Can you demonstrate the process path for enthalpy 2 (going out)? ||!PDF *Elementary Principles of Chemic X PDF *Elementary Principles of Chemic X +O56°FClearFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDraw(T) Read aloudQ Searchout3. Choose reference states for acetone and nitrogen.Substance428 of 695The reference states may be chosen for computational convenience, since the choice has noeffect on the calculated value of AH. We will arbitrarily choose the inlet stream condition fornitrogen (65°C, 1 atm) as the reference state for this species and one of the two outlet streamconditions for acetone (1, 20°C, 5 atm) as the reference state for acetone, which will enable us to setthe corresponding values for Ĥ in the enthalpy table equal to zero instead of having to calculate them.4. Construct an inlet-outlet enthalpy table.We first write the chosen reference states, then construct the table shown below:References: Ac(1, 20°C, 5 atm), N₂(g, 65°C, 1 atm)ninĤ in(mol/s) (kJ/mol)66.9 Ĥ₁Ac(v)Ac(1)N₂33.1in0nout(mol/s)3.3563.5533.1Ĥ out(kJ/mol)Note the following points about the table:• Nitrogen has only one inlet state (gas, 65°C, 1 atm) and one outlet state (gas, 20°C, 5 atm), so weneed only one row in the table for N₂. Acetone has one inlet state (vapor, 65°C, 1 atm) but twooutlet states (vapor and liquid, each at 20°C and 5 atm), so we need two rows for this species.■Ĥ₂0Ĥ3• We mark out (using dashes) the two cells corresponding to in and Hin for liquid acetone, since noliquid acetone enters the system.• The è values are obtained from the flowchart. The flow rate of acetone vapor at the inlet, forexample, is determined as (100 mol/s)[0.669 mol Ac(v)/mol] = 66.9 mol Ac(v)/s.• Since the nitrogen entering the system and the liquid acetone leaving the system area are at theirreference states, we set their specific enthalpies equal to zero.H• Three unknown specific enthalpies have been labeled and must be determined in Step 5.5. Calculate all unknown specific enthalpies.To calculate the three unknown specific enthalpies in the table, we construct hypotheticalprocess paths from the reference states to the states of the species in the process and evaluate AĤ foreach path. This is the part of the calculation you have not yet learned to do. We will show you thecalculation of H₁ to illustrate the method, give the results of the other calculations, and go into detailabout the required procedures in Sections 8.2-8.5.Ĥ₁ = specific enthalpy of Ac(v, 65°C, 1 atm) relative to Ac(1, 20°C, 5 atm)= AĤ for Ac(1, 20°C, 5 atm) → Ac(v, 65°C, 1 atm)When choosing a process path for the determination of AH, it helps to know that formulas anddata are given in this chapter for enthalpy changes corresponding to certain types of processes:• Section 8.2 gives the formula AĤ = VAP for a change in pressure (AP) undergone by a liquid orsolid with constant specific volume V. The value of V for liquid acetone may be determined as0.0734 L/mol from the specific gravity (0.791) given in Table B.1.{"@683ENGSign in00:11:16 PM5/23/2023 64°FClearElementary Principles of Chemica X +File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDraw(T)Read aloudExample 8.1-1Equipment Encyclopediacondenserwww.wiley.com/college/feldernixturecompoundsprodiems.(8.15408 CHAPTER 8 Balances on Nonreactive ProcessesQ Search428 of 695Energy Balance on a CondenserAcetone (denoted as Ac is partially condensed out of a gas stream containing 66.9 mole% acetone vapor andthe balance nitrogen. Process specifications and material balance calculations lead to the flowchart shownbelow.100 mol/s0.669 mol Ac(v)/mol0.331 mol N₂/mol65°C, 1 atmO(J/s)CONDENSERSolution We will follow the procedure given preceding this example.36.45 mol/s0.092 mol Ac(v)/mol0.908 mol N₂/mol20°C, 5 atm63.55 mol Ac(1)/s20°C, 5 atmThe process operates at steady state. Calculate the required cooling rate.2 Use Ĥ; instead of Û, for a closed constant-pressure system, since Q=AH for such systems.sta-tePerform required material balance calculations. None are required in this example.Write and simplify the energy balance.For this open steady-state system + W₁ = AH+AE+AEp. There are no moving parts inthe system and no energy is transferred by electricity or radiation, so W₁ = 0. No significant verticaldistance separates the inlet and outlet ports, so AEp 0. Phase changes and nonnegligibletemperature changes occur, so AE 0 (relative to AH). The energy balance reduces toΟ =ΔΗ - ΣΗ-ΣΗ13. Choose reference states for acetone and nitrogen.The reference states may be chosen for computational convenience, since the choice has noeffect on the calculated value of AH. We will arbitrarily choose the inlet stream condition fornitrogen (65°C, 1 atm) as the reference state for this species and one of the two outlet streamConditions for acetone (1, 20°C, 5 atm) as the reference state for acetone, which will enable us to setthe corresponding values for H in the enthalpy table equal to zero instead of having to calculate them.{"@683DENGSign in0060:8:46 PM5/23/2023

Question

Can you demonstrate the process path for enthalpy 2 (going out)? ||!PDF *Elementary Principles of Chemic X PDF *Elementary Principles of Chemic X +O56°FClearFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDraw(T) Read aloudQ Searchout3. Choose reference states for acetone and nitrogen.Substance428 of 695The reference states may be chosen for computational convenience, since the choice has noeffect on the calculated value of AH. We will arbitrarily choose the inlet stream condition fornitrogen (65°C, 1 atm) as the reference state for this species and one of the two outlet streamconditions for acetone (1, 20°C, 5 atm) as the reference state for acetone, which will enable us to setthe corresponding values for Ĥ in the enthalpy table equal to zero instead of having to calculate them.4. Construct an inlet-outlet enthalpy table.We first write the chosen reference states, then construct the table shown below:References: Ac(1, 20°C, 5 atm), N₂(g, 65°C, 1 atm)ninĤ in(mol/s) (kJ/mol)66.9 Ĥ₁Ac(v)Ac(1)N₂33.1in0nout(mol/s)3.3563.5533.1Ĥ out(kJ/mol)Note the following points about the table:• Nitrogen has only one inlet state (gas, 65°C, 1 atm) and one outlet state (gas, 20°C, 5 atm), so weneed only one row in the table for N₂. Acetone has one inlet state (vapor, 65°C, 1 atm) but twooutlet states (vapor and liquid, each at 20°C and 5 atm), so we need two rows for this species.■Ĥ₂0Ĥ3• We mark out (using dashes) the two cells corresponding to in and Hin for liquid acetone, since noliquid acetone enters the system.• The è values are obtained from the flowchart. The flow rate of acetone vapor at the inlet, forexample, is determined as (100 mol/s)[0.669 mol Ac(v)/mol] = 66.9 mol Ac(v)/s.• Since the nitrogen entering the system and the liquid acetone leaving the system area are at theirreference states, we set their specific enthalpies equal to zero.H• Three unknown specific enthalpies have been labeled and must be determined in Step 5.5. Calculate all unknown specific enthalpies.To calculate the three unknown specific enthalpies in the table, we construct hypotheticalprocess paths from the reference states to the states of the species in the process and evaluate AĤ foreach path. This is the part of the calculation you have not yet learned to do. We will show you thecalculation of H₁ to illustrate the method, give the results of the other calculations, and go into detailabout the required procedures in Sections 8.2-8.5.Ĥ₁ = specific enthalpy of Ac(v, 65°C, 1 atm) relative to Ac(1, 20°C, 5 atm)= AĤ for Ac(1, 20°C, 5 atm) → Ac(v, 65°C, 1 atm)When choosing a process path for the determination of AH, it helps to know that formulas anddata are given in this chapter for enthalpy changes corresponding to certain types of processes:• Section 8.2 gives the formula AĤ = VAP for a change in pressure (AP) undergone by a liquid orsolid with constant specific volume V. The value of V for liquid acetone may be determined as0.0734 L/mol from the specific gravity (0.791) given in Table B.1.{&#34;@683ENGSign in00:11:16 PM5/23/2023 64°FClearElementary Principles of Chemica X +File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDraw(T)Read aloudExample 8.1-1Equipment Encyclopediacondenserwww.wiley.com/college/feldernixturecompoundsprodiems.(8.15408 CHAPTER 8 Balances on Nonreactive ProcessesQ Search428 of 695Energy Balance on a CondenserAcetone (denoted as Ac is partially condensed out of a gas stream containing 66.9 mole% acetone vapor andthe balance nitrogen. Process specifications and material balance calculations lead to the flowchart shownbelow.100 mol/s0.669 mol Ac(v)/mol0.331 mol N₂/mol65°C, 1 atmO(J/s)CONDENSERSolution We will follow the procedure given preceding this example.36.45 mol/s0.092 mol Ac(v)/mol0.908 mol N₂/mol20°C, 5 atm63.55 mol Ac(1)/s20°C, 5 atmThe process operates at steady state. Calculate the required cooling rate.2 Use Ĥ; instead of Û, for a closed constant-pressure system, since Q=AH for such systems.sta-tePerform required material balance calculations. None are required in this example.Write and simplify the energy balance.For this open steady-state system + W₁ = AH+AE+AEp. There are no moving parts inthe system and no energy is transferred by electricity or radiation, so W₁ = 0. No significant verticaldistance separates the inlet and outlet ports, so AEp 0. Phase changes and nonnegligibletemperature changes occur, so AE 0 (relative to AH). The energy balance reduces toΟ =ΔΗ - ΣΗ-ΣΗ13. Choose reference states for acetone and nitrogen.The reference states may be chosen for computational convenience, since the choice has noeffect on the calculated value of AH. We will arbitrarily choose the inlet stream condition fornitrogen (65°C, 1 atm) as the reference state for this species and one of the two outlet streamConditions for acetone (1, 20°C, 5 atm) as the reference state for acetone, which will enable us to setthe corresponding values for H in the enthalpy table equal to zero instead of having to calculate them.{&#34;@683DENGSign in0060:8:46 PM5/23/2023

Accepted Answer

The enthalpy change of the acetone vapor leaving the condenser is to be determined.