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i MANUFACTURE OF PHENOL FORMALDEHYDE RESIN A PROJECT REPORT Submitted by URMILA.K (41502203018) VARUN RATHI (41502203019) in partial fulfillment for the award of the degree of BACHELOR OF ENGINEERING in CHEMICAL ENGINEERING S.R.M ENGINEERING COLLEGE, KANCHEEPURAM ANNA UNIVERSITY:: CHENNAI 600 025 MAY 2006
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MANUFACTURE OF PHENOL FORMALDEHYDE RESIN

A PROJECT REPORT

Submitted by

URMILA.K (41502203018) VARUN RATHI (41502203019)

in partial fulfillment for the award of the degree

of

BACHELOR OF ENGINEERING

in

CHEMICAL ENGINEERING

S.R.M ENGINEERING COLLEGE, KANCHEEPURAM

ANNA UNIVERSITY:: CHENNAI 600 025

MAY 2006

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ANNA UNIVERSITY: CHENNAI

BONAFIDE CERTIFICATE

Certified that this project report “MANUFACTURE OF PHENOL FORMALDEHYDE RESIN” Is the bonafide work of “URMILA.K (41502203018) and VARUN RATHI (41502203019)” who carried out the project work under my supervision. SIGNATURE SIGNATURE Dr.R.KARTHIKEYAN Dr.R.KARTHIKEYAN HEAD OF THE DEPARTMENT Professor and Head

&

Dr.B.S.M. KUMAR Professor

CHEMICAL ENGNEERING CHEMICAL ENGINEERING S.R.M.Engineering College S.R.M.Engineering College Kattankulathur-603203 Kattankulathur-603203 Kancheepuram District Kancheepuram District

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ACKNOWLEDGEMENT

It is pleasure and privilege for us to present this project report, before which we would like

to thank all those who supported and guided us at the various stages of this project.

We express our sincere thanks to our guides DR.R. Karthikeyan B.E., Ph.D, Professor and

Head of the Department of Chemical Engineering , and Dr.B.S.M.Kumar, M.sc.,

M.Tech.,Ph.D., Professor, Department of Chemical Engineering, S.R.M Engineering

College, for their outstanding guidance, constant encouragement and support, apart from

their ideas and approach which has helped us complete this project .

We would like to mention special thanks to Dr.V.E.Annamalai, Dr.I.A.P.S Murthy, of

Carborundum Universal Ltd., For giving us opportunity in gaining practical knowledge in

recent industry.

We would like to thank all the staff members of our department for their endless suggestions

and guidance towards the completion of this project.

ABSTRACT

Phenol-formaldehyde resins belong to the class of thermo set resins. These are known for

their outstanding heat resistance. PF resins are of two types-resoles and novolaks –

depending on the phenol-formaldehyde ratio. They can be manufactured in both liquid and

powder form. The raw materials which are charged in the reactor at room temperature

undergo an exothermic reaction for two hours. Continuous vacuum distillation takes place

for about 6 hours , till the required viscosity is attained. Thus the phenol formaldehyde resin,

of resole type is manufactured, as proposed .

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TABLE OF CONTENTS

CHAPTERS TITLE PAGE NO. ABSTRACT iv LIST OF TABLES vii LIST OF FIGURES viii LIST OF SYMBOLS ix

1 INTRODUCTION 1

2 PROPERTIES 3

2.1 PHYSICAL PROPERTIES 3

2.2 CHEMICAL PROPERTIES 4

3 APPLICATION 6

4 LITERATURE SURVEY 8

4.1 PROCESS SELECTION 10

5 PROCESS DESCRIPTION 12

5.1 EFFLUENT TREATMENT 16

6 MATERIAL BALANCE 22

7 ENERGY BALANCE 26

8 DESIGN 29

9 PROCESS CONTROL 40

10 PLANT LAYOUT 42

11 COST ESTIMATION 52

12 SAFETY 60

13 STORAGE AND TRANSPORTATION 64

14 CONCLUSION 65

BIBLOGRAPHY 66

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LIST OF TABLES TABLE DESCRIPTION PAGE NO. NUMBER 2.1 Physical properties 3

5.1 Viscosity test 13

7.1 calculation of heat content 26

8.1 Heat transfer data 34

11.1 Delivered cost of equipments 52

11.2 Direct cost factor 53

11.3 Indirect cost factor 53

11.4 Auxillary cost factor 54

LIST OF FIGURES

FIGURE 5.1 FLOW SHEET 15

FIGURE 5.2 FLOW SHEET 21

FIGURE 6.1 REACTOR BALANCE 22

FIGURE 6.2 CONDENSER BALANCE 24

FIGURE 7.1 ENERGY BALANCE FOR 26

A REACTOR

FIGURE 7.2 ENERGY BALANCE FOR 28

A CONDENSER

FIGURE 10.1 PLANT LAYOUT 51

LIST OF SYMBOLS

A Area(m2)

D,d Diameter(m)

L Length (m)

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H Height (m)

m Mass (kg)

Nu Nusselt number

n Number of tubes

P Pressure

Pr Prandtle number

Re Reynolds nymber

V Volume

T Temperature

U Overall heat transfer

Coefficient (W/m2ºC)

Cp Specific heat capacity

(KJ/KgK)

K Thermal conductivity

(W/Mk)

f Shear stress

tsk Skirt thickness (mm)

W Weight of the reactor (N)

Cv Correction factor

GREEK LETTERS

∆T Temperature difference(ºC)

µ viscosity of liquid

λ Latent heat of vapourisation (KJ/Kg)

ρ

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1.INTRODUCTION HISTORY Leo H.Bakeland applied for his famous “heat and pressure” patent for the processing of

phenol formaldehyde resins. This technique made possible the worldwide application of the

first wholly synthetic polymer material. Even from his first patent application of feb 18,

1907, it was clear that baekland , more than his predecessors was fully aware of the value of

the phenolic resins.

So that when bakelite started with phenolic resins the following were already know.

Phenols and formaldehyde are converted to resinous products in the presence of acidic and

alkaline catalysts. These may be permanently fusible and soluble in organic solvents or heat

curable depending upon the preparation conditions. Phenolic resins were already being sold

as substitutes for shellac, ebonite, horn and celluloid. These are colorable , can be mixed

with fillers and under the influence of heart shaped in molds into solid parts.

However , economic of molded parts are not yet possible. The “heat and pressure”

patent became the turning point , indicating clearly the importance of economic processing

techniques for market acceptance. Phenolic resins mixed with fillers could be hardened in a

press or autoclave, which was called bakelistor, under pressure at temperature below 100 * c

in a considerably short time and without the formation of blisters. According to the first

bakelite patent phenol and formaldehyde, catalyst and fibrous cellulosic material were

reacted at elevated temperature. The impregnation of the fibrous material can be improved

by application of vacuum and pressure, infusible products being obtained only if the

formaldehyde was used in excess. Soon afterwards he recommended the impregnation of the

cellulosic fibers with liquid phenolic resins, acid catalyzed resins were being used at this

stage. According to a patent application by lebach in February 1907 insoluble and infusible

condensation products, useful as plastic materials, could be obtained if phenol is reacted

with surplus formaldehyde using neutral or basic salts as catalysts. In the same year bakeland

also patented a process for the preparation of phenolic resins using alkaline catalyst,

preferably ammonia, NaOH and Na2CO3. Henschke granted a patent to him in the USA

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but not in Germany because of the lack of inventive steps considering previous publications.

It was in this patent, however, that resin manufacture was described for the first time just as

it is carried out today.

⇒ The reaction is performed in a closed vessel with a reflux condenser to prevent loss

of volatile material.

⇒ The reaction is interrupted when the desire viscosity is obtained.

⇒ Distillation is performed in a vacuum and can be continued until a solid product,

which is still soluble in alcohols is obtained.

Today, the most important fields of application are the wood industry,

molding and insulation compounds. More than 2/3 of all phenolic resins are used in

these three fields. But also all classic application established by bakeland could maintain

their position.

2. PROPERTIES

2.1 Physical properties Physical Properties Metric Density 1.39 - 1.51 g/ccApparent Bulk Density 0.64 - 0.68 g/ccWater Absorption 0.36 - 0.54 %Linear Mold Shrinkage 0.003 - 0.0065 cm/cmHardness, Rockwell E 75 - 83Tensile Strength, Ultimate 50.5 - 59.7 MPaElongation at Break 0.7 - 0.9 %Tensile Modulus 7.22 - 9.13 GPaFlexural Modulus 7.07 - 8.3 GPaFlexural Yield Strength 80.1 - 95.6 MPaCompressive Yield Strength 187.2 - 198.5 MPaPoisson's Ratio 0.36 Charpy Impact, Notched 0.19 - 0.2 J/cm²

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Electrical Resistivity 4.14e+012 - 3.58139e+013 ohm-cm

Dielectric Constant 5.2 - 5.9Dielectric Strength 10.2 - 13.7 kV/mmDissipation Factor 0.032 - 0.054Arc Resistance 80 - 150 secComparative Tracking Index 175 VCTE, linear 20°C 53 µm/m-°CMaximum Service Temperature, Air

182 - 205 °C

Flammability, UL94 HB Phenol formaldehyde resin is hard, scratch resistant, infusible, and water resistant.

2.2 Chemical Properties

1) Overview of PF Cure Cure behavior is one of the most important characteristics of thermosetting adhesives.

Understanding adhesive cure behavior and its dependence on the temperature and chemical

conversion is important for predicting processing windows and the properties of cured bond

lines.

Thermo set cure usually involves polymerization and cross linking, as it passes through two

stages: gelation and vitrification. Gelation occurs when a three dimensional network

structure with infinite viscosity is formed. It marks the transition between the liquid and gel

state. Vitrification occurs when the glass transition temperature of the thermosetting (pf)

material rises and equals the cure temperature. Vitrification marks the transition from a

liquid or rubber to a glass. Before gelation, thermoset cure is a kinetically controlled process

while after vitrification it is a diffusion-controlled process and the reaction rate decreases

dramatically.

2)ACTION OF HEAT

The base catalyzed reaction of phenol with formaldehyde produces

Intermediates which condense into branched polymers (resoles) at temperatures of between

60 and 100 ‘C . An investigation of

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The degradation properties of PF resins was conducted by Cordey . He

Concluded that the primary degradation pathway for PF resins is oxidative in nature even in

an oxygen deficient atmosphere and that thermal processes only begin to compete at higher

temperatures. The presence of CO is first detected at about 350 ‘C, while CH4 which is the

major volatile product from the thermal degradation of the resin, is evident only at

temperatures above 550° C.

3) Action of acids:

Phenol formaldehyde is resistant to non-oxidizing acids, salts and many organic solvents.

4) Stability:

Phenol formaldehyde is very stable. No decomposition at ordinary temperatures.

5) Toxicity :

Oral LD50 : 9200 mg/kg (rat)

6) Ecological effects:

Can be separated mechanically in water treatment plants.

7) Flammability: Phenol formaldehyde is generally un flammable.

3. APPLICATIONS

Phenolics are little used in general consumer products today due to

the cost and complexity of production and their brittle nature. An exception to the overall

decline is the use in small precision-shaped components where their specific properties are

required, such as molded disc brake cylinders, saucepan handles, electrical plugs and

switches, and electrical iron parts. Today, Bakelite is manufactured under various commercial

brand names such as Micarta. Micarta is produced in sheets, rods and tubes for hundreds of

industrial applications in the electronics, power generation and aerospace industries.

Major use categories of phenolic resins are,

Molding materials. The discovery by bakeland that wood flour compounded

with phenolic resins could be molded under heat and pressure to give a strong

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heat resistant part that would not crack or split apart on aging, was the start of

phenolic resin industry.

Laminates. Liquid one step resins and solvent solutions of one step resins

are used to make laminated structures. Two general classes are recognized:

Industrial and decorative.

BONDING RESINS: This market area includes the use of phenolic resins to bond

friction materials, abrasives, wood particles, and inorganic fibers for insulation.

Friction materials. Phenol Formaldehyde resins is the principal bonding

agent for the asbestos used in friction materials. The major categories are

automotive brake linings, clutch facings, and automatic –transmission discs,

but a wide variety of other products are made, e.g. brakes for oil well

drilling rigs, power derricks, and rail road cars. Bonded abrasives. About half of oil grinding wheel tonnage is resin

bonded, the phenolic resins being used almost exclusively. Resins have

replaced the various ceramic bonds because resinoid wheels can withstand

more mechanical and thermal shock. Coated abrasives. Phenolic resins have replaced hide glue for industrial

grades of “sand paper” where heat is generated in dry grinding or where

water-cooling is required. Insulation. Phenolic resins are used to bond glass and rock wool fibers for

thermal and acoustic insulation. Plywood. Phenol formaldehyde resins for plywood glues are alkaline –

catalyzed liquid one step resins. Foundry use. Phenolic resins are employed in several metal casting

applications. Coatings. Phenolic resins are used in coatings both as the sole film former

and to fortify drying oils. Resins used as the sole reactive ingredient are

alkaline catalyzed one step phenol formaldehyde resin.

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4. LITERATURE SURVEY

Phenol-formaldehyde (PF)

Overview

Phenol-formaldehyde (PF) resin was the first wholly synthetic polymer to be commercialized

(1). It has become one of the most widely utilized synthetic polymers since Baekeland

developed a commercial manufacturing process in 1907. Phenol-formaldehyde resin can be

tailored to different properties suitable for various applications such as molding compounds,

paper impregnates, adhesives, coatings, etc. By varying the catalyst type and the

formaldehyde (F) and phenol (P) molar ratio, two classes of PF resin can be synthesized:

resoles (resols) and novolaks (novolacs). Resoles are synthesized under basic conditions with

excess formaldehyde (i.e. F/P>1); novolaks are synthesized under acidic conditions with

excess phenol (i.e. F/P<1). Resoles and novolaks are inherently different: resoles are heat

curable while novolaks require addition of a cross linking agent such as

hexamethylenetetramine (HMTA) to cure. For most novolaks, this additional step results in

slower cure rates and lower cross linking than resoles .

PF resins were first introduced as binders for particleboard and plywood in the mid 1930’s;

they have since become one of the most important thermosetting adhesives in the wood

composites industry, especially for exterior applications. In 1998, PF resins comprised

approximately 32 percent of the total 1.78 million metric tons of resin solids consumed in

the North American wood products industry. Almost all PF resins currently used in wood

bonding applications are resoles. PF resoles are desirable for exterior applications due to

their rigidity, weather resistance, chemical resistance and dimensional stability. PF resoles, in

either a liquid or a spray-dried form, are currently used as binders for the manufacture of an

important structural wood panel, oriented strand board (OSB). Compared to polymeric

diphenylmethane diisocyanate (PMDI), the only other binder currently used in OSB

manufacturing in North America, PF resoles have the advantage of low cost, good thermal

stability and reasonably fast cure.

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PF Resole Synthesis

PF Resoles are polycondensation products of phenol (P) and formaldehyde (F) in an alkaline

aqueous medium with excess formaldehyde. Formaldehyde is often used in the form of an

aqueous solution during commercial production of PF resoles. The polymeric form of

formaldehyde, Para formaldehyde, is rarely used in industrial processes due to its high cost.

PF resoles used as wood binders are typically synthesized under 100oC with a

formaldehyde/phenol (F/P) ratio of 2 to 1 . The most commonly used catalyst in

commercial resole preparation is sodium hydroxide (NaOH). Besides its catalytic effect,

sodium hydroxide also improves the solubility of PF resoles in aqueous solution, which

allows resoles to be synthesized with a high degree of advancement for fast curing, while

maintaining good process ability.

4.1 PROCESS SELECTION

Process selection is an important criteria for any manufacturing unit. This selection gives

direction to obtain the required product with high efficiency , quality and within the cost to

be produced. The applications of the product defines the condition and changes required for

manufacturing. Importance of process selection has been the key tool for many of the

manufacturing units.

Phenol formaldehyde resin is been manufactured , mainly by two process. Depending on the

application of resin the required process can be chosen.

The two process are explained in brief below,

Manufacture of phenol formaldehyde resin using alkaline catalyst.

PF resins are manufactured in batch process. Phenol and formaldehyde

are charged into the kettle in specified quantaties. The kettle is kept under continuous

agitation. An alkaline catalyst is added to initiate the reaction. Exotherms are controlled and

cooking temperature is maintained by circulating cooling water and by cooling oil within the

pipe and the outer jacket respectively. After 2 hours of reaction continuous distillation takes

place for 6 hours . The whole manufacturing process takes place under vaccum. Once the

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distillation starts heating oil is circulated in place of cooling oil. After achieving the

viscosity/water tolerance , vaccum distillation is stopped and the reactor is cooled to below

40ºC. then the contents are discharged into specific containers.

The product is kept in a cold room at below 10ºC till the time of dispatch.

Manufacture of phenol formaldehyde resin using acid catalyst

Novolak resins are ordinarily manufactured by batch process in a jacketed acid resistant

stainless steel kettles equipped with shell and tube vapour condensers and heavy duty achor

or turbine blade agitator. In a typical reaction reaction cycle molten phenol at 60-65ºC and

warm 37-40% formaldehyde are charged to the kettle from weigh tanks. Agitation is started

and is continuous throughout the cycle.the acid catalyst is then added and the batch is tested

for pH.steam heat is applied to raise the temperature .this heating is necessary for 3-6

hours.at the end of reflux period the condensate is re routed to a reciever and the water is

distilled from the kettle.vaccum is applied when temperature reaches to 120-150ºC.melting

point or solution viscosity is used to test the sample for checking its completion. When the

resin is completed , it is discharged .

5. PROCESS DESCRIPTION

This is a batch process, which takes place for about eight hours. Phenol and formaldehyde

are taken from the raw material storage room. Vacuum is first created in the reactor kettle,

and then charging of phenol is done. Before adding phenol, vacuum pressure is created and

cooling water supply is started. After charging phenol, formaldehyde is charged into the

kettle. The molar ratio of phenol to formaldehyde is of 1:1.5. Now, sodium hydroxide,

which is the catalyst, is added. It is mixed with necessary amount of water. Charging of raw

materials in the reactor kettle takes place at 30°C. Reactor consists of an outer jacket and a

coil around its circumference. The outer jacket carries the cooling oil for the first two hours

of the reaction and the cooling water is circulated in the coil within the reactor for the same

time.

Now, after the entire charging section is complete, condenser valve is opened. As the stirring

continuously takes place, the reaction temperature increases to about 102º C, the reaction

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being an exothermic one. The cooling oil and cooling water helps to control the reaction

temperature at about 60-70ºC. Now, the reaction continues for about 2 hours at the same

temperature. The extent of the reaction or amount reacted is tested by WATER

TOLERANCE TEST also known as GEL TIME TEST. The water tolerance reduces from

infinity to 600, as the reaction continues.

STEP: 1

C6H5OH + 2CH2O → C8H10 O3

STEP: 2

2n C8 H10O3 → [C8H8O2] n + n H2O

OVERALL REACTION

2n C6H5OH + 4nCH2O → n C8H7O2Na + n H2O {naoh}

Now, after two hours of reaction, the reactor behaves as a distillation column and

continuous condensation takes place. Cooling is cut off and hot water and oil is circulated

through the coils and outer jacket respectively. Distillation continues for about 6 hours at

about 60 - 70°C. The distilled water is collected in the receiver. Condensation takes place

in the condenser; thereby changing the phase of vapour to liquid and directing it towards

the distillation receiver. As the condensation takes place, the resin is checked for its

viscosity periodically. The viscosity check is done in FORKED VISCOMETER. This is

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one of the widely used viscometer known for its accuracy and efficiency. The following

table shows the values of the viscosity test.

TABLE NO 5.1

TIME SAMPLE 15 SEC

FIRST SAMPLE

25 SEC SECOND SAMPLE

37 SEC THIRD SAMPLE

54 SEC FOURTH SAMPLE

Finally, the viscosity of resin is measured as 3000 cpi from BROOKFIELD

VISCOMETER, the distillation is stopped and the discharging is done at about 40-55°C.

The discharged phenol formaldehyde resole contains about 20% water. This is taken and

stored in the PVC containers and is stored in cool room at temperatures below 15°C.

Thus semi solid resin, which may be dissolved in organic solvents

such as alcohols and used as varnish or coating or it, may be applied to sheets for subsequent

lamination.

The water from the distillation receiver tank, which contains some amount of phenol goes to

the EFFLUENT TREATMENT PLANT.

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RAWMATERIAL

1RAW

MATERIAL2

BALANCE

REA

CTO

RW

ITHA

GITA

TOR

CONDENSER

DISTILA

TION

REC

EIVER

VAC

UU

MTA

NK

ETP

RESINRECEIVER

15

FIGURE 5.1

5.1

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5.1 PROCESS DESCRIPTION OF EFFLUENT TREATMENT SCREENING AND EQUALISATON The effluent is screened in the bar screen and taken to the equalisation tank where the flow

and parameters are equalised. Oil is separated by belt oil and skimmer mechanism . the

equalisation tank is designed to hold 24 hours retention of effluent.

CHEMICAL TREATMENT Then the equalised and neuralised effluent is pumped to the reaction cum settling tank

where alum,lime and polyelectrolyte solutions are added. The reacted effluent is allowed

settled and the clear effluent is taken on furthur treatment to bio filter and where as a

setteled sludge is applied on the sludge drying beds for disposal.

BIO FILTER A population of micro-organism attached to the filter media degrades the organic matters in

the waste water. Organic matter from liquid is adsorbed on to the biological film or, shine

longer. In the outer positions of the biological slime layer , the organic matter is degraded by

aerobic micro organisms. As the micro- organisms grow , the thickness of the slime layer

increases and the diffused oxygen is consumed before it can penetrate the full depth of the

biological slime layer.

Thus anaerobic environment is established at the surface of the

media, as the slime layer increases in thickness the adsorbed organic matter is metabolised

before it can reach the microorganisms near the media surface. As a result of having no

external organic source available for cell carbon , the microorganisms near the media surface

enter into an endogenous phase of growth and loose their ability to cling to the media

surface.The liquid then washes the slime of media and the new slime layer is called

SLOUGHING and is primarly a function of organic and hydraulic loading on the filter.The

hydraulic loading accounts for the sheer velocities and the organic loading accounts for the

rate of metabolism in the shine layer.

PROCESS MICROBIOLOGY AND ANALYSIS The biological community in the filter consists primarly of protests including aerobic ,

anaerobic and facultative bacteria ,fungi,algae and protozoa. Higher animals such as norms

insects, larve and snails are also present .

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Facultative bacteria are the predominating microorganisms in the bio filter. Fungi

present are also responsible for the waste stabilization but their contribution is usually

important only under low pH Conditions or, with certain induatrial wastes . the protozoan

are predominating of ciliate group and their function is to control the bacteria population.

In predicting the performance of bio filter the organic and hydraulic loading and the

degree of purification required are the most important factors to be considered. Due to the

unstable characteristics of the biological slime layer and the unpredictable hydraulic

characteristics , a generalised kinetic model of the bio filter is very difficult to develop.

The main problem encountered is the design of bio filter is the dtermination of macimum

organic material that can be applied to the filter before oxygen becomes a limiting variable.

Two stage bio filter is the envisaged for the treatment process.

AEROBIC PROCESS The clear overflow of grvitates to the aeration tank for biological degradaiton. A single stage

extended aeration activated sludge system has been adopted for treatment of organics. The process of ACTIVATED SLUDGE PROCESS is to remove organics that

ecape from the primary treatment. “ ACTIVATED SLUDGE “ describes a continous flow ,

biological treatment system characterized by a suspension of aerobic microorganisms

maintained in a realtively homogenious by mixing and turbulence induced in conjuction of

aeration process. Waste water is received in aeration tank where aerobic microorganism is

maintained in suspension. Suface aerators are provided to supply oxygen to the

microorganisms , to completely mixed conditions. The aerobic micoorganisms degrade the

solube and suspended organics in the effluent.

The mixed liquor flows from the aeration tank to settling tank where the

activated sludge is settled. A portion of the settled sludge is returned to aeration tank to

maintain proper microorganisms (MLSS) concentration in aeration tank to permit rapid bio

– degradation of organic matter . the excess sludge is wasted.

Basically the Activated Sludge Process uses aerobic mocrorganisms in

suspension to oxidise soluble and colloidal organics in the presence of molecular oxygen.

During the oxidation process , a portion of the organic material is synthesised into new cells.

A part of the synthesised cells then undergo auto oxidation ( self oxidation or endogenous

respiration) in the Aeration Tank, Oxygen is required to support the synthesis and

endogenous respiration reactions.

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Sufficient numbers of aerators shall be installed in the aeration tank to

transfer , required oxygen necessary to sustain the activity of the microrganisms. In addition

to the oxygen requirements the aerobic microbes require macro nutrients, nitrogen and

phosphorus to sustain the microbial activity. Nitrogen being avaliable in the form of

Ammonia would be readily utilised by the microbes. The aerobic microbes are capable of

utilising about 65-70% of the nitrogen in the feed. Phosphorus on the other has to be

supplemented with phosphorus salts.

The overflow ffrom the aeration tank will contain a high concentration of

solids. A secondary clarifier helps in separating the microbes from the liquid stream to

produce a high quality effluent . the secondary clarifier also aids in maintaining a thick

undeflow sludge concentrtion , crucial to the effective operation of the activated sludge

process.

The aeration tanks would be equiped with diffused aeration system to transfer

oxygen to sustain the activity of microbes. The overflow from the aeration tank shall be

settled in in secondary settling tank. A portion of the settled sludge shall be recycled to

maintain the desired mixed liquor suspended solids in the aeration tank. The overflow from

the secondary settling tanks shall be collected in a treated effulent sump to be taken up for

furthur treatment and disposal.

SLUDGE TREATMENT AND DISPOSAL The sludge from the waste activated sludge from the extended aeration activated sludge

plands shall be drained to sludge drying beds to dewater the sludge. The sludge drained to

the sludge drying beds shall be allowed to dry for a period of about 7 days. The dried sludge

would be scrapped from the sludge drying beds and used as manure, since this sludge, which

only comprises of biological solids is rich in nitrogen and phosphorus. The filtrate from the

sludge drying beds shall be taken up for furthur treatment and disposal.

The solids settled in the primary settling tanks following neutralisation treatment shall be

dried to the sludge drying beds and stored for safe land fillings.

TERITARY TREATMENT PLANT After secondary clarifiaction the efluent is subject to filtration followed by activated carbon

filtration. Pressure land filter comprises of a mild steel pressure vessel containing the media,

provided externally with valves and piping to direct and control flow of water during

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treatment and for cleaning. The media is supported by layers of crushed gravel and graded

pebbles of specific sizes.

And inlet distributor in the form of inverted bell-mouth funnel directs the

inflow of raw water upwards towards thew top dished ends to ensure even distributon across

the surface area of filter beds. Filtered water leaves the filter uniformly by means of a bottom

collecting system which also serves to distribute evenly the flow of water used to xlean the

filter. The bottom collecting system is either a false bottom type or either a header with

perforated laterals depending on the type of filter and diameters. The internal syrface of sand

media filter is painted with anti corrsosive bituminous paint.

After filtration the water is passed throught activated carbon filter for odour

removal also excess chlorine removal. Activated carbon filter comprises of a mild steel

pressure vessel containing the media, provided externally with valves and piping direct and

control flow of water during treatment for cleaning. The media is supported by layers of

crushed gravel and graded pebbles of specific sizes.

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FLOW SHEET

Raw Effluent

Fig 5.2 flow sheet

Raw effluent collection tank

Batch settler , flash mixer

Recycle tank

Bio Filter - 1

Bar Screen

sludge

FeSO4,Lime

Aeration tank

Settling Tank

Collection Tank

Dual Media Filter

Sludge

Filtrate to recycle tank

Bio Filter - 2

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6. MATERIAL BALANCE

STEP: 1

C6H5OH + 2CH2O → C8H10 O3

STEP: 2

2n C8 H10O3 → [C8H8O2] n + n H2O

OVERALL REACTION

2n C6H5OH + 4nCH2O → n C8H7O2Na + n H2O {naoh}

Material balance for the reactor

Basis: 1000 kg of phenol input.

Molecular weight of resin: 295

Molar ratio of phenol to formaldehyde = 1:2

FIG 6.1 REACTOR BALANCE

Reactant side :

Reactor

Phenol = 1000kg

CH2O solution = 1704 CH2O = 37% H2O = 63%

NaOH solution = 132kg NaOH = 31.8% H2O = 68.2%

Resin = 1413kg Unreacted phenol = 94kg Unreacted CH2O= 60kg Unreacted water= 1164 kg

Resin

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Total amount of phenol added = 1000kg

Mole basis = 1000/94

= 10.5 kmoles

Total amount of formalin solution added in mole basis = 21kmoles

Total amount of formalin solution added by weight = 21*30

= 630 kg.

formalin solution contains 37% of formaldehyde = 630 kg

amount of formalin solution = 630 /.37

= 1704kg

amount of water in formalin solution = 1704-630

= 1074kg

NaOH catalyst solution : 10% of phenol in kmoles

NaOH =42kg

Water content in NaOH =90kg

Total amount of reactant water = 1074+90

= 1164 kg

total weight of reactant = 2836 kg

Product ( 90% conversion)

Total resin = 295*.45*10.5

= 1413 kg

total reaction water = .45*18*10.5

= 86.17 kg

water from NaOH = 1.05 kmoles

= 1.05*18 = 18.9 kg

total unreacted water = 1164 kg

total unreacted formaldehyde =30*2

= 60 kg

total unreacted phenol = 94 kg

total weight of the product = 2836 kg

reactor outlet:

unreacted phenol: 1kmole = 94 kg

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unreacted formaldehyde: 2 kmoles = 60kg

total unreacted water : 1269kg

total resin produced : 1413kg

Material balance for condenser

Total feed entering the condenser = 2836 kg

Total amount of water in the mixture=1269kg

20% of water remains in the resin product=1269*0.2=254kg

therefore, total liquid resin =254+1413

=1667kg

total amount of vapour coming out =(1269*0.8)+94+60

=1169kg

condenser balance is given as:

feed = vapour + liquid

2836 = 1667 + 1169

=2836kg.

CONDENSER

Resin+ water = 2836 kg

Total distillate = 1169kg

Resin = 1667kg

FIG 6.2 CONDENSER BALANCE

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7. ENERGY BALANCE

Energy balance for the reactor

Amount of water circulated in the reactor =200 kg

DATA TABLE 7.1:

Compound Mass ,kg Specific heat capacity, Cp MCp∆ T

Phenol 1000 2.34 11700

Formaldehyde 1704 0.5 4262.5

Resin 2836 1.18 117126.8

Water at 55 C 200 4.175 -

Standard heat of reaction:

FIG 7.1

ISOTHERMAL REACTOR

Phenol = 1000kg CH2O solution = 1704 NaOH solution = 132kg (30ºC)

Resin = 1413kg Unreacted phenol = 94kg (60ºC)

Unreacted CH2O= 60kg Unreacted water= 1164 kg(60ºC)

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∆H (reactants)=( mCp∆ T)phenol+ (mCp∆ T)formaldehyde

=(1000*2.34*5)+(1705*0.5*5)

=11700+4262.5

=15962.5

∆H (products)=( mCp∆ T)products

= ((60-25)*1.18*2836)

= 117126.8

∆H =∆H°+ (∑∆H) products - (∑∆H) reactants

= 163.05+117126.8-15962.5

= 101327.35

Mass of oil to be circulated:

∆H =(∑∆H) oil+(∑∆H) water

101325.35 = (m*1.70*(70-20))+(200*4.175*(60-25))

m = 848

Mass of oil circulated in the reactor is: 848 kg

CONDENSER:

H2O (20ºC)

Vapour (60ºC) H2O (60ºC)

H2O(60ºC)

Fig 7.2

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Enthalpy balance:

mλ = mCp∆ T

Mass of vapour =1169 kg

λ at 60ºC = 2358.4 kj/kg

Cp of water = 4.182 kj/kgºC

∆T= 35º C

mλ = mCp∆ T

1169*2858.4 = m*4.182

m = 18835.6 kg

Therefore, mass of cooling water in the condenser = 18,835.6 kg

8. DESIGN OF EQUIPMENTS

REACTOR

Data required for design of reactor:

Density of formaldehyde : 815.3Kg/m3

Density of phenol : 1056.93kg/m3

Density of water : 1000 kg/m3

Mass flow rate of the reactants : 2836 kg/batch

Hours of operation : 8 hours

H/D ratio : 1.5

Average density : : 957.41 kg/m3

To find the volume of the vessel :

Volume to be handled = m/ρ

= 2836*8/957.41

=23.69 m3/batch

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To find the volume of reaction vessel:

Assuming 10% excess volume

Volume = (23.69*(10/100))+23.69

= 26.059 m3/batch

total volume = volume of the cylindrical portion

V = (πD2 H)/ 4

= π D2 (1.5D)/4

26.059 = π D3 *1.5/4

D = {(26.059*4)/ π *1.5}1/3

= 2.8 m

H = 1.5D

= 1.5*2.8

= 4.2 m

The diameter of the vessel is 2.8m .

Height of the reactor is 4.2m.

AGITATOR Type : turbine type

Turbine diameter,Da :30-50% Dt

Diameter of the tank,Dt: 2.8m

Peripheral speed : 200-250 m/min

Viscosity of resin: 3010-3 kg/ms

Specific gravity of resin:1.6

Density of water: 1000kg/m3

Density of resin : 1.6*1000

= 1600 kg/m3

tube turbine diameter = 0.4(Dt)

= 0.4*2.8

=1.12m

tube peripheral speed = 225 m/min

therefore,

πDaN=225

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225/( π*1.12)=N

N=64 rpm

Consider the ratio of the reduction gear available 1/5,1/10,1/15…….

Take 1/15 as the ratio of the reduction gear.

Therefore speed of the motor = 64/(1/15)

= 960 rpm

Power consumption:

Reynolds number, Nre = DVρ/µ

V = N*D

= (64/60)*1.12

= 1.2 m/s

now,

Nre = (1.12*1.194*160)/30*10-3

= 71,321.6

From Np vs Nre graph,

Np = 6

Np = ρgc / ρN 3 Da5

P = (6*1600*1.063*1.125 )/9.8

= 2053.988

Power consumption = 2053.988/75

= 27.38 HP

Now take transmission and other losses as 20%.

Then actual power requirement is,

= 27.38*1.2

= 32.856 HP

Power at the start, P max = P*1.5

= 32.856*1.5

= 49.284 HP

To calculate diameter of shaft:

P= 2 π NT/75 . . . . . . . . . . . . . . .(1)

Torque T is given by,

T = π*Dsh3 Fsh/16

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Dsh - diameter of the shaft

Fsh - shear stress of the shaft.

N - motor speed

N = 960/60

= 16 rps

from (1)

T = P*75/ (2* π*N)

= (49.284*75)/(2* π*16)

= 36.78

T = π* Dsh3 *Fsh/16

Dsh3 = 36.78*16/π*9800000

Dsh = 0.0267m

Shape factors:

S1 = 0.33

S2 = 0.33

S3 =0.25

S4=0.2

S5=0.1

S6=1.0

S3=L/Da=0.25

L =0.25*1.12

= 0.28m

S4 = W/Da =0.2

W = .2*1.12 =.224m

W=0.224m

S5 = J/Dt = 0.1

J = 0.1*2.8 = 0.28m

Agitator dimensions

Diameter of the turbine: 1.12m

Length of the blade : 0.28m

Thickness of the blade : 0.28m

Width of the blade : 0.224m

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Diameter of the shaft: 0.026m

HEAT EXCHANGER 1,1 SHELL AND TUBE HEAT EXCHANGER

Tube side - cooling water

Shell side - vapour

Water inlet temperature = 30ºC

Water outlet temperature = 55ºC

Vapour inlet temperature = 60ºC

Vapour outlet temperature = 60ºC

Mass of the steam entering = 1169 kg

Inside diameter ,Di =0.0225 m

Out side diameter,Do =0.025m

Length of the condenser = 4m

DATA TABLE 8.1 HEAT TRANSFER PROPERTIES

Sl no. PROPERTY WATER at 42.5ºC VAPOUR at 60ºC

1. Specific heat capacity ,Cp 4.179 kj/kg k 1.920 kj/kg k

2. Viscosity, µ 631*10-6 kg/m sec 1.49*10-6 kg/m sec

3. Thermal conductivity ,K 634*10-3 w/m k 22.0*10-3 w/m k

4. Prandtl number, Pr 4.16 0.916

5. Latent heat of vaporization,

λs

- 2354 kj/kg

LMTD = ∆T1-∆T2/ ln(∆T1/ ∆T2)

=(5-30)/ln(5/30)

=13.95

Mass of cooling water =18835kg

Per hour operation =18835/6

Mass flow rate = 3139.6kg/hr

Volumetric flow rate = mass flow rate/density of water

=3139.6/1000

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=3.1396 m3/ hr

velocity = volumetric flow rate/ area

=3.139*4/ (π*0.02252 *3600 )

= 2.19 m/sec

Reynolds number, Nre = DVρ/µ

= 0.0225*.19*1000/(631*10-6)

=79,675.11

Prandtl number, Pr = 4.16

Nusselt number , Nu = 0.023*(Nre)0.8 (Pr)0.4

=0.023*(79,675.11) 0.8 (4.16)0.4

=339.175

Nu = Hi D/K

Hi =9557.186 kj /hr m2 ºC

Shell side steam coefficient, Ho= 8518.35 w/ m2 ºC

Heat lost by steam = heat gained by cooling water

Q= m Cp ∆T

=1169*4.179*30

=146557.53

U = 1/{ (1/Hi)*(Do/Di) + ln(Ro/Ri)*(Ro/k) + 1/Ho}

= 3847.72

Q = UA ∆T

A = 146557.53/(3847.72*5)

= 7.617 m2

A = πDoLN

N = 24 TUBES.

Number of tubes = 24

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SKIRT SUPPORT

To know the weight of the reactor:

Wv = 240CvDm(Hv+ 0.8Dm)t

Assume thicknes of vessel to be 3mm.

Cv = 1.08

Dm = 2.8m

H v = 4.2m

Wv = 240*1.08*2.8(4.2+0.8*2.8)*3*10-3

= 14.02 KN

Approximate weight =п * 2.8 2*4.2*1000*9.81

= 253.7*103 KN

total weight = 14.02+253.7*103

= 253.71*103 KN

(a) stress due to dead weight

fdw = ЄW/ п Dsk Tsk

= 28.48*103 / Tsk

Assume height of the skirt = 1m

Msb = 2/3 * ЄW*Ht*0.08

= 2/3*253.7*103*5.2*0.08

= 70.35*103 Nm

Stress due to seismic load

fsb = 4*Msb*1000/п*Ds*Tsk

= 4*70.35*103*1000/п*2.82*Tsk

= 11.42*106/Tsk

Maximum tensile stress in the skirt support

ft = fsb- fdw

= 11.42*106/Tsk - 28.48*103 / Tsk

Permissible tensile stress of the material = 61.3 * 106 N/m2

ft = 11.39*106 / tsk

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tsk = 185mm

Maximum compressive stress in the skirt

fc = fsb + fdw

= 11.42*106/Tsk +28.48*103 / tsk

Permissible compressive stress of the material = 82.2 N/mm2

Therefore,

tsk = 138.27 mm

Design of the skirt bearing plate

Approximate bolt circle diameter = 2.2m

No of bolts = 2200п/600

= 11.5

no of bolts = 12

take bolt design stress = 125 N/mm2

Ms = 3919 KNm

take W = operating value

= 14.06 KN

fb = maximum allowable bolt stress

Area of cross section Ab = 1/Nb*fb ( 4*Ms/Db – W)

= (1/125*106*12)(4*3919/(2.2) – 14.02*103)

= 4740.9 mm2

total compressive load on the base ring is given by

fb = 4*Ms/п Ds2 + W/ п Ds

= 4*3919000/п 2.8*2.8 + 1402000/п*2.8

= 638.95 KN/nm

taking the bearing pressure as 5 n/mm2

Lb = fb/fc

= 638.05/5

= 127.61mm

actual width = Lb + tsk + 50

= 127.61+138.27+50

= 315.88 mm

Lb = fb/fc’

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fc’ = 638.05/127.61

= 5 N/mm2

Tb = LbҐ(3*fc’/fb)

Tb = 19.56 mm

bolt circle diameter = 2.2m

bearing thickness = 19.56 mm

bearing length = 127.61 mm

Area of cross section = 4740.9 mm2

Skirt thickness = 185 mm

9. PROCESS CONTROL Temperature control - The capability of the cooling system to remove the heat generated

by the reaction is critical to the safe operation of an exothermic process. Facilities should

evaluate capacity of cooling system with respect to controlling unexpected exotherms.

Condensation

cooling of reflux is commonly used to cool exothermic reactions that generate vapor as a

byproduct, but has several limitations to

Control unexpected exotherms. Reflux cooling is limited until the reaction mass reaches the

boiling point of the liquid and cannot control exotherms that begin while the reaction

temperature is below the liquid’s boiling point. As a runaway reaction proceeds, the

increased generation rate of

vapor increases the vapor velocity, the mass flow rate, and the inlet temperature in the

overhead condenser. The increased heat load on the condenser results in only partial

condensation and reflux of water.

Addition of raw materials - Frequently, the reaction rate is controlled by the addition rate

of one reactant or the catalyst and should be determined based on chemistry studies.

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Facilities must pay attention to the order of ingredients, the addition rates, under- or over-

charging, and loss of agitation.

Administrative controls- If administrative controls, such as training and standard operating

procedures, are used as a safeguard against process deviation and accidental release,

consideration must be given to human factors to ensure reliability, especially if an

administrative control is the

Sole layer of protection. Humans make mistakes; the consequences of a human error should

not lead to a catastrophic release. Processes, equipment and procedures must be designed

with potential for human error in

mind. For manual operations, preventive measures should be considered to minimize the

likelihood of human error, for example, interlocks. SOP’s must be understandable,

periodically reviewed, and kept up-to-date. Employees must be trained on the SOP’s and

mechanisms set up to ensure that SOP’s are followed at all times. The consequences of

deviation from SOP’s must be well understood by all employees.

QUALITY CONTROL Quality control of phenolic resins begins with control of the raw materials. Phenol is

generally USP grade. Specifications for phenol include fp , 40.9º C min ; bp, 181.8ºC min ;

and sp gravity , 1.0563 (45/20ºC) . Formaldehyde, also USP grade, is monitored for assay by

the hydroxylamine hydrochloric test, nominal 37% grade should contain 36.8% - 37.2%

CH2O and 1% methanol wax.

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10. PLANT LAYOUT

INTRODUCTION

The economic construction and efficient operation of a process unit will depend upon

how well the plant and equipment specified on the process flow sheet is laid out and on the

profitability of the project with it scope for future expansion. Plant location and site

selection should be made before the plant layout.

Plant location and site selection:

The location of the plant has a crucial effect on the profitability of the project. The

important factors that are to be considered while selecting a site are:

1. Location, with respect to market area

2. Raw material supply

3. Transport facilities

4. Availability of labour

5. Availability of utilities

6. Availability of suitable land

7. Environmental impact and effluent disposal 8. Local community considerations

8. Climate

9.Political and strategic considerations

1. Marketing area

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For materials that are produced in bulk quantities, such as cement, mineral acids, and

fertilizers where the cost of product per tone is relatively low and the cost of transport a

significant fraction of the sales price, the plant should be located close to the primary

product. This consideration will be less important for low volume production, high priced

products, such as pharmaceutical.

2. Raw materials

The availability and price of suitable raw materials will often determine the site location.

Plants producing bulk chemicals are best located close to the source of major raw material,

where this is also close to the marketing area. For the production of formaldehyde the site

should be preferably near a methanol plant.

3. Transport

Transport of raw materials and products is an important factor to be

Considered. Transport of products can be in any of the four modes of

Transport.

4. Availability of labour

Labour will be needed for construction of the plant and its operation. Skilled

construction workers will usually be brought in from outside the site area, but there should

be an adequate pool of unskilled labours available locally; and labour suitable for training to

operate the plant. Skilled tradesman will be needed for plant maintenance. Local trade union

customs and restrictive practices will have to be considered when assessing the availability

and suitability of the local labour for recruitment and training.

5. Environmental impact and effluent disposal

All industrial processes produce waste products, and full consideration must be given to

the difficulties and cost of their disposal. . The disposal of toxic and harmful effluents will be

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covered by the local regulations and the appropriate authorities must be consulted during the

initial survey to determine the standards that must be met.

6. Local community consideration

The proposed plant must fit n with and be acceptable to the local community. Full

consideration must be given to the safe location of the plant so that it does not impose a

significant additional risk to the community on a new site, the local community must be able

to provide adequate facilities for the plant personnel.

7. Land

Sufficient suitable land must be available for the proposed plant and for future expansion.

The land should ideally be flat, well drained and have suitable load-bearing characteristics full

site evaluation should be made to determine the need for piling or other special foundations.

8. Climate

Adverse climatic conditions, at a site will increase costs. Abnormally low temperatures

will require the provision of additional insulation and special heating for equipment and pipe

runs.

9. 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 overriding consideration in the site selection.

After considering the location of the site the plant layout, is completed. It involves

placing of equipment so that the following are minimized:

The various units that should be laid out include:

1. Main processing unit

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2. Storage for raw materials and products

3. Maintenance workshops

4. Laboratories for process control.

5. Fire stations and other emergency services

6. Utilities: steam, boilers, compressed air, power, generation, and refrigeration.

7. Effluent disposal plant

8. Offices for general administration

9. Canteens and other amenity buildings, such as medical centers

10.Car parks

1. Processing area

Processing area also known as plant area is the main part of the

plant where the actual production takes place. There are two ways of

laying out the processing area

1.) Grouped layout

2.) Flow line layout

Grouped layout

Grouped layout places all similar pieces of equipment adjacent. This provides for ease of

operation and switching from one unit to another. This is suitable for all plants.

Flow line layout

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Flow line layout uses the line system, which locates all the

equipment in the order in which it occurs on the flow sheet. This minimizes the length of

transfer lines and therefore reduces the energy needed to transport materials. This is used

mainly for- small volume products.

2. Storage house

The main stage areas should be placed between the loading and unloading facilities and the

process they serve. The amount of space required for storage is determined from how much

is to be stored in what containers. In raw material storage, liquids are stored in small

containers or in a pile on the ground. Automatic storage and retrieving equipment can be

substantially cut down storage.

3. Laboratories

Quality control laboratories are a necessary part of any plant and must be included in all

cost estimates. Adequate space must be provided in them for performing all tests, and for

clearing and storing laboratory sampling and testing containers.

4. Transport

The transport of materials and products to and from the plant will be an overriding

consideration in site selection. If practicable, a site should be selected that is close to at least

two major forms of transport: road, rail, waterway or a seaport.

Rail transport will be cheaper for long distance transport of bulk chemicals. Road

transport is being increasingly used and is suitable for local distribution. Road area also used

for fire fighting equipment and other emergency vehicles and for maintenance equipment.

This means that there should be a road around the perimeter of the site. No roads should be

a dead end. All major traffic should be kept away from the processing areas. It is wise to

locate all loading and unloading facilities as well as, plant offices, personnel facilities near the

main road to minimize traffic congestion within the plant and to reduce danger.

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5. Utilities

The word "Utilities" is now generally used for ancillary services needed in the operation

of any production process. These services will normally be supplied from a central site

facility and will include: Electricity, Steam for process heating, Cooling water, Water for

general use, Inert gas supplies.

Electricity:

Electrical power will be needed at all the sites. Electrochemical processes that require

large quantities of power need to be located close to a cheap source of power. Transformers

will be used to step down the supply voltage to the voltages used on the purpose.

Steam for process heating:

The steam for process heating is usually generated in water tube boilers using the most

economical fuel available. The process temperature can be obtained with low-pressure

steam. A competitively priced fuel must be available on site for steam generation.

Cooling water:

Chemical processes invariably require large quantities of water for cooling. The cooling

water required can be taken from a river or lake or from the sea.

Water for general use:

Water is needed in large quantities for general purpose and the plant must be located near

the sources of water of suitable quality, process water may be drawn from river from wells or

purchased from a local authority.

Offices:

The location of this building should be arranged so as to minimize the time spent by

personnel in traveling between buildings. Administration offices in which a relatively large

number of people working should be located well from potentially hazardous process.

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Canteen:

Canteen should be spacious and large enough for the workers with good and hygienic food.

Fire station:

Fire station should be located adjacent to the plant area, so that in case of fire or

emergency, the service can be put into action.

Medical facilities:

Medical facilities should be provided with at least basic facilities giving first aid to the

injured workers. Provision must be made for the environmentally acceptable disposal of

effluent.

The layout of the plant can be made effective by:

1) Adopting the shortest run of connecting pipe between equipments and the least amount

of structural steel work and thereby reducing the cost.

2. Equipment that need frequent operator attention should be located convenient to control

rooms.

3. Locating the vessels that require frequent replacement of packing or catalyst outside the

building

4. Providing at least two escape routes for operators from each level in process buildings.

5. Convenient location of the equipment so that it can be tied with any future expansion of

the process.

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FINISHEDPRODUCT COLD

STORAGE EXTENSIONAREA

SCRAPYARD

PROCESSINGAREA

SAFETY &HEALTH

CAREDEPARTMEN

T

ADMINISTRATIVEOFFICE

SECURITYOFFICE

ETP

FIRE STATION

PARKINGAREA

ENTRY EXIT

CANTEEN

WORKSHOP

51

FIGURE 10.1

11. COST ESTIMATION

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ESTIMATION OF THE TOTAL CAPITAL INVESTMENT The total capital investment “I” involves the following:

A. The fixed capital investment in the process area, IF.

B. The capital investment in the auxiliary services, IA.

C. The capital investment as working capital, IW. i.e., I = IF + IA + IW

A. FIXED CAPITAL INVESTMENT IN THE PROCESS AREA, IF. This is the investment in all processing equipment within the processing area. Fixed capital

investment in the process area, IF = Direct plant cost + Indirect plant cost. The

approximate delivered cost of major equipments used in the proposed manufacturing plant

are furnished below:

DATA TABLE 11.1

S.No. Equipment Units Cost in lakhs/unit Cost (in lakhs Rs) 1 Reactor 1 30 30 2 Condenser 1 20 20 3 Distillate collection tank 1 10 10 4 Storage tank-phenol 1 10 10

5 Storage tank-formaldehyde

1 10 10

6 Storage tank-resin 1 10 10 7 Bio filters 3 15 45 8 Storage tanks- ETP 5 10 50 9 Flash mixer 1 10 10 10 Vacuum tank 1 10 10 11 Cooling tower 1 20 20 12 Miscellaneous 125

TOTAL 350

CALCULATION OF FIXED CAPITAL INVESTMENT

1) Direct cost factor: DATA TANLE 11.2 DIRECT COST FACTOR

S.No Items Direct cost factor 1 Delivered cost of major

equipments 50

2 Equipment installation 15 3 Insulation 15

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4 Instrumentation 15 5 Piping 25 6 Land & building 30 7 Foundation 10 8 Electrical 15 9 Clean up 5

Total direct cost factor 180 Direct plant cost = (Delivered cost of major equipments)(Total direct cost factor) / 100

Direct plant cost = (350 * 180) / 100

= 630 lakhs

2) Indirect cost factor: DATA TABLE 11.3 INDIRECT COST FACTOR

S.No. Item Indirect cost factor 1 Overhead contractor etc. 30 2 Engineering fee 13 3 Contingency 13 Total indirect cost factor 56

Indirect plant cost = (Direct plant cost)(Total indirect cost factor) / 100

= (630 * 56) / 100

= 352.8 lakhs

Fixed capital investment in the process area, IF = Direct + Indirect plant cost

= 630 + 352.8

= 982.8 lakhs

B. THE CAPITAL INVESTMENT IN THE AUXILLARY SERVICES, IA.

Such items as steam generators, fuel stations and fire protection facilities are commonly

stationed outside the process area and serve the system under consideration.

DATA TABLE 11.4 AUXILLARY COST FACTOR

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S.No. Items Auxiliary services cost factor 1 Auxiliary buildings 5 2 Water supply 2 3 Electric Main Sub station 1.5 4 Process waste system 1 5 Raw material storage 1 6 Fire protection system 0.7 7 Roads 0.5 8 Sanitary and waste disposal 0.2 9 Communication 0.2 10 Yard and fence lighting 0.2 Total 12.3

Capital investment in the auxiliary services = (Fixed capital investment in

process area)*( Auxillary

services cost factor) / 100

= (982.8* 12.3) / 100

= 120.8 lakhs

Installed cost = Fixed capital investment in the process area + Capital

Investment in the auxiliary services

= 982.8 + 120.8

= 1103.6 lakhs

C. THE CAPITAL INVESTMENT AS WORKING CAPITAL, IW. This is the capital invested in the form of cash to meet day-to-day operational expenses,

inventories of raw materials and products. The working capital may be assumed as 15% of

the total capital investment made in the plant ( I ).

Capital investment as working capital, IW = ((982.8+120.8)* 15)/85

= (1103.6* 15) / 85

= 194.75lakhs

Total capital investment I = IF+ IA+ IW

= 982.8 + 194.75+ 120.8

= 1298.35lakhs

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ESTIMATION OF MANUFACTURING COST The manufacturing cost may be divided into three items, as follows:

A. Cost Proportional to total investment

B. Cost proportional to production rate

C. Cost proportional to labour requirement

A. COST PROPORTIONAL TO TOTAL INVESTMENT This includes the factors, which are independent of production rate and

proportional to the fixed investment such as

- Maintenance-labour and material

- Property taxes

- Insurance

- Safety expenses

- Protection, security and first aid

- General services, laboratory, roads, etc.

- Administrative services

For this purpose we shall charge 15% of the installed cost of the plant

= (Installed cost * 15) / 100

= (1103.6* 15) / 100

= 165.54 lakhs

B. COST PROPORTIONAL TO PRODUCTION RATE

The factors proportional to production rate are

- Raw material costs

- Utilities cost – power, fuel, water. Steam, etc.

- Maintenance cost

- Chemical, warehouse, shipping expenses

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Assuming that the cost proportional to production rate is nearly 60% of total capital

investment,

Cost proportional to production rate = (Total capital investment * 60) / 100

= (1298.35 * .6)

= 779.01 lakhs

C. COST PROPORTIONAL TO LABOUR REQUIREMENT The cost proportional to labour requirement might amount to 10% of total

manufacturing cost.

Cost proportional to labour requirement = (165.54 + 779.01)*(0.1) / (0.9)

= 104.95 lakhs

Therefore, manufacturing cost = (165.54 +779.01+ 104.95)

= 1049.5 lakhs

SALES PRICE OF PRODUCT

Market price of = Rs. 100/kg

Production rate =1460000 Kg PA

Total sales income =1460000*100

= 1460 lakhs

PROFITABILITY ANALYSIS

A. DEPRECIATION According to sinking fund method:

R = (V-VS) I / (1+ I)n

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R = Uniform annual payments made at the end of each year

V = Installed cost of the plant

VS = Salvage value of the plant after n years

N = life period (assumed to be 15 years)

I = Annual interest rate (taken as 15%)

R = (1103.6 * .15) / (1+0.15)15-1

= 23.19 lakhs

B. GROSS PROFIT Gross profit = Total sales income - manufacturing cost

= 1460 - 1049.5

=410.5 lakhs

C. NET PROFIT It is defined as the annual return on the investment made after deducting

depreciation and taxes. Tax rate is assumed to be 40%.

Net profit = Gross profit-Depreciation-(Gross profit*Tax rate)

= 410.5-23.19-(410.5*0.4)

= 223.11 lakhs

D. ANNUAL RATE OF RETURN Rate of return = (100*Net profit/Installed cost)

= (100*223.11) /1103.6) = 20.2%

E. PAYOUT PERIOD Payout period = Depreciable fixed investment/((profit)+(depreciation))

= 1103.6 / (223.11 + 23.19)

= 4.48 years

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12. SAFETY

INTRODUCTION In recent years there has been an increased emphasis on process safety as a result of number

of serious accidents. This is due in part to the worldwide attention to issues in the chemical

industry brought on by several dramatic accidents involving gas releases, major explosions

and several environmental accidents. Public awareness of these and other accidents has

provided a driving force for industry to improve its safety record. Local and national

governments are taking a hard look at safety in the industry as a whole and the chemical

industry in particular. There has been an increasing amount of government regulations.

For many reasons, the public often associates chemical industry with environmental and

safety problems. It is vital for the future of the chemical industry that process safety has a

higher priority in the design and operation of chemical process facilities.

INDUSTRIAL ACCIDENTS

An accident has been defined as an unplanned or unexpected event, which causes or is likely to cause an injury. An accident occurs as a result of unsafe actions or exposure to an unsafe environment.

Unsafe actions or unsafe mechanical or physical conditions exist only because of faults of a

particular person.

Faults of persons are inherited from the environment and reasons for the faults are:

i. Improper attitude

ii. Lack of knowledge or skill

iii. Physical unsuitability

iv. Improper mechanical or physical environment

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ACCIDENT PREVENTION

From the foregoing, it will be seen that the occurrence of an injury is the culmination of

a series of events or circumstances that invariably occur in a fused and logical order.

Knowledge of the factors in the accident sequence guides and assists in selecting the

point of attack in prevention work. It permits simplification without sacrifice of

effectiveness. The most important point is that unsafe conditions or actions are the

immediate cause of accidents. The supervision and management can control the actions

of employed persons and so prevent unsafe acts and also guard or remove unsafe

conditions, even though previous events or circumstances in the sequence are

unfavorable.

The four factors that converge to cause accidents are:

i. Personal factor

ii. Hazard factor

iii. Unsafe factor

iv. Proximate casual factor

The solution under the four factors would also lead to two steps. These are planning and

organizing to:

i. Prevent unsafe mechanical or physical conditions

ii. Prevent unsafe action being committed.

HANDLING GUIDELINES

1) Always handle with rubber gloves.

2) Avoid direct skin contact.

3) Wear EYE GOGGLES while handling product.

4) Use a breathing mask while in close proximity to product

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5) Wear an apron while handling product. DO NOT allow product to come

in direct contact with clothes

6) Avoid any direct contact with skin.

7) All employees working inside factory should wear safety shoes.

8) Every employee inside the factory should wear safety helmet to avoid

head injuries.

9) Company should have well equipped medical center.

FIRST AID MEASURES

1) GENERAL INFORMATION: Instantly remove any clothing soiled by the product.

2) After inhalation supply fresh air, consult doctor in case of symptoms.

3) After skin contact instantly wash with water and soap and rinse thoroughly

4) After eye contact rinse opened eye for several minutes under running water.

5) After swallowing rinse mouth and drink plenty of water.

FIRE FIGHTING MEASURES

1) Carbon dioxide, extinguishing powder or water jet is normally used in fires.

2) For large fires water jet or alcohol-resistant foam is used.

3) Collect contaminated fire fighting water separately,it must not enter drains.

SPILLAGE

1) Spray material with water to prevent air pollution through dispersion of particulate

matter.

2) Collect the spilled material using a scrapper.

3) Avoid exposure of spilled material to a direct flame or heat source.

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13. STORAGE AND TRANSPORTATION

Phenol formaldehyde resin in usually stored in cold storage at about 15 degree Celsius.

Storage guidelines

1) The product must be stored in cold storage room.

2) The storage area must be free from moisture.

3) The storage area must be well insulated from any heat source (direct flame).

4) The shelf life for the product which is stored in cold storage room is only 6 months.

5) Advisable to use the product immediately or before 6 months

6) The storage facility must have good ventilation.

7) Clean water must be available in plenty in the vicinity in the event of emergency.

8) The storage area must be designed to avoid direct exposure of the product to the

atmosphere.

9) The containers used for storage should be well sealed containers.

TRANSPORTATION:

Phenol formaldehyde resin is stored in air tight containers and is transported from one place

to other by: lorries, trucks, ships.

DISPOSAL:

After the shelf life period of Phenol formaldehyde resin must be disposed as solid waste

disposal techniques outlined by the pollution control board of the local government.

14. CONCLUSION

The phenol formaldehyde is the largest used resin in the world. Today it is used in all

common places like wood working industry, abrasives, molding and insulation compounds.

The market demand for this resin is always high. The economic importance of phenolic

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lvi

resins today rather proves that they are irreplaceable in the various engineering fields and

distinct areas of daily life.

This project report deals with the manufacturing process, mass, energy, balance and design

aspects. The feasibility of the project and the cost estimation details has also been discussed.

BIBLIOGRAPHY

A.Knop ,W.Scheib ,“Chemistry and Application of Phenolic Resins”.

Kirk and Othmer , “ENCYCLOPEDIA OF CHEMICAL TECHNOLOGY” 4th edition,

volume 18.

Coulson and Richardson, “Chemical Engineering”, 3rd edition, 6th volume.

Joshi and Sharma, “Process Equipment Design “, Khanna Publications.

Donald .Q.Kern, “Process Heat Transfer”, McGraw Hill, International student edition, 3rd

edition.

Dryden’s , “Outlines of Chemical Technology”, East-West press ,3rd edition.

Robert.H.Perry and Don Green, “Perry’s Chemical Engineers Hnadbook,7th edition.

B.I.Bhatt and S.M.Vora, “Stoichiomtery”,4th edition, Tata McGraw Hill publication.

Warren L.McCabe, Julian C.Smith, Peter Harriott , “Unit Operations in Chemical

Engineering”,6th edition, McGraw Hill International Edition.


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