1

1. INDUSTRY PROFILE

THE DCW STORY GOES back to 1925 when the foundation stone of India's first Soda

Ash factory at Dhrangadhra - a small principality in Gujarat in West India - was laid. The

plant was taken over in 1939 and run under the name Dhrangadhra Chemical Works. To

wish the venture luck, the company adopted the horse shoe as their corporate logo, which

stands till today, and is widely recognized as a symbol of excellence.

It is said that every oak begins as an acorn. The process of growth received a major

impetus in 1959 with the commissioning of the chlor-alkali plant at Sahupuram in the

southern state of Tamilnadu. At that time, the area was completely barren. Today this

complex has made its mark on the chemical map of India as well as the world.

Growth at the chlor-alkali complex was rapid as between 1965 and 1970 three plants

were erected that turned the co-product chlorine into a money spinner; a liquid chlorine

plant in 1965, the country's first Trichloroethylene plant in 1968 and an integrated PVC

resin plant in 1970-making the company one of the first in the nascent petrochemicals

field.

In the same year, 1970, the company set up a plant to manufacture upgraded ilmenite,

the first of its kind in Asia, and even today, one of the few of its kind in the world. In 1986,

to reflect the expanded activity spectrum, and its emergence as a multi-product and multi-

locational company, the corporate name was changed simply to DCW Ltd.

PRODUCTS INSTALLED

CAPACITY MT PRODUCTION MT

CAUSTIC SODA DIVISION

CAUSTIC SODA LYE 100000 77612

CAUSTIC SODA SOLID 155

CAUSTIC SODA FLAKES 25910

LIQUID CHLORINE 36000 18963

TRICHLOROETHYLENE 7200 4715

UPGR & W.G.ILM 48000 36394

UTOX 600 1612

IRON OXIDE 297

PVC DIVISION 90000 85758

Table 1.1 Industry Products

2

2. INTRODUCTION

2.1. HISTORY OF TRICHLOROETHYLENE

By 1896, work in Germany led to chlorinated solvents by partial or complete chlorination

of acetylene, and in 1908 to a full-scale plant producing 1,1,2-trichloroethene. Pioneered by

Imperial Chemical Industries in Britain, its development was hailed as an anesthetic

revolution. Originally thought to possess less hepatotoxicity than chloroform, and without

the unpleasant pungency and flammability of ether, TCE use was nonetheless soon found

to have several pitfalls. These included promotion of cardiac arrhythmias, too low volatility

for quick anesthetic induction, reactions with soda lime used in carbon dioxide absorbing

systems, prolonged neurological dysfunction when used with soda lime, and evidence of

hepatotoxicity as had been found with chloroform.

The introduction of halothane in 1956 greatly diminished the use of TCE as a general

anesthetic. TCE was still used as an inhalation analgesic in childbirth given by self-

administration. Fetal toxicity and concerns for carcinogenic potential of TCE led to its

abandonment in the 1980s.

Due to concerns about its toxicity, the use of trichloroethylene in the food and

pharmaceutical industries has been banned in much of the world since the 1970s.

Legislation has forced the substitution of trichloroethylene in many processes in Europe as

the chemical was classified as a carcinogen carrying an R45 risk phrase.

Fig 2.1 Chemical Structure of TCE

3

2.2. PHYSICAL PROPERTIES

CHEMICAL FORMULA CHCI = CCI2

PHYSICAL STATE AT ROOM TEMP liquid

APPEARANCE ALMOST colorless

FLASH POINT nonflammable

VAPOR PRESSURE AT 20°C 68mmHg

BOILING POINT @760mmHg 86.7°C

FREEZING POINT -87.1°C

MOLECULAR.WT 131.4

Sp.GRAVITY @ 20°C 1.464

Sp.HEAT @ 80°C 0.225cal/gm°C

VISCOSITY @ 20°C 0.58cP

REFRACTIVE INDEX@20°C 1.4782

Table 2.2.1 Physical Properties of TCE

4

2.3. CHEMICAL PROPERTIES

TCE is highly soluble in ether and alcohol and is miscible with most organic solvents while

being practically insoluble in water

Compound (1) decomposes to form dichloroacetyl chloride, which in the presence

of water decomposes to dichloroacetic acid and hydrochloric acid (HCl) with

consequent increases in the corrosive action of the solvent on metal surfaces.

Compound (2) decomposes to yield phosgene, carbon monoxide, and hydrogen

chloride with an increase in the corrosive action on metal surfaces.

It is not readily hydrolyzed by water. Under pressure at 150◦C, it gives glycolic acid

with alkaline hydroxides.

Liquid trichloroethylene has been polymerized by irradiation with 20-keV x-rays

Fig 2.3.1 Decomposed Product

5

2.4. PRODUCTION METHODS

Prior to the early 1970s, most trichloroethylene was produced in a two-step process from

acetylene. First, acetylene was treated with chlorine using a ferric chloride catalyst at 90 °C

to produce 1,1,2,2-tetrachloroethane according to the chemical equation

HC≡CH + 2 Cl2 → Cl2CHCHCl2

The 1,1,2,2-tetrachloroethane is then dehydrochlorinated to give trichloroethylene. This can

either be accomplished with an aqueous solution of calcium hydroxide

2Cl2CHCHCl2 + Ca(OH)2 → 2ClCH=CCl2 + CaCl2 + 2 H2O

or in the vapor phase by heating it to 300-500°C on a barium chloride or calcium chloride

catalyst

Cl2CHCHCl2 → ClCH = CCl2 + HCl

Today, however, most trichloroethylene is produced from ethylene. First, ethylene is

chlorinated over a ferric chloride catalyst to produce 1,2-dichloroethane.

CH2 = CH2 + Cl2 → ClCH2CH2Cl

When heated to around 400 °C with additional chlorine, 1,2-dichloroethane is converted to

trichloroethylene

ClCH2CH2Cl + 2 Cl2 → ClCH = CCl2 + 3 HCl

This reaction can be catalyzed by a variety of substances. The most commonly used

catalyst is a mixture of potassium chloride and aluminum chloride. However, various forms

of porous carbon can also be used. This reaction produces tetrachloroethylene as a

byproduct, and depending on the amount of chlorine fed to the reaction, tetrachloroethylene

can even be the major product. Typically, trichloroethylene and tetrachloroethylene are

collected together and then separated by distillation.

6

3. CHEMICAL PROCESS

3.1. VARIOUS PROCESSES

Chlorination of Acetylene

Chlorination of Ethylene

Oxychlorination of Ethylene or Dichloroethane

3.2. CHLORINATION OF ACETYLENE

Reaction involved

Step1.Chlorination

Step2. Dehydrochlorination

The above both reactions are exothermic reactions.

C2H

2 + 2Cl

2 C

2H

2 Cl4 ΔH= -402kJ/kmol

90°c

FeCl3

2C2H

2Cl

4 + Ca(0H)

2 2C

2HCl

3 + CaCl

2 + 2H

2O ΔH= -150kJ/kmol

80-85°c

7

3.3. REASONS FOR THIS PROCESS SELECTION

Selected process - Chlorination of Acetylene

Acetylene generated from CaC2 which easily produced from lime and Coal

Acetylene purity became 99.5% v/v

Higher yield of Tetrachloroethane which is feed for TCE

Lime slurry from Acetylene generation is reused for Dehydrochlorination

Higher equilibrium conversion of 95% without catalyst

Yield of TCE is 90% with 99.6% purity

3.4. PROCESS DESCRIPTION

3.4.1. RAW MATERIALS

Acetylene gas (C2H2)

Chlorine gas (Cl2)

Catalyst (FeCl3)

Stabilizers (Phenol, Thymol)

3.4.2. PROCESS PARAMETERS

Step1. Chlorination of Acetylene

Temperature : 80 to 90°C

Feed molar ratio : Acetylene to Chlorine, 1 : 2

Pressure : 1.1 bar

Step2. Dehydrochlorination of Tetrachloroethane

Temperature : 90°C

Pressure : 1.1 bar Feed

Molar ratio (lime to Tetrachloroethane): 1 : 2

8 Fig 3.4.1 PFD for TCE

9

3.4.3. STEPS INVOLVED IN PROCESS

The acetylene-based process consists of two steps

Step1. First acetylene is chlorinated to 1, 1, 2, 2-tetrachloroethane. The reaction is

exothermic (402 kJ/mol = 96 kcal/mol but is maintained at 80–90°C by the vaporization of

solvent and product. Catalysts include ferric chloride.

Here, Acetylene gas (C2H2) is dried (<40ppm water) with diluted sulphuric acid

(80%) and sent to the bottom of synthesis column with two molar times of Chlorine gas

(Cl2). But it‟s indirectly contacted with teterachloroethane medium that in liquid phase. So,

the reaction followed by absorption of gases with tetrachloroethane. The unreacted gases

synthesized in recovery column. Product is continuously discharged from the overflow line

which attached in the middle of reactor.

This reaction is highly exothermic in nature. So, the reaction mixture continuously

cooled by the open type heat exchanger upto 85°C. Finally product tetrachloroethane stored

in closed container because of their higher volatility and presents of free chlorine.

Step2. The product is then dehydrochlorinated to trichloroethylene at 96–100°C in aqueous

bases such as Ca(OH)2. The yield of trichloroethylene is about 94% based on acetylene. A

significant disadvantage of the alkaline process is the loss of chlorine as calcium chloride

(CaCl2).

In this section aqueous lime with 18% concentration that preheated to 80°C. After

preheating another reactant tetrachloroethane was fed continuously as per the theoretical

requirement. Here Tetrachloroethane (C2H2Cl4) is the limiting reactant. So lime is added

excess quantity.The crude Trichloroethylene obtained from the reactor top in vapor phase,

with the application of total condenser. Also crude product analyzed for it acidity nature

because of the dehydrochlorination reaction takes place.

Finally, the crude Trichloroethylene is charged in the fractionating column and the

product pure Trichloroethylene separated at the top of the column with 99.5% purity.

10

3.5 APPLICATIONS

o Metal cleaning, finishing and vapor degreasing

Cold, ultrasonic and vapor degreasing of metal parts between fabrication steps and

before finishing or assembly, surface preparation of metal parts for galvanizing,

anodizing, electro plating, painting and bonding.

For quick removal of oils and waxes in dip coating of rust preventing formulations,

the oil and grease covered parts are placed in an atmosphere of TCE vapor. The hot

vapor condenses immediately on the cold metal surface, dissolving the oils and

greases and flushing them. The work heats to the vapor temperature in minutes,

leaving the work clean, dry and ready for further treatment.

o Solvent extraction

Extraction of vegetable oils, waxes, animal fats, caffeine, cocoa butter, essential

oils, pharmaceuticals, for recovery of oils and grease from waste, rags and paper,

turnings etc.

Chemical Processes

In dehydration and purification of many chemicals, as a reactant for manufacture of various

halohydrocarbons and other chemicals, as an additive for promoting polymerization of

monomers etc…

Other uses

TCE is a good heat transfer medium in the temperature range of -73°C to +120°C

and is used in simulating out door conditions in material testing.

Used as a coolant for cutting tantalum and other special metals.

Used as a freezing point depressant, especially in fire extinguisher fluids.

Used in making rubber cements, bone glue and adhesive activators.

As a solvent for paints, lacquers, dyes, oils, fats, waxes, resins, halohydrocarbons

and many other chemicals.

11

4. MATERIAL BALANCE

4.1. BATCH REACTOR

Basis: 4380 kg of

Crude TCE (3000 liters)

2(167.8) 74.1 2(131) 111 2(18)

As per the reaction

335.6 kg of Tetra required to produce 262.7 kg of TCE

So for the production of 4380 Kg of TCE = (4380*335.6)/262.7

= 5595.46 kg

(Density of Tetra = 1.58) = 3.542 m3

But in actual process, they are fed 3600 liters of Tetrachloroethane (limiting reactant) for

3000 liters of crude TCE production in one Batch

Reactor conversion = (3.542 / 3.6)*100 = 98.4%

Amount of solid CaCl2 formed = (111/262.7)*4380

= 1850.7 kg

BATCH

REACTOR

4380 Kg of crude TCE

6864 Kg of 18% lime

5595 Kg of Tetra

15% of CaCl2 6230 Kg

2C2H

2Cl

4 + Ca(0H)

2 2C

2HCl

3 + CaCl

2 + 2H

2O ΔH= -150kJ/kmol 80-85°c

Fig 4.1.1 Batch Reactor

12

4.2. DISTILLATION

Here the crude Trichloroethylene contains the following components,

1. Trichloroethylene (TCE)

2. Dichloroethylene (vinyl dine chloride)-ligherends

3. Perchloroethylene (PERC)

Basis: 21 ton per day of pure TCE,

TCE required = 21000/24

= 875 kg/hr

In Fractionator 2

F1

F

F2

B

D

Fractionator 1

Freactionator 2

Fig 4.2.1 Fractionator column

13

Top product contains 99.4% TCE

Bottom contains 3% TCE

Feed contains 93.0% TCE

7.0% Perchloroethylene

TCE balance

0.93F2 = 875 + 0.03B ---------> 1

Top product obtains

0.994D = 875

D = 880.28

Therefore PERC in Distillate = 5.28 kg/hr

PERC balance

0.07F2 = 5.28 + 0.97B ----------->2

From equation 1 and 2

F2 = 942.87 kg/hr

B = 62.59 kg/hr

Fractionator 1

We assume,

All of Perchloroethylene goes in the bottom

Feed composition

TCE = 91.74%

14

Dichloroethylene = 1.65%

PERC = 6.60%

PERC balance

0.066 F = 0.07 F2

F = 1000 kg/hr

Product (TCE) Balance

0.917 F = 0.93 F2 + 0.1 F1 ---------3

From overall balance

F1 = 401.22 kg/hr

RESULTS FROM F1 and F2

Fractionators resulting the following composition

TCE from F1 = 40.12 kg/hr

Dichloroethylene from F1 = 361.1 kg/hr

Product (TCE) from F2 top = 875 kg/hr with 99.5% purity

Heavier ends (Perchloroethylene) = 1000 – 875

= 275 kg/hr

15

5. ENERGY BALANCE

5.1. BATCH REACTOR

Energy balance across a reactor

Basis: 8 hrs- batch

Heat capacity of reactants + Heat supplied form of steam + Heat of reaction (Exothermic)

= Heat capacity of products + Heat loses in reactor

Sensible Heat = m Cp ΔT

Latent Heat = m λ

(In reactor conversion is 98% only)

Take a basis for 8 hours for the time of one batch completion

3600 liters of Tetrachloroethane is converted in 3000 liters of Trichloroethylene

The respective density of the above compounds likely 1.58 and 1.48 gm/cc

ENERGY IN INPUT

Reactants Mass kmol Cp kJ/kmolK Heat Q, kJ

Tetrachloroethane 5760 34.28 0.97 27936

Lime 18% solution 6864 244.27 103.16 2.52×106

Steam supplied 532 28.5 40626 (λ) 1.20×106

Heat of

Reaction(Exothermic)

5760 34.28 150000 98%

conversion

5042

Total energy 3.75×106

Table 5.1.1 Reactor Energy Input

16

ENERGY OUT

Products MASS

kg

kmol Cp

kJ/kmolK

HEAT

Q, kJ

Crude Trichloroethylene 4440 33.4 121.62 1.4×106

Calcium chloride 23% solution 8079 205 137.9 2.12×106

Heat loss

(vapor)

102 5.6 40626 (λ) 0.23×106

Total

energy

3.75×106

5.2. ENREGY BALANCE ACROSS THE CONDENSER

Vapor from the reactor which enters the condenser at 90°C with 1 bar pressure. At the more

mole fraction X=0.91 of condensate we get both liquid and vapor in T-x-y diagram

Liquid Temperature = 85°C,

We take cooling water enters at 30°C and leaving at 42°C

Table 5.1.2 Reactor Energy Output

Fig 5.2.1 Vertical Condenser

17

Product mixture Mol Percentage

%

Heat Capacity Cp

kJ/kmolK

Trichloroethylene 92.0 120.1

Dichlroethylene 2.63 112.1

Tetrachloroethylene 5.27 140

Heat capacity of mixture

Cp(mix)

121.62

Basis: 1 batch = 4 hours

Amount of Heat removed = Heat loss to reduction vapor temperature

from 90°c to 85°c

+

Heat loss due to condensation (latent heat removal)

Q1 = m Cp (mix) ΔT

= 33.4 x 121.62 x (90-85)

= 20311.7 KJ/batch

= 5077.9 kJ/hr

Q2 = m λ

= 4440 x 0.236

= 1047.84 kJ/batch

= 261.96 kJ/hr

Total Heat to be removed = Q1+ Q2 = 5339.86 kJ/hr

Table 5.2.1 Heat Capacity of Vapor

18

TUBE SIDE - COOLING MEDIA

Cold water available at 30°c and we assume it leaves at 42°c

Heat removed = mass flow rate of water x Heat capacity

Mass flow rate of cold water (m) = Q / (Cp ΔT)

= 5339.86x103 / [4.18 x (42-30)]

= 29.57 kg/hr

5.3. ENERGY BALANCE FOR COOLER

30°C

46°C

40°C

85°C

Fig 5.2.2 Temperature Profile

Fig 5.3.1 Cooler

19

The Crude Trichloroethylene condensate from condenser which enters a cooler at 85°C

(358K) and leaves at 40°C (313K).Here the cool water passed in counter current direction

is available at 30°C (303K) and leaves 46°C (319K)

Heat Duty

Q = m Cp ΔT

m – Molar flow rate of fluid (kmol/hr)

Cp- Specific Heat capacity of mixer (kJ/kmolK)

ΔT- Temperature differences (K)

Qh = mh Cph ΔT

= 8.349 x 121.627 x (358-313)

= 45695.89 kJ/hr

The required cool water mass flow rate

Qc = mc Cpc ΔT

We have to remove heat using cool water,

So, Qh = Qc

mc = Qh

Cpc ΔT

= 45695.89

4.18 (319-303)

= 683.25 kg/hr

20

5.4. ENERGY BALANCE IN FRACTIONATOR

In Fractionator

F2 feed rate = 942.88kg/hr

Composition (weight) (Mol)

TCE 93% 94.37%

PERC 7% 5.6%

Total = 7.074 kmol

Amount of heat in Feed

F (Hf) = [(6.676×120.1) + (0.398×147)] × (88)

= 74759.5 kJ/hr

88°C

87°C

110°

C

Fig 5.4.1 Distillation Column

21

Condenser section

V (Hv) = D (HD) + L (HL) + Qc

From the above equation,

Qc = V × λ

V = (1+R) ×D

D = 6.686 kmol/hr

V = (1+2.17) × 6.686 we know that (R = 2.17)

= 21.194 kmol/hr

V × λ = 21.194 × 40266

= 861027.44 kJ/hr

Qc = 86.103 × 104 kJ/hr

Cooling water requirement = Qc / [4.18 × (42-30)]

= 1.716 × 104

kg/hr

Reboiler section

Over all heat balance

F (Hf) + Qb = Qc + D (HD) + W (Hw)

Qb = Qc + D (HD) + W (Hw) – F (Hf)

Qb = 15305.79 + 86.103×104

– 74759.5 = 801576.29 kJ/hr

Steam available at 3 kg/cm2

(pressure steam temperature 135°C)

From steam table λV = 2172 kJ/kg

Steam requirement rate ms = Qb/ λV = 801576.29 / 2172

= 359 kg/hr

22

6. DESIGN OF MAJOR EQUIPEMENT

6.1. CONDENSER

PROCESS DESIGN

(I) Preliminary Calculations:

(a) Heat Balance

Vapor flow rate (G) = 8.349 kmoles/hr.

= 1110 kg/hr

= 0.3083 kg/s

Vapor Feed Inlet Temperature =900c.

Let Condensation occur under sub cooling conditions i.e. FT = 0.8

Condensate outlet temperature = 850C

Average Temperature = 87.50C

Latent heat of vaporization (λ) = {31.4(0.026) + 33.3(0.92) + 35.0(0.0527)}

= 33.4 kJ/mol

Qh = (molar flow rate of condensing vapor) x (latent heat of vapor)

Qh = heat transfer by the condensing vapor

Qh = 8.349 k-moles/hr x 33.4kJ/mol x (1000/3600)

= 275517.1 kJ/hr

= 77.945 kJ/s

10% overload is taken

Qh = 1.1 x 77.945 = 85.74 kJ QC = mass flow rate of cold × specific × t

fluid heat

23

QC = heat transfer by the cold fluid.

Assume: Qh = QC.

Inlet temperature of water = 30 0C.

Let the water be untreated water.

Outlet temperature of water (maximum) = 420C

t = 42-30= 120C

Cp = 4.187 kJ/kg K.

mc = 85.74 × 103

= 1 . 7 1 k g / s .

4.187x103x12

(b) LMTD Calculations

Assume, counter current

LMTD = (T1- t2) – (T2 - t1)

ln (T1- t2 )

(T2 - t1)

T1 = 87.5°C, T2 = 87.5°C, t1 = 30°C, t2 = 42°C; LMTD = 51.28°C

(ΔT) LMTD, corrected = 0.8 × 51.28°C = 41.02°C

Fig 6.1.1 Temperature Drop in Condenser

24

(C) Routing of fluids

organic vapors - Shell

side Liquid - Tube side

(D) Heat Transfer Area

(i) Qh = Qc =UA (ΔT) LMTD, corrected.

U= Overall heat transfer coefficient (W/m2

K)

Assume Ua =300W/m2K

A assumed = 85.74 ×103

300 × 41.02

= 6.886 m2

(ii) Select pipe size

Outer diameter of pipe (OD) = 1” = 0.0254 m

Inner diameter of pipe (ID) = 0.022m

Let length of tube = 2.5m

Let allowance = 0.05m

Heat transfer area of each tube (aheat – transfer) = π × OD × (Length – Allowance)

= π × 0.0254 × (2.5 – 0.05)

= 0.1954 m2

Number of tubes (Ntubes) = A assumed 6.886

=

a heat-transfer 0.1954

= 36 tubes

25

(iii) Choose Shell diameter:

Choose TEMA: P or S. 1” OD tubes in 1.25” □lar pitch.

1 - 1 Vertical Condenser

Ntubes (Corrected) = 48

Shell Diameter (Dc) =305 mm.

Acorrected =9.38 m2

Ucorrected = 220.25 W/m2K

(iv) Fluid velocity check:

(a) Vapor side – need not check

(b) Tube side

Flow area (atube) = apipe× Ntubes

Per pass

Ntube passes

a pipe = C.S of pipe = π(ID2)

4

atube = π (0.0254)2

× 48 = 0.0182 m2/pass

4 1

Velocity of fluid (Vpipe) vp = mpipe

in pipe ρpipe x atube

mpipe = mass –flow rate of fluid in pipe.

ρpipe = Density of fluid in pipe (water)

vp = 1.71 = 0.1 m/s

1000 × 0.0182

Fluid velocity check is satisfied

26

(II) Film Transfer Coefficient:

Properties are evaluated at tfilm :

tfilm = tv + tv + (t1+t2) 87.5 + 87.5 + (30+42)

2 = 2 = 74.62°C

2 2

a) Shell side

Reynolds‟s Number (Re) = 4 Г = 4 W µf µf (Ntubes)

2/3×L

= 4 0.3083 ×

0.0002 (48)2/3×2.5

= 185.5 m2

For vertical condenser

ho = 1.51 kf ρf g 0.33

4 W -0.33

µf2

L Nt0.667µf

ho = 1.51 (0.138)3 (1290)

2 (9.81)

0.33 185.5

-0.33

(0.2 × 10-3

)2

= 2510 W/m2

27

b) Tube side

Gt = mpipe

atube

Gt= Superficial mass velocity

Gt = 1.71 = 93.95 kg/m2s

0.0182

Re = (ID) Gt = 0.022 × 93.95 = 2583.6

μ 0.8 ×10 –3

Pr = μ Cp = 0.8 × 10 –3

× 4.187 × 10 3

= 5.79

K 0.578

hi (ID)

K = 0.023 (Re) 0.8 (Pr) 0.3

hi = inside – heat transfer coefficient

hi = 0.023 (2583.6) 0.8

(5.79) 0.3

× 0.578

0.022

hi = 549.20 W/m2K

Fouling factor

(Dirt –coefficient) = 0.003

=5.28 x 10-4 (W/m2K)-1

1 1 (OD) 1

= + + wall resistance + Fouling factor

U0 ho (ID) hi

Uo = overall heat –transfer coefficient

1 1 0.0254 1

= + × + 4.028x10-5 + 5.28x10-4

U0 1154.9 0.022 549.20

U0 = 325.85 W/m2K

U0 > UD

Therefore our assumption in overall heat transfer (UD) co-efficient is sufficient for design

28

(III) Pressure Drop Calculations

a) Tube Side

Re = 2583.6

f = 0.079 (Re)-¼

= 0.079 (2583.6) -¼

= 9.737 x 10 –3

f = friction factor

Pressure Drop along

the pipe length ( P) L = ( H)L × ρ × g

= 4fLVp2

× ρ × g

2g (ID)

= 4 × 9.737 × 10-3

×2.5 × 93.956 2

× 9.81

2 × 9.81 × 0.022 × 1000

= 1.96 kPa

Pressure Drop in the

end zones ( P)e = 2.5 ρ Vp2

= 2.5 x1000 × (93.95×10-2

) 2

=11.3 kPa 2 2

Total pressure drop

in pipe ( P) total = [1.96 + 11.3] 2 = 13.26 kPa < 70 kPa

b) Shell side, Kern‟s method

Baffle spacing (B) = Ds =305 mm

C′ = 3.1 x 10

–2 – 0.0254 = 0.0056m

PT = pitch = 3.1 x 10 –2

m

29

ashell = shell diameter × C′×B = 0.305 × 0.0056 × 0.305

PT 3.1× 10 –2

= 0.0168 m2

De = 4 PT×0.86 PT - 1 π (OD)2 = 4 (31 x 10

–3)

2 ×0.86 - π (0.0254)

2

2 2 4 2 8

(π do) π ( 0.0254)

2 2

= 16.0mm.

Gs= Superficial velocity in shell = mshell = 0.3083 = 18.351 kg/m

2s

ashell 0.0168

(NRe)s = Gs Dc = 18.351× 16 × 10 –3 = 35375

8.3 × 10-6

f = 1.87 (35375) –0.2

= 0.230

Shell side pressure

drop ( P)s = 4 f (Nb + 1) Ds Gs

2 g ] x 0.5

2 g De ρ v

Nb = 0

( P)s = 4 (0.230) (1) (0.305) (18.351)29.81 × 0.5

2 x 9.81 (16 x 10-3) × 4.57

= 6.461 KPa

(It very near to permissible pressure drop).

30

Shell Inner Diameter = 303mm

Shell thickness = 30mm

Inlet line Diameter = 80mm

Outlet Line Diameter = 18mm

Head = Elliptical type

Tube Inner Diameter = 25.4mm

Tube Length = 2500mm

Fig 6.1.2 Model of Vertical Condenser

31

6.2. DISTILLATION COLUMN

87°C D = 6.686 kmoles xD = 0.99

Enrich.

section

F= 7.074k.moles xF = 0.931 TF =90°C Stripping

Section

100°C

Basis: 1 hour operation

Glossary of notations used

Total

Reboiler

W = 0.38kmol Xw= 0.03

F = molar flow rate of feed, kmol/hr D = molar flow rate of distillate, kmol/hr

W = molar flow rate of residue, kmol/hr.

xF = mole fraction of TCE in liquid

xD = mole fraction of TCE in distillate xW = mole fraction of TCE in residue

Fig 6.2.1 Distillation Column

Design

32

Rm = minimum reflux ratio R = actual reflux ratio

L= molar flow rate of liquid in the enriching section, kmol/hr

G = molar flow rate of vapor in the enriching section, kmol/hr

L = molar flow rate of liquid in stripping section, kmol/hr

G = molar flow rate of vapor in stripping section, kmol/hr

M‟ = average molecular weight of feed, kg/kmol

q = Thermal condition of feed

Feed = Saturated liquid at its bubble point, Temperature= 90°C

M‟ = 132.95 kg/kmol

Relative volatility which predicts the order to separate binary mixture using distillation column

process,

Thus the relative volatility (α) of this mixture is obtained from data page (Perry‟s Hand book)

α = 2.67@ 760mmHg, α = y (1-x)/x (1-y)

Vapor liquid equilibrium data

The operating line for the rectification section is constructed as follows. First the desired top

product composition is located on the VLE diagram, and a vertical line produced until it

X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Y 0.228 0.40 0.533 0.64 0.727 0.80 0.86 0.914 0.96 1

Fig 6.2.2 T-x-y Diagram

33

intersects the diagonal line that splits the VLE plot in half. A line with slope R/(R+1) is then

drawn from this intersection point as shown in the diagram below.

R is the ratio of reflux flow (L) to distillate flow (D) and is called the reflux ratio and is a

measure of how much of the material going up the top of the column is returned back to the

column as reflux.

xD Rm+1 = 0.66

Rm+1 = xD 0.994

= = 1.5 0.66 0.66

Rm = 1.5 – 1.00 = 0.5

R= 1.5 Rm = 0.75

xD 0.994

= = 0.568

R+1 0.75 +1

Number of trays from graph =12

L = RD = 0.568x6.686 = 3.797 k-moles

Fig 6.2.3 V-L-E Diagram

34

G =L+D = 1.838 + 6.686

= 10.483 k-moles q=1 (Feed is saturated liquid)

L = L+qf = 3.797 + 1(7.074)

= 10.871 k-moles

G = G+ (q –1) F = 8.475 +0

= 10.483 k-moles

Fig 6.2.4 McCabe Thiele Diagram

35

PROPERTIES

Enriching section Stripping section

Top Bottom Top Bottom

Liquid (k.moles/hr)

Liquid (kg/hr)

Vapor

(k.moles/hr)

Vapor (kg/hr)

x

y

T liquid (oC)

T vapor(oC)

ρvapor(kg/m3)

ρliquid (kg/m3)

(L/G)(ρg/ρL)0.5

liq (dyn/cm)

µ vapor

µ liq

Dvapor (m2/s)

Dliquid (m2/s)

3.797

499.30

10.483

1378.52

0.994

0.99

87

87.6

4.57

1460

0.0202

28.7

00038

0.27

0.0599

1.688x10-9

3.797

504.81

10.483

1393.71

0.952

0.95

89

89.7

4.51

1480

0.0199

28.7

0.0038

0.271

0.0059

1.688x10-9

10.871

1467.58

10.483

1415.2

0.93

0.932

89.5

90.3

4.46

1515

0.056

26.3

0.0038

0.449

0.148

1.952x10-9

10.871

1804.58

10.483

1740.18

0.036

0.036

100

101.5

5.1

1546

0.0595

26.3

0.0038

0.4491

0.149

1.952x10-9

Table 6.2.1 Properties of Vapor and Liquid in Distillation

column

36

AVERAGE CONDITIONS AND PROPERTIES

Properties Enriching section Stripping section

Liquid (k-moles/hr) (kg/hr)

3.797 502.05

10.871 1636.08

Vapor ( k-moles/hr)

(kg/hr)

10.483

1386.115

10.483

1577.69

Tliq (°c )

88

94.75

Tvapor (°c )

88.65 95.9

ρliq (kg/m3)

1470 1530.5

ρvapor (kg/m3)

4.54 2.483

ENRICHING SECTION

VOLUMETRIC FLOW RATE OF VAPOUR

Assume ideal gas behavior

V = nRT/P

= 3.787 × 0.08206 x 361.5/1 × 3600

V=0.313 m3/s

Assuming vapor velocity = 0.5 m/s

Net cross sectional area of top column =Vol .flow rate / vap. Velocity

= 0.313/0.5

An = 0.624 m2

Table 6.2.2 Average Conditions and Properties

37

Area of the column, Ac=An/0.88

=0.624/0.88

=0.81m2

Ac= (/4) x Dc2

0.81=0.785Dc2

Dc = 1.015 m

STRIPPING SECTION

VOLUMETRIC FLOW RATE OF VAPOUR:

Assume ideal gas behavior

V = nRT/P

= 10.871 x 0.08206 × 367.75/1 × 3600

V= 0.278 m3/s

Vapor velocity =0.5 m/s

Net cross sectional area of top column =Vol .flow rate / vap. Velocity

=0.278/0.5

An = 0.556 m2

Area of the column, Ac=An/0.88

=0.556/0.88

=0.632 m2

Ac= (/4) xDc2

0.632=0.785Dc2

Dc = 0.897 m

38

CROSS SECTIONAL AREA OF DOWN COMER

Ad = Ac - An

=0.81 – 0.556

Ad =0.254 m2

TOWER HEIGHT:

= ((Actual trays – tray for reboiler + tray for condenser) +2) x Tray spacing

Actual trays = (NTP/Tray overall efficiency)

= (12/ 0.17 – 0.616 log (μavg))

= (12 / 0.474) since, μavg = 0.32 cP

= 25.27 = 26

= ((26-1+0) +2) × Tray spacing

= 27 × Tray spacing

=27 × 0.5

Column height =13.5 m

FEED ENTERING ZONE

Feed enter in between the enriching to stripping section. From the above parameters feed

should enter in the 5th

tray which find out from the actual tray calculation and help of graph

chart.

39

TRAY HYDRAULICS (TOP & BOTTOM)

Sieve tray are used

1) Plate spacing = 0.5 m

2) Hole diameter = 0.006 m

3) Hole pitch = 3 x Hole diameter =0.0018 m

4) Tray thickness =0.6 × 0.006=0.0036 m

Ratio of hole area to perforated area = (Ah / Ap)

= 0.9 × (6/18) 2

= 0.1

WEIR HEIGHT

For normal pressure weir height lies between

40-50 mm

Let weir height = 0.05 m

WEIR LENGTH

Weir length = 0.75 × Dc

= 0.75 × 0.879

Weir length = 0.66 m

40

7. COST ESTIMATION AND ECONOMICS

Cost of TCE plant of capacity 5430 TPY in 1968 Rs. 3.26×108

Therefore cost of 7602 TPY in 2011 is: C1 = C2 (Q1/Q2)0.6

= 3.26×108 (7602/5430)

0.6

= Rs.3.99 × 108 Chemical Engineering Plant Cost Index

Cost index in 1968 = 112

Cost index in 2011 = 511 (from chemical magazines) Thus, Present cost of Plant = (original cost) × (present cost index)/(past cost index)

= (Rs.3.99 × 108) × (511/112) = Rs. 18.20×108

i.e., Fixed Capital Cost (FCI) = Rs. 18.20×108 Estimation of Capital Investment Cost I. Direct Costs: material and labor involved in actual installation of complete facility (70-

85% of fixed-capital investment)

a) Equipment + installation + instrumentation + piping + electrical + insulation

+ painting (50-60% of Fixed-capital investment) 1. Purchased equipment cost (PEC): (15-40% of Fixed-capital investment)

Consider purchased equipment cost = 25% of Fixed-capital investment

i.e., PEC = 25% of 18.20×108

= 0.25 × 18.20×108

= Rs. 4.551×108

2. Installation, including insulation and painting: (25-55% of purchased equipment

cost.)

Consider the Installation cost = 40% of Purchased equipment cost

= 40% of 4.551×108

= 0.40 ×4.551×108

= Rs.1.820×108

41

3. Instrumentation and controls, installed: (6-30% of Purchased equipment cost.) Consider the installation cost = 20% of Purchased equipment cost

= 20% of ×4.551x108

= 0.20 ×4.551×108

= Rs. 0.9102×108 4. Piping installed: (10-80% of Purchased equipment cost)

Consider the piping cost = 40% Purchased equipment cost

= 40% of Purchased equipment cost

= 0.40 ×4.551×108

= Rs. 1.8204×108

5. Electrical, installed: (10-40% of Purchased equipment cost)

Consider Electrical cost = 25% of Purchased equipment cost

= 25% of 4.551×108

= 0.25 ×4.551×108

= Rs. 1.1377×108

B. Buildings, process and Auxiliary: (10-70% of Purchased equipment

cost)

Consider Buildings, process and auxiliary cost = 40% of PEC

= 40% of 4.551×108

= 0.40 ×4.551×108

= Rs. 1.8104×108

C. Service facilities and yard improvements: (40-100% of Purchased equipment

42

cost)

Consider the cost of service facilities and yard improvement = 60% of PEC

= 60% of 4.551×108

= Rs. 2.7306×108

D. Land: (1-2% of fixed capital investment or 4-8% of Purchased equipment cost)

Consider the cost of land = 6% PEC

= 6% of 4.551×108

= 0.06 ×4.551×108

= Rs. 0.2730×108

Thus, Direct cost = Rs. 15.0586×108 --------- (82.74% of FCI)

II. Indirect costs: expenses which are not directly involved with material and labor of actual

installation of complete facility (15-30% of Fixed-capital investment)

A. Engineering and Supervision: (5-30% of direct costs) Consider the cost of engineering and supervision = 10% of Direct costs

i.e., cost of engineering and supervision = 10% of 15.0519 ×108

= Rs. 1.5058×108

B. Construction Expense and Contractor‟s fee: (6-30% of direct costs)

Consider the construction expense and contractor‟s fee = 10% of Direct costs

i.e., construction expense and contractor‟s fee = 10% of 15.0586×108

= 1.5059×108

C. Contingency: (5-15% of Fixed-capital investment) Consider the contingency cost = 10% of Fixed-capital investment

i.e., Contingency cost = 10% of 18.20×108

= 0.12× 18.20×108

= Rs. 1.820×108

Thus, Indirect Costs = Rs. 5.1942×108 --------- (28.54% of FCI)

43

III. Fixed Capital Investment: Fixed capital investment = Direct costs + Indirect costs

= (15.0586×108) + (5.1942×108)

i.e., Fixed capital investment = Rs. 20.253×108 IV. Working Capital: (10-20% of Fixed-capital investment)

Consider the Working Capital = 15% of Fixed-capital investment

i.e., Working capital = 15% of 20.253×108

= 0.15 × 20.253×108

= Rs. 3.037×108

V. Total Capital Investment (TCI): Total capital investment = Fixed capital investment + Working capital

= (20.253×108) + (3.037×108)

i.e., Total capital investment = Rs. 23.291×108

Estimation of Total Product cost I. Manufacturing Cost = Direct production cost + Fixed charges + Plant overhead

cost.

A. Fixed Charges: (10-20% total product cost) i. Depreciation: (depends on life period, salvage value and method of calculation-

about 13% of FCI for machinery and equipment and 2-3% for Building Value for Buildings)

Consider depreciation = 13% of FCI for machinery and equipment and 3% for Building

Value for Buildings)

i.e., Depreciation = (0.13×20.253×108) + (0.03×1.8104×108) = Rs. 2.687×108

ii. Local Taxes: (1-4% of fixed capital investment)

Consider the local taxes = 3% of fixed capital investment

i.e. Local Taxes = 0.03×20.253×108

44

= Rs. 0.607×108

iii. Insurances: (0.4-1% of fixed capital investment)

Consider the Insurance = 0.7% of fixed capital investment

i.e. Insurance = 0.007×20.253×108

= Rs. 0.1417×108

iv. Rent: (8-12% of value of rented land and buildings)

Consider rent = 10% of value of rented land and buildings

= 10% of ((0.2730×108) + (1.8104×108))

Rent = Rs. 0.2083x108

Thus, Fixed Charges = Rs. 3.644×108

B. Direct Production Cost: (about 60% of total product cost) Now we have Fixed charges = 10-20% of total product charges – (given)

Consider the Fixed charges = 15% of total product cost

Total product charge = fixed charges/15%

Total product charge = 3.644×108/15%

Total product charge = 3.644×108/0.15

Total product charge (TPC) = Rs. 24.29×108 i. Raw Materials: (10-50% of total product cost) Consider the cost of raw materials = 25% of total product cost Raw material cost = 25% of TPC

= 0.25×24.29×108

Raw material cost = Rs. 6.0725×108 ii. Operating Labor (OL): (10-20% of total product cost)

Consider the cost of operating labor = 12% of total product cost

Operating labor cost = 12% of 24.29×108 = 0.12×24.29×10

8

Operating labor cost = Rs. 2.914×108 iii. Direct Supervisory and Clerical Labor (DS & CL): (10-25% of OL) Consider the cost

45

for Direct supervisory and clerical labor = 12% of OL

Direct supervisory and clerical labor cost = 12% of 2.914×108

= 0.12×2.914×108

Direct supervisory and clerical labor cost = Rs. 0.3497×108

iv. Utilities: (10-20% of total product cost)

Consider the cost of Utilities = 12% of total product cost

Utilities cost = 12% of 24.29×108

= 0.12×20.253×108

Utilities cost = Rs. 2.43×108 v. Maintenance and repairs (M & R): (2-10% of fixed capital investment)

Consider the maintenance and repair cost = 5% of fixed capital investment

i.e. Maintenance and repair cost = 0.05×20.253×108

= Rs.1.0126×108

vi. Operating Supplies: (10-20% of M & R or 0.5-1% of FCI)

Consider the cost of Operating supplies = 15% of M & R Operating supplies cost

= 15% of 1.085×108

= 0.15×1.0126×108

Operating supplies cost = Rs. 0.1518×108 vii. Laboratory Charges: (10-20% of OL)

Consider the Laboratory charges = 15% of OL

Laboratory charges = 15% of 2.914×108 = 0.15×2.914×108

Laboratory charges = Rs. 0.6204×108 viii. Patent and Royalties: (0-6% of total product cost) Consider the cost of Patent and royalties = 4% of total product cost

Patent and Royalties = 4% of 24.29×108

= 0.03×24.29×108

Patent and Royalties cost = Rs. 0.7287×108

46

Thus, Direct Production Cost = Rs. 14.279×108 -------- (59% of TPC)

C. Plant overhead Costs (50-70% of Operating labor, supervision, and

maintenance or 5-15% of total product cost); includes for the following: general plant

upkeep and over head , payroll overhead, packaging, medical services, safety and

protection, restaurants, recreation, salvage, laboratories, and storage facilities.

Consider the plant overhead cost = 60% of OL, DS & CL, and M & R

Plant overhead cost = 60% of [(2.914×108) + (0.3497×108) + (1.0126×108)]

Plant overhead cost = Rs. 2.565×108

Thus, Manufacture cost = Direct production cost + Fixed charges + Plant overhead costs.

Manufacture cost = (24.29×108) + (3.644×108) + (2.565×108) Manufacture cost = Rs.30.499×10

8

II. General Expenses = [Administrative costs + distribution and selling costs + Research and development costs] A. Administrative costs: (2-6% of total product cost) Consider the Administrative costs = 5% of total product cost

Administrative costs = 0.05 × 24.29×108

Administrative costs = Rs. 1.2145×108 B. Distribution and Selling costs: (2-20% of total product cost); includes costs for sales

offices, salesmen, shipping, and advertising.

Consider the Distribution and selling costs = 15% of total product cost

Distribution and selling costs = 15% of 24.29×108

Distribution and selling costs = 0.15 × 24.29×108

Distribution and Selling costs = Rs. 3.464×108

C. Research and Development costs: (about 5% of total product cost)

Consider the Research and development costs = 5% of total product cost

Research and Development costs = 5% of 24.29×108

Research and development costs = 0.05 × 24.29×108

47

Research and Development costs = Rs. 1.2145×108 D. Financing (interest): (0-10% of total capital investment)

Consider interest = 5% of total capital investment

i.e. interest = 5% of 23.291×108 = 0.05×23.291×108

Interest = Rs. 1.1645×108

Thus, General Expenses = Rs. 7.0575×108

IV. Total Product cost = Manufacture cost + General Expenses

= (30.499×108) + (7.0575×108)

Total product cost = Rs. 37.5565×108

V. Gross Earnings/Income: Wholesale Selling Price of TCE per kg = Rs.60 Total Income = Selling price × Quantity of product manufactured

= 60 x 0.7602 x 108

Total Income = Rs. 45.612×108 Gross income = Total Income – Total Product Cost

= (45.612×108) – (37.5565×108)

Gross Income = Rs. 8.056×108

Let the Tax rate be 45% (common)

Net Profit = Gross income - Taxes

= Gross income× (1- Tax rate)

Net profit = 8.056×108(1- 0.45)

= Rs. 4.431×108

Rate of Return Rate of return = Net profit×100/Total Capital Investment

Rate of Return = 4.431×108×100/ (23.291×108)

Rate of Return = 19.02%

48

Break-even Analysis Data available

Annual Direct Production Cost = Rs. 24.29×108

Annual Fixed charges, overhead and general expenses = Rs. 3.644×108

Total Annual sales = Rs. 148.24×108 Wholesale Selling Price MEK per ton. = Rs. 60000

Direct production cost per ton of MEK = (3.644×108)/ (148.24×108/60000) = Rs. 1474.9 per ton Let „n‟ TPA be the break even production rate. Number of tons needed for a break-even point is given by

(3.644×108) + (1474.9 ×n) = (60000×n) => n = 6226.39 tons/year n = 1.7199 tons/day

Hence, the break-even production rate is 1.7199 TPD or 8 .2% of the considered plant

capacity.

49

8. PLANT LAYOUT

The location of the plant can have a crucial effect on the overall profitability of a project,

and the scope for future expansion. Many factors must be considered when selecting a suitable

plant site. The principal factors are:

• Location, with respect to the marketing area

• Raw material supply

• Transport facilities

• Availability of labor

• Availability of suitable land

• Environmental impact and effluent disposal

• Local community consideration

• Climate

• Political and strategic consideration

Raw material availability

The source of raw material is one of the most important factors influencing the selection of a

plant site. This is particularly true if large volumes of raw material are consumed, because

location near the raw material source permits considerable reduction in transportation and

storage charge. Attention should be given to the purchased price of the raw material , distance

from the source of supply , fright or transportation expenses availability and reliability of supply

purity of the material and storage requirements.

Markets

The location of markets or intermediate centers affects the cost of product distribution and

the time required for shipping. Proximity to the major markets is an important consideration in

the selection of plant site , because the buyer usually finds it advantages to purchase from nearby

sources. It should be noted that markets are needed for by products as well as major final

products.

50

Energy availability

Power and steam requirements are high in most industrial plants , and fuel is ordinary

required to supply these utilities. Consequently power and fuel can be one major factor in the

choice of a plant site. Electrolytic processes require a cheap source of electricity and plants using

electrolytic processes are often located near hydroelectric installations. If the plant requires large

quantities of coal or oil, location near a source of fuel supply may be essential for economic

operation. The local cost of power can help determine whether power should be purchased or

self-generated.

Climate

If the plants is located in a cold climate, costs may be increased by the necessity for

construction of protective shelters around the process equipment, and special cooling towers or

air –conditioning equipment may be required if the prevailing temperature are high. Excessive

humidity or extremes of hot or cold weather can have a serious effect on the economic operation

of plant and these factors should be examined when selecting a plant site.

Transportation facilities

Water, railroads and high and ways are the common means of transportation used by major

industrial concerns. The kind and amount of products and raw materials determine the most

suitable type of transportation facilities. In any case, careful attention should be given to local

freight rates and existing railroad lines. The proximity to railroad centers and possibility of canal,

river, lake or ocean transport must be considered. Motor trucking facilities are widely used and

serve as a useful supplement to rail and water facilities. If possible ,the plant site should have

access to all three types of transportation, and certainly, at least two types should be available .

There is usually need for convenient air and rail transportation facilities between the plant and

the main company headquarters, and effective transportation facilities for the plant personnel are

necessary.

Water supply

The process industries use large quantities of water for cooling, washing steam generation and

as raw material. The plant therefore must be located where a dependable supply of water is

51

available. A large river must be located where a dependable supply of water may be satisfactory

if the amount of water required is not too great. The level of the existing water table can be

checked by constancy of the water table and the year-round capacity of local rivers or lakes

should be obtain. It the water supply shows seasonal fluctuations, it may be to construct a

reservoir or to several standby wells. The temperature, mineral content, silt or sand content,

bacteriological content, and cost for supply and purification treatment must also be considered

when choosing a water supply.

Waste disposal

In recent years, many legal restrictions have been placed on the methods for disposing of waste

material from the process industries. The site selected for disposing of a plant should have

adequate capacity and facilities for correct waste disposal. Even though a given area has

minimal restrictions on pollution. It should not be assumed that this condition will continue to

exist. In choosing a plant sit, the permissible tolerance levels for various methods of waste

disposal should be considered carefully and attention should be given to potential for additional

waste-treatment facilities.

Labor supply

The type and supply of labor available in the vicinity of a proposed plant site must be

examined. Consideration should be given to prevailing pay scales, restrictions on number of

hours worked per week, competing industries that can cause dissatisfaction or high turnover rates

among the workers, and variations in the skill and productivity of the workers.

Taxation and legal restrictions

State and local tax rates on property income, unemployment insurance and similar items

vary from form one location to another. Similarly, local regulations on zoning, building codes,

nuisance aspects and transportation can have a major influence on the final choice of a plant site.

In fact, zoning difficulties and obtaining the many required permits can often be much more

important in terms of cost and time delays than many of factors discussed in the preceding

sections.

52

Site characteristics

The characteristics of the land at a proposed plant site should be examined carefully. The

topography of the tract of land and the soil structure must be considered, since either of both may

have a pronounced effect on construction cost or living conditions. Future changes may make it

desirable or necessary to expand the plant facilities. Therefore , even through no immediate

expansion is planned, a new plant should be constructed at a location where additional space is

available.

Flood and fire protection

Many industrial plants are located along rivers near large bodies of water, and risks of flood

of flood or hurricane damage. Before selecting a plant site, the regional history of natural events

of this type should be examined and the consequences of such occurrence considered. Protection

from losses by fire is another important factor in selecting a plant location. In case of major fire,

assistance from outside fire department should be available. Fire hazards in the immediate area

surrounding the plant site must not be overlooked.

Community factors

The character and facilities of a community can have quite an effect on the location of

plant. If a certain minimum number of facilities for satisfactory living of plant personnel do not

exist, it often becomes a burden for the plant to subsidize such facilities. Cultural facilities of the

community are important to sound growth. Churches libraries, schools, civic theaters, concert

associations and other similar groups, if active and dynamic, do much to make a community

progressive. The problem of recreation deserves special consideration. The efficiency, character,

and history of both state and local government should be evaluated. The existence of low taxes is

not in itself a favorable situation unless the community is already well developed and relatively

free of debt.

53

PLANT LAYOUT

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

plant equipment specified on the process flow sheet and laid out.

The principal factors to be considered are:

1. Economic consideration: construction and operation cost.

2. The process requirement

3. Convenience of operation

4. Convenience of maintenance

5. Safety

6. Future expansion

7. Modular construction

COSTS

The cost of construction can be minimized by adopting a layout that gives shortest run of

connecting pipes between equipment, and adopting the least amount of structural steel work.

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

PROCESS REQUIREMENT

All the required equipments have to be placed properly within process. Even the

installation of the auxiliaries should be done in such a way that it will occupy the least space.

OPERATION

Equipment that needs to have frequent operation should be located convenient to the

control room. Valves, sample points, and instruments should be located at convenient position

and height. Sufficient working space and headroom must be provided to allow easy access to

equipment.

54

MAINTENANCE

Heat exchangers need to be sited so that the tube bundles can be easily withdrawn for

cleaning and tube replacement. Vessels that require frequent replacement of catalyst or packing

should be located on the outside of buildings. Equipment that requires dismantling for

maintenance, such as compressors and large pumps, should be placed under cover.

SAFETY

Blast walls may be needed to isolate potentially hazardous equipment, and confine the

effects of an explosion. At least two escape routes for operator must be provided from each level

in the process building.

PLANT EXPANSION

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

expansion of the process. Space should be left on pipe alleys for future needs, service pipes

oversized to allow for future requirements.

MODULAR CONSTRUCTION

In recent years, there has been a move to assemble sections of the plant at the

manufacturer site. These modules will include the equipment, structural steel, piping and

instrumentation. The modules then transported to the plant site, by road or sea.

55

H e a l t h 2

F i r e 1

R e a c t i v i t y 0

P e r s o n a l H

P r o t e c t i o n

9. Material Safety Data Sheet

Trichloroethylene MSDS

2 0

Product Name: Trichloroethylene

Catalog Codes: SLT3310, SLT2590

CAS#: 79-01-6

RTECS: KX4560000

TSCA: TSCA 8(b) inventory: Trichloroethylene

CI#: Not available.

Synonym:

Chemical Formula: C2HCl3

Composition and Information on Ingredients

Composition

CAS #

% by Weight

Name

Trichloroethylene

79-01-6

100

Hazards Identification

Potential Acute Health Effects: Hazardous in case of skin contact (irritant, permeator), of

eye contact (irritant), of ingestion, of inhalation.

Potential Chronic Health Effects:

CARCINOGENIC EFFECTS: Classified + (PROVEN) by OSHA. Classified A5 (Not

suspected for human.) by ACGIH. MUTAGENIC EFFECTS: Not available. TERATOGENIC

EFFECTS: Not available. DEVELOPMENTAL TOXICITY: Not available. The substance is

toxic to kidneys, the nervous system, liver, heart, upper respiratory tract. Repeated or

prolonged exposure to the substance can produce target organs damage.

Fig 9.1 Hazard Identity

Symbol

56

First Aid Measures

Eye Contact

Check for and remove any contact lenses. Immediately flush eyes with running water for at

least 15 minutes, keeping eyelids open. Cold water may be used. Do not use an eye

ointment. Seek medical attention.

Skin Contact

After contact with skin, wash immediately with plenty of water. Gently and thoroughly wash

the contaminated skin with running water and non-abrasive soap. Be particularly careful to

clean folds, crevices, creases and groin. Cover the irritated skin with an emollient. If irritation

persists, seek medical attention. Wash contaminated clothing before reusing.

Serious Skin Contact

Wash with a disinfectant soap and cover the contaminated skin with an anti-bacterial cream.

Seek medical attention.

Inhalation

Allow the victim to rest in a well ventilated area. Seek immediate medical attention.

Serious Inhalation

Evacuate the victim to a safe area as soon as possible. Loosen tight clothing such as a

collar, tie, belt or waistband. If breathing is difficult, administer oxygen. If the victim is

not breathing, perform mouth-to-mouth resuscitation. Seek medical attention.

Ingestion

Do not induce vomiting. Loosen tight clothing such as a collar, tie, belt or waistband. If

the victim is not breathing, perform mouth-to-mouth resuscitation. Seek immediate

medical attention.

Serious Ingestion Not available.

57

Fire and Explosion Data

Flammability of the Product: May be combustible at high temperature.

Auto-Ignition Temperature: 420°C (788°F)

Flash Points: Not available.

Flammable Limits: LOWER: 8% UPPER: 10.5%

Products of Combustion: These products are carbon oxides (CO, CO2), halogenated

compounds.

Fire Hazards in Presence of Various Substances: Not available.

Explosion Hazards in Presence of Various Substances:

Risks of explosion of the product in presence of mechanical impact: Not available.

Risks of explosion of the product in presence of static discharge: Not available.

Fire Fighting Media and Instructions:

SMALL FIRE: Use DRY chemical powder.

LARGE FIRE: Use water spray, fog or foam. Do not use water jet.

Special Remarks on Fire Hazards: Not available.

Special Remarks on Explosion Hazards: Not available

Accidental Release Measures

Small Spill: Absorb with an inert material and put the spilled material in an appropriate waste

disposal.

Large Spill: Absorb with an inert material and put the spilled material in an appropriate waste

disposal. Be careful that the product is not present at a concentration level above TLV. Check

TLV on the MSDS and with local authorities.

58

Handling and Storage

Precautions

Keep locked up Keep away from heat. Keep away from sources of ignition. Empty

containers pose a fire risk; evaporate the residue under a fume hood. Ground all

equipment containing material. Do not ingest. Do not breathe gas/fumes/ vapor/spray.

Wear suitable protective clothing In case of insufficient ventilation, wear suitable

respiratory equipment if ingested, seek medical advice immediately and show the

container or the label. Avoid contact with skin and eyes

Storage

Keep container dry. Keep in a cool place. Ground all equipment containing material.

Carcinogenic, teratogenic or mutagenic materials should be stored in a separate locked safety

storage cabinet or room.

Exposure Controls/Personal Protection

Engineering Controls:

Provide exhaust ventilation or other engineering controls to keep the airborne concentrations of

vapors below their respective threshold limit value. Ensure that eyewash stations and safety

showers are proximal to the work-station location.

Personal Protection:

Splash goggles. Lab coat. Vapor respirator. Be sure to use an approved/certified respirator or

equivalent. Gloves.

Personal Protection in Case of a Large Spill:

Splash goggles, Full suit, Vapor respirator, Boots, Gloves. A self contained breathing

apparatus should be used to avoid inhalation of the product. Suggested protective

clothing might not be sufficient; consult a specialist BEFORE handling this product.

Exposure Limits: TWA: 50 STEL: 200 (ppm) from ACGIH (TLV) TWA: 269 STEL: 1070

(mg/m3) from ACGIH Consult local authorities for acceptable exposure limits.

59

Stability and Reactivity Data

Stability: The product is stable.

Instability Temperature: Not available.

Conditions of Instability: Not available.

Incompatibility with various substances: Not available.

Corrosivity: Extremely corrosive in presence of aluminum. Non-corrosive in presence of glass.

Special Remarks on Reactivity: Not available.

Special Remarks on Corrosivity: Not available.

Polymerization: No

Transport Information

DOT Classification: CLASS 6.1: Poisonous material.

Identification: Trichloroethylene : UN1710 PG: III

Special Provisions for Transport: Not available

60

10. CONCLUSION

In Trichloroethylene manufacturing process, the above explained dehydrochlorination with

Calcium Hydroxide is the energy efficient technique that going in DCW limited. Because the

pyrolysis of Tetrachloroethane requires huge energy. They are only one Industry producing TCE

with 7600 TPY the way of lime process. Also according to the demand of TCE, they are

expanding their production to 10860 TPY. Because, the modern chemical derived from TCE for

the Refrigerant production instead of CFC. An industry, they are performing the condensation of

crude Trichloroethylene vapor using open type co-current vertical condenser with shell side

condensation that vapor product resulting from dehydrochlorination of Tetrachloroethane. I

suggested that present condenser can be replaced by closed type counter current vertical

condenser. This kind of modification will minimize the cold water requirement to 60% of current

status. This will reduce the cooling tower load. Thus above modification will feasible and

process becomes more effective.

61

11. BIBLIOGRAPHY

11.1. BOOKS CONSULTED

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

3rd Ed., East-West press.[1990]

(2) Kirk-Othmer, “Encyclopaedia of Chemical Technology”, 5th Ed, Volume-1, John

Wiley & Sons Inc..[1971]

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

MIR publishers.[1987]

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

7th Ed. Mc-Graw Hill International edition,[1989, 1993]

(5) OLEF A. Hougen, Kenneth M. Watson, Roland A. Ragatz “Chemical Process Principles”,

2nd

edition., CBS Publishers & Distributors, New Delhi.[1986]

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

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

(7) Joshi M. V., “Process Equipment Design”, 2nd Ed., Mc-Millan India Ltd,[2000]

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

Engineers”, 3rd Ed., Mc-Graw Hill Book Company, Page No: 207-208, 484-485[1996].

(9) B.C Bhattacharya, “Chemical equipment Design”, Chemical Engineering Education

Development Centre.[1998]

(10) L.E. Brownell and E.H. Young, “Process Equipment Design”, John Wiley & SonsInc. New

York,[2001]

62

11.2. WEB LINKS

www.wikipedia.com/1,1,2trichloroethyene+(data page)&Wikipedia%20the+free

encyclopedia.htm

http://www.cheresources.com/invision/topic/941-vertical-condenser-design/

http://www.cambridgesoft.com/databases/details/?fid=140

www.chemcad.com

http://www.freepatentsonline.com/3949009.html

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