This preview shows page 1. Sign up to view the full content.
Unformatted text preview: cen84959_ch14.qxd 9/15/06 6:05 AM Page 737 Chapter 14
GAS–VAPOR MIXTURES AND AIRCONDITIONING A t temperatures below the critical temperature, the gas phase of a substance is frequently referred to as a vapor. The term vapor implies a gaseous state that is close to the saturation region of the substance, raising the possibility of condensation during a process. In Chap. 13, we discussed mixtures of gases that are usually above their critical temperatures. Therefore, we were not concerned about any of the gases condensing during a process. Not having to deal with two phases greatly simplified the analysis. When we are dealing with a gas–vapor mixture, however, the vapor may condense out of the mixture during a process, forming a twophase mixture. This may complicate the analysis considerably. Therefore, a gas–vapor mixture needs to be treated differently from an ordinary gas mixture. Several gas–vapor mixtures are encountered in engineering. In this chapter, we consider the air–watervapor mixture, which is the most commonly encountered gas–vapor mixture in practice. We also discuss airconditioning, which is the primary application area of air–watervapor mixtures. Objectives
The objectives of Chapter 14 are to: • Differentiate between dry air and atmospheric air. • Define and calculate the specific and relative humidity of atmospheric air. • Calculate the dewpoint temperature of atmospheric air. • Relate the adiabatic saturation temperature and wetbulb temperatures of atmospheric air. • Use the psychrometric chart as a tool to determine the properties of atmospheric air. • Apply the principles of the conservation of mass and energy to various airconditioning processes.  737 cen84959_ch14.qxd 9/15/06 6:05 AM Page 738 738  Thermodynamics 14–1 DRY AND ATMOSPHERIC AIR DRY AIR T,°C –10 0 10 20 30 40 50 cp ,kJ/kg ·°C 1.0038 1.0041 1.0045 1.0049 1.0054 1.0059 1.0065 Air is a mixture of nitrogen, oxygen, and small amounts of some other gases. Air in the atmosphere normally contains some water vapor (or moisture) and is referred to as atmospheric air. By contrast, air that contains no water vapor is called dry air. It is often convenient to treat air as a mixture of water vapor and dry air since the composition of dry air remains relatively constant, but the amount of water vapor changes as a result of condensation and evaporation from oceans, lakes, rivers, showers, and even the human body. Although the amount of water vapor in the air is small, it plays a major role in human comfort. Therefore, it is an important consideration in airconditioning applications. The temperature of air in airconditioning applications ranges from about 10 to about 50°C. In this range, dry air can be treated as an ideal gas with a constant cp value of 1.005 kJ/kg · K [0.240 Btu/lbm · R] with negligible error (under 0.2 percent), as illustrated in Fig. 14–1. Taking 0°C as the reference temperature, the enthalpy and enthalpy change of dry air can be determined from
hdry air cpT 1 1.005 kJ> kg # °C 2 T 1 kJ> kg 2
(14–1a) FIGURE 14–1 The cp of air can be assumed to be constant at 1.005 kJ/kg · °C in the temperature range 10 to 50°C with an error under 0.2 percent. and
¢ h dry air cp ¢ T 1 1.005 kJ> kg # °C 2 ¢ T 1 kJ> kg 2 (14–1b) T, °C where T is the air temperature in °C and T is the change in temperature. In airconditioning processes we are concerned with the changes in enthalpy h, which is independent of the reference point selected. It certainly would be very convenient to also treat the water vapor in the air as an ideal gas and you would probably be willing to sacrifice some accuracy for such convenience. Well, it turns out that we can have the convenience without much sacrifice. At 50°C, the saturation pressure of water is 12.3 kPa. At pressures below this value, water vapor can be treated as an ideal gas with negligible error (under 0.2 percent), even when it is a saturated vapor. Therefore, water vapor in air behaves as if it existed alone and obeys the idealgas relation Pv RT. Then the atmospheric air can be treated as an idealgas mixture whose pressure is the sum of the partial pressure of dry air* Pa and that of water vapor Pv:
P Pa Pv 1 kPa 2
(14–2) 50 h = const. s The partial pressure of water vapor is usually referred to as the vapor pressure. It is the pressure water vapor would exert if it existed alone at the temperature and volume of atmospheric air. Since water vapor is an ideal gas, the enthalpy of water vapor is a function of temperature only, that is, h h(T ). This can also be observed from the Ts diagram of water given in Fig. A–9 and Fig. 14–2 where the constantenthalpy lines coincide with constanttemperature lines at temperatures FIGURE 14–2 At temperatures below 50°C, the h constant lines coincide with the T constant lines in the superheated vapor region of water. *Throughout this chapter, the subscript a denotes dry air and the subscript v denotes water vapor. cen84959_ch14.qxd 9/15/06 6:05 AM Page 739 Chapter 14
below 50°C. Therefore, the enthalpy of water vapor in air can be taken to be equal to the enthalpy of saturated vapor at the same temperature. That is,
hv 1 T, low P 2 hg 1 T 2
(14–3) WATER VAPOR hg ,kJ/kg T,°C –10 0 10 20 30 40 50 Table A4 2482.1 2500.9 2519.2 2537.4 2555.6 2573.5 2591.3 Eq. 144 2482.7 2500.9 2519.1 2537.3 2555.5 2573.7 2591.9  739 The enthalpy of water vapor at 0°C is 2500.9 kJ/kg. The average cp value of water vapor in the temperature range 10 to 50°C can be taken to be 1.82 kJ/kg · °C. Then the enthalpy of water vapor can be determined approximately from
hg 1 T 2 hg 1 T 2 2500.9 1.82T 1 kJ> kg 2 T in °C
(14–4) Difference, kJ/kg –0.6 0.0 0.1 0.1 0.1 –0.2 –0.6 or 1060.9 0.435T 1 Btu> lbm 2 T in °F (14–5) in the temperature range 10 to 50°C (or 15 to 120°F), with negligible error, as shown in Fig. 14–3. 14–2 SPECIFIC AND RELATIVE HUMIDITY OF AIR
FIGURE 14–3 In the temperature range 10 to 50°C, the hg of water can be determined from Eq. 14–4 with negligible error. The amount of water vapor in the air can be specified in various ways. Probably the most logical way is to specify directly the mass of water vapor present in a unit mass of dry air. This is called absolute or specific humidity (also called humidity ratio) and is denoted by v:
v mv 1 kg water vapor> kg dry air 2 ma PvV> R vT Pv > R v
(14–6) The specific humidity can also be expressed as
v mv ma PaV> R aT Pa > R a 0.622 Pv Pa
(14–7) or
v 0.622Pv 1 kg water vapor> kg dry air 2 P Pv
(14–8) where P is the total pressure. Consider 1 kg of dry air. By definition, dry air contains no water vapor, and thus its specific humidity is zero. Now let us add some water vapor to this dry air. The specific humidity will increase. As more vapor or moisture is added, the specific humidity will keep increasing until the air can hold no more moisture. At this point, the air is said to be saturated with moisture, and it is called saturated air. Any moisture introduced into saturated air will condense. The amount of water vapor in saturated air at a specified temperature and pressure can be determined from Eq. 14–8 by replacing Pv by Pg, the saturation pressure of water at that temperature (Fig. 14–4). The amount of moisture in the air has a definite effect on how comfortable we feel in an environment. However, the comfort level depends more on the amount of moisture the air holds (mv) relative to the maximum amount of moisture the air can hold at the same temperature (mg). The ratio of these two quantities is called the relative humidity f (Fig. 14–5)
f mv mg PgV> R vT PvV> R vT Pv Pg
(14–9) AIR 25 25°C,100 k Pa (Psat,H 2 O @ 25°C = 3.1698 k Pa) Pv = 0 Pv < 3.1698 k Pa Pv = 3.1698 k Pa dry air unsaturated air saturated air FIGURE 14–4 For saturated air, the vapor pressure is equal to the saturation pressure of water. cen84959_ch14.qxd 9/15/06 6:05 AM Page 740 740  Thermodynamics
where
AIR 25°C,1 atm ma = 1 kg kg mv = 0.01 kg 0.01 mv, max = 0.02 kg max Pg Psat @ T (14–10) Combining Eqs. 14–8 and 14–9, we can also express the relative humidity as
f
kg H 2 O kg kg dry air vP and v 1 0.622 v 2 Pg 0.622fPg P fPg (14–11a, b) Specific Specific humidity: ω = 0.01 Relative Relative humidity: φ = 50% FIGURE 14–5 Specific humidity is the actual amount of water vapor in 1 kg of dry air, whereas relative humidity is the ratio of the actual amount of moisture in the air at a given temperature to the maximum amount of moisture air can hold at the same temperature.
(1 + ω) kg of moist air The relative humidity ranges from 0 for dry air to 1 for saturated air. Note that the amount of moisture air can hold depends on its temperature. Therefore, the relative humidity of air changes with temperature even when its specific humidity remains constant. Atmospheric air is a mixture of dry air and water vapor, and thus the enthalpy of air is expressed in terms of the enthalpies of the dry air and the water vapor. In most practical applications, the amount of dry air in the air–watervapor mixture remains constant, but the amount of water vapor changes. Therefore, the enthalpy of atmospheric air is expressed per unit mass of dry air instead of per unit mass of the air–water vapor mixture. The total enthalpy (an extensive property) of atmospheric air is the sum of the enthalpies of dry air and the water vapor:
H Ha Hv m ah a mv h ma v mv hv Dividing by ma gives
h H ma ha ha vh v Dry air 1 kg ha or
moisture ω kg hg h ha vhg 1 kJ> kg dry air 2 (14–12) h = ha + ω hg,kJ/kg dry air since hv hg (Fig. 14–6). Also note that the ordinary temperature of atmospheric air is frequently referred to as the drybulb temperature to differentiate it from other forms of temperatures that shall be discussed. FIGURE 14–6 The enthalpy of moist (atmospheric) air is expressed per unit mass of dry air, not per unit mass of moist air. EXAMPLE 14–1 The Amount of Water Vapor in Room Air A 5m 5m 3m room shown in Fig. 14–7 contains air at 25°C and 100 kPa at a relative humidity of 75 percent. Determine (a) the partial pressure of dry air, (b) the specific humidity, (c) the enthalpy per unit mass of the dry air, and (d ) the masses of the dry air and water vapor in the room.
ROOM 5m×5m×3m T = 25°C P = 100 kPa φ = 75% Solution The relative humidity of air in a room is given. The dry air pressure, specific humidity, enthalpy, and the masses of dry air and water vapor in the room are to be determined. Assumptions The dry air and the water vapor in the room are ideal gases. Properties The constantpressure specific heat of air at room temperature is cp 1.005 kJ/kg · K (Table A–2a). For water at 25°C, we have Tsat 3.1698 kPa and hg 2546.5 kJ/kg (Table A–4). FIGURE 14–7 Schematic for Example 14–1. cen84959_ch14.qxd 9/15/06 6:05 AM Page 741 Chapter 14
Analysis (a) The partial pressure of dry air can be determined from Eq. 14–2:  741 Pa
where P fPg 1 100 Pv fPsat @ 25°C 2.38 2 kPa 1 0.75 2 1 3.1698 kPa 2 97.62 kPa 2.38 kPa Pv
Thus, Pa (b) The specific humidity of air is determined from Eq. 14–8: v 0.622Pv P Pv 1 0.622 2 1 2.38 kPa 2 1 100 2.38 2 kPa 0.0152 kg H2O/kg dry air (c) The enthalpy of air per unit mass of dry air is determined from Eq. 14–12: h ha 1 1.005 kJ> kg # °C 2 1 25°C 2 vh v cpT vh g 1 0.0152 2 1 2546.5 kJ> kg 2 63.8 kJ/kg dry air
The enthalpy of water vapor (2546.5 kJ/kg) could also be determined from the approximation given by Eq. 14–4: hg @ 25°C 2500.9 1.82 1 25 2 2546.4 kJ> kg which is almost identical to the value obtained from Table A–4. (d ) Both the dry air and the water vapor fill the entire room completely. Therefore, the volume of each gas is equal to the volume of the room: Va Vv Vroom 15 m 2 15 m 2 13 m 2 75 m3 The masses of the dry air and the water vapor are determined from the idealgas relation applied to each gas separately: ma mv PaVa R aT PvVv RvT 1 0.287 kPa # m3> kg # K 2 1 298 K 2 1 0.4615 kPa # m3> kg # K 2 1 298 K 2 1 0.0152 2 1 85.61 kg 2 1 2.38 kPa 2 1 75 m3 2 1 97.62 kPa 2 1 75 m3 2 85.61 kg 1.30 kg The mass of the water vapor in the air could also be determined from Eq. 14–6: mv vma 1.30 kg 14–3 DEWPOINT TEMPERATURE If you live in a humid area, you are probably used to waking up most summer mornings and finding the grass wet. You know it did not rain the night before. So what happened? Well, the excess moisture in the air simply condensed on the cool surfaces, forming what we call dew. In summer, a considerable amount of water vaporizes during the day. As the temperature falls during the cen84959_ch14.qxd 9/15/06 6:05 AM Page 742 742
T  Thermodynamics
night, so does the “moisture capacity” of air, which is the maximum amount of moisture air can hold. (What happens to the relative humidity during this process?) After a while, the moisture capacity of air equals its moisture content. At this point, air is saturated, and its relative humidity is 100 percent. Any further drop in temperature results in the condensation of some of the moisture, and this is the beginning of dew formation. The dewpoint temperature Tdp is defined as the temperature at which condensation begins when the air is cooled at constant pressure. In other words, Tdp is the saturation temperature of water corresponding to the vapor pressure:
s
ons t. T1 Tdp 2 P v =c 1 Tdp Tsat @ Pv (14–13) FIGURE 14–8 Constantpresssure cooling of moist air and the dewpoint temperature on the Ts diagram of water. MOIST AIR T < Tdp Liquid water droplets (dew) FIGURE 14–9 When the temperature of a cold drink is below the dewpoint temperature of the surrounding air, it “sweats.” This is also illustrated in Fig. 14–8. As the air cools at constant pressure, the vapor pressure Pv remains constant. Therefore, the vapor in the air (state 1) undergoes a constantpressure cooling process until it strikes the saturated vapor line (state 2). The temperature at this point is Tdp, and if the temperature drops any further, some vapor condenses out. As a result, the amount of vapor in the air decreases, which results in a decrease in Pv. The air remains saturated during the condensation process and thus follows a path of 100 percent relative humidity (the saturated vapor line). The ordinary temperature and the dewpoint temperature of saturated air are identical. You have probably noticed that when you buy a cold canned drink from a vending machine on a hot and humid day, dew forms on the can. The formation of dew on the can indicates that the temperature of the drink is below the dewpoint temperature of the surrounding air (Fig. 14–9). The dewpoint temperature of room air can be determined easily by cooling some water in a metal cup by adding small amounts of ice and stirring. The temperature of the outer surface of the cup when dew starts to form on the surface is the dewpoint temperature of the air. EXAMPLE 14–2
COLD OUTDOORS 10°C AIR 20°C, 75% Typical temperature distribution 18°C 16°C 20°C 20°C 20°C 18°C 16°C Fogging of the Windows in a House In cold weather, condensation frequently occurs on the inner surfaces of the windows due to the lower air temperatures near the window surface. Consider a house, shown in Fig. 14–10, that contains air at 20°C and 75 percent relative humidity. At what window temperature will the moisture in the air start condensing on the inner surfaces of the windows? Solution The interior of a house is maintained at a specified temperature
and humidity. The window temperature at which fogging starts is to be determined. Properties The saturation pressure of water at 20°C is Psat 2.3392 kPa (Table A–4). Analysis The temperature distribution in a house, in general, is not uniform. When the outdoor temperature drops in winter, so does the indoor temperature near the walls and the windows. Therefore, the air near the walls and the windows remains at a lower temperature than at the inner parts of a house even though the total pressure and the vapor pressure remain constant throughout the house. As a result, the air near the walls and the windows undergoes a Pv constant cooling process until the moisture in the air FIGURE 14–10 Schematic for Example 14–2. cen84959_ch14.qxd 9/15/06 6:05 AM Page 743 Chapter 14
starts condensing. This happens when the air reaches its dewpoint temperature Tdp, which is determined from Eq. 14–13 to be  743 Tdp
where Tsat @ Pv 1.754 kPa Pv
Thus, fPg @ 20°C 1 0.75 2 1 2.3392 kPa 2 Tsat @ 1.754 kPa 15.4 °C Tdp Discussion Note that the inner surface of the window should be maintained above 15.4°C if condensation on the window surfaces is to be avoided. 14–4 ADIABATIC SATURATION AND WETBULB TEMPERATURES
Unsaturated air T1, ω1 f1 1 Liquid water Saturated air T2, ω2 f2 100% 2 Relative humidity and specific humidity are frequently used in engineering and atmospheric sciences, and it is desirable to relate them to easily measurable quantities such as temperature and pressure. One way of determining the relative humidity is to determine the dewpoint temperature of air, as discussed in the last section. Knowing the dewpoint temperature, we can determine the vapor pressure Pv and thus the relative humidity. This approach is simple, but not quite practical. Another way of determining the absolute or relative humidity is related to an adiabatic saturation process, shown schematically and on a Ts diagram in Fig. 14–11. The system consists of a long insulated channel that contains a pool of water. A steady stream of unsaturated air that has a specific humidity of v1 (unknown) and a temperature of T1 is passed through this channel. As the air flows over the water, some water evaporates and mixes with the airstream. The moisture content of air increases during this process, and its temperature decreases, since part of the latent heat of vaporization of the water that evaporates comes from the air. If the channel is long enough, the airstream exits as saturated air (f 100 percent) at temperature T2, which is called the adiabatic saturation temperature. If makeup water is supplied to the channel at the rate of evaporation at temperature T2, the adiabatic saturation process described above can be analyzed as a steadyflow process. The process involves no heat or work interactions, and the kinetic and potential energy changes can be neglected. Then the conservation of mass and conservation of energy relations for this twoinlet, oneexit steadyflow system reduces to the following: Mass balance:
# ma1 # m w1 # m a2 # mf # ma # m w2 (The mass flow rate of dry air remains constant) (The mass flow rate of vapor in the air increases by an amount equal . to the rate of evaporation mf ) # mf # mav2 Liquid water at T2 T Adiabatic saturation temperature
Pv1 1 2 Dewpoint temperature s FIGURE 14–11 The adiabatic saturation process and its representation on a Ts diagram of water. or
# mav1 cen84959_ch14.qxd 9/15/06 6:05 AM Page 744 744  Thermodynamics
Thus,
# mf # ma 1 v2 v1 2 # 0 and W 02 Energy balance:
# Ein # ma h1 # mf hf2 # ma h1 # # Eout 1 since Q # ma h2 # ma 1 v2 1 v2 1 v2 cp 1 T2 v1 2 h f2 v1 2 hf2 v1 2 hf2 T1 2 h g1 h f2 or . Dividing by ma gives
h1 # m a h2 h2 1 cpT2 v2hfg2 v2hg2 2 or which yields 1 cpT1 v1hg1 2 v1 (14–14) where, from Eq. 14–11b,
v2 0.622Pg2 P2 Pg2
(14–15) Ordinary thermometer Wetbulb thermometer Air flow Wick Liquid water FIGURE 14–12 A simple arrangement to measure the wetbulb temperature. since f2 100 percent. Thus we conclude that the specific humidity (and relative humidity) of air can be determined from Eqs. 14–14 and 14–15 by measuring the pressure and temperature of air at the inlet and the exit of an adiabatic saturator. If the air entering the channel is already saturated, then the adiabatic saturation temperature T2 will be identical to the inlet temperature T1, in which case Eq. 14–14 yields v1 v2. In general, the adiabatic saturation temperature is between the inlet and dewpoint temperatures. The adiabatic saturation process discussed above provides a means of determining the absolute or relative humidity of air, but it requires a long channel or a spray mechanism to achieve saturation conditions at the exit. A more practical approach is to use a thermometer whose bulb is covered with a cotton wick saturated with water and to blow air over the wick, as shown in Fig. 14–12. The temperature measured in this manner is called the wetbulb temperature Twb, and it is commonly used in airconditioning applications. The basic principle involved is similar to that in adiabatic saturation. When unsaturated air passes over the wet wick, some of the water in the wick evaporates. As a result, the temperature of the water drops, creating a temperature difference (which is the driving force for heat transfer) between the air and the water. After a while, the heat loss from the water by evaporation equals the heat gain from the air, and the water temperature stabilizes. The thermometer reading at this point is the wetbulb temperature. The wetbulb temperature can also be measured by placing the wetwicked thermometer in a holder attached to a handle and rotating the holder rapidly, that is, by moving the thermometer instead of the air. A device that works cen84959_ch14.qxd 9/15/06 6:05 AM Page 745 Chapter 14
on this principle is called a sling psychrometer and is shown in Fig. 14–13. Usually a drybulb thermometer is also mounted on the frame of this device so that both the wet and drybulb temperatures can be read simultaneously. Advances in electronics made it possible to measure humidity directly in a fast and reliable way. It appears that sling psychrometers and wetwicked thermometers are about to become things of the past. Today, handheld electronic humidity measurement devices based on the capacitance change in a thin polymer film as it absorbs water vapor are capable of sensing and digitally displaying the relative humidity within 1 percent accuracy in a matter of seconds. In general, the adiabatic saturation temperature and the wetbulb temperature are not the same. However, for air–water vapor mixtures at atmospheric pressure, the wetbulb temperature happens to be approximately equal to the adiabatic saturation temperature. Therefore, the wetbulb temperature Twb can be used in Eq. 14–14 in place of T2 to determine the specific humidity of air.  745 Wetbulb thermometer Drybulb thermometer EXAMPLE 14–3 The Specific and Relative Humidity of Air FIGURE 14–13 Sling psychrometer. Wetbulb thermometer wick The dry and the wetbulb temperatures of atmospheric air at 1 atm (101.325 kPa) pressure are measured with a sling psychrometer and determined to be 25 and 15°C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air. Solution Dry and wetbulb temperatures are given. The specific humidity,
relative humidity, and enthalpy are to be determined. Properties The saturation pressure of water is 1.7057 kPa at 15°C, and 3.1698 kPa at 25°C (Table A–4). The constantpressure specific heat of air at room temperature is cp 1.005 kJ/kg · K (Table A–2a). Analysis (a) The specific humidity v1 is determined from Eq. 14–14, v1 cp 1 T2 T1 2 v2hfg2 hf2 hg1 where T2 is the wetbulb temperature and v2 is v2 0.622Pg2 P2 Pg2 0.01065 kg H2O> kg dry air
Thus, 1 101.325 1 0.622 2 1 1.7057 kPa 2 1.7057 2 kPa v1 1 1.005 kJ> kg # °C 2 3 1 15 1 2546.5 25 2 °C 4 62.982 2 kJ> kg 1 0.01065 2 1 2465.4 kJ> kg 2 0.00653 kg H2O/kg dry air
(b) The relative humidity f1 is determined from Eq. 14–11a to be f1 v1P2 1 0.622 v1 2 Pg1 1 0.622 1 0.00653 2 1 101.325 kPa 2 0.00653 2 1 3.1698 kPa 2 0.332 or 33.2% cen84959_ch14.qxd 9/15/06 6:05 AM Page 746 746  Thermodynamics
(c) The enthalpy of air per unit mass of dry air is determined from Eq. 14–12: h1 h a1 1 1.005 kJ> kg # °C 2 1 25°C 2 v1h v1 cpT1 v1hg1 1 0.00653 2 1 2546.5 kJ> kg 2 41.8 kJ/kg dry air
Discussion The previous property calculations can be performed easily using EES or other programs with builtin psychrometric functions. 14–5
10 0% THE PSYCHROMETRIC CHART Drybulb temperature FIGURE 14–14 Schematic for a psychrometric chart. Saturation line 15°C
T wb Tdp = 15°C 15°C FIGURE 14–15 For saturated air, the drybulb, wetbulb, and dewpoint temperatures are identical. The state of the atmospheric air at a specified pressure is completely specified by two independent intensive properties. The rest of the properties can be calculated easily from the previous relations. The sizing of a typical airconditioning system involves numerous such calculations, which may eventually get on the nerves of even the most patient engineers. Therefore, there is clear motivation to computerize calculations or to do these calculations once and to present the data in the form of easily readable charts. Such charts are called psychrometric charts, and they are used extensively in airconditioning applications. A psychrometric chart for a pressure of 1 atm (101.325 kPa or 14.696 psia) is given in Fig. A–31 in SI units and in Fig. A–31E in English units. Psychrometric charts at other pressures (for use at considerably higher elevations than sea level) are also available. The basic features of the psychrometric chart are illustrated in Fig. 14–14. The drybulb temperatures are shown on the horizontal axis, and the specific humidity is shown on the vertical axis. (Some charts also show the vapor pressure on the vertical axis since at a fixed total pressure P there is a onetoone correspondence between the specific humidity v and the vapor pressure Pv , as can be seen from Eq. 14–8.) On the left end of the chart, there is a curve (called the saturation line) instead of a straight line. All the saturated air states are located on this curve. Therefore, it is also the curve of 100 percent relative humidity. Other constant relativehumidity curves have the same general shape. Lines of constant wetbulb temperature have a downhill appearance to the right. Lines of constant specific volume (in m3/kg dry air) look similar, except they are steeper. Lines of constant enthalpy (in kJ/kg dry air) lie very nearly parallel to the lines of constant wetbulb temperature. Therefore, the constantwetbulbtemperature lines are used as constantenthalpy lines in some charts. For saturated air, the drybulb, wetbulb, and dewpoint temperatures are identical (Fig. 14–15). Therefore, the dewpoint temperature of atmospheric air at any point on the chart can be determined by drawing a horizontal line (a line of v constant or Pv constant) from the point to the saturated curve. The temperature value at the intersection point is the dewpoint temperature. The psychrometric chart also serves as a valuable aid in visualizing the airconditioning processes. An ordinary heating or cooling process, for example, appears as a horizontal line on this chart if no humidification or dehumidification is involved (that is, v constant). Any deviation from a horizontal line indicates that moisture is added or removed from the air during the process. φ= co ns t. Tdb = 15°C Specific humidity, ω lin Sa tur ati on e, φ=
Tw
b = co h v=c
onst. ns t. = co
=1
5°C ns t. cen84959_ch14.qxd 9/15/06 6:05 AM Page 747 Chapter 14
EXAMPLE 14–4 The Use of the Psychrometric Chart  747 Consider a room that contains air at 1 atm, 35°C, and 40 percent relative humidity. Using the psychrometric chart, determine (a) the specific humidity, (b) the enthalpy, (c) the wetbulb temperature, (d ) the dewpoint temperature, and (e) the specific volume of the air. Solution The relative humidity of air in a room is given. The specific humidity, enthalpy, wetbulb temperature, dewpoint temperature, and specific volume of the air are to be determined using the psychrometric chart. Analysis At a given total pressure, the state of atmospheric air is completely specified by two independent properties such as the drybulb temperature and the relative humidity. Other properties are determined by directly reading their values at the specified state.
(a) The specific humidity is determined by drawing a horizontal line from the specified state to the right until it intersects with the v axis, as shown in Fig. 14–16. At the intersection point we read
h v 0.0142 kg H2O/kg dry air
Tdp (b) The enthalpy of air per unit mass of dry air is determined by drawing a line parallel to the h constant lines from the specific state until it intersects the enthalpy scale, giving φ= 40 Twb % ω v h 71.5 kJ/kg dry air (c) The wetbulb temperature is determined by drawing a line parallel to the Twb constant lines from the specified state until it intersects the saturation line, giving T = 35°C FIGURE 14–16 Schematic for Example 14–4. Twb 24°C (d ) The dewpoint temperature is determined by drawing a horizontal line from the specified state to the left until it intersects the saturation line, giving Tdp 19.4°C (e) The specific volume per unit mass of dry air is determined by noting the distances between the specified state and the v constant lines on both sides of the point. The specific volume is determined by visual interpolation to be v 0.893 m3/kg dry air Discussion Values read from the psychrometric chart inevitably involve reading errors, and thus are of limited accuracy. 14–6 HUMAN COMFORT AND AIRCONDITIONING Human beings have an inherent weakness—they want to feel comfortable. They want to live in an environment that is neither hot nor cold, neither humid nor dry. However, comfort does not come easily since the desires of the human body and the weather usually are not quite compatible. Achieving comfort requires a constant struggle against the factors that cause discomfort, such as high or low temperatures and high or low humidity. As engineers, it is our duty to help people feel comfortable. (Besides, it keeps us employed.) cen84959_ch14.qxd 9/15/06 6:05 AM Page 748 748  Thermodynamics
It did not take long for people to realize that they could not change the weather in an area. All they can do is change it in a confined space such as a house or a workplace (Fig. 14–17). In the past, this was partially accomplished by fire and simple indoor heating systems. Today, modern airconditioning systems can heat, cool, humidify, dehumidify, clean, and even deodorize the air–in other words, condition the air to peoples’ desires. Airconditioning systems are designed to satisfy the needs of the human body; therefore, it is essential that we understand the thermodynamic aspects of the body. The human body can be viewed as a heat engine whose energy input is food. As with any other heat engine, the human body generates waste heat that must be rejected to the environment if the body is to continue operating. The rate of heat generation depends on the level of the activity. For an average adult male, it is about 87 W when sleeping, 115 W when resting or doing office work, 230 W when bowling, and 440 W when doing heavy physical work. The corresponding numbers for an adult female are about 15 percent less. (This difference is due to the body size, not the body temperature. The deepbody temperature of a healthy person is maintained constant at about 37°C.) A body will feel comfortable in environments in which it can dissipate this waste heat comfortably (Fig. 14–18). Heat transfer is proportional to the temperature difference. Therefore in cold environments, a body loses more heat than it normally generates, which results in a feeling of discomfort. The body tries to minimize the energy deficit by cutting down the blood circulation near the skin (causing a pale look). This lowers the skin temperature, which is about 34°C for an average person, and thus the heat transfer rate. A low skin temperature causes discomfort. The hands, for example, feel painfully cold when the skin temperature reaches 10°C (50°F). We can also reduce the heat loss from the body either by putting barriers (additional clothes, blankets, etc.) in the path of heat or by increasing the rate of heat generation within the body by exercising. For example, the comfort level of a resting person dressed in warm winter clothing in a room at 10°C (50°F) is roughly equal to the comfort level of an identical person doing moderate work in a room at about 23°C ( 10°F). Or we can just cuddle up and put our hands between our legs to reduce the surface area through which heat flows. In hot environments, we have the opposite problem—we do not seem to be dissipating enough heat from our bodies, and we feel as if we are going to burst. We dress lightly to make it easier for heat to get away from our bodies, and we reduce the level of activity to minimize the rate of waste heat generation in the body. We also turn on the fan to continuously replace the warmer air layer that forms around our bodies as a result of body heat by the cooler air in other parts of the room. When doing light work or walking slowly, about half of the rejected body heat is dissipated through perspiration as latent heat while the other half is dissipated through convection and radiation as sensible heat. When resting or doing office work, most of the heat (about 70 percent) is dissipated in the form of sensible heat whereas when doing heavy physical work, most of the heat (about 60 percent) is dissipated in the form of latent heat. The body helps out by perspiring or sweating more. As this sweat evaporates, it absorbs latent heat from the body and cools it. Perspiration is not much help, however, if the relative humidity of FIGURE 14–17 We cannot change the weather, but we can change the climate in a confined space by airconditioning.
© Vol. 77/PhotoDisc 23°C Waste heat 37°C FIGURE 14–18 A body feels comfortable when it can freely dissipate its waste heat, and no more. cen84959_ch14.qxd 9/15/06 6:05 AM Page 749 Chapter 14
the environment is close to 100 percent. Prolonged sweating without any fluid intake causes dehydration and reduced sweating, which may lead to a rise in body temperature and a heat stroke. Another important factor that affects human comfort is heat transfer by radiation between the body and the surrounding surfaces such as walls and windows. The sun’s rays travel through space by radiation. You warm up in front of a fire even if the air between you and the fire is quite cold. Likewise, in a warm room you feel chilly if the ceiling or the wall surfaces are at a considerably lower temperature. This is due to direct heat transfer between your body and the surrounding surfaces by radiation. Radiant heaters are commonly used for heating hardtoheat places such as car repair shops. The comfort of the human body depends primarily on three factors: the (drybulb) temperature, relative humidity, and air motion (Fig. 14–19). The temperature of the environment is the single most important index of comfort. Most people feel comfortable when the environment temperature is between 22 and 27°C (72 and 80°F). The relative humidity also has a considerable effect on comfort since it affects the amount of heat a body can dissipate through evaporation. Relative humidity is a measure of air’s ability to absorb more moisture. High relative humidity slows down heat rejection by evaporation, and low relative humidity speeds it up. Most people prefer a relative humidity of 40 to 60 percent. Air motion also plays an important role in human comfort. It removes the warm, moist air that builds up around the body and replaces it with fresh air. Therefore, air motion improves heat rejection by both convection and evaporation. Air motion should be strong enough to remove heat and moisture from the vicinity of the body, but gentle enough to be unnoticed. Most people feel comfortable at an airspeed of about 15 m/min. Veryhighspeed air motion causes discomfort instead of comfort. For example, an environment at 10°C (50°F) with 48 km/h winds feels as cold as an environment at 7°C (20°F) with 3 km/h winds as a result of the bodychilling effect of the air motion (the windchill factor). Other factors that affect comfort are air cleanliness, odor, noise, and radiation effect.  749 23°C f = 50% Air motion 15 m/min FIGURE 14–19 A comfortable environment.
© Reprinted with special permission of King Features Syndicate. 14–7 AIRCONDITIONING PROCESSES
Humidifying H Dehumidifying e hu ati m ng id ify and in g Heating Maintaining a living space or an industrial facility at the desired temperature and humidity requires some processes called airconditioning processes. These processes include simple heating (raising the temperature), simple cooling (lowering the temperature), humidifying (adding moisture), and dehumidifying (removing moisture). Sometimes two or more of these processes are needed to bring the air to a desired temperature and humidity level. Various airconditioning processes are illustrated on the psychrometric chart in Fig. 14–20. Notice that simple heating and cooling processes appear as horizontal lines on this chart since the moisture content of the air remains constant (v constant) during these processes. Air is commonly heated and humidified in winter and cooled and dehumidified in summer. Notice how these processes appear on the psychrometric chart. Cooling in ol Co de FIGURE 14–20 Various airconditioning processes. hu m id g an ify d in g cen84959_ch14.qxd 9/15/06 6:05 AM Page 750 750  Thermodynamics
Most airconditioning processes can be modeled as steadyflow processes, . . mout can be expressed for dry air and thus the mass balance relation min and water as
Mass balance for dry air: Mass balance for water: # a ma
in # a m a 1 kg> s 2
out (14–16) # a mw
in # # a mw or a m a v
out in # a ma v
out (14–17) Disregarding the kinetic and . potential energy changes, the steadyflow . energy balance relation Ein Eout can be expressed in this case as
# Q in # Win # a mh
in # Q out # Wout # a mh
out (14–18) The work term usually consists of the fan work input, which is small relative to the other terms in the energy balance relation. Next we examine some commonly encountered processes in airconditioning. Simple Heating and Cooling (V
Heating coils Air T1, ω1, f1 Heat T2 constant) ω2 = ω1 f2 < f1 FIGURE 14–21 During simple heating, specific humidity remains constant, but relative humidity decreases. f2 = 80% 2 v = constant Cooling 12°C f1 = 30% 1 30°C Many residential heating systems consist of a stove, a heat pump, or an electric resistance heater. The air in these systems is heated by circulating it through a duct that contains the tubing for the hot gases or the electric resistance wires, as shown in Fig. 14–21. The amount of moisture in the air remains constant during this process since no moisture is added to or removed from the air. That is, the specific humidity of the air remains constant (v constant) during a heating (or cooling) process with no humidification or dehumidification. Such a heating process proceeds in the direction of increasing drybulb temperature following a line of constant specific humidity on the psychrometric chart, which appears as a horizontal line. Notice that the relative humidity of air decreases during a heating process even if the specific humidity v remains constant. This is because the relative humidity is the ratio of the moisture content to the moisture capacity of air at the same temperature, and moisture capacity increases with temperature. Therefore, the relative humidity of heated air may be well below comfortable levels, causing dry skin, respiratory difficulties, and an increase in static electricity. A cooling process at constant specific humidity is similar to the heating process discussed above, except the drybulb temperature decreases and the relative humidity increases during such a process, as shown in Fig. 14–22. Cooling can be accomplished by passing the air over some coils through which a refrigerant or chilled water flows. The conservation of mass equations for a heating or cooling process that . . . involves no humidification or dehumidification reduce to ma1 ma2 ma for dry air and v1 v2 for water. Neglecting any fan work that may be present, the conservation of energy equation in this case reduces to
# Q # ma 1 h2 h1 2 or q h2 h1 FIGURE 14–22 During simple cooling, specific humidity remains constant, but relative humidity increases. where h1 and h2 are enthalpies per unit mass of dry air at the inlet and the exit of the heating or cooling section, respectively. cen84959_ch14.qxd 9/15/06 6:05 AM Page 751 Chapter 14 
Humidifier 751 Heating with Humidification
Problems associated with the low relative humidity resulting from simple heating can be eliminated by humidifying the heated air. This is accomplished by passing the air first through a heating section (process 12) and then through a humidifying section (process 23), as shown in Fig. 14–23. The location of state 3 depends on how the humidification is accomplished. If steam is introduced in the humidification section, this will result T2). If humidification is in humidification with additional heating (T3 accomplished by spraying water into the airstream instead, part of the latent heat of vaporization comes from the air, which results in the cooling of the heated airstream (T3 T2). Air should be heated to a higher temperature in the heating section in this case to make up for the cooling effect during the humidification process.
EXAMPLE 14–5 Heating and Humidification of Air
1 Air Heating coils 2 3 ω2 = ω1
Heating section ω3 > ω2
Humidifying section FIGURE 14–23 Heating with humidification. An airconditioning system is to take in outdoor air at 10°C and 30 percent relative humidity at a steady rate of 45 m3/min and to condition it to 25°C and 60 percent relative humidity. The outdoor air is first heated to 22°C in the heating section and then humidified by the injection of hot steam in the humidifying section. Assuming the entire process takes place at a pressure of 100 kPa, determine (a) the rate of heat supply in the heating section and (b) the mass flow rate of the steam required in the humidifying section. Solution Outdoor air is first heated and then humidified by steam injection. The rate of heat transfer and the mass flow rate of steam are to be determined. Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The constantpressure specific heat of air at room temperature is cp 1.005 kJ/kg · K, and its gas constant is Ra 0.287 kJ/kg · K (Table A–2a). The saturation pressure of water is 1.2281 kPa at 10°C, and 3.1698 kPa at 25°C. The enthalpy of saturated water vapor is 2519.2 kJ/kg at 10°C, and 2541.0 kJ/kg at 22°C (Table A–4). Analysis We take the system to be the heating or the humidifying section, as appropriate. The schematic of the system and the psychrometric chart of the process are shown in Fig. 14–24. We note that the amount of water vapor in the air remains constant in the heating section (v1 v2) but increases in the humidifying section (v3 v2). (a) Applying the mass and energy balances on the heating section gives f
= f1 3 0% 3 = 60 % 3 1 10°C 2 22°C 25°C Heating coils T1 = 10°C f1 = 30% Air · V1 = 45 m3/min T2 = 22°C 1 2 Humidifier T3 = 25°C f3 = 60% 3 Dry air mass balance: Water mass balance: Energy balance: # Q in # m a1 # ma1v1 # ma h1 # # m a2 ma # m a2v2 S v1 v2 # # # ma h 2 S Q in m a 1 h2 h1 2 The psychrometric chart offers great convenience in determining the properties of moist air. However, its use is limited to a specified pressure only, which is 1 atm (101.325 kPa) for the one given in the appendix. At pressures other than FIGURE 14–24 Schematic and psychrometric chart for Example 14–5. cen84959_ch14.qxd 9/15/06 6:05 AM Page 752 752  Thermodynamics
1 atm, either other charts for that pressure or the relations developed earlier should be used. In our case, the choice is clear: Pv1 Pa1 v1 # ma v1 h1 f1Pg1 P1 Pv1 fPsat @ 10°C 1 100 1 0.287 kPa # m3> kg # K 2 1 283 K 2 R aT1 Pa 99.632 kPa # 45 m3> min V1 55.2 kg> min v1 0.815 m3> kg 0.622Pv1 P1 Pv1 v1hg1 1 100 0.622 1 0.368 kPa 2 0.368 2 kPa 0.368 2 kPa 1 0.3 2 1 1.2281 kPa 2 0.368 kPa 99.632 kPa 0.815 m3> kg dry air 0.0023 kg H 2O> kg dry air 1 0.0023 2 1 2519.2 kJ> kg 2 1 0.0023 2 1 2541.0 kJ> kg 2 15.8 kJ> kg dry air cpT2 v2hg2 cpT1 1 1.005 kJ> kg # °C 2 1 10°C 2 1 1.005 kJ> kg # °C 2 1 22°C 2 h2 28.0 kJ> kg dry air # ma 1 h2 since v2 becomes v1. Then the rate of heat transfer to air in the heating section # Q in h1 2 1 55.2 kg> min 2 3 1 28.0 15.8 2 kJ> kg 4 673 kJ/min
(b) The mass balance for water in the humidifying section can be expressed as # m a2v2
or # mw # ma 1 v3 3 100 # ma3v3 v2 2 1 0.60 2 1 3.1698 2 4 kPa # mw
where v3 0.622f3Pg3 P3 f3Pg3 0.622 1 0.60 2 1 3.1698 kPa 2 0.01206 kg H2O> kg dry air
Thus, # mw 1 55.2 kg> min 2 1 0.01206 0.539 kg/min 0.0023 2 Discussion The result 0.539 kg/min corresponds to a water requirement of close to one ton a day, which is significant. Cooling with Dehumidification
The specific humidity of air remains constant during a simple cooling process, but its relative humidity increases. If the relative humidity reaches undesirably high levels, it may be necessary to remove some moisture from the air, that is, to dehumidify it. This requires cooling the air below its dewpoint temperature. cen84959_ch14.qxd 9/15/06 6:05 AM Page 753 Chapter 14
The cooling process with dehumidifying is illustrated schematically and on the psychrometric chart in Fig. 14–25 in conjunction with Example 14–6. Hot, moist air enters the cooling section at state 1. As it passes through the cooling coils, its temperature decreases and its relative humidity increases at constant specific humidity. If the cooling section is sufficiently long, air reaches its dew point (state x, saturated air). Further cooling of air results in the condensation of part of the moisture in the air. Air remains saturated during the entire condensation process, which follows a line of 100 percent relative humidity until the final state (state 2) is reached. The water vapor that condenses out of the air during this process is removed from the cooling section through a separate channel. The condensate is usually assumed to leave the cooling section at T2. The cool, saturated air at state 2 is usually routed directly to the room, where it mixes with the room air. In some cases, however, the air at state 2 may be at the right specific humidity but at a very low temperature. In such cases, air is passed through a heating section where its temperature is raised to a more comfortable level before it is routed to the room.  753 f1 = 80% x f2 = 100% 1 2 14°C 30°C Cooling coils EXAMPLE 14–6 Cooling and Dehumidification of Air
2 T2 = 14°C f2 = 100% Air Condensate 1 T1 = 30°C f1 = 80% · V1 = 10 m3/min Air enters a window air conditioner at 1 atm, 30°C, and 80 percent relative humidity at a rate of 10 m3/min, and it leaves as saturated air at 14°C. Part of the moisture in the air that condenses during the process is also removed at 14°C. Determine the rates of heat and moisture removal from the air. 14°C Condensate removal Solution Air is cooled and dehumidified by a window air conditioner. The
rates of heat and moisture removal are to be determined. Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and the water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The enthalpy of saturated liquid water at 14°C is 58.8 kJ/kg (Table A–4). Also, the inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. Therefore, we can determine the properties of the air at both states from the psychrometric chart to be FIGURE 14–25 Schematic and psychrometric chart for Example 14–6. h1 v1 v1 85.4 kJ/kg dry air 0.0216 kg H2O/kg dry air 0.889 m3/kg dry air and h2 v2 39.3 kJ/kg dry air 0.0100 kg H2O/kg dry air Analysis We take the cooling section to be the system. The schematic of the system and the psychrometric chart of the process are shown in Fig. 14–25. We note that the amount of water vapor in the air decreases during the process (v2 v1) due to dehumidification. Applying the mass and energy balances on the cooling and dehumidification section gives Dry air mass balance: Water mass balance:
in # ma1 # ma1v1 # # Energy balance: a mh Q out # # ma2 ma # # # # ma2v2 mw S mw ma 1 v1 v2 2 # # # # a mh S Q out m 1 h 1 h 2 2 mw hw
out cen84959_ch14.qxd 9/15/06 6:05 AM Page 754 754  Thermodynamics
Then, # ma # mw # Q out # V1 v1 1 11.25 kg> min 2 1 0.0216 1 11.25 kg> min 2 3 1 85.4 0.889 m3> kg dry air 10 m3> min 11.25 kg> min 39.3 2 kJ> kg 4 0.0100 2 0.131 kg/min 1 0.131 kg> min 2 1 58.8 kJ> kg 2 511 kJ/min
Therefore, this airconditioning unit removes moisture and heat from the air at rates of 0.131 kg/min and 511 kJ/min, respectively.
Water that leaks out Hot, dry air Evaporative Cooling
Conventional cooling systems operate on a refrigeration cycle, and they can be used in any part of the world. But they have a high initial and operating cost. In desert (hot and dry) climates, we can avoid the high cost of cooling by using evaporative coolers, also known as swamp coolers. Evaporative cooling is based on a simple principle: As water evaporates, the latent heat of vaporization is absorbed from the water body and the surrounding air. As a result, both the water and the air are cooled during the process. This approach has been used for thousands of years to cool water. A porous jug or pitcher filled with water is left in an open, shaded area. A small amount of water leaks out through the porous holes, and the pitcher “sweats.” In a dry environment, this water evaporates and cools the remaining water in the pitcher (Fig. 14–26). You have probably noticed that on a hot, dry day the air feels a lot cooler when the yard is watered. This is because water absorbs heat from the air as it evaporates. An evaporative cooler works on the same principle. The evaporative cooling process is shown schematically and on a psychrometric chart in Fig. 14–27. Hot, dry air at state 1 enters the evaporative cooler, where it is sprayed with liquid water. Part of the water evaporates during this process by absorbing heat from the airstream. As a result, the temperature of the airstream decreases and its humidity increases (state 2). In the limiting case, the air leaves the evaporative cooler saturated at state 2 . This is the lowest temperature that can be achieved by this process. The evaporative cooling process is essentially identical to the adiabatic saturation process since the heat transfer between the airstream and the surroundings is usually negligible. Therefore, the evaporative cooling process follows a line of constant wetbulb temperature on the psychrometric chart. (Note that this will not exactly be the case if the liquid water is supplied at a temperature different from the exit temperature of the airstream.) Since the constantwetbulbtemperature lines almost coincide with the constantenthalpy lines, the enthalpy of the airstream can also be assumed to remain constant. That is,
Twb constant constant
(14–19) FIGURE 14–26 Water in a porous jug left in an open, breezy area cools as a result of evaporative cooling. 2' 2 ~ Twb = const. ~ h = const. 1 Liquid water COOL, MOIST AIR 2 1 HOT, DRY AIR and
h
(14–20) FIGURE 14–27 Evaporative cooling. during an evaporative cooling process. This is a reasonably accurate approximation, and it is commonly used in airconditioning calculations. cen84959_ch14.qxd 9/15/06 6:05 AM Page 755 Chapter 14
EXAMPLE 14–7 Evaporative Cooling of Air by a Swamp Cooler  755 of discharged air and the lowest temperature to which the air can be cooled are to be determined. Analysis The schematic of the evaporative cooler and the psychrometric chart of the process are shown in Fig. 14–28. (a) If we assume the liquid water is supplied at a temperature not much different from the exit temperature of the airstream, the evaporative cooling process follows a line of constant wetbulb temperature on the psychrometric chart. That is,
Tmin 1 T2 95°F Twb constant
AIR 2' 2 1 T1 = 95°F f = 20% P = 14.7 psia The wetbulb temperature at 95°F and 20 percent relative humidity is determined from the psychrometric chart to be 66.0°F. The intersection point of the Twb 66.0°F and the f 80 percent lines is the exit state of the air. The temperature at this point is the exit temperature of the air, and it is determined from the psychrometric chart to be T2 70.4°F (b) In the limiting case, air leaves the evaporative cooler saturated (f 100 percent), and the exit state of the air in this case is the state where the Twb 66.0°F line intersects the saturation line. For saturated air, the dry and the wetbulb temperatures are identical. Therefore, the lowest temperature to which air can be cooled is the wetbulb temperature, which is FIGURE 14–28 Schematic and psychrometric chart for Example 14–7. Tmin T2¿ 66.0°F Discussion Note that the temperature of air drops by as much as 30°F in this case by evaporative cooling. Adiabatic Mixing of Airstreams
Many airconditioning applications require the mixing of two airstreams. This is particularly true for large buildings, most production and process plants, and hospitals, which require that the conditioned air be mixed with a certain fraction of fresh outside air before it is routed into the living space. The mixing is accomplished by simply merging the two airstreams, as shown in Fig. 14–29. The heat transfer with the surroundings is usually small, and thus the mixing processes can be assumed to be adiabatic. Mixing processes normally involve no work interactions, and the changes in kinetic and potential energies, if any, are negligible. Then the mass and energy balances for the adiabatic mixing of two airstreams reduce to
Mass of dry air: Mass of water vapor: Energy: # ma1 # v1ma1 # ma1h1 # ma2 # v2ma2 # ma2h2 # ma3 # v3ma3 # ma3h3
(14–21) (14–22) (14–23) f 1 =2 0% 2 f Solution Air is cooled steadily by an evaporative cooler. The temperature 2 2' = 80 % Air enters an evaporative (or swamp) cooler at 14.7 psi, 95°F, and 20 percent relative humidity, and it exits at 80 percent relative humidity. Determine (a) the exit temperature of the air and (b) the lowest temperature to which the air can be cooled by this evaporative cooler. cen84959_ch14.qxd 9/15/06 6:05 AM Page 756 756
ω1 h1  Thermodynamics
. Eliminating ma3 from the relations above, we obtain
# m a1 # ma2
Mixing section 3 1 v2 v3 v3 v1 h2 h3 h3 h1 (14–24) ω3 h3 2 ω2 h2 h2 h3 h1 h3 – h1 C h2 – h3 2 A 1 3 D ω2 – ω3 ω3 – ω1 B ω2 ω3 ω1 This equation has an instructive geometric interpretation on the psychrov3 to v3 v1 is equal to the metric chart. It shows that the ratio of v2 . . ratio of ma1 to ma2. The states that satisfy this condition are indicated by the dashed line AB. The ratio of h2 h3 to h3 h1 is also equal to the ratio of . . ma1 to ma2, and the states that satisfy this condition are indicated by the dashed line CD. The only state that satisfies both conditions is the intersection point of these two dashed lines, which is located on the straight line connecting states 1 and 2. Thus we conclude that when two airstreams at two different states (states 1 and 2) are mixed adiabatically, the state of the mixture (state 3) lies on the straight line connecting states 1 and 2 on the psychrometric chart, and the ratio of the distances 23 and 31 is equal to the ratio of mass flow . . rates ma1 and ma2. The concave nature of the saturation curve and the conclusion above lead to an interesting possibility. When states 1 and 2 are located close to the saturation curve, the straight line connecting the two states will cross the saturation curve, and state 3 may lie to the left of the saturation curve. In this case, some water will inevitably condense during the mixing process.
EXAMPLE 14–8 Mixing of Conditioned Air with Outdoor Air FIGURE 14–29 When two airstreams at states 1 and 2 are mixed adiabatically, the state of the mixture lies on the straight line connecting the two states. Saturated air leaving the cooling section of an airconditioning system at 14°C at a rate of 50 m3/min is mixed adiabatically with the outside air at 32°C and 60 percent relative humidity at a rate of 20 m3/min. Assuming that the mixing process occurs at a pressure of 1 atm, determine the specific humidity, the relative humidity, the drybulb temperature, and the volume flow rate of the mixture. Solution Conditioned air is mixed with outside air at specified rates. The
specific and relative humidities, drybulb temperature, and the flow rate of the mixture are to be determined. Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties The properties of each inlet stream are determined from the psychrometric chart to be h1 v1 v1
and 39.4 kJ> kg dry air 0.010 kg H2O> kg dry air 0.826 m3> kg dry air 79.0 kJ> kg dry air h2 v2 v2 0.0182 kg H2O> kg dry air 0.889 m3> kg dry air cen84959_ch14.qxd 9/15/06 6:05 AM Page 757 Chapter 14
Analysis We take the mixing section of the streams as the system. The schematic of the system and the psychrometric chart of the process are shown in Fig. 14–30. We note that this is a steadyflow mixing process. The mass flow rates of dry air in each stream are
Saturated air T1 = 14°C · V1 = 50 m3/min 1 Mixing section P = 1 atm T2 = 32°C f2 = 60% 2  757 # ma1 # ma2 # V1 v1 # V2 v2 # ma2 0.826 m3> kg dry air 0.889 m3> kg dry air 1 60.5 20 m3> min 50 m > min
3 60.5 kg> min 22.5 kg> min 83 kg> min 3 V3 v3 f3 T3 · From the mass balance of dry air, # ma3 # ma1 22.5 2 kg> min V2 = 20 m3/min · # ma1 # ma2 60.5 22.5
which yield v2 v3 v3 v1 h2 h3 h3 h1 79.0 h3 h3 39.4 10 = f 2 3 1 0.0182 v3 v3 0.010 v3 h3 0.0122 kg H2O/kg dry air 50.1 kJ> kg dry air 14°C f 1 32°C These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: T3 f3 v3 # V3 # ma3v3 19.0°C 89% 0.844 m3> kg dry air 70.1 m3/min FIGURE 14–30 Schematic and psychrometric chart for Example 14–8. Finally, the volume flow rate of the mixture is determined from 1 83 kg> min 2 1 0.844 m3> kg 2 Discussion Notice that the volume flow rate of the mixture is approximately equal to the sum of the volume flow rates of the two incoming streams. This is typical in airconditioning applications. Wet Cooling Towers
Power plants, large airconditioning systems, and some industries generate large quantities of waste heat that is often rejected to cooling water from nearby lakes or rivers. In some cases, however, the cooling water supply is limited or thermal pollution is a serious concern. In such cases, the waste heat must be rejected to the atmosphere, with cooling water recirculating and serving as a transport medium for heat transfer between the source and the sink (the atmosphere). One way of achieving this is through the use of wet cooling towers. A wet cooling tower is essentially a semienclosed evaporative cooler. An induceddraft counterflow wet cooling tower is shown schematically in =6 0% The specific humidity and the enthalpy of the mixture can be determined from Eq. 14–24, 0% 2 cen84959_ch14.qxd 9/15/06 6:05 AM Page 758 758  Thermodynamics
AIR EXIT WARM WATER FAN AIR INLET COOL WATER FIGURE 14–31 An induceddraft counterflow cooling tower. WARM WATER COOL WATER AIR INLET FIGURE 14–32 A naturaldraft cooling tower. Fig. 14–31. Air is drawn into the tower from the bottom and leaves through the top. Warm water from the condenser is pumped to the top of the tower and is sprayed into this airstream. The purpose of spraying is to expose a large surface area of water to the air. As the water droplets fall under the influence of gravity, a small fraction of water (usually a few percent) evaporates and cools the remaining water. The temperature and the moisture content of the air increase during this process. The cooled water collects at the bottom of the tower and is pumped back to the condenser to absorb additional waste heat. Makeup water must be added to the cycle to replace the water lost by evaporation and air draft. To minimize water carried away by the air, drift eliminators are installed in the wet cooling towers above the spray section. The air circulation in the cooling tower described is provided by a fan, and therefore it is classified as a forceddraft cooling tower. Another popular type of cooling tower is the naturaldraft cooling tower, which looks like a large chimney and works like an ordinary chimney. The air in the tower has a high watervapor content, and thus it is lighter than the outside air. Consequently, the light air in the tower rises, and the heavier outside air fills the vacant space, creating an airflow from the bottom of the tower to the top. The flow rate of air is controlled by the conditions of the atmospheric air. Naturaldraft cooling towers do not require any external power to induce the air, but they cost a lot more to build than forceddraft cooling towers. The naturaldraft cooling towers are hyperbolic in profile, as shown in Fig. 14–32, and some are over 100 m high. The hyperbolic profile is for greater structural strength, not for any thermodynamic reason. The idea of a cooling tower started with the spray pond, where the warm water is sprayed into the air and is cooled by the air as it falls into the pond, as shown in Fig. 14–33. Some spray ponds are still in use today. However, they require 25 to 50 times the area of a cooling tower, water loss due to air drift is high, and they are unprotected against dust and dirt. We could also dump the waste heat into a still cooling pond, which is basically a large artificial lake open to the atmosphere. Heat transfer from the pond surface to the atmosphere is very slow, however, and we would need about 20 times the area of a spray pond in this case to achieve the same cooling.
EXAMPLE 14–9 Cooling of a Power Plant by a Cooling Tower Cooling water leaves the condenser of a power plant and enters a wet cooling tower at 35°C at a rate of 100 kg/s. Water is cooled to 22°C in the cooling tower by air that enters the tower at 1 atm, 20°C, and 60 percent relative humidity and leaves saturated at 30°C. Neglecting the power input to the fan, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water. FIGURE 14–33 A spray pond.
Photo by Yunus Çengel. Solution Warm cooling water from a power plant is cooled in a wet cooling
tower. The flow rates of makeup water and air are to be determined. Assumptions 1 Steady operating conditions exist and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and the water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. cen84959_ch14.qxd 9/15/06 6:05 AM Page 759 Chapter 14
Properties The enthalpy of saturated liquid water is 92.28 kJ/kg at 22°C and 146.64 kJ/kg at 35°C (Table A–4). From the psychrometric chart,
2 WARM WATER 35°C 100 kg/s 3 30°C f2 = 100%  759 h1 v1 v1 42.2 kJ/kg dry air 0.0087 kg H2O/kg dry air 0.842 m3/kg dry air h2 v2 100.0 kJ/kg dry air 0.0273 kg H2O/kg dry air Analysis We take the entire cooling tower to be the system, which is shown schematically in Fig. 14–34. We note that the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. (a) Applying the mass and energy balances on the cooling tower gives System boundary 1 AIR 1 atm 20°C f1 = 60% · V1 22°C 100 kg/s COOL WATER 4 # # m a2 m a # # # # m3 m a1v1 m4 m a2v2 Water mass balance: # # # # m3 m4 ma 1 v 2 v1 2 mmakeup or # # # # # # Energy balance: a m h a m h S ma1h1 m3 h 3 m a2h2 m4 h4 Dry air mass balance:
in out # m a1 Makeup water or # m3 h3 · Solving for ma gives # ma 1 h2 h1 2 # m3 1 h3 # 1 m3 h4 2 # m makeup 2 h4 FIGURE 14–34 Schematic for Example 14–9. # ma
Substituting, 1 h2 h1 2 1 v2 v1 2 h4 96.9 kg> s # ma 3 1 100.0 # V1 # m makeup 42.2 2 kJ> kg 4 # mav1 1 100 kg> s 2 3 1 146.64 3 1 0.0273 92.28 2 kJ> kg 4 0.0087 2 1 92.28 2 kJ> kg 4 81.6 m3/s 0.0087 2 Then the volume flow rate of air into the cooling tower becomes 1 96.9 kg> s 2 1 0.842 m3> kg 2 v1 2 1 96.9 kg> s 2 1 0.0273 (b) The mass flow rate of the required makeup water is determined from # m a 1 v2 1.80 kg/s Discussion Note that over 98 percent of the cooling water is saved and recirculated in this case. SUMMARY
In this chapter we discussed the air–watervapor mixture, which is the most commonly encountered gas–vapor mixture in practice. The air in the atmosphere normally contains some water vapor, and it is referred to as atmospheric air. By contrast, air that contains no water vapor is called dry air. In the temperature range encountered in airconditioning applications, both the dry air and the water vapor can be treated as ideal gases. The enthalpy change of dry air during a process can be determined from ¢ h dry air cp ¢ T 1 1.005 kJ> kg # °C 2 ¢ T The atmospheric air can be treated as an idealgas mixture whose pressure is the sum of the partial pressure of dry air Pa and that of the water vapor Pv , P Pa Pv cen84959_ch14.qxd 9/15/06 6:05 AM Page 760 760  Thermodynamics
where v2 0.622Pg2 P2 Pg2 The enthalpy of water vapor in the air can be taken to be equal to the enthalpy of the saturated vapor at the same temperature: hv(T, low P) hg(T) 2500.9 1060.9 1.82T (kJ/kg) T in °C 0.435T (Btu/lbm) T in °F in the temperature range 10 to 50°C (15 to 120°F). The mass of water vapor present per unit mass of dry air is called the specific or absolute humidity v, v mv ma 0.622Pv 1 kg H 2O> kg dry air 2 P Pv where P is the total pressure of air and Pv is the vapor pressure. There is a limit on the amount of vapor the air can hold at a given temperature. Air that is holding as much moisture as it can at a given temperature is called saturated air. The ratio of the amount of moisture air holds (mv) to the maximum amount of moisture air can hold at the same temperature (mg) is called the relative humidity f, f mv mg Pv Pg Psat @ T. The relative and specific humidities can where Pg also be expressed as f vP and v 1 0.622 v 2 Pg 0.622fPg P fPg Relative humidity ranges from 0 for dry air to 1 for saturated air. The enthalpy of atmospheric air is expressed per unit mass of dry air, instead of per unit mass of the air–watervapor mixture, as h ha vhg 1 kJ> kg dry air 2 and T2 is the adiabatic saturation temperature. A more practical approach in airconditioning applications is to use a thermometer whose bulb is covered with a cotton wick saturated with water and to blow air over the wick. The temperature measured in this manner is called the wetbulb temperature Twb, and it is used in place of the adiabatic saturation temperature. The properties of atmospheric air at a specified total pressure are presented in the form of easily readable charts, called psychrometric charts. The lines of constant enthalpy and the lines of constant wetbulb temperature are very nearly parallel on these charts. The needs of the human body and the conditions of the environment are not quite compatible. Therefore, it often becomes necessary to change the conditions of a living space to make it more comfortable. Maintaining a living space or an industrial facility at the desired temperature and humidity may require simple heating (raising the temperature), simple cooling (lowering the temperature), humidifying (adding moisture), or dehumidifying (removing moisture). Sometimes two or more of these processes are needed to bring the air to the desired temperature and humidity level. Most airconditioning processes can be modeled as steadyflow processes, and therefore they can be analyzed by applying the steadyflow mass (for both dry air and water) and energy balances, Dry air mass: Water mass: Energy: # a mw
in # a ma
in # a ma
out # # a m w or a m av
out in # a m av
out The ordinary temperature of atmospheric air is referred to as the drybulb temperature to differentiate it from other forms of temperatures. The temperature at which condensation begins if the air is cooled at constant pressure is called the dewpoint temperature Tdp: Tdp Tsat @
Pv # Q in # Win # a mh
in # Q out # Wout # a mh
out Relative humidity and specific humidity of air can be determined by measuring the adiabatic saturation temperature of air, which is the temperature air attains after flowing over water in a long adiabatic channel until it is saturated, v1 cp 1 T2 T1 2 v2hfg2 hf2 hg1 The changes in kinetic and potential energies are assumed to be negligible. During a simple heating or cooling process, the specific humidity remains constant, but the temperature and the relative humidity change. Sometimes air is humidified after it is heated, and some cooling processes include dehumidification. In dry climates, air can be cooled via evaporative cooling by passing it through a section where it is sprayed with water. In locations with limited cooling water supply, large amounts of waste heat can be rejected to the atmosphere with minimum water loss through the use of cooling towers. cen84959_ch14.qxd 9/15/06 6:05 AM Page 761 Chapter 14 REFERENCES AND SUGGESTED READINGS
1. ASHRAE. 1981 Handbook of Fundamentals. Atlanta, GA: American Society of Heating, Refrigerating, and AirConditioning Engineers, 1981. 2. S. M. Elonka. “Cooling Towers.” Power, March 1963. 3. W. F. Stoecker and J. W. Jones. Refrigeration and Air Conditioning. 2nd ed. New York: McGrawHill, 1982.  761 4. L. D. Winiarski and B. A. Tichenor. “Model of Natural Draft Cooling Tower Performance.” Journal of the Sanitary Engineering Division, Proceedings of the American Society of Civil Engineers, August 1970. PROBLEMS*
Dry and Atmospheric Air: Specific and Relative Humidity
14–1C Is it possible to obtain saturated air from unsaturated air without adding any moisture? Explain. 14–2C Is the relative humidity of saturated air necessarily 100 percent? 14–3C Moist air is passed through a cooling section where it is cooled and dehumidified. How do (a) the specific humidity and (b) the relative humidity of air change during this process? 14–4C What is the difference between dry air and atmospheric air? 14–5C Can the water vapor in air be treated as an ideal gas? Explain. 14–6C What is vapor pressure? 14–7C How would you compare the enthalpy of water vapor at 20°C and 2 kPa with the enthalpy of water vapor at 20°C and 0.5 kPa? 14–8C What is the difference between the specific humidity and the relative humidity? 14–9C How will (a) the specific humidity and (b) the relative humidity of the air contained in a wellsealed room change as it is heated? 14–10C How will (a) the specific humidity and (b) the relative humidity of the air contained in a wellsealed room change as it is cooled? 14–11C Consider a tank that contains moist air at 3 atm and whose walls are permeable to water vapor. The surrounding air at 1 atm pressure also contains some moisture. Is it possible for the water vapor to flow into the tank from surroundings? Explain. 14–12C Why are the chilled water lines always wrapped with vapor barrier jackets? 14–13C Explain how vapor pressure of the ambient air is determined when the temperature, total pressure, and the relative humidity of air are given. 14–14E Humid air at 100 psia and 400 F and a humidity ratio of 0.025 lbm H2O/lbm dry air is expanded to 15 psia in an isentropic nozzle. How much of the initial water vapor has been converted to liquid water at the nozzle outlet? 14–15 Humid air at 100 kPa, 20 C, and 90 percent relative humidity is compressed in a steadyflow, isentropic compressor to 800 kPa. What is the relative humidity of the air at the compressor outlet? Answer: 0.4%
800 kPa Humid air *Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with the icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with the icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 100 kPa 20 C, 90% FIGURE P14–15 cen84959_ch14.qxd 9/15/06 6:05 AM Page 762 762  Thermodynamics
14–32 The dry and wetbulb temperatures of atmospheric air at 95 kPa are 25 and 17°C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air, in kJ/kg dry air. 14–33 The air in a room has a drybulb temperature of 22°C and a wetbulb temperature of 16°C. Assuming a pressure of 100 kPa, determine (a) the specific humidity, (b) the relative humidity, and (c) the dewpoint temperature. Answers:
(a) 0.0090 kg H2O/kg dry air, (b) 54.1 percent, (c) 12.3°C 14–16 A tank contains 21 kg of dry air and 0.3 kg of water vapor at 30°C and 100 kPa total pressure. Determine (a) the specific humidity, (b) the relative humidity, and (c) the volume of the tank. 14–17 Repeat Prob. 14–16 for a temperature of 24°C. 14–18 A room contains air at 20°C and 98 kPa at a relative humidity of 85 percent. Determine (a) the partial pressure of dry air, (b) the specific humidity of the air, and (c) the enthalpy per unit mass of dry air. 14–19 Repeat Prob. 14–18 for a pressure of 85 kPa. 14–20E A room contains air at 70°F and 14.6 psia at a relative humidity of 85 percent. Determine (a) the partial pressure of dry air, (b) the specific humidity, and (c) the enthalpy per unit mass of dry air. Answers: (a) 14.291 psia,
(b) 0.0134 lbm H2O/lbm dry air, (c) 31.43 Btu/lbm dry air 14–34 Reconsider Prob. 14–33. Determine the required properties using EES (or other) software. What would the property values be at a pressure of 300 kPa? 14–35E The air in a room has a drybulb temperature of 80°F and a wetbulb temperature of 65°F. Assuming a pressure of 14.7 psia, determine (a) the specific humidity, (b) the relative humidity, and (c) the dewpoint temperature.
Answers: (a) 0.0097 lbm H2O/lbm dry air, (b) 44.7 percent, (c) 56.6°F 14–21 Determine the masses of dry air and the water vapor contained in a 240m3 room at 98 kPa, 23°C, and 50 percent relative humidity. Answers: 273 kg, 2.5 kg DewPoint, Adiabatic Saturation, and WetBulb Temperatures
14–22C What is the dewpoint temperature? 14–23C Andy and Wendy both wear glasses. On a cold winter day, Andy comes from the cold outside and enters the warm house while Wendy leaves the house and goes outside. Whose glasses are more likely to be fogged? Explain. 14–24C In summer, the outer surface of a glass filled with iced water frequently “sweats.” How can you explain this sweating? 14–25C In some climates, cleaning the ice off the windshield of a car is a common chore on winter mornings. Explain how ice forms on the windshield during some nights even when there is no rain or snow. 14–26C When are the drybulb and dewpoint temperatures identical? 14–27C When are the adiabatic saturation and wetbulb temperatures equivalent for atmospheric air? 14–28 A house contains air at 25°C and 65 percent relative humidity. Will any moisture condense on the inner surfaces of the windows when the temperature of the window drops to 10°C? 14–29 After a long walk in the 8°C outdoors, a person wearing glasses enters a room at 25°C and 40 percent relative humidity. Determine whether the glasses will become fogged. 14–30 Repeat Prob. 14–29 for a relative humidity of 30 percent. 14–31E A thirsty woman opens the refrigerator and picks up a cool canned drink at 40°F. Do you think the can will “sweat” as she enjoys the drink in a room at 80°F and 50 percent relative humidity? 14–36 Atmospheric air at 35°C flows steadily into an adiabatic saturation device and leaves as a saturated mixture at 25°C. Makeup water is supplied to the device at 25°C. Atmospheric pressure is 98 kPa. Determine the relative humidity and specific humidity of the air. Psychrometric Chart
14–37C How do constantenthalpy and constantwetbulbtemperature lines compare on the psychrometric chart? 14–38C At what states on the psychrometric chart are the drybulb, wetbulb, and dewpoint temperatures identical? 14–39C How is the dewpoint temperature at a specified state determined on the psychrometric chart? 14–40C Can the enthalpy values determined from a psychrometric chart at sea level be used at higher elevations? 14–41 The air in a room is at 1 atm, 32°C, and 60 percent relative humidity. Determine (a) the specific humidity, (b) the enthalpy, in kJ/kg dry air, (c) the wetbulb temperature, (d ) the dewpoint temperature, and (e) the specific volume of the air, in m3/kg dry air. Use the psychrometric chart or available software. 14–42 A room contains air at 1 atm, 26°C, and 70 percent relative humidity. Using the psychrometric chart, determine (a) the specific humidity, (b) the enthalpy, in kJ/kg dry air, (c) the wetbulb temperature, (d ) the dewpoint temperature, and (e) the specific volume of the air, in m3/kg dry air. 14–43 Reconsider Prob. 14–42. Determine the required properties using EES (or other) software instead of the psychrometric chart. What would the property values be at a location at 2000 m altitude? 14–44 The air in a room has a pressure of 1 atm, a drybulb temperature of 24°C, and a wetbulb temperature of 17°C. Using the psychrometric chart, determine (a) the specific cen84959_ch14.qxd 9/15/06 6:05 AM Page 763 Chapter 14
humidity, (b) the enthalpy, in kJ/kg dry air, (c) the relative humidity, (d ) the dewpoint temperature, and (e) the specific volume of the air, in m3/kg dry air. 14–45 Reconsider Prob. 14–44. Determine the required properties using EES (or other) software instead of the psychrometric chart. What would the property values be at a location at 3000 m altitude? 14–46 Atmospheric air at a pressure of 1 atm and drybulb temperature of 30 C has a relative humidity of 80 percent. Using the psychrometric chart, determine (a) the wetbulb temperature, (b) the humidity ratio, (c) the enthalpy, (d) the dewpoint temperature, and (e) the water vapor pressure. 14–47E Atmospheric air at a pressure of 1 atm and drybulb temperature of 90 F has a wetbulb temperature of 85 F. Using the psychrometric chart, determine (a) the relative humidity, (b) the humidity ratio, (c) the enthalpy, (d) the dewpoint temperature, and (e) the water vapor pressure.  763 14–52C How does the human body respond to (a) hot weather, (b) cold weather, and (c) hot and humid weather? 14–53C What is the radiation effect? How does it affect human comfort? 14–54C How does the air motion in the vicinity of the human body affect human comfort? 14–55C Consider a tennis match in cold weather where both players and spectators wear the same clothes. Which group of people will feel colder? Why? 14–56C Why do you think little babies are more susceptible to cold? 14–57C How does humidity affect human comfort? 14–58C What are humidification and dehumidification? 14–59C What is metabolism? What is the range of metabolic rate for an average man? Why are we interested in the metabolic rate of the occupants of a building when we deal with heating and airconditioning? 14–60C Why is the metabolic rate of women, in general, lower than that of men? What is the effect of clothing on the environmental temperature that feels comfortable? 14–61C What is sensible heat? How is the sensible heat loss from a human body affected by the (a) skin temperature, (b) environment temperature, and (c) air motion? 14–62C What is latent heat? How is the latent heat loss from the human body affected by the (a) skin wettedness and (b) relative humidity of the environment? How is the rate of evaporation from the body related to the rate of latent heat loss? 14–63 An average person produces 0.25 kg of moisture while taking a shower and 0.05 kg while bathing in a tub. Consider a family of four who each shower once a day in a bathroom that is not ventilated. Taking the heat of vaporization of water to be 2450 kJ/kg, determine the contribution of showers to the latent heat load of the air conditioner per day in summer. 14–64 An average (1.82 kg or 4.0 lbm) chicken has a basal metabolic rate of 5.47 W and an average metabolic rate of 10.2 W (3.78 W sensible and 6.42 W latent) during normal activity. If there are 100 chickens in a breeding room, determine the rate of total heat generation and the rate of moisture production in the room. Take the heat of vaporization of water to be 2430 kJ/kg. 14–65 A department store expects to have 120 customers and 15 employees at peak times in summer. Determine the contribution of people to the total cooling load of the store. 14–66E In a movie theater in winter, 500 people, each generating sensible heat at a rate of 70 W, are watching a movie. The heat losses through the walls, windows, and the roof are estimated to be 130,000 Btu/h. Determine if the theater needs to be heated or cooled. Air 1 atm 90 F Twb = 85 F FIGURE P14–47E
14–48E Determine the adiabatic saturation temperature of the humid air in Prob. 14–47E. Answer: 85 F 14–49 Atmospheric air at a pressure of 1 atm and drybulb temperature of 28 C has a dewpoint temperature of 20 C. Using the psychrometric chart, determine (a) the relative humidity, (b) the humidity ratio, (c) the enthalpy, (d) the wetbulb temperature, and (e) the water vapor pressure. 14–50 Determine the adiabatic saturation temperature of the humid air in Prob. 14–49.
Water Humidifier 1 atm 28 C Tdp = 20 C AIR 100% FIGURE P14–50 Human Comfort and AirConditioning
14–51C What does a modern airconditioning system do besides heating or cooling the air? cen84959_ch14.qxd 9/15/06 6:05 AM Page 764 764  Thermodynamics
Heating coils 15 C 100% 20 m/s AIR 200 kPa 14–67 For an infiltration rate of 1.2 air changes per hour (ACH), determine sensible, latent, and total infiltration heat load of a building at sea level, in kW, that is 20 m long, 13 m wide, and 3 m high when the outdoor air is at 32°C and 50 percent relative humidity. The building is maintained at 24°C and 50 percent relative humidity at all times. 14–68 Repeat Prob. 14–67 for an infiltration rate of 1.8 ACH. 30 C FIGURE P14–75
14–76 Calculate the rate at which the exergy of the humid air of Prob. 14–75 is increased. Take T0 15 C. Simple Heating and Cooling
14–69C How do relative and specific humidities change during a simple heating process? Answer the same question for a simple cooling process. 14–70C Why does a simple heating or cooling process appear as a horizontal line on the psychrometric chart? 14–71 Air enters a 40cmdiameter cooling section at 1 atm, 32°C, and 30 percent relative humidity at 18 m/s. Heat is removed from the air at a rate of 1200 kJ/min. Determine (a) the exit temperature, (b) the exit relative humidity of the air, and (c) the exit velocity. Answers: (a) 24.4°C, (b) 46.6
percent, (c) 17.6 m/s Heating with Humidification
14–77C Why is heated air sometimes humidified? 14–78 Air at 1 atm, 15°C, and 60 percent relative humidity is first heated to 20°C in a heating section and then humidified by introducing water vapor. The air leaves the humidifying section at 25°C and 65 percent relative humidity. Determine (a) the amount of steam added to the air, and (b) the amount of heat transfer to the air in the heating section. Answers: (a) 0.0065 kg H2O/kg dry air, (b) 5.1 kJ/kg dry air 14–79E Air at 14.7 psia, 50°F, and 60 percent relative humidity is first heated to 72°F in a heating section and then humidified by introducing water vapor. The air leaves the humidifying section at 75°F and 55 percent relative humidity. Determine (a) the amount of steam added to the air, in lbm H2O/lbm dry air, and (b) the amount of heat transfer to the air in the heating section, in Btu/lbm dry air. 14–80 An airconditioning system operates at a total pressure of 1 atm and consists of a heating section and a humidifier that supplies wet steam (saturated water vapor) at 100°C. Air enters the heating section at 10°C and 70 percent relative humidity at a rate of 35 m3/min, and it leaves the humidifying section at 20°C and 60 percent relative humidity. Determine (a) the temperature and relative humidity of air when it leaves the heating section, (b) the rate of heat transfer in the heating section, and (c) the rate at which water is added to the air in the humidifying section.
Sat. vapor 100°C Heating coils 10°C 70% 35 m3/min AIR P = 1 atm Humidifier 20°C 60% 1200 kJ/min 32°C, 30% AIR 18 m/s 1 atm FIGURE P14–71
14–72 Repeat Prob. 14–70 for a heat removal rate of 800 kJ/min. 14–73E Humid air at 1 atm, 100 F, and 70 percent relative humidity is cooled at constant pressure to the dewpoint temperature. Determine the cooling, in Btu/lbm dry air, required for this process. 14–74 Humid air at 150 kPa, 40 C, and 70 percent relative humidity is cooled at constant pressure in a pipe to its dewpoint temperature. Calculate the heat transfer, in kJ/kg dry air, required for this process. Answer: 6.8 kJ/kg dry air 14–75 Saturated humid air at 200 kPa and 15 C is heated to 30 C as it flows through a 4cmdiameter pipe with a velocity of 20 m/s. Disregarding pressure losses, calculate the relative humidity at the pipe outlet and the rate of heat transfer, in kW, to the air. FIGURE P14–80 cen84959_ch14.qxd 9/15/06 6:05 AM Page 765 Chapter 14
14–81 Repeat Prob. 14–75 for a total pressure of 95 kPa for the airstream. Answers: (a) 19.5°C, 37.7 percent, (b) 391 kJ/min,
(c) 0.147 kg/min  765 Cooling with Dehumidification
14–82C Why is cooled air sometimes reheated in summer before it is discharged to a room? 14–83 An airconditioning system is to take in air at 1 atm, 34°C, and 70 percent relative humidity and deliver it at 22°C and 50 percent relative humidity. The air flows first over the cooling coils, where it is cooled and dehumidified, and then over the resistance heating wires, where it is heated to the desired temperature. Assuming that the condensate is removed from the cooling section at 10°C, determine (a) the temperature of air before it enters the heating section, (b) the amount of heat removed in the cooling section, and (c) the amount of heat transferred in the heating section, both in kJ/kg dry air. 14–84 Air enters a 30cmdiameter cooling section at 1 atm, 35°C, and 60 percent relative humidity at 120 m/min. The air is cooled by passing it over a cooling coil through which cold water flows. The water experiences a temperature rise of 8°C. The air leaves the cooling section saturated at 20°C. Determine (a) the rate of heat transfer, (b) the mass flow rate of the water, and (c) the exit velocity of the airstream.
Water T Cooling coils 14–88E Ten thousand cubic feet per hour of atmospheric air at 1 atm and 85 F with a dewpoint temperature of 70 F are to be cooled to 60 F. Determine the rate at which condensate leaves this system and the cooling rate when the condensate leaves the system at 65 F. 14–89 On a summer day in New Orleans, Louisiana, the pressure is 1 atm; the temperature is 32 C; and the relative humidity is 95 percent. This air is to be conditioned to 24 C and 60 percent relative humidity. Determine the amount of cooling, in kJ, required and water removed, in kg, per 1000 m 3 of dry air processed at the entrance to the system.
Cooling coils T2 = 24 C f2 = 60% 1 atm Condensate T1 = 32 C f1 = 95% FIGURE P14–89
14–90 How far will the temperature of the humid air of Prob. 14–89 have to be reduced to produce the desired dehumidification? Answer: 15.8 C 14–91 A simple ideal vaporcompression refrigeration system using refrigerant134a as the working fluid is used to provide the cooling required in Prob. 14–89. It operates its evaporator at 4 C and its condenser at a saturation temperature of 39.4 C. The condenser rejects its heat to the New Orleans summer air. Calculate the exergy destruction, in kJ, in the total system per 1000 m 3 of dry air processed.
Condenser T + 8°C 35°C 60% 120 m/min AIR 20°C Saturated FIGURE P14–84
14–85 Reconsider Prob. 14–84. Using EES (or other) software, develop a general solution of the problem in which the input variables may be supplied and parametric studies performed. For each set of input variables for which the pressure is atmospheric, show the process on the psychrometric chart. 14–86 Repeat Prob. 14–84 for a total pressure of 95 kPa for air. Answers: (a) 293.2 kJ/min, (b) 8.77 kg/min, (c) 113 m/min 14–87 Atmospheric air at 1 atm, 30 C, and 80 percent relative humidity is cooled to 20 C while the mixture pressure remains constant. Calculate the amount of water, in kg/kg dry air, removed from the air and the cooling requirement, in kJ/kg dry air, when the liquid water leaves the system at 22 C. Answers: 0.0069 kg H2O/kg dry air, 27.3 kJ/kg dry air 3 Expansion valve 4 Evaporator 2 Compressor 1 24 C 60% 1 atm Condensate 32 C 95% FIGURE P14–91E cen84959_ch14.qxd 9/15/06 6:05 AM Page 766 766  Thermodynamics
98 kPa. Determine the heat transfer rate from the air, in kW, and the mass flow rate of condensate water, if any, in kg/h. 14–92 Atmospheric air from the inside of an automobile enters the evaporator section of the air conditioner at 1 atm, 27°C and 50 percent relative humidity. The air returns to the automobile at 10°C and 90 percent relative humidity. The passenger compartment has a volume of 2 m3 and 5 air changes per minute are required to maintain the inside of the automobile at the desired comfort level. Sketch the psychrometric diagram for the atmospheric air flowing through the air conditioning process. Determine the dew point and wet bulb temperatures at the inlet to the evaporator section, in °C. Determine the required heat transfer rate from the atmospheric air to the evaporator fluid, in kW. Determine the rate of condensation of water vapor in the evaporator section, in kg/min. Evaporative Cooling
14–96C Does an evaporation process have to involve heat transfer? Describe a process that involves both heat and mass transfer. 14–97C During evaporation from a water body to air, under what conditions will the latent heat of vaporization be equal to the heat transfer from the air? 14–98C What is evaporative cooling? Will it work in humid climates? 14–99 Air enters an evaporative cooler at 1 atm, 36°C, and 20 percent relative humidity at a rate of 4 m3/min, and it leaves with a relative humidity of 90 percent. Determine (a) the exit temperature of the air and (b) the required rate of water supply to the evaporative cooler.
Water mω Cooling coils AIR Condensate Humidifier 1 atm 36°C f1 = 20% AIR f2 = 90% FIGURE P14–92
14–93 Two thousand cubic meters per hour of atmospheric air at 28°C with a dew point temperature of 25°C flows into an air conditioner that uses chilled water as the cooling fluid. The atmospheric air is to be cooled to 18°C. Sketch the system hardware and the psychrometric diagram for the process. Determine the mass flow rate of the condensate water, if any, leaving the air conditioner, in kg/h. If the cooling water has a 10°C temperature rise while flowing through the air conditioner, determine the volume flow rate of chilled water supplied to the air conditioner heat exchanger, in m3/min. The air conditioning process takes place at 100 kPa. 14–94 An automobile air conditioner uses refrigerant134a as the cooling fluid. The evaporator operates at 275 kPa gage and the condenser operates at 1.7 MPa gage. The compressor requires a power input of 6 kW and has an isentropic efficiency of 85 percent. Atmospheric air at 22°C and 50 percent relative humidity enters the evaporator and leaves at 8°C and 90 percent relative humidity. Determine the volume flow rate of the atmospheric air entering the evaporator of the air conditioner, in m3/min. 14–95 Air from a workspace enters an air conditioner unit at 30°C dry bulb and 25°C wet bulb. The air leaves the air conditioner and returns to the space at 25°C drybulb and 6.5°C dewpoint temperature. If there is any, the condensate leaves the air conditioner at the temperature of the air leaving the cooling coils. The volume flow rate of the air returned to the workspace is 1000 m3/min. Atmospheric pressure is FIGURE P14–99
14–100E Air enters an evaporative cooler at 14.7 psia, 90°F, and 20 percent relative humidity at a rate of 150 ft3/min, and it leaves with a relative humidity of 90 percent. Determine (a) the exit temperature of air and (b) the required rate of water supply to the evaporative cooler. Answers: (a) 65°F,
(b) 0.06 lbm/min 14–101 Air enters an evaporative cooler at 1 atm, 32°C, and 30 percent relative humidity at a rate of 5 m3/min and leaves at 22°C. Determine (a) the final relative humidity and (b) the amount of water added to air. 14–102 What is the lowest temperature that air can attain in an evaporative cooler if it enters at 1 atm, 29°C, and 40 percent relative humidity? Answer: 19.3°C 14–103 Air at 1 atm, 15°C, and 60 percent relative humidity is first heated to 30°C in a heating section and then passed through an evaporative cooler where its temperature drops to 25°C. Determine (a) the exit relative humidity and (b) the amount of water added to air, in kg H2O/kg dry air. 14–104E Desert dwellers often wrap their heads with a watersoaked porous cloth. On a desert where the pressure is 1 atm, temperature is 120 F, and relative humidity is 10 percent, what is the temperature of this cloth? cen84959_ch14.qxd 9/15/06 6:05 AM Page 767 Chapter 14
Adiabatic Mixing of Airstreams
14–105C Two unsaturated airstreams are mixed adiabatically. It is observed that some moisture condenses during the mixing process. Under what conditions will this be the case? 14–106C Consider the adiabatic mixing of two airstreams. Does the state of the mixture on the psychrometric chart have to be on the straight line connecting the two states? 14–107 Two airstreams are mixed steadily and adiabatically. The first stream enters at 32°C and 40 percent relative humidity at a rate of 20 m3/min, while the second stream enters at 12°C and 90 percent relative humidity at a rate of 25 m3/min. Assuming that the mixing process occurs at a pressure of 1 atm, determine the specific humidity, the relative humidity, the drybulb temperature, and the volume flow rate of the mixture. Answers: 0.0096 kg H2O/kg dry air, 63.4
percent, 20.6°C, 45.0 m3/min  767 1 32°C 40% P = 1 atm AIR 12°C 90% 3 v3 f3 T3 14–110E Calculate the rate of increase for the total entropy of Prob. 14–109E. Answer: 3.40 10 4 Btu/s # R 14–111 Saturated humid air at 1 atm and 10 C is to be mixed with atmospheric air at 1 atm, 32 C, and 80 percent relative humidity, to form air of 70 percent relative humidity. Determine the proportions at which these two streams are to be mixed and the temperature of the resulting air. 14–112 A stream of warm air with a drybulb temperature of 40°C and a wetbulb temperature of 32°C is mixed adiabatically with a stream of saturated cool air at 18°C. The dry air mass flow rates of the warm and cool airstreams are 8 and 6 kg/s, respectively. Assuming a total pressure of 1 atm, determine (a) the temperature, (b) the specific humidity, and (c) the relative humidity of the mixture. 14–113 Reconsider Prob. 14–112. Using EES (or other) software, determine the effect of the mass flow rate of saturated cool air stream on the mixture temperature, specific humidity, and relative humidity. Vary the mass flow rate of saturated cool air from 0 to 16 kg/s while maintaining the mass flow rate of warm air constant at 8 kg/s. Plot the mixture temperature, specific humidity, and relative humidity as functions of the mass flow rate of cool air, and discuss the results. Wet Cooling Towers
14–114C How does a naturaldraft wet cooling tower work? 14–115C What is a spray pond? How does its performance compare to the performance of a wet cooling tower? 14–116 A wet cooling tower is to cool 60 kg/s of water from 40 to 26°C. Atmospheric air enters the tower at 1 atm with dry and wetbulb temperatures of 22 and 16°C, respectively, and leaves at 34°C with a relative humidity of 90 percent. Using the psychrometric chart, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water. Answers:
(a) 44.9 m3/s, (b) 1.16 kg/s
AIR EXIT 34°C 90% 2 FIGURE P14–107
14–108 Repeat Prob. 14–107 for a total mixingchamber pressure of 90 kPa. 14–109E Two humid air streams are adiabatically mixed at 1 atm pressure to form a third stream. The first stream has a temperature of 100 F, a relative humidity of 90 percent, and a volume flow rate of 3 ft 3/s, while the second stream has a temperature of 50 F, a relative humidity of 30 percent, and a volume flow rate of 1 ft 3/s. Calculate the third stream’s temperature and relative humidity. WARM WATER 60 kg/s 40°C 1 100 F 90% 3 ft3/s P = 1 atm AIR 1 ft3/s 50 F 30% 2 3 T3 f3 AIR INLET 1 atm Tdb = 22°C Twb = 16°C 26°C COOL WATER Makeup water FIGURE P14–109E FIGURE P14–116 cen84959_ch14.qxd 9/15/06 6:05 AM Page 768 768  Thermodynamics
flows through the compressedair lines. Disregarding any pressure losses, determine if there will be any condensation in the compressedair lines on a day when the ambient air is at 20°C and 50 percent relative humidity. 14–123E Determine the mole fraction of the water vapor at the surface of a lake whose surface temperature is 60°F, and compare it to the mole fraction of water in the lake, which is very nearly 1.0. The air at the lake surface is saturated, and the atmospheric pressure at lake level can be taken to be 13.8 psia. 14–124 Determine the mole fraction of dry air at the surface of a lake whose temperature is 18°C. The air at the lake surface is saturated, and the atmospheric pressure at lake level can be taken to be 100 kPa. 14–125E Consider a room that is cooled adequately by an air conditioner whose cooling capacity is 7500 Btu/h. If the room is to be cooled by an evaporative cooler that removes heat at the same rate by evaporation, determine how much water needs to be supplied to the cooler per hour at design conditions. 14–126E The capacity of evaporative coolers is usually expressed in terms of the flow rate of air in ft3/min (or cfm), and a practical way of determining the required size of an evaporative cooler for an 8fthigh house is to multiply the floor area of the house by 4 (by 3 in dry climates and by 5 in humid climates). For example, the capacity of an evaporative cooler for a 30ftlong, 40ftwide house is 1200 4 4800 cfm. Develop an equivalent rule of thumb for the selection of an evaporative cooler in SI units for 2.4mhigh houses whose floor areas are given in m2. 14–127 A cooling tower with a cooling capacity of 100 tons (440 kW) is claimed to evaporate 15,800 kg of water per day. Is this a reasonable claim? 14–128 During a summer day in El Paso, Texas, outdoor air is at 1 atm, 40 C, and 20 percent relative humidity. Water at 20 C is evaporated into this air to produce air at 25 C and 80 percent relative humidity. How much water, in kg H2O/kg dry air, is required and how much cooling, in kJ/kg dry air, has been produced?
Water 20 C 14–117 Water at 30 C is to be cooled to 22 C in a cooling tower which it enters at a rate of 5 kg/s. Humid air enters this tower at 1 atm and 15 C with a relative humidity of 25 percent and leaves at 18 C with a relative humidity of 95 percent. Determine the mass flow rate of dry air through this tower. Answer: 6.29 kg/s 14–118 How much work potential, in kJ/kg dry air, is lost in the cooling tower of Prob. 14–117? Take T0 15 C. 14–119E Water at 100 F is to be cooled in a cooling tower which it enters at a rate of 10,000 lbm/h. Humid air enters this tower at 1 atm, 60 F, and 20 percent relative humidity with a dry air flow rate of 7000 lbm/h and leaves at 75 F and 0.018 lbm H2O/lbm dry air. Determine the relative humidity at which the air leaves the tower and the water’s exit temperature. 14–120 A wet cooling tower is to cool 25 kg/s of cooling water from 40 to 30°C at a location where the atmospheric pressure is 96 kPa. Atmospheric air enters the tower at 20°C and 70 percent relative humidity and leaves saturated at 35°C. Neglecting the power input to the fan, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water.
Answers: (a) 11.2 m3/s, (b) 0.35 kg/s 14–121 A naturaldraft cooling tower is to remove waste heat from the cooling water flowing through the condenser of a steam power plant. The turbine in the steam power plant receives 42 kg/s of steam from the steam generator. Eighteen percent of the steam entering the turbine is extracted for various feedwater heaters. The condensate of the higher pressure feedwater heaters is trapped to the next lowest pressure feedwater heater. The last feedwater heater operates at 0.2 MPa and all of the steam extracted for the feedwater heaters is throttled from the last feedwater heater exit to the condenser operating at a pressure of 10 kPa. The remainder of the steam produces work in the turbine and leaves the lowest pressure stage of the turbine at 10 kPa with an entropy of 7.962 kJ/kg K. The cooling tower supplies the cooling water at 26°C to the condenser, and cooling water returns from the condenser to the cooling tower at 40°C. Atmospheric air enters the tower at 1 atm with dry and wetbulb temperatures of 23 and 18°C, respectively, and leaves saturated at 37°C. Determine (a) the mass flow rate of the cooling water, (b) the volume flow rate of air into the cooling tower, and (c) the mass flow rate of the required makeup water. Review Problems
14–122 The condensation of the water vapor in compressedair lines is a major concern in industrial facilities, and the compressed air is often dehumidified to avoid the problems associated with condensation. Consider a compressor that compresses ambient air from the local atmospheric pressure of 92 kPa to a pressure of 800 kPa (absolute). The compressed air is then cooled to the ambient temperature as it 1 atm 40 C 20% AIR 25 C 80% FIGURE P14–128
14–129 If the system of Prob. 14–128 is operated as an adiabatic system and the air produced by this system has a rela cen84959_ch14.qxd 9/15/06 6:05 AM Page 769 Chapter 14
tive humidity of 80 percent, what is the temperature of the air produced? Answer: 24.6 C 14–130E Ambient conditions for a summer day in Atlanta, Georgia are 1 atm, 90 F, and 85 percent relative humidity. Determine the cooling required to condition 1000 ft 3 of this air to 75 F and 50 percent relative humidity. What is the minimum humid air temperature required to meet this cooling requirement? 14–131E A simple ideal vaporcompression refrigeration system using refrigerant134a provides the cooling of the Prob. 14–130E. This system operates its condenser such that the saturation temperature is 19.5 F higher than the summer air to which it rejects heat and its evaporator is 10 F cooler than the lowest humid air temperature in the system. What is the system’s COP? Answer: 6.40 14–132E A typical winter day in Fairbanks, Alaska has a pressure of 1 atm, a temperature of 32 F, and a relative humidity of 60 percent. What is the relative humidity inside a home where this air has been heated to 70 F? 14–137  769 Reconsider Prob. 14–136. Using EES (or other) software, determine the properties of the air at the initial state. Study the effect of heating the air at constant volume until the pressure is 110 kPa. Plot the required heat transfer, in kJ, as a function of pressure. 14–138E Air at 15 psia, 60°F, and 50 percent relative humidity flows in an 8in diameter duct at a velocity of 50 ft/s. Determine (a) the dewpoint temperature, (b) the volume flow rate of air, and (c) the mass flow rate of dry air. 14–139 Air enters a cooling section at 97 kPa, 35°C, and 30 percent relative humidity at a rate of 6 m3/min, where it is cooled until the moisture in the air starts condensing. Determine (a) the temperature of the air at the exit and (b) the rate of heat transfer in the cooling section. 14–140 Outdoor air enters an airconditioning system at 10°C and 40 percent relative humidity at a steady rate of 22 m3/min, and it leaves at 25°C and 55 percent relative humidity. The outdoor air is first heated to 22°C in the heating section and then humidified by the injection of hot steam in the humidifying section. Assuming the entire process takes place at a pressure of 1 atm, determine (a) the rate of heat supply in the heating section and (b) the mass flow rate of steam required in the humidifying section. 14–141 Air enters an airconditioning system that uses refrigerant134a at 30°C and 70 percent relative humidity at a rate of 4 m3/min. The refrigerant enters the cooling section at 700 kPa with a quality of 20 percent and leaves as saturated vapor. The air is cooled to 20°C at a pressure of 1 atm. Determine (a) the rate of dehumidification, (b) the rate of heat transfer, and (c) the mass flow rate of the refrigerant. 14–142 for air. Repeat Prob. 14–141 for a total pressure of 95 kPa Heating coils AIR 1 atm 32 F 60% 70 F FIGURE P14–132E
14–133E The relative humidity of the air in a home discussed in Prob. 14–132E is to be restored to 60 percent by evaporating 60 F water into the air. How much heat, in Btu, is required to do this in a home of 16,000 ft 3 volume? 14–134E The U.S. Department of Energy estimates that 190,000 barrels of oil would be saved per day if every household in the United States raised the thermostat setting in summer by 6°F (3.3°C). Assuming the average cooling season to be 120 days and the cost of oil to be $70/barrel, determine how much money would be saved per year. 14–135 The airconditioning costs of a house can be reduced by up to 10 percent by installing the outdoor unit (the condenser) of the air conditioner at a location shaded by trees and shrubs. If the airconditioning costs of a house are $500 a year, determine how much the trees will save the home owner in the 20year life of the system. 14–136 A 3m3 tank contains saturated air at 25°C and 97 kPa. Determine (a) the mass of the dry air, (b) the specific humidity, and (c) the enthalpy of the air per unit mass of the dry air. Answers: (a) 3.29 kg, (b) 0.0210 kg H2O/kg dry air,
(c) 78.6 kJ/kg dry air 14–143 An airconditioning system operates at a total pressure of 1 atm and consists of a heating section and an evaporative cooler. Air enters the heating section at 10°C and 70 percent relative humidity at a rate of 30 m3/min, and it leaves the evaporative cooler at 20°C and 60 percent relatively humidity. Determine (a) the temperature and relative humidity of the air when it leaves the heating section, (b) the rate of heat transfer in the heating section, and (c) the rate of water added to air in the evaporative cooler. Answers:
(a) 28.3°C, 22.3 percent, (b) 696 kJ/min, (c) 0.13 kg/min 14–144 Reconsider Prob. 14–143. Using EES (or other) software, study the effect of total pressure in the range 94 to 100 kPa on the results required in the problem. Plot the results as functions of total pressure. 14–145 14–146 Repeat Prob. 14–143 for a total pressure of 96 kPa. A naturaldraft cooling tower is to remove 50 MW of waste heat from the cooling water that enters the tower at 42°C and leaves at 27°C. Atmospheric air enters the tower at 1 atm with dry and wetbulb cen84959_ch14.qxd 9/15/06 6:05 AM Page 770 770  Thermodynamics
14–151 A room contains 50 kg of dry air and 0.6 kg of water vapor at 25°C and 95 kPa total pressure. The relative humidity of air in the room is (a) 1.2% (d ) 65.2% (b) 18.4% (e) 78.0% (c) 56.7% temperatures of 23 and 18°C, respectively, and leaves saturated at 37°C. Determine (a) the mass flow rate of the cooling water, (b) the volume flow rate of air into the cooling tower, and (c) the mass flow rate of the required makeup water. 14–147 Reconsider Prob. 14–146. Using EES (or other) software, investigate the effect of air inlet wetbulb temperature on the required air volume flow rate and the makeup water flow rate when the other input data are the stated values. Plot the results as functions of wetbulb temperature. 14–148 Atmospheric air enters an airconditioning system at 30°C and 70 percent relative humidity with a volume flow rate of 4 m3/min and is cooled to 20°C and 20 percent relative humidity at a pressure of 1 atm. The system uses refrigerant134a as the cooling fluid that enters the cooling section at 350 kPa with a quality of 20 percent and leaves as a saturated vapor. Show the process on the psychrometric chart. What is the heat transfer from the air to the cooling coils, in kW? If any water is condensed from the air, how much water will be condensed from the atmospheric air per min? Determine the mass flow rate of the refrigerant, in kg/min. 14–152 A 40m3 room contains air at 30°C and a total pressure of 90 kPa with a relative humidity of 75 percent. The mass of dry air in the room is (a) 24.7 kg (d ) 41.4 kg (b) 29.9 kg (e) 52.3 kg (c) 39.9 kg 14–153 A room contains air at 30°C and a total pressure of 96.0 kPa with a relative humidity of 75 percent. The partial pressure of dry air is (a) 82.0 kPa (d ) 90.6 kPa (b) 85.8 kPa (e) 72.0 kPa (c) 92.8 kPa 14–154 The air in a house is at 20°C and 50 percent relative humidity. Now the air is cooled at constant pressure. The temperature at which the moisture in the air will start condensing is (a) 8.7°C (d ) 9.3°C (b) 11.3°C (e) 10.0°C (c) 13.8°C Cooling coils 14–155 On the psychrometric chart, a cooling and dehumidification process appears as a line that is (a) horizontal to the left (b) vertical downward (c) diagonal upwards to the right (NE direction) (d ) diagonal upwards to the left (NW direction) (e) diagonal downwards to the left (SW direction) AIR Condensate 14–156 On the psychrometric chart, a heating and humidification process appears as a line that is (a) horizontal to the right (b) vertical upward (c) diagonal upwards to the right (NE direction) (d ) diagonal upwards to the left (NW direction) (e) diagonal downwards to the right (SE direction) 14–157 An air stream at a specified temperature and relative humidity undergoes evaporative cooling by spraying water into it at about the same temperature. The lowest temperature the air stream can be cooled to is (a) the dry bulb temperature at the given state (b) the wet bulb temperature at the given state (c) the dew point temperature at the given state (d ) the saturation temperature corresponding to the humidity ratio at the given state (e) the triple point temperature of water 14–158 Air is cooled and dehumidified as it flows over the coils of a refrigeration system at 85 kPa from 30°C and a humidity ratio of 0.023 kg/kg dry air to 15°C and a humidity FIGURE P14–148
14–149 An uninsulated tank having a volume of 0.5 m3 contains air at 35°C, 130 kPa, and 20 percent relative humidity. The tank is connected to a water supply line in which water flows at 50°C. Water is sprayed into the tank until the relative humidity of the air–vapor mixture is 90 percent. Determine the amount of water supplied to the tank, in kg, the final pressure of the air–vapor mixture in the tank, in kPa, and the heat transfer required during the process to maintain the air– vapor mixture in the tank at 35°C. Fundamentals of Engineering (FE) Exam Problems
14–150 A room is filled with saturated moist air at 25°C and a total pressure of 100 kPa. If the mass of dry air in the room is 100 kg, the mass of water vapor is (a) 0.52 kg (d ) 2.04 kg (b) 1.97 kg (e) 3.17 kg (c) 2.96 kg cen84959_ch14.qxd 9/15/06 6:05 AM Page 771 Chapter 14
ratio of 0.015 kg/kg dry air. If the mass flow rate of dry air is 0.7 kg/s, the rate of heat removal from the air is (a) 5 kJ/s (d ) 20 kJ/s (b) 10 kJ/s (e) 25 kJ/s (c) 15 kJ/s  771 14–159 Air at a total pressure of 90 kPa, 15°C, and 75 percent relative humidity is heated and humidified to 25°C and 75 percent relative humidity by introducing water vapor. If the mass flow rate of dry air is 4 kg/s, the rate at which steam is added to the air is (a) 0.032 kg/s (d ) 0.0079 kg/s (b) 0.013 kg/s (e) 0 kg/s (c) 0.019 kg/s 14–162 Design an inexpensive evaporative cooling system suitable for use in your house. Show how you would obtain a water spray, how you would provide airflow, and how you would prevent water droplets from drifting into the living space. 14–163 The operation of a cooling tower is governed by the principles of fluid mechanics, heat transfer, and mass transfer, as well as thermodynamics. The laws of thermodynamics do place bounds on the conditions under which satisfactory operation may be expected, while the other sciences determine equipment sizes and other factors. Use the second law as expressed by the increase in entropy or other appropriate principle and the first law to place bounds on the humid air at its inlet in comparison to the conditions at the liquidwater inlet. Do the same for the humidair outlet conditions as compared to the liquidwater outlet conditions. 14–164 A hurricane is a large heat engine driven by the exchange of water with humid air. Evaporation of ocean water occurs as the air approaches the eye of the storm, and condensation occurs as rain near the eye of the storm. Develop a plot of the wind speed near the eye of the storm as a function of the amount of water released from the air as rain. On this plot, indicate the minimum air temperature and relative humidity necessary to sustain each wind speed. Hint: As an upper bound, all of the energy released by the condensing water would be converted into kinetic energy. Design and Essay Problems
14–160 The airconditioning needs of a large building can be met by a single central system or by several individual window units. Considering that both approaches are commonly used in practice, the right choice depends on the situation on hand. Identify the important factors that need to be considered in decision making, and discuss the conditions under which an airconditioning system that consists of several window units is preferable over a large single central system, and vice versa. 14–161 Identify the major sources of heat gain in your house in summer, and propose ways of minimizing them and thus reducing the cooling load. cen84959_ch14.qxd 9/15/06 6:05 AM Page 772 ...
View
Full
Document
This note was uploaded on 03/09/2009 for the course ME 430 taught by Professor Y during the Spring '09 term at CUNY City.
 Spring '09
 Y

Click to edit the document details