For point 1, I cannot locate the humidity and enthalpy. For point 2, I cannot locate the humidity and enthalpy. Where do you locate these on the graph. The graph is provided for this. ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +58°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloudWet Bulb,Dew Point orSaturationTemperatureDry Bulb 2025==iQ Search223=S5524454 of 6958%60A365RR#9:2Dry Bulb Temperature, F14$.5.2.7CDEnthalpy at saturation,of dry airper85Humid Volume, ft/lb Dry Air95Grains of Pounds ofmoisturemaismoistureper pound.per poundof dry airof dry air1051100251200179002401470010006002FIGURE 8.4-2 Psychrometric chart-U.S. customary units. Reference states: H₂O (liquid, 32°F, 1 atm), dry air (0°F, 1 atm). (Data obtained from Carrier Corporation.){"ƠⓇ63+60DENGSign in(0)+O7:38 AM5/22/2023 ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +58°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloud8.70+458 of 695 (2) ODNote: In labeling the outlet gas stream, we have implicitly written a balance on dry air.Degree-of-Freedom Analysis:Point 1:Point 2:7 unknowns (m₁, m₂, m3, Ĥ₁, Ĥ₂, Ĥ3, 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)Balance on H₂O:-1 enthalpy of condensate (from known heat capacity of liquid water)-1 energy balance= 0 degrees of freedomFraction H₂0 Condensed:Q Search80°F80% RHm₁ =Figure 8.4-21.0 lbm DASaturatedm₂ =51°F Figure 8.4-21.0 lbm DAm₁ = m₂ + m30.018 lbm H₂Olbm DAha = 0.018 lb H₂O/lbm DAĤ₁ = 38.8 Btu/lbm DA0.010 lbm condensed0.018 lbm fedRHirelativeharidity0.0079 lbm H₂Olbm DAm₁ = 0.018 lbmm₂ = 0.0079 lbmm3 = 0.010 lb, H₂O condensedHh₁ = 0.0079 lbm H₂O/lbm DAĤ₂ = 20.9 Btu/lbm DA0.555= 0.018 lbm H₂O12 In assuming this basis, we are temporarily ignoring the specification of the volumetric flow rate at the outlet. After theprocess is balanced for the assumed basis, we will scale up to an outlet flow rate of 1000 ft³/min.= 0.0079 lbm H₂OpaEnthalpy of CondensateSince the reference condition for water on Figure 8.4-2 is liquid water at 32°F, we must use the same condition8.4 Phase-Change Operations 439{"ƠⓇJ6360DENGSign in(0)+O7:39 AM5/22/2023

Question

For point 1, I cannot locate the humidity and enthalpy.  For point 2, I cannot locate the humidity and enthalpy.  Where do you locate these on the graph.  The graph is provided for this. ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +58°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloudWet Bulb,Dew Point orSaturationTemperatureDry Bulb 2025==iQ Search223=S5524454 of 6958%60A365RR#9:2Dry Bulb Temperature, F14$.5.2.7CDEnthalpy at saturation,of dry airper85Humid Volume, ft/lb Dry Air95Grains of Pounds ofmoisturemaismoistureper pound.per poundof dry airof dry air1051100251200179002401470010006002FIGURE 8.4-2 Psychrometric chart-U.S. customary units. Reference states: H₂O (liquid, 32°F, 1 atm), dry air (0°F, 1 atm). (Data obtained from Carrier Corporation.){&#34;ƠⓇ63+60DENGSign in(0)+O7:38 AM5/22/2023 ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +58°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloud8.70+458 of 695 (2) ODNote: In labeling the outlet gas stream, we have implicitly written a balance on dry air.Degree-of-Freedom Analysis:Point 1:Point 2:7 unknowns (m₁, m₂, m3, Ĥ₁, Ĥ₂, Ĥ3, 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)Balance on H₂O:-1 enthalpy of condensate (from known heat capacity of liquid water)-1 energy balance= 0 degrees of freedomFraction H₂0 Condensed:Q Search80°F80% RHm₁ =Figure 8.4-21.0 lbm DASaturatedm₂ =51°F Figure 8.4-21.0 lbm DAm₁ = m₂ + m30.018 lbm H₂Olbm DAha = 0.018 lb H₂O/lbm DAĤ₁ = 38.8 Btu/lbm DA0.010 lbm condensed0.018 lbm fedRHirelativeharidity0.0079 lbm H₂Olbm DAm₁ = 0.018 lbmm₂ = 0.0079 lbmm3 = 0.010 lb, H₂O condensedHh₁ = 0.0079 lbm H₂O/lbm DAĤ₂ = 20.9 Btu/lbm DA0.555= 0.018 lbm H₂O12 In assuming this basis, we are temporarily ignoring the specification of the volumetric flow rate at the outlet. After theprocess is balanced for the assumed basis, we will scale up to an outlet flow rate of 1000 ft³/min.= 0.0079 lbm H₂OpaEnthalpy of CondensateSince the reference condition for water on Figure 8.4-2 is liquid water at 32°F, we must use the same condition8.4 Phase-Change Operations 439{&#34;ƠⓇJ6360DENGSign in(0)+O7:39 AM5/22/2023

Accepted Answer

we have to find the enthalpy and humidity ratio from the given graph by given vales