A Project Report onDesign A Plant To Manufacture

50000 TPA OfStyrene Oxide

ByUJJWAL BAJPAI (1071110009)

ARITRA MUKHERJEE (1071110029)Under the guidance of

Dr. K. TamilarasanProfessor, Chemical Engineering

Production of Styrene Oxide

Stream in Kmoles/hr Temperature(K) ⌠CpdT Unit Enthalpy in

(KJ/hr)

Styrene 66.172 298.15 0.00E+00 KJ/Kmole 0

Ethyl

Benzene

0.326 298.15 0.00E+00 KJ/Kmole 0

66.498

Energy Balance for Pre-heater

Component A B D

Styrene 18.9578 -2.2959 1.01872

Ethyl Benzene 4.19845 -0.3127 0.17860

Cp/R=A+B(T/100)+C(T/100)2

Stream out Kmoles/hr Temperature(K) ⌠Cpdt Unit Enthalpy

out (KJ/hr)

Styrene 66.172 343.15 5102.19 KJ/Kmole 3.38E+05

Ethyl Benzene 0.326 343.15 8717.91 KJ/Kmole 2.84E+03

66.498 3.40E+05

Q(heat transfer rate)= Enthalpy out- Enthalpy in= 3.40E+05 KJ/hr

Mass of Steam=Q/ λs at 1 atm= 127.23 Kg/hr

Energy Balance for Reactor

Component A B D

Styrene 18.9578 -2.2959 1.01872

Ethyl Benzene 4.19845 -0.3127 0.17860

Reactants Kmoles/hr Mass Flow rate

(Kg/hr)

Temperature

(in Kelvin)

⌠CpdT unit Enthalpy in

KJ/hr

Styrene

9.93 1032.28 343.15 5102.19

KJ/Kmole

5.06E+04

Ethyl Benzene

0.33 34.58 343.15 8717.92

KJ/Kmole

2.84E+03

Styrene

dichloride 26.47 4632.04 343.15 10785.30

KJ/Kmole

2.85E+05

Styrene

Chlorohydrin 29.78 4660.16 343.15 12907.10

KJ/Kmole

3.84E+05

NaOH solut 248.72 6198.71 298.15 142.38 KJ/Kg 8.83E+05

Enthalpy

of

reactants 1.606E+06

Products Kmoles/hr Mass Flow

rate

(Kg/hr)

Temperature (in

Kelvin)

⌠Cpdt unit Enthalpy

out KJ/hr

Styrene 9.926 1032.283 343.150 5102.19 KJ/Kmole 5.06E+04

Ethyl Benzene 0.326 34.582 343.150 8717.92 KJ/Kmole 2.84E+03

Styrene

dichloride 1.429 250.147 343.150 10785.3 KJ/Kmole 1.54E+04

Styrene

Chlorohydrin 1.597 250.000 343.150 12907.1 KJ/Kmole 2.06E+04

Styrene Oxide 53.219 6386.316 343.150 8687.23 KJ/Kmole 4.62E+05

NaCl aq 245.740 5316.593 343.150 147.78 KJ/Kg 7.8E+05

NaCl solid 56.203 3287.846 343.150 2250.00 KJ/Kmole 1.26E+05

Enthalpy

of

Products

1.46E+06

Calculating Heat of FormationStyrene

Oxide

Contribution Heat of

formation

(KJ/mole)

Heat of

formation

contributed

(KJ/mole)

Hvap

(KJ/mole)

Hvap

contributed(KJ/mole)

-O-

(ring) 1 -138.16 -138.16 4.682 4.682

-CH2- 1 -26.8 -26.8 2.398 2.398

>CH- 1 8.67 8.67 1.942 1.942

=C< 1 46.43 46.43 3.06 3.059

=CH- 5 2.09 10.45 2.54 12.72

-99.41 24.801

=-31.12 KJ/mole

Heat of Formation (Ideal Gas, 298 K)

Heat of Vaporization at Normal Boiling Point=40.101 KJ/mole

Heat of Formation at liquid state = Heat of Formation- Heat of Vaporization at Normal Boiling Point=-71.221 KJ/mole

Reactants Heat of

Formation

(KJ/mole)

Products Heat of Formation

(KJ/mole)

Styrene Chlorohydrin -201.504 Styrene Oxide 71.221

NaOH -469.15 Water -285

NaCl -407

-670.654 -620.779

ΔH°rxn1=Heat of formation of Products - Heat of formation of reactants = 49.875 KJ/mole

Calculating Standard Heat of Reaction

Reactant

Joule/mole

Product

Joule/mole

Styrene

Chlorohydrin

12907.1 Styrene Oxide

8687.23

NaOH 2684.7 NaCl 3394.71

Water2250

-15591.8 14431.94

Considering stotiomety

=48.615 KJ/mole

Similarly ΔHrxn2=-35.187 KJ/mole

ΔHrxn1 48.62 KJ/mole

Heat released by reaction 1 1.37E+06 KJ/hr

ΔHrxn2 -35.19 KJ/mole

Heat released by reaction 2 -8.81E+05 KJ/hr

Heat released by reaction = Moles Of Products formed*ΔHrxn

Q(heat transfer rate)=Enthalpy of outlet Stream - Enthalpy of inlet Stream

+ Heat released by reaction 1 + Heat released by reaction 2

= 1.606E+06 -1.46E+06 +1.37E+06 -8.81E+05

= 3.47E+5 KJ/hr (heat lost)

Thus We need to use a steam Jacket

Steam at 1 atm enters the jacket and leaves out of the jacket as water(liquid) at 100°C

Latent Heat of Vaporization of steam(λs)= 2676 KJ/Kg

Mass flowrate of Steam =(Q/ λs)= 96.21 Kg/hr

Energy Balance for Distillation Column

Mole fraction of low boiler in feed =0.1542 Mole fraction of low boiler in Top Product =0.97Mole fraction of low boiler in Bottom Product =0.01

Pressure 60 mm of HgTemperature 90.5°C

60.8°C Styrene

101.8°C Styrene Oxide

Using Clayeperon equation find x,y

Temperature °C P1sat mm of Hg P2sat mm of Hg x1 y160.80 60.00 12.37 1.00 1.0063.73 66.95 14.02 0.87 0.9766.66 74.56 15.86 0.75 0.9369.58 82.89 17.91 0.65 0.8972.51 91.99 20.18 0.55 0.8575.44 101.90 22.69 0.47 0.8078.37 112.69 25.46 0.40 0.7481.30 124.41 28.51 0.33 0.6884.23 137.13 31.88 0.27 0.6187.16 150.91 35.57 0.21 0.5390.08 165.82 39.63 0.16 0.4593.01 181.93 44.07 0.12 0.3595.94 199.31 48.92 0.07 0.2498.87 218.03 54.22 0.04 0.13

101.80 238.19 60.00 0.00 0.00

Condenser DutyVapor Formed=(R+1)*D=34.40 Kmoles/hr

Since q=1 feed is saturated

λ Styrene 35008.00 KJ/Kmole

λ Styrene Oxide 40101.00 KJ/Kmole

mole fraction of top product 0.97

λ Effective 35160.79 KJ/Kmole

Where λ is latent heat of vaporization

Energy Required= V * λ Effective= 1.210E+06 KJ/hr

Specific heat capacity of water=4.18KJ/KgKInlet and outlet temperature of condenser cooling water=25°C and 50°CQ=mCpΔTMass flowrate of water= 5252.63 Kg/hr

Reboiler Duty

Vapor Formed in reboiler(V’)=Vapor from the top of the column= D=34.40 Kmoles/hr

Since q=1 feed is saturated

λ Styrene 35008.00 KJ/Kmole

λ Styrene Oxide 40101.00 KJ/Kmole

mole fraction of top product 0.01

λ Effective 40050.07 KJ/Kmole

Where λ is latent heat of vaporization

Energy Required= V’ * λ Effective= 1.378E+06 KJ/hr

Latent heat of Steam= λs=2676 KJ/hr

Mass flowrate of steam required = (V’ * λ Effective)/ λs=514.874 Kg/hr

Reactor Design

Let 1-Styrene Chlorohydrin2-Styrene Dichloride

Order 2 Order 2

Since pH has to be maintained 12 thus OH- will be in excess and both of the reactions will behave as pseudo order 1 reaction

Where F1°= moles of feed Kmole/hr

V=Volume of reactor m3

X1

0.000 0.087

0.050 0.091

0.100 0.097

0.150 0.102

0.200 0.109

0.250 0.116

0.300 0.124

0.350 0.134

0.400 0.145

0.450 0.158

0.500 0.174

0.550 0.193

0.600 0.217

0.650 0.248

0.700 0.290

0.750 0.348

0.800 0.435

0.850 0.579

0.900 0.869

0.947 1.624

0.000

0.200

0.400

0.600

0.800

1.000

1.200

1.400

1.600

1.800

0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000

`

X 1 (Conversion )

Reactor Design

Area Under the curve = 0.264 m3/kmol

Residence time: 2.5 hours (From Literature)

Area Under the curve =

k (Rate Constant) = (Area Under the curve * C1°)/ Residence time

= 0.638 hr-1

Volume of liquid system = (Residence time * F1°)/C1°= 12.328 m3

Actual Volume of liquid system will be calculated by giving 10% allowance=1.3893 m3

Weight of Catalyst =

=1929.405 kg /h

Actual Volume of Catalyst=1.1(Mass of Catalyst/ Density Of Catalyst)=1.3893 m3

Total Volume of reactor = Volume of Catalyst+Volume of liquid system= 14.950 m3

HEIGHT AND DIAMETER OF THE REACTOR:Let us assume an ellipsoidal head at the top and at the bottom of the vessel

Taking H/D = 2,

Diameter of the reactor D: 2.013 mHeight of the reactor H =2* D

=4.026 m

Reactor Design

Distillation Column Design

Mole fraction of low boiler in feed =0.1542 Mole fraction of low boiler in Top Product =0.97Mole fraction of low boiler in Bottom Product =0.01

Pressure 60 mm of HgTemperature 90.5°C

60.8°C Styrene

101.8°C Styrene Oxide

Using Clayeperon equation find x,y

Temperature °C P1sat mm of Hg P2sat mm of Hg x1 y160.80 60.00 12.37 1.00 1.0063.73 66.95 14.02 0.87 0.9766.66 74.56 15.86 0.75 0.9369.58 82.89 17.91 0.65 0.8972.51 91.99 20.18 0.55 0.8575.44 101.90 22.69 0.47 0.8078.37 112.69 25.46 0.40 0.7481.30 124.41 28.51 0.33 0.6884.23 137.13 31.88 0.27 0.6187.16 150.91 35.57 0.21 0.5390.08 165.82 39.63 0.16 0.4593.01 181.93 44.07 0.12 0.3595.94 199.31 48.92 0.07 0.2498.87 218.03 54.22 0.04 0.13

101.80 238.19 60.00 0.00 0.00

x-y diagram

Rmin was calculated from this graph R=1.2*R

Distillation Column Design Plate Spacing (lt): 0.6 m (assumed)

ρl = Density of styrene in liquid phase = 859.41 kg/ m3 at 70oC and 60mm of Hgρv = Density of styrene in gaseous phase =0.2916 Kg/ m3 70oC and 60mm of Hg

Vapor Velocity

Molar flowrate of Vapors Formed=34.402 kmole/hrMass of Vapors formed Vm=

kmole/hr vapor formed*Molecular wt of styrene * /3600

=0.9938 Kg/s

Column Diameter=

Diameter of column=1.2231 m

From Graph,Nm+ 1 = 12

Where Nm = Number of theoretical plates Nm = 11Tray Efficiency = 0.6Actual Number of trays = 11 / 0.6 =19

COLUMN HEIGHT = (Actual number of trays – 1 )*Tray Spacing + Top and bottom allowance = (19 – 1)* 0.6 + 2 * 0.6

= 12 m

COLUMN HEIGHT: 12 mCOLUMN DIAMETER: 1.223 m

Reference

I.L. Finar Organic Chemistry Volume 1 Sixth Edition Pearson page

380,616,89,569

David M.Himmelblau, Basic principles and calculations in chemical

engineering, edition 6, Prentice Hall of India, New Delhi, 1998.

George T. Austin Shreve’s Chemical Process Industries Fifth Edition page

661-687.

Citation No:US2582114 ,Filing Date: JULY 07,1949,Publishing,Date: JAN

08,1952,Title: U.S. Rubber Co.,(Styrene Oxide Synthesis).

M. Gopala Rao Marshall Sittig Dryden’s Outline of Chemical Technology

Third Edition East West Press page 518-525