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Point 1: Point 2: -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 Balance on H₂O: Fraction H₂0 Condensed: 80°F 80% RH 1.0 51°F) Saturated Figure 1.4-2 h = 0.018 lb H₂O/lb DA H₁ 38.8 Btu/lb, DA DA 842 0.018 lb H₂O -0.018b H₂O lb DA 1.0lb DA 0.0079 lb H₂O Ib,,, ᎠᎪ forate on the us psychso. 0.018 -0.0079 my=0.010 lb H₂O condensed 0.010 lb condensed 0.018 lb, fed k₂ = 0.0079 lb H₂O/lb DAN H₂= 20.9 Btu/lb DA -0.0079 lb H₂O Chart "Docate on us psychro. chart =0.555

Point 1: Point 2: -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 Balance on H₂O: Fraction H₂0 Condensed: 80°F 80% RH 1.0 51°F) Saturated Figure 1.4-2 h = 0.018 lb H₂O/lb DA H₁ 38.8 Btu/lb, DA DA 842 0.018 lb H₂O -0.018b H₂O lb DA 1.0lb DA 0.0079 lb H₂O Ib,,, ᎠᎪ forate on the us psychso. 0.018 -0.0079 my=0.010 lb H₂O condensed 0.010 lb condensed 0.018 lb, fed k₂ = 0.0079 lb H₂O/lb DAN H₂= 20.9 Btu/lb DA -0.0079 lb H₂O Chart "Docate on us psychro. chart =0.555

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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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°F Mostly sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw (T) Read aloud Test Yourself (Answers, p. 659) Q Search Energy Balance: The open-system energy balance with Ws, AEk, and AEp set equal to zero is m₂Ĥ₁ Q=AH = = Σm.#₁ - Σm out (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 humid air streams are obtained from the psychrometric chart in Btu/lb dry air, and (2) the mass units of m; and Ĥ₂ must cancel when the two are multiplied in the energy balance, the tabulated values of m; for these streams must be in lbm dry air. locate a the us psychrumetric chart 459 of 695 Humid air H₂O(1) Q=AH = on H₂O(1, 32°F)→ H₂O(1,51°F) Btu AHH₂= 1.0; (51°F 32°F) = 19.0 Btu/lbm H₂O lb °F References: Dry air (DA) (g. 0°F, 1 atm), H₂O (1, 32°F, 1 atm) Substance min 1.0 lbm DA The references were of necessity chosen to be the ones used to generate the psychrometric chart. Substituting the values in the table into the energy balance yields 1.0 lbm DA ■ Vbasis = 38.8 Btu/lbm DA Q= 20.9 Btu 0.010 lbm H₂O(1) + lbm DA = -17.7 Btu To calculate the cooling requirement for 1000 ft³/min of delivered air, we must first determine the volume 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°F VH = 13.0 ft³/lbm DA 1.0 lbm DA -17.7 Btu mout 1.0lbm DA 0.010 lbm 13.0 ft³ = 13.0 ft³ lbm DA 1000 ft³/min 13.0 ft³ Ĥ out 20.9 Btu/lbm DA 19 Btu/lbm 19 Btu lbm H₂O 1.0 lbm DA 38.8 Btu lbm DA -1360 Btu/min Air at 25°C and 1 atm has a relative humidity of 20%. Use the psychrometric chart to estimate the absolute humidity, wet-bulb temperature, dew point, humid volume, and specific enthalpy of the air. o {" @ 683 ENG Sign in 00 : 6:15 PM 5/28/2023
Transcribed Image Text:||! PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X + 76°F Mostly sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw (T) Read aloud Test Yourself (Answers, p. 659) Q Search Energy Balance: The open-system energy balance with Ws, AEk, and AEp set equal to zero is m₂Ĥ₁ Q=AH = = Σm.#₁ - Σm out (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 humid air streams are obtained from the psychrometric chart in Btu/lb dry air, and (2) the mass units of m; and Ĥ₂ must cancel when the two are multiplied in the energy balance, the tabulated values of m; for these streams must be in lbm dry air. locate a the us psychrumetric chart 459 of 695 Humid air H₂O(1) Q=AH = on H₂O(1, 32°F)→ H₂O(1,51°F) Btu AHH₂= 1.0; (51°F 32°F) = 19.0 Btu/lbm H₂O lb °F References: Dry air (DA) (g. 0°F, 1 atm), H₂O (1, 32°F, 1 atm) Substance min 1.0 lbm DA The references were of necessity chosen to be the ones used to generate the psychrometric chart. Substituting the values in the table into the energy balance yields 1.0 lbm DA ■ Vbasis = 38.8 Btu/lbm DA Q= 20.9 Btu 0.010 lbm H₂O(1) + lbm DA = -17.7 Btu To calculate the cooling requirement for 1000 ft³/min of delivered air, we must first determine the volume 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°F VH = 13.0 ft³/lbm DA 1.0 lbm DA -17.7 Btu mout 1.0lbm DA 0.010 lbm 13.0 ft³ = 13.0 ft³ lbm DA 1000 ft³/min 13.0 ft³ Ĥ out 20.9 Btu/lbm DA 19 Btu/lbm 19 Btu lbm H₂O 1.0 lbm DA 38.8 Btu lbm DA -1360 Btu/min Air at 25°C and 1 atm has a relative humidity of 20%. Use the psychrometric chart to estimate the absolute humidity, wet-bulb temperature, dew point, humid volume, and specific enthalpy of the air. o {" @ 683 ENG Sign in 00 : 6:15 PM 5/28/2023
||! PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X + 76°F Mostly 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 Q Search 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. 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 1: Point 2: 458 of 695 3 1 lbm DA mi(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 freedom 80% RH m₁ = m₂ = ■ AIR COOLER 80°F) Figure 8.4-2 51°F Saturated 1.0 lbm DA 1 lb DA, 51°F m₂(lb H₂O(v)) H₂(Btu/lb DA) 1.0 lbm DA m3(lb H₂O(1)), 51°F Å3(Btu/lb.) Figure 8.4-2 m₁ = m₂ + m3 H 0.018 lbm H₂O lbm DA h₂0.018 lbm H₂O/1bm DA Ĥ₁ = 38.8 Btu/lbm DA 0.0079 lbm H₂O lbm DA m₁ = 0.018 lbm m₂ = 0.00791 m3 = 0.010 lbm H₂O condensed 0.010 lbm condensed 0.018 lbm fed forate on the us psychso. h₂ = 0.0079 lbm H₂O/lb DAN Ĥ₂ = 20.9 Btu/lbm DA 0.555 = 0.018 lbm H₂O chart DADocate on is psychro. chart 12 In assuming this basis, we are temporarily ignoring the specification of the volumetric flow rate at the outlet. After the process balanced for the assumed basis, we will scale up to an outlet flow rate of 1000 ft³/min. o = 0.0079 lbm H₂O {" @ 683 ENG Sign in 00 : 6:15 PM 5/28/2023
Transcribed Image Text:||! PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X + 76°F Mostly 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 Q Search 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. 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 1: Point 2: 458 of 695 3 1 lbm DA mi(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 freedom 80% RH m₁ = m₂ = ■ AIR COOLER 80°F) Figure 8.4-2 51°F Saturated 1.0 lbm DA 1 lb DA, 51°F m₂(lb H₂O(v)) H₂(Btu/lb DA) 1.0 lbm DA m3(lb H₂O(1)), 51°F Å3(Btu/lb.) Figure 8.4-2 m₁ = m₂ + m3 H 0.018 lbm H₂O lbm DA h₂0.018 lbm H₂O/1bm DA Ĥ₁ = 38.8 Btu/lbm DA 0.0079 lbm H₂O lbm DA m₁ = 0.018 lbm m₂ = 0.00791 m3 = 0.010 lbm H₂O condensed 0.010 lbm condensed 0.018 lbm fed forate on the us psychso. h₂ = 0.0079 lbm H₂O/lb DAN Ĥ₂ = 20.9 Btu/lbm DA 0.555 = 0.018 lbm H₂O chart DADocate on is psychro. chart 12 In assuming this basis, we are temporarily ignoring the specification of the volumetric flow rate at the outlet. After the process balanced for the assumed basis, we will scale up to an outlet flow rate of 1000 ft³/min. o = 0.0079 lbm H₂O {" @ 683 ENG Sign in 00 : 6:15 PM 5/28/2023
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