THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
8th Edition
ISBN: 9781307434316
Author: CENGEL
Publisher: INTER MCG
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Chapter 14.7, Problem 113RP

(a)

To determine

The molar analysis of the saturated air.

(a)

Expert Solution
Check Mark

Answer to Problem 113RP

The molar analysis of water is 0.0313, nitrogen is 0.7566, oxygen is 0.2025 and argon is 0.0097.

Explanation of Solution

Express the pressure of air.

Pair=PPv=PPsat@25°C (I)

Here, the saturation pressure at temperature of 25°C is Psat@25°C, vapor pressure is Pv and atmospheric pressure is P.

Express the molar fraction of water.

yH2O=PvP=Psat@25°CP (II)

Here, pressure of water is PH2O.

Express the molar fraction of nitrogen.

yN2=PN2P=yN2,dryPairP (III)

Here, pressure of nitrogen is PN2 and mass fraction of dry nitrogen is yN2,dry.

Express the molar fraction of oxygen.

yN2=PO2P=yO2,dryPairP (IV)

Here, pressure of oxygen is PO2 and mass fraction of dry oxygen is yO2,dry.

Express the molar fraction of argon.

yAr=PArP=yAr,dryPairP (V)

Here, pressure of oxygen is PO2 and mass fraction of dry oxygen is yO2,dry.

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure at temperature of 25°C.

Psat@25°C=3.1698kPa

Perform unit conversion of atmospheric pressure from atmtokPa.

P=1atm=1atm[101.325kPaatm]=101.325kPa

Substitute 101.325kPa for P and 3.1698kPa for Psat@25°C in Equation (I).

Pair=101.325kPa3.1698kPa=98.155kPa

Substitute 101.325kPa for P and 3.1698kPa for Psat@25°C in Equation (II).

yH2O=3.1698kPa101.325kPa=0.0313

Hence, the molar analysis of water is 0.0313.

Substitute 0.781 for yN2,dry, 98.155kPa for Pair and 101.325kPa for P in Equation (III).

yN2=(0.781)(98.155kPa)101.325kPa=0.7566

Hence, the molar analysis of nitrogen is 0.7566.

Substitute 0.209 for yO2,dry, 98.155kPa for Pair and 101.325kPa for P in Equation (IV).

yO2=(0.209)(98.155kPa)101.325kPa=0.2025

Hence, the molar analysis of oxygen is 0.2025.

Substitute 0.01 for yAr,dry, 98.155kPa for Pair and 101.325kPa for P in Equation (V).

yAr=(0.01)(98.155kPa)101.325kPa=0.0097

Hence, the molar analysis of argon is 0.0097.

(b)

To determine

The density of air before and after the process.

(b)

Expert Solution
Check Mark

Answer to Problem 113RP

The density of air before and after the process is 1.186kg/m3and1.170kg/m3 respectively.

Explanation of Solution

Express the molar mass of dry air.

Mdryair=yiMi=yN2,dryMN2+yO2,dryMO2+yAr,dryMAr (VI)

Here, total molar fraction is yi, total molar mass is Mi, molar mass of nitrogen is MN2, molar mass of oxygen is MO2 and molar mass of argon is MAr.

Express the molar mass of saturated air.

Msatair=yiMi=yN2MN2+yO2MO2+yArMAr+yH2OMH2O (VII)

Here, molar mass of water vapor is H2O.

Express the density of air before the process.

ρdryair=P(Ra/Mdryair)T (VIII)

Here, gas constant of air is Ra.

Express the density of air after the process.

ρsatair=P(Ra/Msatair)T (IX)

Here, gas constant of air is Ra.

Conclusion:

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses.

MN2=28kg/kmolMAr=39.9kg/kmolMO2=32kg/kmolMH2O=18kg/kmol

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the universal gas constant of air.

Ra=8.314kPam/kmolK

Substitute 0.781 for yN2,dry, 28kg/kmol for MN2, 39.9kg/kmol for MAr, 32kg/kmol for MO2, 0.209 for yO2,dry and 0.01 for yAr,dry in Equation (VI).

Mdryair=(0.781)(28)+(0.209)(32)+(0.01)(39.9)=29kg/kmol

Substitute 0.7566 for yN2, 28kg/kmol for MN2, 39.9kg/kmol for MAr, 32kg/kmol for MO2, 0.2025 for yO2, 0.0097 for yAr, 18kg/kmol for MH2O and 0.0313 for yH2O in Equation (VII).

Msatair=(0.7566)(28)+(0.2025)(32)+(0.0097)(39.9)+(0.0313)(18)=28.62kg/kmol

Perform the unit conversion of temperature from °CtoK.

T=25°C=(25+273)K=298K

Substitute 101.325kPa for P, 8.314kPam/kmolK for Ra, 29kg/kmol for Mdryair and 298K for T in Equation (VIII).

ρdryair=101.325kPa(8.314kPam/kmolK/29kg/kmol)(298K)=1.186kg/m3 (X)

Hence, the density of air before the process is 1.186kg/m3.

Substitute 101.325kPa for P, 8.314kPam/kmolK for Ra, 28.62kg/kmol for Msatair and 298K for T in Equation (IX).

ρsatair=101.325kPa(8.314kPam/kmolK/28.62kg/kmol)(298K)=1.170kg/m3 (XI)

Hence, the density of air after the process is 1.170kg/m3.

From the result obtained in Equations (X) and (XI), It is obtained that the density of dry air is larger than that of saturated air as the molar mass of dry air being larger than that of water.

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Chapter 14 Solutions

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

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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