Phthalic anhydride

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CONTENTS

S.No. CHAPTER NAME

1. INTRODUCTION

2. PROPERTIES

3. USES

4. SELECTION OF RAW MATERIAL

5. MATERIAL BALANCES

6. ENERGY BALANCES

7. EQUIPMENT DESIGN 8. COST ESTIMATION

9. PLANT LOCATION 10. PLANT LAYOUT 11. REFERENCES & BIBILIOGRAPHY

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INTRODUCTION

Phthalic Anhydride is an important aromatic di-carboxylic acid anhydride.It is an ortho-derivative of phthalic acid.Its abbreviation is PAN.

Preparation Methods:

It is normally prepared either by oxidation of Naphthalene or ortho-xylene.

a) From Naphthalene:

C10H8 + 4.5 O2 C8H4O3 + 2 CO2 + 2 H2O (g) b) From o-xylene: C8H10 + 3O2 C8H4O3 + 3 H2O

PROPERTIES

a. Physical Properties:

Molecular Formula: C8H4O3

Molecular weight : 148.1

Form: Flakes

Molten State: Clear liquid that resembles water

Odor: Mild Characteristic

Color: White

Melting Point: 130.6◦ C

Boiling Point: 284.5◦ C Density g/ml at 20◦C: 1.527

Vapor Pressure mm Hg at 25◦ C: 0.00052

Solubility : Slightly soluble in hot water and ether; sublimes below the Melting point

Latent Heat of Vaporization: 11850 cal/mol

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b. Chemical Properties:

PAN is an aromatic di-carboxylic acid. So it exhibits the reaction characteristics of dicarboxylic acid.

Some of the typical chemical reactions are:

Polymerization Reactions:

C8H4O3 + CH2OH – CHOH - CH2OH Glyptal (an alkyd resin) (PAN) (Glycerol) (Polyester)

Alcoholysis (conversion into esters):

C8H4O3 + CH3CH2CHOHCH3 sec-Butyl hydrogen phthalate

(PAN) (Sec-Butyl alcohol)

Friedel- Crafts acylation(Formation of Ketones): C8H4O3 + C6H6 o-Benzoylbenzoic acid

(PAN) (Benzene)

USES

PAN has various applications and a few of the areas are as follows:

Insecticides Pharma Products Reinforced Plastics Paints Plasticizers for PVC, Cellulose, Resins & Cellophane Dyes, Plastics, Printing Inks, Varnishes Printing Inks Dyes

SELECTION OF RAW MATERIAL

1. PAN can be manufactured from o-xylene, n-pentane and naphthalene.

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-H2O

AlCl3 0◦C

2. Ortho-xylene is clearly preferred modern day feed stock for PAN manufacture.3. Compared to n-pentane and naphthalene, o-xylene produces higher yields (all carbon atoms appear in the product), it is cheaper than naphthalene and provides a more efficient process.4. Converting an existing plant to o-xylene can reduce raw material costs by 25%.5. Advances are also being made in catalyst selectivity to improve the quality of the product and eliminate side reactions.

SELECTION OF REACTOR

1. Phthalic anhydride processes can be classified according to the type of reactor used.2. Three reactor configurations have been commercially developed. (a) Fixed-bed vapor phase reactor (b) Fluidized bed vapor phase reactor and (c) Liquid phase reactor 3. The fixed-bed vapor phase reactors have proven to be superior to fluidized-bed and liquid phase reactors.4. Fluidized bed processes have proved difficult to maintain and have suffered from erosion problems and excessive catalyst losses.5. The construction costs of liquid-phase processes have proven prohibitive. METHOD OF PRODUCTION

a) Raw Materials: Air and o-xylene

b) Basis: 60,000 metric tons/year of PAN product, 98 wt% purity (or) 8400 kg/hr (approx.) of PAN product, 98 wt% purity

c) Feed Requirements:o-xylene: 8480 kg/hrAir: o-xylene (wt ratio): 10:1Air: 84800 kg/hr

d) Chemical Reactions:

Major reaction:

C8H10 + 3O2 C8H4O3 + 3 H2O

Major side reaction:

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V2O5

355-375 °C

C8H10 + 10.5O2 5 H2O + 8CO2

Minor side reaction:C8H10 + 7.5 O2 C4H2O3 + 4 H2O + 4CO2

e) Process Flow Sheet:

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f) Process Description:

The raw materials are air and o-xylene. The o-xylene feed, which may be considered pure and at 0.75 atm, is pumped to 3 atm and then vaporized in a

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fired heater. Air, which may be assumed to contain only O2 & N2 is compressed to 3atm and heated in a heat exchanger.

The hot air and vaporized o-xylene are mixed and sent to a packed bed reactor. Packed bed reactor is multitubular reactor filled with supported V2O5. 100% of o-xylene is reacted in this reactor. The selectivity data is given in the following table:

Fractional Conversion of o-Xylene into (Yield of) Indicated Product:

T(°C) Maleic anhydride CO2 Phthalic anhydride300 1.00 0.00 0.00320 0.536 0.0339 0.425340 0.215 0.102 0.683360 0.100 0.200 0.700380 0.0463 0.356 0.598400 0.0215 0.602 0.377420 0.00 1.00 0.00

In the selectivity data table we observe the fractional conversion of o-xylene into Phthalic anhydride and maleic anhydride are 0.7 and 0.1 at 360◦ C respectively. Also, the selectivity for complete combustion reaction is 0.2. Therefore, the reactor is maintained at 355-365◦ C. Also, the reactor is maintained at 3 atm and a contact time within the reactor is about 0.1-0.4 seconds. Since all the three reactions taking place in a reactor are highly exothermic, the temperature is controlled around 355-365◦ C using Molten salt (High Heat Transfer Salt). The reactor effluent, which is at 2 atm enters a complex series of devices known as switch condensers. The feed to switch condensers may be no higher than 180 ◦C, hence reactor effluent must be cooled. The net result of switch condensers is that all light gases and water leave the top stream of reactor with small amounts of both anhydrides (PAN & MAN), while large amounts of PAN and MAN leave in stream which is feed to distillation column. The stream containing large amounts of PAN is fed to distillation column which results in formation of bottom product of quality 98% PAN.

g) Catalyst: It is a combination of Vanadium Pentoxide(V2O5) and Titanium dioxide(Ti2O5).It is prepared by heating a suspension containing compounds and spray coating the suspension onto a support which can be either 6mm porcelain spheres. Surface Area available for reaction is about 10-12 m2/g.

h) Major Engineering Problems:

1. Explosion Hazards-minimized by adding excess air to stay below the lower explosive limit.

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2. Fixed bed tubular reactor design-tube size and heat transfer must be considered to avoid too high a center temperature within each catalytic tube.

3. Catalyst development for high specificity of oxidation.

4. Choice of coolant –mercury and diphenyl are also used. If molten salt is used, then corrosion will be a problem.

MATERIAL BALANCE

Let us assume that 10 kmol of o-xylene is fed to a reactor. Selectivity for PAN reaction= 0.7

C8H10 + 3O2 C8H4O3 + 3 H2O

o-xylene reacted = 7 kmol

Oxygen reacted = 21 kmol

PAN formed = 7 kmol

Water (g) formed = 21 kmol

Selectivity for complete combustion reaction=0.2

C8H10 + 10.5O2 5 H2O + 8CO2

o-xylene reacted = 2 kmol

Oxygen reacted = 21 kmol Carbon dioxide formed = 16 kmol

Water (g) formed = 10 kmol

Selectivity for MAN reaction=0.1

C8H10 + 7.5 O2 C4H2O3 + 4 H2O + 4CO2

o-xylene reacted= 1 kmol

Oxygen reacted =7.5 kmol Carbon dioxide formed =4 kmol

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Water(g) formed= 4 kmol

Maleic anhydride = 1 kmol

Molecular weight of o-xylene =106

Therefore 10 kmol of o-xylene = 1060 kg

On weight basis, Air : o-xylene = 10:1

Air fed =10,600 kg = 367.55 kmol

Oxygen, Nitrogen in air are 77.18 kmol, 290.37 kmol.

Excess Oxygen = 77.18 – 49.5 = 27.68 kmol

Nitrogen fed = 290.37 kmol

Material Balance around the reactor :

Reactants entering: o-xylene = 10 kmol = 10*106 = 1060 kg

Oxygen fed = 77.18 kmol = 77.18*32 = 2469.76 kg

Nitrogen fed = 290.37 kmol = 290.37*28 = 8130.36 kg

Total weight = 1060+10600=11660 kg

Products leaving:

PAN = 7 kmol = 7*148 = 1036 kg

H2O(g) = 35 kmol=35*18 = 630 kg

CO2 = 20 kmol =20*44 = 880 kg

MAN = 1 kmol = 1*98 = 98 kg

O2 = 27.68 kmol = 27.68*32 = 885.76 kg

N2 = 290.37 kmol = 290.37 *28 = 8130.36 kg

Total weight = 11660.12 kg

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Law of Conservation of mass is satisfied.

Material Balance around the Switch Condenser:

Feed to switch condenser:

PAN = 7 kmol

H2O (g) = 35 kmol

CO2 = 20 kmol

MAN = 1 kmol

O2 = 27.68 kmol

N2 = 290.37 kmol

Assume that 2 mol% and 0.2 mol% of MAN and PAN present in the feed leave in the top stream.

Top Stream Leaving the Condenser:

PAN = (0.2/100)*7 = 0.014 kmol

H2O (g) = 35 kmol

CO2 = 20 kmol

MAN = (2/100)*1= 0.02 kmol

O2 = 27.68 kmol

N2 = 290.37 kmol

Bottom stream Leaving the Condenser:

MAN = 0.98 kmol = 0.98*98 = 96kg

PA N = 6.986 kmol = 6.986*148 = 1034 kg

Material Balance around the Distillation Column:

Feed to Distillation Column:

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MAN = 96 kg

PA N = 1034 kgTotal (F) = 1130 kg

PAN composition (weight basis) in Distillate (D) and Residue (W) are 2% and 98% respectivelyOverall Material Balance: F = D + W where F= 1130 kg (wt. basis)PAN balance (wt. basis) : xF*F = xD*D + x W* W Where xF*F = 1034 kg xD = 0.98 xW = 0.02

Solving above two equations, we get D = 76.5 kg; W = 1053.5 kg

Aim: To produce PAN product, 60000 metric tons/year, 98 wt% purity.

Here 1 year = 300 days

60,000 tons/yr = 200 tons/day=25/3 tons/hr=25000/3 kg/hr;

On similar basis of Material Balance using 10 kmol of o-xylene, inorder to produce 60,000 tons/yr of bottom product we have D = 605.1 kg/hr

PAN and MAN in D are 12.1 kg and 593 kg; respectively.

PAN = 24500/3 kg/hr;

MAN = 500/3 kg/hr;

Total PAN to distillation column = 24562.1/3 = 55.26 kmol/hrTotal PAN to distillation column = 2279/3 = 7.75 kmol/hr

We know that 2% and 0.2% of MAN and PAN present in the stream leaving the reactor is discharged to atmosphere.

Total PAN leaving the reactor = 55.26 /0.998 = 55.37 kmol/hrTotal MAN leaving the reactor = 7.75/0.98 = 7.91 kmol/hr

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Selectivity of MAN reaction is 0.1.Therefore, o-xylene to be charged to reactor = 7.91/0.1 = 79.1 kmol/hr

Start: o-xylene fed = 80 kmol/hr

Material Balance around the reactor :

Reactants entering: o-xylene = 80 kmol/hr = 80*106 =8480 kg/hr

On weight basis,Air: o-xylene = 10:1

Air fed = 84,800 kg/hr = 2940.4 kmol/hr

Oxygen fed = 617.5 kmol/hr = 617.5*32 = 19760 kg/hr

Nitrogen fed = 2322.9 kmol/hr = 22322.9*28 = 65040 kg/hr

Reactants kmol/hr mol.wt kg/hro-xylene 80 106 8480

O2 617.5 32 19760

N2 2322.9 28 65041.2 Total 93281.2 Selectivity for PAN reaction= 0.7

C8H10 + 3O2 C8H4O3 + 3 H2O

o-xylene reacted = 80*0.7 = 56 kmol/hr

Oxygen reacted = 56*3 = 168 kmol/hr PAN formed = 56*1 = 56 kmol/hr Water(g) formed = 56*3 =168 kmol /hr

Selectivity for complete combustion reaction=0.2

C8H10 + 10.5O2 5 H2O + 8CO2

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o-xylene reacted = 80*0.2 =16 kmol/hr

Oxygen reacted =10.5*16 = 168 kmol/hr Carbon dioxide formed = 8*16 = 128 kmol/hr

Water (g) formed = 5*16 = 80 kmol/hr

Selectivity for MAN reaction=0.1

C8H10 + 7.5 O2 C4H2O3 + 4 H2O + 4CO2

o-xylene reacted = 80*0.1 = 8 kmol/hr

Oxygen reacted = 7.5 *8 = 60 kmol/hr Carbon dioxide formed = 4*8 = 32 kmol/hr

Water (g) formed = 4*8 = 32 kmol/hr

Maleic anhydride = 8*1 = 8 kmol/hr

Excess oxygen = 617.5-(168+168+60) = 221.5 kmol/hr

Products leaving:

Products kmol/hr mol.wt kg/hrPAN 56 148 8288H2O (g) 280 18 5040CO2 160 44 7040MAN 8 98 784o-xylene 0 106 0O2 221.5 32 7088N2 2322.9 28 65041.2 Total 93281.2 Law of Conservation of mass is satisfied.

Material Balance around the Switch Condenser:

Feed to switch condenser:

PAN = 56 kmo/hr

H2O = 280 kmol/hr

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CO2 = 160 kmol/hr

MAN = 8 kmol/hr

O2 = 221.5 kmol /hr

N2 = 322.9 kmol/hr

Assume that 2 mol% and 0.2 mol% of MAN and PAN are present in the feed leaving the top stream.

Top Stream Leaving the Condenser:

PAN = (0.2/100)*56 = 0.112 kmol /hr

H2O(g) = 280 kmol/hr

CO2 = 160 kmol/hr

MAN = (2/100)*8 = 0.16 kmol/hr

O2 = 221.5 kmol /hr

N2 = 2322.9 kmol/hr

Bottom stream Leaving the Condenser:

MAN = 7.84 kmol/hr = 7.84*98 = 768.3 kg/hr

PA N = 55.888 kmol/hr = 55.888*148 = 8271.4 kg/hr Total = 9039.7 kg/hr

Material Balance around the Distillation Column:

Feed to Distillation Column:

MAN = 768.3 kg/hr

PA N = 8271.4 kg/hr----------------------------------- Total (F) = 9039.7 kg/hr------------------------------------

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PAN composition (weight basis) in Distillate(D) and Residue(W) are 2% and 98% respectively.

Overall Material Balance: F = D + W , where F= 9039.7 kg/hr (wt basis)PAN balance (wt basis) : xF*F = xD*D + x W* W Where xF*F=8271.44 kg xD= 0.98 xW= 0.02Solving above two equations, we get D= 611.98 kg/hr; W= 8427.71 kg/hrRounding Off: Phthalic anhydride Product (W) = 8400 kg/hr

Distillate product (D) = 640 kg/hrENERGY BALANCE

Since the gases are at high temperature and low pressure, the gases can be assumed to be ideal . Compound mole fractiono-xylene 0.0265air 0.9735

Specific heat of reacting mixture (air and o-xylene)

(Cp[M1]) = 6.386 + 4.885 *10-3 *T -2.179 *10-6 *T2 + 4.765*10-9 *T3 -3093.39 T -2

∆H net = 3020.4*1000 ∫(6.386 + 4.885*10-3*T – 2.179*10-6*T2 + 4.765*10-9*T3

– 3093.39*T-2) dT where T is from 298 K to 593 K = 3020.4*1000*2526.38 = 0.763*1010 Cal/hr = 31940.46 MJ/hr

∆H1= o-xylene stream;

∆H2= air stream;

∆H1+∆H2 =∆H net

2.508 *10-3 *T2 + 6.56 *T -4945.12 +3177.6* T -1 = 0Solving T = 610.5 K = 337.5◦ C

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Now, ∆H1= 80*1000*∫(-3.786 + 0.1424*T – 8.224*10-5*T2 + 1.798*10-7*T3)dT where T is from 423 K to 610.5 K

∆H2 = 27555.47 MJ/hr o-xylene stream:

Assuming 1% losses,

Heat transfer rate supplied by Natural gas to o-xylene stream=4385/0.99 = 4429.3 MJ/hr

Calorific Value of Natural gas = 54 kJ/g

Amount of Natural gas: m(54) = 4429.3 *1000 kJ/hr m=82 kg/hr

Air stream:

Assuming 1% losses,

Heat transfer rate supplied by steam to air stream = 27555.47/0.99 = 27833.81 MJ/hr

Energy Balance around the Reactor:

Heat transfer rate required to raise the reacting mixture from 320◦ C to 360◦ C = 3020.4*1000* 386.08 = 0.1166 *1010 cal/hr = 4881.1 MJ/hrPAN reaction: C8H10 + 3O2 C8H4O3 + 3 H2O

∆H◦ 298 K = -371.79 + 3(-242) -19.01 = -1116.8 KJ/mol ∆H◦ 633K = -1116.8 +11.814 = -1104.98 kJ/mol

∆H net(1)= n(∆H◦ 633K ) = 56 *1000*(-1104.98)*1000=-61878.88MJ/hrMAN reaction:C8H10 + 7.5 O2 C4H2O3 + 4 H2O + 4CO2

∆H◦ 298 K = -469.65 + 4(-242) + 4(-393.77) -19.01 = -3031.74 kJ/mol∆H◦ 633K= -3031.74 -7.69 =-3039.43 kJ/mol

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∆H net(2)= n(∆H◦ 633K ) = 8*1000*-3039.43*1000 = -24315.44 MJ/hr

Complete Combustion:C8H10 + 10.5O2 5 H2O + 8CO2

∆H◦ 298 K = 8(-393.77)+5(-242)-19.01 = -4379.17 kJ/mol∆H◦ 633K = -4379.17 + 174.785= -4204.38kJ/mol

∆H net(3) = n(∆H◦ 633K ) =16*1000*-4204.38*1000 = -67,270.16 MJ/hrOverall Heat transfer rate liberated = -61878.88 -24315.44-67270.16 = -153464.48 MJ/hrLet us allow the Products to leave the reactor at 400◦ C.∆HO2 = 288.68 MJ/hr∆HN2=(23 22.9*1000)(1218.53)=2830.51 MJ/hr ∆HMAN= (8000)*5392.32= 43.14 MJ/hr∆HCO2 = (160*1000)*1945.38 =(160*1000) * 1945.38 =311.26 MJ/hr∆HH2O = (280*1000)*1479.97=414.39MJ/hr∆HPAN= 600.41 MJ/hrOverall Heat utilized for raising products from 360◦ C to 400◦ C = 4488.4MJ/hr Net enthalpy heat rate change with in the reactor =4881.1-153464.5+4488.4 = -144095.01 MJ/hr144095.01 MJ/hr is to be utilized by molten salt. Energy balance around the salt cooler: Q = m Cp ∆ T 144095.01 *106 = m *1560*(395-150) m= 377 tons/hr =104.72 kg/sEnergy balance around the heat exchanger – 2(HE-2):

Boiler feed water (bfw) available at 549 kPa,90◦ C

It is to be delivered at high pressure (4300 kPa)Saturated steam enthalpy (at 4300 kPa and 254◦ C)=2799.4 kJ/kgAt heat exchanger-2, 144095.01*106= m(2799.4*1000) m = 51473.5 kg/hr Flow rate of boiler feed water = 51473.5 kg/hr

Heat possessed by products leaving reactor at 400◦ C :∆HO2=221.5 *1000*11633.92 = 2576.9 MJ/hr

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∆HN2 = (2322.9*1000)*11120.96 = 25832.9 MJ/hr ∆HMAN = 8000*40662.54 = 325.3 MJ/hrBoiling point of water = 406.7 K at 3 atm∆HH2O = 16403.4 MJ/hr∆HPAN=56*1000*82634.21= 4627.52 MJ/hr∆HCO2 =160*1000*16441.09 =2630.58 MJ/hr

Overall Heat possessed by products leaving the reactor = 2576.9 + 25832.9 +325.3 +2630.58 +16403.4+4627.52 = 52396.6 MJ/hrEnergy balance around the heat exchanger – 3(HE-3):

Energy possessed by stream after heat exchanger-3 :∆HO2 = 221.5 *1000* (3735.32)=827.37 MJ/hr∆HN2 =2322.9 *1000 *3651.46=8481.97 MJ/hr∆HMAN = 8*1000*8713.958=69711664 cal/hr=291.8 MJ/hr∆HH2O =13929.75 MJ/hr∆HPAN=56*1000*8713.958=487981648 cal/hr=2042.59 MJ/hr∆HCO2 = 794.86 MJ/hrTotal Heat = 26368.3 MJ/hr52397-26368=26029 MJ/hr is being utilized by heat exchanger – 3,Boiling Point of water = 427 K at 525 kPa Saturated steam enthalpy =2749.7 kJ/hrm (2749.7*1000) =26029 *106

m= 9466 kg/hrFlow rate of cool water = 9466 kg/hr

Energy balance around the Switch Condenser:Energy possessed by effluent stream:

Effluents leave the heat exchanger – 3 at 140◦ C∆HO2 = 221.5 *1000* (3431.55) = 760.09 MJ/hr∆HN2 =2322.9 *1000 *3358.4=7801.2 MJ/hr

∆HMAN= 160*9797==1.57 MJ/hr∆HH2O =13832.94 MJ/hr(since B.P of water = 393 K at 2 atm)∆HPAN= 112 * 20480 = 2.29 MJ/hr∆HCO2 =160*1000*4548.34= 727.73 MJ/hrTotal energy possessed by effluent stream =23125.76 MJ/hrSince Cp (MAN) =Cp (PAN) in liquid state, we have

Cp (MAN) =Cp (PAN) = Cp (Feed)Specific enthalpy = 7972.15 Cal/mol = 33369.8 kJ/kmolTotal enthalpy of feed = (55.888+7.84) (33,369.8) = 63.728 *33,369.8 = 2126.6 MJ/hr

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Energy remaining =26368.3 – (2126.64+23,125.8) = 1115.9 MJ/hrLet cool water be condensing medium and its temperature rise by 10◦Ci.e., from 30 ◦C to 40◦C 1115.9 *1000 =m (4.18)(40-30) m=26696.2 kg/hr=26.7 T/hr26.7 T/hr of cooling water is being used.

Energy balance around the Distillation Column:

In W, 55.62 and 1.71 kmol/hr of PAN and MAN are present.In D, 0.086 and 6.4 kmol/hr of PAN and MAN are present.Solidification point of PAN = 130.8◦CBottom product is at 150◦CSpecific enthalpy of bottom product (hW ) =8713.96 cal /mol =36474.9 kJ/kmolTotal Enthalpy of bottom product = (55.62+1.71) (36474.9) = 57.33 * 36474.9 = 2091.1 MJ/hrTop Product is at 60◦C (333K)Specific enthalpy of top product (hD) = 2317.48 cal/mol =9700 .5 kJ/kmol Total enthalpy of Top product =(6.4+0.086) (9700.5) =6.486 *9700.5 -62.92 MJ/hr

Composition of MAN and PAN in D are 0.987, 0.013Assuming PAN and MAN obey’s Raoults law, y = ax/ (1+(a-1)x); where a=sqrt( atop* a bottom)Here MAN is more volatile component.Using Antonie’s equation,Vapor pressure of MAN at 333 K, 423 K are 3.536 mm Hg, 186.2 mm Hg.

Vapor pressure of PAN at 333 K, 423 K are 0.153 mm Hg, 17.36 mm Hg.atop =3.536/0.153 =23.11; a bottom =186.2/17.36= 10.73

a=sqrt( atop* a bottom) =sqrt(23.11*10.73) = 15.75y = 15.75x/(1+14.75x) is equilibrium relation.

Let us assume that boiling point of feed varies linearly with compositionBoiling point of PAN and MAN are 560 K and 473 KFor feed composition, Boiling point of feed =(55.888*560+7.84*473)/(55.888+7.84) = 549.3 KSpecific enthalpy of feed if feed is saturated liquid =18,752.08 cal/mol = 78492.5 kJ/kmolSpecific enthalpy of feed = 7972.15 cal/mol =33369.8 kJ/kmol Normal Heats of vaporization

PAN 11850 cal/mol = 49601.73 kJ/kmolMAN 5850 cal/mol = 24486.93 kJ/kmol

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Let us assume that latent heat of vaporization of feed varies linearly with compositionλ feed = (55.888*49601.73 +7.84 *24486.93)/(55.888+7.84) = 46511.66 kJ/kmol

Specific enthalpy of feed if feed is saturated vapor = Specific enthalpy of saturated liquid + Latent heat of vaporization of feed =78492.5 +46511.66 =125004.2 kJ/kmolq= (Hv - hF )/(Hv– hL ) = (125004.2-33369.8)/(125004.2- 78492.5) = 1.97Assume that constant Molar overflow rate is prevailing.Feed line is y = qx/(q-1) – xF /(q-1) = 2.03 x -0.127Point of intersection of feed line and equilibrium line is 2.03 x -0.127 = 15.75x / (1+14.75x) x = 0.53 => y = 0.95For minimum reflux (Rm) Top section operating line passes through pinch point (0.53,0.95) and (0.98,0.98)Slope = (0.98- 0.95)/(0.98- 0.53) = 1/15 = Rm / ( Rm + 1) Rm =1/14Ropt = 1.5 Rm =1.5/14 =0.107Mole fraction of MAN in top product =(627.2/98)/[(627.2/98) +(12.8/98)] = 0.987Mole fraction of PAN in top product =1- 0.987 =0.013Assuming boiling point varies linearly, Boiling point of Top product = 0.987 * 473 + 0.013 *560 =474.13 Kλ top product = 0.987*24486.93 +0.013 *49601.73 = 24813.42 kJ/kmolSpecific enthalpy of top product if it is saturated vapor = 12628.32 +24 813.42 =37441.74 kJ/kmol

Energy balance around Total Condenser:

V*HV = D*hD + Lo* hLo + Qc

Using V=(R+1)D, Lo=R*D, hLo= hD

Q c = (R+!) D [HV –hD] =(1+0.107)(6.4+0.086 )[37441.74-9700.5] = 199.2 MJ/hrCooling water is available at 30◦C and let its temperature rise by 10◦C mCp∆T = 199182.16 kJ/hr m (4.18)(40-30) = 199182.16 kJ/hr m= 4765.12 kg/hr = 4.765 T/hr

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Flow rate of cooling water = 4765.12 kg/hr = 4.765 T/hr

Overall Energy balance around Distillation Column: F*hF + QB = Q c + D*hD + W* hW 2126.6 + QB =199.2 +62.92 +2091.1 QB = 226.62 MJ/hrReboiler load :Electrical Power to be supplied to Reboiler at a rate of 226.62 MJ/hr

DESIGN OF TUBULAR REACTOR

Estimation of volume of Reacting mixture:

1 kmol – 22.414 m3( at STP)P1V1/T1 =P2V2/T2

1*22.414/273= 3* V2/593V2=16.23 m3

1 kmol of o-xylene at 3 atm and 320 C =16.23 m31 mol of reactant mixture =30.88 gm93,281.2 kg/hr =3020.4 kmol/hr =49021.09 m3/hr=13.62 m3/s

Estimation of volume of products mixture:

P1V1/T1 =P2V2/T2

1*22.414/273 =3*V2/673 V2=18.42 m3

1 kmol of products at 2 atm and 400 C =18.42 m3

1 mol of product mixture =30.6gm93,281.2 kg/hr =3048.4 kmol/hr = 56151.53 m3/hr=15.6 m3/s

Basis: 16 m3/s of gasUse 1.5” tubes of outside diameter (BWG 10no) and length =4mWall thickness for BWG 10 no tube =0.134”Therefore,Inside diameter of tube =1.5-2*0.134=1.232”=31.3 mmVolume of each tube = π/4 * d2 *L = π/4 (31.2928*10-3)2*4 =3.0764*10-3 m3

Residence time with in reactor =0.4 sec T =volume of reactor/volumetric flow rate of gas through reactor 0.4=16/Q

Q = 40 m3/sec

Let ‘N” be no.of tubes N(3.0764*10-3) = 40

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N = 13002 = 13000 tubes (approx.)Heat transfer area =N(π*do*h) = 13000*π*38.1*10-3* 4 = 6224 m2

Triangular pitch is used to accommodate 13000 no of tubes.Triangular tube pitch ,Pt=1.25do=1.25 *38.1=47.625 mm

Bundle diameter:Db =do(Nt/K1)1/n1

=38.1(13000/0.319)1/2.142 =5409.9 =5410 mm

Graph: Shell inside diameter-bundle diameter vs bundle diameter By extrapolating the bundle diameter to 5.41 m, We have Shell ID –Bundle dia = 20 +(20-10)/(1.2-0.2)*(5.41-1.2) = 20+(4.21)10=62.1 mm Therefore shell ID =Bundle dia + 62.1 mm = 5410 +62.1=5472.1 mm =5474 mmHeight of a reactor = 10mVolume of a reactor =(π/4)*D2*H = (π/4) (5.474)2 *10=235.34 m3

ShellMaterial of Construction: PAN and MAN vapors are not particularly corrosive to most steels. The shell is susceptible to attack from cooling salt only. The preferred construction material for reactor is low alloy steel (IS:3609)

From Code IS 2825, f=12.6 kgf/mm2

=1260 kgf/cm2

This is a Class-I vessel. Therefore, J = weld joint efficiency factor = 1Here t = PDi/(200fJ-P) +CA where P=3 atm = (3.04*10-2*5474)/(200*1*12.6-3.04 *10-2) + 1 =1.066 mm Since t< 12 mm, Use t =12 mm (including corrosion allowance) in order to withstand its own weight.2:1 ellipsoidal dome end :

Calculation of Dome end Thickness: t = PDoC/200f J +CA +TAAssume t=10 mm hi=Di/4 =5474/4 =1368.5 mm ho =hi + t =1368.5 +10 =1378.5 mm

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Ro =0.82Di + t =0.82 *5474 +10 =4498.68 mm ro= 0.15 Di+T =0.15*5474 +10=831.1 mm Do =Di +2t =5474 +2*12 =5498 mmDo

2/4R0 = 54982/(4*4498.68) = 1679.82mSqrt(Doro/2)=Sqrt (5498*831.1/2)=1511.52ho = 1378.5 mmho is least. Therefore hE=ho=1378.5 mm

hE/Do=1378.5/5498 =0.251;t/Do =10/5498 =1.819 *10-3

From graph hE/Do vs C, we have C=1.35t = PDoC/200f J = (3.04*10-2*5498*1.35)/(200*1*12.6) = 0.089 mmSo, thickness of dome = 10mmShell side fluid is molten salt.Total Heat removed = 144095 MJ/hrHeat removed per tube = (144095.01*10-6)/(3600*13000) = 3079 Wq = UA∆TSubstituting A=6224 m2, ∆T=395-150 = 245°C, q = 144095.01 MJ/hrWe have U = 26.25 W/m2°CInternal diameter of vessel = 5474 mmWe know that Ds/5 < baffle spacing < Ds

i.e., 1094.8< baffle spacing<5474Use Baffle spacing = 1360 mmNumber of baffles = 4

Design Specifications:Tubes:

Number of tubes: 13000Length: 4mCatalyst filled height: 3.8 mInside diameter: 31.3 mmOutside diameter: 38.1 mmGauge: BWG 10 numberHeat transfer area: 6224m2

Passes: 1 (flow in upward direction)Tube Pattern: TriangularTube Pitch: 47.6 mmInlet temperature: 320° COutlet temperature: 400° C

Shell:

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Fluid: Molten SaltComposition of fluid: 40% NaNO2, 7% NaNO3, 50% KNO3

Flow: 377 Tons/hrInlet temperature: 150 °COutlet temperature: 395 °CHeat removed per tube: 3079 WHeat transfer coefficient: 26.25 W/m2 °CInside Diameter : 5475 mmPasses: 1Baffles: 4Baffle spacing: 1360 mm

Details about construction material:Type of material: Low alloy steelMaterial specification: IS:3609Grade or designation: 2.5% Cr, 1% MoTensile strength (min): 49 Kg/mm2

Yield stress (min): 25 Kg/mm2

Reactor specifications:Orientation: VerticalOperation mode: ContinuousHeight: 10 mOutside diameter: 5498 mmClass: IDesign Pressure: 3 atmWall thickness: 12 mmType of Ends: Ellipsoid, 2.5:1End thickness: 10 mmMan holes: 2 * 450 mm * 400 mm

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ECONOMIC EVALUATION

Equipment Cost:Type No. Reqd. Unit price (Rs.) Total (Rs.)Pump 2 50,000 1,00,000Reactor 1 15,00,000 15,00,000Fired Heater 1 2,00,000 2,00,000Compressor 1 1,00,000 1,00,000Heat Exchanger 1 1,00,000 1,00,000Switch Condenser 1 1,00,000 1,00,000Distillation Column 1 2,00,000 2,00,000

Equipment installation cost, Pi = 1.43 Pd = 32,89,000

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Piping = 20% of Pi = 4,60,000Instrumentation = 5% of Pi = 1,15,000Building & site development = 5% of Pi = 1,15,000Auxiliaries = 5% of Pi = 1,15,000Outside lines = 10% of Pi = 2,30,000Total physical plant cost = T.P.C = 63,94,000Contingencies = 10% of T.P.C= 19,18,200Size factor = 5% of T.P.C = 1,27,880Total fixed capital cost = = 84,40,080

MANUFACTURING COSTS:

1. Raw Material

Cost of 1 Kg of o-xylene = Rs. 20Total cost per year = 20 * 160.56 * 300= Rs. 9,63,360Hence total cost for raw material = Rs. 9,63,360

2. Utilities

Total cost of power requirement = 75,00,000 / yr

3. Labor & supervision:

Designation No. reqd. Salary/month Salary P.a.Managing Director 1 25,000/- 3,00,000/-Plant Manager 2 20,000/- 4,80,000/-Plant Engineers 2 15,000/- 3,60,000/-Plant Operators 4 15,000/- 5,20,000/-Skilled labor 4 5,000/- 2,40,000/-Unskilled labor 10 2,500/- 3,00,000/-Watchman 4 1,500/- 72,000/-

Total Annual Salary 21,36,000/-

Perks = 25% of total 5,34,000/-Total 26,70,000/-

Designation No. reqd. Salary/month Salary P.a.Managing (Finance & Sales) 1 15,000/- 1,80,000/-Officer (Finan. & Persna Sales)

1 10,000/- 1,20,000/-

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Accountants 2 6,000/- 2,88,000/-Typists 5 3,000/- 1,80,000/-Clerks 3 2,500/- 1,30,000/-Drivers 10 2,000/- 2,40,000/-Attendants 8 1,500/- 96,000/- Total Annual Salary 12,24,000/- Perks = 25% of total 5,34,000/-

Total 14,07,600/-Therefore, total annual wages = 14,07,600 + 26,70,000 = Rs. 40,77,600/-4. Repair and maintenance of Equipment: = 0.2% of Pi = 0.002*230000 = Rs. 4600A. Total direct conversion expenses = 1 + 2 + 3 + 4 = Rs. 1,00,45,560B. Indirect conversion expenses a. Depreciation on equip (1% of EPC) = Rs.2300b. Depreciation on buildg. (5% of BC) = Rs. 57500c. Insurance (3% of TPC) = Rs. 1,91,820Therefore, total conversion expenses = Rs. 2,51,620Therefore, total manufacturing costs or cost of sales, C.P. = A + B = Rs. 98,18,965Selling price for 1 ton of phthalic anhydride = Rs. 185So, for production of 60000 tons, the total selling price = Rs. 1,11,00,000Therefore, gross profit = S.P. – C.P. = Rs. 12,81,035Income tax = 38% of G.P. = Rs. 4,86,793.3Therefore, net profit = Rs. 794241.7Capital earning rate = net profit / total investment * 100

= 42.997Capital payout time = total capital investment / property earnings + Depreciation

= 2.323 yearsRate of returns = Net profit * 100 / total capital investment

= 21.24 %PLANT LOCATION

Plant Location and Site Selection:The major requirements for an Ethylene Glycol plant are ethylene oxide and water. Glycol plants are almost always located very close to Ethylene oxide plants to reduce transportation expenses, as the transportation of ethylene oxide is expensive due to its explosive tendencies. The plant considered here is located adjacent to an Ethylene oxide plant. The most optimum location would be in a petrochemical industrial area where there is a market for fiber grade glycol.

The other considerations are as follows:

1. Raw Materials Availability:

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Probably the location of the raw materials of an industry contributes more toward the choice of plant site than any other factor in most chemical operations low delivered cost of raw materials must be weighed up against other operating costs.This is especially noticeable in those industries in which the raw materials areinexpensive and bulky and is made more compact and obtain a high bulk value during the process of manufacture. The supply of basic raw materials should be controlled directly be user. Physical distance is not the only controlling factor in source of raw materials, for purchase price and buying expense, base point procuring, reserve stock and reliability of supply are also determinants.

2. Markets and Transportation:The existence of transportation facilities has given too many of the greatest track centers of world. A location should be chosen, if possible which has several competitions will help to maintain low rates and give better service. Often times, a location is selected outside the city in order to have a rail road siding available and thus eliminate trucking costs to freight years from excessive costs of transportation. There will be more long distance water transportation used in the future to reduce the cost of freight years from excessive costs of transportation. There will be long-distance water transportation used in the future to reduce the cost of freight, with the spread between production cost and sales cost constantly narrowing. We would see that the product has a ready market at a close distance from the plant site so that transportation will not become a big problem. Also we have to see that the product has as a ready market so that there will be demand throughout the year for the product.

3. ClimateThe plant site should be at a place where the climate is mild. Excessive cold, torrid heat and excessive humidity should not be present where the plant is situated for this will reduce the productivity part of the workmen. Also if excessive of conditions is present then the air conditioning and other facilities also will increase the expenditure.

4. Power supply The Chemical Engineering industries are the largest users of electric power equipment among the industries today because the modern demand is for extreme flexibility that sometimes errors on the side of too many individual drives. Power for chemical industry is primarily from coal water and oil: in as much as they provide for the generation of steam both for processing and of electricity production. A plant should establish near a hydraulic power generated project. By keeping the availability of power and deep water transportation overweight all other considerations including that of extremely severe winter weather making for difficult operating conditions.

5. Water supply

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Water for industrial purpose can be obtained from one of two general sources: the plant’s own source of municipal supply, if the demand for water is large it is more economical for the industry to supply its own water. Such supply may be obtained from drilled wells, rivers, lakes, dammed steams, or other impounded supplier, before a company enters upon any project, it must ensure itself if a sufficient supply of water for all industrial, sanitary and fire demand, both present and future. Data on temperature of water and on maximum, minimum and average rain fall can be obtained from governmental agencies if surface water is to be pounded or the date on stream flow of reverse can be acquired likewise if wells are to be relied on, Geologists and practical well drillers should be consulted.

6. Labor supplyA certain careful study of the supply of a cheap labor should be made. Factors to be considered in labor studies are supply, kind diversity, intelligence, wage scales regulation efficiency and costs. The success of many of organizations depends upon the means by which its labor gets to and from their works. A cheap site may have to be avoided if the laborers cone a long distance they will be tired in coming to the plant. Also technical skill should be given due importance.

7. Community and Site characters The nature of the sub soil is very important while considering the plant of the industry. Also due consideration should be given for the expansion of the plant. The cost of the land is important, as well as local building costs and living conditions. Also even if there is no immediate plan for expanding, a new plant should be constructed at a location where additional space is available.

PLANT LAYOUTPlant layout in its broadest sense is a part of the overall system. It includes everything from the original of the building to the location and movement of a small component. It is an integral part of: PRODUCTION PLANNING: It allows, promotes and aids the creation of utility. MAINTENANCE: It affects the amount, difficulty and time required for it. MATERIAL HANDLING: This is necessitated by the design & layout of the plant. ORGANIZATION: Physical layout often determines areas of authority, spheres of personnel influence.

Obviously machines, equipment, materials, employees, fixtures and all the necessary facilities for engaging in an activity must be given a place of work. How they are located and where they are located, may well determine the firm’s efficiency, its profit potential and its existence. Noise, color, tight and dictate work

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environment. Proper work environment will increase the productivity to optimum levels, boost the morale and job satisfaction.

Good layout design requires through knowledge of work flow, product flow and information flow. Engineering, management & future expansion are to be imbibed into the layout design. Technology is continuously upgrading making better manufacturing techniques available and correct layout will accommodate these challenges. Consideration is to be given for backward integration and forward integration of the product.

Further the arrangement of the equipment and facilities specified in the process flow sheet is a necessary requirement for accurate pre-construction cost estimation of future detailed design involving piping, structural and electrical facilities. Careful attention to the development of the plot and the elevation plans will point out unusual plant requirements and therefore, give reliable information on building site costs required for precise pre-construction cost accounting.The following list will suggest some of the reasons for what good layout is about:1. Reduce manufacturing costs.2. Increase employee safety.3. Better service to the customer.4. Reduce capital investment.5. Increase flexibility.6. Improve employee morale through improved employee comforts and conveniences in work area.7. Better quality of the product.8. Effective utilization of floor space.9. Reduce work in process to a minimum.10. Reduce work delays and stoppages.11. Better work methods and utilization of labor.12. Improve control and supervision.13. Easier maintenance.14. Reduce manufacturing cycle.15. Better utilization of equipment and facilities.16. Eliminate congestion points.

In developing an effective layout for an enterprise, we should in mind several fundamentals, which exert a significant influence in achieving a good and workable arrangement.

The following are among the major fundamentals most often citied:

Storage layout:Storage facilities for raw materials, intermediate and finished products may be located in isolated areas or in adjoining areas. Hazardous materials become a menace to life and property when stored in large quantities and should be kept

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isolated. Arranging storage of materials so as to facilitate or simplify handling is a point to be considered in design. Where it is possible to pump a single material to an elevation so that subsequent handling can be accomplished by gravity into intermediate reaction storage units.

Equipment layout:In making layout, ample space should be assigned to each piece of equipment accessibility is an important factor for maintenance. Unless a process is well seasoned, it is not always possible to predict just how its various units may have to be changed in order to be in harmony with each other. It is extremely poor economy to put the equipment layout too closely to a building. A slightly larger building will cost little more than that is crowded. The extra cost will indeed be small in comparison with the penalties that will be extracted if the building was to be extracted. The relative levels of the several pieces of equipment and their accessories determine their placement. Although gravity flow is usually preferable, it is not altogether.

Necessary because liquids can be transported by blowing or by pumping and solids can be moved by mechanical means. Access for initial construction and maintenance is a necessary part of planning, for example, over head equipment must have space for lowering into place, and heat exchange equipment should be located near access areas here trucks or hoists can be placed for pulling and replacing tube bundles. Thus space should be provided for movement of cranes and fork trucks as well as access way around doors and underground hatches. Therefore each plant presents its own challenges that need to be incorporated in the layout.

Plant expansion:Expansion must always be kept in mind. The question of multiplying the number of units or increasing the size of the prevailing unit or units merit more study than it can be given here. Suffice it to say that one must exercise engineering judgment. Correcting inconsiderate layout plan may involve scrapping the serviceable equipment or shut down the running equipment. Nevertheless the cost of change must be borne for economics of large units and in the end make replacement inevitable.

Floor space:Floor space mayor may not be major factor in the design of a particular plant. The value of land may be considerable item. The engineer should, however, follow the rule of practicing economy of floor space, consistent with good house keeping in the plant and with proper consideration given to line flow of materials, space to permit working on parts of equipment that need servicing , safety and comfort to the operators.

Utilities servicing:

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The distribution of gas, air, water, steam, power and electricity is not always a major item of consideration but flexibility of designing these items should permit to meet almost any condition. Regard to the proper placement of each of these services practicing good design reduces the cost of maintenance.

Workshop:A workshop is also provided to supply tools on demand from laboratory and process. Therefore, this is laid out nearer to the process area.

Safety units:These are located to the processing area, because probably accidents occur at the processing. Thus they can be easily controlled.

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REFERENCES AND BIBLIOGRAPHY

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1. Chemical Engineering Design (4th edition)by R K Sinnot

2. Process Heat Transfer by Donald Q.Kern

3. Perry’s Chemical Engineer’s Hand Book (7th edition)by Robert H.Perryby Don W. Green

4. Chemical Engineering Thermodynamics(6th edition)by JM Smith ,HC VanNess ,MM.Abott

5. Stoichiometry(4th edition)by BI Bhatt and SM Vora

6. Unit Operations of Chemical Engineering(7th edition)by McCabe,Smith,Harriot

7. Mass Transfer Operations(3rd edition)by Robert E Treybal

8. Chemical Reaction Engineering (3rd edition)by Octave Levenspiel

9. Outlines of Chemical Technology(3rd edition)by Charles E dryden

10. Code for Unfired Pressure Vessels IS:282511. Organic Chemistry (6th edition)

by Robert Morrison and Robert Boyd

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