Phthalic anhydride

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

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[...] Overall Material Balance: F = D + W , where 9039.7 kg/hr (wt basis) PAN balance (wt basis) : xF*F = xD*D + x W Where xF*F= 8271.44 kg xD= 0.98 xW= 0.02 Solving above two equations, we get 611.98 kg/hr; 8427.71 kg/hr Rounding Off: Phthalic anhydride Product = 8400 kg/hr Distillate product = 640 kg/hr ENERGY BALANCE Since the gases are at high temperature and low pressure, the gases can be assumed to be ideal . Compound mole fraction o-xylene 0.0265 air 0.9735 Specific heat of reacting mixture (air and o-xylene) = 6.386 + 4.885 *10-3 - 2.179 *10-6 *T2 + 4.765 *10-9 *T3 - 3093.39 T net = 3020.4 *1000 6.386 + 4.885 *10-3*T 2.179 *10-6*T2 + 4.765 9*T3 3093.39 dT where T is from 298 K to 593 K = 3020.4 *1000* 2526.38 = 0.763 *1010 Cal/hr = 31940.46 MJ/hr o-xylene stream; air stream; net 2.508 *10-3 *T2 + 6.56 - 4945.12 + 3177.6 * T = 0 Solving T = 610.5 K = 337.5 C Now, 3.786 + 0.1424 8.224 *10-5*T2 + 1.798 *10-7*T3)dT where T is from 423 K to 610.5 K = 27555.47 MJ/hr o-xylene stream: Assuming 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 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 C to C = 3020.4 *1000* 386.08 = 0.1166 *1010 cal/hr = 4881.1 MJ/hr PAN reaction: C8H10 + 3O2 C8H4O3 + 3 H2O 298 K = - 371.79 + 3(-242) - 19.01 = - 1116.8 KJ/mol 633K = - 1116.8 + 11.814 = - 1104.98 kJ/mol net(1)= 633K ) = 56 1104.98 61878.88 MJ/hr MAN reaction: C8H10 + 7.5 O2 C4H2O3 + 4 H2O + 4CO2 298 K = - 469.65 + 4(-242) + 393.77 ) - 19.01 = - 3031.74 kJ/mol 633K= - 3031.74 - 7.69 3039.43 kJ/mol net(2)= 633K ) = 8*1000*- 3039.43 *1000 = - 24315.44 MJ/hr Complete Combustion: C8H10 + 10.5 O H2O + 8CO2 298 K = 393.77 19.01 = - 4379.17 kJ/mol 633K = - 4379.17 + 174.785 = - 4204.38 kJ/mol net(3) = 633K ) =16*1000*- 4204.38 *1000 = - MJ/hr Overall Heat transfer rate liberated = - 61878.88 - 24315.44 - 67270.16 = - 153464.48 MJ/hr Let us allow the Products to leave the reactor at C. [...]