| Case StudyProduction Of Phthalic Anhydride | Semester Project | | | 1/6/2012 |

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Table of Contents 1. INTRODUCTION 2 1.1Physical Properties: 2 1.2 Applications: 2 1.3 PAN Producers in Pakistan: 3 1.4 End Users of PAN: 3 2. PROCESS OVERVIEW 4 3. REACTOR CALCULATIONS 5 4. SALT CIRCULATION RATE 6 5. SWITCH CONDENSER CALCULATIONS 7 6. AFTER COOLER CALCULATIONS 8 7. VAPORIZER CALCULATIONS 9 8. PRE HEATER CALCULATIONS 10 9. GAS COOLER CALCULATIONS 11 10. STRIPPING COLUMN CALCULATIONS 12 11. RECTIFICATION COLUMN CALCULATIONS 13 Bibliography 14

Table 1-Physical properties of Phthalic Anhydride 2 Table 2-Heats of Formation (1) 4

Figure 1-Detailed Process Flow Diagram (PFD) of the Phthalic Anhydride production process 3 Figure 2-LMTD across the switch condenser unit 6

1. INTRODUCTION
Phthalic Anhydride (PAN) is one of the first cyclic anhydrides and is a white crystalline solid. It is an organic compound with the general formula C6H4(CO)2O and is an anhydride of Phthalic Acid.
1.1Physical Properties:
These are summarized in the table below: Molecular Formula | C8H4O3 | Molar Mass | 148.1 g/mol. | Appearance | White Flakes | Density | 1.53 g/cm3 | Melting Point | 131oC | Boiling Point | 280OC | Flash Point | 152oC |
Table [ 1 ]-Physical properties of Phthalic Anhydride 1.2 Applications:
One of the main uses of PAN is in the manufacture of plasticizers for PVC processing. It is also used in the synthesis of: * Phthalate Esters * Unsaturated Polyester Resins * Alkyl Resins * Dyes * Anthraquinone (Used for H2O2 production)
1.3 PAN Producers in Pakistan:
Phthalic Anhydride production rates have been on a rise across countries like Australia, Indonesia, Taiwan, Japan and Korea. Recently, it has found its branches in Pakistan too. Companies producing PAN in Pakistan are: * Nimir Chemicals Pakistan Limited * Resources Group Pakistan
1.4 End Users of PAN:
End users of PAN can be divided into three broad categories. Companies that manufacture phthalate plasticizers for flexible PVC and other resins, those that manufacture unsaturated polyesters for fiber-reinforced plastics and those that are involved in the manufacture of alkyl resins for oil based coatings.

2. PROCESS OVERVIEW
Phthalic Anhydride is produced by the partial oxidation of o-xylene. The chemical reaction governing the whole process is as follows:
C6H4(CH3)2+ 3O2→ C6H4(CO)2O+3H2O

O-xylene is the preferred raw material and the reactants are introduced in a fixed bed, catalytic, tubular reactor. The LAR (Low Air Ratio) process is the most favored process as it uses a catalyst which allows a low air to o-xylene ratio.

Naphthalene can also be used as a raw material but ortho-xylene is more desirable due to the higher production rates, higher efficiency and cost effectiveness.

There are three principal steps in PAN production: * Reaction * Condensation * Purification
The reaction is highly exothermic and requires heat removal through molten salt circulation.

The following figure gives a detailed Process Flow Diagram (PFD) of the whole production process:

Figure [ 1 ]-Detailed Process Flow Diagram (PFD) of the Phthalic Anhydride production process

3. REACTOR CALCULATIONS
PAN Yield = 1.10 kg PAN/kg o-xylene
PAN Rate= 5000 kg/hr.
Therefore: o-xylene feed rate = 5000/1.10 = 4545.45 kg/hr.
Hydrocarbon Loading (o-xylene in each tube) = 0.34 kg/hr.

So,
Number of Tubes = 4545.45/0.34 =13368.97 = 13400

Now; for heat generated, we go about the following procedure;
Given the reaction;
C6H4(CH3)2+ 3O2→ C6H4(CO)2O+3H2O

Moles of o-xylene = 4545.45/106 = 42.88 kmol/hr.
Moles of PAN = 5000/148 = 33.78 kmol/hr.
Moles of water = (42.88 X 3) X 33.78/42.88 = 101.34 kmol/hr.

Component | Molar Mass | Heat of Formation (J/mol.) | O-xylene | 106 | 24400 | Water | 18 | 285800 | PAN | 148 | 1361000 |
Table [ 2 ]-Heats of Formation (1)
Heat of reaction liberated is assumed to be heat generated.
Heat of Reaction = ∆Hfoproducts- ∆Hfo(reactants) = {(33.78 X 103 X 1361) + (101.34 X103 X 285.8) } – {42.88 X 103 X 24.4} = 7.38 X 107 KJ/hr = 20525.4 KJ/s. = KW * Heat Generated = 20.53 MW = 21 MW
Finally, for Heat Transfer Area;
A = Number of Tubes X Surface Area of each Tube
A = 13400 X 2π X (25 X 10-3 /2) X 3 * Heat Transfer Area = 3157.3 m2

4. SALT CIRCULATION RATE

Having found the heat generated in the reactor, we then assume that there is no heat loss as the salt circulates through the reactor.
Assuming the specific heat, CP, of molten salt = 2.5 KJ/kg/oC
Assuming the temperature difference (T – TO) = 100 OC

For this purpose;

Q = mCP∆T
20525.4 KJ/s. = m X 2.5 X 100 m = 82.1 kg/s. m= 82.1 X 3600/1000

Mass Flow Rate of Salt = 295.6 Tonnes/hr.

5. SWITCH CONDENSER CALCULATIONS

Now if we move forward with the objective of subliming 50 % PAN from vapor to solid;

Air Inlet Flow rate (taken from the PFD) = 45000 kg/hr.
PAN inlet flow rate = 0.5 X 5000 = 2500 kg/hr.
Total Gas Inlet Flow Rate = 45000 + 2500 = 47500 kg/hr.

After the sublimation process, the only gas leaving the condenser will be the air.
Total Gas Outlet Flow Rate = 45000 kg/hr.

Outlet liquid will consist solely of PAN.
Total Liquid Outlet Flow Rate = 2500 kg/hr.

To find the Heat Duty, given that it is a sublimation process, we need to have the Enthalpy of Sublimation for PAN.
Enthalpy of Sublimation of PAN = 597.297 KJ/kg (2)

Duty = Q =ml
Q = 2500 X 597.297
Q = 1493242.5 KJ/hr.
Duty = 414.8 KW

For Heat Transfer Area; we estimated the heat transfer coefficient.
Heat Transfer Coefficient for flowing liquids in Condensers = 600 W/m2/oC (3) For the Log Mean Temperature Difference (LMTD);

Figure [ 2 ]-LMTD across the switch condenser unit
LMTD = (T1 –T3) –(T2-T4)ln{T1-T3T2-T4}
LMTD = 35-20ln⁡{3520} = 26.8 0C
Therefore; A = Q/ (U X T)
Heat Transfer Area = 25.79 m2

6. AFTER COOLER CALCULATIONS
The After-Cooler serves the objective of condensing 50% of PAN vapors to liquid. As the after-cooler precedes the switch condenser, PAN will be at a higher temperature at the after-cooler inlet than it is at the switch condenser inlet. Hence, it is assumed that the LMTD for this section will be greater than the condensers due to high temperature gradient.

Assuming LMTD = 35OC
Total Gas Inlet Flow Rate = 45000 + 2500 = 47500 kg/hr.
Total Gas Outlet Flow Rate = 45000 kg/hr.
Total Liquid Outlet Flow Rate = 2500 kg/hr.
Heat Transfer Coefficient for flowing liquids in Coolers = 600 W/m2/oC (3)

Duty = Q =ml
Q = 2500 X 597.297
Q = 1493242.5 KJ/hr.
Duty = 414.8 KW

A = Q/ (U X T)
A= 414.8/(0.6 X 35)

Heat Transfer Area = 19.75 m2

7. VAPORIZER CALCULATIONS

The reason for installing the vaporizer is to vaporize the o-xylene feed for the reactor up to a temperature of 300-350oC.
For this section, we require an estimate of the latent heat of vaporization and the specific heat capacity of o-xylene.

Latent heat of vaporization of o-xylene = 347 kJ/kg (4).
Specific Heat Capacity of o-xylene = 0.7 kJ/kgoC (4).
Flow Rate of o-xylene = 4545.45 kg/hr.
Assuming o-xylene enters at room temperature i.e. 25oC
Assuming o-xylene is heated upto maximum temperature i.e. 350oC

As the vaporizer comes before the after-cooler, another higher estimate of the LMTD is taken.
Estimated LMTD = 45OC

Heat duty will therefore be,
Heat Duty = ml + mCP∆T
Heat Duty =4545.45 X {347 + 0.7 X (350 -25)} = 2611361.025 kJ/hr. Duty= 725.378kW

As for the Heat Transfer Area;
A = Q/ (U X T)
A = 725.378/ (0.6 X 45)

Heat Transfer Area = 26.9 m2

8. PRE HEATER CALCULATIONS The purpose is to heat the reactor inlet air up to a temperature of 350oC.

Given air flow-rate = 45,000 kg/hr.
Specific Heat capacity of air = 1.0035 kJ/kgoC (5).
Assuming air enters at room temperature i.e. 25oC

Heat Transfer Coefficient for an Air-Preheater = 80-100 W/m2/oC (6).

Therefore,
Heat Duty = mCP∆T = 45000 X 1.0035{350-25} = 14676187.5 KJ/hr. = 4076.71 KW

Duty = 4.076 MW Assuming U = 100 W/m2.oC and LMTD = 50oC,
A = Q/ (U X T)
A= 4076.71 / (0.1 X 50)

Heat Transfer Area = 815.342 m2

9. GAS COOLER CALCULATIONS The gas cooler is used to cool the reactor product gases consisting of PAN vapor and air to the after cooler inlet temperature of 205oC. As stated in the literature, air and o-xylene can reach upto a maximum temperature of 500oC. Therefore ∆T for both PAN and air = 295oC m for PAN = 5000 kg/hr. m for Air = 45000 kg/hr. Cp PAN = 1.08 KJ/kg/oC Cp Air = 1.0035 KJ/kg/oC Finally; Heat Duty = mCP∆T + mCP∆T = {5000 X 295 X 1.08} + {45000 X 295 X 1.0035} Duty = 4142.9 KJ/s. Duty = 4.142 MW Now, estimating U = 100 W/m2.oC and LMTD = 50 oC,
A = Q/ {U X T}
Heat Transfer Area = 828.58 m2

10. STRIPPING COLUMN CALCULATIONS
We use Fenske equation to calculate the minimum number of theoretical plates.
Reference to the PFD, we have:
Xd = 0.99
Xb = 0.01
Fenske Equation:

Vapor Pressure of PAN = 1.333 Pa
Vapor Pressure of Maleic Anhydride = 21.33 Pa

α= 1.00 for PAN α = 21.33/1.333 = 16 for Maleic Anhydride αavg = (16 X 1 )0.5 = 4
Finally;
N = log {(0.99/1-0.99)(1-0.01/0.01)} / log4
N = 6.63
Therefore 5-7 minimum number of stages.

11. RECTIFICATION COLUMN CALCULATIONS
Unlike the stripping column, here PAN has to be separated from the waste residue. All other assumptions are taken to be the same.
Xd = 0.998
Xb = 0.02
Assuming that the vapor pressure of the residue is very low and has a value = 0.005 Pa α= 1.00 for Residue α = 1.33/0.005 = 26600 for PAN αavg = (26600 X 1 )0.5 =163
Finally;
N = log {(0.998/1-0.998)(1-0.02/0.02)} / log163
N= 2.63
Therefore minimum number of stages = 2-4

Bibliography
1. MDPI. 2007, International Journal of Molecular Sciences.
2. National Institute of Standard and Technology. s.l. : Material Measurement Laboratory.
3. Perry, R.H and D. Green. Perry's Chemical Engineering Handbook-6th edition. New York : Mc Graw Hill.
4. Cameo Chemicals. Datasheet.
5. Engineering Tool Box. [Online] http//engineeringtoolbox.com.
6. Power Plant Engineering. s.l. : McGraw-Hill.