kg) Components
Input
(kg) Black cake Filtrate solution
NaCl 1508.4785 75.4239 1433.0545
H2O 962.2416 96.2242 866.0174
Cystine 120.0000 108.0000 12.0000
Impurities 8.1920
Other amino acids 409.6007 401.4087
287.8402 2712.4806
3000.3208 3000.3208
6.4 DECOLOURIZATION TANK1 10% HCl is required to dissolve cystine cake.
HCl required to dissolve 287.8402 kg of cystine cake =287.8402 x 0.1 = 28.7840 kg
Water required to dissolve 287.8402 kg of cystine cake =287.8402 x 0.9 = 259.0562 kg
7% of carbon is required to decolorize cystine cake.
Carbon required= 287.8402 x 0.07 = 20.1488 kg
HCl=28.7840 kg Water=259.0562 kg Carbon=20.1488 kg
DECOLORIZATION TANK 1
Cystine=108 kg NaCl=75.4239 kg Water=96.2242 kg Impurities=8.192kg
Cystine=108kg H2O=355.2804kg NaCl=75.4239kg HCl=28.7840kg Carbon=20.1488kg Impurities=8.192kg
Components Input (kg) Output (kg)
Cystine 108.0000 108.0000
NaCl 75.4239 75.4239
H2O 355.2804 355.2804
HCl 28.7840 28.7840
Carbon 20.1488 20.1488
Impurities 8.1920 8.1920
595.8291 595.8291
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6.5 FILTER PRESS 2 8% of cystine loss. 3% of NaCl loss. 3% of HCl loss.98% of carbon separated. 6% of
water loss. Waste cake consists of 80% of total impurities.
FILTER PRESS2
Waste cake
Cystine=108 kg H2O=355.2804 kg Imp. =8.192kg NaCl=75.4239 kg HCl=28.7840 kg Carbon=20.1488kg
Filtrate Cystine=99.36kg H2O=333.9636kg Imp. =1.6384kg NaCl=73.1612kg
Cystine=0.08 x108 =8.64 kg HCl=0.03 x28.7840 =0.8635 kg H2O=0.06 x355.2804 =21.3168 kg NaCl=0.03 x75.4239 =2.2627 kg Carbon=0.98 x20.1488 =19.7458 kg Imp. =0.8 x8.192kg =6.5536 kg
HCl=27.9205kg
Output (kg) Components Input
(kg) Waste cake Filtrate
Cystine 108.0000 8.6400 99.3600
NaCl 75.4239 2.2627 73.1612
HO 355.2804 21.3168 333.9636
HCl 28.7840 0.8635 27.9205
Carbon 20.1488 19.7458 0.4030
Impurities 8.1920 6.5536 1.6384
59.3824 536.4467
595.8291 595.8291
6.6 NEUTRALIZATION TANK 2
NaOH + HCl → NaCl + H2O
Amount of HCl reacted = 27.9205/36.46 = 0.7657 moles
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Amount of NaOH required = 0.7657 x 40 = 30.6280 kg
5% NaOH solution is required.
Water required to neutralize 536.4467 kg of filtrate = 536.4467 x 0.95 = 509.6244 kg
Components Input (kg) Output (kg)
Cystine 99.3600 99.3600
NaCl 73.1612 117.9271
H2O 843.5880 857.3706
HCl 27.9205
Carbon 0.4030 0.4030
Impurities 1.6384 1.6384
NaOH 30.6289
1076.6991 1076.6991
6.7 FILTER PRESS 3 Loss of cystine is 10%. Brown cystine cake consists of 5% of NaCl. 5% of moisture.
50% loss of impurities.
NEUTRALIZATION TANK 2
Brown cake
Filtrate solution
Cystine=0.9x99.36 =89.4240 kg NaCl=0.05x117.9271 =5.8964 kg H2O=0.05x857.3706 =42.8685 kg Imp. =0.50x1.638 =0.8192 kg
Cystine=99.36 kg NaCl=117.9271 kg H2O=857.3706kg Carbon=0.403 kg Imp. =1.6384 kg
FILTER PRESS 3
Water=509.6244 kgNaOH=30.6280 kg
Cystine=9.936 kg H2O=814.5021 kg NaCl=112.0307 kgCarbon=0.403 kg Imp. =0.8192 kg
Cystine=99.36 kg HCl=27.9205 kg NaCl=73.1612 kg Carbon=0.403 kg Water=333.9636 kg Imp.=1.6384 kg
Cystine=99.36kg NaCl=73.1612 +44.7658 =117.9271kg H2O =333.9636 +509.6244+3.7826 =857.3706 kg Carbon= 0.403 kg Imp. = 1.6384 kg
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Output (kg) Components Input
(kg) Filtrate Brown cake
NaCl 117.9271 112.0307 5.8964
H2O 857.3706 814.5021 42.8685
Cystine 99.3600 9.9360 89.4240
Carbon 0.4030 0.4030
Impurities 1.6384 0.8192 0.8192
937.691 139.0081
1076.6991 1076.6991
6.8 DECOLORIZATION TANK 2 20% of HCl is required to dissolve Brown cystine cake.
HCl required = 139.0081 x 0.2 = 27.8016 kg
H2O required = 80% of 139.0081 = 111.2065 kg
10% of carbon is required to decolorize Brown Cystine cake.
Carbon required = 139.0081 x 0.1 = 13.9008 kg
Components Input (kg) Output (kg)
Cystine 89.4240 89.4240
NaCl 5.8964 5.8964
H2O 154.0750 154.0750
HCl 27.8016 27.8016
Impurities 0.8192 0.8192
Carbon 13.9008 13.9008
291.9170 291.9170
DDEECCOOLLOORRIIZZAATTIIOONN TTAANNKK 22
Cystine =89.4240 kg NaCl=5.8964 kg Water=42.8685 kg Impurities=0.8192 kg
Cystine =89.4240kg
HCl=27.8016 kg Water=111.2065 kgCarbon=13.9008 kg
NaCl=5.8964 kg H2O=154.0750 kg HCl=27.8016 kg Imp. =0.8192 kg Carbon=13.9008 kg
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6.9 FILTER PRESS 4 Cystine loss-10%;NaCl loss-10%; HCl loss-5%
In waste cake: Impurities-5% from its weight in feed
Carbon-98% from its weight in feed.
Output (kg) Components Input
(kg) Waste cake Filtrate
Cystine 89.4240 8.9424 80.4816
NaCl 5.8964 0.5896 5.3068
HCl 27.8016 1.3901 26.4115
Impurities 0.8192 0.0410 0.7782
H2O 154.0750 7.7038 146.3712
Carbon 13.9008 13.6228 0.2780
32.2897 259.6273
291.9170 291.9170
FILTER PRESS 4
Cystine =89.4240 kgNaCl=5.8964 kg Carbon=13.9008 kg H2O=154.0750 kg HCl=27.8016 kg Imp. =0.8192 kg
Waste cakeCystine=8.9424kg NaCl=0.5896 kg H2O=7.7038 kg HCl=1.3901 kg Imp. =0.0410 kg Carbon=13.6228kg
Cystine=80.4816kg Filtrate
NaCl=5.3068 kg H2O=146.3712 kg HCl=26.4115 kg Imp. =0.7782 kg Carbon=0.2780 kg
6.10 FINAL NEUTRALIZATION TANK
NaOH + HCl → NaCl + H2O
Amount of HCl required = 26.4115/36.46 = 0.7244 moles
Amount of NaOH required = 0.7244 x 40 = 28.9760 kg
10% NaOH solution is required to neutralize cystine solution.
H2O required to neutralize 259.6273 kg of cystine solution=259.6273 x0.9=233.6646 kg
32
Amount of NaCl formed = 0.7244 x 40 = 28.9760 kg
Amount of H2O formed= 0.7244 x 18 =13.0392 kg
Components Input (kg) Output (kg)
Cystine 80.4816 80.4816
NaOH 28.9760 28.9760
Water 380.0358 380.0358
HCl 26.4115 26.4115
Impurities 0.7782 0.7782
Carbon 0.2780 0.2780
NaCl 5.3068 5.3068
522.2679 522.2679
FINAL NEUTRALIZATION
TANK
NaOH=28.9760 kgWater=233.6646 kg
Cystine=80.4816 kg NaCl=5.3068 kg Water=146.3712 kg HCl=26.4115 kg Impurities=0.7782 kg Carbon=0.2780 kg
Cystine=80.4816kgNaCl=47.6551kg H2O=393.075kg Imp. =0.7782 kg Carbon=0.2780kg
6.11 CENTRIFUGE 15% Cystine loss. Impurities are reduced to 50%. Carbon is completely removed.
98% water loss. 98% NaCl loss. Cystine cake
Decanted water
CENTRIFUGECystine=80.4816 kg NaCl=47.6551 kg Water=393.0750 kg Imp. =0.7782 kg Carbon=0.2780 kg
Cystine=68.4094kgNaCl=0.9531 kg Water=7.8615 kg Imp. =0.3891kg
Cystine=12.0722kgImp. =0.3891kg NaCl=46.7020kg Water=385.2135kgCarbon=0.2780kg
33
Output (kg) Components Input
(kg) Decanted water Cystine cake
Cystine 80.4816 12.0722 68.4094
Water 393.0750 385.2135 7.8615
NaCl 47.6551 46.7020 0.9531
Impurities 0.7782 0.3891 0.3891
Carbon 0.2780 0.2780
444.6548 77.6131
522.2679 522.2679
6.12 TRAY DRYER 98% of the total moisture is removed in this process.
Cystine=68.4094 kg
TRAY DRYERNaCl=0.9531 kg H2O=7.8615 kg Imp. =0.3891 kg
Cystine=68.4094 kgNaCl=0.9531 kg H2O=0.1572 kg Imp. =0.3891kg
H2O=7.7043kg Output (kg) Components Input
(kg) Water removed Cystine
Cystine 68.4094 68.4094
Water 7.8615 7.7043 0.1572
NaCl 0.9531 0.9531
Impurities 0.3891 0.3891
7.7043
77.6131 77.6131
PRODUCT L-cystine = 68.4094 kg /batch = 136.8188 kg/day (Two batches in a day)
Thus, L-cystine produced in an yr (300 working days) = 41tpa
34
ENERGY BALANCE
ENERGY BALANCE
35
Reference temperature: 25 °C = 298.15 K
Room temperature = 30 °C =303.15 K
Heat of formation (ΔHf cal/kmol) Specific heat constant (CP cal/mol K)
Na2CO3 (s) = -269.46 x 10³ Na2CO3 (s) = 28.9
HCl (l) = -39.85 x 10³ HCl (l) = 6.7 + 0.00084T
NaOH (l) = -112.193 x 10³ NaOH (l) = 9.4373
NaCl (s) = -98.321 x 10³ NaCl (s) = 10.79 + 0.0042T
H2O (l) = -68.3164 x 10³ H2O (l) = 18
CO2 (g) = -94.052 x 10³ CO2 (g) = 10.34 + 0.00274T – 195500/T²
C (s) = 2.673 + 0.002617T – 116900/T²
H2O (g) = 8.22 + 0.00015T + 0.00000134 T²
7.1 GLASS LINED REACTOR 110°C
Hair + H2O + HCl Hydrolyzed hair
HEAT ASSOCIATED WITH REACTANTS
303.15
QHCL = 480 x 10³/36.46 ʃ (6.7 + 0.00084T) dT cal = 0.4576 x 10³ kcal
298.15
303.15
QH2O = 320 x 10³/18 ʃ (18) dT cal = 0.0889 x 10³ kcal
298.15
ΔQ reactant = 0.5465 x 10³ kcal
HEAT ASSOCIATED WITH PRODUCTS
383.15
QHCL = 470.4 x 10³/36.46 ʃ (6.7 + 0.00084T) dT cal = 7.6614 x 10³ kcal
298.15
383.15
QH2O = 313.6 x10³/18 ʃ (18) dT cal =26.656 x 10³ kcal
36
298.15
373.15
Evaporated acid: QH2O = 6.4 x10³ /18 ʃ (8.22+ 1.5xE-4 T + 1.34xE-6 T² ) dT
298.15
+ 6.4 x10³/18 x (9729) = 3.6838 x 10³ kcal
383.15
QHCL = 9.6 x10³/36.46 ʃ (6.7 + 0.00084T) dT + 9.6 x10³/18 x (3860)
298.15
= 1.1720 x 10³ kcal
ΔQ product = 39.1732 x 10³ kcal
CHANGE IN ENERGY
ΔQ required = ΔQ product – ΔQ reactant + ΔH° R
= 39.1732 – 0.5465 + 0 = 38.6267 x 10³ kcal
Components Input (kcal) x 10³ Output (kcal) x 10³
HCl 0.4576 7.6614
H2O(l) 0.0889 26.6560
H2O(g) 1.1720
Heat of reaction 0
Heat supplied 38.6267
39.1732 39.1732
7.2 NEUTRALIZATION TANK 1 50°C
Na2CO3 + 2HCl 2NaCl + CO2 + H2O
STANDARD HEAT OF REACTION AT 298.15K
ΔHR° = (ΔHf) product – (ΔHf) reactant
For 1mole of Na2CO3 :
ΔHR°= {[2(-98.321) + (-94.052) + (-68.3164)] – [(-269.46) + 2(-.85)]} x 10³
= – 9.8504 x 10³ kcal
37
For 1367.5908/106 moles of Na2CO3 :
ΔHR°= (- 9.8504 x 10³) x 1367.5908/106
= -127.0879 x 10³ kcal
HEAT ASSOCIATED WITH REACTANTS
383.15
QHCL = 470.4 x10³/36.46 ʃ (6.7 + 0.00084T) dT cal = 7.6614 x 10³ kcal
298.15
383.15
QH2O = 730.0092 x10³/18 ʃ (18) dT cal = 62.0508 x10³ kcal
298.15
303.15
QNa2CO3=1367.5908 x10³/106 ʃ (28.9) dT cal=18.6431 x10³ kcal
298.15
ΔQ reactant = 88.3553 x10³ kcal
HEAT ASSOCIATED WITH PRODUCTS
323.15
QNaCl = 1508.4785 x10³ /58.46 ʃ (10.79 + 0.0042T) dT cal =7.8022 x10³ kcal
298.15
323.15
QH2O = 962.2416 x10³ /18 ʃ (18) dT cal = 24.056 x10³ kcal
298.15
323.15
Qcystine = 120 x10³ /240.31 ʃ (141.2) dT cal = 1.7627 x10³ kcal
298.15
323.15
QCO2=567.6792 x10³/44 ʃ (10.34+0.00274T–195500/T²)dT cal = 2.9552 x10³ kcal
298.15
38
ΔQ product = 36.5761 x10³ kcal
CHANGE IN ENERGY
ΔQ required = ΔQ product – ΔQ reactant + ΔH° R
= (36.5761 – 88.3553 – 127.0879) x10³= - 178.8671 x10³ kcal
Components Input (kcal) x10³ Output (kcal) x10³
HCl 7.6614
H2O(l) 62.0508 24.0560
Na2CO3 18.6431
NaCl 7.8022
Cystine 1.7627
CO2 2.9552
Heat of reaction 127.0879
Heat released 178.8671
215.4432 215.4432
7.3 NEUTRALIZATION TANK 2
50°C
NaOH + HCl NaCl + H2O
STANDARD HEAT OF REACTION AT 298.15K
ΔHR° = (ΔHf) product – (ΔHf) reactant
For 1mole of NaOH
ΔHR° = {[(-98.321) + (-68.3164)] – [(-112.193) + (-39.85)]} X10³ = – 14.5944 x10³ kcal
For 30.628/40 moles of NaOH
ΔHR° = (-14.5944 x10³) x 30.628/40 = -11.1749 x10³ kcal
HEAT ASSOCIATED WITH REACTANTS
303.15
Qcystine = 99.36 x10³/240.31 ʃ (141.2) dT cal = 0.2919x10³ kcal
39
298.15
303.15
QNaCl = 73.1612 x10³/58.46 ʃ (10.79 + 0.0042T) dT cal = 0.0754 x10³ kcal
298.15
303.15
QNaOH = 30.628 x10³ /40 ʃ (9.4373) dT cal = 0.0361 x10³ kcal
298.15
303.15
QHCL = 27.9205 x10³ /36.46 ʃ (6.7 + 0.00084T) dT cal = 0.0266 x10³ kcal
298.15
303.15
QH2O = 843.588 x10³/18 ʃ (18) dT cal = 4.2179 x10³ kcal
298.15
303.15
Qcarbon = 0.403 x10³/12 ʃ (2.673+0.002617T–116900/T²)dT cal = 0.0003 x10³ kcal
298.15
ΔQ reactant = 4.6482 x10³ kcal
HEAT ASSOCIATED WITH PRODUCTS
323.15
Qcystine = 99.36 x10³/240.31 ʃ (141.2)dT cal = 1.4595 x10³ kcal
298.15
323.15
QNaCl = 117.9271 x10³/58.46 ʃ (10.79 + 0.0042T) dT cal =0.6099 x10³kcal
298.15
323.15
QH2O = 857.3706 x10³/18 ʃ (18) dT cal = 21.4343 x10³ kcal
298.15
40
323.15
Qcarbon = 0.403 x10³/12 ʃ (2.673+0.002617T–116900/T²)dT cal = 0.0019 x10³ kcal
298.15
ΔQ product = 23.5056 x10³kcal
CHANGE IN ENERGY
ΔQ required = ΔQ product – ΔQ reactant + ΔH° R
= (23.5056 – 4.6482 – 11.1749) x10³ = 7.6825 x10³ kcal
Components Input (kcal) x10³ Output(kcal) x10³
Cystine 0.2919 1.4595
NaCl 0.0754 0.6099
H2O 4.2179 21.4343
HCl 0.0266
Carbon 0.0003 0.0019
NaOH 0.0361
Heat of reaction 11.1749
Heat supplied 7.6845
23.5056 23.5056
7.4 FINAL NEUTRALIZATION TANK 50°C
NaOH + HCl NaCl + H2O
STANDARD HEAT OF REACTION AT 298.15K
ΔHR° = (ΔHf) product – (ΔHf) reactant
For 1mole of NaOH:
ΔHR° = {[(-98.321) + (-68.3164)] – [(-112.193) + (-39.85)]} x10³= – 14.5944 x10³ kcal
For 28.976/40 moles of NaOH:
ΔHR°= (-14.5944 x10³) x 28.976/40 = -10.5722 x10³ kcal
HEAT ASSOCIATED WITH REACTANTS
303.15
Qcystine = 80.4816 x10³ /240.31 ʃ (141.2) dT cal = 0.2364 x10³ kcal
41
298.15
303.15
QNaCl = 5.3068 x10³ /58.46 ʃ (10.79 + 0.0042T) dT cal = 0.0055 x10³ kcal
298.15
303.15
QNaOH = 28.976 x10³ /40 ʃ (9.4373) dT cal = 0.0342 x10³ kcal
298.15
303.15
QHCL = 26.4115 x10³ /36.46 ʃ (6.7 + 0.00084T) dT cal = 0.0334 x10³ kcal
298.15
303.15
QH2O = 380.0358 x10³ /18 ʃ (18) dT cal = 1.9002 x10³ kcal
298.15
303.15
Qcarbon = 0.278 x10³ /12 ʃ (2.673+0.002617T–116900/T²)dT cal = 0.0002 x10³ kcal
298.15
ΔQ reactant = 2.2099 x10³ kcal
HEAT ASSOCIATED WITH PRODUCTS
323.15
Qcystine =80.4816x10³/240.31 ʃ (141.2) dT cal=1.1822 x10³kcal
298.15
323.15
QNaCl = 47.6551 x10³/58.46 ʃ (10.79 + 0.0042T) dT cal = 0.2464 x10³kcal
298.15
323.15
QH2O = 393.075 x10³/18 ʃ (18) dT cal = 9.8269 x10³ kcal
298.15
42
323.15
Qcarbon = 0.278 x10³/12 ʃ (2.673+0.002617T–116900/T²)dT cal = 0.0013 x10³ kcal
298.15
ΔQ product = 11.2568 x10³kcal
CHANGE IN ENERGY
ΔQ required = ΔQ product – ΔQ reactant + ΔH° R
= 11.2568 – 2.2099 – 10.5722 = -1.5253 x10³ kcal
Components Input(kcal) x10³ Output(kcal) x10³
Cystine 0.2364 1.1822
NaOH 0.0342
H2O 1.9002 9.8269
HCl 0.0334
Carbon 0.0002 0.0013
NaCl 0.0055 0.2464
Heat of reaction 10.5722
Heat released 1.5253
12.7821 12.7821
7.5 TRAY DRYER HEAT ASSOCIATED WITH REACTANTS
303.15
Qcystine = 68.4094 x10³ /240.31 ʃ (141.2) dT cal = 0.2010 x10³ kcal
298.15
303.15
QNaCl = 0.9531 x10³ /58.46 ʃ (10.79 + 0.0042T) dT cal = 0.0001 x10³kcal
298.15
303.15
QH2O = 7.8615 x10³/18 ʃ (18) dT cal = 0.0393 x10³ kcal
298.15
43
ΔQ reactant = 0.2404 x10³kcal
HEAT ASSOCIATED WITH PRODUCTS
393.15
Qcystine = 68.4094 x10³/240.31 ʃ (141.2) dT cal =3.8186 x10³ kcal
298.15
393.15
QNaCl = 0.9531 x10³/58.46 ʃ (10.79 + 0.0042T)dT cal = 0.0189 x10³kcal
298.15
393.15
QH2O = 0.1572 x10³/18 ʃ (18) dT cal = 0.0149 x10³kcal
298.15 393.15
Evaporated water: QH2O = 7.7043 x10³/18 ʃ (8.22+ 1.5xE-4 T + 1.34xE-6 T² ) dT
298.15
+ 7.7043 x10³/18 x (9729) = 4.4345 x10³kcal
ΔQ product = 8.2869 x10³kcal
CHANGE IN ENERGY
ΔQ required = ΔH product – ΔH reactant = (8.2869 – 0.2404 ) x10³ = 8.0465 x10³kcal
Components Input(kcal) x10³ Output(kcal) x10³
Cystine 0.2010 3.8186
H2O(l) 0.0393 0.0149
NaCl 0.0001 0.0189
H2O(g) 4.5071
Heat supplied 8.0465
8.2869 8.2869
44
DESIGN
DESIGN
8.1 DESIGN OF REACTOR VESSEL Data:
Density of water = 1000kg/m³
45
Density of HCl = 84.3681 kg/m³
Density of Hair = 237 kg/m³
Amount of water = 320 kg
Amount of HCL = 480 kg
Amount of Hair = 800 kg
Volume of water = 320/1000 = 0.32 m³
Volume of HCL = 480/845.3681 = 0.5678 m³
Volume of Hair = 800/237 = 3.3755 m³
Total volume = 4.2633 m³
Let us take 10% excess volume
Working volume = 4.2633(1+0.1) = 4.6896 m³
Working volume = volume of cylindrical portion + Volume of conical portion
= {π x (Dt/2)² x Hcy} + {π /3x (Dt/2)² x Hconi}
Where, Dt – Diameter of Vessel (m)
Hcy – Height of Cylindrical Portion (m)
Hconi – Height of Conical portion (m)
Ht – Height of Reactor Vessel (m)
∴ {π x (Dt/2)² x Hcy} + {π /3x (Dt/2)² x Hconi} = 4.6896
Hcy = 2Dt & Hconi = Dt /2
{π x (Dt/2)² x 2Dt } + {π /3x (Dt/2)² x Dt /2} = 4.6896
π /2 x Dt³ + π /24 x Dt³ = 4.6896
13π /24 x Dt³ = 4.6896
Dt = 1.402 m
Hcy = 2Dt = 2.804 m
Hconi = Dt /2 = 0.701 m
Ht = Hcy + Hconi = 3.505 m
Thickness of cylindrical portion = pDt = ρgHcy x Dt = 7861.093 x 9.81x 2.804 x 1.402
2f 2fs 2x (5 x106)
46
= 0.0303 m = 3.03 cm
Thickness of lining = 20% of Thickness of cylindrical portion = 0.2 x 3.03 = 0.06 cm
Thickness of conical head = pDt/2f cosα = ρgHconi x Dt = 7861.093 x 9.81x 0.701x 1.402
2fs cosα 2x (9.8 x106) xcos45°
= 0.0055 m = 5.5 mm
Where, fs – Shear Stress (kg/cm²)
P – Pressure exerted (N/m²)
AGITATOR
Da =Diameter of Agitator = 1/3 Dt = 0.4673 m
Let us assume the speed of agitator ‘u’ as 350 rpm
π DaN = u
π x 0.4673 x N = 350
N = speed of motor = 238.4089 rpm = 3.9734 rps
Reynolds number
NRe = NDa²ρ/µ
ρ = density = 237 kg/m³
µ = viscosity = 6.002 x 10-4 kg/ms
NRe = 342614 > 104, thus turbulent region
From NRe vs. power number (Np) table : Np = 5
Power number
Np = P/( ρ x N³ x Da5)
P = Np x ρ x N³ x Da5 = 5 x 237 x (3.9734) ³ x (0.46735) = 1656.4721 watts
= 1.6565 kw = 2.2214 HP
25% excess power
P = 2.2214 (1+0.25) = 2.7768 HP
47
Shaft diameter
Tc = Torque = HP x 75 x 60/ 2π N = 2.7768 x 75 x 60 = 8.3417 Nm
2π x 238.4089
Tmax = c x Tc = 1.9 x 8.3417 = 15.8492 Nm
ZP = Polar Modulus of shaft = Tmax/fs = 15.8492/ (5 x 106) = 3.1698 x 10-6 m³
π ds³/16 = ZP
ds = 0.0253 m
Some standard ratios
Da/ Dt = 1/3
E/Dt = 1/3
L/Da = ¼
W/Da = 1/5
H/Dt = 1
J/Dt = 1/12
E = height of impellor above vessel floor = 0.4673 m
L = length of impellor blade = 0.1168 m
W = width of impellor = 0.0935 m
H =depth of liquid in vessel = 1.402 m
J = width of baffle = 0.1168 m
DESIGN SUMMARY
Diameter of Vessel (Dt) = 1.402 m
Height of Cylindrical Portion (Hcy) = 2.804 m
Height of Conical portion (Hconi) = 0.701 m
Height of Reactor Vessel (Ht) = 3.505 m
Thickness of cylindrical portion = 3.03 cm
48
Thickness of lining = 0.06 cm
Thickness of conical head = 0.55 cm
Diameter of Agitator (Da) =0.4673 m
Shaft diameter (ds) = 0.0253 m
Height of impellor above vessel floor (E) = 0.4673 m
Length of impellor blade (L) = 0.1168 m
Width of impellor (W) = 0.0935 m
Depth of liquid in vessel (H) = 1.402 m
Width of baffle (J) = 0.1168 m
49
FIGURE 8.1 REACTOR
8.2 DESIGN OF FILTER PRESS
50
Basis: Carbon – water slurry: 13.91kg of C /154.08 kg H2O = 0.09 kg of C /kg of H2O
Specific gravity of carbon = 1.8
Assumptions: A plate and frame filter with frame 0.3 m²
The press takes 120s to dismantle.
120s to reassemble.
120s to remove cake from each layer.
Pressure = 275 KN/ m² = 275 x 10³ N/m²
Cake porosity = ε = 0.5
Let ‘n’ be the no. of frames and ‘d’ thickness of frame.
Total time for one complete cycle = Tf + 120n +240 sec
Overall rate of filtration = Vf
Tf + 120n +240
For constant rate of filtration Vf ² = ΔP A² Tf / µ c α
Where, Vf = volume of filtrate (m³)
Tf = time for filtration (s)
ΔP= pressure difference = (275-101.3) x 10³ N/ m²
A = area = 2n x 0.09 = 0.18n m²
µ = viscosity = 10-3 kgm/s
c =mass of solid deposited on filter per unit volume of filtrate
=13.6228 = 93 kg/m³
(146.3712/1000)
α = specific cake resistance = 8.8 x 107 x [1+ 1.64 x 10³x (ΔP)0.86] m/kg
Vf = volume of frames = 0.3² nd = 0.9nd
Volume of cake per unit volume of filtrate 0.09/ (0.5x1.8)
Vf ² = ΔP A² Tf
µ c α
(0.9 nd) ² = (275-101.3) x 10³ x (0.18n) ² x Tf
93 x 10-3 x α
(0.9 nd) ² = 1.28 n² Tf x 10-5
51
Tf = 6.328 x 104 d²
Let us take overall rate of filtration as 1.25 x 10-4 m³/s
1.25 x 10-4 = Vf
Tf + 120n +240
1.25 x 10-4 = 0.9nd / (6.328 x 104 d² +120n + 240)
7.91d² + 0.0015 n + 0.03 = 0.9nd
n = (7.91d² + 0.03) / (0.9d - 0.0015)
To find minimum number of trays: dn/dd = 0
(0.9d - 0.0015) (15.82d) - (7.91d² + 0.03)0.9 = 0
14.238 d²– 0.0237d – 7.119 d² – 0.027 = 0
7.119 d²– 0.0237d – 0.027 = 0
d = 0.0633 m = 63.3 mm
n = (7.91d² + 0.03) / (0.9d - 0.0015) ~ 2
Tf = 253.56 sec
Time for one complete cycle = Tf + 120n +240 = 733.56 sec
Overall rate of filtration = 0.9 nd = 1.55 x 10-4 m³/s
Tf + 120n +240
DESIGN SUMMARY
Number of frames (n) = 2
Thickness of frame (d) = 63.3 mm
Time of filtration (Tf) = 253.56 s
Time for one cycle = 733.56 s
Overall rate of filtration = 1.55 x 10-4 m³/s
Pressure = 275 KN/m²
Area of filter (A) = 0.9 m²
52
FIGURE 8.2 FILTER PRESS
8.3 DESIGN OF STORAGE TANK
53
Basis: Storage for one month
Material construction – stainless steel
Data:
Density of stainless steel = 7700 kg/m³
Density of NaCl = 2160 kg/m³
Density of H2O = 1000 kg/m³
Density of cystine = 1671 kg/m³
Density of carbon = 1800 kg/m³
Amount of NaCl = 224.0651 kg
Amount of H2O = 1629.0068 kg
Amount of cystine = 19.872 kg
Amount of carbon = 0.806 kg
Volume of NaCl = 0.1037 m³
Volume of H2O = 1.629 m³
Volume of cystine = 0.0119 m³
Volume of carbon = 0.0005 m³
Total volume = 1.7451 m³
Average density = total mass/ total volume = 1073.72 kg/m³
Volume per batch = 1.7451 m³
2 batches per day volume per day = 3.5 m³
25 working days in a month volume per month = 87.5 m³
Assume H/D ratio as 0.5
Volume V = π /4 D²H
87.5 = π /4 D² (0.5 D)
87.5 = π /8 D3
D = 6.06 m
H = 0.5 D = 3.03 m
Giving 10% allowance
54
D = 6.06(1+0.1) = 6.666 m
H = 3.03(1+0.1) = 3.333 m
Pressure p = avg. density x g x H
= 1073.72 x 9.81 x 3.333
= 35107.133 N/m² = 35.1071 KN/m²
Design pressure p = 35.1071 (1+0.1) = 38.6178 kN/m²
As the diameter is less than 15m, assume minimum steel thickness as 5mm.
DESIGN SUMMARY
Diameter of tank (D) = 6.666 m
Height of tank (H) = 3.333 m
Design pressure (p) = 38.6178 kN/m²
Plate thickness = 5 mm
55
FIGURE 8.3 STORAGE TANK
A -Storage tank; B - solution; C - Sludge; D - Tank roof; E - Loading valve; F - Riser; G - Pump hose ; H - Bottom hose; J - Main valve.
56
COST ESTIMATION
COST ESTIMATION
Number of working days per year = 300
Cost of 1kg of L-Cystine = Rs 1,10,000
Production of L-Cystine = 136.8188 kg/day
Gross sales for 1 yr or total income =110000x136.8188x 300 = Rs 450,00,00,000
TURN OVER RATIO:
57
It can be defined as the ratio of total income to fixed capital investment.
Turn over ratio = Total Income
Fixed capital Investment
For chemical industries the turn over ratio is one.
Thus, Fixed capital investment = Gross Annual Sales = Rs 450,00,00,000
But, Fixed capital investment = Direct cost + Indirect cost
DIRECT COST: It is taken as 70% of the fixed capital investment= 0.7 x 4500000000 = Rs 3,37,50,00,000
The costs involved in the direct cost are,
i. Equipment cost
ii. Installation & Painting cost
iii. Instrumentation Cost
iv. Electrical cost
v. Piping Cost
vi. Building, process and auxiliary cost
vii. Service facilities & yard improvement cost
viii. Land cost
Equipment cost
It is taken as 24% of fixed capital investment = 0.24 x 4500000000 = Rs 108,00,00,000
Painting and installation cost
It is taken as 40% of the equipment cost = 0.4 x 1080000000 = Rs 43,20,00,000
Instrumentation cost
It can be taken as 10% of equipment cost = 0.1 x 1080000000 = Rs 10,80,00,000
58
Piping cost
It is 25% of the equipment cost = 0.25 x 108000000 = Rs 27,00,00,000
Electrical cost
It can be taken as 25% of equipment cost = 0.25 x 108000000 = Rs 27,00,00,000
Building, process and auxiliary cost
It is 39.1677% of equipment cost = 0.391677 x 108000000 = Rs 42,30,00,400
Service facilities & yard improvement cost
It can be taken as 40% of equipment cost = 0.1 x 1080000000 = Rs 10,80,00,000
Land cost
It is usually taken as 1% of fixed capital cost = 0.01 x 4500000000 = Rs 4,50,00,000
INDIRECT COST:
Indirect cost = Fixed Capital Investment – Direct Cost
= 4500000000 – 3375000000 = Rs 1,12,50,00,000
It consists of the following items
i. Engineering and supervision cost
ii. Contingency
iii. Working capital
Engineering and supervision cost
It can be taken as 10% of equipment cost = 0.1 x 1080000000 = Rs 10,80,00,000
Contingency
It can be taken as 3.7% of fixed capital = 0.037 x 4500000000 = Rs 16,65,00,000
Working capital
It is 20% of total capital investment
Total capital investment = fixed capital + working capital
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= 4500000000 + (0.2 x 4500000000) = Rs 5,40,00,00,000
Working capital = 0.2 x 5400000000 = Rs 1,08,00,00,000
Estimation of total product cost
Annual income = Rs 45,00,00,00,00
Gross earning is 10% of annual income = 0.1 x 4500000000 = Rs 45,00,00,000
Product cost = Total annual income – Gross earnings
= 4500000000 – 450000000 = Rs 4,05,00,00,000
Direct production cost
It can be taken as 60% of the total product cost = 0.6 x 4050000000 = Rs 2,43,00,00,000
Raw materials cost
It is 2% of the total product cost = 0.02 x 4050000000 = Rs 8,10,00,000
Operating labor cost
It can be taken as 15% of total product cost = 0.15 x 4050000000 = Rs 60,75,00,000
Direct supervisory & clinical labor cost
It is 20% of operating labor cost = 0.2 x 607500000 = Rs 12,15,00,000
Utilities
It can be taken as 15% of total product cost = 0.15 x 4050000000 = Rs 60,75,00,000
Maintenance & repair cost
It is 3.6% of fixed capital investment cost = 0.036 x 4500000000 = Rs 16,20,00,000
Laboratory charges
It is taken as 6.67% of operating labor cost = 0.0667 x 607500000 = Rs 4,05,20,300
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Royalties
It is taken as 1.45% of the fixed capital cost = 0.0145 x 4500000000 = Rs 6,52,50,000
Fixed charges
It can be taken as 20% of product cost = 0.2 x 4050000000 = Rs 81,00,00,000
Plant overheads
This includes cost for general upkeep and overhead packaging, medical services, safety
and protection, recreation, sewage, laboratories, and storage facilities.
It is 5% of the total product cost = 0.05 x 4050000000 = Rs 20,25,00,000
Depreciation
Depreciation for machinery is 10% of fixed capital cost
= 0.1 x 4500000000 = Rs 45,00,00,000
Depreciation of building is 3% of the land cost
= 0.03 x 45000000 = Rs 13,50,000
Total depreciation value = 4500000000 – 1350000 = Rs 44,86,50,000
Insurance
It is 1% of the fixed capital cost = 0.01 x 4500000000 = Rs 45000000
Rent value
It is 3.033% of the total product cost = 0.03033 x 4050000000 = Rs 12,28,36,500
General expenses
Administrative cost includes cost for officer, legal fees, office supplier and
communication.
It is 5% of the total; product cost = 0.05 x 4050000000 = Rs 20,25,00,000
Distribution and selling cost
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It accounts for 7% of the total product cost = 0.07 x 4050000000 = Rs 28,35,00,000
Research and development cost
It is 1% of the total product cost = 0.01x 4050000000 = Rs 4,05,00,000
Financing
It is 2% of the total product cost = 0.02 x 4050000000 = Rs 8,10,00,000
Net profit
It is obtained after deduction of taxes from the Gross Earnings.
Net profit is 40% of the Gross Earnings = 0.4 x 4500000000 = Rs 1,80,00,00,000
Determination of Pay-Back period (without interest charges)
= Depreciable fixed capital investment
(Average profit + average depreciation)/yr
= 4500000000
(1800000000 + 450000000)
= 2 yrs.
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PLANT LAYOUT
AND
LOCATION
PLANT LAYOUT AND LOCATION
A suitable site must be found for a new project, and the site and equipment layout
planned.
10.1 PLANT LOCATION AND SITE SELECTION The geographical location of the final plant can have strong influence on the success of
the industrial venture. Considerable care must be exercised in selecting the plant site, and
many different factors must be considered. The location of the plant can also have a
crucial effect on the profitability of a project.
The choice of the final site should first be based on a complete survey of the
advantages and disadvantages of various geographical areas and ultimately, on the
advantages and disadvantages of the available real estate. The various principal factors
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that must be considered while selecting a suitable plant site are briefly discussed in this
section. The factors to be considered are:
1. Raw material availability.
2. Location, with respect to the marketing area.
3. Availability of suitable land.
4. Transport facilities.
5. Availability of labors.
6. Availability of utilities (Water, Electricity).
7. Environmental impact and effluent disposal.
8. Local community considerations.
9. Climate.
10. Political strategic considerations.
11. Taxations and legal restrictions
10.1.1 RAW MATERIALS AVAILABILITY:
The source of raw materials is one of the most important factors influencing the selection
of a plant site. Attention should be given to the purchased price of the raw materials,
distance from the source of supply, freight and transportation expenses, availability and
reliability of supply, purity of raw materials and storage requirements.
10.1.2 LOCATION:
The location of markets or intermediate distribution centers affects the cost of product
distribution and time required for shipping. Proximity to the major markets is an
important consideration in the selection of the plant site, because the buyer usually finds
advantageous to purchase from near-by sources.
10.1.3 AVAILABILITY OF SUITABLE LAND:
The characteristics of the land at the proposed plant site should be examined carefully.
The topography of the tract of land structure must be considered, since either or both may
have a pronounced effect on the construction costs. The cost of the land is important, as
well as local building costs and living conditions. Future changes may make it desirable
or necessary to expand the plant facilities.
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10.1.4 TRANSPORT
The transport of materials and products to and from plant will be an overriding
consideration in site selection. If practicable, a site should be selected so that it is close to
at least two major forms of transport: road, rail, waterway or a seaport. 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. If possible the plant site
should have access to all three types of transportation.
10.1.5 AVAILABILITY OF LABORS:
Labors will be needed for construction of the plant and its operation. Skilled construction
workers will usually be brought in from outside the site, but there should be an adequate
pool of unskilled labors available locally; and labors suitable for training to operate the
plant. Skilled tradesmen will be needed for plant maintenance.
10.1.6 AVAILABILITY OF UTILITIES:
The word “utilities” is generally used for the ancillary services needed in the operation of
any production process. These services will normally be supplied from a central facility
and includes Water, Fuel and Electricity which are briefly described as follows:
Water: -
The water is required for large industrial as well as general purposes, starting with water
for cooling, washing, steam generation and as a raw material. The plant therefore must be
located where a dependable water supply is available namely lakes, rivers, wells, seas. If
the water supply shows seasonal fluctuations, it’s desirable to construct a reservoir or to
drill several standby wells
Electricity: -
Power and steam requirements are high in most industrial plants and fuel is ordinarily
required to supply these utilities. Power, fuel and steam are required for running the
various equipments like generators, motors, turbines, plant lightings and general use and
thus be considered as one major factor is choice of plant site.
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10.1.7 ENVIRONMENTAL IMPACT AND EFFLUENT DISPOSAL:
Facilities must be provided for the effective disposal of the effluent without any public
nuisance. In choosing a plant site, the permissible tolerance levels for various effluents
should be considered and attention should be given to potential requirements for
additional waste treatment facilities. 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
10.1.8 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.
10.1.9 CLIMATE:
Adverse climatic conditions at site will increase costs. Extremes of low temperatures will
require the provision of additional insulation and special heating for equipment and
piping. Similarly, excessive humidity and hot temperatures pose serious problems and
must be considered for selecting a site for the plant. Stronger structures will be needed at
locations subject to high wind loads or earthquakes.
10.1.10 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.
10.1.11 TAXATION AND LEGAL RESTRICTIONS:
State and local tax rates on property income, unemployment insurance, and similar items
vary from one location to another. Similarly, local regulations on zoning, building codes,
nuisance aspects and others facilities can have a major influence on the final choice of the
plant site.
10.2 THE SITE LAYOUT
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The process units and ancillary buildings should be laid out to give the most economical
flow of materials and personnel around the site. Hazardous processes must be located at a
safe distance from other buildings. Consideration must also be given to the future
expansion of the site. The ancillary buildings and services required on a site, in addition
to the main processing units will include:
1. Raw material and Product Storage.
2. Maintenance Workshop.
3. Stores for maintenance and operating supplies.
4. Laboratories for process control.
5. Fire Station and other emergency services.
6. Utilities: steam boilers, compressed air, power generation, refrigeration, transformers.
7. Effluent disposal plant.
8. Offices for general administration.
9. Canteens and other amenity buildings, such as medical Centre.
10. Car parks.
10.3 PLANT LAY OUT After the flow process 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 unit 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.
Thus the economic construction and efficient operation of a process unit will depend on
how well the plant and equipment specified on the process flow sheet is laid out. The
principal factors that are considered are listed below:
1. Economic considerations: construction and operating costs.
2. Process requirements.
3. Convenience of operation.
4. Convenience of maintenance.
5. Health and Safety considerations.
6. Future plant expansion.
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7. Modular construction.
10.3.1 COSTS:
The coat of construction can be minimized by adopting a layout that gives the shortest
run of connecting pipe between equipment, and least amount of structural steel work.
However, this will not necessarily be the best arrangement.
10.3.2 PROCESS REQUIREMENTS:
An example of the need to take into account process consideration is the need to elevate
the base of columns to provide the necessary net positive suction head to a pump.
10.3.3 CONVENIENCE OF OPERATION:
Equipment that needs to have frequent attention should be located convenient to the
control room. Valves, sample points, and instruments should be located at convenient
positions and heights. Sufficient working space and headroom must be provided
10.3.4 CONVENIENCE OF MAINTENANCE:
Heat exchangers need to be sited so that the tube bundles can be easily withdrawn for
cleaning and tube replacement. Vessels that require frequent replacement of catalyst or
packing should be located on the out side of buildings. Equipment that requires
dismantling for maintenance, such as compressors and large pumps, should be places
under cover.
10.3.5 HEALTH AND SAFETY CONSIDERATIONS:
Blast walls may be needed to isolate potentially hazardous equipment, and confine the
effects of an explosion. At least two escape routes for operators must be provided from
each level in process buildings.
10.3.6 FUTURE PLANT EXPANSION:
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Equipment should be located so that it can be conveniently tied in with any future
expansion of the process. Space should be left on pipe alleys for future needs, and service
pipes over-sized to allow for future requirements.
10.3.7 MODULAR CONSTRUCTION:
In recent years there has been a move to assemble sections of plant at the plant
manufacturer’s site. These modules will include the equipment, structural steel, piping
and instrumentation. The modules are then transported to the plant site, by road or sea.
The advantages of modular construction are:
1. Improved quality control.
2. Reduced construction cost.
3. Less need for skilled labors on site.
The disadvantages of modular construction are:
1. Higher design costs & more structural steel work.
2. More flanged constructions & possible problems with assembly, on site.
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FIGURE 10.1 SITE LAYOUT
70
SAFETY
AND
LOSS PREVENTION
SAFETY AND LOSS PREVENTION
Any organization has a legal and moral obligation to safeguard the health and welfare of
its employees and the general public.
Safety and loss prevention in process design can be considered under the following broad
headings.
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• Identification and assessment of hazards.
• Control of the hazards; for example, by containment of flammable and toxic
materials.
• Control of the process.
Process can be divided into those that are intrinsically safe and those for which the safety
has to be engineered in. an intrinsically safe process is one in which safe operation is
inherent in the nature of the process.
The designer should always select a process that is inherently safe whenever it is practical
and economical. The process that we are employing to manufacture l-cystine is inherently
safe. However in most chemical manufacturing processes, dangerous situations can
develop if the process conditions deviate from the design values.
Some of the process hazards are listed.
1. Toxic materials
2. Dust explosion.
3. low temperature
4. Flammable materials
5. Corrosion and erosion
6. Leakage joints and packing
PREVENTIVE AND PROTECTIVE MEASURES The basic safety and fire protective measures that should be included in all chemical
process designs are listed below.
1. Adequate and secure water supplies for fire fighting.
2. Correct structural design of vessels, piping and steel work.
3. Pressure relief devices.
4. Earthing of electrical equipments.
5. Adequate separation of hazardous equipments.
6. Safe design and location of control rooms.
PREVENTIVE MEASURES - CATEGORIES
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1. Those that reduce the number of incidents such as, sound mechanical design of
equipment and piping, operating and maintenance procedures and operator
training.
2. Those that reduce the scale of a potential incident such as, measures for fire
protection and fixed fire fighting equipments.
The other common safety measures followed in the process industry are:
1. Compulsory wearing of helmets.
2. Wearing goggles while working in the furnace, any other fired equipment.
3. Wearing gloves while handling chemicals.
4. Wearing leather shoes in order to protect the legs from heavy materials.
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MATERIALS
OF
CONSTRUCTION
MATERIALS OF CONSTRUCTION
The most important characteristics to be considered when selecting a material of
construction are:
1. Mechanical properties.
(a) Strength – tensile strength
(b) Stiffness – elastic modulus (young’s modulus)
(c) Toughness – fracture resistance
(d) Hardness – wear resistance
(e) Fatigue and creep resistance
2. The effect of high and low temperatures on the mechanical properties.
3. Corrosion resistance.
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4. Any special properties required such as, thermal conductivity, electrical
resistance, magnetic properties.
5. Ease of fabrication – forming, welding, casting.
6. Availability in standard sizes – plates, sections, tubes.
7. Cost.
12.1 REACTOR The reactor is a glass-lined equipment. Its main body is made of high quality carbon steel
lined with special silicate glass by firing them at high temperature. The glass-lined
equipment, therefore, has high mechanical features and corrosion resistance
12.2 NEUTRALIZATION TANK Neutralization tank is cylindrical and have bolted down and gasketed covers. All
hardware for bolting are stainless steel, gaskets are neoprene.
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12.3 DECOLORIZATION TANK Decolorization tank is cylindrical in shape. It is made up of stainless steel as it is fairly
corrosion resistant and has good mechanical properties.
INSTRUMENTATION
76
AND CONTROL
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
13.2.1 SAFE PLANT OPERATION
77
1) To keep the process variables within known safe limits.
2) To detect dangerous situations as they develop and to provide alarms and automatic
shut down systems.
3) To provide inter locks and alarms to prevent dangerous operating procedures.
13.2.2 PRODUCTION RATE
To achieve the design product output.
13.2.3 PRODUCT QUALITY
To maintain the product composition within the specified quality standards.
13.2.4 COST
To operate the lowest production cost, commensurate with the other objectives.
13.3 TYPICAL CONTROL SYSTEMS 13.3.1 LEVEL CONTROL
In any equipment where an interface exists between two phases (liquid-vapor), some
means of maintaining 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.
13.3.2 PRESSURE CONTROL
Pressure control will be necessary for most systems
handling vapor or gas. The method of control will depend on the nature of the process.
13.3.3 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
78
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.
13.3.4 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.
13.3.5 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.
79
LC
FC
FC
TC
FC
FIGURE 13.1 STIRRED TANK REACTOR CONTROL SCHEME, Temperature: cascade control, and Reagent: flow control
80
CONCLUSION
CONCLUSION
In this project the dominant route for the manufacture of L-Cystine from Human Hair is
discussed. The process is an acid hydrolysis extraction process and is done in batch
operation. The yield obtained is 98% pure. This chapter provides a brief idea about the
material and energy balance calculations, design of a reactor vessel, the layout of the
plant and the pay back period in cost estimation. This process consists of raw materials of
low cost and finally yields a product of commercially high cost.
81
82
NOMENCLATURE
NOMENCLATURE
A – Area of filter (m²)
CP – Specific Heat Constant (cal/mol K)
c – Mass of solid deposited on filter per unit volume of filtrate
ds – Diameter of shaft (m)
Dt – Diameter of Vessel (m)
Da – Diameter of Agitator (m)
d – Thickness of frame (m)
D – Diameter of storage tank (m)
E – Height of Agitator above vessel floor (m)
fs – Shear Stress (kg/cm²)
g – Gravitational acceleration (m/s²)
ΔHf – Heat of Formation (kcal/mol)
H – Height of storage tank (m)
Hcy – Height of Cylindrical Portion (m)
83
Hconi – Height of Conical portion (m)
Ht – Height of Reactor Vessel (m)
ΔH°R – Heat of reaction (kcal/mole)
H – Depth of liquid in Vessel (m)
J – Width of Baffle (m)
L – Length of Blade (m)
L – Length of Agitator Blade (m)
Np – Power number
N – Rotational speed of Motor (rpm)
NRe – Reynolds number
n – Number of Frames in filter press
P – Power requirement for motor (Watts)
ΔP – Pressure difference (N/m²)
p – Design pressure for storage tank (N/m²)
Q – Heat required (kcal)
T – Thickness of Vessel (m)
Tc – Torque (Nm)
Tmax – Maximum Torque (Nm)
Tf – Time for filtration (s)
U – Speed of Agitator (rpm)
Vf – Volume of filtrate (m³)
W – Width of Blade (m)
Zp – Polar Modulus of shaft
Greek letters:
µ – Viscosity (kg/ms)
ρ – Density (kg/m³)
α – specific cake resistance (m/kg)
ε – Cake porosity
84
BIBLIOGRAPHY
85
BIBLIOGRAPHY
Douglas D. Schoon , John Halal
“Hair Structure & Chemistry”
Milady Publishing Company 1993
David Mautner Himmelblau
“Basic principles & Calculations in chemical engineering”
Prentice Hall PTR Publications 1996
Robert H Perry, Don W Green & James O Maloney
“Chemical Engineers Hand Book”
Mc Graw Hill Publications 1999
John Metcalfe Coulson, John Francis Richardson
“Plant Design & Economics for Chemical Engineers”
Butterworth and Heinemann Publications 2002
Warren L. McCabe, Julian C. Smith, Peter Harriot
“Unit Operations of Chemical Engineering”
Mc Graw Hill Publications 2004
Robert Ewald Treybal
“Mass Transfer operations’
Mc Graw Hill Publications 2000
M V Joshi, V V Mahajani
“Process Equipment Design”
Mac Millan Publications 2003
Max S Peters & Klaus D. Timmerhaus
“Plant Design & Economics for Chemical Engineers”
Mc Graw Hill Publications 2004
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