Calculate the temperature of the benzene stream fed to the reactor and the required rate of heat addition to or removal from the reactor. Use the following approximate heat capacities in your calculations: Cp[Btu/(lbm °F)] = 0.57 for propylene, 0.55 for butane, 0.45 for benzene, and 0.40 for cumene. ### Cumene Production Process**Chemical Reaction**:Cumene (\(C_6H_5C_3H_7\)) is produced by reacting benzene with propylene. The reaction has an enthalpy change (\(\Delta H_r^o\)) of -39,520 Btu.**Process Flow Diagram**:The process involves several key components, including a reactor, heat exchanger, and distillation columns. Here's a detailed explanation:- **Reactor**: The starting point where the chemical reaction occurs.- **Heat Exchanger**: This device heats the benzene and propylene mixture before entering the reactor. The reactor effluent at 350.0°F is cooled to 200.0°F in the exchanger.- **Distillation Columns (T1 and T2)**: These columns separate various components after the reaction. - **T1**: The effluent is fed into this column, where butane and unreacted propylene are removed overhead. - **T2**: The bottoms product from T1, containing cumene and unreacted benzene, is fed into this second column for further separation.- **Recycling**: The unreacted benzene is recycled back into the system. Specifically, 20.0% of the propylene is not reacted and is removed as an overhead product.**Chemical Feed**:- **Propylene and Butane**: The liquid feed consists of 85.0 mole% propylene and 15.0% n-butane, maintained at 77°F.- **Benzene**: Both fresh and recycled benzene are at 77°F and mixed in a 1:2.5 mole ratio (fresh:recycle) before entering the heat exchanger.**Production Rate**:The system produces cumene at a rate of 1200.0 lb\(_m\)/h.This overview provides a conceptual understanding of cumene production, emphasizing heat exchange, separation processes, and recycling within the system. ## Tutorial: Calculating Mass Flow Rates and CompositionsThis section aims to guide you through calculating the mass flow rates of streams fed to a reactor, as well as the molar flow rates and composition of both the reactor effluent and the overhead product from the first distillation column, labeled T1.### Reactor FeedYou will need to determine the following mass flow rates:- **Propylene fed to reactor:** Enter the flow rate in lb\(_m\)/h.- **Butane fed to reactor:** Enter the flow rate in lb\(_m\)/h.- **Benzene fed to reactor:** Enter the flow rate in lb\(_m\)/h.### Reactor EffluentNext, calculate and input the compositions and total molar flow rate of the reactor effluent:- **Total Molar Flow Rate:** Enter the flow rate in lb-mol/h.- **Composition of Butane (\(x_{butane}\)):** Enter the molar fraction.- **Composition of Propylene (\(x_{propylene}\)):** Enter the molar fraction.- **Composition of Benzene (\(x_{benzene}\)):** Enter the molar fraction.- **Composition of Cumene (\(x_{cumene}\)):** Enter the molar fraction.### Overhead from T1Finally, determine the overhead product variables from the first distillation column T1:- **Total Molar Flow Rate:** Enter the flow rate in lb-mol/h.- **Composition of Butane (\(x_{butane}\)):** Enter the molar fraction.- **Composition of Propylene (\(x_{propylene}\)):** Enter the molar fraction.These calculations are essential for understanding the efficiency and product distribution in chemical processes. Make sure you have all necessary data before proceeding with the calculations.

The basis of the given process is taken as 1200 lbm/h of the cumene (C) produced. Convert it into moles of cumene produced as: n˙C=m˙CMC=1200 lbm/h120 lbm/lb-mol=10.0 lb-mol/h The flowchart for the overall process is drawn below.Since there is a reaction taking place, material balance is applied to elements rather than molecules. Apply material balance on Carbon (C) as: 3×0.85n˙1+4×0.15n˙1+6n˙2=3n˙3+4n˙4+9×103.15n˙1+6n˙2=3n˙3+4n˙4+90 ...... (1) Apply material balance on Hydrogen (H) as: 6×0.85n˙1+10×0.15n˙1+6n˙2=6n˙3+10n˙4+120×106.6n˙1+6n˙2=6n˙3+10n˙4+120 ...... (2) The equation for unreacted propylene is, n˙3=0.200.85n˙1 ...... (3) The equation for unreacted butane is, 0.15n˙1=n˙4 ...... (4) Substitute equations (3) and (4) in equations (1) and (2) as: 3.15n˙1+6n˙2=30.200.85n˙1+40.15n˙1+902.04n˙1+6n˙2=90 ...... (5)6.6n˙1+6n˙2=60.200.85n˙1+100.15n˙1+1204.08n˙1+6n˙2=120 ...... (6) Simultaneously, solve equations (5) and (6) to calculate the value of n1 and n2 as: n˙1=14.71 lb-mol/hn˙2=10.0 lb-mol/hUse the value of n1 to calculate n3 and n4 from equations (3) and (4) as: n˙3=2.50 lb-mol/hn˙4=2.21 lb-mol/h According to the given values, the pure benzene stream contains fresh benzene and recycled benzene in the ratio of 1:2.5 which is fed to the reactor. Therefore, n˙Benzene=n˙2+2.5n˙2=10+2.510=35 lb-mol/h Benzene in the recycle stream is the benzene that exits the reactor. Thus, n˙Benzenereactor=2.5n˙2=2.510=25 lb-mol/h Now, prepare a flowchart for the reactor as: The molar flow rate of the propylene and butane stream to the reactor is: n˙1C3H6=0.85×14.71=12.50 lb-mol/hn˙1C4H10=0.15×14.71=2.21 lb-mol/h The total molar flow rate of the reactor effluent is: n˙effluent=2.50+2.21+25+10=39.71 lb-mol/h Calculate the mole fraction of each of the species in the effluent stream as: xC6H6=n˙C6H6n˙effluent=2539.71=0.630 lb-mol C6H6/lb-molxC3H6=n˙C3H6n˙effluent=2.5039.71=0.063 lb-mol C3H6/lb-molxC4H10=n˙C4H10n˙effluent=2.2139.71=0.056 lb-mol C4H10/lb-molxC9H12=n˙C9H12n˙effluent=1039.71=0.252 lb-mol C9H12/lb-molAll the propylene and butane present in the reactor effluent are separated from the mixture in the first distillation column, T1. Thus, the moles of propylene and butane present in the overhead product of T1 are: n˙Overhead T1=2.50 lb-mol C3H6/h+2.21 lb-mol C4H10/h=4.71 lb-mol/h The composition of the overhead product from T1 is calculated as: xC3H6=n˙C3H6n˙Overhead T1=2.504.71=0.531 lb-mol C3H6/lb-molxC4H10=n˙C4H10n˙Overhead T1=2.214.71=0.469 lb-mol C4H10/lb-mol The flowchart for the heat exchanger is drawn below. Assume the heat exchanger to work adiabatically, so the energy balance equation becomes: ΔH˙=0 This equation can be written as: 0=∑n˙iH^i, out−H^i, in0=∑n˙iCpiTi, out−Ti, inNow, substitute the values for all the species in the above equation and calculate the value of as: 0=nC9H12MC9H12CpC9H12TC9H12,out−TC9H12,in+nC3H6MC3H6CpC3H6TC3H6,out−TC3H6,innC4H10MC4H10CpC4H10TC4H10,out−TC4H10,in+nC6H6MC6H6CpC6H6TC6H6,out−TC6H6,in+nPure benzeneMC6H6CpC6H6TPure benzene,out−TPure benzene,in0=10 lb-mol/h120 lbm/lb-mol0.40 Btu/lbm⋅∘F200−350∘F+2.50 lb-mol/h42.08 lbm/lb-mol0.57 Btu/lbm⋅∘F200−350∘F+2.21 lb-mol/h58.12 lbm/lb-mol0.55 Btu/lbm⋅∘F200−350∘F+25 lb-mol/h78.11 lbm/lb-mol0.45 Btu/lbm⋅∘F200−350∘F+35 lb-mol/h78.11 lbm/lb-mol0.45 Btu/lbm⋅∘FT−77∘FT=258.6∘FTo calculate the heating requirement of the reactor, take the reference for energy balance as C6H6 (l), C3H6(l), C4H10(l), and C9H12(l) at 77°F. Since, the reference is taken as molecules and not elements, the heat of the reaction is to be included in the energy balance equation. The inlet-outlet enthalpy table for the reactor is: Specific enthalpies are calculated as: H^1=MWC6H6×CpC6H6×Treactor,in−77=78.11×0.45×258.6−77=6383.15 Btu/lb-molH^2=MWC6H6×CpC6H6×Treactor,out−77=78.11×0.45×350−77=9595.81 Btu/lb-molH^3=MWC3H6×CpC3H6×Treactor,out−77=42.08×0.57×350−77=6548.07 Btu/lb-molH^4=MWC4H10×CpC4H10×Treactor,out−77=58.11×0.55×350−77=8725.22 Btu/lb-molH^5=MWC9H12×CpC9H12×Treactor,out−77=120×0.40×350−77=13104 Btu/lb-molNow, apply energy balance to the reactor. The potential energy of the system is zero and any kinetic change in the system is assumed to be zero. Also, there is no shaft work done on the system. Thus, ΔH˙=Q˙ It can be written as: Q˙=n˙C9H12ΔH^∘r77∘CvC9H12+∑outn˙iH^i−∑inn˙iH^iQ˙=10.01−39520+259595.81+2.506548.07+2.218725.22+1013104−356383.15Q˙=−212022.09 Btu/hQ˙=−2.12×105 Btu/h