EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220100257056
Author: CENGEL
Publisher: YUZU
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Chapter 14.7, Problem 112P
To determine

The exergy destruction per unit mass of dry air.

Expert Solution & Answer
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Answer to Problem 112P

The exergy destruction per unit mass of dry air is 0.726Btu/bm dry air.

Explanation of Solution

Apply the dry air mass balance on the cooling tower.

m˙a1=m˙a2=m˙a

Here, mass flow rate of air at inlet and outlet is m˙a1andm˙a2, and mass flow rate of liquid water is m˙a.

Apply the water mass balance on the cooling tower.

m˙3+m˙a1ω1=m˙4+m˙a2ω2m˙3m˙4=m˙a(ω2ω1)m˙3m˙4=m˙makeup (I)

Here, mass flow rate of water at state 3 and 4 is m˙3 and m˙4, and mass flow rate of required makeup water is m˙makeup.

Apply the energy balance on the cooling tower.

inm˙h=outm˙hm˙a1h1+m˙3h3=m˙a1h1+m˙4h4m˙3h3=m˙a(h2h1)+(m˙3m˙makeup)h4m˙a=m˙3(h3h4)(h2h1)(ω2ω1)h4 (II)

Write the expression to obtain the mass of water stream at state 3 (m3).

m3=m˙3m˙a (III)

Write the expression to obtain the mass of water stream at state 4 (m4).

m4=m3(ω2ω1) (IV)

Write the expression to obtain the entropy change of water stream (Δswater).

 Δswater=m4s4m3s3 (V)

Write the expression to obtain the entropy change of water vapor in the air stream (Δsvapor).

 Δsvapor=ω2sg2ω1sg1 (VI)

Write the expression to obtain the vapor pressure at inlet conditions (Pv1).

Pv1=ϕ1Pg1=ϕ1Psat@65°F (VII)

Here, saturation pressure of water at 65°F is Pg1 and relative humidity at state 1 is ϕ1.

Write the expression of the atmospheric pressure of an ideal gas mixture (P1).

P1=Pa1+Pv1 (VIII)

Here, dry air partial pressure at state 1 is Pa1.

Write the expression to obtain the vapor pressure at second inlet conditions (Pv2).

Pv2=ϕ2Pg2=ϕ2Psat@75°F (IX)

Here, saturation pressure of water at 75°F is Pg2 and relative humidity at state 2 is ϕ2.

Write the expression of the atmospheric pressure of an ideal gas mixture (P2).

P1=Pa1+Pv1 (X)

Here, dry air partial pressure at state 2 is Pa2.

Write the expression to obtain the entropy change of dry air (Δsa)

Δsa=s2s1

Δsa=cpln(T2T1)Rln(Pa2Pa1) (XI)

Here, temperature at state 2 is T2 and temperature at state 1 is T1.

Write the expression for entropy generation in the cooling tower (sgen).

sgen=Δswater+Δsvapor+Δsa (XII)

Write the exergy destruction per unit mass of dry air (xdest).

xdest=T0sgen (XIII)

Conclusion:

Refer Table A-4E, “Saturated water – Temperature table”, obtain the enthalpy (hf@95°F=h3) as 63.04Btu/lbm and entropy (sf@95°F=s3) as 0.12065Btu/lbmR at a temperature of 95°F.

Refer Table A-4E, “Saturated water – Temperature table”, obtain the enthalpy (hf@80°F=h4) as 48.07Btu/lbm and entropy (sf@80°F=s4) as 0.09328Btu/lbmR at a temperature of 80°F.

Refer Fig A-31E, “Psychrometric chart at 1 atm total pressure”, at inlet temperature (T1=65°F), and relative humidity (ϕ1=30%), read the value of inlet enthalpy (h1) as 19.9Btu/lbmdry air, inlet specific humidity (ω1) as 0.00391lbmH2O/lbmdryair, and inlet specific volume of dry air (v1) as 13.31ft3/lbmdryair.

Refer Fig A-31E, “Psychrometric chart at 1 atm total pressure”, at outlet temperature (T2=75°F) and relative humidity (ϕ2=80%), read the value of exit enthalpy (h2) as 34.3Btu/lbmdry air, and exit specific humidity (ω2) as 0.0149lbmH2O/lbmdryair.

Substitute 3lbm/s for m˙3, 63.04Btu/lbm for h3, 48.07Btu/lbm for h4, 19.9Btu/lbmdry air for h1, 34.3Btu/lbmdry air for h2, 0.00391lbmH2O/lbmdryair for ω1, and 0.0149lbmH2O/lbmdryair for ω2 in Equation (II).

m˙a=3lbm/s(63.04Btu/lbm48.07Btu/lbm)(34.3Btu/lbmdry air19.9Btu/lbmdry air)[(0.0149lbmH2O/lbmdryair0.00391lbmH2O/lbmdryair)48.07Btu/lbm]=3.22lbm/s

Substitute 3 lbm water/s for m˙3 and 3.22 lbm dry air/s for m˙a in Equation (III).

m3=3 lbm water/s3.22 lbm dry air/s=0.9317 lbm water/lbm dry air

Substitute 0.9317 lbm water/lbm dry air for m3, 0.00391lbmH2O/lbmdryair for ω1, and 0.0149lbmH2O/lbmdryair for ω2 in Equation (IV).

m4=0.9317 lbm water/lbm dry air(0.01490.00391)lbmH2O/lbmdryair=0.9207lbmwater/lbmdryair

Substitute 0.9317 lbm water/lbm dry air for m3, 0.9207 lbm water/lbm dry air for m4, 0.12065Btu/lbmR for s3 and 0.09328Btu/lbmR for s4 in Equation (V).

 Δswater={(0.9207lbmwater/lbmdryair×0.09328Btu/lbmR)(0.9317lbmwater/lbmdryair×0.12065Btu/lbmR)}=0.02653Btu/Rlbmdryair

Refer Table A-4E, “Saturated water – Temperature table”, obtain the entropy (sg@65°F=sg1) as 2.0788Btu/lbmR at a temperature of 65°F.

Refer Table A-4E, “Saturated water – Temperature table”, obtain the entropy (sg@80°F=sg2) as 2.0352Btu/lbmR at a temperature of 80°F.

Substitute 2.0352Btu/lbmR for sg2, 2.0788Btu/lbmR for sg1, 0.00391lbmH2O/lbmdryair for ω1, and 0.0149lbmH2O/lbmdryair for ω2 in Equation (VI).

 Δsvapor={(0.0149lbmH2O/lbmdryair×2.0352Btu/lbmR)(0.00391lbmH2O/lbmdryair×2.0788Btu/lbmR)}=0.0222 Btu/Rlbm dry air

Refer Table A-4E, “Saturated water – Temperature table”, obtain the properties of water at a temperature of 65°F.

Psat@65°F=0.30578 psia

Substitute 0.30578 psia for Psat@65°F and 0.3 for ϕ1 in Equation (VII).

Pv1=(0.3)(0.30578 psia)=0.0917 psia

Rewrite Equation (VIII) and substitute 14.696 psia for P1 and 0.0917 psia for Pv1.

Pa1=P1Pv1=14.696 psia0.0917 psia=14.60 psia

Refer Table A-4E, “Saturated water – Temperature table”, obtain the properties of water at a temperature of 75°F.

Psat@75°F=0.43016 psia

Substitute 0.43016 psia for Psat@75°F and 0.80 for ϕ2 in Equation (IX).

Pv2=(0.80)(0.43016 psia)=0.3441 psia

Rewrite Equation (X) and substitute 14.696 psia for P2 and 0.3441 psia for Pv2.

Pa2=P2Pv2=14.696 psia0.3441 psia=14.35 psia

Refer Table A-2E, “Ideal gas specific heats of various common gases”, obtain the value of cp and R for air as 0.240Btu/lbmR and 0.06855Btu/lbmR respectively.

Substitute 0.240Btu/lbmR for cp, 0.06855Btu/lbmR for R, 75°F for T2, 65°F for T1, 14.35 psia for Pa2 and 14.60 psia for Pa1 in Equation (XI).

Δsa={0.240Btu/lbmRln((75°F+460)R(65°F+460)R)0.06855Btu/lbmRln(14.35 psia14.60 psia)}=0.005712Btu/lbmR dry air

Substitute 0.005712Btu/Rlbm dry air for Δsa, 0.0222 Btu/Rlbm dry air for Δsvapor and 0.02653Btu/Rlbmdryair for Δswater in Equation (XII).

sgen={0.02653Btu/Rlbm dry air+0.02220Btu/Rlbm dry air+0.005712Btu/Rlbm dry air}=0.001382Btu/Rlbm dry air

Substitute 65°F for T0, and 0.001382Btu/Rlbm dry air for sgen in Equation (XIII).

xdest=(65°F+460)R×0.001382Btu/Rlbm dry air=0.726Btu/bm dry air

Thus the exergy destruction per unit mass of dry air is 0.726Btu/bm dry air.

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Chapter 14 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 14.7 - A tank contains 15 kg of dry air and 0.17 kg of...Ch. 14.7 - Prob. 12PCh. 14.7 - Prob. 13PCh. 14.7 - 14–13 A room contains air at 20°C and 98 kPa at a...Ch. 14.7 - A room contains air at 85F and 13.5 psia at a...Ch. 14.7 - An 8-m3 tank contains saturated air at 30C, 105...Ch. 14.7 - Prob. 17PCh. 14.7 - Prob. 18PCh. 14.7 - Prob. 19PCh. 14.7 - Andy and Wendy both wear glasses. On a cold winter...Ch. 14.7 - In summer, the outer surface of a glass filled...Ch. 14.7 - In some climates, cleaning the ice off the...Ch. 14.7 - Prob. 23PCh. 14.7 - Prob. 24PCh. 14.7 - Prob. 25PCh. 14.7 - Prob. 26PCh. 14.7 - A thirsty woman opens the refrigerator and picks...Ch. 14.7 - Prob. 28PCh. 14.7 - The air in a room has a dry-bulb temperature of...Ch. 14.7 - Prob. 31PCh. 14.7 - Prob. 32PCh. 14.7 - How do constant-enthalpy and...Ch. 14.7 - At what states on the psychrometric chart are the...Ch. 14.7 - How is the dew-point temperature at a specified...Ch. 14.7 - Can the enthalpy values determined from a...Ch. 14.7 - Prob. 37PCh. 14.7 - Prob. 39PCh. 14.7 - Prob. 41PCh. 14.7 - Prob. 42PCh. 14.7 - Prob. 43PCh. 14.7 - Prob. 44PCh. 14.7 - What does a modern air-conditioning system do...Ch. 14.7 - How does the human body respond to (a) hot...Ch. 14.7 - Prob. 47PCh. 14.7 - How does the air motion in the vicinity of the...Ch. 14.7 - Consider a tennis match in cold weather where both...Ch. 14.7 - Prob. 50PCh. 14.7 - Prob. 51PCh. 14.7 - Prob. 52PCh. 14.7 - What is metabolism? What is the range of metabolic...Ch. 14.7 - What is sensible heat? How is the sensible heat...Ch. 14.7 - Prob. 55PCh. 14.7 - Prob. 56PCh. 14.7 - Prob. 57PCh. 14.7 - Prob. 58PCh. 14.7 - Repeat Prob. 1459 for an infiltration rate of 1.8...Ch. 14.7 - An average person produces 0.25 kg of moisture...Ch. 14.7 - An average (1.82 kg or 4.0 lbm) chicken has a...Ch. 14.7 - How do relative and specific humidities change...Ch. 14.7 - Prob. 63PCh. 14.7 - Prob. 64PCh. 14.7 - Prob. 65PCh. 14.7 - Humid air at 40 psia, 50F, and 90 percent relative...Ch. 14.7 - Air enters a 30-cm-diameter cooling section at 1...Ch. 14.7 - Prob. 68PCh. 14.7 - Prob. 69PCh. 14.7 - Why is heated air sometimes humidified?Ch. 14.7 - Air at 1 atm, 15C, and 60 percent relative...Ch. 14.7 - Prob. 72PCh. 14.7 - An air-conditioning system operates at a total...Ch. 14.7 - Prob. 74PCh. 14.7 - Why is cooled air sometimes reheated in summer...Ch. 14.7 - Prob. 76PCh. 14.7 - Prob. 77PCh. 14.7 - Air enters a 40-cm-diameter cooling section at 1...Ch. 14.7 - Repeat Prob. 1479 for a total pressure of 88 kPa...Ch. 14.7 - Prob. 81PCh. 14.7 - Prob. 83PCh. 14.7 - Prob. 84PCh. 14.7 - Prob. 85PCh. 14.7 - Atmospheric air at 1 atm, 32C, and 95 percent...Ch. 14.7 - Prob. 88PCh. 14.7 - Prob. 89PCh. 14.7 - Does an evaporation process have to involve heat...Ch. 14.7 - Prob. 93PCh. 14.7 - Prob. 94PCh. 14.7 - Air at 1 atm, 20C, and 70 percent relative...Ch. 14.7 - Two unsaturated airstreams are mixed...Ch. 14.7 - Consider the adiabatic mixing of two airstreams....Ch. 14.7 - Prob. 98PCh. 14.7 - Two airstreams are mixed steadily and...Ch. 14.7 - A stream of warm air with a dry-bulb temperature...Ch. 14.7 - Prob. 104PCh. 14.7 - How does a natural-draft wet cooling tower work?Ch. 14.7 - What is a spray pond? How does its performance...Ch. 14.7 - The cooling water from the condenser of a power...Ch. 14.7 - Prob. 108PCh. 14.7 - A wet cooling tower is to cool 60 kg/s of water...Ch. 14.7 - Prob. 110PCh. 14.7 - Prob. 111PCh. 14.7 - Prob. 112PCh. 14.7 - Prob. 113RPCh. 14.7 - Prob. 114RPCh. 14.7 - Prob. 115RPCh. 14.7 - Prob. 116RPCh. 14.7 - Prob. 117RPCh. 14.7 - Prob. 118RPCh. 14.7 - Prob. 119RPCh. 14.7 - Prob. 120RPCh. 14.7 - 14–121 The relative humidity inside dacha of Prob....Ch. 14.7 - Prob. 122RPCh. 14.7 - Prob. 124RPCh. 14.7 - 14–126E Air at 15 psia, 60°F, and 70 percent...Ch. 14.7 - Prob. 127RPCh. 14.7 - Air enters a cooling section at 97 kPa, 35C, and...Ch. 14.7 - Prob. 129RPCh. 14.7 - Humid air at 101.3 kPa, 36C dry bulb and 65...Ch. 14.7 - 14–131 Air enters an air-conditioning system that...Ch. 14.7 - Prob. 132RPCh. 14.7 - Prob. 133RPCh. 14.7 - Conditioned air at 13C and 90 percent relative...Ch. 14.7 - Prob. 138RPCh. 14.7 - A room is filled with saturated moist air at 25C...Ch. 14.7 - Prob. 141FEPCh. 14.7 - A 40-m3 room contains air at 30C and a total...Ch. 14.7 - Prob. 143FEPCh. 14.7 - The air in a house is at 25C and 65 percent...Ch. 14.7 - On the psychrometric chart, a cooling and...Ch. 14.7 - On the psychrometric chart, a heating and...Ch. 14.7 - An airstream at a specified temperature and...Ch. 14.7 - Prob. 148FEPCh. 14.7 - Air at a total pressure of 90 kPa, 15C, and 75...
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