INTEGRATED PROJECTPRODUCTION OF BENZENEMATERIAL AND ENERGY BALANCE

Material and Energy BalanceIn our production of benzene project, we have conducted the material and energy balance of the process in the reactor and cooler.In the cooler, the process that takes place is a non-reactive process.

Non-reactive process ( Cooler )0.041 C7H8 (v)0.495 H2 (v)0.337 CH4 (v)0.127 C6H6 (v)0.041 C7H8 (l)0.495 H2 (v)0.337 CH4 (v)0.127 C6H6 (l)H2O ( l,1250C) H2O (l, 25oC )

Cooler2 671.015OC, 24,811 atm3 38OC, 24,466 atm

Based on the PFD diagram of the cooler, there are 4 unknowns which we need to find which are 2, 3, 4 and 5.The value of 2 can be determined by the mass balance at the reactor as 2 is at the outlet of the reactor. When 2 is known, 3 can be determine by mass balance of hydrogen.This is a cooler, so there are also inlet and outlet for water .

At the inlet of the cooler, all of the components are in vapour state as they are the products from the reactor.At the outlet, some of the components remain as vapour and some as liquid because they are being cooled at the cooler. Toluene and benzene are in liquid state while methane and hydrogen are in vapour.

Mass Balance of CoolerMass Balance :Since 2 = 9.11 x 10^6 kg/hr2 = 3Balance for hydrogen , H2 , 0.4952 = 0.4953 0.495 ( 9.11 x 10^6 ) = 0.4953 3 = 9.11 x 10^6 kg/hr

Energy Balance of CoolerEnergy Balance : Reference state : H2 ( g, 250C, 1 atm ) ; C7H8, CH4, C6H6 ( g, 671.0150C, 24.811 atm )

Componentmin (kg/hr) Hin (kJ/kg) mout ( kg/hr) Hout ( kJ/kg)C7H83.74x105 03.74x105 H2H24.51x106 H14.51x106 H3CH43.07x106 03.07x106 H4C6H61.16x106 01.16x106 H5

In the energy balance, there are 5 unknowns which are H1, H2, H3, H4 and H5. All of these are the enthaply of components in the process.All of the enthalpies are determined based on the reference state of the components.Because the components are in different states at the outlet, calculation of enthalpy is different between them.

Find H1 : assuming that hydrogen, H2, is an ideal gas

From table B.8 Using interpolation : 671.015 - 600 = H1 - 16.81 700 - 600 19.81 - 16.81 71.015 = H1 - 16.81 100 3 2.1305 = H1 - 16.81 H1 = 18.941 kJ 1 mol 1000g mol 2g 1 kg = 9470.5 kJ/kg

T ( 0C )H ( kJ/mol )60016.81671.015H170019.81

Find H2 : from table B.2

State 1 : C7H8( v, 672.0150C, 24.811 atm )

State 2 : C7H8( v, 110.620C, 1 atm )

State 3 : C7H8( l, 110.620C, 1 atm )

State 4 : C7H8( l, 380C, 24.466 atm )

Cp : C7H8 = 0.09418 + 3.8 x 10^-4 T - 2.786 x 10^-7 T2 + 8.033 x 10^-11 T3 HAHBHC

HA = Cp ( C7H8 , v ) dT = 0.09418 + 3.8 x 10^-4 T - 2.786 x 10^-7 T2 + 8.033 x 10^-11 T3 dT = 0.09418 T + 3.8 x 10^-4 T2/2 - 2.786 x 10^-7 T3/3 + 8.033 x 10^-11 T4/4 = ( 10.4182 + 2.32499 - 0.12771 + 3.00713 x 10^-3) - ( 63.1962 + 85.5496 - 28.05799 + 4.07143 ) = 12.6205 - 124.75924 = -112.14 kJ 1 mol 1000g mol 92.13g 1 kg = -1217.193 kJ/kg

HB = Hv C7H8(110.620C, 1 atm) = 33.47 kJ 1 mol 1000g mol 92.13g 1 kg = 363.291 kJ/kgHC = P + Cp( C7H8( l) ) dTCp : C7H8(l) = 0.1488 + 3.24 x 10^-4 TCp( C7H8( l) ) dT =

= ( 5.6544 + 0.23393 ) - ( 16.4603 + 1.9824 ) 0.1488 + 3.24 x 10^-4 T dT = 0.1488 T + 3.24 x 10^-4 T^2/2

= 5.888 - 18.443 = -12.555 kJ 1 mol 1000g mol 92.13g 1 kgCp( C7H8( l) ) dT = - 136.275 kJ/kgSG for C6H6 = 0.866 Density = 0.866 kg/L P = 1 L ( 1-24.446) atm 92.13g 1kg 0.008314 kJ/mol.K 0.866 kg 1 mol 1000g 0.08314 L.atm/mol.K= -0.2494 kJ 1 mol 1000g mol 92.13g 1 kg= -2.70704 kJ/kg

HC = -2.70704 - 136.275 = - 138.98 kJ/kgH2 = HA + HB + HC= -1217.193 + 363.291 - 138.98= - 99.88 kJ/kgFind H3 : assuming that hydrogen, H2, is an ideal gasFrom table B.8

T ( 0C )H ( kJ/mol )25038H31002.16

Using interpolation : 38 - 25 = H3 - 0 100 - 25 2.16 - 0 13 = H3 75 2.16 H3 = 0.3744 kJ 1 mol 1000g mol 2g 1 kg = 187.2 kJ/kg Find H4 : From table B.2

Cp : CH4 = 0.03431 + 5.469 x 10^-5 T + 3.661 x 10^-9 T2 - 1.1 x 10^-11 T3

H4 = Cp dT

= 0.03431 + 5.469 x 10^-5 T + 3.661 x 10^-9 T2 - 1.1 x 10^-11 T3 dT = 0.03431 T + 5.469 x 10^-5 T2/2 + 3.661 x 10^-9 T3/3 - 1.1 x 10^-11 T4/4 = ( 1.30378 + 0.039486 + 6.6962 x 10^-5 - 5.7341 x 10^-6 ) - ( 23.0225 + 12.3124 + 0.3687 - 0.55752 ) = 1.34333 - 35.1461 = - 33.803 kJ 1 mol 1000g mol 16.04g 1 kgH4 = -2107.42 kJ/kg

Find H5 : From table B.2

State 1 : C6H6( v, 672.0150C, 24.811 atm ) HA

State 2 : C6H6( v, 80.100C, 1 atm ) HB

State 3 : C6H6( l, 80.100C, 1 atm ) HC

State 4 : C6H6( l, 380C, 24.466 atm )Cp : C6H6 = 0.07406 + 3.295 x 10^-4 T - 2.52 x 10^-7 T2 + 7.757 x 10^-11 T3

HA = Cp dT = 0.07406 + 3.295 x 10^-4 T - 2.52 x 10^-7 T2 + 7.757 x 10^-11 T3 dT = 0.07406 T + 3.295 x 10^-4 T2/2 - 2.52 x 10^-7 T3/3 + 7.757 x 10^-11 T4/4 = (5.932 + 1.05704 - 0.044026 + 7.98296 x 10^-4) - ( 49.6954 + 74.1805 - 25.3791 + 3.93154 ) = 6.9458 - 102.42834 = - 95.48 kJ 1 mol 1000g mol 78.11g 1 kgHA = -1222.38 kJ/kgHB = Hv C6H6(80.100C, 1 atm) = 30.765 kJ 1 mol 1000g mol 78.11g 1 kg = 393.87 kJ/kg

HC = P + Cp( C6H6( l) ) dTCp : C6H6 (l) = 0.1265 + 2.34 x 10^-4 TCp( C6H6( l) ) dT = 0.1265 + 2.34 x 10^-4 T d = 0.1265 T + 2.34 x 10^-4 T^2/2 = ( 4.807 + 0.16895 ) - ( 10.133 + 0.7507 ) = 4.97595 - 10.884 = - 5.908 kJ 1 mol 1000g mol 78.11g 1 kg = -75.64 kJ/kg

SG of C6H6 = 0.879Density = 0.879 kg/L P = 1 L ( 1-24.446) atm 78.11g 1kg 0.008314 kJ/mol.K 0.879 kg 1 mol 1000g 0.08314 L.atm/mol.K = - 0.2083 kJ 1 mol 1000g mol 78.11g 1 kg = -2.667 kJ/kg HC = -2.667 - 75.64 = -78.307 kJ/kgH5 = HA + HB + HC = -1222.38 + 393.87 - 78.307 = -906.817 kJ/kg

Find QQ=Q=I HI -i HiQ=(3.74x105)( - 99.88)+( 4.51x106 )( 187.2)+( 3.07x106)( -2107.42)+( 1.16x106)( -906.817)-( 4.51 x106 )( 9470.5)Q= (-3.47x107) + (8.44x108)+(-6.47 x109)+(-6.47x109) - (4.27 x1010) Q= -1.213 x1010-(4.27 x1010)Q= -5.48 x1010