Locate the two points and humid volume on the US psychrometric chart. ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +76°FMostly sunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDraw(T)Read aloudTest Yourself(Answers, p. 659)Q SearchEnergy Balance:The open-system energy balance with Ws, AEk, and AEp set equal to zero ism₂Ĥ₁Q=AH == Σm.#₁ - Σmout(There are no dots over the extensive variables in this equation because the basis of calculation is an amount,not a flow rate.) The enthalpy table for the process is shown below. Since (1) the enthalpies (H₂) of the humidair streams are obtained from the psychrometric chart in Btu/lb dry air, and (2) the mass units of m; and Ĥ₂must cancel when the two are multiplied in the energy balance, the tabulated values of m; for these streamsmust be in lbm dry air.locate athe uspsychrumetricchart459 of 695Humid airH₂O(1)Q=AH =onH₂O(1, 32°F)→ H₂O(1,51°F)BtuAHH₂= 1.0; (51°F 32°F) = 19.0 Btu/lbm H₂Olb °FReferences: Dry air (DA) (g. 0°F, 1 atm), H₂O (1, 32°F, 1 atm)Substancemin1.0 lbm DAThe references were of necessity chosen to be the ones used to generate the psychrometric chart. Substitutingthe values in the table into the energy balance yields1.0 lbm DA■Vbasis =38.8 Btu/lbm DAQ=20.9 Btu 0.010 lbm H₂O(1)+lbm DA= -17.7 BtuTo calculate the cooling requirement for 1000 ft³/min of delivered air, we must first determine thevolume of delivered air corresponding to our assumed basis and scale the calculated value of Q by the ratio(1000 ft³/min)/(Vbasis). From the psychrometric chart, for humid air saturated at 51°FVH = 13.0 ft³/lbm DA1.0 lbm DA-17.7 Btumout1.0lbm DA0.010 lbm13.0 ft³= 13.0 ft³lbm DA1000 ft³/min13.0 ft³Ĥ out20.9 Btu/lbm DA19 Btu/lbm19 Btulbm H₂O1.0 lbm DA38.8 Btulbm DA-1360 Btu/minAir at 25°C and 1 atm has a relative humidity of 20%. Use the psychrometric chart to estimatethe absolute humidity, wet-bulb temperature, dew point, humid volume, and specific enthalpyof the air.o{"@683ENGSign in00:6:15 PM5/28/2023 ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +76°FMostly sunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDraw(T)Read aloudExample 8.4-6Equipment Encyclopediaheat exchangerwww.wiley.com/college/felderSolutionQ SearchMaterial and Energy Balances on an Air ConditionerAir at 80°F and 80% relative humidity is cooled to 51°F at a constant pressure of 1 atm. Use the psychrometricchart to calculate the fraction of the water that condenses and the rate at which heat must be removed todeliver 1000 ft³/min of humid air at the final condition.Basis: 1 lbm Dry Air¹2A 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 1:Point 2:458 of 695 31 lbm DAmi(lbm H₂O(v))80°F, h, = 80%Ĥ₁ (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:Balance on H₂O0:Fraction H₂O Condensed: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 freedom80% RHm₁ =m₂ =■AIRCOOLER80°F) Figure 8.4-251°FSaturated1.0 lbm DA1 lb DA, 51°Fm₂(lb H₂O(v))H₂(Btu/lb DA)1.0 lbm DAm3(lb H₂O(1)), 51°FÅ3(Btu/lb.)Figure 8.4-2m₁ = m₂ + m3H0.018 lbm H₂Olbm DAh₂0.018 lbm H₂O/1bm DAĤ₁ = 38.8 Btu/lbm DA0.0079 lbm H₂Olbm DAm₁ = 0.018 lbmm₂ = 0.00791m3 = 0.010 lbm H₂O condensed0.010 lbm condensed0.018 lbm fedforate onthe uspsychso.h₂ = 0.0079 lbm H₂O/lb DANĤ₂ = 20.9 Btu/lbm DA0.555= 0.018 lbm H₂OchartDADocate onis psychro.chart12 In assuming this basis, we are temporarily ignoring the specification of the volumetric flow rate at the outlet. After theprocess balanced for the assumed basis, we will scale up to an outlet flow rate of 1000 ft³/min.o= 0.0079 lbm H₂O{"@683ENGSign in00:6:15 PM5/28/2023

Question

Locate the two points and humid volume on the US psychrometric chart. ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +76°FMostly sunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDraw(T)Read aloudTest Yourself(Answers, p. 659)Q SearchEnergy Balance:The open-system energy balance with Ws, AEk, and AEp set equal to zero ism₂Ĥ₁Q=AH == Σm.#₁ - Σmout(There are no dots over the extensive variables in this equation because the basis of calculation is an amount,not a flow rate.) The enthalpy table for the process is shown below. Since (1) the enthalpies (H₂) of the humidair streams are obtained from the psychrometric chart in Btu/lb dry air, and (2) the mass units of m; and Ĥ₂must cancel when the two are multiplied in the energy balance, the tabulated values of m; for these streamsmust be in lbm dry air.locate athe uspsychrumetricchart459 of 695Humid airH₂O(1)Q=AH =onH₂O(1, 32°F)→ H₂O(1,51°F)BtuAHH₂= 1.0; (51°F 32°F) = 19.0 Btu/lbm H₂Olb °FReferences: Dry air (DA) (g. 0°F, 1 atm), H₂O (1, 32°F, 1 atm)Substancemin1.0 lbm DAThe references were of necessity chosen to be the ones used to generate the psychrometric chart. Substitutingthe values in the table into the energy balance yields1.0 lbm DA■Vbasis =38.8 Btu/lbm DAQ=20.9 Btu 0.010 lbm H₂O(1)+lbm DA= -17.7 BtuTo calculate the cooling requirement for 1000 ft³/min of delivered air, we must first determine thevolume of delivered air corresponding to our assumed basis and scale the calculated value of Q by the ratio(1000 ft³/min)/(Vbasis). From the psychrometric chart, for humid air saturated at 51°FVH = 13.0 ft³/lbm DA1.0 lbm DA-17.7 Btumout1.0lbm DA0.010 lbm13.0 ft³= 13.0 ft³lbm DA1000 ft³/min13.0 ft³Ĥ out20.9 Btu/lbm DA19 Btu/lbm19 Btulbm H₂O1.0 lbm DA38.8 Btulbm DA-1360 Btu/minAir at 25°C and 1 atm has a relative humidity of 20%. Use the psychrometric chart to estimatethe absolute humidity, wet-bulb temperature, dew point, humid volume, and specific enthalpyof the air.o{&#34;@683ENGSign in00:6:15 PM5/28/2023 ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +76°FMostly sunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDraw(T)Read aloudExample 8.4-6Equipment Encyclopediaheat exchangerwww.wiley.com/college/felderSolutionQ SearchMaterial and Energy Balances on an Air ConditionerAir at 80°F and 80% relative humidity is cooled to 51°F at a constant pressure of 1 atm. Use the psychrometricchart to calculate the fraction of the water that condenses and the rate at which heat must be removed todeliver 1000 ft³/min of humid air at the final condition.Basis: 1 lbm Dry Air¹2A 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 1:Point 2:458 of 695 31 lbm DAmi(lbm H₂O(v))80°F, h, = 80%Ĥ₁ (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:Balance on H₂O0:Fraction H₂O Condensed: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 freedom80% RHm₁ =m₂ =■AIRCOOLER80°F) Figure 8.4-251°FSaturated1.0 lbm DA1 lb DA, 51°Fm₂(lb H₂O(v))H₂(Btu/lb DA)1.0 lbm DAm3(lb H₂O(1)), 51°FÅ3(Btu/lb.)Figure 8.4-2m₁ = m₂ + m3H0.018 lbm H₂Olbm DAh₂0.018 lbm H₂O/1bm DAĤ₁ = 38.8 Btu/lbm DA0.0079 lbm H₂Olbm DAm₁ = 0.018 lbmm₂ = 0.00791m3 = 0.010 lbm H₂O condensed0.010 lbm condensed0.018 lbm fedforate onthe uspsychso.h₂ = 0.0079 lbm H₂O/lb DANĤ₂ = 20.9 Btu/lbm DA0.555= 0.018 lbm H₂OchartDADocate onis psychro.chart12 In assuming this basis, we are temporarily ignoring the specification of the volumetric flow rate at the outlet. After theprocess balanced for the assumed basis, we will scale up to an outlet flow rate of 1000 ft³/min.o= 0.0079 lbm H₂O{&#34;@683ENGSign in00:6:15 PM5/28/2023

Accepted Answer

As two indepent properties are defined we can determine the whole system in the humid(psychrometric…