ppt final.pptx

DEPARTMENT OF CHEMICAL ENGINEERING JAYPEE UNIVERSITY OF ENGINEERING & TECHNOLOGY, A-B ROAD, RAGHOGARH, DT. GUNA – 473226, M.P., INDIA

PRODUCTION OF PHENOL FROM CHLOROBENZENE

Shikhar Prakash (101446)Sumit Verma (101449)

1Production of Phenol from Chlorobenzene

Natural occurrences Natural occurrences:

During fractionation of coal tar, carbonic oil fraction is

separated containing 30-35wt% phenolic components and liquid

byproducts from gasification process.

Applications: Used as an disinfectant in household cleaners and in mouthwash,

used forsynthesis of dyes, aspirin (C9H8O4). Used for making picric acid , Used in the manufacturing of

Bisphenol - A Phenol is also used to produce Aniline

Product specifications & standards

Phenol is classified as a Class B poison.The US Pharmacopeia (USP) specifications for phenol

include:

a) Purity is to be no less than 98 wt%.

b) Clear solubility of 1 part of phenol in 15 parts of water.

c) A congealing temperature to be not lower than 39°C.

d) A content of non volatiles is no more than 0.05 wt%.

Phenol demand forecast in India

Graph 1: Production / Import of phenol in metric ton per year.

2011 2012 2013 2014 20150

50,000

100,000

150,000

200,000

250,000

300,000

DemandProductionImport

Year

Metric tonnes

Plant capacity on demand - supply basis (for 2014)

Demand: 240,000 TPA

Production: 80,000 TPA

Import: 160,000 TPA

Plant Capacity: 200,000 TPA (To meet the demand).

S. No.

Specification UnitsProcesses

Cumene Oxidation Toluene Benzoic Acid

Sulpho-nation

Chloro Benzene

Rasching

Feed Stocks1. Toluene T/T - 1.45 - - -2. Benzene T/T 0.67 - 0.95 1.01 0.9253. Chlorine T/T - - - 1.04 -4. Propylene T/T 0.38 - - - -

Utilities

(Consumption per tonne of phenol)

Steam Tonne 3.5 9.0 2.7 9.0 16.5Cooling water m3 35 65 80 250 50Process water m3 0.02 - 6.5 1.0 -Electricity KWh 350 650 100 3900 350Fuel Oil 106 KJ 4.5 3.5 15 5.5 14Inert gas(High Pressure)

SCF 80 40 - - -

Natural Gas BTU - - 2100 10000 4000

The most widely accepted technology for producing phenol is cumene oxidation process because of better economics due to by-product acetone. If one wants to avoid acetone, the choice will be toluene oxidation process.

Selected process

Phenol from chlorobenzeneThere aretwo reactions to convert chlorobenzene to phenol using chlorination route:

Causticization

C6H5 Cl+ NaOH → C6H5ONa

NaOH is in aqueous media (Operating conditions: 425°C and 350 atm, Exothermic reaction)

Hydrolysis C6H5ONa + HCl (aq) → C6H5 Cl + NaCl (aq)

(Operating conditions: 1 atm & 60 ° C)

Fig1. PFD for production of Phenol from Chlorobenzene

Material Balance

Phenol production with a purity of 99% = 2,00,000 TPA.

All calculations are done on per hour operation.

Working days in a year = 330 days.

Molecular weight of phenol = 94.

Per hour production = 25252.53 kg/hr.

= 268.644 kg mol/hr.

Various assumptions and values taken from literature:

Overall material balance

Component In (Kg/h) Out (Kg/h)

Chlorobenzene 32113.143 1605.668

Sodium hydroxide 14263.810 3423.303

Diphenyl oxide 2575.277 2575.277

Phenol -------- 25507.566

Sodium chloride -------- 15839.168

Total 48952.230 48950.982

Energy balance ΔH 298 = -114.567 KJ/mol

ΔH C6H5Cl = 285.297(230.0238) (373-303) = 17718836.42 J/mol

ΔH NaOH = 356.622 (86.0937) (573-303) = 3243363.955 J/mol

ΔHC6H5OH =271.033 (239.334)(353-303) = 3243363.955 J/mol

ΔH NaCl = 271.033(84.4090) (353-303) = 1143881.74 J/mol

Heat of reaction = Σ Heat of product – Σ Heat of reactant+ ΔH 298

Heat of reaction = -21735942.74 J/mol

Production of Phenol from Chlorobenzene 11

Heat balance across Heat balance across heat exchanger

heat exchanger inlet outlet reactor inlet

Tref = 303 K Tref = 303K,

Tin = 313 K T= 573K

QNaOH = 310595.5465 J Q NaOH = 8342126.471 J

QC6H5Cl = 5449157.38 J QC6H5Cl = 14019901.52 J

Q C12H10O = 45211.0613 J QC12H10O = 1358477.399 J

Heat balance reactor outlet Heat balance neutralizer in

Tref = 303 K, T = 698 K T ref = 303K, T= 333 K

QHCl = 19671615.81 J QHCl = 309069.2717 J

Q C6H5Cl = 1190158.848 J QNaOH = 223550.9719 J

QNaOH =2920034.506 J Q C12H10O = 136762.8128 J

Q C12H10O = 2147280.632 J Production of Phenol from Chlorobenzene 12

QC6H5OH = 3116121.024 J

QNaCl = 115353535.308 J

QC6H5Cl = 111834.3687 J

QNaOH = 372452.6559 J

Q C12H10O = 229770.867 J

Energy balance across distillation column

Assumptions:Temperature of distillate = 460 K

Boiling point of Diphenyl oxide = 531.46 K

Boiling point of phenol =455 K

Feed enters at =353 K



Utility required in reactor

Heat of reaction = -21735942.74 J/mol

Cp of water = -75.327 J/mol K

ΔH = n*Cp*dT

-21735942.74 = n*(-75.327)*(40)

n = 7213.861 moles

Weight of water required = 7213.861*18 = 129849.5126 g/hr

Process & mechanical design of neutralizing tank C6H5ONa + HCl C6H5Cl + NaCl EQUIPMENT - Cylindrical vessel with a drain valve and a turbine agitator

Total weight of the reacting mass = 48950.587 kg/h Average density at 60oC= 1.4006 kg/litre Total volume of reacting mass= Total weight of the reacting mixture

Average Density

Let residence time (T) = 2 hours

Vo= mav/dav =34.949 m3/hr

T = V / Vo

V = 69.898 m3

Where V = Volume of reactor

L/D = 1.25

D= 4.14 m, H=5.18 m

Mechanical design The maximum pressure in the tank is 14.70 psi (1

atm)

ts=[{(14.70 x 4.14)/(2 x 20000 x 0.8)}+0.005] x 2.54 =5.13 mm

Take shell thickness= 5mm

Design pressure is above 100 Kpa, ellipsoidal head would be an ideal choice

Major axis: Minor axis=2:1

Inside depth (B=A/2), A=D/2, B=D/4

B=1.035 cm

Volume V of the ellipsoidal head

V = (3.14 x 4.142 x 5.18/4) + (3.14 x 4.143/12) = 88.306 m3

Design of agitator

Let diameter of the agitator= 0.6 x D

Da= 0.6 x 4.14= 2.484 m

Speed of the agitator=75 rpm = 1.2 rps

Average Viscosity =3.6 cp

Average density of the slurry=1.1747 kg/lt

Calculating Reynolds Number Nre

Nre = 112584.00

Calculating Froude number Nfr

Nfr = n2 x Da/g = 1.252 x 2.484/9.81 = 0.3956

1 cm=1/30.8 ft

1 kg/lt= 62 lb/ft3

9.81 m/s2=32.17 ft/s2

Power=[(0.123)-0.0935 x 1.253 x (8.1710)5 x 1.4006 x 62]/32.17

Power=331.52 ft/lbf

Power=0.513 HP

Assuming frictional losses

Power= 1.5 HP

Torque in the shaft=2 x 550/2 π n

Torque in the shaft= 140 ft/lbf

Distillation Column: Process design Feed components:

Phenol – 271.0333 kmol/hrDiphenyl oxide – 15.13 kmol/hrTotal feed = 286.163 kmol/hr

Distillate components:

Phenol – 265.64 kmol/hrDiphenyl oxide – 2.683 kmol/hrTotal = 268.323 kmol/hr

Bottom components:

Phenol – 5.393 kmol/hrDiphenyl oxide – 12.447 kmol/hrTotal = 17.84 kmol/hr

Component Feed, xf Distillate, xd Bottom, xb

Phenol 0.947 0.99 0.302

Diphenyl oxide 0.053 0.009 0.698

Using Antoine Equation, we find vapor pressure:

P = A-(B/ (T+C)) T in Kelvin, Pressure in bar.

Component A B C

Phenol 4.24688 1509.677 -98.949

Diphenyl oxide 4.13678 1800.415 -95.324

Top temp at 123°C Bottom temp at 150°C

Phenol 0.14599 0.3948

Diphenyl oxide 0.01409 0.04487

Calculated Vapour Pressure in bar:

Heavy key component – Diphenyl oxideLight key component – Phenol

Relative volatility calculation :α top = 10.36α bottom = 8.798

α avg = (α top* α bottom)0.5

= (10.36*8.798)0.5

= 9.5475.

Fenske equation :

Nm = log [(XLK/ XHK)d*( XHK/ XLK)b] / log αavg

(XHK) d = 0.302

(XLK) d = 0.698

(XHK) b = 0.990

(XLK) b = 0.009

Nm = log [(0.698/ 0.302)*( 0.990/ 0.009)] / log 9.5475

= 2.52 Nm ≈ 3

Underwood’s Method:

∑ (αi*xif)/ (αi-θ) = 1-q

We get, θ =1.4960

∑ (αi*xif)/(αi-θ) = Rm+1

We get, Rm = 0.138

Gilliland’s correlation:

On solving, we get –N = 8.983 ≈ 9Assuming, tray efficiency = 0.5Therefore, actual no. of trays = 9/0.5 = 18

Tower diameter required at top:

Distillation: Vacuum distillation at 100mm Hg

Molar flow rate of vapor & liquid at top in enriching section –

L = R*D = 0.2074*268.323 = 55.650 kmol/hr

V = (R+1)*D = 1.2074*268.323 = 323.973 kmol/hr

L/V = 55.650/323.973 = 0.172

Mavg = ∑xiMi = 94.7621

Assuming tray spacing = 0.6 m

From Figure 11.27 of R.K Sinnott (Coulson & Richardsons) page-568

K1 = 0.08

uf = flooding velocity

uf = 3.6 m/sec

Now, Actual velocity = 0.85 * uf = 2.88 m/sec

Volumetric flow rate of vapor at the top, Qv = (V+Mavg) / ρv = 2.168 m3/s

An = net area required at the top

An = Qv/V = 0.753 m2

Down comer area, Ad = 0.12*Ac (Reference: Sinnott)

An = Ac – Ad

On solving, Ac = 0.856 m2

Di = 0.98 m

Tower diameter required at bottom:Similar procedure was followed to calculate

bottom diameter:Bottom diameter comes out to be 1.02 m

Height of Distillation Column:Height of column, Hc = no. of trays * tray

spacingActual no. of plates = 18Tray spacing = 0.6 mSo, height of column = 18 * 0.6 =10.8 m

Mechanical designCalculation of Thickness:Allowable design stress, f= 1.18MN/m2 Welding joint efficiency factor, J=0.85

t = [PDi/ (2*f*J-P)]

= [(13.33*103*1.02)/ (2*1.18x106*0.85)-(13.33*103)]

= 0.06 m

Economic Analysis

Costing of distillation columnThickness of shell = 0.06 m

Weight of vessel = = x 10.8 x 0.06 x 7850 = 16292.003 Kg Where r is the internal radius of distillation

column, L is the length of column, t is the thickness of shell.Density of carbon steel = 7850 Kg/m3

Weight of head, skirt = 10% weight of vessel.Total weight of column without tray = 1.1 x 16292.003 = 17921.23 Kg



Therefore, Purchased cost of distillation column without trays (1990) = $ 105

Purchased cost of tray = $ 300 per/trayTotal cost of trays = 18 x $ 300 = $ 5400Quantity factor of each tray = 1.10Total cost of the trays = 1.10 x $ 5400 = $ 5940Total cost of distillation column in 1990 = $ (105 + 5940) =$

105940Cost index in 1990 (C1) = 356 (Chemical engineering plant

cost index refer Klaus D.Timmerhaus Page 163)

Cost index in 1990 (C2) = 588.6 P1 = Purchased cost of distillation column in 1990P2 = Purchased cost of distillation column in 2014 Present Cost = Original Cost P2 =105940 = $ 17518 = Rs 17518 59 = Rs 10,334,327


Costing of various equipments1. Purchased cost of distillation column = Rs

10,334,3272. Purchased cost of heat exchanger = Rs.

82837363. Purchased cost of Neutralizer = Rs 3217984. Purchased cost of reactor = Rs 29966100


S No Raw material Cost ($/kg)

Total Consumption

in Kg/Year

Rs/Year

01 Chlorobenzene 0.5 241619202 7127766459

02 Sodium Hydroxide 0.3 85856815.44 1519665633

03 Diphenyl ether 3 20396193.84 3610126310

Total 12257558402

Cost of raw materials

Component $/kg Total production (kg/yr) Rs/year

Phenol 1.6 202019923 19120683690

Diphenyl ether 3 20396193.84 3610126310

Total product cost 22730810000

Total product cost


Cash flow statement

Cash flow statement Non-discounted parameters(a) CCP (cumulative cash position) = Rs 38.02 crores

(b) Payback PeriodLand + working capital = 1.12 + 3.67 = 4.8Payback period is the point at which cumulative cash flow is

equal to the sum of land and working capital.Payback period (PBP) = =3+((4.939-4.8)/(4.939-0.77))Therefore, PBP = 3.033 years

(c) Rate of return on investment (ROROI)

Where n is the project life. ROROI = 20.68 %



Plant layout

First Aid safety measuresEyes: Get medical aid immediately. Do NOT allow victim to rub or keep eyes closed. Extensive irrigation with water is required (at least 30 minutes).

Skin: Get medical aid immediately. Immediately flush skin with plenty of soap and water for at least 15 minutes while removing contaminated clothing and shoes. Wash clothing before reuse. Discard contaminated clothing in a manner which limits further exposure. SPEEDY ACTION IS CRITICAL! Destroy contaminated shoes.

Ingestion: Do NOT induce vomiting. If victim is conscious and alert, give 2-4 cupfuls of milk or water. Never give anything by mouth to an unconscious person. Get medical aid immediately.

Inhalation: Get medical aid immediately. Remove from exposure to fresh air immediately. If breathing is difficult, give oxygen. Do NOT use mouth-to-mouth resuscitation. If breathing has ceased apply artificial respiration using oxygen and a suitable mechanical device such as a bag and a mask.

Notes to Physician: Persons with liver or kidney disease should not be exposed to phenol for any length of time.

Antidote: Activated charcoal, followed by cathartic, may be preferred to ipecac induced emesis or lavage in decontamination of the GI tract and preventing systemic absorption of phenol.


Conclusions

After studying the entire process of manufacture of phenol, there are a few recommendations: ◦ Phenol production has a very good scope

in India.◦ For most of the equipments, the fixed

capital investment is higher but the payback period is small.


Thank You..

References [1] http://www.helium.com/items/2025739-medical-uses-for-phenol

[2] http://www.greener-industry.org.uk/pages/phenol/1PhenolAnnualProd.htm

[3] Kirk R. E., Othmer D. F., Encyclopedia of Chemical Technology, John Wiley and Sons, 1994, Volume 18 Page no: 291

[4]http://www.honeywell-pmt.com/sm/chemicalintermediates/phenol-n3/phenol-properties- spec.html?c=21

[5] http://www.icispricing.com/il_shared/Samples/SubPage186.asp

[6] http://en.wikipedia.org/wiki/Phenol#Properties

[7] http://en.wikipedia.org/wiki/Phenol

[8] http://hazard.com/msds/mf/baker/baker/files/p1949.htm

[9] Dryden C. E., Outlines of Chemical Technology, East-West Press, 2008

[10] Kirk R. E., Othmer D. F., Encyclopedia of Chemical Technology, John Wiley and Sons, 1999-2012

[11] Carl L.Yaws, Chemical Properties Handbook, Tata McGraw Hill publications, Third Edition.

[12] Max S.Peters, Klaus D.Timmerhaus, Plant design and economics fro chemical engineers, McGraw-Hill International editions, chemical and petroleum engineering series, fourth edition.

[13] http://www.matche.com/equipcost/

[14] http://avogadro.chem.iastate.edu/MSDS/phenol.htm


ppt final.pptx

Documents

jmol production of phenol

production import of

india production of

phenol demand forecast

molecular weight of

hour production

jmol h naoh

jmol h nacl