INTRODUCTION: Chlorine is the eleventh most abundant element in the lithosphere. It is

highly reactive and hence rarely found in the free state. It exists mainly in

the form of chlorides. In seawater it is present as 2.9 wt% NaCl and 0.3 wt% MgCl2. In salt deposits formed by evaporation of seawater, chlorine is present in large quantities as rock salt (NaCl) and Sylrite (KCl), together with bischofite (MgCl2 . 6 H2O), Carnallite (KCl MgCl2 . 6 H2O), Tachhydrite (CaCl2 . 2MgCl2 . 12 H2O) and others. Occasionally it is also found as heavy metal chlorides, usually in the form of double salts such as

atacamite (CuCl2 .3Cu(OH)2). Plants and animals always contain chlorine in the form of chlorides or free HCl.

Cl2 is formed by oxidation of HCl or chlorides by compounds such as Manganese dioxide, permanganates, dichromate, chlorates or bleaching

powder. Oxygen from the atmosphere acts as an oxidizing agent in the

presence of catalysts.

Initially chlorine was prepared by oxidation of HCl using manganese

dioxide for bleaching purposes in textile industries. This process was

developed by WELDON but the yield obtained was very less and was about 35%. Later DEACON developed a process based on oxidation of HCl gas by atmospheric oxygen in the presence of copper salt( CuCl2) and the yield was 65% of the theoretical value.

In 1800 A.D CRUICKSHANK first prepared chlorine

electrochemically. In this process a diaphragm cell or a mercury cell is used. Currently 95% of worlds production of Cl2 is done by using this chlor- alkali process. Today worlds capacity of production of chlorine is more than 80 x 106 tones/ annum.

CHLORINE IN INDIA:

The birth of chlorine industries in India has been recent and dates back to the 1920s. The plants were set up mainly to meet the requirement of the pulp and bleaching industries. However, with the start of the caustic soda

industries, the chlorine was available in plenty and it was even necessary to dispose it off. The main plants in India that manufactures chlorine by the

electrolysis of brine are: -

1. D.C.M Chemicals 2. Tata Chemicals

3. National rayon corporation 4. Travancore Cochin Chemicals

5. Mettur Chemicals Almost all of the above plants were set up with a short span of years,

with the result that there was a marked increases in the chlorine production.

But soon, with the Government of India following a policy of planned development, it was possible to find channels where chlorine could be

effectively utilized. With the setting up of organic chemical industries for

the manufacture of such diverse products as PVC, DDT, BHC, etc, the demand of chlorine has gone up.

PHYSICAL PROPERTIES OF CHLORINE: Chlorine exists in all three physical states. At NTP it is a greenish

yellow pungent, poisonous gas, which can be liquefied to a mobile yellow

liquid. Solid chlorine forms pale yellow rhombic crystals. Some of the important physical properties of chlorine are listed below:

PROPERTIES

Atomic Number (Z) 17 Atomic Mass (A) 35.453 g

Stable isotope mass 35, 37 g

Melting Point (mp) 172.17 K Boiling Point (bp) 239.10 K

Critical temperature (Tc) 417.15 K Critical Pressure (Pc) 7.71 x 105 N/m2 Critical density (c) 565 Kg / m3

Density at 0C 3.213 Kg / m3

Density relative to air 2.48 Kg / m3

Enthalpy of fusion Hf 90.33 KJ/Kg

Enthalpy of vaporization Hv 287.1 KJ/Kg

Standard Electrode Potential E0 1.359 V

Enthalpy of dissociation Hd 239.44 KJ/ Kmol

Electron affinity 364.25 KJ/Kmol

Enthalpy of hydration 405.7 KJ/ Kmol

USES OF CHLORINE:

During the span of chlorine industry in the world, the largest single use has been in the manufacture of pulp and paper. In the early period of the

industry, the second largest use was in sanitation for sewage treatment and purification of water.

Later on the demand increased because of the invention of new

compounds. Some of the major uses of chlorine are listed below: 1. Benzene hexachloride: Manufacture of insecticides

2. Carbon tetrachloride: Manufacture of fluorocarbons for refrigerants

and propellants. 3. Chloral: insecticides manufacture

4. Chloro paraffins: lubricant additives and paints 5. Dichlorobenzene: organic solvents 6. Chloroform: manufacture of fluorocarbons 7. Ethyl Chloride: antiknock compounds 8. Ethylene oxide and glycol: antifreeze fluids and synthetic fibres

9. Methyl chloride: silicones manufacture; solvent and catalyst carrier. 10. Methylene chloride: paint remover, propellants and general solvent

11. Monochloroacetic acid: herbicides, detergents

12. Perchloroethylene: dry cleaning 13. Tri chloroethylene: metal degreasing

14. Vinyl chloride: plastic and resin products

METHODS OF PRODUCTION AND PROCESS SELECTION

1. KEL CHLORINE PROCESS: This process was developed by KELLOG, which uses HCl as a raw

material from the waste gases of a fluorinated hydrocarbon plant.

Concentrated sulphuric acid and 1% nitrosylsulphuric acid are used as

catalysts.

Sulphuric acid catalyst is fed from the top of the stripper column. The

HCl gas reacts with the catalyst to form nitrosyl chloride. O2, the ultimate

oxidizing agent blows the remaining HCl out of the H2SO4, which becomes more concentrated and then get cooled in the flash vaporizer. This acid is

then fed back into the process. Nitrosyl chloride, HCl, O2 and water vapour flows as gaseous streams into the oxidizer and reacts there, resulting

increase in the temperature. In the absorber rest of the HCl is oxidized.

Concentrated H2SO4 acid is fed at the top, reacts with the oxides and N2 to form nitrosyl sulphuric acid, absorbs the water that has formed and is fed

back into the stripper. The cooled dried chlorine gas still contains 2% HCl and up to 10%

O2. Both are removed by liquefaction.

On the account of the aggressive nature of the chemicals, expensive materials, such as tantalum plated equipment and pipes must be used.

2. OXIDATION OF HCl BY NITRIC ACID:

The nitrosyl chloride route to chlorine is based on the strongly

oxidizing properties of nitric acid. The practical problem lies in the separation of the chlorine from HCl

and other gaseous components. The dilute nitric acid must be concentrated for further use. Corrosion problems are severe.

3. DEACON PROCESS:

Air oxidizes HCl in vapour phase, over a hot copper containing

catalyst (CuCl2). The principal difficulties of fugitive CuCl2, low Cl2 composition in the exit stream, limited available materials of construction

with high maintenance and poor catalyst activity makes this process unsuitable.

4. AIRCO PROCESS:

The Air Reduction Company has made a number of developments in the basic Deacon process in recent years using both air and O2. The net

result indicates that the process is practical today from an industrial by

product containing HCl. HCl is mixed with air or O2 and preheated in a heat exchanger and

sent to the fluidised bed reactor that operates at 300C to 500C.

4 HCl + O2 2 Cl2 + 2 H2O +

The product gases from reactor is cooled to 110C

and fed to the

product gas cooler where substantially all the water from reaction is

condensed. The top stream of the product gas cooler contains mainly Cl2, O2,

N2, H2O and HCl. This stream is sent to product gas absorber where the HCl gas is absorbed with 22% HCl solution and the rest of the gases leave the

absorber from the top. This stream is then taken to liquefaction system from which the pure chlorine is obtained.

This process is chosen for the production since the problem of

catalysis has been solved by the use of improved copper base catalyst bodies with at least one year of useful activity, utilizing rare earth as promoters and

accelerators in a special reverse in flow reactor which makes the reaction

self sustaining without the addition of external heat. New methods, materials of construction and engineering technique make this process economically

feasible.

PROCESS DESCRIPTION HCl is mixed with air, fed into a fluidised bed reactor containing

cupric chloride catalyst, and maintained at a suitable temperature in the

range of 300 - 500C. HCl in the feed reacts with oxygen or oxidizes to give

chlorine and water. The product gas containing chlorine, water, unchanged HCl and inert gases are passed to a packed tower cooler/ scrubber, operating

somewhat above the atmospheric pressure. In the tower the gases comes in contact with 33 36% HCl, thereby cooling the insoluble gases and

absorbing the unchanged HCl. The acid enters the tower at 20C. Most of

the water and some amount of HCl contained in the product gas are

dissolved in the acid. The liquid effluent coming out of the scrubber is split up into two streams. One stream is passed to the top of the scrubber through

a cooler, which lowers the temperature to 20C while the other is sent to the

top of a stripper column (Expeller). In the stripper, the HCl present in the

incoming stream from the product gas cooler is stripped off and hence a gas

containing around 98% HCl (the other constituents being water and chlorine) leaves the stripper from the top that is fed back into the reactor. A mixture of water and HCl containing 20 22% HCl leaves from the base of

the expeller at a temperature of 147C, which is fed to a HCl absorber

through a cooler and the flow rate is maintained around 1000 kg/ hr. In the absorber, almost all of the HCl coming from the product gas cooler/ scrubber

are absorbed. The liquid leaving the base of the HCl absorber contains 33

36% of HCl which is then mixed with fresh 36% HCl and then fed to the top of the expeller. The gaseous chlorine leaving the top of the HCl absorber

column is then dried and sent to liquefaction unit to get pure liquid chlorine.

The workings of various equipments are described as follows:

1. The catalytic reactor consists of a U shaped steel container with Foamsil lined catalyst container. Both feed and

discharge ports are at the bottom. Periodic reversal of flow, controls reaction temperature, preheats the feed. The copper

catalyst acts as an efficient heat regenerator. Gases leaves the

reactor slightly below 200C

2. In the cooler, product gas is scrubbed by cold 33 36% HCl solution. Substantially all the water of reaction condenses to

produce a concentrated solution of HCl, which is stripped of its

HCl content in the expeller described below. The cooler is internally lined with non corrosive material.

3. In the absorber, the cooled reactor gas stripped of its water

content is absorbed in cold 20 22% HCl solution in a falling film unit of karbate and plastic construction. It leaves as 33

35% solution, which is then stripped in the expeller. The absorbing liquid (22% HCl) is the cooled foots of the expeller.

4. The expeller or stripper is a karbate packed tower which is fed with strong & cold HCl solution which is stripped to produce a

product of 98% HCl gas and foots as close as 20% solution of HCl. The water build up is discharged from the system by stripping HCl as gas, in a separate system.

5. The dry gas from the absorption tower is then fed to the liquefaction unit where most of the sniff gases and O2 is removed and chlorine emerges as the pure liquid.

THERMODYNAMIC CONSIDERATIONS

The oxidation of hydrogen chloride is a reversible reaction and hence

the feasibility of the desired forward reaction is highly dependent upon various factors such as temperature, pressure, etc. Arnold and Kobe have

studied the effect of varying these factors. They have calculated the

thermodynamic functions of hydrogen chloride and chlorine from spectroscopic measurements. The reaction is:

4 HCl + O2 2 Cl2 + H2 O ; Hr = - 192 kcal / kg HCl

The equilibrium constant K for this reaction has been calculated from the following equation.

-G / T = 2.303 R log K

where G = change in free energy of the reaction

= - 274460 + 8.65 T 0.00229 T2 + 0.263 x 10-6 T3

T = temperature

R = gas constant

Effect of other variables:

Pressure:

Increase in the pressure gives an increase in the HCl conversion and hence increase in the chlorine content. However, the pressure chosen for

operation is only 1 atm. This is done after considering the economics point of view of the plant. High-pressure operation involves the compression of

gases, which requires considerable amount of money. Secondly, the reactor

will have to be designed to withstand the pressure too. At high pressure, the equipment is highly prone to failure due to stress corrosion cracking.

Dilution of oxygen with inert gases: If pure oxygen is used instead of air, higher content of chlorine is

obtained in the product gas. At 350C, the HCl conversion increases by 2%.

And the chlorine recovery also becomes simpler. But due to the high cost of O2 as compared to the air, one usually goes for air.

Hydrogen chloride / Air Ratio: Increasing the excess oxygen content from 0 to 300% increases the

conversion by 10% at 700K, but the chlorine content decreases by 12%.

Thus the cost of recovery increases.

MATERIAL BALANCE

BALANCE AROUND REACTOR Basis: 1 hour of operation

Assume on stream time = 7500 hrs/ year Chlorine production = (10000 x 1000)/7500 = 1333.33 kg/ hr. = 18.51 kmol/hr From stoichiometry of the reaction:

4 HCl + O2 2 Cl2 + 2 H2O

No. of moles of water produced = 18.51 kmol/ hr

= 333.33 kg/hr

Theoretical O2 required = 18.496/ 2

=9.255 kmol/hr = 296.16 kg/hr

Theoretical amount of HCl required = 18.51 x 2 = 37.02 kmol/hr

= 1350.208 kg/hr Assuming 75% conversion, Actual amount of HCl required = 1350.208/ 0.75 = 1800.25 kg/hr Amount of water in 98% HCl gas = 1800.25/0.98 1800.25 = 36.74 kg/hr Since the feed contains 98% HCl and rest 2% water,

Total flow rate of feed (HCl +Water) = 1800.25 +36.74 = 1836.99 kg/hr Assuming 30% excess oxygen,

Actual amount of oxygen required = 9.251 x 1.3

= 12.026 kmol/ hr = 384.704 kg/hr Assuming 5% inert as nitrogen, Amount of nitrogen in oxygen = 12.022/0.95 12.022 = 0.6327 kmol/ hr = 17.71 kg/hr

Actual molar ratio between HCl and O2

HCl / O2 = 49.322/ 12.655 = 3.897 Unreacted HCl =1800.277 1350.208 = 450.045 kg /hr Unreacted oxygen = 9.251 x 0.3 = 2.7744 kmol/hr

= 88.78 kg/hr

COMPONENT MATERIAL IN (KG)

MATERIAL OUT (KG)

HCl 1800.27 450.045

O2 384.704 88.78

N2 17.71 17.71

H2O 36.74 369.709

Cl2 - 1313.216

TOTAL 2239.461 2239.461

BALANCE AROUND PRODUCT GAS COOLER:

G

Cl2 = 1313.216 kg HCl = 450.045kg I = 2239.48 Y, 36% HCl O2 = 88.78 kg N2 = 17.71 kg

H2O = 369.709 kg Overall balance I = G + Y

HCl balance:

450.045 = Amount of HCl in G + 0.36 x Y Water Balance:

369.709 = Y x (1 0.36 - 0.005) Y = 582.211 kg/hr

Amount of chlorine in Y = 0.005 x 582.211 = 2.911 kg/hr Amount of HCl in G = 450.045 0.36 x 582.11 = 240.485 kg/hr

G = I Y = 2239.481 582.11 = 1657.281 kg/hr

MATERIAL OUT (kg)

COMPONENT MATERIAL IN I (kg)

G Y

HCl 450.045 240.48 209.56 CL2 1313.216 1310.30 2.91 O2 88.78 88.78 - N2 17.71 17.71 -

H2O 369.709 - 369.709

BALANCE AROUND EXPELLER:

B, 98% HCl

Y =582.11 D, 36% HCl 36% HCl C, 22% HCl

Overall balance: Y + D = B + C

98% of B = 1800.277 kg/hr

Total amount of B = 1800.277/ 0.98

= 1837.017 kg/hr. Now, Y = 582.11 kg/hr

The overall equation can be represented as

582.11 + D = 1837.017 + C (1) HCl balance: 582.11 x 0.36 + 0.36 D = 0.98 B + 0.22 C (2) Solving 1 & 2 we get, C = 8132.956 kg/hr D = 9387.8462 kg/ hr

MATERIAL IN (kg) MATERIAL OUT(kg) COMPONENT Y D B C

HCl Cl2 O2 N2

H2O

209.56 2.91

-

-

396.64

3379.62 -

-

-

6008.22

1800 2.91

-

-

33.83

1789.25 -

-

-

3643.706

TOTAL 9969.95 9969.95

BALANCE AROUND ABSORBER:

E

G= 1657.281 kg/hr HCl = 240.4875 C = 1000 O2 = 88.78 G Kg/hr N2 = 17.71 Cl2 = 1310.305

F, 36% HCl Overall balance: G + C = E+ F

E + F = 2657.28 (1) Assuming 0.5% HCl present in the stream E, the HCl balance is: 240.485 + 0.22 x 1000 = 0.005 E + 0.36 F

0.005 E + 0.36 F = 460.485 (2) Solving equation 1 & 2 we get

E = 1397.56 kg/hr. F = 1259.71 kg/hr.

MATERIAL IN (kg) MATERIAL OUT(kg) COMPONENT G C E F

HCl Cl2 O2 N2

H2O

240.485 1310.305

88.78 17.71

-

220.0 -

-

-

780.0

6.98 1284.08 88.78 17.71

-

453.495 26.225

-

-

780.0

TOTAL 2657.28 2657.28

ENERGY BALANCE

EQUATIONS FOR SPECIFIC HEAT: COMPONENT EQUATION

(Kcal/K mol) TEMP. RANGE

(K) Cl2 HCl N2 O2

8.28 +0.00056T 6.7+0.00084T 6.5 + 0.001T

8.27 + 0.000258T- 187700/T2

273 2000 273 2000 300 3000 300 - 5000

BASIS: 1 HOUR OF OPERATION

HEAT BALANCE IN REACTOR:

Heat of reaction at 450C = - 192 Kcal/ kg of HCl reacted

Total heat of reaction Hr = - 192 x 1350.25 x 4.18

= - 1083656 kJ/hr

The feed enters into the reactor at temperature of 20C.

Heat input, Qi = m Cp T =[(384.704/32 x 6.1592) + (17.71/28

x 6.79) + (1800.2/36.5 x 6.946) + (36.74/ 18 x 4.18)] x 4.18 x (20 0) = 35356.635 kJ/hr. Let T be the temperature of the product stream, which leaves the

reactor.

Heat output, Qo = m Cp T = [(1313.216/71 x 8.24) + (88.78/32

x 7.089) + (450.045/36.5 x 7.013) + (17.71/ 28 x 6.87) + (369.71/18 x 8.456)] x 4.18 x (T 273) + (369.7 x 2253.27)

Qo = [1844.04 x (T 273)] + 833047.443 At steady state;

Input + Generation = Output

Qi + Hr = Qo 35356.635 + 1083456 = 1844.04 X (T 273 ) + 833047.44 Solving the equation we get

T = 428.2 K or T = 155.2 C

HEAT BALANCE AROUND PRODUCT GAS COOLER:

Product from the converter leaves at 155.2 C. This stream is then

cooled to 110C in a reverse flow heat exchanger and then fed at the bottom

of the product gas cooler.

Substantially all the water present in the product gas condenses in the

cooler. And hence the temperature of the streams M & Y are assumed to be

60C.

I = 2239.46 G = 1657, Tg

X = 250, 20C M = 824.11, 60C

Heat input:

Qi = mi Cp ( T TR) The reference temperature TR= 0C.

Qi = 2239.46 x 0.6553 x 110 = 161427 kJ/hr Qx = 250 x 2.92 x 20 = 14600 kJ/hr Heat output:

Qm = 824.11 x 2.98 x 60 = 147350.87 kJ/hr Qg = 1657 x 0.567 x Tg = 939.52 Tg Heat input = Heat output

161427 + 14600 = 147350.87 + 939.52 Tg Tg = 30C

HEAT BALANCE AROUND EXPELLER:

B = 1837, 20C

Y= 582.11, D = 9387.84, 30C

60C

Qs

C = 8132.95, 147C

The feed is fed to the expeller from HCl tank. This feed temperature is

assumed to be 30 C.

Tc = 147 C

Heat input:

QY = 852.11 x 2.94 x 60 = 102684.2 kJ/hr QD = 9387.84 x 2.52 x 30 = 709872.8 kJ/hr

Heat output:

QB = 1837 x 0.497 x 20 = 18277.2 kJ/hr QC = 8132.95 x 2.926 x 147 = 3298160.7 kJ/hr Heat input = heat output

QS + 102684.2 + 709872.8 = 18277.2 + 3498160.7 QS = 2703880.9 kJ/hr = 751078105 J/sec. Assuming process steam is available at 5 atm.

5 atm = 2109.687 kJ/kg. Steam flow rate ms = 751078.03/ 2109687 = 0.356 kg/sec.

HEAT BALANCE AROUND ABSORBER: E= 1397.56 C = 1000

G = 1657.281 F = 1259.71

Let QAB be the heat of absorption of HCl at 30C. QAB = 420 kcal / kg HCl. HCl absorbed = 453.49 220 = 233.49 kg/hr.

Total heat of absorption = 420 x 4.18 x 233.49

Qab = 409915.04 kJ/hr. Qab + Q + QG +QC = QE + QF Qg = 939.52 x 30 = 28185.6 kJ/hr. Qc = 1000 x 0.7 x 4.18 x 30 = 87780 kJ/hr. QF = 1259.56 x 2.94 x 30 = 111093.2 kJ/hr. QE = 1397.56 x 0.54 x 30 = 22640.472 kJ/hr. Substituting all the values in the above equation we get Q = 392146.97 kJ/hr.

DESIGN OF HEAT EXCHANGER:

147C 30C

45C 25

Hot HCl solution at 147 C is to be cooled from that temperature to 30C.

Let the cooling water is available at 25C and it is heated to 45C. Let mc be the mass flow rate of water. PROPERTIES:

COMPONENT Tb

C Cp

kJ/kg C

c.p

K

W/m C

kg/m3

Aq HCl 88.5 2.926 0.95 0.69 1072.4

Water 37.5 4.18 0.75 0.628 993.148

HEAT BALANCE: mh = 8132.95 / 3600 = 2.26 kg/sec

Qh = mh Cp T = 2.26 x 2.926 x (147 30) = 773.69 kW

Qc = mc Cp T = mc x 4.18 x ( 20) mc = 9.255 kg/sec LMTD = [ (147 45) (45 25)] / ln [(147 45) / (45 25)] = 32.16 C

R = (147 30) / (45 25) = 5.8 S = (45 25)/ (147 25) = 0.164 From the graph; Ft = 0.9

Actual LMTD = 32.16 x 0.9 = 28.94 C

HEAT TRANSFER CALCULATION:

Let us assume HCl in shell side; Water in tube side. Choose UD= 500 W/m2 C

Cross Sectional Area = Q/ UD T = 773.69 x 103/ 500 x 28.94 = 53.46 m2 Let us choose OD 16 BWG tubes.

at= 0.3048 x 0.1963 = 0.0598 m2/m.

Let us choose 12 ft pipe.

L = 3.66 m

Heat transfer area per tube = 3.66 x 0.0598

= 0.219 m2

Number of tubes = 53.46/0.219 = 245 tubes

Now, let us choose TEMA P or S, 1 4 type of heat exchanger with

1 triangular pitch.

Nt = 290

Actual area = 290 x 0.219

= 63.51 m2

(UD)corrected = 773.69 x 103 / (63.51 x 28.94) = 441 W/m2C

TUBE SIDE VELOCITY:

At = nt / np x /4 x di2

= 290 / 4 x /4 x (0.015748)2 = 0.014 m2

vt = m/ ( At) = 9.255/(993.148 x 0.014) = 0.6656 m/s

SHELL SIDE VELOCITY: Sm = (P d0) Ls x Ds/ P Ds = 540 mm Sm = [(25.4 19.05) x 10-3 x (540 x 10-3) x 0.3] / 2.54 x 10-2 = 0.022 m2

vs = m/ a = 2.26 / (1072.4 x .022) = 0.9636 m/s

HEAT TRANSFER COEFFICIENT CALCULATION:

SHELL SIDE: d0 = 19.05 x 10-3 m di = 15.75 x 10-3 m

Reynolds Number = v do / = (1072.4 x 0.9636 x 19.05 x 10-3)/0.95 x 10-3) =20721.68

Prandtls Number = Cp / K = 2.926 x 103 x 0.95 x 10-3 / 0.69 = 4.028 For Reynolds number above 10000, Sieder Tate equation holds

good.

ho do/ K = 0.023 x NRe0.8 x NPr0.33 (b / w)0.14 b = 1.2 c.p

w = 1.7 c.p

hodo/K = 0.023 x (20721.68)0.8 (4.028)0.33 x (1.2/ 1.7)0.14

= 98.89

ho = 3582.1 W/m2 C

TUBE SIDE:

Reynolds Number = v di/ = 993.19 x 0.6656 x 15.75 x 10-3/ = 13882.41

Prandtls Number = Cp / K = 4.18 x 103 x 0.75 x 10-3/ 0.69 = 4.54 For this value of Reynolds number Dittus Bolter equation holds

good.

hidi/K = 0.023 x NRe0.8 x NPr0.33

= 0.023 x (13882.41)0.8 x (4.54)0.33

= 78.078

hi = 3108 W/m2 K

Assuming dirt coefficient hd = 1000 W/ m2C

We have,

1/ U = 1/ ho + Do/Di x 1/ hi + 1/ hd + [Do ln (Do / Di)] / 2 x Kw 1/U= 1/ 3582.1 + (1/3108 x 19.05/15.78) + 1/ 1000 + (19.05 x 10-3

ln(19.05/15.78) x 1/32) = 1.99 x 10-3

U= 502.43 W/m2C

PRESSURE DROP CALCULATION: TUBE SIDE:

Friction factor, f = 0.079 NRe-0.25 = 0.079 x (13882.41)-0.25

= 7.278 x 10-3

h = 4f L Vt2 / 2g Di

= 4 x 7.278 x 10-3 x 3.66 x 0.66562/ (15.75 x 10-3 x 2 x 9.8) = 0.153 m

PL = g h = 993.18 x 9.8 x 0.153 = 1.49 kPa

Pc = 2.5( Vt2/2) = 2.5 ( 993.18 x 0.66562/2) = 0.55 kPa

PT = [Pc + PL] x Np = [1.49 + 0.55] x 4 = 8.16 kPa

SHELL SIDE:

From Perry,

Pc = bfk w2 Nc / Sm2 ( w/b)0.14

= 2 x 10-3 x 0.08 x 2.262 x Nc x 1.420.14 / (1072.4 x0.0222) = 1.6537 x 10-3 Nc Nc = Ds[ 1 2(lc/Ds)] / Pp = 540 x 10-3 [1 2(0.25)] / 2.54 x 10-2

= 10.63 11

Pc = 11 x 1.6537 x 10-3

= 0.018 kPa

MECHANICAL DESIGN:

Hastalloy is the material of construction, whose density is 8600 kg/m3 and the permissible stress is ft = 860 kg/cm2 SHELL SIDE:

1. SHELL THICKNESS: ts = [P D/ (2fJ + P)] + C P = 5 kg/cm2 ; D = 540 mm; f = 860 kg/cm2; J = 85%

ts = [5 x 540/ (2 x 860 x .85 + 5)] + 2 = 3.84 mm

4 mm

2. NOZZLE DIAMETER:

Mass flow rate, m = A v

A = m/v = 2.26/1072.4 x 0.9626 = 5.18 x 10-3

Diameter of the nozzle = [5.18 x 10-3 x 4/]1/2

= 0.0812 m

= 8.12 cm 3. NOZZLE THICKNESS:

tn = [PD/(2 fJ p)] + C = [5 x 81.2/ {(2 x 860 x 0.85) 5}] + 2 = 2.2786 mm

3 mm

4. HEAD THICKNESS:

Assuming a torrispherical head, the thickness can be calculated as: th = PRcW/2fJ

P = 5

W = ( 3 + Rc/Rk) = 1.34

th = (5 x 540 x 1.34)/(2 x 860 x 0.85) = 2.47 mm

Taking 2mm corrosion allowance we have,

th = 2.47 + 2

= 4.47 mm

5 mm

TUBE SIDE: 5. TUBE SHEET THICKNESS:

tts = FG (0.25 P/f)0.5

= 1 x 400 (0.25 x 5/860)0.5

= 15 mm 6. CHANNEL DESIGN:

a. Channel Length = 1.3 x (cross sectional area of tube/pass)/Ds = 1.3 x 290/4 x /4 x (1.5748)2 / 54 = 3.39 mm

4 mm

b. Channel thickness:

tc = Gc (kP/f)1/2

= 400 x (0.3 x 5/860)1/2

= 16.70 mm 7. NOZZLE DESIGN:

m = A v

A = m/v = 9.255/(993.18 x 0.6656)

= 0.014 m2

Diameter of the nozzle = d = (0.014 x 4/)1/2

= 0.1335 m = 13.35 cm SUPPORT DESIGN:

For this shell and tube heat exchanger we use a saddle type of support.

1) SHELL WEIGHT: Ws = (r02 ri2) x h x = (0.5462 0.542) x 3.7 x 8600 = 651.37 kg

2) TUBE WEIGHT: Wt = /4 x [ro2 ri2] h nt

= /4 x [(17.4 x 10-3)2 (15.75 x 10-3)2] x 3.7 x 290 x 8600 = 396.4 kg

3) WEIGHT OF LIQUID: Wl = 700 kg

4) WEIGHT OF TIE ROD, END COVERS, BAFFLES: WA = 500 kg

Total weight = 652 + 397 + 700 + 500

= 2249 kg Assume total weight to be equal to 2500 kg.

5) LONGITUDINAL BENDING MOMENT: M1 = QA[ 1 {(1 A/L + (R2 H2)/2AL)/(1 + 4/3 H/L)}] Q = W/2 [L + 4/3 H] L = 3.7m, H = 0.2 m Q = 2500/4 x 2 [3.7 + 4/3 x 0.2] = 1239.6 kg

Let us take A = 0.25 m

M1 = 1239.6 x 25 [1 {1 25/366 + (542 202)/(2 x 25 x 366)}/{1 + 4/3 x 20/366}]

= 1239.6 x 25 [1 (1.069/1.0728)] = 104.56 kg cm M2 = QL/4 [ {1 + 2(R2 H2)/L2}/{1 + 4/3 H/L} 4A/L] = 1239 x 366/4 [{1+2(542 202)/3662}/{1+4/3x 20/366} 25 x

4/366] = 78702 kg cm

6) STRESSES IN SHELL AT THE SADDLE: f1 = M1 / k1R2t

= 104.56/(0.107 x x 542 x 0.5) = 0.213 kg/cm2

f2 = M1/(k2R2t) = 104.56/(0.192 x x 542 x 0.5) = 0.118 kg/cm2

The stresses are well within the permissible limit. 7) STRESS IN THE SHELL AT MID SPAN:

f3 = M2 / R2 t = 789702/( x 542 x 0.5) = 17.18 kg/cm2

Axial stress due to internal pressure:

Fp = pD/4t = 5 x 540/4 x (4-2) = 337.5 kg/cm2

The combined stresses (fp + f1), (fp f2) and (fp + f3) are well within limit. Hence the design can withstand the load.

DESIGN OF ABSORBER

E = 1397.56 kg/hr C = 1000 kg/hr Cl2 = 1284.08 HCl = 220

HCl = 6.98 H2O = 780 O2 = 88.78 Yt Xt N2 = 17.71

G = 1657.28

F = 1259.72 O2 = 88.78 HCl = 453.495 N2 = 17.71 Yb Xb H2O = 780 Cl2 = 1310.305 Cl2 = 26.225

Gm = Molar flow rate of inert in the feed

Lm = Molar flow rate of solvent (36 % HCl) Y = Mole ratio of HCl in gas phase X = Mole ratio of HCl in the liquid phase

Yb = 6.588/ (18.455 + 0.6325 + 2.77) = 0.3014 kmol of HCl / kmol of inert in the feed.

At the top,

Yt = 0.19123 / ( 18.455 +2.77 + 0.6325) = 0.00889 kmol HCl / kmol inert

Gas flow rate at the bottom of the tower, Gm = 1657.288 kg/hr = 0.46 kg/sec Average molecular weight of the feed gas Mf = 0.231 x 36.5 + 0.648 x 71 + 0.0973 x 32 + 0.0222 x28 = 58.175 kg / kmol

Gm = 0.46 / 58.78 = 0.0079 kmol/ sec

Inert in the feed (Gm) = Gm x mole fraction of inert = 0.0079 x 0.7684 = 0.00607 kmol/sec Xt = mol of HCl /mol of inert in the solvent

Xt = 6.0274 / 43.33 = 0.139 [ Lm / Gm]min = ( Yb Yt) / ( Xb* - Xt )

where Xb* is the equilibrium composition of liquid stream which is

obtained from the X-Y plot. The equilibrium data is given below:

X Y X Y

0.01 1.105 x 10-7 0.139 1.343 x 10-3

0.02

0.0315 0.043

0.055 0.067 0.08

0.094 0.1074

0.123

1.013 x 10 -6

2.98 x 10-6

6.78 x 10-6

1.46 x 10-5

3.08 x 10-5

6.58 x 10-5

1.39 x 10-4

3.0 x 10-4

6.319 x 10-4

0.156 0.1733 0.192 0.211 0.232

0.254 0.277 0.302

0.329

2.868 x 10-3

6.032 x 10-3

0.01317

0.0288 0.0622 0.1376 0.328 0.901 4.71

From the graph we get, Xb* = 0.276

[ Lm / Gm]min = ( 0.3014 0.00889) / (0.276 0.139) = 2.135 Taking [Lm/ Gm] = 2.5 x [ Lm/ Gm] = 2.5 x 2.135 = 5.3375

Lm = Gm x 5.3375

= 0.00607 x 5.3375 = 0.0324 kmol/sec

Average molecular weight of the solvent;

Ms = 0.122 x 36.5 + 0.8779 x 18 = 20.255 kg/ kmol

Mass flow rate of inert,

L = Lm x Ms

= 0.0324 x 20.255 = 0.656 kg/sec

Gas flow rate at the bottom of the tower,

Gb = G + (Gm Yb) x mol wt of HCl = 0.00607 x 58.175 + (0.00607 x 0.3014 x 36.5 ) = 0.4197 kg/sec Liquid flow rate at the bottom of the tower,

Lb = L + Gm ( Yb - Yt ) x 36.5 = 0.656 + 0.00607 x ( 0.3014 0.00889) x 36.5 = 0.656 + 0.0648 = 0.7208 kg/ sec

Density of the feed gas,

g = (M x 273) / ( 22.7 x 303 ) = (58.28 x 273) x ( 303 x 22.7) = 2.3 kg/ m3

Density of the solvent at 30C = 1103.4 kg / m3

To specify flooding conditions,

Choose 1 ceramic rasching rings as packing material with diameter d = 25.4 mm

From perry,

Surface area a = 190 m2/ m3

= 0.74

Fb = 510 m-1

Calculating [ Lb / Gb] [l / g ]1/2 = (0.7208/0.4197) x [2.3/ 1103.4]1/2

= 0.07842

For the above value from perry,

Gf2Fb0.2 gl g = 0.17 ( A)

= water / solvent = 992.2 / 1103.4 = 0.889

l = 1.6 x 10-3

Substituting all the values in the equation (A) , [Gf2 x 510 x 0.889 x (1.6 x 10-3)0.2 ] / [2.3 x 1103.4 x9.81] = 0.17 Gf2 = 33.45

Gf = 5.78 kg/m2 sec

= 20822.1 kg/ m2 hr

Cross sectional area of the tower (A) = Gb/ 80% of Gf = 1511 / 0.8 x 20822.1

A = 0.0907 m2

Diameter of the tower, Dc = [ 0.0907 x 4 / ]1/2

= 0.340 m Diameter of tower / Diameter of the packing = 0.340/0.0254 = 13.3

Which is greater than 10

PRESSURE DROP CALCULATION: LEVAS CORRELATION:

P = C2 10 3 tl x g Utg2 (B)

P = Pressure drop in Inch H2O / ft packing

g = Density of gas in lb/ft3

Utg , Utl = Velocity of gases in ft/sec C2 & C3 are constants

Superficial liquid flow rate, L = L / cross sectional area of tower

= 0.656 / 0.0907

= 7.2326 kg/m2sec = 5333.8 lb / ft2 hr For this liquid flow rate constants C2 & C3 are obtained from the perry

for the rasching rings,

C2 = 0.8 & C3 = 0.0348

Superficial gas flow rate

G = Gb / g = 0.4197 / 0.0907 = 4.627 kg/m2 sec

Utg = G / a = 1.2095 / 2.3 = 2.01 m/sec = 6.595 ft/sec

Utl = L /l = 7.2326/ 1103.45 = 6.554 x 10-3 m/sec = 0.0215 ft/sec

g = 0.143 lb/ft3

Substituting the values in equation (B) we get, P = 5 inch of water / Ft of packing

= 415.2 mm of H2O/ m of packing DEGREE OF WETTING:

Wetting rate Lp = Liquid rate (L) / specific area of packing Specific area = 190 m2/m3

Lp = 7.2326/ (1103.45 x 190) = 3.449 x 10-5 m3/m sec = 1.336 ft3 /ft hr

It is recommended that wetting rate for all packing should be 0.85 ft3/ft hr, except for rings of diameter greater than 3 inch. For all the

remaining packing materials the value is 1.3 ft3/ ft hr.

TOWER HEIGHT CALCULATION:

Z = HOG NOG HOG = Height of overall gas phase transfer unit

NOG = Number of overall gas phase transfer unit

HOG = HG + m [Gm/Lm] x HL CORNELLS RELATION:

HG 0.017 x x D1.24 x Z0.99 x Scg0.5

( L f1 f2 f3 )0.6 D = Diameter of the column, m

Z = Packed height, m

L = Liquid rate kg/m2 sec

F1 = [l / w]0.6 , f2 = [ m/l]1.25 , f3= [w / l ]0.8

Scg = Gas phase Schmidt number = g/ ( g Dg) Dg = Diffusivity of gas

f1 = [ 1.6/ 1.0]0.6 = 1.078 f2 = [1000/ 1103.4]1.25 = 0.8842 Now,

w = 72.8 dyne/cm

l = w xw + m xx ( solvent surface tension) m

1/4 = P ( l)

P = Parachor = Ph + PCl = 15.5 + 55.2 = 70.7

m1/4

= ( 15.5 +55.2 ) (0.03474 ) = 2.456 m = 36.39 dyne/cm

l = 0.122 x 36.39 + 0.8779 x 72.8

= 68.35 dyne/cm f3 = [72.8/68.35]0.8 = 1.0517 Scg = g/ g Dg

Dg = [T1.75 x 10-3 [ {MA + MB} / MA MB]1/2] / [ P { (v)A1/3 + ( v)B1/3}2]

MA = 36.5 ; MB = 71.0

( v)A = 16.5 + 19.5 = 36 ( v )B = 19.5 x 2 = 39 P = 1 atm, T = 303 K

Dg = [10-3 x (303)1.75 { 107.5 / 2591.5}0.5]/ 1 x [361/3+ 391/3]2

= 0.359 x 10-5 m2/sec

g = 0.014 cp

g= 2.3 kg/m3 Scg = 0.014 x 10-3 / (2.3 x 0.359 x 10-5) = 1.695 Corresponding to the 80% flooding for ceramic rasching rings;

= 70 m

HG = [0.017 x 70 x 0.341.24 Z1/3 (1.695)0.5] /[7.2326 x 1.078 x 0.884 x 1.0517 )0.6]

HG = 0.127 Z1/3

HL = [ C/3.28] x [l/ l Dl ]0.5 x [Z / 3.05]0.15

= Correction factor for a given packing in m

C = Correction factor for high gas rate

l = 1103.4 kg/m3

l = 1.6 x 10-3

= 0.07 m

C = 0.6 (for 80% flooding) Dl = 7.4 x 10-8 ( MB)1/2 T/ B VA0.6

VA = molar volume of solute A at its normal boiling temperature

= 30.68 cc/gmol

A = 1.48 gm/cc = 0.0405 gmol/cc MB = 20.25

De = [7.4 x 10-8 ( M)0.5 T] / [ Vm0.6] = [7.4 x 10-8 (2.6 x 20.255)0.5 x 303] / 1.6 x (30.68)0.6

= 1.3037 x 10-5 cm2/sec

HL = [0.07 x 0.6/3.28] x [1.6 x 10-3 /(1103.4 x 1.303 x 10-9)]0.5 [ Z / 3.05]0.15

HL = 0.3613 Z0.15 HOG = 0.127 Z1/3 + m (Gm/Lm) x 0.3613 Z0.15 m = Slope of equilibrium curve = 1.4106 Lm/Gm = Slope of operating line = 5.3375 HOG = 0.127 Z1/3 + 1.4106 x 1/5.3375 x 0.3613 Z0.15

HOG= 0.127 Z1/3 + 0.09545 Z0.15

Calculation of NOG:

Yt

NOG = dy/(y y*) + ln[(1 + YB) / ( 1 + YT)] Yb

Y* is generated from the x y plot:

Y Y* Y Y* 1/(Y Y*) 0.004 0.001 3 x 10-3 333.33

0.06 0.16 0.18

0.22 0.26 0.30

0.005 0.004 0.006 0.007 0.01

0.014

0.055 0.156 0.174

0.213 0.25

0.286

18.18

6.41 5.747 4.69 4.0

3.49

The plot of 1/ Y Y* vs Y is made and area under the curve is

calculated.

Area under the curve = [ 2 x 0.002] x 2073 = 8.292

NOG = 8.292 ln [(1 + 0.3014) / (1 + 0.00889)] = 8.1647

Z = HOG NOG = 1.0369 Z1/3 + 0.779 Z0.15

Solving the equation by trial and error method we get,

Z = 2.3 m

MECHANICAL DESIGN OF ABSORBER: Material of construction is chosen as Hastalloy.

Density of Hastalloy = 8600 kg/m3 Tensile strength = 860 kg/cm2 Compressive strength = 689 kg/cm2

Shell thickness: ts= Pi ri / ft J +C = 1 x 17.0/800 + 2 mm

= 2.2 mm

COMPRESSIVE STRESSES:

1. Due to shell weight:

Fc1 = weight/ cross sectional area

= X = 8600 X

2. Weight due to insulation: Insulation material chosen is asbestos

Taking insulation thickness ti = 3cm

Density of insulator (i) = 2200 kg/m3 fc2 = i ti/ts X = 2200 x 3/ 0.22 X

= 30000 X 3. Stress due to internals:

fc3 = 125 di2/ ( do2 di2) X di = 0.34 m, do= di + ti + 2ts= 0.3744 m

fcs = [125 x (0.34)2 / [0.37442 0.342] ] X = 588 X

4. Wind load:

Pressure due to wind load

Pw= 0.0024 Vw2

Vw = wind velocity in miles/hour = 501 miles/hr

Pw= 0.0024 (50)2= 6 lb/ft2 = 0.28728 kg/m2 fw = Pw deff X2/ 2 ro2 (ts c) = 0.28728 x 0.3944 x X2 / 2 x

(0.1872)2 x 0.2 x 10-3 = 2572.88 X2

TENSILE STRESSES: 1) With load:

ftensile= fwind + fall - fci (1) ftensile = 860 kg/cm2 fall = Pi ri / 2 (ts C) = 1 x 17/ 2 x 0.02 = 425 kg/cm2 = 425 x 104 kg/m2

fci = 39188 X Substituting all the values in equation (1) 860 x 104 = 2572.88 X2 39188X + 425 x 104 solving the equation we get;

X = 49 m 2) Without load:

Ftensile = fwind - fci 860 x 104 = 2572.88 X2 39188 X Solving the above equation,

X = 65.92 m The height of the tower obtained with and without load is greater than

2.3 m.

The thickness calculated is valid.

COMPRESSIVE STRENGTH: fcompression = 689 kg/m2

1) With load: fcomp = fwind fall + fci 689 x 104 = 2572.88 X2 425 x 104 + 39188 X On solving the equation,

X = 58.62 m 2) Without load:

fcomp = fwind + fci 689 x 104 = 257.88 X2 + 39188 X On solving we get,

X = 44.69 m Since the calculated value is greater than the actual height of the

tower, hence the thickness can withstand the compressive and tensile forces.

SKIRT DESIGN: The cylindrical shell for the skirt is designed for the combination of

the stresses due to vessel dead weight, wind load. The skirt thickness is

uniform and is designed to withstand the maximum values of tensile or

compressive stresses.

1) Due to dead weight: fdb = W/ Do tsk Weight of the shell = 1819 kg Liquid load = 0.46 kg Packing weight = 140 kg

Weight of cooling coils = 40 kg

W = 2000kg Choose skirt thickness to be 5 mm

fd = W/ Do ts Do = 0.3444 m

fd = 2000 / ( x 0.3444 x 5 x10-3) = 33.15 kg/cm2

2) Due to wind load: Plw = K p1 h1 D0

p1 = 50 kg/m2

h1 = 2.3 m D0 = 0.3444 m

Plm = 0.7 x 50 x2.3 x 0.3444 = 30.912 kg/m2 Bending moment due to wind at the base of the tower is determined

by:

Mw = Plw x H/2 = 30.912 x 2.3/2 = 35.548

The stresses fwb = Mw / Z = 4 Mw/ Dok2 tsk fwb = 4 x 35.5488 / x 0.384 x 5 x 10-3

= 23574.03 kg/m2 Maximum compressible stress on the skirt

= fwb + fdb = 23574.03 + 331572.8 = 355146.82 kg/m2 = 35.51 kg/cm2

Since allowable compressible stress for 5mm thick plates is 689 kg/cm2, which is much more than the actual stress, hence the designed vessel

can withstand the load.

GENERAL SAFETY PRECAUTIONS WHILE HANDLING CHLORINE

For safe handling of chlorine, it is necessary to adhere to the

following guidelines: 1. Never apply or expose cylinder to heat for any purposes. Never

place them in a bath of hot water or expose them to heat, to

increase gas discharge rate. 2. Store cylinders and containers far away from flammable and

other materials with which chlorine reacts and in a clean well

ventilated fire resistant area. 3. Never mix chlorine with other gas in a shipping container.

4. Never allow any liquid or moisture to enter the chlorine container.

5. Do not manifold cylinders being emptied in the liquid phase. 6. Permit only reliable and trained personal to handle chlorine. 7. Report all leaks and other irregularities to the supplier.

ECONOMICS Cost of Soda ash plant of capacity 10000 TPA in 1971 is Rs.3 x 106. Chemical Engineering Plant Cost Index: Cost index in 1971 = 132 Cost index in 2002 = 402

Thus, Present cost of Plant = (original cost) (present cost index)/(past cost index)

= (3 x 106) (402/132) = Rs. 8.74106 i.e., Fixed Capital Cost (FCI) = Rs. 8.74106

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 = 30% of Fixed-capital investment

i.e., PEC = 30% of 8.74106= 0.3 8.74106

= Rs. 2.622106 2. Installation, including insulation and painting: (25-55%

of purchased equipment cost.) Consider the Installation cost = 40% of Purchased

equipment cost

= 40% of 1.3325108 = 0.40 2.622106= Rs.1.05 x 106

3. Instrumentation and controls, installed: (6-30% of

Purchased equipment cost.) Consider the installation cost = 15% of Purchased

equipment cost

= 15% of = 0.15 2.622106= Rs. 0.393106

4. Piping installed: (10-80% of Purchased equipment cost)

Consider the piping cost = 40% Purchased equipment

cost

= 40% of Purchased equipment cost = 0.40 2.622106 = Rs. 1.04106

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

Consider Electrical cost = 25% of Purchased equipment cost = 25% of 2.622106= 0.25 2.622106= Rs. 0.655106

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

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

= 40% of 2.622106= 0.40 2.622106= Rs. 1.048108 C. Service facilities and yard improvements: (40-100% of Purchased equipment cost)

Consider the cost of service facilities and yard improvement = 50% of PEC

= 50% of 1.3325 108 = 0.50 2.622106= Rs. 1.311106

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

Consider the cost of land = 6% PEC = 6% of 1.3325 108 = 0.06 2.622106 = Rs 0.1311106

Thus, Direct cost = Rs. 8.5651 x 106 II. Indirect costs: expenses which are not directly involved with material

and labour 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 = 15% of

Direct costs

i.e., cost of engineering and supervision = 15% of 4.41 108 = 0.158.5651 x 106= Rs. 1.275106

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

Consider the construction expense and contractors fee = 10% of

Direct costs

i.e., construction expense and contractors fee = 10% of 8.5651 x 106

= 0.18.5651 x 106= Rs. 0.85106

C. Contingency: (5-15% of Fixed-capital investment) Consider the contingency cost = 5% of Fixed-capital investment i.e., Contingency cost = 5% of 5.33108 = 0.05 8.5651 x 106

= Rs. 0.425 106

Thus, Indirect Costs = Rs. 2.72106

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

= (2.72106) + (8.5651106) i.e., Fixed capital investment = Rs. 11.23106

IV. Working Capital: (10-20% of Fixed-capital investment) Consider the Working Capital = 15% of Fixed-capital

investment

i.e., Working capital = 15% of 5.9316108 = 0.15 11.23106 = Rs. 1.6845108

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

capital = (11.23106) + (1.6845106)

i.e., Total capital investment = Rs. 12.91106

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 = 10% of FCI for machinery and equipment

and 3% for Building Value for Buildings) i.e., Depreciation = (0.1011.23106) + (0.031.048106) = Rs. 1.154106

ii. Local Taxes: (1-4% of fixed capital investment) Consider the local taxes = 4% of fixed capital investment

i.e. Local Taxes = 0.0411.23106= Rs. 0.674106 iii. Insurances: (0.4-1% of fixed capital investment)

Consider the Insurance = 0.6% of fixed capital investment i.e. Insurance = 0.00611.23106= Rs. 0.0674106

iv. Rent: (8-12% of value of fixed capital investment) Consider rent = 10% of value of fixed capital investment

= 0.10 11.23106 Rent = Rs. 1.123x106

Thus, Fixed Charges = Rs. 2.794108

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 = 2.794106/15% Total product charge = 2.794106/0.15 Total product charge (TPC) = Rs. 18.63106

B. Direct Production Cost: (about 60% of total product cost)

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 18.63106 = 0.2518.63106 Raw material cost = Rs. 5.589106

ii. Operating Labour (OL): (10-20% of total product cost) Consider the cost of operating labour = 15% of total product cost Operating labour cost = 15% of 18.63106= 0.1518.63106 Operating labour cost = Rs. 2.79106

iii. Direct Supervisory and Clerical Labour (DS & CL): (10-25% of OL)

Consider the cost for Direct supervisory and clerical labour = 12% of

OL

Direct supervisory and clerical labour cost = 12% of 2.79106 = 0.122.79106 Direct supervisory and clerical labour cost = Rs. 0.335106

iv. Utilities: (10-20% of total product cost) Consider the cost of Utilities = 15% of total product cost Utilities cost= 15% of 18.63106= 0.1218.63106 Utilities cost = Rs. 2.79106

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 = 0.0511.23106 = Rs. 0.5615106

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 0.5615106 = 0.15 0.5615106 Operating supplies cost = Rs. 0.08422106

vii. Laboratory Charges: (10-20% of OL) Consider the Laboratory charges = 15% of OL Laboratory charges = 15% of 2.79106= 0.152.79106 Laboratory charges = Rs. 0.4185106

viii. Patent and Royalties: (0-6% of total product cost) Consider the cost of Patent and royalties = 5% of total product cost Patent and Royalties = 5% of 18.63106 = 0.0518.63106 Patent and Royalties cost = Rs. 0.9315106

Thus, Direct Production Cost = Rs. 13.49106 C. Plant overhead Costs (50-70% of Operating labour, supervision, and maintenance or 5-15% of total product cost); includes for the following: general plant upkeep and overhead, 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 = 0.60 ((2.79106) + (0.335106) +

(0.5615106)) Plant overhead cost = Rs. 2.212106

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

Manufacture cost = (2.794106) + (13.49106) + (2.212106)

Manufacture cost = Rs. 18.496106

II. General Expenses = Administrative costs + distribution and selling costs + research and development costs

Administrative costs :( 2-6% of total product cost) Consider the Administrative costs = 5% of total product cost Administrative costs = 0.05 2.79106 Administrative costs = Rs. 0.1395108

A. 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 = 10% of total product

cost

Distribution and selling costs = 10% of 18.63106 Distribution and selling costs = 0.15 18.63106 Distribution and Selling costs = Rs. 1.863106

C. Research and Development costs: (about 3% of total product cost) Consider the Research and development costs = 3% of total

product cost

Research and Development costs = 3% of 18.63106 Research and development costs = 0.05 18.63106 Research and Development costs = Rs. 0.5589106

Thus, General Expenses = Rs. 2.5614106

IV. Total Product cost = Manufacture cost + General Expenses

= (18.496106) + (2.5615106) Total product cost = Rs. 21.057108

V. Gross Earnings/Income: Wholesale Selling Price of chlorine per ton = $ 60 (USD) Let 1 USD = Rs. 50.00 Hence, Wholesale Selling Price of chlorine per ton. = 60 50 = Rs3000

Total Income = Selling price Quantity of product manufactured = 3000 10000

Total Income = Rs. 30106 Gross income = Total Income Total Product Cost

= (30106) (21.057106) Gross Income = Rs. 8.943106 Let the Tax rate be 45% (common) Net Profit = Gross income - Taxes = Gross income (1- Tax rate) Net profit = 8.943106 (1 - 0.45) = Rs 4.92 x 106

Rate of Return: Rate of return = Net profit100/Total Capital Investment Rate of Return = 4.92 x 106100/ (12.91106) Rate of Return = 38.0%

PLANT LOCATION AND LAYOUT: The success of any industrial venture in terms of returns on

investment depends largely on its location. The most important factors

which influences the location of the plant are:

1. Availability of raw materials

2. Accessibility to market

3. Power and water resources 4. Transportation facilities

5. Labor supply 6. Waste disposal

PLANT LAYOUT: The layout of a plant plays an important role in determining

construction and manufacturing costs. The main factors to be considered are:

1. Operational convenience and accessibility. 2. Future expansion.

3. Proper distribution of utilities and services. 4. Health and safety.

5. Transportation within the plant. 6. Waste disposal.

A detailed drawing showing the plant layout after considering these

factors is attached.

Production of Chlorine

1

REFERENCES

1. Ullmanns Encyclopaedia of industrial chemistry

Wolfgang Gerhard, Volume A6, Fifth edition. 2. Encyclopaedia of Chemical Processing and design

John. J. Mcketta. Volume 8

3. Chlorine Manufacture, Properties and Uses J. S. Sconce Second Edition

4. Perrys Chemical Engineers Handbook

Robert. H. Perry. Sixth Edition. 5. Plant design and Economics for chemical engineers

Max. S. Peters, Klaus. D. Timmerhaus, Third Edition. 6. Chemical Engineering

J.M. Coulson & J. F. Richardson. Volume: 6 7. Process Equipment Design

M. V. Joshi. Second Edition.

Chlorine_Introduction.pdf (p.1-2)Chlorine_Properties&uses.pdf (p.3-4)Chlorine_Methods-2520of-2520Production.pdf (p.5-10)Chlorine_Material-2520Balance.pdf (p.11-15)Chlorine_Energy-2520Balance.pdf (p.16-19)Chlorine_Design-2520of-2520Equipments.pdf (p.20-42)Chlorine_Pollution-2520control&Safety.pdf (p.43)Chlorine_Cost-2520Estimation&Economics.pdf (p.44-52)Chlorine_Plant-2520Location&Layout.pdf (p.53)Chlorine_Bibliography.pdf (p.54)