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Example 8.4-6 Equipment Encyclopedia heat exchanger www.wiley.com/college/felder Solution Material and Energy Balances on an Air Conditioner Air at 80°F and 80% relative humidity is cooled to 51°F at a constant pressure of 1 atm. Use the psychrometric chart to calculate the fraction of the water that condenses and the rate at which heat must be removed to deliver 1000 ft/min of humid air at the final condition. Basis: 1 lbm Dry Air¹2 A flowchart for the process is shown below. By convention we show heat transfer (0) into the process unit, but since the air is being cooled we know that will be negative. E Point 1: Point 2: 1 DA (lb H₂O(v)) 80°F. A,= 80% A(Buth DA) AIR COOLER Q (Btu) Note: In labeling the outlet gas stream, we have implicitly written a balance on dry air. Degree-of-Freedom Analysis: Balance on H₂O: 1 lb DA, 51F H₂O(v)) Bulb, DA) 7 unknowns (my, my, my, A₁, A₂, A3, Q) -1 material balance (H₂O-dry air is already balanced on the chart) -2 absolute humidities from psychrometric chart (for inlet and outlet air) -2 enthalpies from psychrometric chart (for inlet and outlet air) H₂O()), 51°F -1 enthalpy of condensate (from known heat capacity of liquid water) -1 energy balance = 0 degrees of freedom 1.0 lb DA A(Bulb) 80°F Figure 8.4-20.018 lb H₂O/lb DA 80% RH H₁38.8 Btu/lb DA 51°F Figure 8.4-2 Saturated 1.0 lbm DA m₁m₂ + my 0.018 lb H₂O lbm DA = 0.018 lb H₂O h = 0.0079 lb, H₂O/lb DA A₂ = 20.9 Btu/lb DA 0.0079 lb H₂O lb DA = 0.0079 lb H₂O

Example 8.4-6 Equipment Encyclopedia heat exchanger www.wiley.com/college/felder Solution Material and Energy Balances on an Air Conditioner Air at 80°F and 80% relative humidity is cooled to 51°F at a constant pressure of 1 atm. Use the psychrometric chart to calculate the fraction of the water that condenses and the rate at which heat must be removed to deliver 1000 ft/min of humid air at the final condition. Basis: 1 lbm Dry Air¹2 A flowchart for the process is shown below. By convention we show heat transfer (0) into the process unit, but since the air is being cooled we know that will be negative. E Point 1: Point 2: 1 DA (lb H₂O(v)) 80°F. A,= 80% A(Buth DA) AIR COOLER Q (Btu) Note: In labeling the outlet gas stream, we have implicitly written a balance on dry air. Degree-of-Freedom Analysis: Balance on H₂O: 1 lb DA, 51F H₂O(v)) Bulb, DA) 7 unknowns (my, my, my, A₁, A₂, A3, Q) -1 material balance (H₂O-dry air is already balanced on the chart) -2 absolute humidities from psychrometric chart (for inlet and outlet air) -2 enthalpies from psychrometric chart (for inlet and outlet air) H₂O()), 51°F -1 enthalpy of condensate (from known heat capacity of liquid water) -1 energy balance = 0 degrees of freedom 1.0 lb DA A(Bulb) 80°F Figure 8.4-20.018 lb H₂O/lb DA 80% RH H₁38.8 Btu/lb DA 51°F Figure 8.4-2 Saturated 1.0 lbm DA m₁m₂ + my 0.018 lb H₂O lbm DA = 0.018 lb H₂O h = 0.0079 lb, H₂O/lb DA A₂ = 20.9 Btu/lb DA 0.0079 lb H₂O lb DA = 0.0079 lb H₂O

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 need help locating the relative humidities and enthalpies on the psychrometric chart, for points 1 and 2.  

||! PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X + 53°F Partly sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw (T) Read aloud Example 8.4-6 Equipment Encyclopedia heat exchanger www.wiley.com/college/felder Solution Buvuny auga do VL Material and Energy Balances on an Air Conditioner Air at 80°F and 80% relative humidity is cooled to 51°F at a constant pressure of 1 atm. Use the psychrometric chart to calculate the fraction of the water that condenses and the rate at which heat must be removed to deliver 1000 ft³/min of humid air at the final condition. Point 1: Basis: 1 lbm Dry Air¹² A flowchart for the process is shown below. By convention we show heat transfer (Q) into the process unit, but since the air is being cooled we know that Q will be negative. Point 2: Balance on H₂O: Search 458 of 695 on a une chart is tỰ De uova in une 1 lb DA m₁(lbm H₂O(v)) 80°F, h, = 80% A₁ (Btu/lb DA) Q (Btu) Note: In labeling the outlet gas stream, we have implicitly written a balance on dry air. Degree-of-Freedom Analysis: 80% RH m₁ = m₂ = AIR COOLER 7 unknowns (m₁, m2, m3, H₁, H₂, H3, Q) -1 material balance (H₂O-dry air is already balanced on the chart) -2 absolute humidities from psychrometric chart (for inlet and outlet air) -2 enthalpies from psychrometric chart (for inlet and outlet air) -1 enthalpy of condensate (from known heat capacity of liquid water) -1 energy balance = 0 degrees of freedom ■ 80°F Figure 8.4-2 51°F Saturated 4 OD 1.0 lbm DA m₁ = m₂ + m3 Figure 8.4-2 1.0 lbm DA 1 lbm DA, 51°F m₂(lbm H₂O(v)) Ĥ₂(Btu/lb DA) m3(lbm H₂O(1)), 51°F A3(Btu/lb) ha=0.018 lbm H₂O/lbm DA Ĥ₁ = 38.8 Btu/lbm DA 0.018 lbm H₂O lbm DA = 0.018 lbm H₂O ha = 0.0079 lbm H₂O/lbm DA H₂ = 20.9 Btu/lbm DA 0.0079 lbm H₂O lbm DA = 0.0079 lbm H₂O {" J 63 50 ENG Sign in I ● 7:19 AM 5/19/2023 + O {0}
Transcribed Image Text:||! PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X + 53°F Partly sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw (T) Read aloud Example 8.4-6 Equipment Encyclopedia heat exchanger www.wiley.com/college/felder Solution Buvuny auga do VL Material and Energy Balances on an Air Conditioner Air at 80°F and 80% relative humidity is cooled to 51°F at a constant pressure of 1 atm. Use the psychrometric chart to calculate the fraction of the water that condenses and the rate at which heat must be removed to deliver 1000 ft³/min of humid air at the final condition. Point 1: Basis: 1 lbm Dry Air¹² A flowchart for the process is shown below. By convention we show heat transfer (Q) into the process unit, but since the air is being cooled we know that Q will be negative. Point 2: Balance on H₂O: Search 458 of 695 on a une chart is tỰ De uova in une 1 lb DA m₁(lbm H₂O(v)) 80°F, h, = 80% A₁ (Btu/lb DA) Q (Btu) Note: In labeling the outlet gas stream, we have implicitly written a balance on dry air. Degree-of-Freedom Analysis: 80% RH m₁ = m₂ = AIR COOLER 7 unknowns (m₁, m2, m3, H₁, H₂, H3, Q) -1 material balance (H₂O-dry air is already balanced on the chart) -2 absolute humidities from psychrometric chart (for inlet and outlet air) -2 enthalpies from psychrometric chart (for inlet and outlet air) -1 enthalpy of condensate (from known heat capacity of liquid water) -1 energy balance = 0 degrees of freedom ■ 80°F Figure 8.4-2 51°F Saturated 4 OD 1.0 lbm DA m₁ = m₂ + m3 Figure 8.4-2 1.0 lbm DA 1 lbm DA, 51°F m₂(lbm H₂O(v)) Ĥ₂(Btu/lb DA) m3(lbm H₂O(1)), 51°F A3(Btu/lb) ha=0.018 lbm H₂O/lbm DA Ĥ₁ = 38.8 Btu/lbm DA 0.018 lbm H₂O lbm DA = 0.018 lbm H₂O ha = 0.0079 lbm H₂O/lbm DA H₂ = 20.9 Btu/lbm DA 0.0079 lbm H₂O lbm DA = 0.0079 lbm H₂O {" J 63 50 ENG Sign in I ● 7:19 AM 5/19/2023 + O {0}
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