Ideal Gas Mixtures and Psychrometrics

Abstract

This chapter covers the fundamentals and applications of ideal gas mixtures with particular emphasis on air-water vapor mixtures. The basics, key definitions, and laws related to ideal gas mixtures are covered in Sect. 15.1. Section 15.2 focuses exclusively on air-water vapor mixtures, including properties, definitions, and the use of psychrometric charts. Air-conditioning processes, including dehumidification, heating, and cooling, are covered in Sect. 15.3. Section 15.4 addresses the use of mass and energy balances for cooling towers. Mixing of air streams is analyzed in Sect. 15.5. Section 15.6 illustrates the use and applications of psychrometric formulas.

Appendices

Practice Problems

Practice Problem 15.1

2 lbm of carbon dioxide escapes into a lab space with dimensions of 10 ft × 20 ft × 15 ft. The air in the lab space is at 14 psia and 70 °F. Determine:

• A. The mass of air in the lab space before contamination

• B. The mole fraction of carbon dioxide in the lab space

• C. The parts per million (ppm) of carbon dioxide in lab air on a volumetric basis

• D. The final pressure in the chamber assuming the temperature to be constant

Practice Problem 15.2

In atmospheric air at 100 kPa and 30 °C, the partial pressure of water vapor is 1.3 kPa. Calculate:

• A. The partial pressure of dry air

• B. The relative humidity of atmospheric air

Practice Problem 15.3

An air-water vapor mixture at 14 psia and 65 °F has 1% water vapor by mass. Calculate the relative humidity of this mixture.

Practice Problem 15.4

Air is at 30 °C, dry bulb, and 20 °C wet bulb. Determine:

• A. The humidity ratio

• B. The relative humidity

• C. The dew point temperature

• D. The enthalpy per kilogram mass of dry air

• E. The specific volume

Practice Problem 15.5

A cold surface at 45 °F is in contact with air at 65 °F dry bulb and 70% relative humidity. Will there be a condensate film on the cold surface? Justify your answer.

Practice Problem 15.6

Outdoor air flowing at 150 m³/min has 50% relative humidity and is at 5 °C dry bulb temperature. Determine the rate at which moisture needs to be added to outdoor air to make it compatible with air inside a facility where the conditions are 20 °C and 70% relative humidity.

Solutions to Practice Problems

Practice Problem 15.1

• (Solution)

• A. Calculate the volume of the lab space, V = 10 ft × 20 ft × 15 ft = 3000 ft³

Calculate the individual gas constant for air (M_air = 29 lbm/lbmol) using Eq. 13.16.

$$ {R}_{\mathrm{air}}=\frac{\overline{R}}{M_{\mathrm{air}}}=\frac{10.73\frac{\mathrm{psia}-{\mathrm{ft}}^3}{\mathrm{lbm}\mathrm{ol}-{}^{\circ}\mathrm{R}}}{29\frac{\mathrm{lbm}}{\mathrm{lbm}\mathrm{ol}}}=0.37\ \mathrm{psia}-{\mathrm{ft}}^3/\mathrm{lbm}-{}^{\circ}\mathrm{R} $$

Calculate the absolute temperature of air by using Eq. 13.2.

$$ T=460{}^{\circ}+70{}^{\circ}\mathrm{F}=530{}^{\circ}\mathrm{R}=\mathrm{constant} $$

Calculate the mass of air in the lab space prior to contamination using the ideal gas law (Eq. 13.12).

$$ {m}_{\mathrm{air}}=\frac{PV}{R_{\mathrm{air}}T}=\frac{\left(14\ \mathrm{psia}\right)\left(3000\ {\mathrm{ft}}^3\right)}{\left(0.37\frac{\mathrm{psia}-{\mathrm{ft}}^3}{\mathrm{lbm}-{}^{\circ}\mathrm{R}}\right)\left(530{}^{\circ}\mathrm{R}\right)}=214.18\ \mathrm{lbm} $$

• B. Calculate the moles of carbon dioxide by using Eq. 13.13. From Table 13.1, the molecular weight of carbon dioxide is 44 lbm/lbmol.

$$ {N}_{{\mathrm{CO}}_2}=\frac{m_{{\mathrm{CO}}_2}}{M_{{\mathrm{CO}}_2}}=\frac{2\ \mathrm{lbm}}{44\ \frac{\mathrm{lbm}}{\mathrm{lbm}\mathrm{ol}}}=0.0454\ \mathrm{lbm}\mathrm{ol} $$

Calculate the mols of air by using Eq. 13.13. From Table 13.1, the molecular weight of air is 29 lbm/lbmol.

$$ {N}_{\mathrm{air}}=\frac{m_{\mathrm{air}}}{M_{\mathrm{air}}}=\frac{214.18\ \mathrm{lbm}}{29\frac{\mathrm{lbm}}{\mathrm{lbm}\mathrm{ol}}}=7.38\ \mathrm{kmol} $$

Calculate the mole fraction of carbon dioxide in the lab space using Eq. 15.2

$$ {y}_{{\mathrm{CO}}_2}=\frac{N_{{\mathrm{CO}}_2}}{N}=\frac{N_{{\mathrm{CO}}_2}}{N_{{\mathrm{CO}}_2}+{N}_{\mathrm{air}}}=\frac{0.0454\ \mathrm{lbmol}}{0.0454\ \mathrm{lbmol}+7.38\ \mathrm{lbmol}}=0.0061 $$

• C. Parts per million on a volumetric basis is one volume of carbon dioxide per million volumes of air, carbon dioxide mixture. For a mixture of ideal gases, volume fraction = mole fraction (Eq. 15.5). From the results for part B,

$$ {\displaystyle \begin{array}{l}{\mathrm{PPM}}_{{\mathrm{CO}}_2}=\frac{0.00611\ \mathrm{lbmol}\ {\mathrm{CO}}_2}{1\ \mathrm{lbmol}\ {\mathrm{CO}}_2,\mathrm{air}\ \mathrm{mixture}}=\frac{0.00611\ \mathrm{vol}\ {\mathrm{CO}}_2}{1\ \mathrm{vol}\ {\mathrm{CO}}_2,\mathrm{air}\ \mathrm{mixture}}\times \frac{10^6}{10^6}\\ {}\kern14.25em =6110\ \mathrm{ppm}\end{array}} $$

Practice Problem 15.2

• (Solution)

• A. Since atmospheric pressure is the sum of partial pressures of water vapor and dry air (Eq. 15.8), calculate the partial pressure of dry air as shown.

$$ {P}_{\mathrm{a}}=P-{P}_{\mathrm{w}}=100\ \mathrm{kPa}-1.3\ \mathrm{kPa}=98.7\ \mathrm{kPa} $$

• B. From steam tables, at 30 °C, the saturation pressure is P_sat = 0.0042 MPa = 4.2 kPa

Calculate the relative humidity of atmospheric air using Eq. 15.10.

$$ \phi =\frac{P_{\mathrm{w}}}{P_{\mathrm{sat}}\ \mathrm{at}\ {T}_{\mathrm{DB}}\ \mathrm{of}\ \mathrm{mixture}}=\frac{1.3\ \mathrm{kPa}}{4.2\ \mathrm{kPa}}=0.31\ \left(31\%\right) $$

Practice Problem 15.3

• (Solution)

Since the mass fraction of water vapor is 0.01, the mixture will have 0.01 lbm water/lbm air-water vapor mixture. Calculate the humidity ratio using Eq. 15.9.

$$ \omega =\frac{0.01\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{lbm}\ \mathrm{mixture}}=\frac{0.01\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}}{0.90\ \mathrm{lbm}\ \mathrm{dry}\ \mathrm{air}}=0.0111\ \mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}/\mathrm{lbm}\ \mathrm{dry}\ \mathrm{air} $$

Solve Eq. 15.11 for the partial pressure of water vapor, P_w, and substitute the known values.

$$ {P}_{\mathrm{w}}=\frac{\omega P}{0.622+\omega }=\frac{\left(0.0111\frac{\mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{lbm}\ \mathrm{dry}\ \mathrm{air}}\right)\left(14\ \mathrm{psia}\right)}{0.622\frac{\mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{lbm}\ \mathrm{dry}\ \mathrm{air}}+0.0111\frac{\mathrm{lbm}\ {\mathrm{H}}_2\mathrm{O}}{\mathrm{lbm}\ \mathrm{dry}\ \mathrm{air}}}=0.246\ \mathrm{psia} $$

From steam tables, at 65 °F, the saturation pressure is P_sat = 0.315 psia. Calculate the relative humidity of the mixture using Eq. 15.10.

$$ \phi =\frac{P_{\mathrm{w}}}{P_{\mathrm{sat}}\ \mathrm{at}\ {T}_{\mathrm{DB}}\ \mathrm{of}\ \mathrm{mixture}}=\frac{0.246\ \mathrm{psia}}{0.315\ \mathrm{psia}}=0.78\ \left(78\%\right) $$

Practice Problem 15.4

• (Solution)

Locate the state point of moist air at the intersection of 30 °C dry bulb vertical line and 20 °C wet bulb inclined line as shown in the figure. Using the state point as a reference, determine all the properties of moist air as shown.

• A. ω = 0.011 kg water/kg dry air

• B. RH = 40%

• C. T_DP = 15 °C

• D. h = 58 kJ/kg dry air

• E. v = 0.87 m³/kg dry air

Practice Problem 15.5

• (Solution)

The water vapor in the air will condense on the cold surface if the dew point of the moist air is below 45 °F, which is the temperature of cold surface. Using the psychrometric chart, determine the dew point of the moist air as shown in the figure. First, locate the state point at the intersection of 65 °F dry bulb vertical line and the 70% relative humidity curve. Move horizontally left from the state point to the saturation curve to determine the dew point.

From the figure, the dew point is 56 °F. Since the dew point is greater than the surface temperature of 45 °F, there will be no condensation.

Practice Problem 15.6

• (Solution)

Locate the state points of the outdoor air (5 °C, 50% rh) and the inside air (20 °C, 70% rh) as shown on the psychrometric chart. Determine the moisture content at the state points as shown.

Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1.

$$ {\dot{m}}_{\mathrm{a}}=\frac{\dot{V}}{\nu_1}=\frac{150\frac{{\mathrm{m}}^3}{\min }}{0.79\frac{{\mathrm{m}}^3}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}}=190\ \mathrm{kg}/\min $$

Mass balance for water results in the following equations.

$$ \mathrm{Mass}\ \mathrm{flow}\ \mathrm{of}\ \mathrm{water}\ \mathrm{in}=\mathrm{mass}\ \mathrm{flow}\ \mathrm{of}\ \mathrm{water}\ \mathrm{out} $$

$$ {\dot{m}}_{\mathrm{a}}{\omega}_1+{\dot{m}}_{\mathrm{moisture}}={\dot{m}}_{\mathrm{a}}{\omega}_2 $$

Solve the preceding equation for the mass flow rate of moisture to be added, and substitute the known values.

$$ {\displaystyle \begin{array}{l}{\dot{m}}_{\mathrm{moisture}}={\dot{m}}_{\mathrm{a}}\left({\omega}_2-{\omega}_1\right)\\ {}\kern1em =\left(190\frac{\mathrm{kg}}{\min}\right)\left(0.01\frac{\mathrm{kg}\ \mathrm{water}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}-0.003\frac{\mathrm{kg}\ \mathrm{water}}{\mathrm{kg}\ \mathrm{d}\ \mathrm{a}}\right)\\ {}\kern1.25em =1.33\ \mathrm{kg}\ \mathrm{moisture}/\min \end{array}} $$