Problema 1Problema 1: En la figura se muestra un posible diagrama de flujo para la produccin de cido perclrico. La reaccin sigue la estequiometra:

Si el H2SO4 alimentado al reactor es un 20% de exceso sobre la cantidad estequiometrca requerida parala reaccin con la alimentacin fresca de Ba(ClO4)2, y si se alimentan 1000 lb/h de la corriente 1, todaslas variables desconocidas de las corrientes. Suponga que todas las composiciones estn en fraccin masa.CorrienteEspecieFracc. MasaFracc. MolMasamolPM%Msico%Mol1Ba(ClO4)20.90.72859002.6838335.349072.85HClO40.10.271510011001027.15Total1110003.6838-1001002H2SO411315.61883.220698100100Total11315.61883.2206-1001003Ba(ClO4)20.0096350040.00395794912.6758520.0378335.340.963500350.3957949H2SO40.0428022320.06016501956.31080.5746984.280223196.0165019HClO40.4782593090.658820573629.26.29210047.825930965.882057BaSO40.4693034550.277056458617.417642.646233.3446.930345527.705646Total111315.604299.5504-1001004Ba(ClO4)20000335.3400Total0000-005H2SO41156.31080.574698100100Total1156.31080.5746-100100Corriente 11000lb/h6HClO411629.26.292100100100Total11629.26.292-1001007BaSO40.9798825840.985915493617.41762.646233.3497.988258498.591549Ba(ClO4)20.0201174160.01408450712.6758520.0378335.342.011741591.4084507Total11630.0934522.6838-1001008BaSO40.980.986617.4705622.646227233.349898.6Ba(ClO4)20.020.01412.59979690.037573335.3421.4Total11630.0703582.6838-100100900lb1lbmol1 mol H2SO41.2h335.341 mol Ba(ClO4)2=3.22 lbmol/h H2SO4mol H2SO4 entrada - mol H2SO4 salida = mol H2SO4 consumido3.2206 mol entrada - 0.5746 mol salida = 2.646 mol H2SO4 consumidosEcuaciones en funcion de Jn8=2.68382.646 mol H2SO4 cons1 mol Ba(ClO4)2=2.646 mol Ba(ClO4)2 consumidos2.6838 -J = n8*0.014J=2.6461 mol H2SO4J = 0.986*n8HClO4 =2*J+1 = 6.292 mol HClO4 salida2.6838 mol entrada - 2.646 mol consumido = 0.0378 mol Ba(ClO4)2 salidaH2SO4 = 3.2206 - J =0.5746 mol H2SO4 salidaBaSO4 = J =2.646 mol BaSO4 salida

Problema 2CorrienteEspecieFlujo molarFracc molar% molarAlim frescaA91.230.3421082234.2108224B175.440.6578917865.7891776TOTAL266.671100175.44 mol B2 mol A1.3=91.23 mol A alim frescaSalida reactorA24.560.056372945.637294285 mol BB111.110.2550324825.5032479C1000.2295315322.9531526166.67 mol B consum2 mol A=66.67 mol A consumD2000.4590630545.90630525 mol BTOTAL435.671100RecirculacionB102.34110091.23 mol A alimTOTAL102.341100-66.67 mol A consumProductosA24.560.073680747.36807368=24.56 mol salida AB8.770.026310262.63102631111.11 = 102.34 + Bproduc=8.77 mol B producidoC1000.30000330.0003166.67 mol B cons =0.6Bfresca + 102.34 = 277.78=175.44 mol B alim frescaD2000.60000660.0006mol B alimTOTAL333.331100mol B alim al reactor=277.78 mol BConversion de un paso B =0.6100mol/h C100 mol C5 mol B=166.67 mol B consumidosConversion global de B =0.95h3 mol C100 mol C6 mol D= 200 mol D producidosh3 mol C

(277.78)*(1-0.6)=111.11 mol B salida reactor0.95=(277.78 - Br) - (111.11 - Br)=Br=102.34 mol B recirculado277.78 - Br

Hoja3