Question

(25) Reactor Heat Duty Specification The reactor in this process is run isothermally at 350°C. Since the reaction is exothermic, determine the rate (in kJ/h) at which heat must be removed. C3H6 + CH CHı2 or P+BIB $i = 94.0 mol/h C3H6 + CH12 → C12H18 or P+IB → DB $2 = 6.00 mol/h Reactor T = 350°C 210 mol B(g)/h 104 mol P(g) /h T = 350°C 116 mol B(g)/h 4.00 mol P(g)h 88.0 mol IB(g)/h 6.00 mol DB(g)/h

Answer 1

Inlet Mole of C6H6 (B) =210 mol/h Mole of C3H6 (P)=104 mol/h Inlet Temperature=350 °C=350+273=623 K Specific heat of P=64.32 J/mol-K Specific heat of B=82.4 J/mol-K Heat at inlet= [Mole of B*Specific heat of B+ Mole of P*Specific heat of P]*Temperature Inlet= [210 mol/h*82.4 J/mol-K+104 mol/h*64.32 J/mol- K]*623 K=14947813.44 J/hr=14947813.44 /1000=14947.813 KJ/h Outlet Mole of C6H6 (B) =116 mol/h Mole of C3H6 (P) =4 mol/h Mole of C9H12 (B) =88 mol/h Mole of C12H18 (DB) =6 mol/h Specific heat of IB=143.3 J/mol-K Specific heat of DB=184.4 J/mol-K Outlet Temperature=350 °C=350+273=623 K Heat at outlet= [Mole of B*Specific heat of B+ Mole of P*Specific heat of P+ Mole of B*Specific heat of B+ Mole of DB*Specific heat of DB]*Temperature Outlet= [116 mol/h*82.4 J/mol-K+4 mol/h*64.32 J/mol- K+88 mol/h*143.3 J/mol-K+6 mol/h*184.4 J/mol-K]*623 K Heat at outlet=14660735.04 J/hr=14660735.04 /1000=14660.735 KJ/h Rate of heat transfer=Heat at outlet-Heat at inlet=14660.735 KJ/h -14947.813 KJ/h=-287.078 KJ/h

Inlet Mole of C6H6 (B) =210 mol/h Mole of C3H6 (P) =104 mol/h Inlet Temperature=350 °C=350+273=623 K Specific heat of P=64.32 J/mol-K Specific heat of B=82.4 J/mol-K Heat at inlet= [Mole of B*Specific heat of B+ Mole of P*Specific heat of P]*Temperature Inlet= [210 mol/h*82.4 J/mol-K+104 mol/h*64.32 J/mol- K]*673 K=16147477.44 J/hr=16147477.44/1000=16147.477 KJ/h Outlet Mole of C6H6 (B) =116 mol/h Mole of C3H6 (P) =4 mol/h Mole of C9H12 (B)=88 mol/h Mole of C12H18 (DB) =6 mol/h Specific heat of B=143.3 J/mol-K Specific heat of DB=184.4 J/mol-K Outlet Temperature=350 °C=350+273=623 K Heat at outlet=[Mole of B*Specific heat of B+ Mole of P*Specific heat of P+ Mole of B*Specific heat of B+ Mole of DB*Specific heat of DB]*Temperature Outlet= [116 mol/h*82.4 J/mol-K+4 mol/h*64.32 J/mol- K+88 mol/h*143.3 J/mol-K+6 mol/h*184.4 J/mol-K]*673 K Heat at outlet=15837359.04 J/hr=15837359.04 /1000=15837.36 KJ/h Rate of heat transfer=Heat at outlet-Heat at inlet=15837.36 KJ/h-16147.477 KJ/h=-310.117 KJ/h