Production of Carbitol

INTRODUCTION

Carbitol or Diethylene glycol ethyl ether (DEGEE) is colorless liquid with faint, sweet, pleasant odor and bitter taste. Its boiling point is relatively high and vapor pressure and evaporation rate low. Like most glycol ethers, DEGEE has very good solubility and mixes completely with water and with both polar and non-polar solvents.

In 1993 DEGEE was registered as an ingredient in 178 Swedish chemical products, and estimated annual use was just under 500 tons of pure substance. The major area of use was as solvent, but the substance was also used in paint, varnish, cleaners and binders. In Sweden DEGEE is not used in pharmaceuticals or non-prescription diet supplements, but does occur in cosmetics and skin care products (personal communication, Cecilia Ulleryd, Swedish Medical Products Agency, Nov. 15, 1996).

In the United States, DEGEE was reported to occur in 80 cosmetic preparations in 1981. The substance, under the name Transcutol, is used in skin medications and it has also been found in chemical air fresheners for consumer use. The use of DEGEE in Sweden increased rapidly from 1985 to 1992, and in the following year as well. DEGEE, along with mono-(EGBE) and diethylene glycol butyl ether (DEGBE), have been identified as the solvents most widely used in water-based paints and varnishes . Global use in 1993 was estimated to be 31,000 tons.

PROPERTIES AND USES OF CARBITOL

Chemical and physical characteristics of carbitol.

PropertiesValue

Chemical FormulaCH3CH2OCH2CH2OCH2CH2

OH

Molecular weight134.2

Density (20oC)0.99

Boiling point202 oC

Melting point- 76 oC

Vapor pressure (25oC)19 Pa (0.14 mm Hg)

Relative evaporation rate0.02 (n-butyl acetate = 1)

Saturation concentration (25oC)180 ppm

Vapor pressure (25oC)0.13 mm Hg

Vapor density4.62

Flash point96 C

Freezing point-76 C

Refractive index (20o C)1.4300

Latent heat of vaporization85 cal/g

Heat of combustion-6330 cal/g

FLAMMABILITY (FLASH POINT):

This chemical has a flash point of 96 C (205 F). It is combustible. Fires involving this material can be controlled with a dry chemical, carbon dioxide or Halon extinguisher. A water spray may also be used. The auto ignition temperature of this compound is 204 C (400 F).

REACTIVITY:

This chemical can react with oxidizing materials. It is incompatible with strong acids, acid chlorides and acid anhydrides . It is also incompatible with alkalies. It may react with peroxides, oxygen, nitric acid and sulfuric acid.

STABILITY:

This chemical is hygroscopic. ethanol or acetone should be conditions.Solutions of this chemical in water, 95% stable for 24 hours under normal lab

USES:

This compound is used as a solvent for dyes, nitrocellulose and resins.

It is used in non-aqueous stains for wood, for setting the twist and conditioning yarns and cloth, in textile printing, textile soaps, lacquers, cosmetics and quick-drying varnishes and enamels. This compound is also used in brake fluid diluent and in organic synthesis. It used to determine saponification values of oils and as neutral solvent for mineral oil-soap and mineral oil-sulphated oil mixtures (giving fine dispersions in water)

PRODUCTION OF CARBITOL.

Polyethylene glycols as high as the octaethylene glycol have been prepared. The ethers of these polyethylene glycols have attained commercial significance, and this is particularly true of the diethylene glycol ethyl ether, which is sold commercially under the name of carbitol. This product is especially useful in the manufacture of laminated glass, wherein the celluloid interleaf is misted over with a spray of carbitol, which increases the adhesion to the glass. In the printing and dyeing of textile fabrics, it makes possible more economical use of dyes. Carbitol is a solvent with a mild odor, a low rate of evaporation, and a boiling point of 201.9oC. it enters into the manufacture of wood stains and automobile polishes and is used as a lacquer solvent. However, it is particularly valuable in the cosmetic field, where it is employed for creams and hair tonics.Preparation of Glycol Ethers (carbitol)

C2H5OH + CH2CH2OHO

CH2CH2 OC2H5

EthylEthyleneCellosolve

AlcoholOxide

CH2OC2H5CH2CH2OC2H5

+O

CH2

CH2OHCH2

CH2

CH2OH

CARBITOL

There is a large quantity of data available on the German process for the manufacture of the glycol ethers. The monomethyl, ethyl, and n-propyl ethers of ethylene glycol are manufactured by the continuous reaction of ethylene glycol are manufactured by the continuous reaction of ethylene oxide with the anhydrous alcohol at about 200oC and at 25-45 atm pressure. One volume of ethylene oxide and 6 volume of alcohol are fed to a pressure tower packed with iron raschig rings. Excess of alcohol is used to give the required high ratio of glycol ether: diglycol ether, to control the heat librated in the reactor, and to avoid high concentrations of the ethylene oxide and alcohol is exothermic, about 20-25 kg-cal per g mole of ethylene oxide reacted. The reaction product emerges from the base of the pressure tower and is distilled semi continuously. The alcohol is recycled; the pure glycol and diglycol ethers are isolated by batch fractionation. After removal of excess alcohol, the crude product contains about 85 percent glycol ether, 10 percent diglycol ether, and 2-3 percent polygylcol ethers. The yield of ethers is about 90-95 percent on ethylene oxide and alcohol consumed. The reaction is controlled to give complete conversion of ethylene oxide. Conditions in the pressure tower and feed ratios for methyl, ethyl and n-proply alcohols are given in table. The contact time has been calculated on the assumption that the reactor capacity is 3.5 cubic m.

Either acids or bases may be used as catalysts in the reaction between ethylene oxide and alcohol. However, acids are corrosive and must be neutralized before treating the crude reaction product, and alkalies lead to the formation of resins with the acetaldehyde present in the ethylene oxide. Because of the above reasons a noncatalytic process was developed. The Germans also developed a process using aqueous ethyl alcohol, but as this involves a difficult products separation, it is not preferred.

Conditions for Preparation of Glycol Ethers

AlcoholTemp, oCPressureRate, lb/hrContact

atmtime, hr

EthyleneAlcohol

oxide

20040120720About 4

Methyl

Ethyl200-21035100600About 5

220-2302580-90450-540About 6

n-propyl

MATERIAL BALANCE

Basis: 100 Kg moles of C2H4O The conversion is taken as 95%Of the 100 Kg moles of ethylene oxide sent, 95% is reacting.

The crude stream a contain the products and 5 Kg moles of C2H4O Neglecting the 5 Kg moles of C2H4O i.e. assuming that all the 5 Kg moles of C2H4O are going out with the top product of the tower.

On a ethylene oxide free basis.The weight fraction (%) of crude product Stream 'a' is:

85% - Monoglycol ether

10% - Diglycol ether

5% - Triglycol ether

The mole fraction of this stream 'a' on a ethylene oxide free basis is as: 89.885% - Monoglycol ether

7.101% - Diglycol ether

3.011% - Triglycol ether

For a crude product of 100 Kg moles (C2H4O-free),

89.885 Kg moles of monoglycol ether = 89.885 Kg moles of C2H4O 7.101 Kg moles of carbitol = 14.202 Kg moles of C2H4O3.011 Kg moles of triglycol ether = 9.033 Kg moles of C2H4O

Therefore, Total ethylene oxide that has reacted for 100 Kg moles of (C2H4O-free) crude product =113.12 Kg moles of C2H4O.

For 100 Kg moles of C2H4O sent only 95 Kg moles reacts.

Therefore,for 95 Kg moles of C2H4O reacting,the total crude product (C2H4O-free) formed is

(95x 100)/113.12 = 83.981 Kg moles of crude product

The crude product is (83.981+5)Kg moles for 100 Kg moles of C2H4O feed. The total C2H5OH reacting = [(83.981x89.885)/100 + (83.981x7.101)/100] = 81.4504 Kg moles

The recycle must be such that the ratio is 1:6

Total recycle is: (6x100) 81.450 = 518.5 Kg moles The composition of the stream b is

518.55 Kg moles of C2H4O+(83.981+5)Kg moles of crude product The composition of stream a on a C2H4O free basis is known.

For tower (1):

F = 83.981

F = D+W----------(1)

And FxF = DxD + WxW ----------(2)

Now fixing xD as 0.99 and xW =0.05, xF=0.89885 We have solving (1) and (2):

83.981x 0.89885= Dx0.99+(83.981-D)0.05 Solving for D we have

D = 75.837 Kg moles And W = 8.143 Kg moles

Let us take the 0.01 mole fraction lost is distillate is all carbitol(it is lighter then triglycol ether)

Therefore carbitol lost above is 0.75837 Kg moles

Now W= 8.143 Kg moles and has (8.143x0.05)=0.40715 Kg moles of mono-glycol ether

The carbitol in crude stream a on a C2H4O free basis is = (83.981x7.101)/100= 5.963 Kg moles

Of this 5.963 Kg moles of carbitol, 0.75837 Kg moles of carbitol is lost in distillate.

Therefore carbitol present =5.963 0.75837 =5.205 Kg moles

The remaining in triglycol ether =8.143 5.205 0.40715 = 2.5307

For tower (2):

F =8.143 Kg moles

Fixing xD as 0.99 and this is the purity of the product and xW =0.05 We have F =D + WFxF =DxD + WxW

And xF of carbitol =5.205/8.143 =0.639 8.143x0.6391 = Dx0.99 + (8.143-D)x0.05 Solving for D we get

D=5.1040 Kg moles And W =3.038 Kg moles

For 100 Kg moles of feed we have 5.1040 Kg moles of carbitol of 0.99 purity 25 TPD = 186.567 Kg moles/day

The ethylene oxide feed is 3655.309 Kg mole/day

The C2H5OH feed is (81.450x186.567)/5.1040

= 2977.249 Kg mole/dayCrude product(C2H4O-free) is (83.981x3655.309)/100

=3069.765 Kg mole/day Crude product with C2H4O is (88.981x3655.309)/100

= 3252.530 Kg moles/day The C2H4O recycle is (518.55x3655.309)/100

= 18954.558 Kg mole/day

Mono glycol ether,distillate tower (1) =75.837x36.55309 =2772.069 kg moles/dayF of tower (2) is (8.143x3655.309)/100

= 297.6510 Kg moles/day And D is (5.1040x3655.309)/100= 186.566 Kg moles/day

And w is (3.038x3655.309)/100 = 111. 048 Kg moles/day

ENERGY BALANCE

The datum temperature is taken as 30c.

1) ENERGY BALANCE AROUND THE REACTOR

Feed of ethylene oxide and ethyl alcohol can be assumed at room temp (30c).

1 hour of operation is taken as the basis.

Heat input + generation rate heat output - heat supplied externally b. Heat input = (mcpt)alcohol + (mcpt)EtO 0 + 0 ,( treference = tfeed) 0

c. Generation rate

The heat of reaction is 23kcal/gmol of EtO reacted.

Generation rate = +Rfeed rateconversion 234180103152.300.95

13.9100159109J/hr c. Heat output (mcpt)alcohol+ (mcpt)mge+(mcpt)carbitol+(mcpt)tge+(mcpt)EtO

The product outlet temperature is taken as that of reaction temp,200c. (mcpt)alcohol = 789.77175170kJ/hr

(mcpt)mge = 114.97210170 kJ/hr

(mcpt)EtO = 7.614115170 kJ/hr

(mcpt)carbitol = 9.0825481.4170 kJ/hr

(mcpt)tge = 439.6143.85125170 kJ/hr

Therefore heat output = 28.780106 kJ/hr

d. Heat supplied externally = ( 28.780- 13.910 )106 kJ/hr

14.8700109 J/hr 4130.5kW.

2) 2) ENERGY BALANCE AROUND THE ALCOHOL RECOVERY TOWER

Heat input + Reboiler load= heat output + heat load of condenser

a. Heat input

output from the reactor is feed for the tower hence, heat input = 28.780109J/hr.b. Heat output

The products of the tower are at 100c.The output of the tower are distillate and residue.

heat out with distillate = (mcpt)alcohol= 789.2715070 kJ/hr.heat out with residue = (mcpt)carbitol + (mcpt)mge + (mcpt)tge 9.0825343.2570 + 114.9710570 3.8512376.7570 1.16718kJ/hr

The total heat output = 9.453106kJ/hr

c. Condenser load- Reboiler load = (28.780-9.453)109

19.327109J/hr.

3) ENERGY BALANCE AROUND TOWER 1 ( MGE-SEPARATION)

Heat input + Reboiler load= heat output + heat load of condenser

a. Heat input

output from the alcohol recovery tower is feed for the tower hence, heat input = 1.1671109J/hr.b. Heat output

The products of the tower are at 150c.

The output of the tower are distillate and residue.heat out with distillate = (mcpt)mge= 114.97205120 kJ/hr. heat out with residue = (mcpt)carbitol + (mcpt)tge 9.0825418.99120 + 3.8512457.52120 0.667106 kJ/hr

The total heat output = 3.4952106 kJ/hr

c. Reboiler load - Condenser load = (3.4952-1.1671)109 2.3281109J/hr.

4) ENERGY BALANCE AROUND TOWER 2 ( PRODUCT TOWER). Heat input + Reboiler load

heat output + heat load of condenser

b. Heat input

output from tower 1 is feed for the tower hence, heat input = 0.667109J/hr.b. Heat outputThe products of the tower are at 202c.The output of the tower are distillate and residue.

heat out with distillate = (mcpt)carbitol

= 9.0825481.6172 kJ/hr. heat out with residue = (mcpt)tge 3.8512505.6172 0.3349106 kJ/hr

The total heat output = 1.0872106 kJ/hr

c. Reboiler load - Condenser load = (1.0872-0.667)109 0.4202109J/hr

MAJOR EQUIPMENT DESIGN

DISTILLATION COLUMN

Process Design

The feed rate for the tower is 297.6510 kmols/day Therefore F = 297.6510 kmols/day

= 12.4021 kmols/hr The mole fraction of carbitol in the feed is 0.639. i.e Xf = 0.639

Neglecting the residual mono-glycol ether and other impurities, thus the mixture can be treated as binary mixture of carbitol and triglycol ether.

The distillate rate is 186.566 kmols/day

Therefore D = 186.566 kmols/day

= 7.7735 kmols/hr

The mole fraction of carbitol in the distillate is 0.99. i.e xd = 0.99

The residue rate is 111.048 kmols/day

Therefore W = 111.048 kmols/day = 4.627 kmols/hr

The mole fraction of carbitol in the residue is 0.05 i.e xw = 0.05

The feed is taken as saturated liquid, since it is coming from a Reboiler.

Therefore q = ( Hv - Hf )/ ( Hv Hl )

= 1

Therefore the slope of the q-line is q/(q-1) =

Vapor pressure data

TemperatureVapor pressure of MCBVapor pressure of DCB

( K )(mm Hg)(mm Hg)

405700190

408810210

411900240

4151000260

4181050300

4221150330

4261300370

4291400400

4331500450

4371700500

4411800575

4442000625

4482200680

4532400740

Vapor liquid equilibrium data

Temperaturexaya

(K)( mole fraction of( mole fraction of

MCB in liquid )MCB in gas )

4080.920.98

4110.790.94

4150.680.90

4180.610.85

4220.520.79

4260.420.72

4290.360.66

4330.300.59

4370.220.49

4410.150.36

4440.100.26

4480.050.15

4530.010.04

From graph:Minimum reflux ratio ( Rm ) Operating reflux ratio

= 0.17857 = 1.5 x Rm

= 0.2678 Slope of operating line xd/( R+1) = 0.7808 From the graph the number of ideal stages is eight. Therefore total no of trays in the column is seven. Number of stages in enriching section is four. Number of stages in stripping section is three.

Liq flow rate in enriching section L

d. = D x R

7.7735 x 0.2678

2.0817 kmols/hr

Vapor flow rate in the enriching section G

G = L + D

( 1 + R )D

1.2678 x 7.7735

9.855 kmols/hr

Liq flow rate in the stripping section L

d. = L + qF

2.0817 + 12.4021

14.4838 kmols/hr

Vapor flow rate in the stripping section G

e. = G + ( q - 1 )F

= 9.855 kmols/hr

EVALUATION OF AVERAGE PROPERTIES OF THE MIXTURE

DENSITY CALCULATION

Liq phase calculationsCarbitol density at various temperatures

( ! )202c = 5.977kmol/m3. ( ! )213c = 5.857kmol/m3. ( ! )261c = 5.379kmol/m3.Triglycol ether ( TGE ) density at various temperatures( ! )202c = 6.6209kmol/m3. ( ! )213c = 6.5141kmol/m3. ( ! )261c = 6.0870kmol/m3.

Vapor phase calculations Assuming ideal gas

PV = nRT With P = 1 atm

R = 0.082 m3atm/kmol k T = temp in Kelvin

Carbitol density at various temperatures ( ! )204c = 0.025565kmol/m3.

( ! )218c = 0.024835kmol/m3. ( ! )282c = 0.21970kmol/m3.

Triglycol ether density at various temperatures ( ! )204c = 0.25566kmol/m3.

( ! )218c = 0.02483kmol/m3. ( ! )282c = 0.021972kmol/m3.

Mixture properties

( !l ) kg/m3 = avg molecular wt/( x1 !carbitol + x2 !TGE )

( !l )202c = 134.16/(0.99/5.977 + 0.01/6.6209)

= 802.65 kg/m3 ( !l )213c = 838.248 kg/m3( !l )261c = 885.248 kg/m3( !v )204c= 3.4298 kg/m3

( !v )218c= 3.3550 kg/m3

( !v )282c= 3.233 kg/m3

The average properties in the enriching section and stripping section can be summarized into the table as below.

PROPERTYENRICHINGSTRIPPING

SECTIONSECTION

TOPBOTTOMTOPBOTTOM

Liq flow rate2.08172.081714.483814.4838

kmols/hr

Vap flow rate9.8559.8559.8559.855

kmols/hr

Liq mole0.990.70.70.13

fraction x

Vap mole0.990.930.930.18

fraction y

Avg ( liq )134.16138.80138.80147.92

molecular wt

g/mol

Avg ( vap )134.16135.12135.12147.12

molecular wt

g/mol

Temp liq (c)202213213261

Temp vap(c)214218218282

Liq flow rate279.28288.9392010.382142.44

kg/hr (L)

Vap flow rate1322.141331.601331.601449.86

kg/hr(G)

Liq density !l802.65838.32838.32885.248

kg/m3

9DS GHQVLW\ !v3.42983.355023.355023.233

kg/m3

L/G[ !v !l)0.50.013800.0137260.095500.08930

ENRICHING SECTION DESIGN

1.TOWER DIAMETER CALCULATIONThe maximum value of L/G[ !v !l)0.5 is 0.1380 at the top of the section

Choosing Intalox saddles, polypropylene Nominal size = 25mmSurface area = 206 m2/m3.

Fp = 105 per meter

From steam tables the density of water at 202c is 802.65kg/m3.

!water !liq = 862.28/802.65 = 1.0742

l1/3(mix) = x1 11/3 + x2 21/3

0.99(0.41/3) + 0.01(0.921/3)

0.7360

l (mix) = 0.4 cp

From graph (18-38)G2)S% 0.2 !g !l*g) = 0.22

Solving we obtain Gf = 7.954 kg/m2s

Choosing 65% flooding

We get G = 0.65Gf = 5.17027kg/m2s.

Therefore the cross sectional area of the tower

Ac = mass flow rate/G

1322.14/( 3600 x 5.17027)

0.07103 m2

The diameter of the column = {(4xAc)/}0.5

Dc = 0.30073 mTaking the corrected Dc = 0.300 m Therefore Ac = 0.07068m2= 0.76085ft2

tower dia /packing dia (Dc/dp) = 0.30/0.025 = 12 ( > 10 )

WETTING RATE

Lmin = 279.28/(802.65x60x4.546097x10-3)Gallons/min = 1.2756GPM

Degree of wetting Lmin/tower area in ft2

1.2756/0.760855

1.676536 GPM/ft2

Hence acceptable ( greater than the limits of 0.5 GPM/ft2 ).

PRESSURE DROP CALCULATIONS

Pressure drop can be calculated using the equationS &2 *10 (C3*Utl) !g*Utg2- equation 18.48

where

C2, C3 are constants to be evaluated from table 18.7 Utl, Utg are superficial liq and vapor flow rate ( ft/sec ).!g isavg vapor density ( lb/ft3 ).

SSUHVVXUH GUop in inch water/ft packing

)

Utl =L /$F !

l

=284.11/( 3600x0.7068x802.48x0.3048)

= 4.4647x10-3 ft/s.Utg = G $F !g)

= 1326.87/(3600x0.7068x3.39241x0.3048) =5.0431 ft/s.Density of gas !g = 3.3924 Kg/m3 = 0.21177 lb/ft3

C2 = 0.31C3 = 0.0222S&2 *10 (C3*Utl) !g*Utg

0.31*10 (0.0222*0.004464)*0.21177* 5.04312

1.6700 in water/ft packing

139.06 mm water/m packing

STRIPPING SECTION DESIGN

1.TOWER DIAMETER CALCULATIONThe maximum value of L/G[ !v !l)0.5 is 0.09550 at the top of the section

Choosing Intalox saddles, polypropylene Nominal size = 25mmSurface area = 206 m2/m3.

Fp = 105 per meter

From steam tables the density of water at 213c is 848.662kg/m3.

!water !liq = 862.28/838.32 = 1.01233

l1/3(mix) = x1 11/3 + x2 21/3

0.7(0.41/3) + 0.3(0.941/3)

0.7663

l (mix) = 0.45 cpFrom graph (18-38)

G2)S% 0.2 !g !l*g) = 0.15

Solving we obtain Gf = 7.038 kg/m2s

Choosing 65% flooding

We get G = 0.65Gf = 4.574kg/m2s.

Therefore the cross sectional area of the tower

Ac = mass flow rate/G

1331.60/( 3600 x 4.574)

0.0808 m2

The diameter of the column = {(4xAc)/}0.5

Dc = 0.3205 m Taking the corrected Dc = 0.300 m Therefore Ac = 0.07068m2= 0.76085ft2

tower dia /packing dia (Dc/dp) = 0.30/0.025 = 12 ( > 10 )

WETTING RATE

Lmin = 2010.38/(838.32x60x4.546097x10-3)Gallons/min = 8.791GPM

Degree of wetting Lmin/tower area in ft2

8.791/0.760855

11.555 GPM/ft2

Hence acceptable ( greater than the limits of 0.5 GPM/ft2 ).PRESSURE DROP CALCULATIONS

Pressure drop can be calculated using the equationS &2 *10 (C3*Utl) !g*Utg2- equation 18.48

where

C2, C3 are constants to be evaluated from table 18.7 Utl, Utg are superficial liq and vapor flow rate ( ft/sec ).!g isavg vapor density ( lb/ft3 ).

SSUHVVXUH GUop in inch water/ft packing

Utl = L / $F !l)

2076.41/( 3600x0.7068x861.78x0.3048)

0.031067 ft/s.

Utg = G $F !g)= 1390.73/(3600x0.7068x3.29401x0.3048)

=5.443 ft/s.'HQVLW\ RI JDV !g = 3.29401 Kg/m3 = 0.20565 lb/ft3

C2 = 0.31C3 = 0.0222S&2 *10 (C3*Utl) !g*Utg

0.31*10 (0.0222*0.0.31067)*0.20565* 5.4432

1.8917 in water/ft packing

157.60 mm water/m packing

AVERAGE CONDITIONS FOR ENRICHING AND

STRIPPING SECTION

PROPERTYENRICHINGSTRIPPING

SECTONSECTION

Liq flow rate kmols/hr2.081714.4838

Liq flow rate kgs/hr284.102076.41

Vapor flow rate9.8559.855

kmols/hr

Temp of liq c207.5247.5

Temp of vap c211250

'HQVLW\ RI OLT !liq820.485861.784

kg/m3

'HQVLW\ RI YDS !vap3.392413.2940

kg/m3

l cp0.4250.46

vap cp0.0100250.010756

liq dynes/cm12.62215.6123

DlAB cm2/sec16.408x10-617.58x10-6

DVAB cm2/sec0.052780.06082

Liq Schmidt number315.609303.627

Nsl

Vap Schmidt number0.5590.5368

Nsg

TOWER HEIGHT CALCULATION

ENRICHING SECTION

The height of the enriching section can be given by

Z = H og *NogWhere

Hog = Hg + m*Gm/Lm*Hl

Nog = NTln -1)is the stripping factor given by mGm/Lm

Hg% 'c1.11*Z0.33*Scg / (L*f1*f2*f3*f4)0.5

From fig 18.65

Dc = 0.3m

L = 284.10/(0.7068x3600)

= 1.11653 kg/s m2

f1l water )0.16

( 0.425/1.0 )0.16

0.872

f2 !water !l )1.25

(1000/820)1.25

1.0322

f3water/ l)0.8

= ( 72.8/12.622 )0.8= 4.0625

Hg = 0.029*50*(0.3)1.11*(Z)0.33*(0.559)0.5/(1.11653*0.872*1.3022*4.0625)0.5 = 0.5365*(Z)0.33

By Cornell method eqn 18-56Hl !& l !lDl)0.5(z/3.05)0.15

Liquid rate = 5.2147 Kg/m2 s

! P IURP ILJ -60)C = 0.8 (from fig 18-59)

Hl = 0.035*0.8/3.28 * (315.69)0.5 * (Z/3.05)0.15

= 0.1282*(Z)0.15

H og = 0.5365*(Z)0.33 + 0.1214*4.734*0.1282*(Z)0.15 = 0.5365*(Z)0.33 + 0.0737*(Z)0.15Nog = NTln -1)is the stripping factor given by mGm/Lm

m is the slope of equilibrium line in enriching section. m (top) = 0.1

m (bottom) = 0.1428 m (avg) = 0.1214 Gm/Lm = 9.855/2.0817

4.734 = 0.5750 NT = 4

Nog = 4ln0.5750/(0.5750-1)

= 5.2144

Z = H og *Nog {0.5365*(Z)0.33 + 0.0737*(Z)0.15}5.2144

2.796*(Z)0.33 + 0.3844*(Z)0.15 Solving the above equation by trail and error, we get Z = 5.35m Therefore height of enriching section is 5.35mSTRIPPING SECTION

The height of the stripping section can be given by

Z = H og *NogWhere

Hog = Hg + m*Gm/Lm*Hl

Nog = NTln -1)is the stripping factor given by mGm/Lm

Hg% 'c1.11*Z0.33*Scg / (L*f1*f2*f3*f4)0.5

From fig 18.65

Dc = 0.3m

L = 2076.41/(0.7068x3600)

= 8.160 kg/s m2

f1l water )0.16

( 0.46/1.0 )0.16

0.8831

f2 !water !l )1.25

(1000/861.78)1.25

1.204

f3water/ l)0.8

( 72.8/15.612 )0.8

3.427

Hg = 0.029*50*(0.3)1.11*(Z)0.33*(0.559)0.5/(6017.69*0.8831*1.204*3.427)0.5 = 0.052*(Z)0.33

By Cornell method eqn 18-56Hl !& l !lDl)0.5(z/3.05)0.15

Liquid rate = 8.160 Kg/m2 s

! P IURP ILJ -60)C = 0.8 (from fig 18-59)

Hl = 0.07*0.8/3.28 * (303.627)0.5 * (Z/3.05)0.15

= 0.2516*(Z)0.15

H og = 0.0527*(Z)0.33 + 0.86751*0.25167*(Z)0.15

= 0.0527*(Z)0.33 + 0.21826*(Z)0.15

Nog = NTln -1)is the stripping factor given by mGm/Lm

m is the slope of equilibrium line in stripping section. m (top) = 0.3

m (bottom) = 2.25 m (avg) = 0.1.275 Gm/Lm = 0.6804= 0.86751

NT = 3

Nog = 3ln0.8675/(0.8675-1)

= 3.218

Z = H og *Nog {0.0527*(Z)0.33 + 0.21826*(Z)0.15}3.218

0.1673*(Z)0.33 + 0.7023*(Z)0.15

Solving the above equation by trail and error, we get Z = 0.84m

Therefore height of stripping section is 0.84m Total height of the tower is

Z = (5.35+0.84) = 6.19m(B) MECHANICAL DESIGN OF DISTILLATION COLUMN

Diameter of the tower Di = 0.3m

Working pressure = 1 atm=1.0329 kg/m2

Design pressure pd = 1.1362 kg/m2

Shell material Plain Carbon steel

Permissible tensile stress (ft)= 950kg/cm2

Insulation thickness = 50mm

Density of insulation = 770 kg/m3

Top disengaging space = 1m

Bottom separator space = 2m

Skirt height = 2m

Density of material column = 7700 kg/m3

Wind pressure = 130 kg/m2

1) Shell thicknessts= PDi/(2fJ-P) + C

P= design pressure in kg/cm3 f=allowable tensile stress kg/cm3

C= corrosion allowance ( 2 mm )

J=joint factor

ts= (1.1362*300)/(2*950*0.85-1.1362) + C ts = 2.2112 mm

minimum thickness allowable is 6mm ts= 6mm

2) Head Design

Shallow dished and torispherical head

Thickness of head is given by

th= PRCW/2fJ

Rc=crown radius,300mm

W=stress intensification factor

W= 0.25(3 + (RC/RK)

Rk= knuckle radius ,6% of crown radius.

W= 1.7706 mm

th= 1.1362*300*1.7706/2*950*0.85 ts= 0.3737 mmminimum thickness is ts= 6mm

3) Shell thickness at different heights

At a distance X m from the top of the shell the stress are; Axial Stress: (tensile)fap = pi Di___4(ts C)

1.1362*300/4*(6-2)

142.025 kg/m3

4)Compressive stress due to weight of shell up to a distance X fds = /4 * ( Do2 Di2 ) s X/4 * ( Do2 Di2 ) s X

0.77X kg/m3

5) Compressive stress due to weight of insulation fd(ins) = Dins tin ins Dm (ts C )

fd(ins) = 412*50*770*X

306*(6-2) fd(ins) = 1.1015X

6) compressive Stress due to the weight of the liquid and packing

fd = OLT DQG SDFNLQJ ZW XQLW KHLJKW ; Dm(ts-C)For the chosen packing, 25mm plastic intalox saddles

= 0.91

approximate wt/m3 kg/m3 = 76 Number of elements per m3 = 55800

Total volume of the packing = *Dc2*h/4

= 0.4375m3

Total void volume = 0.91*0.4375 = 0.3981m3Total volume of the actual packing = 0.039379m3

Average density of the liq in the column is 841.13kg/m3.

fd = (76+841.13*0.3981)/0.4375*X Dm(ts- c)

24.42X kg/cm2

7) Stress due to the weight of the attachments The total weight of the attachments

The weight of the head is taken as 1020kgs Wa = (1020 + 140X)

Fd(att) = (1020 + 140X)/(*30.6*0.4)

26.525 + 3.640X kg/cm2

8) Total compressive dead weight stress at height X ( sum of 2-7) fds=29.9315X + 26.525 9) Stress due to wind load at distance X fws = 1.4*PwX2/*Do(ts-c)

(1.4*130*X2)/(*31.2*.4)

4.642X2

10) Stress in upwind side

fmax= fws+fap-fdx0.8*950 = 4.642X2+142.025-29.9315X-26.525

4.642X2-29.9315X-644.5 = 0 solving for X

X = 15.2177 m 11) stress in down sidefmax= fws+fap+fdx4.642X2-29.9315X-928.55 = 0 X = 17.730 m

From this, whole tower of 6mm thickness is enough.

12) Skirt design

The material of construction for skirt is carbon steel IS:2062-1962

Minimum weight of vesselWmin= (Di+ts) ts(H-2)s + 2WH

H=11.29 (Total height of tower including skirt height)

s= 7700kg/m3 (specific weight of shell material)

WH= 1020 kg (weight of head)

Wmin= (0.3+0.006) (0.006)(11.29-2)7700 + 2*1020

Wmin= 2171.33 kg

Maximum weight of vesselWmax=Ws+Wi+Wl+WaWs=10800kg(weight of shell during test) Wi= 4200kg (weight of insulation)

Wl= 656.67 kg (weight of water during test) Wa= 4400 kg (weight of attachments)

Wmax=20056.67 kg

Wind loadPW= K1pwHD

For minimum weight of column,D = 0.3m

Pw(min)= 0.7*130*11.29*0.3 = 308.217 kg

Pw(max)= 0.7*130*11.29*0.312 = 320.54 kg

Minimum wind momentMW(min) = PW(min)*H/2 308.217*11.29/2 1739.88 kg m

Maximum wind moment

MW(min) = PW(min)*H/2 320.54*11.29/2 1809.44 kg m

Bending stresses

fb(min)= 4Mw(min) *D2*t

= 4*1739.88 *0.32*t= 2.461/tkg/cm2

fb(max)= 4Mw(max) *D2*t

4*1809.88 *0.32*t

2.5604/t kg/cm2

Minimum dead load stressFds(min)= Wmin/dt 2171.33 /*0.3*t 0.23038/t kg/cm2

Maximum dead load

Fds(max)= Wmax/dt

= 0.21280/t kg/cm2

Maximum tensile stress without any eccentric load fz= fbs(max)-fbs(min)

980*0.8= 0.0994/t t= 0.1449mm

Maximum compressive stress without any eccentric load

fz= fbs(max)-fbs(min)fz= 0.125 E (t/Do) 0.125*2.04*106*t/0.3

850000t

850000t = 2.5604/t+2.461/t t=2.4305 m

Minimum skirt thickness is 7mm,by providing 1mm corrosion allowancets=8mm

DESIGN OF SKIRT BEARING BOLTS

Maximum compressive stress between bearing plate and foundation fc= Wmax/A + Mw/2

A=(Do-l)/2l=outer radius of bearing plate minus outer radius of skrit Z=Rm2 lRm=(Do-l)/2

fc = 20056 /((0.3-l)l) + 1809.44/((0.3-l)2l

The allowable compressive stress of concrete foundation varies from 5.5 to 9.5 MN/m20.55*106 = 20056 /((0.3-l)l) + 1809.44/((0.3-l)2l

l=57mm

As required width of bearing plate is very small a 100 mm width is selected

l=0.1 m

therefore fc= 0.50253*106 kg/m2

thickness of bearing plate

tbp=l(3fc/f)

= 100(3*0.5026*106/96*106) = 12.53 mmBearing plate thickness of 12.53 mm is required

As the plate thickness required is less than 20mm, no reinforcement is required.

fmin= Wmin/A Ms/Z

2171.33/(0.3-0.1)0.1 + 1739.88/(0.3-0.1)20.1 103897kg/m2

The modulus value is taken, less than zero implies that the vessel must be anchored to the concrete foundation by means of anchor bolts toprevent overturning owning to the bending moment induced by the wind or seismic load.

Therefore anchor bolts are to be used Pbolt*n=fmin*AwherePbolt = load on one anchor boltfmin = stress determined by eqn10.211

A = area of the contact between bearing plate and foundation 103897.5*3.14*(0.3-0.1)*0.1

6528.07 kg/m2

For hot rolled carbon steel f=5.73*106 kg/m2 (arn)f=nPbolt

arn=1139.2 mm2 For 12X1.5 ,ar=63mm2

Number of bolts=11392/63=18 bolts

Therefore 18 bolts are to be distributed equally.

MINOR EQUIPMENT DESIGN

CONDENSER

Process Design

The reflux condenser, condenses vapor of the column and send it as the reflux.

The vapor flow rate is 9.855kmols/hr. Tvap = 202c. ( sat vap ).Molecular wt of vap = 134+HDW RI FRQGHQVDWLRQ 202c = 21000Btu/lbmol 364.52kJ/kg.

48846kJ/kmol.

Heat load Q = 9.85548846 481377.33kJ/hr. 133.71kW.

Let the overload be 10%.

Therefore Q = 1.1133.71kW= 147.0875kW.

Let water at a temp of 25c be used to condense the vapor. Fixing the outlet temp of water as 35c.

The water flow rateWc = 147.0875/Cpwater(Tout- Tin) 147.0875/4.18710

3.512kg/s. Tvap = 202c,

7lmtd = {(202-35)-(202-25)}/ln{(202-35)/(202-25)} = 171.95c

Let us assume an overall heat transfer coefficient,( U ) of 567.83J/m2sc. As the heat load is very low, we shall use a DPHE.

Required area for heat transfer is A Q/ 7lmtdU

147.08875/(171.95567.83)

1.369m2. N = A/( Ld02 ) Choosing 2 NPS, 40SCH and 1.25 as the tubes. Inner dia of 2 NPS pipe= 5.25cm

Outer dia of 2NPS pipe= 6.032cm

Inner dia of 1.25 NPS pipe= 3.505cm

Outer dia of 1.25 NPS pipe= 4.216cm

Taking the length of the pipe as 6 ft ( 1.828m) Length available for heat transfer = 1.528m Heat transfer area, outside area of inner pipe = Ld02N where N, is the number of hairpins. Therefore , = 1.369m2/(1.5280.042162) = 3.382 Taking the number of hairpins N as 4. Therefore the corrected heat transfer area = 1.61906m2 Corrected overall heat transfer coefficient = 480W/m2c Location of the fluids , the vapor is taken in the annulus and water in the tube. OVERALL HEAT TRANSFER COEFFICIENT CALCULATION (Ud) The overall heat transfer coefficient is given by the equation 1/Ud = 1/ho+(Do/Di)(1/hi)+ {xwDo/(Dwk)}+ dirt factor where, ho, hi are outside and inside heat transfer coefficients. xw,Dw are wall thickness and mean wall diameter. k is wall material thermal conductivity. 1)ANNULUS SIDE (carbitol vapors) The individual heat transfer coefficient ho = 0.725{[K3!2g @ >' 7@`0.25where,

K is the thermal conductivity of condensate = 0.12461W/mK! LV WKH GHQVLW\ RI FRQGHQVDWHNJ P3.

g is acceleration due to gravity= 9.81m/s2.

LV WKH KHDW RI FRQGHQVDWLRQN- NJ

D is the outside dia of the tube= 0.04216m

LV WKH YLVFRVLW\ RI WKH FRQGHQVDWHFS

7 LV WHPS GLIIHUHQFH RI WKH FRQGHQVDte and the wall = 202-116

= 86c.

Therefore,ho = 959.92W/m2K

2) Inside heat transfer coefficient ( tube side)

hidi/K = 0.023(Re)0.8(NPr)0.4Where,

Re is the Reynolds number. NPr is the prandtl number.

5H'9!

4Mass flow rate/(' 43.512/(0.035050.8510-3) 150092.0

NPr = CP . 4.181030.8510-3/0.16

5.82

Therefore

hidi/K = 0.023(150092)0.8(5.82)0.4 = 643.84

hi = 643.84K/di 643.840.61/0.03505

11205.3W/m2K 3) Wall Resistance can be expressed as

Mean temp of the wall is 116c.

{xwDo/(Dwk)} = 0.003550.04216/(0.0384945)= 8.6410910-5 m2K/W

4) Dirt factor of 0.0005 is assumed.

Therefore,1/Ud = 1/ho+(Do/Di)(1/hi)+ {xwDo/(Dwk)}+ dirt factor

0.0889/(0.0736611205.3) + (1/959.92) + 8.6410910-5 + 0.0005 1.7358810-3. Ud = 576.076W/m2K

Since the Uassumed < Ud, the design is acceptable from heat transfer point of view.

PRESSURE DROP CALCULATIONS

1)TUBE SIDEPT=(4fLvt2/2gDi)tg

Tube side Reynolds number=NRe= 150092.0

Friction factor f = 0.079(NRe)-1/4 =0.079(150092) -1/4 =4.01310-3

Tube side velocity vt = 3.656m/s

PT =

(44.01310-314.6243.6562/(29.80.03505)9969.8

44.54 kPa.

2) ANNULUS SIDE PA= (4fLva2/2gDH)g

Annulus side Reynolds number can be calculated as DH Mass flow rate/( area of annulus)

0.010640.3668/(0.009710-37.68310-4)

508.91103 Friction factor f = 0.079(NRe)-1/4

=0.079(508910) -1/4 =2.95710-4

Annulus side velocity is 138.784m/s at one end negligible at other end. Therefore the annulus side velocity = (138.78 + 0 ) 0.5

= 69.392m/s.

PA =

(42.95710-414.62469.3922/(29.80.01064))3.349.8 13.0728 kPa.

Hence the pressure drops are acceptable.

Mechanical Design

Let the material of construction be 15C-8, carbon steel.

T( tensile strength ) = 410 N/mm2.

y( yield strength ) = 220N/mm2. The pressure in the annulus is taken as 1 atm. The design pressure be taken as

P = 1.251 atm = 1.25atm

1)The total load of the bolt is given by

Fa = Pressure annular cross sectional area

(1.25-1)7.68310-3

19.461N

2)The total load capacity of the bolt is given by

= C(Ar)1.418

= 2.29(Ar)1.418

Therefore , stress area of the bolt Ar = (19.461/4.52268)-1.418

= 4.52268mm2.

3)From table 9.8, for the obtained Ar , bolt dia d = 3mmpitch= o.5mm.

4) Initial tension load in a bolt

Fi = 2805d

4815N

5) Effect of applied load on bolt stress

The final load on the bolt = K Fa + Fi , K for asbestos gasket = 0.6 F = 0.619.461 + 4815

= 4826.67N5) Number of bolts,

An empirical rule for the number of bolts in pipe joints is given by

N = 0.024Dc+ 2 , Dc = dia of cylinder 60.32mm N = 3.44768

As a standard we can provide 6 bolts

6) The maximum spacing of the bolt in any fluid tight joint

s G

s = 6d

s = 18mm

7) The extension at one end is the same as that of the pipe. The bolt circle diameter is given byD2 = D1 + 3.2d

D1 = 1.8D +20 = 128.576mmD2 = 128.57 + 3.23

138.176mm

8) The flange thickness is given by

t = 0.35D + 9 mm

0.3560.32 + 9

31.12mm

POLLUTION AND SAFETY

The raw material ethylene oxide used for the production of carbitol will explode in the presence of common igniters. Pure EtO vapor is difficult to ignite compared to oxide-air or hydrocarbon-air mixtures, requiring spark energies about ten thousand times larger.

The design of the plant should be such that the gas mixtures handled are always outside the explosive limit. The actual safe operating ranges are dependent on operating temperatures, pressures, equipment configurations, gas composition, dynamics of catalyst and instrumentation.

Health affects

Toxic effects

There are no reports on effects of occupational exposure. There is one case report describing a man who drank about 300 ml of DEGEE. He developed severe symptoms of poisoning: CNS effects, breathing difficulty, thirst, acidosis and albuminuria .

An unpublished report (Kligman, 1972) cited by Opdyke describes dermal application of 20% DEGEE in petroleum jelly, under occlusion, to 25 volunteers for 48 hours. The application resulted in no irritation or sensitization. In another sensitization study, pure DEGEE was applied under occlusion to the backs of 98 young men for 7 days, followed by a 3-day application 10 days later. No skin sensitization or edema was observed, but 7 of the men had pronounced skin reddening.

ACUTE/CHRONIC HAZARDS:

When heated to decomposition this compound emits acrid smoke, irritating fumes and toxic fumes of carbon monoxide and carbon dioxidePersonal Protection and Exposure Control

Eye: Wear goggles to avoid splash. Skin: Use protective gloves.Inhalation: Use mask while spraying. Avoid inhalation.

Engineering Controls: Sufficient ventilation to keep within OSHA PEL/TLV limit.The dried film of the product may contain all or some of the following OSHA chemicals And may become a dust nuisance when removed by sanding or grinding.OSHA recommends a PEL/TWA of 15 mg/m3 for the respirable fraction. ACGIH recommends TLV/TWA of 10 mg/m3 for total dust. Use approved mask, eye goggles and gloves while sanding, or grinding. Skin absorption may contribute to the overall exposure of the material.

HEALTH EFFECTS OF OVER EXPOSURE AND FIRST AID

Primary RoutesSymptomsFirst Aid

EyeMay cause burning andFlush with water. Seek

irritation uponmedical attention

direct contact.

SkinDirect skin contact mayWash with soap and

cause skinwater.

irritation and dermatitis.

IngestionSevere oral intoxicationDo not induce

will lead tovomiting. Seek medical

intense burning of theattention.

throat and may

result in drowsiness,

numbness and

headache followed by

weakness &

nausea.

InhalationAcute overexposure inRemove person to fresh

mist form mayair. Apply

result in irritation ofartificial respiration.

throat & lungs.

COST ESTIMATION AND ECONOMICS

Marshall and Shift index in 1992 for equipments is 943.1 Cost estimation based on the equipment

Cost of the bare columns is 3*11000. Saddles cost 720/m3.

Total cost of the packing ,for the 3 columns 3*315. Cost of the reactor is 16000.

There are five heat exchangers, reboilers and condensers, of shell and tube type. The approximate cost of the each exchanger is 3000.

The cost of one DPHE is 1500.

Therefore the total cost of the equipment = 66445. Taking 1 = Rs 63.

Total cost of the equipment = Rs 66445*63*1048/943.1. = Rs 4.65115*106.

ESTIMATION OF DIRECT COST

COMPONENTSCOSTS

Purchased equipment cost ( E )Rs 4.65115*106

Purchased equipment installationRs 1.8139*106

( 39% of E )

Instrumentation (installed cost),Rs 1.3023*106

28%E

Piping installed, 31%ERs 1.4418*106

Electrical installation, 10%ERs 0.465115*106

Yard improvement, 10%ERs 0.465115*106

Service facility, 55%ERs 2.55813*106

Land , 6%ERs 0.279069*106

TOTAL DIRECT COST (D)Rs 13.999*106

ESTIMATION OF INDIRECT COST1.Engg and supervision ( 32% E )= Rs 1.48836*106

2.Construction + contractor fees ( 25% direct cost )

= Rs 3.4997*106

Therefore total indirect cost ( I )= Rs 4.988118*106

DIRECT AND INDIRECT COST (TOTAL) = Rs 18.9871*106

Contingence ( 10%D+I )= Rs 1.89871*106

Fixed capital investment ( FCI ) ,

contingence + D + I= Rs 20.88801*106

Working Capital: (10-20% of Fixed-capital investment)Consider the Working Capital = 15% of Fixed-capital investment i.e., Working capital = 15% of 20.88801*106= 0.15 20.88801*106= Rs 3.13320*106

Total Capital Investment (TCI):

Total capital investment

e. Fixed capital investment + Working capital

f. Rs 24.02121*106

i.e., Total capital investment = Rs 24.02121*106

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 = Rs. 2.71541106

ii. Local Taxes: (1-4% of fixed capital investment) Consider the local taxes = 3% of fixed capital investment i.e. Local Taxes = 0.0320.88801*106

Rs. 0.6266106

iii. Insurances: (0.4-1% of fixed capital investment) Consider the Insurance = 0.7% of fixed capital investment i.e. Insurance = 0.007 Rs 20.88801*106

Rs. 0.14621607106

iv. Rent: (8-12% of value of rented land and buildings) Consider rent = 10% of value of rented land and buildings Rent = Rs. 0.1302322x106

Thus, Fixed Charges = Rs. 3.61844106

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 = 3.61844106/15%

Total product charge = 3.61844106/0.15

Total product charge(TPC) = Rs. 24.1229106

Raw Materials: (10-50% of total product cost)

Consider the cost of raw materials = 25% of total product cost

Raw material cost = 25% of 24.1229106

Raw material cost = Rs. 6.03073106

ii. Operating Labour (OL): (10-20% of total product cost) Consider the cost of operating labour = 12% of total product cost operating labour cost = 12% of 24.1229106

Operating labour cost = Rs 2.89478106

iii. Direct Supervisory and Clerical Labour (DS & CL): iv. (10-25% of OL)Consider the cost for Direct supervisory and clerical labour= 12% of OL

Direct supervisory and clerical labour cost

12% of 2.89478106

0.34736106

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

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

Utilities cost = 12% of 24.1229106

= 0.12 24.1229106

Utilities cost = Rs. 2.89464106

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 = 0.0520.88801106= Rs. 1.0444106

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 1.0444106Operating supplies cost = Rs. 0.15666106

vii. Laboratory Charges: (10-20% of OL) Consider the Laboratory charges = 15% of OL Laboratory charges = 15% of 2.89478106 Laboratory charges = Rs. 0.434212106

Patent and Royalties: (0-6% of total product cost)

Consider the cost of Patent and royalties = 4% of total product cost

d. Patent and Royalties = 4% of 24.1229106

e. Patent and Royalties cost = Rs. 0.964919106

Thus, Direct Production Cost = Rs. 14.7676106

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

= 60% of ((2.8947106) + (0.3473106) + (1.04441106)) Plant overhead cost = Rs. 2.5719106

Thus, Manufacture cost = Direct production cost + Fixed charges + Plant overhead costs.Manufacture cost = Rs. 20.9580106

AI. 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 = Rs. 1.206145106

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 2.41229107

Distribution and Selling costs = Rs. 3.6184106

C. Research and Development costs: (about 5% of total product cost) Consider the Research and development costs = 5% of total productcost

Research and Development costs = 5% of 2.41229107

Research and Development costs = Rs. 1.2010106

D. Financing (interest): (0-10% of total capital investment) Consider interest = 5% of total capital investment i.e. interest = 5% of 2.40212107

Interest = Rs. 1.20106106

Thus, General Expenses = Rs. 7.23178106

III. Total Product cost = Manufacture cost + General Expenses Total product cost = Rs. 28.1897106

V. Gross Earnings/Income:

Wholesale Selling Price of carbitol per ton = 70Hence Wholesale Selling Price of carbitol per ton. = 63 70 = Rs. 4400 Total Income = Selling price Quantity of product manufactured

= 4400 (25 T/day) (330 days/year) Total Income = Rs. 36.3106

Gross income = Total Income Total Product Cost

= 36.3106 24.122106

Gross Income = Rs. 12.1771106

Let the Tax rate be 45% (common) Taxes = 45% of Gross income= 45% of 12.1771106

Taxes = Rs. 5.4796106

Net Profit = Gross income - Taxes = Gross income (1- Tax rate) Net profit = (12.1771106 ) (5.4796106)= Rs. 6.6974106

Rate of Return:

Rate of return = Net profit100/Total Capital Investment Rate of Return = 6.6974106100/ (24.02121106) Rate of Return = 27.881%

PLANT LOCATION AND layout

THE LOCATION OF THE PLANT CAN HAVE A CRUCIAL EFFECT ON THE PROFITABILITY OF A PROJECT, AND THE SCOPE FOR FUTURE EXPANSION. MANY FACTORS MUST BE CONSIDERED WHEN SELECTING A SUITABLE SITE, AND ONLY A BRIEF REVIEW OF THE PRINCIPAL FACTORS WILL BE GIVEN IN THIS SECTION. THE PRINCIPAL FACTORS TO BE CONSIDERED ARE:

g. LOCATION, WITH RESPECT TO THE MARKETING AREA.

h. RAW MATERIAL SUPPLY.

i. TRANSPORT FACILITIES.

j. AVAILABILITY OF LABOUR.

k. AVAILABILITY OF UTILITIES: WATER, FUEL, POWER.

l. AVAILABILITY OF SUITABLE LAND.

m. ENVIRONMENTAL IMPACT, AND EFFLUENT DISPOSAL.

n. LOCAL COMMUNITY CONSIDERATIONS.

o. CLIMATE.

p. POLITICAL STRATEGIC CONSIDERATIONS.

MARKETING AREA

FOR MATERIALS THAT ARE PRODUCED IN BULK QUANTITIES: SUCH AS CEMENT, MINERAL ACIDS AND FERTILIZERS, WHERE THE COST OF THE PRODUCT PER TON IS RELATIVELY LOW AND THE COST OF TRANSPORT A SIGNIFICANT FRACTION OF THE SALES PRICE, THE PLANT SHOULD BE LOCATED CLOSE TO THE PRIMARY MARKET. THIS CONSIDERATION WILL BE LESS IMPORTANT FOR LOW VOLUME PRODUCTION, HIGH-PRICED PRODUCTS; SUCH ASPHARMACEUTICALS. IN AN INTERNATIONAL MARKET, THERE MAY BE AN ADVANTAGE TO BE GAINED BY LOCATING THE PLANT WITHIN AN AREA WITH PREFERENTIAL TARIFF..

RAW MATERIALS

THE AVAILABILITY AND PRICE OF SUITABLE RAW MATERIALS WILL OFTEN DETERMINE THE SITE LOCATION. PLANTS PRODUCING BULK CHEMICALS ARE BEST LOCATED CLOSE TO THE SOURCE OF THE MAJOR RAW MATERIAL; WHERE THIS IS ALSO CLOSE TO THE MARKETING AREA.

TRANSPORT

THE TRANSPORT OF MATERIALS AND PRODUCTS TO AND FROM PLANT WILL BE AN OVERRIDING CONSIDERATION IN SITE SELECTION.

IF PRACTICABLE, A SITE SHOULD BE SELECTED THAT IS CLOSE AT LEAST TWO MAJOR FORMS OF TRANSPORT: ROAD, RAIL, WATERWAY OR A SEAPORT. ROAD TRANSPORT IS BEING INCREASINGLY USED, AND IS SUITABLE FOR LOCAL DISTRIBUTION FROM A CENTRAL WAREHOUSE. RAIL TRANSPORT WILL BE CHEAPER FOR THE LONG-DISTANCE TRANSPORT OF BULK CHEMICALS.

AIR TRANSPORT IS CONVENIENT AND EFFICIENT FOR THE MOVEMENT OF PERSONNEL AND ESSENTIAL EQUIPMENT AND SUPPLIES, AND THE PROXIMITY OF THE SITE TO A MAJOR AIRPORT SHOULD BE CONSIDERED.AVAILABILITY OF LABOUR

LABOUR WILL BE NEEDED FOR CONSTRUCTION OF THE PLANT AND ITS OPERATION. SKILLED CONSTRUCTION WORKERS WILL USUALLY BE BROUGHT IN FROM OUTSIDE THE SITE, BUT THERE SHOULD BE AN ADEQUATE POOL OF UNSKILLED LABOUR AVAILABLE LOCALLY; AND LABOUR SUITABLE FOR TRAINING TO OPERATE THE PLANT. SKILLED TRADESMEN WILL BE NEEDED FOR PLANT MAINTENANCE. LOCAL TRADE UNION CUSTOMS AND RESTRICTIVE PRACTICES WILL HAVE TO BE CONSIDERED WHEN ASSESSING THE AVAILABILITY AND SUITABILITY OF THE LABOUR FOR RECRUITMENT AND TRAINING.

UTILITIES (SERVICES)

THE WORD UTILITIES IS NOW GENERALLY USED FOR THE ANCILLARY SERVICES NEEDED IN THE OPERATION OF ANY PRODUCTION PROCESS. THESE SERVICES WILL NORMALLY BE SUPPLIED FROM A CENTRAL FACILITY; AND WILL INCLUDE:

I. ELECTRICITY: - POWER REQUIRED FOR ELECTROCHEMICAL PROCESSES, MOTORS, LIGHTINGS, AND GENERAL USE J. STEAM FOR PROCESS HEATING: - THE STEAMS REQUIRED FOR THE PROCESS ARE GENERATED IN THE TUBE BOILERS USING MOST ECONOMIC FUEL.

B. COOLING WATER: - NATURAL AND FORCED DRAFT COOLING TOWERS ARE GENERALLY USED TO PROVIDE THE COOLING WATER REQUIRED ON SITE.

C. WATER FOR GENERAL USE: - THE WATER REQUIRED FOR THE GENERAL PURPOSE WILL BE TAKEN FROM LOCAL WATER SUPPLIES LIKE RIVERS, LAKES AND SEAS. BECAUSE OF THIS REASON ALL THE PLANTS LOCATED ON THE BANKS OF RIVER.

D. DEMATERIALIZED WATER: - DEMATERIALIZED WATER, FROM WHICH ALL THE MINERALS HAVE BEEN REMOVED BY ION-EXCHANGE IS USED WHERE PURE WATER IS NEEDED FOR THE PROCESS USE, IN BOILER FEED WATER.

E. REFRIGERATION: - REFRIGERATION IS NEEDED FOR THE PROCESSES, WHICH REQUIRE TEMPERATURES BELOW THAT ARE PROVIDED BY THE COOLING WATER.

F. INERT-GAS SUPPLIES.

G. COMPRESSED AIR: - IN AN ETHYLENE OXIDE PLANT COMPRESSED AIR IS ONE OF THE RAW MATERIALS. IT IS ALSO NEEDED FOR PNEUMATIC CONTROLLERS ETC.

EFFLUENT DISPOSAL FACILITIES: - FACILITIES MUST BE PROVIDED FOR THE EFFECTIVE

DISPOSAL OF THE EFFLUENT WITHOUT ANY PUBLIC NUISANCE.

ENVIRONMENTAL IMPACT, AND EFFLUENT DISPOSAL

ALL INDUSTRIAL PROCESSES PRODUCE WASTE PRODUCTS, AND FULL CONSIDERATION MUST BE GIVEN TO THE DIFFICULTIES AND COAT OF THEIR DISPOSAL. THE DISPOSAL OF TOXIC AND HARMFUL EFFLUENTS WILL BE COVERED BY LOCAL REGULATIONS, AND THE APPROPRIATE AUTHORITIES MUST BE CONSULTED DURING THE INITIAL SITE SURVEY TO DETERMINE THE STANDARDS THAT MUST BE MET.

LOCAL COMMUNITY CONSIDERATIONS

THE PROPOSED PLANT MUST FIT IN WITH AND BE ACCEPTABLE TO THE LOCAL COMMUNITY. FULL CONSIDERATION MUST BE GIVEN TO THE SAFE LOCATION OF THE PLANT SO THAT IT DOES NOT IMPOSE A SIGNIFICANT ADDITIONAL RISK TO THE COMMUNITY.

LAND (SITE CONSIDERATIONS)

SUFFICIENT SUITABLE LAND MUST BE AVAILABLE FOR THE PROPOSED PLANT AND FUTURE EXPANSION. THE LAND SHOULD BE IDEALLY FLAT, WELL DRAINED AND HAVE LOAD-BEARING CHARACTERISTICS. A FULL SITE EVALUATION SHOULD BE MADE TO DETERMINE THE NEED FOR PILING OR OTHER FOUNDATIONS.CLIMATE

ADVERSE CLIMATIC CONDITIONS AT SITE WILL INCREASE COSTS. ABNORMALLY LOW TEMPERATURES WILL REQUIRE THE PROVISION OF ADDITIONAL INSULATION AND SPECIAL HEATING FOR EQUIPMENT AND PIPING. STRONGER LOCATIONS WILL BE NEEDED AT LOCATIONS SUBJECT TO HIGH WIND LOADS OR EARTHQUAKES.

POLITICAL AND STRATEGIC CONSIDERATIONS

CAPITAL GRANTS, TAX CONCESSIONS, AND OTHER INDUCEMENTS ARE OFTEN GIVEN BY GOVERNMENTS TO DIRECT NEW INVESTMENT TO PREFERRED LOCATIONS; SUCH AS AREAS OF HIGH UNEMPLOYMENT. THE AVAILABILITY OF SUCH GRANTS CAN BE THE OVERRIDING CONSIDERATION IN SITE SELECTION.

PLANT LAY OUT

The economic construction and efficient operation of a process unit will depend on how well the plant and equipment specified on the process flow sheet is laid out. The principal factors are considered are:

Economic considerations: construction and operating costs.

The process requirements.

Convenience of operation.

Convenience of maintenance.

Safety.

Future expansion. Modular construction. Costs The cost of construction can be minimized by adopting a layout that gives the shortest run of connecting pipe between equipment, and at least amount of structural steel work. However, this will not necessarily be the best arrangement for operation and maintenance. Process requirements An example of the need to take into account process consideration is the need to elevate the base of columns to provide the necessary net positive suction head to a pump or the operating head for a thermosyphon reboiler. Operations Equipment that needs to have frequent attention should be located convenient to the control room. Valves, sample points, and instruments should be located at convenient positions and heights. Sufficient working space and headroom must be provided to allow easy access to equipment. Maintenance Heat exchangers need to be sited so that the tube bundles can be easily withdrawn for cleaning and tube replacement. Vessels that require frequent replacement of catalyst or packing should be located on the out side of buildings. Equipment that requires dismantling formaintenance, such as compressors and large pumps, should be places 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 operators must be provided from each level in process buildings.

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, and service pipes over-sized to allow for future requirements.

Modular construction

In recent years there has been a move to assemble sections of plant at the plant manufacturers site. These modules will include the equipment, structural steel, piping and instrumentation. The modules are then transported to the plant site, by road or sea. The advantages of modular construction are:

f. Improved quality control.

g. Reduced construction cost.

h. Less need for skilled labour on site. Some of the disadvantages are; Higher design costs & more structural steel work.

More flanged constructions & Possible problems with assembly, on site.

THE PLANT LAYOUT KEYWORDS

Raw material Storage

Product Storage

Process Site

Laboratories

Workshop

Canteen & Change house

Fire Brigade

Central Control Room

Security office 10.Administrative Building Site for Expansion Project.

Effluent treatment plant 13.Power house

Emergency water storage 15.Plant utilities

A detailed plant layout is drawn and submitted with this thesis report. This plant layout is just a reference plant layout. There may be a lot of changes in actual plant layout.

BIBLIOGRAPHY

(1)

a. R. H. PERRY AND DON W. GREEN, PERRYS CHEMICAL ENGINEERS HAND BOOK, 6TH ED. MC-GRAW HILL INTERNATIONAL EDITION,

b. H.SAWISTOWSKI &W.SMITH, MASS TRANSFER PROCESS CALCULATIONS, INTERSCIENCE PUBLISHERS,

c. R. K. SINNOTT, COULSON AND RICHARDSONS CHEMICAL ENGINEERING SERIES, VOLUME-6, CHEMICAL EQUIPMENT DESIGN 3RD ED., BUTTER WORTH-HEINEMANN, PAGE NO: 828-855, 891-895

d. JOSHI M. V., PROCESS EQUIPMENT DESIGN, 2ND ED., MC-MILLAN INDIA LTD, e. 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. f. (7) B.C BHATTACHARYA, CHEMICAL EQUIPMENT DESIGN, CHEMICAL ENGINEERING EDUCATION DEVELOPMENT CENTRE.g. h. (8) L.E. BROWNELL AND E.H. YOUNG, PROCESS EQUIPMENT DESIGN, JOHN WILEY & SONS INC. NEW YORK,