Ajitha

CONTENTS

List of Tables 5

List of figures 7

List of Plates 8

Nomenclature 9

CHAPTER 1 INTRODUCTION 14

1.1 Introduction 14

1.2 History of the compound 14

1.3 Chemical identity 15

1.4 Physical and Chemical Properties 16

1.5 Uses 19

1.6 List of manufacturers and cost 20

CHAPTER 2 METHODS OF MANUFACTURE 21

2.1 Methods of production 21

2.2 Reasons for selection of process 22

2.3 Raw Materials used 23

2.4 Process description 23

CHAPTER 3 MATERIAL BALANCE 25

3.1 Data 25

3.2 Assumptions 26

1

3.3 Individual mass balance 283.3.1 High pressure reactor 28

3.3.2 Decanter I 30

3.3.3 Distillation Column I 30

3.3.4 Continuous stirred tank reactor 34

3.3.5 Decanter II 36

3.3.5 Distillation Column II 37

CHAPTER 4 ENERGY BALANCE 42

4.1 Data 42

4.2 Assumptions 42

4.3 Individual energy balance 43

4.3.1 High Pressure Reactor 43

4.3.2 Cooler 46

4.3.3 Distillation Column I 48

4.3.4 Continuous stirred tank reactor 52

4.3.5 Preheater 56

4.3.6 Distillation Column II 59

CHAPTER 5 PROCESS EQUIPMENT DESIGN 65

5.1 High pressure reactor 65

5.1.1 Process design 65

5.1.2 Mechanical design 66

5.1.3 Design summary 76

5.2 Distillation Column I 77

2




5.3 Continuous stirred tank reactor 93




5.4 Distillation column II 116




5.5 Preliminary design 144

5.5.1 Cooler 144

CHAPTER 6 COST ANALYSIS 146

6.1 Estimation of equipment cost 146

6.2 Estimation of capital investment 147

6.3 Estimation of raw material cost 149

6.4 Estimation of total product cost 150

6.5 Determination of payback period 155

CHAPTER 7 PROCESS INSTRUMENTATION 157

AND CONTROL

7.1 Introduction 157

7.2 Instruments 157

3

7.3 Aim 158

7.4 Process control 158

CHAPTER 8 SITE AND PLANT LAYOUT 160


8.2 Site layout 160

8.3 Plant layout 161

CHAPTER 9 PROCESS SAFETY 163


9.2 Hazards in industries 164

9.3 Material safety data 164

CONCLUSION 168

REFERENCES 169

4

LIST OF TABLES

TABLE NO. NAME PAGE NO.

1 Property estimation using Aspen Properties 16

2 List of Manufacturers 20

3 Cost of glutaraldehyde 20

4 Molecular weight of various components 25

5 Antoine constants 25

6 Material balance for high pressure reactor 29

7 Material balance for distillation column I 31

8 Distillation column I inlet and outlet compositions 32

9 Relative volatility determination- Distillation Column I 32

10 Compositions of top product-Distillation Column I 34

11 Material balance for continuous stirred tank reactor 36

12 Material balance for distillation column II 38

13 Distillation column II inlet and outlet compositions 39

14 Relative volatility determination- Distillation Column II 40

15 Compositions of top product-Distillation Column II 41

16 Heat of formation at 298K and Latent Heat of vaporization 42

17 Specific Heat capacity for high 43

pressure reactor components

18 Energy Balance for high pressure reactor 46

19 Specific Heat capacity for condenser components 46

5


20 Energy Balance for Condenser 48

21 Specific heat capacity for distillation column I 48

Components

22 Energy Balance for distillation column I 52

23 Specific Heat capacity for continuous stirred tank reactor 52

24 Energy Balance for continuous stirred tank reactor 56

25 Specific Heat capacity for preheater components 56

26 Energy Balance for Preheater 58

27 Specific heat capacity for distillation column II 59

components

28 Energy Balance for distillation column II 64

29 High pressure reactor composition 65

30 Physical Properties of high pressure reactor components 67

31 Physical properties of distillation column II components 78

32 Properties of distillation column I reboiler components 89

33 Data for continuous stirred reactor process design 94

34 Physical properties of Continuous stirred tank reactor 95

components

35 Nozzle design for continuous stirred tank reactor 111

36 Data for distillation Column II process design 117

6


37 Relative volatility calculations for feed of 118

distillation column II

38 Relative volatility calculations for distillation 118

column II distillate

39 Relative volatility calculations for distillation 118

column II residue

40 Properties of distillation column II components 122

41 Nozzle design for distillation column II 133

42 Properties of distillation column II reboiler components 139

43 Equipment costs 146

44 Direct costs 147

45 Indirect costs 148

46 Raw material costs 149

47 Utilities costs 151

48 Total Direct production costs 151

49 Fixed charges estimation 153

50 General expenses 153

51 Income estimation 154

7

LIST OF FIGURES

FIGURE NO. NAME PAGE NO.

1 VISCOSITY VS TEMPERATURE 17

2 VAPOR PRESSURE VS TEMPERATURE 17

LIST OF PLATES

PLATE NO. NAME

1 PROCESS FLOW DIAGRAM

2 MATERIAL BALANCE FLOW SHEET

3 ENERGY BALANCE FLOW SHEET

4 HIGH PRESSURE REACTOR

5 DISTILLATION COLUMN I

6 CONTINUOUS STIRRED TANK REACTOR

7 DISTILLATION COLUMN II

8 PROCESS INSTRUMENTATION AND CONTROL DIAGRAM

9 PLANT LAYOUT

8

NOMENCLATURE

General Notations :

NOTATION DEFINITION UNITS

A Area of cross-section m2

B.C.D Bolt Circle diameter cm

Cp Specific Heat Capacity kJ/kgK

D Diameter m

G Diameter of gasket load reaction mm

H Hydrostatic end force kg

ΔH Enthalpy change kJ

ΔHf Heat of formation kJ/kgmole

ΔHr Heat of reaction kJ

J Joint efficiency %

L Length m

M Molecular weight kg/kgmole

9


N Impeller speed rpm

K Thermal conductivity W/mK

P Pressure kg/cm2

Q Heat transferred kJ/hr

R Gas constant kJ/kmolK

Rc,Rk Crown radius, Knuckel radius m

ΔT Temperature change K

T Torque Nm

U Heat transfer coefficient W/m2K

V Volume m3

VO Volumetric flow rate m3/hr

Wm Load kg

Y Yield stress kg/cm2

b Gasket seating width mm

f Permissible stress kg/cm2


10

g Acceleration due to gravity m2/s

h Individual heat transfer coefficient W/m2K

ki Volatility of component no unit

m Mass kg

m Gasket factor no unit

t Thickness mm

v Velocity m/s

xd, xr, xf Mole fraction in distillate, residue, feed no unit

Nre Reynolds number no unit

Npr Prandtl number no unit

Np Power number no unit

Greek symbols:

μ Viscosity gm/cm-s

η Efficiency %


11

λv Latent heat of vaporization kJ/kg

α Relative volatility no unit

Ф Underwood’s dimensionless constant no unit

ρ Density gm/cc

Sub scripts:

12

NOTATION DEFINITION

HKf, LKf Heavy key in feed, Light key in feed

HKr,LKr Heavy key in residue, Light key in residue

HKd, LKd Heavy key in distillate, Light key in residue

av Average

c Crown

c Condenser

d Distillate

e Equivalent (diameter)

f Feed

hk Heavy key

i Component

i Internal (diameter)

j Jacket

k Knuckle

l Liquid

lk Light key

o Outer(diameter)

r Residue

s Shell

v Vapor

CHAPTER 1-INTRODUCTION

13

1.1 INTRODUCTION

Glutaraldehyde is a dialdehyde with two carbonyl groups. It is a highly reactive

chemical compound which, because of its particular characteristics, offers considerable

promise as a chemical intermediate for the synthesis of derived chemicals, and in other

fields of use. Glutaraldehyde is widely used in industries like crude oil and natural gas

extraction, beverage manufacturers, hospitals and X-ray processing. Due to its high

reactivity it is available as 25 or 50 wt% aqueous solution.

1.2 HISTORY OF THE COMPOUND [16]

The first report of the synthesis of glutaraldehyde appeared in 1908, but its first

commercial use, as a tanning agent, was not recognised until about 30 years ago. Interest

in glutaraldehyde peaked in the early 1960’s when several investigations founded to have

outstanding disinfection and sterilization capabilities. By 1963 high level disinfectants,

cold chemical sterilants and potent sporicides were marketed with glutaraldehyde as the

active ingredient. Concerns about the health risks associated with the use of

formaldehyde in the early 1970s led to a further impetus in glutaraldehyde use.

Interests have been intense throughout the years for glutaraldehyde, right upto the

present, as its still is essentially the standard for chemical forms of sterilization.

1.3 CHEMICAL IDENTITY

14

Glutaric dialdehyde is a colorless or yellow oily liquid with a pungent odor.

1.3.1 Chemical Name

The IUPAC name is 1, 5-PENTANEDIAL.

The common name is GLUTARALDEHYDE

The Chemical Abstracts Service (CAS) number is 111-30-8.

1.3.2 Molecular and structural formula

The molecular formula is C5H8O2.

The structural formula is OHC-(CH2)3-CHO

1.3.3 Synonyms

1, 3-Diformylpropane; Glutaral; Glutardialdehyde; Glutaric dialdehyde; 1,5-

Pentanedione; Potentiated Acid Glutaraldehyde; Pentanedial; 1,5-Pentanedial; Glutaric

aldehyde; Glutaric acid dialdehyde; Dioxopentane; Gluteraldehyde;

1.3.4 Trade names

Aidal (sterilant), Aldecyde 28, Aldesan, Aldespray, Aldetex, Alhydex, Aqucar, Asep,

Biomate, Cidex, Coldcide-25 microbiocide, Cronex, Derugan, Dioxopentane, Glutaral,

Glutaralum, Glutarol, Glutasept, Glutex, Hospex, Keymix Glutacide, Microcide, Nalco,

Parvocide, Pentanedione, Performax, Piror Slimicide, Protectol GDA, Protosan, Relugan

GT (tanning), Sporicidin, Sonacide, Sterilite, Surflo, Technicide, Ucarcide, Ucarsan,

Uconex, Ultrasan, Wavicide (sterilant), Zenicide Plus, Zexocide.

1.4 PHYSICAL AND CHEMICAL PROPERTIES

15

1.4.1 Physical properties

Glutaraldehyde is a colourless oily liquid. It is commercially available as a clear aqueous

solution at concentrations up to approximately 50% w/w.

Commercial samples may have a slightly coloured tint and an odour of rotten apples.

Appearance at 20°C and 101.3 kPa : light yellow, viscous liquid.

Table 1: Property estimation using Aspen properties [17]

PROPERTY NAME ESTIMATED VALUE

Freezing Point (°C) -77.15

Boiling Point ( °C) 188

Heat of formation for ideal gas at 25°C (Kcal /mole) -73.492

Standard Gibbs free energy of formation for ideal gas at

25°C (Kcal/mole)

-48.581

Critical Pressure (bar) 35.9

Critical Temperature (°C) 386.85

Critical Volume (mole/cc) 347

Compressibility Factor 0.227

Enthalpy of Vaporisation (Kcal/mole) 10.995

Specific Gravity at 60°F 1.0109

Dipole Moment (Debye) 3.327

Standard Liquid Molar Volume at 60°F (mole/cc) 347

Heat of Combustion (Kcal/mole) -613.595

Molecular weight 100.117

1.4.1.1 Property Analysis: [17]

16

Viscosity vs Temperature

-1

-0.2

0.6

1.4

2.2

0 100 200 300

Temperature (Deg C)

Vis

cosi

ty(c

P)

liquid

vapor

Figure 1: Viscosity vs Temperature

Vapour Pressure vs Temperature

0

0.5

1

1.5

0 100 200 300

Temperature(Deg C)

Vap

ou

r P

ress

ure

(b

ar)

vapor

Figure 2: Vapor pressure vs Temperature

1.4.2 Chemical properties

17

1) Glutaraldehyde is an aliphatic dialdehyde that undergoes most of the typical

aldehyde reactions to form acetals, cyanohydrins, oximes, hydrazones and bisulfite

complexes.

2) Glutaraldehyde in solutions is susceptible to aerial oxidation to give the

corresponding carboxylic acid, glutaric acid.

3) Glutaraldehyde reacts with proteins by a cross-linking reaction which is mainly

between the NH2 groups,and which depends upon time, pH and temperature. The

reaction is less efficient under alkaline conditions.

4) Glutaraldehyde polymerises in water to a glassy form which regenerates the

dialdehyde on vacuum distillation. In solution, glutaraldehyde partially polymerises to

oligomers to give a mixture of variable composition. The degree of polymerisation

increases with pH and temperature.

5) When heated to elevated temperatures (> 400°C), glutaraldehyde in aqueous

solution decomposes thermally to form carbon oxides and hydrocarbons

1.5 USES

18

Glutaraldehyde has a wide variety of uses throughout the world with its use spread over a

number of different industries.

It is used primarily as a biocide but it also has wide use as a fixative, and some use as a

therapeutic agent. The main uses of glutaraldehyde are:

as a biocide in water treatment

as a biocide in aquaculture

as a biocide in sanitary solutions for aircraft and portable toilets

in tanning as a fixative

as a cold disinfectant in the health care industry

in animal housing for disinfection

in small quantities as a disinfectant for air ducts

as a preservative in industrial oils

as a hardener in x-ray film processing

an intermediate in the production of pharmaceuticals, pesticides and crop

protection

as a tissue fixative in electron and light microscopy and in histochemistry

as a cross-linking agent for micro encapsulation

in small quantities as an embalming agent

as a water-resistant in the manufacture of wallpaper and paper towelling

as a preservative in cosmetics

in biochemistry applications as an amine-reactive homo-bifunctional cross linker

to examine the oligomeric state of proteins.

19

1.6 LIST OF MANUFACTURERS AND COST OF GLUTARALDEHYDE:

1.6.1 List of manufacturers: [18]

Table 2: List of manufacturers

Company Country

Alfa Aesar USA

Quat-Chem Specialty Chemicals UK

GFS Chemicals USA

Accepta UK

Amsa USA

Advanced Sterilization Products USA

Durotec SOUTH AFRICA

Sigma-Aldrich USA

Mid South Chemicals USA

Dow Chemicals USA

1.6.2 Cost: [18]

Table 3: Cost of glutaraldehyde as on 18/01/2008

Quantity (ml) Cost in Indian Rupees

500 1721.34

1000 5731.48

CHAPTER 2-METHODS OF MANUFACTURE

20

2.1 METHODS OF PRODUCTION [12] [13] [14] [15]

2.1.1 Synthesis from pyridine

In this method, the dihydropyridine obtained from the reduction of pyridine is treated

with hydroxylamine to give glutaric dioxime. The treatment of this oxime with nitric

oxide or with amyl nitrite, gives glutaraldehyde.

2.1.2 Ozonolysis of cyclopentene

This method involves the synthesis of the ozonide of cyclopentene and the decomposition

of this product to form, among other products, varying amounts of glutaraldehyde.

2.1.3 Addition hydrolysis of benzimidazolium salt

Glutaraldehyde can also be prepared by the addition hydrolysis reaction of

benzimidazolium salt with saturated dialdehyde as the di-grignard reagent. This method

involves the separate synthesis of dibenzylbenzimidazolium salt and the grignard reagent

followed by the heating of these reactants for 18 hours.

2.1.4 Thermal hydrolysis of alkoxydihydropyran

This process involves the heating of a derivative of dihydropyran, for example, a 2-

alkoxy-3,4-dihydropyran with water at temperatures of from 100° to 200° C. The thermal

hydrolysis of the alkoxy dihydropyran results in the formation of glutaraldehyde.

2.1.5 Acid hydrolysis of the alkoxydihydropyran

This is a process for the continuous preparation of glutaraldehyde by reaction of

an alkoxydihydropyran, with water at 50°C and atmospheric pressure to form

glutaraldehyde and the alcohol corresponding to the alkoxy group with the help of acid

catalysts.

21

Among the processes listed, the continuous process for the production of glutaraldehyde

by acid hydrolysis of alkoxydihydropyran has been chosen.

2.2 REASONS FOR SELECTION OF THE PROCESS:

In the synthesis of glutaraldehyde from pyridine, the method requires the previous

indirect synthesis of the oxime from chemicals which themselves may be

prepared only with difficulty under considerable expense. However the selected

method does not involve undesirable number of intermediate steps.

Acid hydrolysis of the alkoxydihydropyran does not involve the formation of

highly unstable intermediates as in the case of ozonolysis of cyclopentene in

which the cyclopentene ozonide formed is highly unstable and explosive.

The synthesis using bis-grignard reagent is not preferred because it consumes

more time than the acid hydrolysis method. Also,the latter process being a

continuous one reduces the dimensions of the plant for the same throughput per

unit time compared to the batch wise reaction.

Thermal hydrolysis of alkoxydihydropyrans is feasible only at a temperature

range of 100° to 200°C. Hence acid catalysts are used for lowering the reaction

temperature as given in the selected method of production.

2.3 RAW MATERIALS USED :

The basic raw materials used in the process are:

1) Acrolein

2) Methyl Vinyl Ether

22

3) Water

4) Hyrdoquinone(Inhibitor)

5) Maleic Acid (Catalyst)

2.4 PROCESS DESCRIPTION:

Glutaraldehyde can be continuously prepared by the hydrolysis of the

alkoxydihydropyran of the general formula

or its derivative in the presence of solid acid catalysts. Here, we are using 2 methoxy-

3,4dihydropyran. The reaction is generally carried out at a temperature of 50°C and at

atmospheric pressure.

2.4.1 Preparation of 2 methoxy-3,4 dihydropyran[13]

2 methoxy-3,4dihydropyran is prepared by reacting equimolar amounts of alpha-

beta-olefinic aldehyde such as acrolein with methylvinyl ether in a high pressure reactor

at an elevated temperature of about 180°C and at a high pressure of 30 atm. Sufficient

amount of hydroquinone is added to inhibit the polymerization of acrolein. The product

mixture containing unreacted acrolein and methyl vinyl ether along with the product

formed after condensation through a condenser is fed to a distillation column.2 methoxy

3,4 dihydropyran being less volatile is obtained in the bottom product with traces of

acrolein.

CH2=CH-CHO + CH2=CH-O-CH3 -------------->

Acrolein Vinyl Methyl Ether Methoxydihydropyran

23

2.4.2 Acid Hydrolysis of 2 methoxy-3,4dihydropyran [14]

The 2 methoxy-3,4dihydropyran from the distillate residue is then hydrolysed using water

in a continuous stirred tank reactor. The reaction occurs at 50°C and 1 atm. The molar

ratio of water and 2 methoxy 3,4 dihydropyran is the ratio 8:1.Maleic acid is the catalyst

utilized in this reaction. Glutaraldehyde and methanol are the products formed.

Glutaraldehyde is then separated from methanol by distillation. The distillate is rich in

methanol and part of it is condensed back through a condenser. A product enriched in

glutaraldehyde is taken off at the bottom of the reaction column and sent to storage. Part

of the bottom stream is vaporized again in a reboiler and recirculated to the lower part of

the column.

The bottoms generally comprise of about 50-55% by weight Glutaraldehyde.

+ H2O -------------------CHO-(CH2)3-CHO + CH3OH

2 methoxy-3,4 dihydropyran water Glutaraldehyde

Methanol

CHAPTER 3-MATERIAL BALANCE

24

3.1 DATA:

Table 4: Molecular weight of various components

COMPONENT MOLECULAR WEIGHT

Acrolein 56

Methyl Vinyl Ether 58

2 Methoxy 3, 4 Dihydropyran 114

Glutaraldehyde 100

Water 18

Methanol 32

Maleic acid 116

Hydroquinone 110

Table 5: Antoine Constants

COMPONENT A B C

Acrolein 15.90 2606.53 -45.15

Methyl Vinyl

Ether *

7.02 1016.34 -36.72

2 Methoxy 3, 4

Dihydropyran

17.56 3954.27 -37.28

Glutaraldehyde* 7.03 2120.37 -40.1

Water* 7.17 1715.4 -41.05

Methanol 16.49 3593.39 -35.22

Antoine equation: ln P = A-(B/(T+C)); For * log P = A-(B/(T+C))

3.2 GENERAL ASSUMPTIONS :

1. A recovery of 99% is assumed for Methoxy Dihydropyran.

2. The purity of both Glutaraldehyde and Methoxy Dihydropyran is assumed to be

50% and 99% respectively.

25

3. Purity of acrolein and methyl vinyl ether is assumed to be 100%.

4. As methyl vinyl ether is highly volatile compared to acrolein and methoxy

dihydropyran, it is assumed to have fully vaporized in the distillation column.

5. Owing to its low volatility, glutaraldehyde is assumed to be completely recovered

in the residue.

BASIS: 1 hr operation

It is desired to produce 5 tonnes of glutaraldehyde solution per day. Assuming purity of

50% for pure glutaraldehyde,

Amount of glutaraldehyde to be produced =0.5×5000

= 2500 kg/day

Assuming that the plant operates for 20 hours per day

Amount of glutaraldehyde produced per hour = (2500)/ (20)

= 125 kg

=1.25 kgmoles

For a conversion of 98% for 2 Methoxy 3, 4 Dihydropyran

No. of moles of Methoxy Dihydropyran required to

26

produce 1.25 kgmoles/hr of glutaraldehyde = (1.25 × 100)/98

= 1.275 kgmoles

= 145.40 kg

Recovery of Methoxy Dihydropyran = 99%

Actual amount of Methoxy Dihydropyran to be produced = 145.4/0.99

= 146.86 kg

For a conversion of 96% for Acrolein,

No. of moles of acrolein required

to produce 1.275 kgmoles / hr of MDP = (1.275)/0.96

= 1.342 kgmoles

= 75.15 kg

3.3 INDIVIDUAL MASS BALANCES

Basis: 1 hr operation

27

3.3.1 HIGH PRESSURE REACTOR:

Acrolein and methyl vinyl ether are reacted in the molar ratio of 1:1. 0.4 kg of

Hydroquinone is added to the reaction mixture to prevent the polymerization of acrolein.

Since the reaction requires the reactants to be in their pure form, we assume 100% purity

for the reactants. The conversion for the reaction is 96%. Here acrolein is the limiting

reagent.

The reaction is given by

CH2=CH-CHO + CH2=CH-O-CH3 -------------->

Acrolein Vinyl Methyl Ether Methoxydihydropyran

Amount of acrolein entering the reactor = 1.342 kgmoles

=75.15 kg

Amount of methyl vinyl ether entering the reactor = 1.342 kgmoles

= 77.84 kg

For 96% conversion, amount of Methoxydihydropyran formed = 0.96 × 1.342

28

= 1.288 kgmoles

= 146.87 kg

Amount of acrolein remaining = 1.288 × (1-0.96)

= 0.053 kgmoles

= 3.006 kg

Amount of methyl vinyl ether remaining = 1.288 × (1-0.96)

= 0.053 kgmoles

= 3.11 kg

Table 6: Material balance for high pressure reactor

COMPONENT WEIGHT in kg (INLET) WEIGHT in kg

(OUTLET)

Acrolein 75.15 3.006

Methyl Vinyl Ether 77.84 3.113

2 Methoxy 3, 4

Dihydropyran

0 146.87

Hydroquinone 0.4 0.4

TOTAL 153.39 153.39

3.3.2 DECANTER I:

29

Hydroquinone alone is removed by decanter. It is assumed that hydroquinone is removed

completely.

Amount of stream leaving the decanter = 153.39-0.4

= 152.99 kg

3.3.3 DISTILLATION COLUMN I:

Due to the vast difference in boiling points between acrolein, methyl vinyl ether

and Methoxy Dihydropyran, it is assumed that 99% of Methoxy dihydropyran is

recovered in the bottoms having a purity of 99%.The feed is introduced into the column

at its bubble point ( 98.1°C).The column is maintained at a pressure of 1 atm.

Assuming a recovery of 99% for methoxydihydropyran,

Amount of methoxydihydropyran in residue = 146.87 × 0.99

=145.41 kg

Amount of methoxydihydropyran in distillate =146.87-145.41

=1.46 kg

Amount of methyl vinyl ether in distillate =3.113 kg

Assuming a purity of 99% for methoxydihydropyran

30

Amount of residue formed =145.41/0.99

=146.87 kg

Hence amount of acrolein in residue =146.87×(1-0.99)

=1.46 kg

Amount of acrolein in distillate = 3.006-1.46

=1.55 kg

Table 7: Material balance for distillation column I

COMPONENT FEED (kg) DISTILLATE

WEIGHT

(kg)

RESIDUE

WEIGHT

(kg)

Methyl vinyl ether 3.113 3.113 0

Acrolein 3.006 1.55 1.46

Methoxydihydropyran 146.87 1.46 145.41

TOTAL

6.12 146.87

152.99 152.99

Calculation of reflux ratio by Fenske Underwood-Gilligan Method:[5]

31

Table 8: Distillation column I inlet and outlet composition

COMPONENT FEED MOLE

FRACTION

zi

DISTILLATE MOLE

FRACTION

xdi

RESIDUE

MOLE

FRACTION

xwi

Methyl vinyl ether 0.038 0.570 0

Acrolein 0.038 0.292 0.02

Methoxydihydropyran 0.923 0.137 0.979

Light key- Acrolein

Heavy key- Methoxydihydropyran

ki = Pi / Psat where Pi = vapor pressure, P =Total pressure

αi = ki / khk where αi = relative volatility of each component with respect to the heavy

key.

Table 9: Relative volatility determination-Distillation column I

COMPONENT ki αi ki αi ki αi

FEED DISTILLATE RESIDUE

Acrolein - LK 3.58 8.96 2.02 11.88 6.24 7.01

Methyl Vinyl ether 12.80 32.01 7.86 46.23 20.55 23.08

Methoxydihydropyran- HK 0.4 1 0.17 1 0.89 1

Solving the equation

32

∑ [(αi × zi × F)/(αi – Ф)]= F × (1- q) where q= 1 as feed is at its bubble point

Ф = 6.26

Substituting this value in the below equation gives the minimum reflux ratio

∑ [(αi × xdi × D)/(αi – Ф)] = D × ( Rm + 1)

Rm = 0.65

Actual reflux ratio (L/D) = 1.2 × Rm

= 0.78

No of kgmoles of distillate (D) = 0.092

No of kgmoles of liquid recycled to the column ( L) =0.092 × 0.78

=0.071

No of kgmoles of vapor entering the condenser = L + D

=0.071 + 0.092

=0.163

Table 10: Compositions for top product-Distillation column I

33

DISTILLATE (D) RECYCLE LIQUID (L) VAPOR (G)

Component Mole

fraction

Kgmoles Mass

(kg)

Mole

fraction

Kgmoles Mass

(kg)

Mole

fraction

Kgmoles Mass

(kg)

Acrolein 0.293 0.027 1.55 0.293 0.021 1.17 0.292 0.048 2.72

Methyl

Vinyl Ether

0.57 0.053 3.113 0.57 0.04 2.32 0.57 0.093 5.43

Methoxy

dihydropyran

0.137 0.012 1.46 0.137 0.01 1.14 0.137 0.022 2.6

TOTAL 1 0.092 6.12 1 0.071 4.63 1 0.163 10.75

3.3.5 CONTINUOUS STIRRED TANK REACTOR:

The cooled residue from the distillation column containing 98% Methoxy

dihydropyran is fed to a reactor along with water in the molar ratio of 1:8. The water

solution consists of 0.6% by weight of solid maleic acid catalyst. The conversion for the

reaction is 98%. Here Methoxy dihydropyran is the limiting reagent.

Methoxydihydropyran Glutaraldehyde

+CH3OH

Methanol

34

Amount of Methoxy dihydropyran entering the reactor =45.41kg

= 1.276 kgmoles

Amount of water entering the reactor = 1.276 × 8 kgmoles

= 10.2 kgmoles

= 183.67 kg

Amount of maleic acid catalyst added = (0.006×183.67)/

(1-0.006)

=1.108 kg

For 98% conversion,

Amount of Glutaraldehyde formed = 0.98 ×1.276

= 1.25 kgmoles

= 125 kg

Amount of methanol formed = 0.98 × 1.276

= 1.25 kgmoles

= 40 kg

Amount of Methoxy dihydropyran unreacted = 1.276 × 0.02

35

= 0.026 kgmoles

= 2.91 kg

Amount

Amount of water unreacted = 8.954 kgmoles

= 161.17 kg

Table 11: Material balance for Continuous stirred tank reactor

COMPONENT WEIGHT in Kg (INLET) WEIGHT in Kg

(OUTLET)

Methoxy dihydropyran 145.41 2.91

Water 183.67 161.17

Glutaraldehyde 0 125

Methanol 0 40

Maleic acid 1.108 1.108

Acrolein 1.46 1.46

TOTAL 331.64 331.64

3.3.6 DECANTER II:

Solid maleic acid alone is removed by decanter. It is assumed that maleic acid is removed

completely.

Mass of stream leaving the decanter = 331.648-1.108

= 330.54 kg

36

3.3.7 DISTILLATION COLUMN II:

Light Key: Methanol

Heavy Key: Water

The key components are distributed in both the distillate and the residue. Acrolein

being more volatile than methanol, and Methoxy dihydropyran and Glutaraldehyde, being

less volatile than water, are completely recovered in the distillate and residue

respectively.

Glutaraldehyde is assumed to have a purity of 50%

Amount of Glutaraldehyde in the residue = 125 kg

Amount of residue formed = 125/0.5

= 250 kg

Amount of distillate formed = 330.54-250

= 80.54 kg

Amount of Acrolein in the distillate = 1.46 kg

Amount of Methoxy dihydropyran in the residue = 2.91 kg

Percentage of Water in the distillate = 48.75%

Amount of Water in the residue = 121.89 kg

37

Amount of Water in the distillate = 39.28 kg

Percentage of Methanol in the distillate =0.08%

Amount of Methanol in the residue = 0.2 kg

Amount of Methanol in the distillate = 39.8 kg

Table 12: Material balance for distillation column II

COMPONENT FEED (kg) DISTILLATE

WEIGHT

(kg)

RESIDUE

WEIGHT

(kg)

Glutaraldehyde 3.113 0 125

Methoxy dihydropyran 2.91 0 2.91

Water 161.17 39.28 121.89

Methanol 40 39.8 0.2

Acrolein 1.46 1.46 0

TOTAL

80.54 250

330.54 330.54

38

Table 13: Distillation column II inlet and outlet compositions

COMPONENT FEED MOLE

FRACTION

zi

DISTILLATE MOLE

FRACTION

xdi

RESIDUE

MOLE

FRACTION

xwi

Acrolein 0.0023 0.0076 0

Methanol 0.1086 0.3603 0.0008

Water 0.7782 0.6321 0.8408

MethoxyDihydropyran 0.0022 0 0.0032

Glutaraldehyde 0.1086 0 0.1552

TOTAL 1 1 1

Light key –Methanol

Heavy key – Water


αi = ki / khk where αi = relative volatility of each component with respect to the heavy

key.

39

Table 14: Relative volatility determination-Distillation Column II

COMPONENT ki αi

Acrolein 3.3708 3.9662

Methanol 3.0022 3.5325

Water 0.8499 1

MethoxyDihydropyran 0.3656 0.4301

Glutaraldehyde 0.0369 0.0435

Solving the equation

∑ [(αi × zi × F)/(αi – Ф)]= F × (1- q) where q= 1 as feed is at its bubble point

Ф = 2.69

Substituting this value in the below equation gives the minimum reflux ratio

∑ [(αi × xdi × D)/(αi – Ф)] = D × ( Rm + 1)

Rm = 2.73

Actual reflux ratio (L/D) = 1.2 × Rm

= 3.28

No of kgmoles of distillate ( D ) = 3.452

No of kgmoles of liquid recycled to the column (L) =3.452 × 3.28

40

=11.31

No of kgmoles of vapor entering the condenser = L + D

=11.31 + 3.452

=14.762

Table 15: Compositions for top product-Distillation Column II

DISTILLATE (D) RECYCLE LIQUID (L) VAPOR (G)

Component Mole

fraction

Kgmoles Mass

(kg)

Mole

fraction

Kgmoles Mass

(kg)

Mole

fraction

Kgmoles Mass

(kg)

Acrolein 0.0076 0.006 1.46 0.0076 1.12 4.78 0.0076 0.112 6.24

Methanol 0.3603 0.291 39.8 0.3603 30.49 130.4 0.3603 5.318 170.2

Water 0.6321 0.51 39.28 0.6321 30.09 128.7 0.6321 9.331 168

MethoxyDihydropyran 0 0 0 0 0 0 0 0 0

Glutaraldehyde 1 0.807 80.54 1 61.7 263.9 1 14.76 344.44

TOTAL 1 3.441 79.91 1 18.122 420.85 1 21.563500.76

41

CHAPTER 4 - ENERGY BALANCE

4.1 DATA :

Table 16: Enthalpy of formation at 298K and Latent heat of vaporisation[17]

S.No COMPONENT ΔHformation

(kJ/kgmole)

Latent heat of vaporization λ

(kJ/kg)

1 Acrolein -81.78 1600.98

2 Vinyl Methyl Ether -107.97 1418.54

3 Methoxydihydropyran -301.48 4161.02

4 Water -241.75 1097.856

5 Methanol -200.89 2363.87

6 Glutaraldehyde -307.62 7037.79

4.2 GENERAL ASSUMPTIONS

1. No heat loss in flowing streams and in equipments.

2. The reactors operate under isothermal conditions.

3. Ideal state of gas and liquid phases.

4. Reference Temperature = 25°C

4.3 INDIVIDUAL ENERGY BALANCES:

42

4.3.1 HIGH PRESSURE REACTOR

The vital reaction for the formation of 2 methoxy 3,4 dihydropyran from Acrolein

and Methyl Vinyl ether is carried out in a Reactor vessel at 180°C and 30 atm pressure.

The feed stream enters the reactor at a temperature of 30°C.

Enthalpy of the feed Stream:

Table 17: Specific heat capacity for high pressure reactor components[17]

S.No COMPONENT INLET SPECIFIC

HEAT

CAPACITY(kJ/kgK

)

OUTLET SPECIFIC

HEAT CAPACITY

(kJ/kgK)

1 Acrolein 2.053 3.684

2 Vinyl Methyl Ether 1.344 1.77

3 Hydroquinone 1.247 1.785

4 Methoxydihydropyran - 2.497

Enthalpy of each component in the reactor feed stream at 30°C is calculated as follows:

Enthalpy of Acrolein = 75.15×2.053× (303-298)

= 771.47 kJ

Enthalpy of Methyl Vinyl Ether = 77.84 ×1.344 × (303-298)

=523.08 kJ

Enthalpy of Hydroquinone =0.4 × 1.247 × (303-298)

43

=2.49 kJ

Total Enthalpy of feed stream at 30°C = 1297.04 kJ

Enthalpy of exit Stream:

Enthalpy of each component in the recycle stream at 180°C is calculated as follows:

Enthalpy of Acrolein = 3.006× 3.684 × (453-298)

= 1716.48 kJ

Enthalpy of Methyl Vinyl Ether = 3.113 × 1.77 × (453-298)

= 854.05 kJ

Enthalpy of 2 methoxy 3,4 dihydropyran =146.87× 2.497× (453-298)

=56843.83 kJ

Enthalpy of hydroquinone = 0.4 × 1.785 × (453-298)

= 110.67 kJ

Total Enthalpy of exit stream at 180°C = 59525.03 kJ

Inside the reactor, the reaction of Acrolein and Methyl Vinyl ether occurs as follows:

CH2=CH-CHO + CH2=CH-O-CH3 --------------> Acrolein Vinyl Methyl Ether Methoxydihydropyran

Heat of Reaction at 25°C

∆HR25°C = ∆H products - ∆H reactants

= [-301.48 - (-107.97 -81.78)]

= -111.73 kJ/kgmole

Heat of reaction at 180°C

44

∆HR180°C = ∆HR25°C + Σ Cpi × ∆T

Where Σ Cpi × ∆T = (CpMethoxydihydropyran × ∆T – ( CpMethylvinylether × ∆T + CpAcrolein × ∆T))

= ((284.65 × (453-298)) – (( 102.74 × (453-298))+(206.3×(453-298))))

= -3780.45 kJ/kgmole

Hence Heat of reaction at 180°C = -111.73 + (-3780.45)

= -3892.18 kJ/kgmole

For 1.288 moles reacted, heat of reaction at 180°C = -3892.18 × 1.288

= -5013.12 kJ

Heat to be removed = Total Enthalpy of exit stream + Heat of reaction – Total enthalpy

of inlet stream

Total Enthalpy of inlet stream = 1297.04 kJ

Total Enthalpy of exit stream = 59525.03 kJ

Heat of reaction = -5013.12 kJ

Heat to be removed =59525.03 + (-5013.12)-(1297.04)

=53214.87 kJ

Heat Heat energy is removed from the system using a jacket with water as the cooling

fluid.

The flow rate of the cooling fluid is calculated as follows:

Inlet temperature of the cooling fluid= 50

Outlet temperature of the cooling fluid=30

Mass of the fluid, m= Qr/(Cpw × ΔT)

45

= 53214.5/(4.18 × (323-303))

= 636.5 kg

The operation time is 1hour.

Thus the flow rate for 1 hour =636.5kg/hr

Table 18: Energy balance for high pressure reactor

COMPONENTSENTHALPY OF THE ENTERING STREAM

(kJ)

ENTHALPY OF THELEAVING STREAM

(kJ)

Acrolein 771.47 1716.48Methyl Vinyl Ether 523.08 854.05Methoxydihydropyran 56843.83Hydroquinone 2.49 110.67Heat of reaction 5013.12Heat removed 53214.87

TOTAL 59525.03 59525.03

4.3.2 COOLER:

The entering stream is cooled to the bubble point of the mixture i.e. 98.1°C

Table 19: Specific heat capacity for cooler components [17]


HEAT

CAPACITY(kJ/kgK)

OUTLET

SPECIFIC

HEAT

CAPACITY

(kJ/kgK)

1 Acrolein 1.7 2.53

2 Vinyl Methyl Ether 1.83 3.13

3 Methoxydihydropyran 1.76 1.98

46

Inlet stream

Enthalpy of acrolein = 3.006 × 1.7 × (453-298)

= 792.08 kJ

Enthalpy of Methyl vinyl ether = 3.113 × 1.83 × (453-298)

= 883 kJ

Enthalpy of Methoxydihydropyran = 146.87× 1.76 × (453-298)

= 40066.13 kJ

Total enthalpy of inlet stream = 41741.21 kJ

Outlet stream

Enthalpy of Acrolein = 3.006 ×2.53 ×(371.1-298)

= 555.93 kJ

Enthalpy of Methyl Vinyl Ether = 3.113 ×3.13 ×(371.1-298)

= 712.26 kJ

Enthalpy of 2 methoxy 3,4 dihydropyran =146.87×1.98 ×(371.1-298)

= 21257.67 kJ

Total enthalpy of outlet stream = 22525.86 kJ

Heat removed by the cooler = Enthalpy of inlet stream – Enthalpy of outlet stream

= 41741.21- 22525.86

= 19215.35 kJ

47

Cooling water is supplied at 30°C which attains a final temperature of 60°C

Mass of cooling water required = Qc/(Cpw ×Δ T)

= 19215.35/(4.18 × (333-303))

= 153.23 kg/hr

Table 20: Energy balance for the cooler

COMPONENT ENTHALPY OF THE ENTERING STREAM(kJ)

ENTHALPY OF THE LEAVING STREAM

(kJ)

Acrolein 792.08 555.93Methyl Vinyl Ether 883 712.26Methoxydihydropyran 40066.13 21257.67Heat removed 19215.35Total 41741.21 41741.21

4.3.3 DISTILLATION COLUMN I

The feed enters at its bubble point of 98.1°C at a pressure of 1 atm.

Table 21: Specific heat capacity for distillation column I components [17]

S.No COMPONENT DISTILLATE Cp

(kJ/kgK)

VAPOR Cp

(kJ/kgk)

RESIDUE

Cp

(kJ/kgK)

1 Acrolein 2.05 1.44 2.71

2 Vinyl Methyl Ether 2.45 1.49 3.84

3 Methoxydihydropyran 1.56 1.41 2.11

48

Enthalpy of feed:

Enthalpy of acrolein = 3.006 × 2.53×(371.1-298)

= 555.93 kJ

Enthalpy of Methyl Vinyl Ether = 3.113 × 3.31×(371.1-298)

= 753.22 kJ

Enthalpy of 2 methoxy 3,4 dihydropyran =146.87×1.98×(371.1-298)

=21257.67 kJ

Total enthalpy of feed = 22566.82 kJ

CONDENSER HEAT DUTY:

Enthalpy of vapour:

The vapor is assumed to be at its dew point ( 76.3°C)

Enthalpy of acrolein = 2.72 × 1.44 × (349.3-298)

+ 2.72 × 1600.98

= 4555.59 kJ

Enthalpy of Methyl Vinyl Ether = 5.43 × 1.49 × (349.3-298)

+ 5.43 × 1418.54

= 8117.72 kJ

Enthalpy of 2 methoxy 3,4 dihydropyran =2.6 × 1.41 × (349.3-298)

+ 2.6 × 4161.02

=11006.71 kJ

Total enthalpy of vapour = 23680.02 kJ

49

Enthalpy of distillate:

The distillate is assumed to be at its bubble point ( 30.4°C)

Enthalpy of acrolein = 1.55 × 2.05 × (303.4-298)

= 17.15 kJ

Enthalpy of Methyl Vinyl Ether = 3.113×2.45 × (303.4-298)

= 41.18 kJ

Enthalpy of 2 methoxy 3,4 dihydropyran =1.46 ×1.56 × (303.4-298)

=12.29 kJ

Total enthalpy of distillate = 70.62 kJ

Enthalpy of recycled liquid:

The liquid is recycled at its bubble point ( 30.4°C)

Enthalpy of acrolein = 1.17 ×2.05 × (303.4-298)

= 12.95 kJ

Enthalpy of Methyl Vinyl Ether = 2.32 × 2.45 × (303.4-298)

= 30.69 kJ

Enthalpy of 2 methoxy 3,4 dihydropyran =1.14 × 1.56 × (303.4-298)

=9.6 kJ

Total enthalpy of recycled liquid = 53.24 kJ

50

Heat removed by condenser= Enthalpy of vapour – (Enthalpy of distillate + Enthalpy of recycled liquid)

=23680.02-(70.62 + 53.24)

=23556.16 kJ

Cooling water to condenser enters at 30°C and leaves at 50°C

Mass of cooling water required = Qcd/(Cpw × ΔT)

= 23556.1/(4.18×(323-303)

=281.77 kg/hr

REBOILER HEAT DUTY:

The residue is assumed to be at its bubble point (122.4°C)

Enthalpy of Acrolein in residue =1.46 × 2.71 × (395.4-298)

=385.37 kJ

Enthalpy of methoxydihydropyran in residue =145.41× 2.1 × (395.4-298)

=29827.13 kJ

Total enthalpy of residue = 30212.5 kJ

Heat supplied to the reboiler =Enthalpy of distillate + Heat removed by condenser + Enthalpy of residue-Enthalpy of feed

= 70.62 + 23556.15 + 30212.5 - 22566.82

= 31272.45 kJ

For saturated steam at 7 atm, λ = 2064.9 kJ/kg

Amount of steam supplied = 31272.45/λ

= 31272.45/2064.9

= 15.14 kg/hr

51

Table 22: Energy balance for the distillation column I

COMPONENT ENTHALPY OF THE ENTERING

STREAM(kJ)

ENTHALPY OF THE LEAVING STREAM (kJ)

TOP BOTTOM

Acrolein 555.93 17.15 385.37Methyl Vinyl Ether 753.22 41.18 0Methoxydihydropyran 21257.67 12.29 29827.13Heat supplied 23556.15 -31272.45

TOTAL 22566.82 22566.82

4.3.4 CONTINUOUS STIRRED TANK REACTOR:

Glutaraldehyde is formed from Methoxy dihydropyran and water in a Reaction

vessel at 50ºC and 1atm pressure. Water enters the reactor at a temperature of 30ºC.

Table 23: Specific heat capacity for continuous stirred tank reactor components [17]


HEAT

CAPACITY(kJ/kgK

)

OUTLET SPECIFIC

HEAT CAPACITY

(kJ/kgK)

1 Acrolein 2.71 2.17

2 Water 3.89 4.03

3 Maleic acid 1.18 1.251

4 Methoxydihydropyran 2.105 1.684

5 Methanol - 3.426

6 Glutaraldehyde - 2.057

52

Enthalpy of the feed Stream

Enthalpy of each component in the reactor feed stream at 30ºC is calculated as follows:

Enthalpy of Acrolein = 1.46 × 2.71 × (393-298)

= 375.877 kJ

Enthalpy of Methoxydihydropyran = 145.41× 2.105×(393-298)

= 29078.364 kJ

Enthalpy of Maleic acid = 1.108 × 1.189 ×(303-298)

= 6.587 kJ

Enthalpy of Water = 183.67 ×3.893×(303-298)

=3575.14 kJ

Total Enthalpy at 30°C = 33035.968 kJ

Enthalpy of the Exit Stream

Enthalpy of each component in the reactor outlet at 50ºC is calculated as follows:

Enthalpy of Acrolein = 1.46× 2.175× (323-298)

= 79.386 kJ

53

Enthalpy of Methoxydihydropyran = 2.91×1.684 × (323-298)

= 122.511 kJ

Enthalpy of Glutaraldehyde = 125× 2.057 × (323-298)

= 6428.125 kJ

Enthalpy of Methanol = 40 ×3.426 ×(323-298)

= 3426 kJ

Enthalpy of Maleic acid = 1.108×1.251 ×(323-298)

= 34.653 kJ

Enthalpy of Water = 161.71×4.036× (323-298)

= 16316.539 kJ

Total Enthalpy at 50ºC = 26407.216 kJ

Inside the reactor, the reaction of Methoxydihydropyran and water occurs as

follows:

Methoxydihydropyran Glutaraldehyde +CH3OH

Methanol

54

Heat of Reaction

∆HR = ∆H products - ∆H reactants

∆H50 =∆Hf25 + Cp50 x ∆T

= {(-307.626 +5142.5) + (-200.89 + 2740.8)} – {(-301.484 + 4799.4) +

(-241.75 +1618.2)}

= 1500.418 kJ

Therefore, for 1.25 Kmoles, heat of reaction = 1875.523 kJ

Heat removed or added =Total Enthalpy of exit stream +Heat of reaction –Total enthalpy

of entering stream

= 26407.216 + 1875.523 - 33035.968

= -4753.23 kJ

Heat to be removed = 4753.23 kJ

Heat energy is removed from the system using a jacket with water as the cooling fluid.

Amount of water required, m = Q/ (Cp × ∆T)

= 4753.23 / (4.184 × 30)

= 37.87kg

The operation time is 1hour.

Thus the flow rate for 1 hour =37.87 kg/hr

55

Table 24: Energy balance for continuous stirred tank reactor

COMPONENTS

ENTHALPY OF THE

ENTERING STREAM

(kJ)

ENTHALPY OF THE

LEAVING STREAM

(kJ)

Acrolein 375.877 79.388

Maleic acid 6.587 34.653

Methoxydihydropyran 29078.364 122.511

Water 3575.14 16316.539

Methanol 3426

Glutaraldehyde 6428.125

Heat of reaction 1875.523

Heat removed -4753.23

Total 28282.738 28282.738

4.3.5 PRE-HEATER:

The entering stream is heated to the bubble point of the mixture i.e. 95° C

Table 25: Specific heat capacity for preheater components [17]

S.No COMPONENT OUTLET SPECIFIC HEAT

CAPACITY

(kJ/kgK)

1 Acrolein 2.479

2 Water 4.406

3 Methoxydihydropyran 1.962

4 Methanol 3.83

56

5 Glutaraldehyde 2.233

Inlet stream

Enthalpy of Acrolein = 79.386 kJ

Enthalpy of Glutaraldehyde = 6428.125 kJ

Enthalpy of Methanol = 3426 kJ

Enthalpy of water = 16316.539 kJ

Enthalpy of Methoxydihydropyran = 122.511 kJ

Total enthalpy of inlet stream = 26372.561 kJ

Outlet stream

Enthalpy of Acrolein = 1.46 × 2.479×(368-298)

= 253.35 kJ

Enthalpy of Glutaraldehyde = 125 × 2.233 × (368-298)

= 19538.75 kJ

Enthalpy of Methanol = 40 ×3.831 × (368-298)

= 10726.8 kJ

Enthalpy of water = 161.17× 4.406×(368-298)

57

=49708.05 kJ

Enthalpy of Methoxydihydropyran = 2.91× 1.962× (368-298)

=399.65 kJ

Total enthalpy of outlet stream = 80626.6 kJ

Heat supplied by the pre-heater = Enthalpy of outlet stream – Enthalpy of inlet stream

= 80626.6 - 26372.54

= 54254.06 kJ

Amount of steam required = Qh/ λ

= 54254.06 /2064.9

= 26.27 kg/h

Table 26: Energy balance for the pre-heater

COMPONENTS

ENTHALPY OF THE

ENTERING STREAM

(KJ)

ENTHALPY OF THE

LEAVING STREAM

(KJ)

Acrolein 79.38 253.35

Methoxydihydropyran 122.51 399.65

Water 16316.53 49708.05

Methanol 3426 10726.8

Glutaraldehyde 6428.12 19538.75

58

Heat supplied 54254.06

Total 80626.6 80626.6

4.3.6 DISTILLATION COLUMN II

The feed enters at its bubble point of 95°C at a pressure of 1 atm.

Table 27: Specific heat capacity for distillation column II components [17]

S.No COMPONENT DISTILLATE

Cp

(kJ/kgK)

VAPOR Cp

(kJ/kgk)

RESIDUE

Cp

(kJ/kgK)

1 Acrolein 2.37 1.474 -

2 Methanol 3.73 1.526 3.99

3 Glutaraldehyde - - 2.25

4 Water 4.29 1.88 4.49

5 Methoxydihydropyran - - 1.98

Enthalpy of feed:

Enthalpy of Acrolein = 1.46 × 2.479 × (368-298)

= 253.35 kJ

Enthalpy of Glutaraldehyde = 125× 2.233 × (368-298)

= 19538.75 kJ

Enthalpy of Methanol = 40 × 3.831 × (368-298)

= 10726.8 kJ

59

Enthalpy of water = 161.17×4.406× (368-298)

=49708.051 kJ

Enthalpy of Methoxydihydropyran = 2.91 ×1.962 × (368-298)

=399.659 kJ

Total enthalpy of feed = 80625.86 kJ

Enthalpy of vapour:

The vapor is assumed to be at its dew point (91.8°C)

Enthalpy of Acrolein = 6.24×1.474 × (364.8-298)

+6.249 ×510.742

= 3806.923 kJ

Enthalpy of Methanol = 170.2×1.526×(364.8-298)

+ 170.2 × 1097.856

=204204.73 kJ

Enthalpy of water = 168 ×1.88 ×(364.8-298)

+ 168 × 2363.87

=418228.27 kJ

Total enthalpy of vapour = 626239.93 kJ

60

CONDENSER HEAT DUTY:

Enthalpy of distillate:

The distillate is assumed to be at its bubble point (81.4°C)

Enthalpy of Acrolein = 1.46×2.37×(338-298)

= 138.408 kJ

Enthalpy of Methanol = 39.8×3.73×(338-298)

= 5938.16 kJ

Enthalpy of water = 39.28×4.29×(338-298)

=6740.448 kJ

Total enthalpy of distillate = 12817.016 kJ

Enthalpy of recycled liquid:

The liquid is recycled at its bubble point ( 81.4°C)

Enthalpy of Acrolein = 4.788×2.37×(338-298)

= 453.902 kJ

Enthalpy of Methanol = 130.4×3.73×(338-298)

= 19455.68 kJ

Enthalpy of water = 128.7× 4.29×(338-298)

=22084.92 kJ

61

Total enthalpy of recycled liquid = 41994.502 kJ

Heat removed by condenser = Enthalpy of vapour – (Enthalpy of distillate + Enthalpy of

recycled liquid)

=626239.92 -(12817.016 + 41994.502)

=571428.4 kJ

Cooling water to condenser enters at 30°C and leaves at 60°C

Mass of cooling water

required = Qcd/(Cpw × ΔT)

= 571428.4/(4.18×(333-303))

=4556.8 kg/hr

REBOILER HEAT DUTY:

The residue is assumed to be at its bubble point (104.6°C)

Enthalpy of Glutaraldehyde in the residue = 125× 2.25×(377.6-298)

= 22387.5 kJ

Enthalpy of Methanol in the residue = 0.2 × 3.99 ×(377.6-298)

= 63.521 kJ

Enthalpy of water in the residue = 39.28× 4.49×(377.6-298)

=14038.829 kJ

62

Enthalpy of Methoxydihydropyran in the residue = 2.91 × 1.98 × (377.6-298)

=458.639 kJ

Total enthalpy of residue = 36948.489 kJ

Heat supplied to the reboiler =Enthalpy of distillate + Heat removed by condenser +

Enthalpy of residue- Enthalpy of feed

= 12817.016 + 571428.4 + 36948.489 – 80625.86

= 540568.03 kJ

For saturated steam at 7 atm, λ = 2064.9 kJ/kg

Amount of steam supplied = 540568.03/ λ

= 540568.03 /2064.9

= 261.7 kg/h

63

Table 28: Energy balance for distillation column II

COMPONENT ENTHALPY

OF THE

ENTERING

STREAM(kJ)

ENTHALPY OF THE

LEAVING STREAM (kJ)

TOP BOTTOM

Acrolein 253.35 138.408 0

Methanol 10726.8 5938.16 63.521

Methoxydihydropyran 399.659 0 458.639

Water 49708.051 6740.448 14038.829

Glutaraldehyde 19538.75 0 22387.5

Heat removed from

the condenser

571428.4

Heat supplied to the

reboiler

-

540568.03

Total 80626.61 80626.61

64

CHAPTER 5- PROCESS EQUIPMENT DESIGN

5.1 HIGH PRESSURE REACTOR DESIGN

5.1.1 PROCESS DESIGN

Order of Reaction is assumed to be 2.

Residence time = 2 hours

Table 29: High pressure reactor compositionCOMPONENT MASS

FRACTION(yi)

MASS (kg) MOLE FRACTION (xi)

DENSITY (g/cc)

Acrolein 0.4899 75.15 0.4993 0.6115

Methyl Vinyl Ether 0.5074 77.84 0.4993 0.2852

Hydroquinone 0.0026 0.4 0.0013 1.33

Average density of reaction mixture = 1/(∑ xi / ρi ) =1/ [(0.4993/0.6115) + (0.4993/0.2852) +

(0.0013/1.33)] = 0.3893 gm/cc

Volume of reaction mixture V0 = Total mass of mixture/ Average density

= ((75.15 + 77.84 + 0.4) × 1000)/(0.3893 × 10-6)

= 0.394 m3/hr

Volume of reactor = V0 × t

= 2 × 0.394

= 0.788 m3

65

Assuming an excess volume of 30%

Actual reactor volume = 1.3 × 0.788

= 1.024 m3

Assuming a L/D ratio of 1.5 for the reactor

Volume = (( π × Di2 × L)/4 )+ ((π × Di

3 ) /12) = 1.024 m3

Internal diameter of reactor Di = 0.892 m

Length of reactor L = 0.892 × 1.5

= 1.338 m

5.1.2 MECHANICAL DESIGN: [2]

SHELL DESIGN

The shell material is chosen to be C steel of tensile strength 4921 kg/cm2 with a safety factor of 1.5. Joint efficiency( J ) = 0.85

Internal pressure = 30 atm = 30.99 kg/cm2

Design pressure( P ) = 1.1 × internal pressure

= 1.1 × 30.99

= 34.09 kg/cm2

Working stress fs = 4921/1.5 = 3280.66 kg/cm2

Thickness of shell = (P x Dj )/ (2 × f × J – P)

= (34.09 x 0.892)/ (2 × 3280.66 × 0.85 -34.09)

= 5.48 × 10-3 m

= 5.48 mm

As the components are not corrosive in nature, a corrosion allowance of 0.75 mm in provided.

66

Actual thickness = 5.48 + 0.75

= 6.23 mm

COOLING JACKET DESIGN:

Table 30: Physical Properties of high pressure reactor componentsCOMPONENT VISCOSITY

μ (gm/cm-s)SPECIFIC HEAT CAPACITY Cp (kJ/kgK)

THERMAL CONDUCTIVITY K (W/m-k)

Acrolein 0.0014 3.684 0.0997

Methyl Vinyl Ether 0.00093 1.77 0.0773

Hydroquinone 1.13 1.785 0.0278

Amount of heat to be removed = 53214.87kJ

Average Viscosity of mixture = e∑ xi × ln(μ

i)

= e(0.4993 × ln (0.0014) + 0.4993 × ln(0.00093) +0.0013 × ln(1.13) )

= 1.15 × 10-3 gm/cm-s

Average molecular weight of mixture= ∑ xi × Mi

= (0.4993 × 56 + 0.4993 × 58 + 0.0013 × 110)

= 57.06 kg/kgmole

Average thermal Conductivity of = ∑ xi × Ki

Mixture

= (0.4993 × 0.0997 + 0.4993 × 0.0773 + 0.0013×0.0278)

= 0.088 J/s-m-K

Average specific heat of mixture = ∑ xi × Cpi

67

= (0.4993 × 3.684 + 0.4993 × 1.77 + 0.0013 × 1.785)

= 2.725 kJ/kg-K

Internal diameter of reactor = 0.892 m

Outside diameter of reactor = 0.904 m

Equivalent diameter = (Do2 - Di

2 ) / Di

= (0.9042 – 0.8922 )/0.892 = 0.024 m

Inside coefficients calculation:

Reynolds number (Nre ) = (D2 × N × ρ)/μ

= (0.8922 × 200 × 389.3)/ (1.15 × 10-3 × 60) = 897831.87

Prandtl Number (Npr ) = (Cp × μ)/K = (2.725 × 1.17 × 10-3 × 0.1)/(0.088 × 10-3) = 3.62

Height of impeller ht = Di/3

= 0.892/3

= 0.297 m

Inside heat transfer coefficient hi = 0.74 × ( Nre0.67 × Npr

0.33 × K)/ ht

= 0.74 × (897831.870.67 × 3.620.33 × 0.088)/ 0.297

68

=3265.67 W/m2-K

Outside coefficients calculations:

Reynolds number Nre = (De × v × ρ)/ μ

= (0.024 × 10 × 0.889 ×104)/ 0.0081 = 263407.4

The K value for water is 0.613 W/m2-K, Specific heat = 4.17 kJ/kgK

Prandtl number Npr = (Cp × μ)/ K = (4.17 × 0.0081× 0.1)/( 0.613 × 10-3 ) = 5.51

Centre line diameter of jacket = Dij + (Doj-Dij )/2 = 904 + (1004-904)/2 = 954 mm

Outside coefficient ho = 0.027 × (Nre0.8 × Npr

0.33 × ( 1+ ( 3.5×( De/Dc))) × K)/ ht

=0.027×( 263407.40.8 × 5.51.33 × (1+ ( 3.5×(0.024/0.954))) ×0.613) /0.297

= 2310.99 W/m2-K

Inside fouling factor ffi = 0.0004

Outside fouling factor ffo = 0.0009

69

Wall thickness (x) = 6 mm, Thermal conductivity of material (Km) = 16.02 W/m2-K

Overall Heat Transfer (U ) = [( (1/hi) + ffi + ( 1/ho) + ffo + (x/Km))]-1

coefficient

= [( 1/3265.67) + 0.0004 + (1/2310.99) + 0.0009 + (0.006/16.2)] -1

= 415.05 W/m2-K

Q = U × A × ΔT

Water enters at 50ºC and leaves at 30 ºC ΔT= (50-30)/ln(50/30)

= 39.15ºC

Heat transfer area (A) = 53214.87/(415.05 × 39.15)

= 3.27 m2

Length of jacket tj = 0.75 × L = 0.75 × 1.338

= 1.003 m

DESIGN OF HEAD

As the internal pressure is above 15 kg/cm2, we use an elliptical head.

The head is also constructed from C steel of tensile strength 4921 kg/cm2 with a safety factor of 1.5

Thickness of head = (P × D × V)/( 2 × f × J)

70

where V= (2+ k 2 ) , k- Ratio of major axis: minor axis = 2:1 6

V= (2 + 4)/6 = 1

Thickness of head = (34.09 × 89.2 × 1)/( 2 × 3280.66 × 0.85) = 0.54cm

= 5.4 mm

Corrosion allowance = 0.75 mm

Actual thickness of head = 6.15 mm

GASKET DESIGN

Spiral edge wound metal gasket is used as it can withstand high pressure and temperature

Soft steel is used which has a gasket yield of 316 kg/cm2 and a gasket factor of 3.

Gasket Width Gw:

Go = [Y – m × P]0.5

Gi [Y – (m+1) × P]0.5

= [316- 3 × 34.09]0.5

[316-34.09 × (3+1)]0.5

= 1.0907Assuming Gi = 892 mm

Go= 892 × 1.09 = 972.28 mm

Gw = (Go- Gi)/2

= (972.2-892)/2

= 40.1 mm

71

Basic gasket seating width bo = Gw/2

= 40.1/2 = 20.05 mm

Effective gasket seating width b = 0.5 × b0.5

= 0.5 × 20.050.5

= 2.23 mm

Diameter of gasket load reaction G = Gi + (2 ×Gw)– 2 × b

= 892 + (2 ×40.1) – 2 × 2.23

= 967.74 mm

Bolt load calculations :

Gasket seating load at atmospheric conditions

Wm1=π× b × G × Y

= 3.14 × 2.23 × 967.74 × 316 × 10-2

= 21423.99 kg

Gasket seating load at operating conditions

Wm2 = H + HP

where H= (π × G2 × P)/4

= (π × 967.742 × 34.096 × 10-2 )/4

=250790.27 kg

HP = 2 × π × b × G × M × P

= 2 × 3.14 × 2.23 × 967.74 × 3 × 34.09 × 10-2

= 13867.29 kg

72

Hence Wm2 = 250790.27+13867.29

= 264657.56 kg

DESIGN OF BOLTS

The bolt is made from hot rolled C steel of tensile stress fb= 545.19 kg/cm2

For bolt area the higher value among Wm2 and Wm1 is chosen.

Bolt Area (A) = (Wm2)/ fb

= (264657.56)/ 545.19

= 485.44 cm2

No of bolts (N) = (G in cm)/2.5

= (96.77)/ 2.5 = 38.7

The no of bolts is always a multiple of 4.

Therefore the answer is rounded off to the nearest multiple of 4.

Actual no. of bolts = 40

Bolt diameter bd = [(A × 4 ) /(N × π)]0.5

= [(485.44 × 4)/ (40 × 3.14)] 0.5

= 3.93 cm

Bolt circle diameter (B.C.D) = Outside gasket diameter +12× Diameter of bolt + 12

= 967.74 + 12 × 39.3 + 12

= 1451.34 mm

73

= 145.1 cm

Moment of arm hg = 0.5 × (B.C.D – G)

= 0.5 × (145.1 – 96.77)

= 24.2 cm

DESIGN OF FLANGE

Flange thickness tf = G × ( P / k × f)0.5

Where k = 1/( 0.3 + ( 1.5 × Wm2 × hg)/ (H × G)) =1/(0.3+(1.5×264657.56×24.2)/(250790.27×96.77)) = 1.43 cm

Substituting k, tf = 96.77 × ( 34.09/( 1.43 × 3280.66))0.5

= 8.24 cm


Actual flange thickness ta = 8.24 + 0.075

= 8.31 cm

Bolt spacing = (2 × bd) + [(6 × ta)/( m + 0.5)]

= (2 × 3.93) + [(6 × 8.24)/ (2 + 0.5)]

= 27.63 cm

AGITATOR DESIGN[7]

Anchor agitator is chosen with an impeller speed of 200 rpm.

Di = internal diameter of reactor = 89.2 cm

74

Agitator Diameter (Da) = 0.9 × Di

= 80.28 cm

Agitator blade width (W) = 0.1 × Di

= 8.92 cm

Height of liquid (H) = Di

= 89.2 cm

Agitator submerged height (L) = 0.9 × Di

= 80.28 cm

Clearance between agitator andTank bottom (C) = 0.05 × Di

= 4.46 cm

Impeller Reynolds number Nre = (Da2 × ρ × N)/ μ

= (80.282 × 200 × 0.3893)/( 60 × 1.17 × 10-3)

= 7148122.96For anchor impeller

[7]Kt = 0.35

Np = kt = 0.35

Power required to drive agitator P = (Kt × N3 × Da5 × ρ)/( 75 × gc)

= (0.35 × 2003 × 0.80285 × 389.3)/( 75 × 603 × 9.8)

= 2.28 HP

Shaft design

The standard material used for most shafts are commercial steel with fs = 560 kg/cm2

Torque to be transmitted to shaft (T ) = (P × 4500)/( 2 × π ×100)

75

= 1.86 Nm

For steady load ks = 1

Design value of torque = ks × T

= 1.86 Nm

For solid shaft

fs × (π × d3)/ 16 = ks × T

Solving d = 5.54 mm

5.1.3 DESIGN SUMMARY:

Volume of Reactor =1.024 m3

Area of Heat Transfer =3.27 m2

Shell Internal Diameter =0.892 m

Thickness of Shell =6.23mm

Shell Length =1.338 m

Thickness of Jacket =50 mm

Length of Jacket =1.003 m

Head Thickness =6.15 mm

Pitch Circle Diameter =145.1 cm

Flange Thickness =8.31 cm

76

Diameter of Agitator Shaft =5.54 mm

Power Required =2.28 Hp

Height of Agitator above bottom =4.46 cm

MaximumSpeed =200rpm

5.2 DISTILLATION COLUMN I DESIGN

5.2.1 PROCESS DESIGN:[5]

The feed is at its bubble point temperature of 98.1º C.

Light key –Acrolein

Heavy key – Methoxydihydropyran


αi = ki / khk where αi = relative volatility of each component with respect to the heavy key.

Minimum number of theoretical stages

Nmin + 1 = [log((( xlkD × D)/(xhkD × D)) × (( xhkW ×W)/(xlkW ×W)))]/ log(αlk,av)

From Table 9 αlk, av = (αlkD × αlkW × αlkF)1/3

= (11.56 × 6.99 × 8.96 )1/3

= 8.98From Table 8

Nmin + 1 = [log((0.293/0.137)/(0.979/0.02))]/log(8.99)

= 2.11

77

Nmin = 1.11

Number of stages(N) = ((0.75 × (1- (( R-Rmin))) 0.566) + Nmin) ____________(R + 1)________________ (1-0.75 × ( 1- (( R-Rmin))) 0.566 ) (R + 1)

= (0.75 × ( 1- (( 0.78-0.65))) 0.566) + 1.11) ____________(0.78 + 1)________________ (1-0.75 × ( 1- (( 0.78-0.65))) 0.566 ) (0.78 + 1)

= 5.5

Therefore no of stages = 6

Table 31: Physical properties of distillation column II componentsCOMPONENT VISCOSITY

μ (gm/cm-s)LIQUID DENSITY (kg/cc)

Acrolein 0.0078 783

Methyl Vinyl Ether 0.0066 676

Methoxydihydropyran 0.009 948

Efficiency η = 10(1.67 + 0.3 × log10

(L/V) – 0.25 × log10

(μl

× αF

))

Average viscosity of feed mixture μl = e(0.038× ln(0.0066) + 0.038 × ln(0.0078) + 0.92 × ln(0.009))

= 9.01× 10-3 g/cm-s

Relative volatility of light key in feed αF = 8.96

Assuming equimolal liquid and vapor flow, L=V

Efficiency η = 10(1.67 + 0.3 × log10

(L/V) – 0.25 × log10

(0.009 × 8.96))

=87 % Actual no of stages = 6/0.87

78

=6.8

Therefore actual no of stages = 7

CALCULATION OF FEED STAGE:

ln( Nd/ Nr) = 0.206 × ln [( W × xHKf × (xLKr)2 )/ ( D × xLKf × (xHKd)2 )] Nd + Nr = Actual no of plates = 7

Solving both equations

Nr = 3

Nd = 7 - 4 = 3

The feed plate is located at the 4th plate from the top.

VAPOR LOAD CALCULATION:

Average molecular weight of vapor Mv = ∑ xi ×Mi

= (56 × 0.292) + (58 × 0.57) + (114 × 0.137)

=65.05

Temperature T = 349.3 K

Pressure P =1.03 kg/cm2

R=8.314 J/kg-K

Density of vapor ρv = (P × Mv)/ (R × T)

= (1.03 × 65.05)/ (8.47×10-5× 1000 × 349.3) =2.26 kg/m3

Vapor load V = (D × (R+1) × Mv)/ (ρv)

79

= ( 0.092 × (0.7822+1) ×65.05)/ (2.26)

= 4.71 m3/hr

Density of liquid ρl = 1/(∑ xi / ρi )

= 1/[(0.293/783) + (0.57/676) + (0.137/948)]

= 734.26 m3

COLUMN DESIGN:

Assume a plate spacing ps of 0.9m

Maximum allowable vapor velocity C = ( (-0.171×ps

2) +( 0.271× ps)-0.047)×((ρl- ρv)/ ρv)0.5

= 6.03

Diameter of column Dc = (4 x V)/ (π × C × ρv) = (4 x 4.71)/ (π × 6.03 × 2.26)

= 0.44 m

Height of column Hc = ((Anp -1) +2) × ps

=((7-1) + 2) × 0.9

= 7.2 m

Column area Ac = (π × Dc2)/4

= (π× 0.442)/4 = 0.152 m2

80

Downcomer area Da = 0.12 × Ac

= 0.12 × 0.152

=0.018 m2

Active area Aca = Ac-(2 × Da)

= 0.152 - (2 × 0.018)

=0.116 m2

5.2.2 MECHANICAL DESIGN [2]

SHELL DESIGN:

The shell is made of mild steel of tensile stress f= 931.5 × 105 N/m2

Operating pressure (op) = 1.11 × 105 N/m2

Thickness of shell ts = (op × Dc)/ ((2 × f × 0.85)-op)

=(1.11×105 ×0.44)/((2×931.5×105×0.85)- 1.11× 105)

= 0.3 mm

Shell should have minimum thickness of 2 mm

Hence we take the thickness of shell =2mm


Therefore actual thickness of shell = 2 + 0.75 = 2.75mm

HEAD DESIGN:

A torispherical head is chosen for the given operating conditions

The head is made of mild steel of tensile stress fh = 931.5 × 105 N/m2

81

Crown radius Rc = Dc = 0.44 m

Knuckel radius Rk = 0.08 × Rc = 0.035 m

Thickness of head th =(op × Dc × M)/ (2 × fh × 0.85)

Where M = 0.25 × ( 3 + (Rc/ Rk)0.5)

= 0.25 × ( 3 + (0.44/0.035)0.5)

= 1.63

Therefore substituting in the expression for thickness of head th = 1 mm

Minimum thickness of head should be 2 mm

Hence the thickness of head is taken as 2 mm


Therefore actual thickness of head = 2 + 0.75 = 2.75mm

GASKET DESIGN:

For this operating condition, asbestos is taken as the material.

Asbestos has a gasket yield of 1.11 × 106 N/m2 and a gasket factor of 2.

Gasket Width Gw:

Go /Gi = [Y – m × op]0.5

[Y – (m+1) × op]0.5

= [(1.11 × 10 6 )- 3 ×1.11 × 10 5 ] 0.5

[1.11 × 106 - 1.11 × 105× (3+1)]0.5

= 1.005 Assuming Gi = 0.44 m

Go= 0.44 × 1.005 = 0.442 m

82

Gw = (Go- Gi)/2

= (0.442-0.44)/2

= 0.001 m

Minimum gasket width must be 10 mm

Therefore the gasket width is taken to be 10 mm

Basic gasket seating width bo = Gw/2

= 0.01/2 = 0.005 m

Effective gasket seating width b = 0.5 × b0.5

= 0.5 × 0.0050.5

= 0.035 m

Diameter of gasket load reaction G = Gi + (2 × Gw) – 2 × b

= 0.94 + (2×0.1) – 2 × 0.035

= 0.89 m

BOLT LOAD CALCULATIONS :

Gasket seating load at atmospheric conditions

Wm1= π × b × G × Y

83

= π × 0.035 × 0.89 × 1.11 × 106

= 1086900 N

Gasket seating load at operating conditions

Wm2 = H + HP

where H= (π× G2 × op)/ 4

= π × 0.892 × 1.11 × 105

=69092 N

HP = 2 × π × b × G × M × op

= 2 × π × 0.035 × 0.89 × 2 × 1.11 × 105

= 43913 N

Hence Wm2 = 69092 +43913

= 113005 N

DESIGN OF BOLTS :

The bolt is made from hot rolled C steel of tensile stress fb= 534.6 × 105 N/ m2

For bolt area the higher value among Wm2 and Wm1 is chosen.

Bolt Area (A) = Wm1/ fb

= 1086900/534.6 × 105

= 0.0203 m2

No of bolts (N) = (G in cm)/ 2.5

84

= 89/2.5

= 35.6

The no of bolts is always a multiple of 4.

Therefore the answer is rounded off to the nearest multiple of 4.

Actual no. of bolts = 36

Bolt diameter bd = ( A × 4 )/ (N × π ) = (0.0203 × 4)/( N × π ) = 0.026 m

Bolt circle diameter (B.C.D) = Outside gasket diameter + 12 × Diameter of bolt + 0.0012

= 0.945 + 12 × 0.026+ 0.0012

=1.27 m

Moment of arm hg = 0.5 × (B.C.D – G)

= 0.5 × (1.27 – 0.89)

= 0.19 m

DESIGN OF FLANGE :

Flange thickness tf = G × ( op / k × f)0.5

Where k = 1/( 0.3 + ( 1.5 × Wm1 × hg)/ H × G) = 1/( 0.3 + ( 1.5 × 1086900 × 0.19)/ 69092 × 0.89) = 1.19

85

Substituting the value of k, tf = 0.89 × (1.11 × 105 /( 1.19 ×1.11 × 106 ))0.5

= 0.037 m =3.7 cm

DESIGN OF CONDENSER :

From energy balance heat load of condenser Q = 23556.16 kJ/hr

Water flow rate W = 281.77 kg

Volumetric flow rate of water (V) = W/1000

= 281.77/1000

= 0.281 m3/hr

Let N be the total number of tubes

Assuming two passes on the tube side, no of tubes per pass = N/2

By thumb rule Dic = 0.75 × Doc

Assuming Doc= 0.0195 m, Dic = 0.0146 m

Let velocity of coolant water (VC) be 2 m/s

Cross-sectional area of tube (At ) = (π × Dic2)/4

= (π × 0.01462)/4

= 1.6 × 10-5 m2

No of tubes ( N ) = (V × 2)/ (At × VC × 3600)

= (0.281 × 2)/ ( 1.6 × 10-5 × 2 × 3600)

86

= 5

LMTD (ΔTlm) = (76.3-30.4)/(ln(76.3/30.4))

= 49.87 ºC

Heat transfer coefficient (U) = 850 W/m2 ºC

Heat transfer area (Ah ) = Q/ (U × ΔTlm) = (23556.16)/( 850 × 49.87) = 0.55 m2

Outside tube surface area/metre (As) = π× Doc

= π× 0.0195

= 0.0612 m

Length of each tube (L) = (Ah/( As × N))

= (4.18)/(0.0612 × 5)

= 1.79 m

Tube bundle diameter (Bd) = ( Doc) × (N/k)1/n

For triangular pitch k=0.156,n=2.291 Substituting the values for k and n, Bd = 0.162 m

Shell diameter is 10% excess of tube bundle diameter

Shell diameter (Sdc) = 1.1 × 0.162

87

= 0.178 m

MECHANICAL DESIGN :

Design pressure Pd = 1.13 kg/cm2

fc = 1250 kg/cm2

Shell thickness = (Pd × Sdc)/ ((2 × fc × 0.85)- Pd)

= (1.13× 0.178)/((2 × 1250 × 0.85)- 1.13)

=0.00009 m

Minimum shell thickness should be 2 mm

Hence shell thickness is taken as 2 mm


Therefore actual shell thickness for condenser = 2 + 0.75 = 2.75mm

Baffle spacing = (Shell thickness/5)

= (0.0027/5)

=0.54 mm

REBOILER DESIGN:

Table 32: Properties of distillation column I reboiler componentsComponent Specific heat

(kJ/kgK)Boiling point (ºC)

Latent Heat(kJ/kg)

Critical pressure(bar)

Acrolein 1.55 53 510.72 50

Methoxydihydropyran 2.105 127 323.94 40.1

88

Sensible heat of acrolein (Sha) = Cpa × ΔT

= 1.55 × 53

= 82.55 kJ/kg

Sensible heat of methoxydihydropyran (Shmdp) = Cpmdp × ΔT

= 2.105 × 127

= 267.33 kJ/kg

Total heat load =∑ [(Shi + λi) × mir/3600]

= [((82.55 + 510.74) × 1.46) + ((323.98 + 267.33) × 145.08)]/3600

= 24.12 kW

Maximum heat load = 1.05 × Total heat load

= 1.05 × 24.12

= 25.32 kW

At 5 atm, temperature of saturated steam is 180.9 ºC ( from steam table)

Average temperature difference ΔTr = ∑(Steam temp - Bpi)/2

= [(180.9-52) + (180.9-127)]/2 = 90.9 ºC

89

Outside area required Ao =(Maximum heat load × 1000)/ (U × ΔTr)

=(25.32 × 1000)/(750 × 90.9)

= 3.7 m2

Assuming internal diameter of tube (di) = 8mm , wall thickness (w) = 2 mm outside diameter of tube (do) = 0.012 m, Length (l) = 3 m

No of tubes (N) = Ao/(do × π × l)

=3.7/(0.012 × π × 3)

= 32

By tube layout, tube outside diameter = 0.095 m

Shell diameter = 2 × 0.095

= 0.19 m

Heat flux based on estimated area Qa = Maximum heat load/ Ao

= 25.32/3.7

=6.81 kW

MECHANICAL DESIGN :

Design pressure Pdr = 1.13 kg/cm2

fr = 1250 kg/cm2

Diameter of vessel (Sdr)= 0.19 m

Shell thickness = (Pd × Sdr)/ ((2 × fr × 0.85)- Pd)

90

= (1.13× 0.19)/((2 × 1250 × 0.85)- 1.13)

=0.0001 m




Therefore actual shell thickness for reboiler = 2 + 0.75 = 2.75 mm


= (0.00275/5)

=0.54 mm

5.2.3 DESIGN SUMMARY

Minimum Reflux Ratio =4.3800

Actual Reflux Ratio =5.2560

Number Of Trays =7

Plate Spacing =0.9 m


Shell Length =7.2 m

Crown Radius =0.44 m

Knuckle Radius =0.035 m

Thickness Of Head =2.75 mm

Flange Thickness =3.7 cm

91

Pitch Circle Diameter =127 mm

Number Of Bolts =36

Bolt Diameter =0.026 m

Condenser Tube Number =5

Tube Length =1.79 m

Tube bundle diameter =0.162 m

Shell Thickness =2.75 mm

Shell Diameter =0.178 m

Baffle Spacing =0.54 mm

Reboiler Tube length =3m

No Of Tubes =32

Shell Diameter For Reboiler =0.19 m

Shell Thickness =2.75 mm

Baffle Spacing =0.54 mm

5.3 DESIGN OF CONTINUOUS STIRRED TANK REACTOR

EQUIPMENT DESCRIPTION:

The two types of flow reactors are the continuous stirred tank reactor (CSTR) and

the plug flow reactor (PFR). A CSTR, as name suggests, is a reactor in which its contents

92

are well stirred and uniform throughout. For all design calculations, the exit stream

concentration is taken as the concentration inside the reactor.

For a required conversion, a PFR requires less volume compared to the CSTR.

Yet, the CSTR is preferred to the PFR as uniform removal of heat is difficult in the later.

PURPOSE AT HAND:

The main purpose of this CSTR is to convert the Dihydropyran into

Glutaraldehyde by reacting with excess water.

OBJECTIVES OF DESIGN:

The main objectives of the design are to find out the following:

Volume of the reactor VR

Diameter and Height of the reactor dR, hR

Area of heat transfer for the jacket, Areq

ASSUMPTIONS:

The reaction is assumed to follow first order kinetics with a rate equation as follows:

-rA=k.CA kmol/m3.s

The height to diameter ratio (h/d) of the reactor vessel is assumed to be 2.

93

5.3.1 PROCESS DESIGN:

DATA USED:

Table 33: Data for continuous stirred tank reactor process design

Operating Temperature 50C

Operating Pressure P 760mm Hg

Gas constant R 8.314kJ/K

Conversion XA 0.98

Heat to be removed in the jacket 7253.131kJ

Temperature difference (approach) 25C

DESIGN METHODOLOGY & CALCULATIONS:

Initially, basic calculations to find the average liquid density and average

molecular weight are performed.

Table 34: Physical properties of Continuous stirred tank reactor components

Components xi

Cp,

kJ/kg

,

kg/ms

K,W/

mk , kg/m3 Ln()

Acrolein 0.0023 2.174 0.00026 0.1486 806.7 -8.25

Water 0.8861 4.037 0.00055 0.6374 984.6 -7.50

94

Maleic Acid 0.0008 1.2519 0 0.164 993.4 0

MethoxyDihydrapyran 0.1108 1.684 0.000394 0.1833 965.6 -7.83

1

Average viscosity of the mixture:

Ln(i) = .xi. Ln (i)

=(0.0023×-8.25)+(0.8862×-

7.50)+(0.0008×0)+(0.1108×-7.83)

= -7.8392

i = 0.000532 kg/ms

Average heat capacity of the mixture = xi.Cpi

=(0.0023×2.174)+(0.8861×4.037)+(0.0008×1.251)

+ (0.1108×1.684)

=3.769 kJ/kg

Average thermal conductivity =xi.Ki

=(0.0023×0.148)+(0.8861×0.637)+(0.0008×0.164)

+(0.1108×0.183)

95

=0.5856 W/mk

Average density of the mixture :

1/I =xi / i

=(0.0023×806.7)+(0.8861×984.6)+(0.0008×993.4)

+(0.1108×965.6)

=0.001

I =981.974 kg/m3

The basic design equation of a CSTR is as follows:

= V/VO =CAO. XA /-rA

= XA/k.(1- XA)

= 5 hrs

FO = 331kg/hr

VO = FO/L

= 331/975.535

= 0.339m3/hr

Volume of reactor V = .VO = 5x0.339 = 1.697m3

Allowance volume is calculated as 20% excess to the reactor volume to accommodate the

agitator.

Total volume VR = 1.2×V = 1.2×1.697 = 2.04m3

96

The diameter and height of the reactor are calculated from the reactor volume as follows:

Height-Diameter ratio, h/d = 2

Equation for volume is VR = .d2.h/4

Inner Diameter of reactor, dSi3 = VR.4/

dSi = 3(VR.4/)

= 3(2.04×4/3.14)

= 1.374m

Height of reactor, h = 2xd = 2x1.374 = 2.748m

5.3.2 MECHANICAL DESIGN:[2]

DATA USED:

Shell and jacket

Material of construction = Stainless steel

Shell internal diameter, dSI =1374mm

Shell length, lS =2748mm

Joint efficiency, J =85 %(shell),100%(flange, nozzle)

Permissible stress, fS =120N/mm2

97

Operating pressure, P =0.101325N/mm2

Operating temperature, T =50C

Head-shallow dished head with flange


Head external diameter, do =1374mm

Crown radius, rc= do /2 =687mm

Knuckle radius, rk=6% of do =82.44mm

Flange for head and shell


Nominal diameter, dN =1374mm

Gasket

Material of construction = asbestos

Gasket internal diameter, dGi =1374mm

Gasket yield =260kg/cm2

Agitator

Diameter of agitator, dA =dSi/3

98

=1374/3

=458mm

Height of agitator

above bottom, ha =dSi/3

=1374/3

=458mm

Number of blades =6

Length of blades, lA = dSi/4

=1374/4

=343.5mm

Width of blades, wA = dSi/5

=1374/5

=274.8mm

Maximum speed, NA =220rpm

Permissible stress, fA =55N/mm2

Support skirt:

99

Material of construction = Carbon steel

Skirt height hskirt =1200mm


The mechanical design involves determining the thickness of the vessels and jackets used

taking into account the operating pressure and temperature.

The design pressure and temperature are calculated from the operating conditions to

accommodate any surges in the same during operation.

Design pressure, PD =1.1×P

=1.1×0.101325

=0.11146N/mm2

Design temperature, TD =1.1×T

=1.1×50

=55C

SHELL AND JACKET DESIGN:

100

The length of the jacket used for transferring heat is usually chosen as 75% of the length

of the vessel. This may be appended as per requirements.

Length of the jacket, lJ =0.75×lS

=0.75×2748

=2061mm

Usual practice is to provide an allowance of 10mm

Effective length of jacket, lJeff =lJ+10

=2061+10

=2071mm

Thickness of the shell and jacket are calculated using the formula given below:

t =PD.di/(2.f.j-PD) mm

Thickness of shell, tS =0.11146×1374/((2×120×0.85)-0.11146)

=0.75mm

A minimum thickness of 2mm is a must. In case of stainless steel an additional thickness

of 0.5mm for corrosion allowance is provided.

101

Thickness of shell, tS =2+0.5

=2.5mm

Outer diameter of reactor, dO =dI+tS

=1374+2.5

=1376.5mm

Heat transfer area:

The area required for heat transfer is calculated as below:

Q =Uo.A.ΔT

, Areq =Q/Uo. ΔT

Equivalent diameter, dE =(dSO2-dSI

2)/di2

=(1376.52-13742)/13742

=0.004m

Calculation of Inside Heat transfer coefficient:

NRe=di2.N./ ; N=Agitator speed =220 rpm

NRe =1.3742×220×981.974/(0.000532×60)

=12777128.21

102

NPr=Cp./K

NPr =2.985×0.000532×1000/0.5856

=2.711

Inside heat transfer coefficient is given by the formula,

hi=0.74×NRe0.67×NPr

0.33×K/ha

hi=0.74×(12777128.21) 0.67×(2.711)0.33×585.6/0.458

=75904.9 W/m2K

Calculation of outside heat transfer coefficient:

NRe=de.v./ ; v=1.5m/s

NRe =0.004×1.5×889/0.00081

=6585.19

The K value for water is 0.613 W/m2-K, Specific heat = 4.17 kJ/kgK

Npr = (Cp × μ)/ K

= (4.17 × 0.00081)/( 0.613 × 10-3 )

= 5.51

103

Outside heat transfer coefficient is given by the formula,

ho=0.27×NRe0.8×NPr

0.33×(1+(3.5×(de/di)×(K/ha)))

ho=0.27×6585.190.8×5.510.33 ×

(1+3.5×(0.004/1.374))×(0.000613/0.458)

=727.44 W/m2K

For stainless steel,

Inside Fouling friction factor Ffi=0.00004

Outside Fouling friction factor, Ffo=0.00009

Overall heat transfer coefficient is given by the formula,

U=1/((1/hi)+Ffi+(1/ho)+Ffo+(ts/K)

Wall thickness (x) = 2.5 mm, Thermal conductivity of material (Km) = 16.02 W/m2-K

U=1/((1/75904.9)+0.00004+(1/727.44)+

0.00009+(0.0025/16.02))

=597.405 W/m2K

104

Areq =4753.23/(597.405 x30)

=0.265 m2

Internal jacket diameter, dji =dsi+25

=1399

1400mm

Thickness of jacket, tJ =0.11146×1400/((2×120×0.85)-0.11146)

=0.75mm

A minimum thickness of 2mm is a must. In case of stainless steel an additional

thickness of 0.5mm for corrosion allowance is provided.

Thickness of jacket, tJ =2+0.5

=2.5mm

HEAD-SHALLOW DISHED HEAD WITH FLANGE DESIGN:

A shallow dished head is employed as the pressure conditions are moderate.

Stress intensification factor, W =[3+(rc/rk)0.5]/4

=[3+((687/82.44)0.5]/4

105

=1.471

Thickness of head, th =PD.rc.W/(2.f.J)

=0.11146×687×1.471/(2×120×1)

=1.31mm

A minimum thickness of 2mm is a must. In case of stainless steel an additional thickness

of 0.5mm for corrosion allowance is provided.

Thickness of head, th =2+0.5

=2.5mm

Flange length on the head is usually 3 times the thickness of the head.

Height of flange, lF =3.th

=3×2.5

=7.5mm

Minimum height is 20mm

lF =20mm

Flange for head and shell:

The flange is made up of stainless steel with a steel lining in the form of a ring. The

gasket is made of asbestos. The other data used are obtained from standard tables for the

corresponding material of gasket.

106

Gasket external diameter dgo:

dGO/dGI = (Gasket yield-(m.PD))/(yield-(PD(m+1)))0.5

=(260-(2×0.11146))/(260-(0.11146×(2+1)))0.5

=1.0008

dGO =1374×1.0002

=1374.2mm

Gasket factor, m =2.00

Minimum design seating stress Ya =11.2N/mm2

The following calculations are performed to find out the load on the bolts and hence

determine the number of bolts and bolt dimensions.

Basic gasket seating width, b =(dGo-dGi)/4

=(1374.2-1374)/4

=0.05mm

Effective gasket seating width, b =2.5(bo)

=2.5×0.05

107

=0.559mm

Diameter of gas loading reaction, G =(dGo+dGi)/2=(1374.2+1374)/2=1374.1mm

The minimum bolt load at atmospheric conditions and design pressure and

temperature are calculated as follows:

Bolt load at atmospheric conditions:

Wm1 =.b.G.Ya

=3.14×0.559×1374.1×11.2

=27013.31N

Bolt load at design conditions:

Wm2 =(.2.b.G.m.PD)+( .G2.PD/4)

=(3.14×2×0.599×1374.1×2×0.11146)+

(3.14×1374.12×0.11146/4)

=165743.49N

Permissible stress on bolts, fB =58.70 N/mm2

108

CSA of bolt w.r.t Wm1,

Am1 =Wm1/fB

=27013.31/(58.70×100)

=4.60cm2

CSA of bolt w.r.t Wm2,

Am2 =Wm2/fB

=165743.49/(58.70×100)

=28.32cm2

The number of bolts required, NB =G/(2.5×10)

=1374.1/(2.5×10)

=54.9

55

Diameter of bolt, dB =( Am2.4/. NB)

=(28.32×4/(55×3.14))x10

=8.099mm

Pitch circle diameter of the bolt, B =dGO+2db+12

109

=1374.2+2×12+12

=1402.4mm

Outer diameter of flange, dF =B+2dB

=1402.4+2×18

=1418.6mm

Permissible stress fF =95.2N/mm2

Radial distance from gasket, hg =(B-G)/2=(1402.4-1374.1)/2=14.17mm

Hydrostatic end force, H =.G2.PD/4

=3.14×1374.12×0.11146/4

=165205.83N

Factor k =1/[0.3+(1.5.Wm.hg/H.G)]

=1/[(0.3+(1.5×165743.49×14.17/(165205.83×

1374.1)]

=3.16

A corrosion allowance of about 20% is provided in the thickness calculations.

Flange thickness, tf =(G.PD/K.fF)+c

110

=(1374.1×(0.11146/(3.1.6×95.2))0.5)×1.2

=33.7 mm

Nozzle for head:

The thickness of the nozzles provided on the head is calculated as follows using the

following formula:

tn = PD.di/(2.f.j-PD) mm

Table 35: Nozzle design for continuous stirred tank reactor

Nozzle Type

d N,

(mm) fN,(N/mm2) t N,(mm) tNmin ,(mm) c,(mm) tN,(mm)

Feed-Top 90 130 0.038 2 1 3

Outlet-Bottom 90 130 0.038 2 1 3

Jacket inlet &

outlet 50 130 0.021 2 1 3

Baffle design:

Baffle thickness, tbaffle =da/12

=458/12

=38.2mm

AGITATOR DESIGN:

111

From the data for agitator, Reynolds number is calculated. This is then used to find the

Power number from the plot of Re vs NP. The Power required is estimated as below:

Reynolds number, Re =.N. da2/

=(981.974×220×0.4582/(60×0.000532)

=141968.02

Impeller type: six blade turbine

The corresponding value of power number is found to be 6.2 for six blade turbine.

Power number, NP =P.gc/.N3.da5

=6.2

Power, P =Np.( .N3.da5)/gc

=6.2×(981.974×2203×0.458)/(9.81×603)

=1401.91W

=1.88 hp

A gland loss of 10 % and a transmission system loss of 20% are to determine the total

power requirement.

Total power, Preq =1.2×(1.1×(1.88))

=2.48 hp

112

Based on the power required, the diameter of the shaft is calculated as follows:

Continuous torque, Tc =P×750×60/2..N

=2.48×750×60/(2×3.14/220)

=80.78N.mm

Maximum torque, Tmax =1.5×Tc

=1.5x80.78

=121.18N.mm

Polar modulus, Zp =Tmax×1000/f

=121.18×1000/55

=2203.28mm3

Zp =.dshaft3/16

Diameter of shaft dshaft=3(Zp×16/) =(2203.28×16/3.14)

=22.39mm

22mm

113


Time for Conversion =5 hr

Volume of Reactor =2.0358 m3

Area of Heat Transfer =0.265 m2


Thickness of Shell =2.5mm

Shell External Diameter =1.3765m

Shell Length =2.748m

Jacket Internal Diameter =1.4m

Thickness of Jacket =2.5mm

Length of Jacket =2.061m

Effective Length of Jacket =2.071m

Head External Diameter =1.374m

Crown Radius =0.6870m

Knuckle Radius =0.0824m

Pitch Circle Diameter =1402.4mm

Height of Flange =20 mm

Outside Diameter of Flange =14.186m

Flange Thickness =33.7 mm

Diameter of Agitator Shaft =22mm

Power Required =2.48 hp

Height of Agitator above bottom =0.4580m

Number of Blades =6

114

Length of Blades =0.3435m

Width of Blades =0.2748m

5.4 DESIGN OF DISTILLATION COLUMN II

EQUIPMENT DESCRIPTION:

A multi-component distillation column is used to effectively separate

components, usually liquids in industries, based on the difference in their boiling

points. A minimum temperature difference of 30-50C is essential to carry out the

separation.

115

PURPOSE AT HAND:

The main purpose of the distillation column-2 is to separate Glutaraldehyde

from the lower boiling component viz. Methanol.

OBJECTIVES OF DESIGN:

The main objectives of the design are to find out the following:

Minimum reflux ratio and Actual reflux ratio Rm, R.

Minimum and Theoretical Number of Stages Nm , Ntheo.

Actual number of plates N, from efficiency.

Flooding velocity Uf.

Column diameter Dc.

Column height Hc.

ASSUMPTIONS:

Liquid-Vapor flow rate is assumed to be constant through out the column.

The efficiency of the sieve plates used in the column is assumed to be 85%.

The space between the individual plates inside the column is assumed to be

approximately 8m.

5.4.1 PROCESS DESIGN:[7]

DATA USED:

Table 36: Data for distillation Column II process design

Feed Temperature 95C

Distillate Temperature 91.8C

Residue Temperature 104.6C

116

Heavy Key (Lk) Water

Light Key (Lk) Methanol

Operating Pressure P 760mm Hg

Gas constant R 8.314kJ/K

Plate Efficiency 85%

Plate Spacing 0.8m


The Fenske-Underwood method is employed to design the multi-component distillation

column.

Feed is sent at its bubble point temperature.

Hence, q=1

Relative Volatility calculations for feed at 95C:

Table 37: Relative volatility calculations for feed of distillation column II

compound (pi) (xfi) ki=pi/p α=ki/kc Xf*Ki

Acrolein 2561.842 0.002 3.3708 3.9662 0.007684

Methanol 304.19741 0.108 3.0022 3.5325 0.32616

Water 86.114398 0.779 0.8499 1 0.661398

MDP 277.82546 0.002 0.3656 0.4301 0.00081

GA 3.7437755 0.108 0.0369 0.0435 0.003948

117

0.999 1

Relative Volatility calculations for distillate at 91.8C:

Table 38: Relative volatility calculations for distillate of distillation column II

compound (pi) (ydi) ki=pi/p α=ki/kc Xd/Ki

Acrolein 2323.8 0.0043 3.0576 4.1553 0.0014

Methanol 268.04 0.3615 2.64538 3.5951 0.13664

Water 74.558 0.6343 0.73583 1 0.86196

1 1

Relative Volatility calculations for residue at 104.6C:

Table 39: Relative volatility calculations for residue of distillation column II

compound (pi) (xbi) ki=pi/p α=ki/kc Xb*Ki

Methanol 403.93259 0.00077 3.9865 3.39681 0.0031

Water 118.91517 0.84084 1.1736 1 0.9868

MDP 381.02841 0.00317 0.50135 0.42719 0.0016

GA 5.5664658 0.15521 0.05494 0.04681 0.0085

0.99999 1

Underwood's Equations:

The Underwood's method is used to find out the Minimum Reflux ratio.

∑((αi*xfi)/(αi-Ф))=1-q

{(3.966*0.002)/(3.966-Ф)} + {(3.532*0.108)/(3.532-Ф)} + {(1*0.779)/(1-Ф)} +

{(0.403*0.002)/(0.403-Ф)} + {(0.0434*0.108)/(0.0434-Ф)} = 1-1

118

The value of Ф can be determined by iterating the above equation.

Therefore, Ф = 2.69

Substituting in the equation below to find Minimum Reflux ratio,

∑((αi*xdi)/(αi-Ф))= Rm+1

{(3.966*0.004)/(3.966-2.69)} + {(3.532*0.361)/(3.532-2.69)} + {(1*0.634)/(1-2.69)}

= Rm+1

On solving, we get Rm+1=3.73 Rm=2.73

R= 1.2 x Rm R= 3.28

Fenske’s Equation:

Nm= ln[(xd/xb)lk (xd/xb)hl]/ln[lk/hk] - 1

The Fenske’s equation is used to find the minimum number of theoretical stages

Nm+1 =ln[(0.3603/.0008)(0.6321/0.8408)]/ln[3.532/1]

Nm = 4.098

The number of theoretical stages is found out by the following correlation:

(R-Rm)/(R+1) =(3.28-2.73)/(3.28+1)

119

=0.128

(N-Nm)/(N+1) =0.75×[1-((R-Rmin)/(R+1))0.566]

=0.75×[1-(0.128)0.566]

(N-Nm)/(N+1) =0.516

(N-4.127)/(N+1) =0.516

N =10.882 11

Efficiency η = 85%

Actual no of stages =11/0.85

=12.9

13

Location of feed tray:

Nr/Ns=ln[(xd/xf)lk/(xd/xf)hk]/ln[(xf/xb)lk/(xf/xb)hk]

=ln[(0.3603/0.109)/(0.6321/0.778)]/ln[(0.109/0.0008)/(0.778/0.8408)]

=0.281

Nr =0.281Ns

120

N =Nr+Ns

=13

Ns=4; Nr=9

Thus the feed is introduced into the fourth plate from the top.

Flow rate of the distillate stream D =80.54 kg/hr

Flow rate of the reflux stream L = D × R

=80.54 × 3.28

=263.9 kg/hr

Flow rate of the vapor stream V = D (R+1)

=80.54 × (3.28 +1)

=344.4 kg/hr

The average liquid and vapor densities are calculated for the next stage of calculation as

follows:

Table 40: Properties of distillation column II components

Components xi , kg/m3 Mi

Acrolein 0.002 757.12 56

Methanol 0.109 704 32

121

Water 0.778 961.2 18

MethoxyDihydropyran 0.002 925.566 114

Glutaraldehyde 0.109 1106 100

1 757.12

Average density of Liquid L = xi. i

=(0.002×757.12)+(0.108×704)+(0.779×961.2)

+(0.002×925.566)+(0.108×1106)

=947.62 kg/m3

Average density of Vapor v = P ( xi. Mi)/RT

=101.325×[(0.002×56)+(0.109×32)+(0.779×18)+

(0.002×114)+(0.108×100)]/(8.314×368.6)

=101.325 × 28.618/ (8.314 × 368.6)

=0.946 kg/m3

Now, the flooding velocity is determined in order to calculate the diameter of the

distillation column. The plate spacing can be taken between ranges of 0.5 – 1.5m.

Plate spacing ls =0.8m (assumed)

Flooding velocity, uf =(-0.171ls2 + 0.271ls – 0.047)((l-v)/ v) m/s

=((-0.171× 0.8 2+0.27×0.8–0.047) ((947.414-

0.954/0.954))

122

=1.9092 m/s

Maximum velocity, umax = 0.9 x uf

=0.9 x 1.9092

=1.7183 m/s

Liquid flow rate, VLO =FL/(x3600) m3/s

=501.574/(0.946 x 3600)

=0.1472m3/s

The active area inside the column available for flow of the fluid, down comer area, holes

area, free area and the cross-sectional area are calculated as follows:

Net Area, Anet =VLO/umax

=0.1472/1.7183

=0.0857 m2

Percentage Hole area =10%

Active Area, Aactive = Anet x (1-0.1)

=0.0857 x0.9

=0.0771 m2

123

Free Area, Afree =Anet-Aactive

=0.0866-0.0771

=0.0086m2

Percentage Down comer Area= 10%

Column CSA, Ac = Aactive-2(0.1xAactive)

=0.0779-2(0.1x0.0771)

=0.0617m2

The column diameter and height are calculated from the column CSA as follows:

Using the free area, the number of holes on the sieve trays is calculated as follows:

Number of holes, Nhole =Free area/Area of each hole

Diameter of a hole =5mm

Nhole =0.0.0086/(×(5×103)2/4)

=0.0086/(3.14×(5×103) 2/4)

=436.65

437

124

Column Diameter, dc =(4xAc)/

=(4x0.0617/3.14

=0.3134m

Column Height, hc =[(N-1)+2]×ls

=[(13-1)+2]×0.8

=12.8m

5.4.2 MECHANICAL DESIGN:[2]

DATA USED:

Shell and Jacket:

Material of construction =carbon steel

Shell internal diameter, dsi =313.4mm

125

Shell length, ls =12800mm

Joint efficiency, J =85%(shell),100%(flange)

Permissible stress, fs =120N/mm2

Operating pressure, P =0.101325N/mm2

Operating temperature, T =95C

Head – Elliptical head with flange:

Material of construction =Carbon steel

Head external diameter, do =313.4mm

Crown radius, rc =156.7mm

Knuckle radius, rk =6% of do

=18.806mm

Flange for head and shell:


Nominal diameter, dn =313.4mm

Gasket:

Material of construction =Asbestos

126

Gasket internal diameter, dgi =313.4mm

Gasket yield =260kg/cm2

Trays-sieve type:

Number of Trays Nt =13

Plate spacing, ls =800mm

Support skirt:


Skirt height hskirt =1800mm


The mechanical design involves determining the thickness of the vessels and jackets

used taking into account the operating pressure and temperature.

The procedure and formulae used are narrated below:

The design pressure and temperature are calculated from the operating conditions to

accommodate any surges in the sane, during operation.

127

Design pressure, Pd =1.1×P

=1.1×0.101325

=0.11146N/mm2

Design temperature, Td =1.1×T

=1.1×95.6C

=105.16C

Shell design:

Thickness of the shell is calculated using the formula given below:

Ts =( Pd.ds)/(2.f.j- Pd)

=(0.11146×313.4)/(2×120×0.85-0.11146)

=0.0175mm

A minimum thickness of 2mm is a must.

An additional thickness of 0.5mm is provided for corrosion allowance since the

material of construction is carbon steel.

Thickness of the shell, ts=2+0.5=2.5mm

Head-Shallow dished head with flange design:

128

A shallow dished head is employed as the pressure conditions are moderate.

Stress intensification factor W =[3+(rc/rk)0.5]/4

=[3+(156.7/18.806)0.5]/4

=1.47

Thickness of head, th = Pd ×rc×W/(2×f.j)

=0.11146×156.7×1.47/(2×120×0.85)

=0.126mm

A minimum thickness of 2mm is a must.

An additional thickness of 0.5mm is provided for corrosion allowance.

Thickness of head, th =2.5mm

Height of flange, lf =3× th

=3×2.5=7.5mm

Flange for head and Shell:

129

The flange is made up of stainless steel with a steel lining in the form of a ring.

The gasket is made of asbestos. The other data used are obtained from standard tables for

the corresponding material of the gasket.

Gasket factor, m=2.00

Minimum design seating stress Ya=11.2N/mm2

The following calculations are performed to find out the load on the bolts and hence

determine the number of bolts as well as the bolt dimensions.

Gasket external diameter dgo:

dgo /dgi =(Gasket yield-(m. Pd))/(yield-( Pd (m+1)))0.5

=(260-(2×0.11146))/(260-(0.11146×(2+1)))0.5

=1.0008

dgo =313.4×1.0002

=313.5mm

Basic gasket seating width, bo =( dgo- dgi)/4

=(313.5-313.4)/4

=0.025mm

Effective gasket seating width, b =2.5bo

=2.5×0.025

130

=0.395mm

Diameter of gasket loading reaction =( dgo+d dgi)/2

=(313.5+313.4)/2

=313.45mm

The minimum bolt load at atmospheric conditions and design pressure and temperature

are calculated as follows:

Bolt load at atmospheric conditions:

Wm1 =.b.G.Ya

=3.14×0.395×313.45×11.2

=4357.8N

Bolt load at design conditions:

Wm2 =(.2.b.G.m. Pd)+(.G2. Pd /4)

=(3.14×2×.0.395×313.45×2×0.1114)+

(3.14×313.452×0.1114/4)

=8771.3N

Permissible stress on bolts fB=58.7N/mm2

131

CSA of bolt w.r.t Wm1, Am1=Wm1/fB=4357.8/(58.70)=74.23mm2

CSA of bolt w.r.t Wm2, Am2=Wm2/fB=8771.3/(58.7)=149.42mm2

The number of bolts required, Nb =G/(2.5×10)

=313.45/(2.5×10)

=12.53

13

Diameter of bolt, db =(Am2×4/(Nb×)

=((149.42 ×4)/(13×3.14))

=3.862mm

Calculation for pitch circle diameter is as follows:

B= dgo +2db+12=313.5+2×3.862+12=333.164mm

Outer diameter of flange, dF= B+2db=333.164+2×10.23=340.817mm

The next step is to estimate the thickness of the flange.

Permissible stress, fF=95.2N/mm2

132

Radial distance from gasket, hg =(B-G)/2=(333.164-313.45)/2=9.8434mm

=10mm

Hydrostatic end force, H =.G2. Pd /4

=3.14×313.452×2×0.11146/4

=8597.9N

Factor K =1/[0.3+(1.5.Wm.hg/H.G)]

=1/[0.003+(1.5×8771.3×10/(8597.9×313.45))]

=2.873

A corrosion allowance of about 20% is provided in the thickness calculation.

Flange thickness tF =G.( Pd /K.fF) + c

=313.45×(0.11146/(2.873×95.2)) × 1.2

=2.064mm

Nozzles:

The thickness of the nozzles provided is calculated using the formula below:

133

tn= Pd.di/(2.f.j- Pd) mm

Table 41: Nozzle design for distillation column II

Nozzle Type

dN

(mm) fN (N/mm2) tN (mm) tNmin (mm) c (mm)

tN

(mm)

Feed 90 95 0.52 2 1 3

Vapor Outlet 90 95 0.52 2 1 3

Reflux 90 95 0.52 2 1 3

Reboiler Feed 90 95 0.52 2 1 3

Liquid Outlet 90 95 0.52 2 1 3

Trays sieve type:

Usually a hole diameter of 5mm and a plate thickness of 3mm is chosen in the

case of stainless steel. In this case, the specifications of the sieve trays and hence the

down comer area as well as the number of holes is calculated as follows:

Hole diameter, dhole =5mm

Plate thickness, tp =5mm

Downcomer area, Adown =(Column CSA-Active Area)/2

=(0.0617-0.0771)/2=0.077m2

134

Free Area, Afree=0.0866m2

Number of holes, Nhole =Free area/Area of each hole

=Afree/(.dhole2/4)

=4×0.0866×106/(3.14×52)

=437

CONDENSER:

Heat duty on Condenser, Qc =571428.412 kJ/hr

Water flow rate,W = 455.248 kg/hr

Volumetric flow rate of water (V) = W/1000

= 455.248 /1000

= 0.455 m3/hr

Let N be the total number of tubes

Assuming two passes on the tube side, no of tubes per pass = N/2

By thumb rule, Dic = 0.75 × Doc

Assuming, Doc = 0.019 m

Dic = 0.0143 m

135

Let velocity of coolant water (VC) be 2 m/s

Cross-sectional area of tube (At ) =(×Dic2)/4;

=(3.14×0.01432)/4

=0.000160m2

Number of tubes = (V×2)/(At×Vc)

=(0.455×2)/(0.000160×2)

=28.43

=29

LMTD (ΔTlm) = (91.8-81.4)/(ln(91.8/81.4))

= 86.50 ºC

Heat transfer coefficient (U) = 850 W/m2 ºC

Heat transfer area (Ah ) =Qc/(U×ΔTlm)

=571428.412 /(850×86.50)

=0.78m2

136

Outside tube surface area/metre (As) = × Doc

=× 0.019

= 0.0596 m

Length of each tube (L) = (Ah/( As × N))

=(0.78)/(0.0596×29)

= 0.451 m

Tube bundle diameter (Bd) = ( Doc) × (N/k)1/n

For triangular pitch k=0.156,n= 2.291

Substituting the values for k and n, Bd = 0.183 m

Shell diameter is 10% excess of tube bundle diameter

Shell diameter (Sdc) = 1.1 × 0.183

= 0.2014 m

fc = 55 N/mm2

137

Shell thickness = Pd×Sdc / ((2 × fc × 0.85)- Pd)

= (0.11146×173)/((2×95×0.85) – 0.11146)

=0.206 mm




Therefore actual shell thickness for condenser = 2 + 0.5 = 2.5mm


= (0.0025/5)

=0.5 m

REBOILER:

Heat duty on Reboiler, Qr= 540567.305 kJ/hr

Table 42: Properties of distillation column II reboiler components

Components M, kg

Cp,

kJ/kgk Tb, C v, kJ/kg Pc, bar Sh,kJ/kg

Methanol 0.2 1.5988 65 1097.856 80.84 103.922

138

Water 121.891 1.8984 100 2363.87 220.55 189.84

MethoxyDihydropyra

n 2.91 2.1085 127 323.94 40.108 267.7795

Glutaraldehyde 125 2.339 188 460.245 35.9 439.732

Sensible heat (Sh) of Mathanol = Cpa × ΔT

= 1.598 × 65

= 103.87kJ/kg

Similarly, Sensible heats for all components are found out.

Total heat load =∑( Sensible heat + λi) × mir / 3600

=((103.922+1097.85)×0.2)+((189.84+2363.87)×

121.891) +((267.77+323.94))×2.91)+

((439.732+460.245)×125))/3600

= 118.126 kW

Maximum heat load = 1.05 × Total heat load

= 1.05 × 118.126

= 124.032 kW

139

At 6 atm, temperature of saturated steam is 219.95 ºC ( from steam table)

Average temperature difference ΔTr = ∑(Steam temp - Bpi)/2

= [(219.95-65) + (219.95-100)+( 219.95-127)+

( 219.95-188)]/2

= 199.89 ºC

Outside area required Ao =(Maximum heat load × 1000)/ 850 × ΔTr

=(124.032× 1000)/ 850 ×199.89

= 0.73 m2

Assuming internal diameter of tube (di) = 8mm , wall thickness (w) = 2 mm

outside diameter of tube (do) = 0.012 m, Length (l) = 3 m

No of tubes (N) = Ao/(do × × l)

=0.73/(0.012 × 3.14× 3)

= 6.54

=7

140

Shell diameter = 2 × 0.012

= 0.024 m

Shell thickness = Pd×Sdr / ((2 × fc × 0.85)- Pd)

= (0.11146×24)/((2×95×0.85) – 0.11146)

=0.165 mm




Therefore actual shell thickness for condenser = 2 + 0.5 = 2.5 mm


= (0.0025/5)

=0.5 m


Flooding Velocity =1.9092m/s

Minimum Reflux Ratio =4.3800

Actual Reflux Ratio =5.2560

Number Of Trays =13

141

Plate Spacing =0.8000m

Plate Thickness =5mm

Hole Diameter =5

Number Of Holes =437

Shell Internal Diameter =0.3134m

Shell Length =9.6000m

Head External Diameter =0.3134m

Crown Radius =0.1567m

Knuckle Radius =0.0188m

Thickness Of Head =2.5mm

Flange Thickness =2.5mm

Height Of Flange =7.5mm

Flange Outside Diameter =340.817mm

Pitch Circle Diameter =333.1640mm

Number Of Bolts =13

Bolt Diameter =3.8265mm

Condenser Tube Number =29

Tube Length =0.451 m

Tube Internal Diameter =0.0143m

Tube External Diameter =0.0190m

Tube bundle diameter =0.183m

Shell Thickness =2.5mm

Shell Diameter =0.2014m

Baffle Thickness =0.5000m

142

Reboiler Tube length =3m

Tube Internal Diameter =0.0080m

Tube External Diameter =0.0120m

No Of Tubes =7

Shell Diameter For Reboiler =0.0240m

Shell Thickness =2.5mm

Baffle Spacing =0.5000m

5.5 PRELIMINARY DESIGN :

5.5.1 COOLER

The cooler inlet temperature is 180ºC. Its outlet temperature is the bubble point of the

mixture i.e. 98.1ºC.Water is used as the coolant which enters at 30ºC and leaves at 60ºC

143

Heat removed by cooler (Q) =19215.35kJ/hr

Amount of water required (m) = 153.23 kg/hr

Logarthmic mean temperature differenceΔTlm = ((180-30)-(98.1-60))/ Log((180-

30)/(98.1-60))

= 81.67 ºC

Assuming overall heat transfer coefficient of 700 W/m2 ºC

Heat transfer area = Q/(U × ΔTlm)

= (19215.35)/ (700 × 81.67)

=0.33 m2

Inside diameter of tube = 0.014 m

Outside diameter of tube = 0.016 m

Length of tube =1.95 m

Surface area of one tube = π × do × L

144

= π × 0.016 × 1.95

= 0.098 m2

No of tubes N = (Heat transfer area)/(Surface area of one tube)

= 0.33/0.098=4

CHAPTER 6 -COST ESTIMATION

6.1 ESTIMATION OF EQUIPMENT COST : [19] [3] [11]

The details of the price of the equipment purchased for the process are tabulated below:

145

Table 43: Equipment cost

S.No Equipment Quantity

Cost of

Equipment(Rs.) Total Cost(Rs.)

1 Distillation Column 1 1 17,33,000 17,33,000

2 Distillation Column 2 1 20,45,000 20,45,000

3 High Pressure Reactor 1 14,48,000 14,48,000

4 CSTR 1 9,76,000 9,76,000

5 Cooler 1 2,00,000 2,00,000

6 Storage Tank 5 1,00,000 5,00,000

Total 65,02,000 69,02,000

The total purchase equipment cost PEC =Rs. 69,02,000

The delivered cost of the purchased equipment is calculated as 10% of the purchased

equipment cost as follows:

Purchased equipment delivered cost PED = 1.1*PEC

= Rs.75,92,200

The purchased equipment delivered essentially consists of the following

1. Fabricated equipment

2. Process machinery

3. Pumps

4. Valves

146

6.2 ESTIMATION OF CAPITAL INVESTMENT

The various categories contributing to the direct costs are calculated based on the

purchased equipment delivered.

The corresponding factors for Liquid Gas operations and the direct costs are tabulated as

follows:

Table 44: Direct costs

Category Total Cost

(Rs.)1 Purchased equipment delivered 75,92,2002 Purchased equipment installation [47% of PED] 35,68,334

3 Instrumentation and controls (installed) [36% of PED]

27,33,192

4 Piping (installed) [68% of PED] 51,62,696

5 Electrical system (installed) [11% of PED] 8,35,142

6 Buildings(including services) [18% of PED] 13,66,596

7 Yard improvement [10% of PED] 7,59,220

8 Service facilities (installed) [7% of PED] 5,31,454

Total 2,25,48,834

The indirect costs are calculated and listed as follows:

Table 45: Indirect costs

S.No Category Total Cost

(Rs.)

1 Engineering and supervision [32% of PED] 24,29,504

147

2 Construction expenses [34% of PED] 25,81,348

3 Legal expenses (installed) [4% of PED] 3,03,688

4 Contractor’s fees [19% of PED] 14,42,518

5 Contingency [37% of PED] 28,09,114

Total 95,66,172

Fixed Capital Investment FCI = Direct plant costs + Indirect plant costs

=2,25,48,834+95,66,172

=Rs.3,21,15,006

Working Capital Investment WCI = 15% of Total Capital investment

=3,21,15,006×0.15/(1-0.15)

=Rs.56,67,354

Total Capital Investment TCI = Fixed capital investment + Working capital

investment

=3,21,15,006 + 56,67,354

=Rs. 3,77,82,360

148

6.3 ESTIMATION OF RAW MATERIALS COSTS

The quantity of raw materials used and the corresponding costs have to be calculated in

order to proceed with the calculations to determine the total product cost.

Estimation of total raw materials cost:

Table 46: Raw material costs

S.No Utility Quantity/Year Cost/kg (Rs.) Total Cost

(Rs.)

1 Acrolein 450900 1540 69,43,86,000

2 VinylMethylEther 467040 2010 93,87,50,400

3 Hydroquinone 1000 1132 11,32,000

4 Water 1104000 0.06 66,240

5 Maleic Acid 1000 1432 14,32,000

TOTAL 163,57,66,640

6.4 ESTIMATION OF TOTAL PRODUCT COSTS:

The total product cost is composed of the following heads:

Manufacturing Costs:

149

1. Direct production costs

2. Fixed Charges

3. Plant overheads

General expenses

1. Administrative costs

2. Distribution and selling costs

3. Research and development costs

The direct product cost consists of several heads of which the operating labor and

utilities are calculated as follows:

Estimation of operating labor:

Man-hours requirements per 1000 kg of product manufactured = 5

Annual production of product =15,00,000 kg

Therefore, Man-hours required annually =3000

Total cost per Man-hour =Rs.10000

Therefore, Cost of operating labor = Rs.3,00,00,000

Estimation of Utilities

Table 47: Utilities cost

S.No Component Quantity/Year Cost/1000kg

(Rs.)

Total

Cost(Rs.)

150

1 Water – Cooling 21107200 8 1,69,000

2 Water – Process 1118080 15 16000

3 Electricity 100000 7 7,00,000

4 Waste Disposal 12800000 25 3,20,000

Total 12,05,000

Now the remaining heads are calculated as percentages. The Total direct production

costs are determined as follows on an annual basis:

Table 48: Total direct production costs


(Rs.)

1 Raw material cost 163,57,66,640

2 Operating labor 3,00,00,000

3 Operating supervision [15% of operating cost] 45,00,000

4 Utilities cost 12,05,000

5 Maintenance and Repairs [6% FCI] 19,26,900

6 Operating supplies [15% of maintenance cost] 2,89,035

7 Laboratory charges 48,00,000

8 Patents and Royalties 1,00,000

Total 167,85,87,575

Estimation of Fixed Charges

Table 49: Fixed charges estimation

151


(Rs.)

1 Local Taxes [4% of FCI] 12,84,600

2 Insurance [1% of FCI] 3,21,150

Total 16,05,750

The plant overhead is calculated based on the operating labor as 50% of the operating

labor costs, operating supervision and maintenance costs as shown:

Plant Overhead costs = 0.5 × (3,00,00,000 + 45,00,000+19,26,900)

=Rs.3,64,26,900

Therefore, Manufacturing cost = Direct production costs + Fixed charges + Plant

overhead costs

Manufacturing cost =167,85,87,575+16,05,750+3,64,26,900

=Rs.171,66,20,225

Therefore, The Total product cost TPC = Manufacturing cost/0.8

=Rs.214,57,75,281

The General expenses are calculated as follows:

Table 50: General expenses

152


(Rs.)

1 Administrative costs [20% of Operating labor] 60,00,000

2 Distribution and Selling costs 1,58,16,000

3 Research and Development costs 2,37,35,000

Total 2,41,89,500

The total income is calculated based on the sales of the manufactured product and the

revenue obtained from sale of by-products recovered from the process.

Quantity of Glutaraldehyde produced annually = 15,00,000 kg

Selling price of Glutaraldehyde acid = Rs.1500/kg

Revenue from sale of product annually = 15,00,000 ×1500

= Rs. 225,00,00,000

Income from sale of by-products and recovered raw materials are tabulated below:

Table 51: Income estimation

153

S.No Component Quantity/Year Cost/1000kg

(Rs.)

Total

Cost(Rs.)

1 Water – Cooling 1203200 2.5 30,08,000

2 Water – Process 1118080 2 22,36,160

3 Methanol 240000 30 72,00,000

Total 1,24,44,160

Total annual income = 225,00,00,000+12444160

= Rs.2262444160

Annual Gross earnings = Total income – Total Product cost

= 2262444160-214,57,75,281

= Rs.116668879

Considering and accommodating an income tax of 35%, we can estimate the annual

gross earnings after taxes as follows:

Annual gross earnings =Annual gross earnings×(1-0.35)

=116668879×(1-0.35)

=Rs.7,58,34,771

6.5 DETERMINATION OF PAYBACK PERIOD :

For an evaluation period of ten years, the net profit i.e., net profit after tax is

calculated:

154

Net profit = Annual gross earnings – ((FCI/10)*(1-0.35))

=7,58,34,771-((32115006/10*(1-0.35))

=Rs.73747295

The annual depreciation in the fixed capital invested is estimated by using

conventional straight line method of depreciation. The straight line method

accommodates for the deprecation uniformly through out the service life of the

commodity.

The depreciation is estimated as follows assuming the salvage value is zero at the end

of service life, as follows:

Annual depreciation =FCI/service life

=32115006/10

= Rs.3211500

The payback period of the capital invested is estimated using a simple formula as

follows:

Payback period = Total direct expenditure(FCI)/(Net profit - Annual depreciation)

= 3778236/(73747295 – 3211500)

=1.53 years

155

CHAPTER 7- PROCESS INSTRUMENTATION AND CONTROL

7.1 INTRODUCTION

156

The piping and instrumentation (P & I) diagram is a pictorial representation of a process

plant depicting all the equipments along with the piping, valves, insulation and

instrumentation. There are standard symbols for most items in a P & I diagram. All

piping must be marked with line designations, symbols showing service, pipe size and

pipe specification. The P & I diagram should include:

1) Mechanical equipments with names and numbers

2) All valves and their identification

3) Process piping, sizes and identification.

4) Miscellaneous vents, drains, special fittings etc.

5) Flow directions.

6) Computer control system.

7.2 INSTRUMENTS

The different instruments used in the P & I diagram generally are Flow meters, Level

meters, Thermometers, Quality Analysis, Radiation measurement and Weight calculation.

The instruments are used for indicating, recording and controlling purposes. The

instruments are all identified by a code number. The first letter of the code refers to the

property measured. For Example, F for Flow meters, T for thermometers and L for Level

meters. The second letter is either I, R or C Which indicates to indicating, recording and

controlling respectively. Then the letters are followed by a number used to identify the

instrument uniquely amidst a number of similar instruments.

7.3 AIM:

1. To control the process variables so that they are within known safe operating

limits.

2. To maintain the product composition within the specified quality standards.

157

3. To detect dangerous situations and develop alarm and automatic shut-down

systems.

4. To operate at lowest possible production cost.

5. To achieve the desired product output.

7.4 PROCESS CONTROL [8] [4]

The variables that need to be controlled in chemical processes are temperature, pressure,

liquid level, flow rate, composition etc. Temperature is usually controlled by heat

exchange with a heat transfer medium. Pressure is usually controlled by regulating the

flow of effluent from the equipment vessel. Control of the effluent flow rate is the most

common method to regulate the liquid level.

7.4.1 CONTROL OF EQUIPMENTS

REACTORS

In reactors, as reaction rates are highly sensitive to temperature changes, temperature

control often dominates the design of the reactor. Temperature is controlled by means of

an external jacket and regulating the flow of the heating or cooling medium. A primary

requirement for the effective control of the reactor is to provide the reactants in the

appropriate ratio to get the desired composition of the product. This is achieved by means

of flow control of the entering reactants. Pressure controller maintains the pressure

constant inside the reactor.

DISTILLATION COLUMNS

Distillation columns have a large number of closed control loops and these are highly

interactive and depend on each other. For controlling the quality of one specified product,

the reflux ratio should be maximized and this is done by controlling the flow rates of the

158

reflux or vapor. The pressure of the tower is controlled by changing the amount of vapor

in the overhead.

CHAPTER 8 - SITE AND PLANT LAYOUT

8.1 INTRODUCTION: [8]

159

A suitable site layout has been designed for the manufacture of Glutaraldehyde.

Provision has been made for ancillary buildings and services needed for efficient plant

operation; and for the environmentally acceptable treatment and disposal of the effluent.

8.2 SITE LAYOUT:

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 future

expansion of the site. The ancillary buildings and services required on a site, in addition

to the main processing units will include:

Storages for raw materials and products; tank farms and warehouses

Maintenance workshops

Stores for maintenance and operating supplies

Laboratories for process control

Fire stations and other emergency services

Utilities: steam boilers, compressed air, water, power generation, transformer

stations

Effluent disposal

Offices for general administration

Canteens and other amenity buildings, such as medical centers

Car and heavy vehicle parking

When sketching the preliminary site layout, the process units will normally be sited

first and arranged to give a smooth flow of materials through the various processing steps

from raw material and final product stage. The location of the principal ancillary

buildings should then be decided. They should be arranged so as to minimize the time

spent by personnel in traveling in traveling between buildings.

160

Administration offices and laboratories, in which relatively large number of people will

be working, should be located well away from potentially hazardous processes. Control

rooms will normally be located adjacent to the processing units, but with potentially

hazardous processes may have to be sited at a safer distance. Utility buildings should be

sited to give the most economical run of pipes to and from the process units.

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

the process units they serve. Storage tanks containing hazardous materials should be sited

at least 70m from site boundary.

8.3 PLANT LAYOUT:

The economic construction of a process unit will depend on how well the plant

and equipment specified on the process flow sheet is laid out. The principal

considerations to be considered are:

Economic considerations: construction and operating costs

The process requirements

Convenience of operation

Convenience of maintenance

Safety

Future expansion

Modular construction

Costs:

Adopting a layout that gives the shortest run of connecting pipe between

equipment, and the least amount of structural steel work can minimize the cost of

construction. However, this will not necessarily be the best arrangement for operation and

maintenance.

161

Operation:

Equipment that needs to have frequent operator attention should be located

convenient to the control room. Valves, sample points and instruments should be located

at convenient positions and heights. Sufficient working place and headroom must be

provided to allow easy access to equipment.

Maintenance:

Heat exchangers need to be sited so that they can be easily cleaned. Equipment

that requires dismantling for maintenance, such as compressors and large pumps should

be placed under cover.

Safety:

Potentially hazardous equipment needs to be isolated and steps to onfine the

effects of explosion must be taken. At least two escape routes for operators must be

provided from each level in process buildimgs.

Plant expansion:

Equipment should be located so that it can be conveniently tied with any future

expansion of the process. Space should be left on pipe allays for future needs and service

pipes oversized to allow for future requirements.

CHAPTER 9- PROCESS SAFETY

9.1 INTRODUCTION

162

Process Safety generally refers to the prevention of unintentional releases of

chemicals, energy, or other potentially dangerous materials (including steam) during the

course of Chemical processes that can have a serious effect. Process safety involves, for

example, the prevention of leaks, spills, equipment malfunction, over-pressures, over-

temperatures, corrosion, metal fatigue and other similar conditions.

In most industries, the main concern is to ensure worker safety by machine

guards, moving load warnings and electrical isolation. Accidents rarely have any effect

on members of the public. However in process industries accidents can result in the

release of toxic materials or large amounts of energy with disastrous consequences for

workers and third parties. Releases from a chemical plant can go well beyond the site

boundary and can cause both long-term and short-term effects. However even in process

industries handling very dangerous materials, the majority of accidents are not related to

processes- they are largely trips, falls and dropped loads.

Much can be done to ensure safety by application of common sense and basic

engineering skills. As processes become more hazardous, however, the problems of

ensuring safe operations become even more complex, requiring the application of

specialist safety analysis methods. Such techniques can only be acquired by specific

training and experience. Process safety programs focus on design and engineering of

facilities, maintenance of equipment, effective alarms, effective control points,

procedures and training.

9.2 HAZARDS IN INDUSTRIES

A personal injury or accident occurs as a result of an accident. An accident occurs

as a result of unsafe actions or exposure to unsafe mechanical conditions. There are ways

of preventing such accidents from ever taking place.

163

Modern technology has been quite successful in developing tailor made

chemicals. However this effort has also introduced some additional problems since

manufacturing and handling experience is frequently inadequate to deal with hazards.

The ever-increasing production of flammable organics, the rush to bring in new products,

all extend the probability of hazards.

Toxic and corrosive chemicals, fire, explosion and fully mechanized equipment

are major hazards encountered in the operation of plants in chemical industries. The

design engineer must be aware of these hazards and must make every attempt to present

designs, which provide maximum protection for the plant personnel and minimum chance

for occurrence of accidents.

9.3 MATERIAL SAFETY DATA :

HEALTH EFFECTS:

Glutaraldehyde is a transparent colorless liquid. It causes irreversible eye damage and

skin burns. It may be fatal if swallowed. Some of its health effects include:

Liquid causes severe conjunctivitis and corneal injury which can permanently

impair vision if prompt first-aid is not provided. Vapor causes stinging sensations

in the eye with excess tear production.

Skin contact causes itching and brown coloration. Prolonged contact may lead to

severe pain followed by swelling with ulceration leading to tissue destruction. It

might also cause absorption of harmful amount of the material.

It is moderately toxic and its ingestion causes chemical burns in the mouth,

esophagus, throat and stomach. This further leads to nausea, dizziness, pain in the

chest and abdomen and might even cause coma.

164

Its vapor is an irritant to the respiratory tract causing stinging sensations in the

nose and throat, bleeding from the nose, coughing etc.

Chronic exposure results in cumulative dermatitis.

FIRST AID MEASURES:

Eye contact

Immediately flush eyes with water and continue washing for at least 15 minutes. Do not

remove contact lenses, if worn. Obtain medical attention without delay, preferably from

an ophthalmologist.

Skin contact

Immediately remove contaminated clothing and shoes. Wash skin with soap and water.

Obtain medical attention. Wash clothing before reuse. Discard contaminated leather

articles such as shoes and belt.

Inhalation

Move to fresh air. Give artificial respiration if not breathing. If breathing is difficult,

oxygen may be given by qualified professionals. Obtain medical attention.

Ingestion

Do not induce vomiting. Do not give anything to drink. Obtain medical attention

immediately.

FIRE EXTINGUISHING MEDIA:

165

Glutaraldehyde solution is non flammable. However once water evaporates, the

remaining material will burn. For large fires use of alcohol type fire extinguishers is

recommended. Use of carbon dioxide in sufficient concentrations can act as an

asphyxiant.

ACCIDENTAL RELEASE MEASURES:

Low concentrations of glutaraldehdye ( 5 ppm or less) can be degraded in biological

waste water treatment system. Small spills are washed with large amounts of water. In

case of large spills the material should be collected for disposal.Spilled material is

decontaminated by careful application of sodium hydroxide, ammonium or sodium

bisulfate.

HANDLING AND STORAGE:

Must not be used in the form of spray or aerosol.

Avoid breathing vapors and do not handle or empty in presence of flammable

vapor.

Wear goggles, protective clothing and gloves.

Wash thoroughly with soap and water after handling.

Remove contaminated clothing and wash before reuse.

Keep container closed and use adequate ventilation.

PERSONAL PROTECTION EQUIPMENT:

RESPIRATORY PROTECTION:

Self-contained breathing apparatus in high vapor concentrations.

166

PROTECTIVE GLOVES:

Polyethylene, Nitrile (NBR) or Butyl gloves

SKIN PROTECTION:

Protective chemical apron and rubber boots.

EYE PROTECTION:

Splash proof mono-goggles or safety glasses with side shields in conjunction with face

shield.

CONCLUSION

Glutaraldehyde is a highly reactive compound which finds great application as an

intermediate in the production of various important derived chemicals. It is primarily

167

used as a disinfectant and is also used as a fixative in tanning industries.

Glutaraldehyde is manufactured by various leading chemical companies like Sigma

Aldrich and Alfa Aesar, USA and Durotec, South Africa to name a few.

The Continuous production of Glutaraldehyde by the acid hydrolysis of 2 Methoxy 3,

4 Dihydropyran that is produced by the reaction between acrolein and Methyl vinyl

ether, has been dealt in detail. The material and energy balance calculations have

been performed to verify and account for the transfer of mass and energy throughout

the process.

The detailed process and mechanical design of four representative equipments

namely, High pressure reactor, two multi component distillation column and a

continuous stirred tank reactor has been dealt with extensively. The diagrams of the

equipments designed have also been presented.

The instrumentation employed in the operation and the control methodologies have

been outlined. This is followed by a site layout that has been proposed based on

several conditions to economize the manufacturing process. The economics of the

process has been worked and the project is considered feasible.

An overview of the safety aspect of the process has been made providing vital

material safety information on the chemicals handled in the process.

The design project for the manufacture of glutaraldehyde has been undertaken subject

to several assumptions. These must be looked into critically in order to implement he

process design proposed in this report.

REFERENCES

1. Bhatt, B.I. and Vora, Stoichometry, Tata Mcgrawhill publications.

168

2. Dawande, S.D., Process design of equipments, Second edition

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4. Perry, R.H. and D.W.Green, Perry’s Chemical Engineering Handbook,

Sixth edition Tata Mcgrawhill publications.

5 Robert E.Treybal , Mass-Transfer Operations, Chemical Engineering series,

Third Edition, Mcgraw hill-International editions.

6. Smith J.M., Van Hess H.C., Abbott, Michael M.,Introduction to Chemical

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14. US Patent 6559346 Jorg Therre, Worms (DE); Carsten Oost, Bad Durkheim

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16. www.nicnas.gov.au

17. ASPEN PLUS

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