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1-1 sedia Telder Energy Balance on a Condenser Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole % acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below. 100 mol/s 0.669 mol Ac(v/mol 0.331 mol Ny/mol 65°C, 1 atm (1/) CONDENSER 36.45 mol/s 0.092 mol Ac(v)/mol 0.908 mol Nymol 20°C, 5 atm 63.55 mol Ac(1/s 20°C, 5 atm The process operates at steady state. Calculate the required cooling rate.

1-1 sedia Telder Energy Balance on a Condenser Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole % acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below. 100 mol/s 0.669 mol Ac(v/mol 0.331 mol Ny/mol 65°C, 1 atm (1/) CONDENSER 36.45 mol/s 0.092 mol Ac(v)/mol 0.908 mol Nymol 20°C, 5 atm 63.55 mol Ac(1/s 20°C, 5 atm The process operates at steady state. Calculate the required cooling rate.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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I know how enthalpy 1 was calculated.  I do not know how enthalpies 2 and 3 were calculated.  Please help me with this problem.  

||! PDF *Elementary Principles of Chemic X + 46°F Sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdf Draw (T) Read aloud Example 8.1-1 Equipment Encyclopedia condenser www.wiley.com/college/felder 427 of 695 HA Te Canice w UC for you to understand une rest of the chapter and to solve the cha-of-chapter problems. Energy Balance on a Condenser Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole% acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below. 100 mol/s 0.669 mol Ac(v)/mol 0.331 mol N₂/mol 65°C, 1 atm 408 CHAPTER 8 Balances on Nonreactive Processes Ô(J/s) 63.55 mol Ac(1)/s 20°C, 5 atm The process operates at steady state. Calculate the required cooling rate. Q Search CONDENSER 2 Use Ĥ, instead of U; for a closed constant-pressure system, since Q=AH for such systems. Solution We will follow the procedure given preceding this example. 36.45 mol/s 0.092 mol Ac(v)/mol 0.908 mol N₂/mol 20°C, 5 atm ■ Perform required material balance calculations. None are required in this example. Write and simplify the energy balance. For this open steady-state system + W, = AH + AÈk +AEp. There are no moving parts in the system and no energy is transferred by electricity or radiation, so W, = 0. No significant vertical distance separates the inlet and outlet ports, so AĖp ~0. Phase changes and nonnegligible temperature changes occur, so AE 0 (relative to AH). The energy balance reduces to r P < Ơ J I ENG Sign in 100 7:07 AM 5/6/2023 • la + O
Transcribed Image Text:||! PDF *Elementary Principles of Chemic X + 46°F Sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdf Draw (T) Read aloud Example 8.1-1 Equipment Encyclopedia condenser www.wiley.com/college/felder 427 of 695 HA Te Canice w UC for you to understand une rest of the chapter and to solve the cha-of-chapter problems. Energy Balance on a Condenser Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole% acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below. 100 mol/s 0.669 mol Ac(v)/mol 0.331 mol N₂/mol 65°C, 1 atm 408 CHAPTER 8 Balances on Nonreactive Processes Ô(J/s) 63.55 mol Ac(1)/s 20°C, 5 atm The process operates at steady state. Calculate the required cooling rate. Q Search CONDENSER 2 Use Ĥ, instead of U; for a closed constant-pressure system, since Q=AH for such systems. Solution We will follow the procedure given preceding this example. 36.45 mol/s 0.092 mol Ac(v)/mol 0.908 mol N₂/mol 20°C, 5 atm ■ Perform required material balance calculations. None are required in this example. Write and simplify the energy balance. For this open steady-state system + W, = AH + AÈk +AEp. There are no moving parts in the system and no energy is transferred by electricity or radiation, so W, = 0. No significant vertical distance separates the inlet and outlet ports, so AĖp ~0. Phase changes and nonnegligible temperature changes occur, so AE 0 (relative to AH). The energy balance reduces to r P < Ơ J I ENG Sign in 100 7:07 AM 5/6/2023 • la + O
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