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14 14-1 Chapter GAS-VAPOR MIXTURES AND AIR CONDITIONING Dry and Atmospheric Air, Specific and Relative Humidity 14-1C Yes; by cooling the air at constant pressure. 14-2C Yes. 14-3C Specific humidity will decrease but relative humidity will increase. 14-4C Dry air does not contain any water vapor, but atmospheric air does. 14-5C Yes, the water vapor in the air can be treated as an ideal gas because of its very low partial pressure. 14-6C The partial pressure of the water vapor in atmospheric air is called vapor pressure. 14-7C The same. This is because water vapor behaves as an ideal gas at low pressures, and the enthalpy of an ideal gas depends on temperature only. 14-8C Specific humidity is the amount of water vapor present in a unit mass of dry air. Relative humidity is the ratio of the actual amount of vapor in the air at a given temperature to the maximum amount of vapor air can hold at that temperature. 14-9C The specific humidity will remain constant, but the relative humidity will decrease as the temperature rises in a well-sealed room. 14-10C The specific humidity will remain constant, but the relative humidity will decrease as the temperature drops in a well-sealed room. 14-11C A tank that contains moist air at 3 atm is located in moist air that is at 1 atm. The driving force for moisture transfer is the vapor pressure difference, and thus it is possible for the water vapor to flow into the tank from surroundings if the vapor pressure in the surroundings is greater than the vapor pressure in the tank. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-2 14-12C Insulations on chilled water lines are always wrapped with vapor barrier jackets to eliminate the possibility of vapor entering the insulation. This is because moisture that migrates through the insulation to the cold surface will condense and remain there indefinitely with no possibility of vaporizing and moving back to the outside. 14-13C When the temperature, total pressure, and the relative humidity are given, the vapor pressure can be determined from the psychrometric chart or the relation Pv = Psat where Psat is the saturation (or boiling) pressure of water at the specified temperature and is the relative humidity. 14-14E Humid air is expanded in an isentropic nozzle. The amount of water vapor that has condensed during the process is to be determined. Assumptions The air and the water vapor are ideal gases. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2a). The saturation properties of water are to be obtained from water tables. Analysis Since the mole fraction of the water vapor in this mixture is very small, P T2 = T1 2 P 1 ( k -1) / k 15 psia = (860 R ) 100 psia 0.4/1.4 = 500 R We will assume that the air leaves the nozzle at a relative humidity of 100% (will be verified later). The vapor pressure and specific humidity at the outlet are then Pv , 2 = 2 Pg , 2 = 2 Psat @ 40F = (1.0)(0.12173 psia) = 0.1217 psia 100 psia 400F 1=0.025 AIR 15 psia 2 = 0.622 Pv , 2 P - Pv , 2 = (0.622)(0.1217 psia) = 0.00509 lbm H 2 O/lbm dry air (15 - 0.1217) psia This is less than the inlet specific humidity (0.025 lbm/lbm dry air), the relative humidity at the outlet must be 100% as originally assumed. The amount of liquid formation is then = 1 - 2 = 0.025 - 0.00509 = 0.0199 lbm H 2 O/lbm dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-3 14-15 Humid air is compressed in an isentropic compressor. The relative humidity of the air at the compressor outlet is to be determined. Assumptions The air and the water vapor are ideal gases. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2a). The saturation properties of water are to be obtained from water tables. Analysis At the inlet, Pv ,1 = 1 Pg ,1 = 1 Psat @ 20C = (0.90)(2.3392 kPa) = 2.105 kPa 800 kPa 2 = 1 = 0.622 Pv,1 P - Pv ,1 = (0.622)(2.105 kPa) = 0.0134 kg H 2 O/kg dry air (100 - 2.105) kPa Humid air Since the mole fraction of the water vapor in this mixture is very small, P T2 = T1 2 P 1 ( k -1) / k 800 kPa = (293 K ) 100 kPa 0.4/1.4 = 531 K The saturation pressure at this temperature is Pg , 2 = Psat @ 258C = 4542 kPa (from EES) 100 kPa 20C 90% RH The vapor pressure at the exit is Pv , 2 = 2 P2 2 + 0.622 = (0.0134)(800) = 16.87 kPa 0.0134 + 0.622 The relative humidity at the exit is then 2 = Pv ,2 Pg , 2 = 16.87 = 0.0037 = 0.37% 4542 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-4 14-16 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity, and the volume of the tank are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity can be determined form its definition, = mv 0.3 kg = = 0.0143 kg H 2 O/kg dry air ma 21 kg (b) The saturation pressure of water at 30C is Pg = Psat @ 30C = 4.2469 kPa 21 kg dry air 0.3 kg H2O vapor 30C 100 kPa Then the relative humidity can be determined from = (0.0143)(100 kPa) P = = 52.9% (0.622 + ) Pg (0.622 + 0.0143)(4.2469 kPa) (c) The volume of the tank can be determined from the ideal gas relation for the dry air, Pv = Pg = (0.529)(4.2469 kPa) = 2.245 kPa Pa = P - Pv = 100 - 2.245 = 97.755 kPa m a R a T (21 kg)(0.287 kJ/kg K)(303 K) = = 18.7 m 3 97.755 kPa Pa V = 14-17 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity, and the volume of the tank are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity can be determined form its definition, = mv 0.3 kg = = 0.0143 kg H 2 O/kg dry air 21 kg ma (b) The saturation pressure of water at 24C is Pg = Psat @24C = 2.986 kPa 21 kg dry air 0.3 kg H2O vapor 24C 100 kPa Then the relative humidity can be determined from = (0.0143)(100 kPa) P = = 75.2% (0.622 + ) Pg (0.622 + 0.0143)2.986 kPa (c) The volume of the tank can be determined from the ideal gas relation for the dry air, Pv = Pg = (0.752)(2.986 kPa) = 2.245 kPa Pa = P - Pv = 100 - 2.245 = 97.755 kPa m a R a T (21 kg)(0.287 kJ/kg K)(297 K) = = 18.3 m 3 97.755 kPa Pa V = PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-5 14-18 A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The partial pressure of dry air can be determined from Pv = Pg = Psat @ 20C = (0.85)(2.3392 kPa) = 1.988 kPa Pa = P - Pv = 98 - 1.988 = 96.01 kPa (b) The specific humidity of air is determined from AIR 20C 98 kPa 85% RH = 0.622 Pv (0.622)(1.988 kPa) = = 0.0129 kg H 2 O/kg dry air (98 - 1.988) kPa P - Pv (c) The enthalpy of air per unit mass of dry air is determined from h = ha + hv c p T + h g = (1.005 kJ/kg C)(20C) + (0.0129)(2537.4 kJ/kg) = 52.78 kJ/kg dry air 14-19 A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The partial pressure of dry air can be determined from Pv = Pg = Psat @ 20C = (0.85)(2.3392 kPa) = 1.988 kPa Pa = P - Pv = 85 - 1.988 = 83.01 kPa (b) The specific humidity of air is determined from AIR 20C 85 kPa 85% RH = 0.622 Pv (0.622)(1.988 kPa) = = 0.0149 kg H 2 O/kg dry air (85 - 1.988) kPa P - Pv (c) The enthalpy of air per unit mass of dry air is determined from h = ha + hv c p T + h g = (1.005 kJ/kg C)(20C) + (0.0149)(2537.4 kJ/kg) = 57.90 kJ/kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-6 14-20E A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The partial pressure of dry air can be determined from Pv = Pg = Psat @ 70 F = (0.85)(0.36334 psia) = 0.309 psia Pa = P - Pv = 14.6 - 0.309 = 14.291 psia (b) The specific humidity of air is determined from AIR 70F 14.6 psia 85% RH = 0.622 Pv (0.622)(0.309 psia) = = 0.0134 lbm H 2 O/lbm dry air (14.6 - 0.309) psia P - Pv (c) The enthalpy of air per unit mass of dry air is determined from h = ha + hv c p T + h g = (0.24 Btu/lbm F)(70F) + (0.0134)(1091.8 Btu/lbm) = 31.43 Btu/lbm dry air 14-21 The masses of dry air and the water vapor contained in a room at specified conditions and relative humidity are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis The partial pressure of water vapor and dry air are determined to be Pv = Pg = Psat @ 23C = (0.50)(2.811 kPa) = 1.41 kPa Pa = P - Pv = 98 - 1.41 = 96.59 kPa The masses are determined to be ma = mv = PaV (96.59 kPa)(240 m 3 ) = = 272.9 kg R a T (0.287 kPa m 3 /kg K)(296 K) PvV (1.41 kPa)(240 m 3 ) = = 2.47 kg Rv T (0.4615 kPa m 3 /kg K)(296 K) ROOM 240 m3 23C 98 kPa 50% RH PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-7 Dew-point, Adiabatic Saturation, and Wet-bulb Temperatures 14-22C Dew-point temperature is the temperature at which condensation begins when air is cooled at constant pressure. 14-23C Andy's. The temperature of his glasses may be below the dew-point temperature of the room, causing condensation on the surface of the glasses. 14-24C The outer surface temperature of the glass may drop below the dew-point temperature of the surrounding air, causing the moisture in the vicinity of the glass to condense. After a while, the condensate may start dripping down because of gravity. 14-25C When the temperature falls below the dew-point temperature, dew forms on the outer surfaces of the car. If the temperature is below 0C, the dew will freeze. At very low temperatures, the moisture in the air will freeze directly on the car windows. 14-26C When the air is saturated (100% relative humidity). 14-27C These two are approximately equal at atmospheric temperatures and pressure. 14-28 A house contains air at a specified temperature and relative humidity. It is to be determined whether any moisture will condense on the inner surfaces of the windows when the temperature of the window drops to a specified value. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv is uniform throughout the house, and its value can be determined from Pv = Pg @ 25C = (0.65)(3.1698 kPa) = 2.06 kPa = 65% 10C 25C The dew-point temperature of the air in the house is Tdp = Tsat @ Pv = Tsat @ 2.06 kPa = 18.0C That is, the moisture in the house air will start condensing when the temperature drops below 18.0C. Since the windows are at a lower temperature than the dew-point temperature, some moisture will condense on the window surfaces. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-8 14-29 A person wearing glasses enters a warm room at a specified temperature and relative humidity from the cold outdoors. It is to be determined whether the glasses will get fogged. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from Pv = Pg @ 25C = (0.40)(3.1698 kPa) = 1.268 kPa The dew-point temperature of the air in the house is Tdp = Tsat @ Pv = Tsat @ 1.268 kPa = 10.5C (from EES) 25C = 40% 8C That is, the moisture in the house air will start condensing when the air temperature drops below 10.5C. Since the glasses are at a lower temperature than the dew-point temperature, some moisture will condense on the glasses, and thus they will get fogged. 14-30 A person wearing glasses enters a warm room at a specified temperature and relative humidity from the cold outdoors. It is to be determined whether the glasses will get fogged. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from Pv = Pg @ 25C = (0.30)(3.1698 kPa) = 0.95 kPa The dew-point temperature of the air in the house is Tdp = Tsat @ Pv = Tsat @ 0.95 kPa = 6.2C (from EES) 25C = 30% 8C That is, the moisture in the house air will start condensing when the air temperature drops below 6.2C. Since the glasses are at a higher temperature than the dew-point temperature, moisture will not condense on the glasses, and thus they will not get fogged. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-9 14-31E A woman drinks a cool canned soda in a room at a specified temperature and relative humidity. It is to be determined whether the can will sweat. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from Pv = Pg @ 80 F = (0.50)(0.50745 psia) = 0.254 psia 80F 50% RH Cola 40F The dew-point temperature of the air in the house is Tdp = Tsat @ Pv = Tsat @ 0.254 psia = 59.7F (from EES) That is, the moisture in the house air will start condensing when the air temperature drops below 59.7C. Since the canned drink is at a lower temperature than the dew-point temperature, some moisture will condense on the can, and thus it will sweat. 14-32 The dry- and wet-bulb temperatures of atmospheric air at a specified pressure are given. The specific humidity, the relative humidity, and the enthalpy of air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) We obtain the properties of water vapor from EES. The specific humidity 1 is determined from 1 = c p (T2 - T1 ) + 2 h fg 2 h g1 - h f 2 where T2 is the wet-bulb temperature, and 2 is determined from 95 kPa 25C Twb = 17C 2 = Thus, 0.622 Pg 2 P2 - Pg 2 = (0.622)(1.938 kPa) = 0.01295 kg H 2 O/kg dry air (95 - 1.938) kPa 1 = (1.005 kJ/kg C)(17 - 25)C + (0.01295)(2460.6 kJ/kg) = 0.00963 kg H 2 O/kg dry air (2546.5 - 71.36) kJ/kg (b) The relative humidity 1 is determined from 1 = 1 P1 (0.00963)(95 kPa) = = 0.457 or 45.7% (0.622 + 1 ) Pg1 (0.622 + 0.00963)(3.1698 kPa) (c) The enthalpy of air per unit mass of dry air is determined from h1 = ha1 + 1 hv1 c p T1 + 1 h g1 = (1.005 kJ/kg C)(25C) + (0.00963)(2546.5 kJ/kg) = 49.65 kJ/kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-10 14-33 The dry- and wet-bulb temperatures of air in room at a specified pressure are given. The specific humidity, the relative humidity, and the dew-point temperature are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) We obtain the properties of water vapor from EES. The specific humidity 1 is determined from 1 = c p (T2 - T1 ) + 2 h fg 2 h g1 - h f 2 where T2 is the wet-bulb temperature, and 2 is determined from 100 kPa 22C Twb = 16C 2 = Thus, 0.622 Pg 2 P2 - Pg 2 = (0.622)(1.819 kPa) = 0.01152 kg H 2 O/kg dry air (100 - 1.819) kPa 1 = (1.005 kJ/kg C)(16 - 22)C + (0.01152)(2463.0 kJ/kg) = 0.00903 kg H 2O/kg dry air (2541.1 - 67.17) kJ/kg (b) The relative humidity 1 is determined from 1 = 1 P1 (0.00903)(100 kPa) = = 0.541 or 54.1% (0.622 + 1 ) Pg1 (0.622 + 0.0091)(2.6452 kPa) (c) The vapor pressure at the inlet conditions is Pv1 = 1 Pg1 = 1 Psat @ 22C = (0.541)(2.6452 kPa) = 1.432 kPa Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 1.432 kPa = 12.3C PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-11 14-34 EES Problem 14-33 is reconsidered. The required properties are to be determined using EES at 100 and 300 kPa pressures. Analysis The problem is solved using EES, and the solution is given below. Tdb=22 [C] Twb=16 [C] P1=100 [kPa] P2=300 [kPa] h1=enthalpy(AirH2O;T=Tdb;P=P1;B=Twb) v1=volume(AirH2O;T=Tdb;P=P1;B=Twb) Tdp1=dewpoint(AirH2O;T=Tdb;P=P1;B=Twb) w1=humrat(AirH2O;T=Tdb;P=P1;B=Twb) Rh1=relhum(AirH2O;T=Tdb;P=P1;B=Twb) h2=enthalpy(AirH2O;T=Tdb;P=P2;B=Twb) v2=volume(AirH2O;T=Tdb;P=P2;B=Twb) Tdp2=dewpoint(AirH2O;T=Tdb;P=P2;B=Twb) w2=humrat(AirH2O;T=Tdb;P=P2;B=Twb) Rh2=relhum(AirH2O;T=Tdb;P=P2;B=Twb) SOLUTION h1=45.09 [kJ/kga] h2=25.54 [kJ/kga] P1=100 [kPa] P2=300 [kPa] Rh1=0.541 Rh2=0.243 Tdb=22 [C] Tdp1=12.3 [C] Tdp2=0.6964 [C] Twb=16 [C] v1=0.8595 [m^3/kga] v2=0.283 [m^3/kga] w1=0.009029 [kgv/kga] w2=0.001336 [kgv/kga] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-12 14-35E The dry- and wet-bulb temperatures of air in room at a specified pressure are given. The specific humidity, the relative humidity, and the dew-point temperature are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity 1 is determined from c p (T2 - T1 ) + 2 h fg 2 h g1 - h f 2 1 = 14.7 psia 80F Twb = 65F where T2 is the wet-bulb temperature, and 2 is determined from 2 = Thus, 0.622 Pg 2 P2 - Pg 2 = (0.622)(0.30578 psia) = 0.01321 lbm H 2 O/lbm dry air (14.7 - 0.30578) psia 1 = (0.24 Btu/lbm F)(65 - 80)F + (0.01321)(1056.5 Btu/lbm) = 0.00974 lbm H 2O/lbm dry air (1096.1 - 33.08) Btu/lbm (b) The relative humidity 1 is determined from 1 = 1 P1 (0.00974)(14.7 psia) = = 0.447 or 44.7% (0.622 + 1 ) Pg1 (0.622 + 0.00974)(0.50745 psia) (c) The vapor pressure at the inlet conditions is Pv1 = 1 Pg1 = 1 Psat @ 70F = (0.447)(0.50745 psia) = 0.2268 psia Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 0.2268 psia = 56.6F (from EES) PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-13 14-36 Atmospheric air flows steadily into an adiabatic saturation device and leaves as a saturated vapor. The relative humidity and specific humidity of air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The exit state of the air is completely specified, and the total pressure is 98 kPa. The properties of the moist air at the exit state may be determined from EES to be h2 = 78.11 kJ/kg dry air 2 = 0.02079 kg H 2 O/kg dry air The enthalpy of makeup water is h w 2 = h f@ 25C = 104.83 kJ/kg (Table A - 4) Water 25C Humidifier 35C 98 kPa AIR 25C 98 kPa 100% An energy balance on the control volume gives h1 + ( 2 - 1 )h w = h2 h1 + (0.02079 - 1 )(104.83 kJ/kg) = 78.11 kJ/kg Pressure and temperature are known for inlet air. Other properties may be determined from this equation using EES. A hand solution would require a trial-error approach. The results are h1 = 77.66 kJ/kg dry air 1 = 0.01654 kg H 2 O/kg dry air 1 = 0.4511 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-14 Psychrometric Chart 14-37C They are very nearly parallel to each other. 14-38C The saturation states (located on the saturation curve). 14-39C By drawing a horizontal line until it intersects with the saturation curve. The corresponding temperature is the dew-point temperature. 14-40C No, they cannot. The enthalpy of moist air depends on , which depends on the total pressure. 14-41 [Also solved by EES on enclosed CD] The pressure, temperature, and relative humidity of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the wet-bulb temperature, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read (a) = 0.0181 kg H 2 O / kg dry air (b) h = 78.4 kJ / kg dry air (c) Twb = 25.5C (d) Tdp = 23.3C (e) v = 0.890 m 3 / kg dry air 14-42 The pressure, temperature, and relative humidity of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the wet-bulb temperature, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read (a) = 0.0148 kg H 2 O / kg dry air (b) h = 63.9 kJ / kg dry air (c) Twb = 21.9C (d) Tdp = 20.1C (e) v = 0.868 m 3 / kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-15 14-43 EES Problem 14-42 is reconsidered. The required properties are to be determined using EES. Also, the properties are to be obtained at an altitude of 2000 m. Analysis The problem is solved using EES, and the solution is given below. Tdb=26 [C] Rh=0.70 P1=101.325 [kPa] Z = 2000 [m] P2=101.325*(1-0.02256*Z*convert(m,km))^5.256 "Relation giving P as a function of altitude" h1=enthalpy(AirH2O,T=Tdb,P=P1,R=Rh) v1=volume(AirH2O,T=Tdb,P=P1,R=Rh) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,R=Rh) w1=humrat(AirH2O,T=Tdb,P=P1,R=Rh) Twb1=wetbulb(AirH2O,T=Tdb,P=P1,R=Rh) h2=enthalpy(AirH2O,T=Tdb,P=P2,R=Rh) v2=volume(AirH2O,T=Tdb,P=P2,R=Rh) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,R=Rh) w2=humrat(AirH2O,T=Tdb,P=P2,R=Rh) Twb2=wetbulb(AirH2O,T=Tdb,P=P2,R=Rh) SOLUTION h1=63.88 [kJ/kg] h2=74.55 [kJ/kg] P1=101.3 [kPa] P2=79.49 [kPa] Rh=0.7 Tdb=26 [C] Tdp1=20.11 [C] Tdp2=20.11 [C] Twb1=21.87 [C] Twb2=21.59 [C] v1=0.8676 [m^3/kg] v2=1.113 [m^3/kg] w1=0.0148 [kg/kg] w2=0.01899 [kg/kg] Z=2000 [m] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-16 14-44 The pressure and the dry- and wet-bulb temperatures of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the relative humidity, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read (a) = 0.0092 kg H 2 O / kg dry air (b) h = 47.6 kJ / kg dry air (c) = 49.6% (d) Tdp = 12.8C (e) v = 0.855 m3 / kg dry air 14-45 EES Problem 14-44 is reconsidered. The required properties are to be determined using EES. Also, the properties are to be obtained at an altitude of 3000 m. Analysis The problem is solved using EES, and the solution is given below. Tdb=24 [C] Twb=17 [C] P1=101.325 [kPa] Z = 3000 [m] P2=101.325*(1-0.02256*Z*convert(m,km))^5.256 "Relation giving P as function of altitude" h1=enthalpy(AirH2O,T=Tdb,P=P1,B=Twb) v1=volume(AirH2O,T=Tdb,P=P1,B=Twb) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,B=Twb) w1=humrat(AirH2O,T=Tdb,P=P1,B=Twb) Rh1=relhum(AirH2O,T=Tdb,P=P1,B=Twb) h2=enthalpy(AirH2O,T=Tdb,P=P2,B=Twb) v2=volume(AirH2O,T=Tdb,P=P2,B=Twb) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,B=Twb) w2=humrat(AirH2O,T=Tdb,P=P2,B=Twb) Rh2=relhum(AirH2O,T=Tdb,P=P2,B=Twb) SOLUTION h1=47.61 [kJ/kg] P1=101.3 [kPa] Rh1=0.4956 Tdb=24 [C] Tdp2=14.24 [C] v1=0.8542 [m^3/kg] w1=0.009219 [kg/kg] Z=3000 [m] h2=61.68 [kJ/kg] P2=70.11 [kPa] Rh2=0.5438 Tdp1=12.81 [C] Twb=17 [C] v2=1.245 [m^3/kg] w2=0.01475 [kg/kg] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-17 14-46 The pressure, temperature, and relative humidity of air are specified. Using the psychrometric chart, the wet-bulb temperature, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be determined. Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain (a) Twb = 27.1C (b) = 0.0217 kg H 2 O / kg dry air (c) h = 85.5 kJ/kg dry air (d) Tdp = 26.2C (e) Pv = Pg = Psat @ 30C = (0.80)(4.2469 kPa) = 3.40 kPa Air 1 atm 30C 80% RH 14-47E The pressure, temperature, and wet-bulb temperature of air are specified. Using the psychrometric chart, the relative humidity, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be determined. Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain (a) = 0.816 = 81.6% (b) = 0.0252 lbm H 2 O / lbm dry air (c) h = 49.4 Btu/lbm dry air (d) Tdp = 83.7F (e) Pv = Pg = Psat @ 90F = (0.816)(0.69904 psia) = 0.570 psia Air 1 atm 90F Twb=85F PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-18 14-48E The pressure, temperature, and wet-bulb temperature of air are specified. The adiabatic saturation temperature is to be determined. Analysis For an adiabatic saturation process, we obtained Eq. 14-14 in the text, c p (T2 - T1 ) + 2 h fg 2 h g1 - h f 2 1 = Water Humidifier 1 atm 90F Twb=85F AIR 100% This requires a trial-error solution for the adiabatic saturation temperature, T2. The inlet state properties are 1 = 0.0252 lbm H 2 O / lbm dry air h g1 = h g @ 90F = 1100.4 Btu/lbm As a first estimate, let us take T2 =85F (the inlet wet-bulb temperature). Also, at the exit, the relative humidity is 100% ( 2 = 1 ) and the pressure is 1 atm. Other properties at the exit state are 2 = 0.0264 lbm H 2 O / lbm dry air h f 2 = h f @ 85F = 53.06 Btu/lbm (Table A - 4E) h fg 2 = h fg @ 85F = 1045.2 Btu/lbm (Table A - 4E) Substituting, 1 = c p (T2 - T1 ) + 2 h fg 2 h g1 - h f 2 = (0.240)(85 - 90) + (0.0264)(1045.2) = 0.0252 lbm H 2 O / lbm dry air 1100.4 - 53.06 which is equal to the inlet specific humidity. Therefore, the adiabatic saturation temperature is T2 = 85F Discussion This result is not surprising since the wet-bulb and adiabatic saturation temperatures are approximately equal to each other for air-water mixtures at atmospheric pressure. 14-49 The pressure, temperature, and wet-bulb temperature of air are specified. Using the psychrometric chart, the relative humidity, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be determined. Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain (a) = 0.618 = 61.8% (b) = 0.0148 kg H 2 O / kg dry air (c) h = 65.8 kJ/kg dry air (d) Twb = 22.4C (e) Pv = Pg = Psat @ 28C = (0.618)(3.780 kPa) = 2.34 kPa Air 1 atm 28C Tdp=20C PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-19 14-50 The pressure, temperature, and wet-bulb temperature of air are specified. The adiabatic saturation temperature is to be determined. Analysis For an adiabatic saturation process, we obtained Eq. 14-14 in the text, c p (T2 - T1 ) + 2 h fg 2 h g1 - h f 2 1 = Water Humidifier 1 atm 28C Tdp=20C AIR 100% This requires a trial-error solution for the adiabatic saturation temperature, T2. The inlet state properties are 1 = 0.0148 kg H 2 O / kg dry air (Fig. A-31) h g1 = h g @ 28C = 2551.9 kJ/kg (Table A-4) As a first estimate, let us take T2 =22C (the inlet wet-bulb temperature). Also, at the exit, the relative humidity is 100% ( 2 = 1 ) and the pressure is 1 atm. Other properties at the exit state are 2 = 0.0167 kg H 2 O / kg dry air h f 2 = h f @ 22C = 92.28 kJ/kg (Table A - 4) h fg 2 = h fg @ 22C = 2448.8 kJ/kg (Table A - 4) Substituting, 1 = c p (T2 - T1 ) + 2 h fg 2 h g1 - h f 2 = (1.005)(22 - 28) + (0.0167)(2448.8) = 0.0142 kg H 2 O / kg dry air 2551.9 - 92.28 which is sufficiently close to the inlet specific humidity (0.0148). Therefore, the adiabatic saturation temperature is T2 22C Discussion This result is not surprising since the wet-bulb and adiabatic saturation temperatures are approximately equal to each other for air-water mixtures at atmospheric pressure. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-20 Human Comfort and Air-Conditioning 14-51C It humidifies, dehumidifies, cleans and even deodorizes the air. 14-52C (a) Perspires more, (b) cuts the blood circulation near the skin, and (c) sweats excessively. 14-53C It is the direct heat exchange between the body and the surrounding surfaces. It can make a person feel chilly in winter, and hot in summer. 14-54C It affects by removing the warm, moist air that builds up around the body and replacing it with fresh air. 14-55C The spectators. Because they have a lower level of activity, and thus a lower level of heat generation within their bodies. 14-56C Because they have a large skin area to volume ratio. That is, they have a smaller volume to generate heat but a larger area to lose it from. 14-57C It affects a body's ability to perspire, and thus the amount of heat a body can dissipate through evaporation. 14-58C Humidification is to add moisture into an environment, dehumidification is to remove it. 14-59C The metabolism refers to the burning of foods such as carbohydrates, fat, and protein in order to perform the necessary bodily functions. The metabolic rate for an average man ranges from 108 W while reading, writing, typing, or listening to a lecture in a classroom in a seated position to 1250 W at age 20 (730 at age 70) during strenuous exercise. The corresponding rates for women are about 30 percent lower. Maximum metabolic rates of trained athletes can exceed 2000 W. We are interested in metabolic rate of the occupants of a building when we deal with heating and air conditioning because the metabolic rate represents the rate at which a body generates heat and dissipates it to the room. This body heat contributes to the heating in winter, but it adds to the cooling load of the building in summer. 14-60C The metabolic rate is proportional to the size of the body, and the metabolic rate of women, in general, is lower than that of men because of their smaller size. Clothing serves as insulation, and the thicker the clothing, the lower the environmental temperature that feels comfortable. 14-61C Sensible heat is the energy associated with a temperature change. The sensible heat loss from a human body increases as (a) the skin temperature increases, (b) the environment temperature decreases, and (c) the air motion (and thus the convection heat transfer coefficient) increases. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-21 14-62C Latent heat is the energy released as water vapor condenses on cold surfaces, or the energy absorbed from a warm surface as liquid water evaporates. The latent heat loss from a human body increases as (a) the skin wetness increases and (b) the relative humidity of the environment decreases. The & & rate of evaporation from the body is related to the rate of latent heat loss by Qlatent = mvapor h fg where hfg is the latent heat of vaporization of water at the skin temperature. 14-63 An average person produces 0.25 kg of moisture while taking a shower. The contribution of showers of a family of four to the latent heat load of the air-conditioner per day is to be determined. Assumptions All the water vapor from the shower is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 2450 kJ/kg. Analysis The amount of moisture produced per day is & mvapor = ( Moisture produced per person)(No. of persons) = (0.25 kg / person)(4 persons / day) = 1 kg / day Then the latent heat load due to showers becomes & & Qlatent = mvapor h fg = (1 kg / day)(2450 kJ / kg) = 2450 kJ / day 14-64 There are 100 chickens in a breeding room. The rate of total heat generation and the rate of moisture production in the room are to be determined. Assumptions All the moisture from the chickens is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 2430 kJ/kg. The average metabolic rate of chicken during normal activity is 10.2 W (3.78 W sensible and 6.42 W latent). Analysis The total rate of heat generation of the chickens in the breeding room is & & Qgen, total = q gen, total (No. of chickens) = (10.2 W / chicken)(100 chickens) = 1020 W The latent heat generated by the chicken and the rate of moisture production are & & Qgen, latent = q gen, latent (No. of chickens) = (6.42 W/chicken)(100 chickens) = 642 W = 0.642 kW & mmoisture = & Qgen, latent h fg = 0.642 kJ / s = 0.000264 kg / s = 0.264 g / s 2430 kJ / kg PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-22 14-65 A department store expects to have a specified number of people at peak times in summer. The contribution of people to the sensible, latent, and total cooling load of the store is to be determined. Assumptions There is a mix of men, women, and children in the classroom. Properties The average rate of heat generation from people doing light work is 115 W, and 70% of is in sensible form (see Sec. 14-6). Analysis The contribution of people to the sensible, latent, and total cooling load of the store are & & Qpeople, total = (No. of people) Qperson, total = 135 (115 W) = 15,525 W & & Qpeople, sensible = (No. of people) Qperson, sensible = 135 (0.7 115 W) = 10,868 W & & Qpeople, latent = (No. of people) Qperson, latent = 135 (0.3 115 W) = 4658 W 14-66E There are a specified number of people in a movie theater in winter. It is to be determined if the theater needs to be heated or cooled. Assumptions There is a mix of men, women, and children in the classroom. Properties The average rate of heat generation from people in a movie theater is 105 W, and 70 W of it is in sensible form and 35 W in latent form. Analysis Noting that only the sensible heat from a person contributes to the heating load of a building, the contribution of people to the heating of the building is & & Qpeople, sensible = (No. of people) Qperson, sensible = 500 (70 W) = 35,000 W = 119,420 Btu/h since 1 W = 3.412 Btu/h. The building needs to be heated since the heat gain from people is less than the rate of heat loss of 130,000 Btu/h from the building. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-23 14-67 The infiltration rate of a building is estimated to be 1.2 ACH. The sensible, latent, and total infiltration heat loads of the building at sea level are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air infiltrates at the outdoor conditions, and exfiltrates at the indoor conditions. 3 Excess moisture condenses at room temperature of 24C. 4 The effect of water vapor on air density is negligible. Properties The gas constant and the specific heat of air are R = 0.287 kPa.m3/kg.K and cp = 1.005 kJ/kgC (Table A-2). The heat of vaporization of water at 24C is h fg = h fg @ 24C = 2444.1 kJ/kg (Table A-4). The properties of the ambient and room air are determined from the psychrometric chart (Fig. A-31) to be Tambient = 32 C = 0.0150 kg/kg dryair w ambient = 50% ambient room = 50% Troom = 24 C wroom = 0.0093 kg/kg dryair Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 1.2 times every hour, the air will enter the room at a mass flow rate of ambient = P0 101.325 kPa = = 1.158 kg/m 3 RT0 (0.287 kPa.m 3 /kg.K)(32 + 273 K) & m air = ambientV room ACH = (1.158 kg/m 3 )(20 13 3 m 3 )(1.2 h -1 ) = 1084 kg/h = 0.301 kg/s Then the sensible, latent, and total infiltration heat loads of the room are determined to be & & Qinfiltration, sensible = m air c p (Tambient - Troom ) = (0.301 kg/s)(1.005 kJ/kg.C)(32 - 24)C = 2.42 kW & & Qinfiltration, latent = m air ( wambient - wroom )h fg = (0.301 kg/s)(0.0150 - 0.0093)(2444.1 kJ/kg) = 4.16 kW & & & Qinfiltration, total = Qinfiltration, sensible + Qinfiltration, latent = 2.42 + 4.16 = 6.58 kW Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-24 14-68 The infiltration rate of a building is estimated to be 1.8 ACH. The sensible, latent, and total infiltration heat loads of the building at sea level are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air infiltrates at the outdoor conditions, and exfiltrates at the indoor conditions. 3 Excess moisture condenses at room temperature of 24C. 4 The effect of water vapor on air density is negligible. Properties The gas constant and the specific heat of air are R = 0.287 kPa.m3/kg.K and cp = 1.005 kJ/kgC (Table A-2). The heat of vaporization of water at 24C is h fg = h fg @ 24C = 2444.1 kJ/kg (Table A-4). The properties of the ambient and room air are determined from the psychrometric chart (Fig. A-31) to be Tambient = 32 C = 0.0150 kg/kg dryair w ambient = 50% ambient room = 50% Troom = 24 C wroom = 0.0093 kg/kg dryair Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 1.8 times every hour, the air will enter the room at a mass flow rate of ambient = P0 101.325 kPa = = 1.158 kg/m 3 RT0 (0.287 kPa.m 3 /kg.K)(32 + 273 K) & m air = ambientV room ACH = (1.158 kg/m 3 )(20 13 3 m 3 )(1.8 h -1 ) = 1084 kg/h = 0.4514 kg/s Then the sensible, latent, and total infiltration heat loads of the room are determined to be & & Qinfiltration, sensible = m air c p (Tambient - Troom ) = (0.4514 kg/s)(1.005 kJ/kg.C)(32 - 24)C = 3.63 kW & & Qinfiltration, latent = m air ( wambient - wroom )h fg = (0.4514 kg/s)(0.0150 - 0.0093)(2444.1 kJ/kg) = 6.24 kW & & & Qinfiltration, total = Qinfiltration, sensible + Qinfiltration, latent = 3.63 + 6.24 = 9.87 kW Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-25 Simple Heating and cooling 14-69C Relative humidity decreases during a simple heating process and increases during a simple cooling process. Specific humidity, on the other hand, remains constant in both cases. 14-70C Because a horizontal line on the psychrometric chart represents a = constant process, and the moisture content of air remains constant during these processes. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-26 14-71 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air remains constant ( 1 = 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31) to be h1 = 55.0 kJ/kg dry air 1 = 0.0089 kg H 2 O/kg dry air (= 2 ) 1200 v 1 = 0.877 m 3 / kg dry air The mass flow rate of dry air through the cooling section is & ma = = 1 1 32C 30% 18 m/s 2 1 atm AIR v1 V1 A1 1 3 (18 m/s)( 0.4 2 /4 m 2 ) (0.877 m / kg) = 2.58 kg/s From the energy balance on air in the cooling section, & & - Qout = ma ( h2 - h1 ) -1200 / 60 kJ / s = (2.58 kg / s)( h2 - 55.0) kJ / kg h2 = 47.2 kJ / kg dry air The exit state of the air is fixed now since we know both h2 and 2. From the psychrometric chart at this state we read T2 = 24.4C (b) 2 = 46.6% v 2 = 0.856 m 3 / kg dry air (c) The exit velocity is determined from the conservation of mass of dry air, & & m a1 = m a 2 V2 = V A V A V&1 V&2 = 1 = 2 v1 v 2 v1 v2 v2 0.856 V1 = (18 m/s) = 17.6 m/s v1 0.877 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-27 14-72 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air remains constant ( 1 = 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31) to be h1 = 55.0 kJ/kg dry air 1 = 0.0089 kg H 2 O/kg dry air (= 2 ) v 1 = 0.877 m 3 / kg dry air The mass flow rate of dry air through the cooling section is & ma = = 1 800 kJ/min 1 v1 V1 A1 1 3 32C 30% 18 m/s 2 1 atm AIR (18 m/s)( 0.4 2 /4 m 2 ) (0.877 m / kg) = 2.58 kg/s From the energy balance on air in the cooling section, & & - Qout = ma ( h2 - h1 ) -800 / 60 kJ / s = (2.58 kg / s)( h2 - 55.0) kJ / kg h2 = 49.8 kJ / kg dry air The exit state of the air is fixed now since we know both h2 and 2. From the psychrometric chart at this state we read T2 = 26.9C (b) 2 = 40.0% v 2 = 0.862 m 3 / kg dry air (c) The exit velocity is determined from the conservation of mass of dry air, & & m a1 = m a 2 V2 = V A V A V&1 V&2 = 1 = 2 v1 v 2 v1 v2 v2 0.862 V1 = (18 m/s) = 17.7 m/s v1 0.877 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-28 14-73E Humid air at a specified state is cooled at constant pressure to the dew-point temperature. The cooling required for this process is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The amount of moisture in the air remains constant ( 1 = 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet and exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31E) to be h1 = 56.7 Btu/lbm dry air 1 = 0.0296 lbm H 2 O/lbm dry air (= 2 ) Tdp,1 = 88.4F The exit state enthalpy is P = 1 atm T2 = Tdp,1 = 88.4F 1 = 1 h2 = 53.8 Btu/lbm dry air 1 100F 70% RH 1 atm 100% RH 2 AIR From the energy balance on air in the cooling section, q out = h1 - h2 = 56.7 - 53.8 = 2.9 Btu/lbm dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-29 14-74 Humid air at a specified state is cooled at constant pressure to the dew-point temperature. The cooling required for this process is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The amount of moisture in the air remains constant ( 1 = 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 150 kPa. The properties of the air at the inlet and exit states are determined to be Pv1 = 1 Pg1 = 1 Psat @ 40C = (0.70)(7.3851 kPa) = 5.1696 kPa h g1 = h g @ 40C = 2573.5 kJ/kg 1 = 0.622 Pv1 0.622(5.1696 kPa) = = 0.02220 kg H 2 O/kg dry air (150 - 5.1696) kPa P1 - Pv1 h1 = c p T1 + 1 h g1 = (1.005 kJ/kg C)(40C) + (0.02220)(2573.5 kJ/kg) = 97.33 kJ/kg dry air Pv 2 = Pv1 = 5.1696 kPa Pg 2 = Pv 2 2 = 5.1696 kPa = 5.1696 kPa 1 T2 = Tsat @ 5.1695 kPa = 33.5C h g 2 = h g @ 33.5C = 2561.9 kJ/kg 1 40C 70% RH 150 kPa 100% RH 2 AIR 2 = 1 h2 = c p T2 + 2 h g 2 = (1.005 kJ/kg C)(33.5C) + (0.02220)(2561.9 kJ/kg) = 90.55 kJ/kg dry air From the energy balance on air in the cooling section, q out = h1 - h2 = 97.33 - 90.55 = 6.78 kJ/kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-30 14-75 Saturated humid air at a specified state is heated to a specified temperature. The relative humidity at the exit and the rate of heat transfer are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The amount of moisture in the air remains constant ( 1 = 2) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 200 kPa. The properties of the air at the inlet and exit states are determined to be Pv1 = 1 Pg1 = 1 Psat @ 15C = (1.0)(1.7057 kPa) = 1.7057 kPa h g1 = h g @ 15C = 2528.3 kJ/kg Pa1 = P1 - Pv1 = 200 - 1.7057 = 198.29 kPa R a T1 Pa1 (0.287 kPa m 3 / kg K)(288 K) 198.29 kPa Heating coils 1 15C 100% RH 20 m/s 30C 2 v1 = = 200 kPa AIR = 0.4168 m 3 / kg dry air 1 = 0.622 Pv1 0.622(1.7057 kPa) = = 0.005350 kg H 2 O/kg dry air P1 - Pv1 (200 - 1.7057) kPa h1 = c p T1 + 1 h g1 = (1.005 kJ/kg C)(15C) + (0.005350)(2528.3 kJ/kg) = 28.60 kJ/kg dry air Pv 2 = Pv1 = 1.7057 kPa Pg 2 = Psat @ 30C = 4.2469 kPa 2 = Pv 2 1.7057 kPa = = 0.402 = 40.2% Pg 2 4.2469 kPa h g 2 = h g @ 30C = 2555.6 kJ/kg 2 = 1 h2 = c p T2 + 2 h g 2 = (1.005 kJ/kg C)(30C) + (0.005350)(2555.6 kJ/kg) = 43.82 kJ/kg dry air Then, V&1 = V1 A1 = V1 & ma = D2 4 (0.04 m) 2 = (20 m/s) 4 = 0.02513 m 3 /s V&1 0.02513 m 3 / s = = 0.06029 kg/s v 1 0.4168 m 3 / kg dry air From the energy balance on air in the heating section, & & Qin = m a (h2 - h1 ) = (0.06029 kg/s)(43.82 - 28.60)kJ/kg = 0.918 kW PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-31 14-76 Saturated humid air at a specified state is heated to a specified temperature. The rate at which the exergy of the humid air is increased is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The amount of moisture in the air remains constant ( 1 = 2) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 200 kPa. The properties of the air at the inlet and exit states are determined to be Pv1 = 1 Pg1 = 1 Psat @ 15C = (1.0)(1.7057 kPa) = 1.7057 kPa h g1 = h g @ 15C = 2528.3 kJ/kg s g1 = s g @ 15C = 8.7803 kJ/kg K Pa1 = P1 - Pv1 = 200 - 1.7057 = 198.29 kPa 1 15C 100% RH 200 kPa 30C 2 AIR v1 = 1 = R a T1 (0.287 kPa m 3 / kg K)(288 K) = = 0.4168 m 3 / kg dry air Pa1 198.29 kPa 0.622 Pv1 0.622(1.7057 kPa) = = 0.005350 kg H 2 O/kg dry air P1 - Pv1 (200 - 1.7057) kPa Pv 2 = Pv1 = 1.7057 kPa Pg 2 = Psat @ 30C = 4.2469 kPa h1 = c p T1 + 1 h g1 = (1.005 kJ/kg C)(15C) + (0.005350)(2528.3 kJ/kg) = 28.60 kJ/kg dry air 2 = Pv 2 1.7057 kPa = = 0.402 = 40.2% Pg 2 4.2469 kPa Pa 2 = P2 - Pv 2 = 200 - 1.7057 = 198.29 kPa h g 2 = h g @ 30C = 2555.6 kJ/kg s g 2 = s g @ 30C = 8.4520 kJ/kg K 2 = 1 h2 = c p T2 + 2 h g 2 = (1.005 kJ/kg C)(30C) + (0.005350)(2555.6 kJ/kg) = 43.82 kJ/kg dry air The entropy change of the dry air is P T 303 198.29 ( s 2 - s1 ) dry air = c p ln 2 - R ln a 2 = (1.005) ln - (0.287) ln = 0.05103 kJ/kg K 288 198.29 T1 Pa1 The entropy change of the air-water mixture is s 2 - s1 = ( s 2 - s1 ) dry air + ( s 2 - s1 ) water vapor = 0.05103 + (0.005350) (8.4520 - 8.7803) = 0.04927 kJ/kg K The mass flow rate of the dry air is V&1 = V1 A1 = V1 & ma = D2 4 (0.04 m) 2 = (20 m/s) 4 = 0.02513 m 3 /s V&1 0.02513 m 3 / s = = 0.06029 kg/s v 1 0.4168 m 3 / kg dry air The exergy increase of the humid air during this process is then, & & = m a ( 2 - 1 ) = m a [(h2 - h1 ) - T0 ( s 2 - s1 )] = (0.06029 kg/s)[(43.82 - 28.60)kJ/kg - (288 K)(0.04927 kJ/kg K)] = 0.062 kW/K PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-32 Heating with Humidification 14-77C To achieve a higher level of comfort. Very dry air can cause dry skin, respiratory difficulties, and increased static electricity. 14-78 Air is first heated and then humidified by water vapor. The amount of steam added to the air and the amount of heat transfer to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 311 kJ / kg dry air . 1 = 0.0064 kg H 2 O / kg dry air ( = 2 ) h2 = 36.2 kJ / kg dry air h3 = 581 kJ / kg dry air . Heating coils T1 = 15C 1 = 60% 1 1 atm AIR T2 = 20C 2 3 T3 = 25C 3 = 65% 3 = 0.0129 kg H 2 O / kg dry air Analysis (a) The amount of moisture in the air remains constant it flows through the heating section ( 1 = 2), but increases in the humidifying section ( 3 > 2). The amount of steam added to the air in the heating section is = 3 - 2 = 0.0129 - 0.0064 = 0.0065 kg H 2 O / kg dry air (b) The heat transfer to the air in the heating section per unit mass of air is qin = h2 - h1 = 36.2 - 311 = 5.1 kJ / kg dry air . PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-33 14-79E Air is first heated and then humidified by water vapor. The amount of steam added to the air and the amount of heat transfer to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31E) to be h1 = 17.0 Btu/lbm dry air 1 = 0.0046 lbm H 2 O/lbm dry air h2 = 22.3 Btu/lbm dry air 2 = 1 = 0.0046 lbm H 2 O/lbm dry air h3 = 29.2 Btu/lbm dry air Heating coils T1 = 50F 1 = 60% 1 14.7 psia AIR T2 = 72F 2 3 T3 = 75F 3 = 55% 3 = 0.0102 lbm H 2 O/lbm dry air Analysis (a) The amount of moisture in the air remains constant it flows through the heating section (1 = 2), but increases in the humidifying section ( 3 > 2). The amount of steam added to the air in the heating section is = 3 - 2 = 0.0102 - 0.0046 = 0.0056 lbm H 2 O/lbm dry air (b) The heat transfer to the air in the heating section per unit mass of air is q in = h2 - h1 = 22.3 - 17.0 = 5.3 Btu/lbm dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-34 14-80 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 23.5 kJ/kg dry air 1 = 0.0053 kg H 2 O/kg dry air (= 2 ) v 1 = 0.809 m 3 /kg dry air h3 = 42.3 kJ/kg dry air Heating coils 10C 70% 35 m3/min 1 AIR 1 atm 2 Sat. vapor 100C Humidifier 20C 60% 3 3 = 0.0087 kg H 2 O/kg dry air Analysis (a) The amount of moisture in the air remains constant it flows through the heating section ( 1 = 2), but increases in the humidifying section ( 3 > 2). The mass flow rate of dry air is & ma = 35 m3 / min V&1 = = 43.3 kg/min v1 0.809 m3 / kg Noting that Q = W =0, the energy balance on the humidifying section can be expressed as & & & E in - E out = E system & & E in = E out & & mi hi = me he & & & mw hw + ma 2 h2 = ma h3 ( 3 - 2 )hw + h2 = h3 0 (steady) =0 Solving for h2, h2 = h3 - ( 3 - 2 )h g @ 100C = 42.3 - (0.0087 - 0.0053)(2675.6) = 33.2 kJ/kg dry air Thus at the exit of the heating section we have 2 = 0.0053 kg H2O dry air and h2 = 33.2 kJ/kg dry air, which completely fixes the state. Then from the psychrometric chart we read T2 = 19.5C 2 = 37.8% (b) The rate of heat transfer to the air in the heating section is & & Qin = m a (h2 - h1 ) = (43.3 kg/min)(33.2 - 23.5) kJ/kg = 420 kJ/min (c) The amount of water added to the air in the humidifying section is determined from the conservation of mass equation of water in the humidifying section, & & m w = m a ( 3 - 2 ) = (43.3 kg/min)(0.0087 - 0.0053) = 0.15 kg/min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-35 14-81 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Sat. vapor Analysis (a) The amount of moisture in the 100C Heating air also remains constant it flows through coils Humidifier the heating section ( 1 = 2), but increases in the humidifying section ( 3 > AIR 10C 2). The inlet and the exit states of the air 20C 70% are completely specified, and the total 60% 3 95 kPa 35 m /min pressure is 95 kPa. The properties of the air at various states are determined to be 1 2 3 Pv1 = 1 Pg1 = 1 Psat @ 10C = (0.70)(1.2281 kPa) = 0.860 kPa ( = Pv 2 ) Pa1 = P1 - Pv1 = 95 - 0.860 = 94.14 kPa v1 = R a T1 (0.287 kPa m 3 / kg K)(283 K) = = 0.863 m 3 / kg dry air Pa1 94.14 kPa 0.622 Pv1 0.622(0.86 kPa) = = 0.00568 kg H 2 O/kg dry air (= 2 ) P1 - Pv1 (95 - 0.86) kPa 1 = h1 = c p T1 + 1 h g1 = (1.005 kJ/kg C)(10C) + (0.00568)(2519.2 kJ/kg) = 24.36 kJ/kg dry air Pv 3 = 3 Pg 3 = 3 Psat @ 20C = (0.60)(2.3392 kPa) = 1.40 kPa 3 = 0.622 Pv3 0.622(1.40 kPa) = = 0.00930 kg H 2O/kg dry air P3 - Pv 3 (95 - 1.40) kPa h3 = c pT3 + 3hg 3 = (1.005 kJ/kg C)(20C) + (0.0093)(2537.4 kJ/kg) = 43.70 kJ/kg dry air Also, & ma = V&1 35 m 3 / min = = 40.6 kg/min v 1 0.863 m 3 / kg Noting that Q = W = 0, the energy balance on the humidifying section gives & & & E in - E out = E system 0 (steady) & & =0 E in = E out & & & & & m e h e = m i hi m w hw + m a 2 h2 = m a h3 ( 3 - 2 )h w + h2 = h3 h2 = h3 - ( 3 - 2 )h g @ 100C = 43.7 - (0.0093 - 0.00568) 2675.6 = 34.0 kJ/kg dry air Thus at the exit of the heating section we have = 0.00568 kg H2O dry air and h2 = 34.0 kJ/kg dry air, which completely fixes the state. The temperature of air at the exit of the heating section is determined from the definition of enthalpy, h2 = c p T2 + 2 h g 2 c p T2 + 2 (2500.9 + 1.82T2 ) 34.0 = (1.005)T2 + (0.00568)(2500.9 + 1.82T2 ) PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-36 Solving for h2, yields T2 = 19.5 C The relative humidity at this state is 2 = Pv 2 Pv 2 0.859 kPa = = = 0.377 or 37.7% Pg 2 Psat @ 19.5C 2.2759 kPa (b) The rate of heat transfer to the air in the heating section becomes & & Qin = m a (h2 - h1 ) = (40.6 kg/min)(34.0 - 24.36) kJ/kg = 391 kJ/min (c) The amount of water added to the air in the humidifying section is determined from the conservation of mass equation of water in the humidifying section, & & mw = ma ( 3 - 2 ) = ( 40.6 kg / min)( 0.0093 - 0.00568) = 0.147 kg / min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-37 Cooling with Dehumidification 14-82C To drop its relative humidity to more desirable levels. 14-83 Air is first cooled, then dehumidified, and finally heated. The temperature of air before it enters the heating section, the amount of heat removed in the cooling section, and the amount of heat supplied in the heating section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air decreases due to dehumidification ( 3 < 1), and remains constant during heating ( 3 = 2). The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The intermediate state (state 2) is also known since 2 = 100% and 2 = 3. Therefore, we can determine the properties of the air at all three states from the psychrometric chart (Fig. A-31) to be h1 = 95.2 kJ / kg dry air 1 = 0.0238 kg H 2 O / kg dry air Cooling section T1 = 34C 1 = 70% 1 w 10C T2 2 Heating section T3 = 22C and h3 = 431 kJ / kg dry air . 3 = 0.0082 kg H 2 O / kg dry air ( = 2 ) 1 atm AIR 3 3 = 50% Also, hw h f @ 10C = 42.02 kJ/kg (Table A - 4) h2 = 31.8 kJ/kg dry air T2 = 11.1C (b) The amount of heat removed in the cooling section is determined from the energy balance equation applied to the cooling section, & & & E in - E out = E system & & E in = E out & & & mi hi = me he + Qout,cooling & & & & & & Qout,cooling = ma1h1 - (ma 2 h2 + mw hw ) = ma (h1 - h2 ) - mw hw 0 (steady) =0 or, per unit mass of dry air, q out,cooling = (h1 - h2 ) - ( 1 - 2 )hw = (95.2 - 31.8) - (0.0238 - 0.0082)42.02 = 62.7 kJ/kg dry air (c) The amount of heat supplied in the heating section per unit mass of dry air is qin,heating = h3 - h2 = 431 - 31.8 = 11.3 kJ / kg dry air . PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-38 14-84 [Also solved by EES on enclosed CD] Air is cooled by passing it over a cooling coil through which chilled water flows. The rate of heat transfer, the mass flow rate of water, and the exit velocity of airstream are to be determined. Assumptions 1 This is a steady-flow 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. Analysis (a) The saturation pressure of water at 35C is 5. 6291 kPa (Table A-4). Then the dew point temperature of the incoming air stream at 35C becomes Tdp = Tsat @ Pv = Tsat @ 0.65.6291 kPa = 26C (Table A-5) since air is cooled to 20C, which is below its dew point temperature, some of the moisture in the air will condense. The amount of moisture in the air decreases due to dehumidification ( 2 < 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. Then the properties of the air at both states are determined from the psychrometric chart (Fig. A-31) to be h1 = 90.3 kJ/kg dry air Water 1 = 0.0215 kg H 2 O/kg dry air T + 8C T v 1 = 0.904 m 3 /kg dry air and Cooling coils h2 = 57.5 kJ/kg dry air 2 = 0.0147 kg H 2 O/kg dry air v 2 = 0.851 m 3 /kg dry air Also, Then, hw h f @ 20C = 83.93 kJ/kg (Table A-4) 1 35C 60% 120 m/min = 8.48 m 3 / min AIR 20C 2 Saturated V&1 = V1 A1 = V1 & m a1 = D2 4 (0.3 m) 2 = (120 m/min) 4 V&1 8.48 m 3 / min = = 9.38 kg/min v 1 0.904 m 3 / kg dry air Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the water), & & & & & Water Mass Balance: mw,i = mw,e ma1 1 = ma 2 2 + mw & & m w = ma ( 1 - 2 ) = ( 9.38 kg/min)(0.0215 - 0.0147 ) = 0.064 kg/min Energy Balance: 0 (steady) & & & & & = 0 E = E E - E = E in out system in out & & & & & & & & mi hi = me he + Qout Qout = ma1h1 - (ma 2 h2 + mw hw ) = ma (h1 - h2 ) - mw hw & Qout = (9.38 kg/min)(90.3 - 57.5)kJ/kg - (0.064 kg/min)(83.93 kJ/kg) = 302.3 kJ/min (b) Noting that the heat lost by the air is gained by the cooling water, the mass flow rate of the cooling water is determined from & & & Qcooling water = m cooling water h = m cooling water c p T & m cooling water = & Qw 302.3 kJ/min = = 9.04 kg/min c p T (4.18 kJ/kg C)(8C) (c) The exit velocity is determined from the conservation of mass of dry air, V A V A V& V& & & m a1 = m a 2 1 = 2 1 = 2 v1 v2 v1 v2 V2 = v2 0.851 V1 = (120 m/min) = 113 m/min 0.904 v1 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-39 14-85 EES Problem 14-84 is reconsidered. A general solution of the problem in which the input variables may be supplied and parametric studies performed is to be developed and the process is to be shown in the psychrometric chart for each set of input variables. Analysis The problem is solved using EES, and the solution is given below. "Input Data from the Diagram Window" {D=0.3 P[1] =101.32 [kPa] T[1] = 35 [C] RH[1] = 60/100 "%, relative humidity" Vel[1] = 120/60 "[m/s]" DELTAT_cw =8 [C] P[2] = 101.32 [kPa] T[2] = 20 [C]} RH[2] = 100/100 "%" "Dry air flow rate, m_dot_a, is constant" Vol_dot[1]= (pi * D^2)/4*Vel[1] v[1]=VOLUME(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_a = Vol_dot[1]/v[1] "Exit vleocity" Vol_dot[2]= (pi * D^2)/4*Vel[2] v[2]=VOLUME(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_a = Vol_dot[2]/v[2] "Mass flow rate of the condensed water" m_dot_v[1]=m_dot_v[2]+m_dot_w w[1]=HUMRAT(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_v[1] = m_dot_a*w[1] w[2]=HUMRAT(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_v[2] = m_dot_a*w[2] "SSSF conservation of energy for the air" m_dot_a *(h[1] + (1+w[1])*Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)) + Q_dot = m_dot_a*(h[2] +(1+w[2])*Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)) +m_dot_w*h_liq_2 h[1]=ENTHALPY(AirH2O,T=T[1],P=P[1],w=w[1]) h[2]=ENTHALPY(AirH2O,T=T[2],P=P[2],w=w[2]) h_liq_2=ENTHALPY(Water,T=T[2],P=P[2]) "SSSF conservation of energy for the cooling water" -Q_dot =m_dot_cw*Cp_cw*DELTAT_cw "Note: Q_netwater=-Q_netair" Cp_cw = SpecHeat(water,T=10,P=P[2])"kJ/kg-K" PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-40 RH1 0.5 0.6 0.7 0.8 0.9 ma 0.1574 0.1565 0.1556 0.1547 0.1538 mw 0.0004834 0.001056 0.001629 0.002201 0.002774 mcw 0.1085 0.1505 0.1926 0.2346 0.2766 Q [kW] -3.632 -5.039 -6.445 -7.852 -9.258 Vel1 [m/s] 2 2 2 2 2 Vel2 [m/s] 1.894 1.883 1.872 1.861 1.85 T1 [C] 35 35 35 35 35 T2 [C] 20 20 20 20 20 w1 0.01777 0.02144 0.02516 0.02892 0.03273 w2 0.0147 0.0147 0.0147 0.0147 0.0147 0.050 0.045 0.040 0.035 0.8 Pressure = 101.0 [kPa] Humidity Ratio 0.030 0.025 0.020 0.015 0.010 0.005 0.000 -10 -5 0C 10 C 20 C 30 C 0.6 0.4 0.2 -0 5 10 15 20 25 30 35 40 T [C] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-41 14-86 Air is cooled by passing it over a cooling coil. The rate of heat transfer, the mass flow rate of water, and the exit velocity of airstream are to be determined. Assumptions 1 This is a steady-flow 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. Analysis (a) The dew point temperature of the incoming air stream at 35C is Pv1 = 1 Pg1 = 1 Psat @ 35C Tdp = (0.6)(5.6291 kPa) = 3.38 kPa = Tsat @ Pv = Tsat @ 3.38 kPa = 25.9C Water, T T + 8C Cooling coils 35C 60% 120 m/min Since air is cooled to 20C, which is below its dew point temperature, some of the moisture in the air will condense. 1 AIR 20C 2 Saturated The amount of moisture in the air decreases due to dehumidification ( 2 < 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 95 kPa. Then the properties of the air at both states are determined to be Pa1 = P1 - Pv1 = 95 - 3.38 = 91.62 kPa v1 = 1 = Ra T1 (0.287 kPa m 3 / kg K)(308 K) = = 0.965 m 3 / kg dry air 91.62 kPa Pa1 0.622 Pv1 0.622(3.38 kPa) = = 0.0229 kg H 2 O/kg dry air (95 - 3.38) kPa P1 - Pv1 h1 = c p T1 + 1hg1 = (1.005 kJ/kg C)(35C) + (0.0229)(2564.6 kJ/kg) = 93.90 kJ/kg dry air and Pv 2 = 2 Pg 2 = (1.00) Psat @ 20C = 2.3392 kPa v2 = Ra T2 (0.287 kPa m 3 / kg K)(293 K) = = 0.908 m 3 / kg dry air (95 - 2.339) kPa Pa 2 0.622 Pv 2 0.622(2.339 kPa) = = 0.0157 kg H 2 O/kg dry air (95 - 2.339) kPa P2 - Pv 2 2 = h2 = c p T2 + 2 hg 2 = (1.005 kJ/kg C)(20C) + (0.0157)(2537.4 kJ/kg) = 59.95 kJ/kg dry air Also, hw h f @ 20C = 83.915 kJ/kg (Table A-4) Then, V&1 = V1 A1 = V1 & m a1 = D2 4 (0.3 m) 2 = (120 m/min) 4 = 8.48 m 3 / min V&1 8.48 m 3 / min = = 8.79 kg/min v 1 0.965 m 3 / kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-42 Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section (excluding the water), Water Mass Balance: & & mw,i = mw ,e & & & ma1 1 = ma 2 2 + mw & & mw = ma ( 1 - 2 ) = (8.79 kg / min)(0.0229 - 0.0157) = 0.0633 kg / min Energy Balance: & & & E in - E out = E system & & E in = E out & & & & & & & & & mi hi = me he + Qout Qout = ma1h1 - (ma 2 h2 + mw hw ) = ma (h1 - h2 ) - mw hw & Qout = (8.79 kg/min)(93.90 - 59.94)kJ/kg - (0.0633 kg/min)(83.915 kJ/kg) = 293.2 kJ/min 0 (steady) =0 (b) Noting that the heat lost by the air is gained by the cooling water, the mass flow rate of the cooling water is determined from & & & Qcooling water = mcooling water h = mcooling water c p T & mcooling water = & Qw 293.2 kJ/min = = 8.77 kg/min c p T (4.18 kJ/kg C)(8C) (c) The exit velocity is determined from the conservation of mass of dry air, & & m a1 = m a 2 V2 = V A V A V&1 V&2 = 1 = 2 v1 v 2 v1 v2 v2 0.908 V1 = (120 m/min) = 113 m/min 0.965 v1 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-43 14-87 Air is cooled and dehumidified at constant pressure. The amount of water removed from the air and the cooling requirement are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 85.5 kJ/kg dry air 1 = 0.0217 kg H 2 O/kg dry air and Cooling coils 2 = 1.0 h2 = 57.6 kJ/kg dry air T2 = 20C 2 =100% 2 22C (Table A-4) 1 atm Condensate 1 Condensate removal T1 = 30C 1 = 80% 2 = 0.0148 kg H 2 O/kg dry air Also, h w h f @ 22C = 92.28 kJ/kg Analysis The amount of moisture in the air decreases due to dehumidification ( 2 < 1). Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: & & & & & m w, i = m w, e m a11 = m a 2 2 + m w = 1 - 2 = 0.0217 - 0.0148 = 0.0069 kg H 2 O/kg dry air Energy Balance: & & & E in - E out = E system 0 (steady) =0 & & E in = E out & & & mi hi = Qout + m e he & & & & & & Qout = m a1 h1 - (m a 2 h2 + m w h w ) = m a (h1 - h2 ) - m w hw q out = h1 - h2 - (1 - 2 )hw = (85.5 - 57.6)kJ/kg - (0.0069)(92.28) = 27.3 kJ/kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-44 14-88E Air is cooled and dehumidified at constant pressure. The amount of water removed from the air and the rate of cooling are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31E) to be h1 = 37.8 Btu/lbm dry air 1 = 0.0158 lbm H 2 O/lbm dry air v 1 = 14.08 ft 3 /lbm dry air and Cooling coils 2 = 1.0 h2 = 26.5 Btu/lbm dry air T2 = 60F 2 =100% 2 65F (Table A-4E) 1 atm Condensate 1 Condensate removal T1 = 85F Tdp1=70F 2 = 0.0111 lbm H 2 O/lbm dry air Also, hw h f @ 65 F = 33.08 Btu/lbm Analysis The amount of moisture in the air decreases due to dehumidification ( 2 < 1). The mass flow rate of air is & m a1 = V&1 (10,000 / 3600) ft 3 / s = = 0.1973 lbm/s v 1 14.08 ft 3 / lbm dry air Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: & & & & & m w, i = m w, e m a11 = m a 2 2 + m w & & m w = m a (1 - 2 ) = (0.1973 lbm/s)(0.0158 - 0.0111) = 0.000927 lbm/s Energy Balance: & & & E in - E out = E system 0 (steady) =0 & & E in = E out & & & mi hi = Qout + m e he & = m h - (m h + m h ) = m (h - h ) - m h & a2 2 & w w &a 1 &w w Qout & a1 1 2 & Qout = (0.1973 lbm/s)(37.8 - 26.5)Btu/lbm - (0.000927 lbm/s)(33.08 Btu/lbm) = 2.20 Btu/s PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-45 14-89 Air is cooled and dehumidified at constant pressure. The amount of water removed from the air and the rate of cooling are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 106.8 kJ/kg dry air 1 = 0.0292 kg H 2 O/kg dry air v 1 = 0.905 m 3 /kg dry air and h2 = 52.7 kJ/kg dry air Cooling coils T2 = 24C 2 =60% 2 28C 1 atm Condensate 1 Condensate removal T1 = 32C 1 =95% 2 = 0.0112 kg H 2 O/kg dry air We assume that the condensate leaves this system at the average temperature of the air inlet and exit. Then, hw h f @ 28C = 117.4 kJ/kg (Table A-4) Analysis The amount of moisture in the air decreases due to dehumidification ( 2 < 1). The mass of air is ma = V1 1000 m 3 = = 1105 kg v 1 0.905 m 3 / kg dry air Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: & & & & mw,i = mw,e ma1 1 = ma 2 2 + mw & m w = m a (1 - 2 ) = (1105 kg)(0.0292 - 0.0112) = 19.89 kg Energy Balance: & & & E in - E out = E system 0 (steady) =0 & & E in = E out & & & mi hi = Qout + m e he & & & & & & Qout = m a1 h1 - (m a 2 h2 + m w h w ) = m a (h1 - h2 ) - m w h w Qout = m a (h1 - h2 ) - m w hw Qout = (1105 kg)(106.8 - 52.7)kJ/kg - (19.89 kg)(117.4 kJ/kg) = 57,450 kJ PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-46 14-90 The humid air of the previous problem is reconsidered. The exit temperature of the air to produce the desired dehumidification is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be (from the data of the previous problem) h1 = 106.8 kJ/kg dry air 1 = 0.0292 kg H 2 O/kg dry air v 1 = 0.905 m 3 /kg dry air and h2 = 52.7 kJ/kg dry air Cooling coils T2 = 24C 2 =60% 2 28C 1 atm Condensate 1 Condensate removal T1 = 32C 1 =95% 2 = 0.0112 kg H 2 O/kg dry air Analysis For the desired dehumidification, the air at the exit should be saturated with a specific humidity of 0.0112 kg water/kg dry air. That is, 2 = 1.0 2 = 0.0112 kg H 2 O/kg dry air The temperature of the air at this state is T2 = 15.8C PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-47 14-91 Air is cooled and dehumidified at constant pressure. The cooling required is provided by a simple ideal vapor-compression refrigeration system using refrigerant-134a as the working fluid. The exergy destruction in the total system per 1000 m3 of dry air is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 106.8 kJ/kg dry air 1 = 0.0292 kg H 2 O/kg dry air v 1 = 0.905 m 3 /kg dry air and h2 = 52.7 kJ/kg dry air Condenser 3 E x 4 Evaporator Compressor 1 2 2 = 0.0112 kg H 2 O/kg dry air We assume that the condensate leaves this system at the average temperature of the air inlet and exit. Then, from Table A-4, hw h f @ 28C = 117.4 kJ/kg T2 = 24C 2 = 60% 1 atm T1 = 32C 1 = 95% Analysis The amount of moisture in the air decreases due to dehumidification ( 2 < 1). The mass of air is ma = Condensate V1 1000 m 3 = = 1105 kg v 1 0.905 m 3 / kg dry air Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: & & & & mw,i = mw,e ma1 1 = ma 2 2 + mw & m w = m a (1 - 2 ) = (1105 kg)(0.0292 - 0.0112) = 19.89 kg Energy Balance: & & & E in - E out = E system 0 (steady) =0 & & E in = E out & & & mi hi = Qout + m e he & = m h - (m h + m h ) = m (h - h ) - m h & a2 2 & w w &a 1 &w w Qout & a1 1 2 Qout = m a (h1 - h2 ) - m w hw Qout = (1105 kg)(106.8 - 52.7)kJ/kg - (19.89 kg)(117.4 kJ/kg) = 57,450 kJ We obtain the properties for the vapor-compression refrigeration cycle as follows (Tables A-11,through A13): PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-48 T1 = 4C h1 = h g @ 4C = 252.77 kJ/kg sat. vapor s1 = s g @ 4C = 0.92927 kJ/kg K P2 = Psat @ 39.4C = 1 MPa h2 = 275.29 kJ/kg s 2 = s1 P3 = 1 MPa h3 = h f @ 1 MPa = 107.32 kJ/kg sat. liquid s 3 = s f @ 1 MPa = 0.39189 kJ/kg K h4 h3 = 107.32 kJ/kg ( throttling) T4 = 4C x 4 = 0.2561 h4 = 107.32 kJ/kg s 4 = 0.4045 kJ/kg K T QH 3 39.4C 2 Win 4 C 4s 4 QL 1 The mass flow rate of refrigerant-134a is mR = QL 57,450 kJ = = 395.0 kg h1 - h4 (252.77 - 107.32)kJ/kg s The amount of heat rejected from the condenser is Q H = m R (h2 - h3 ) = (395.0 kg)(275.29 - 107.32) kJ/kg = 66,350 kg Next, we calculate the exergy destruction in the components of the refrigeration cycle: X destroyed,12 = m R T0 ( s 2 - s1 ) = 0 (since the process is isentropic) Q X destroyed, 23 = T0 m R ( s 3 - s 2 ) + H TH 66,350 kJ = (305 K ) (395 kg)(0.39189 - 0.92927) kJ/kg K + = 1609 kJ 305 K X destroyed, 34 = m R T0 ( s 4 - s 3 ) = (395 kg)(305 K )(0.4045 - 0.39189) kJ/kg K = 1519 kJ The entropies of water vapor in the air stream are s g1 = s g @ 32C = 8.4114 kJ/kg K s g 2 = s g @ 24C = 8.5782 kJ/kg K The entropy change of water vapor in the air stream is S vapor = m a ( 2 s g 2 - 1 s g1 ) = (1105)(0.0112 8.5782 - 0.0292 8.4114) = -165.2 kJ/K The entropy of water leaving the cooling section is S w = m w s f @ 28C = (19.89 kg )(0.4091 kJ/kg K) = 8.14 kJ/K The partial pressures of water vapor and dry air for air streams are Pv1 = 1 Pg1 = 1 Psat @ 32C = (0.95)(4.760 kPa) = 4.522 kPa Pa1 = P1 - Pv1 = 101.325 - 4.522 = 96.80 kPa Pv 2 = 2 Pg 2 = 2 Psat @ 24C = (0.60)(2.986 kPa) = 1.792 kPa Pa 2 = P2 - Pv 2 = 101.325 - 1.792 = 99.53 kPa The entropy change of dry air is PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-49 P T S a = m a ( s 2 - s1 ) = m a c p ln 2 - R ln a 2 T1 Pa1 297 99.53 = (1105) (1.005) ln - (0.287) ln = -38.34 kJ/kg dry air 305 96.80 The entropy change of R-134a in the evaporator is S R, 41 = m R ( s1 - s 4 ) = (395 kg )(0.92927 - 0.4045) = 207.3 kJ/K An entropy balance on the evaporator gives S gen,evaporator = S R,41 + S vapor + S a + S w = 207.3 + (-165.2) + (-38.34) + 8.14 = 11.90 kJ/K Then, the exergy destruction in the evaporator is X dest = T0 S gen, evaporator = (305 K)(11.90 kJ/K) = 3630 kJ Finally the total exergy destruction is X dest, total = X dest, compressor + X dest, condenser + X dest, throttle + X dest, evaporator = 0 + 1609 + 1519 + 3630 = 6758 kJ The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from humid air and rejected to the ambient air at 32C (305 K), which is also taken as the dead state temperature. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-50 14-92 Atmospheric air enters the evaporator of an automobile air conditioner at a specified pressure, temperature, and relative humidity. The dew point and wet bulb temperatures at the inlet to the evaporator section, the required heat transfer rate from the atmospheric air to the evaporator fluid, and the rate of condensation of water vapor in the evaporator section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at the inlet and exit states may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES) Tdp1 = 15.7C Twb1 = 19.5C h1 = 55.60 kJ/kg dry air Cooling coils 1 = 0.01115 kg H 2O/kg dry air v1 = 0.8655 m3 / kg dry air h2 = 27.35 kJ/kg dry air T2 =10C 2 = 90% 2 10C 1 atm Condensate 1 Condensate removal T1 =27C 1 = 50% 2 = 0.00686 kg H 2O/kg dry air The mass flow rate of dry air is & ma = V&1 V car ACH (2 m 3 /change)(5 changes/min) = = = 11.55 kg/min v1 v1 0.8655 m 3 The mass flow rates of vapor at the inlet and exit are & & m v1 = 1 m a = (0.01115)(11.55 kg/min) = 0.1288 kg/min & & m v 2 = 2 m a = (0.00686)(11.55 kg/min) = 0.07926 kg/min An energy balance on the control volume gives & & & & m a h1 = Qout + m a h2 + m w h w2 where the the enthalpy of condensate water is hw 2 = h f@ 10C = 42.02 kJ/kg (Table A - 4) and the rate of condensation of water vapor is & & & m w = m v1 - m v 2 = 0.1288 - 0.07926 = 0.0495 kg/min Substituting, & & & & m a h1 = Qout + m a h2 + m w hw 2 & + (11.55 kg/min)(27.35 kJ/kg) + (0.0495 kg/min)(42.02 kJ/kg) (11.55 kg/min)(55.60 kJ/kg) = Qout & Q = 324.4 kJ/min = 5.41 kW out PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-51 0,050 Pressure = 101.3 [kPa] 0,045 0,040 AirH2O 35C 0,035 0.8 Humidity Ratio 0,030 30C 0,025 0,020 0,015 0,010 10C 15C 20C 25C 0.6 0.4 1 0.2 0,005 0,000 0 2 5 10 15 20 25 30 35 40 T [C] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-52 14-93 Atmospheric air flows into an air conditioner that uses chilled water as the cooling fluid. The mass flow rate of the condensate water and the volume flow rate of chilled water supplied to the air conditioner are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis We may assume that the exit relative humidity is 100 percent since the exit temperature (18C) is below the dew-point temperature of the inlet air (25C). The properties of the air at the exit state may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES) h2 = 51.34 kJ/kg dry air 2 = 0.01311 kg H 2 O/kg dry air The partial pressure of water vapor at the inlet state is (Table A-4) Pv1 = Psat@ 25C = 3.17 kPa Cooling coils T2 = 18C 100% RH 100 kPa 2 Condensate 1 18C Condensate removal T1 = 28C Tdp1 = 25C 2000 m3/h The saturation pressure at the inlet state is Pg1 = Psat@ 28C = 3.783 kPa (Table A - 4) Then, the relative humidity at the inlet state becomes 1 = Pv1 3.17 = = 0.8379 Pg1 3.783 Now, the inlet state is also fixed. The properties are obtained from EES to be h1 = 80.14 kJ/kg dry air 1 = 0.02036 kg H 2 O/kg dry air v 1 = 0.8927 m 3 /kg The mass flow rate of dry air is & ma = V&1 (2000 / 60) m3/h = = 37.34 kg/min v1 0.8927 m3/kg The mass flow rate of condensate water is & & m w = m a (1 - 2 ) = (37.34 kg/min)(0.02036 - 0.01311) = 0.2707 kg/min = 16.24 kg/h The enthalpy of condensate water is hw 2 = h f@ 18C = 75.54 kJ/kg (Table A - 4) An energy balance on the control volume gives PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-53 & & & & m a h1 = Qout + m a h2 + m w hw2 & (37.34 kg/min)(80.14 kJ/kg) = Qout + (37.34 kg/min)(51.34 kJ/kg) + (0.2707 kg/min)(75.54 kJ/kg) & Q = 1055 kJ/min = 17.59 kW out Noting that the rate of heat lost from the air is received by the cooling water, the mass flow rate of the cooling water is determined from & & & Qin = mcw c p Tcw mcw = & Qin 1055 kJ/min = = 25.24 kg/min c p Tcw (4.18 kJ/kg.C)(10C) where we used the specific heat of water value at room temperature. Assuming a density of 1000 kg/m3 for water, the volume flow rate is determined to be V&cw = & mcw cw = 25.24 kg/min = 0.0252 m3 /min 1000 kg/m3 0,050 Pressure = 100.0 [kPa] 0,045 0,040 AirH2O 35C Humidity Ratio 0,035 0,030 30C 0. 92 5 0.8 0,025 0. 0.6 0,020 0,015 0,010 0,005 0 .8 25C 9 1 5 0.4 20C 0. 2 15C 0 .8 87 10C 0 .8 25 0.2 g 5m 3 /k 0,000 0 5 10 15 20 25 30 35 40 T [C] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-54 14-94 An automobile air conditioner using refrigerant 134a as the cooling fluid is considered. The inlet and exit states of moist air in the evaporator are specified. The volume flow rate of the air entering the evaporator of the air conditioner is to be determined. Assumptions 1 All processes are steady flow and the mass flow rate of dry air remains constant during the & & & entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis We assume that the total pressure of moist air is 100 kPa. Then, the inlet and exit states of the moist air for the evaporator are completely specified. The properties may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES) h1 = 43.33 kJ/kg dry air 1 = 0.008337 kg H 2 O/kg dry air v 1 = 0.8585 m 3 /kg dry air h2 = 23.31 kJ/kg dry air R-134a 375 kPa Cooling coils 2 = 0.006065 kg H 2 O/kg dry air The mass flow rate of dry air is given by T2 = 8C 2 =90% 2 8 C AIR Condensate 1 Condensate removal T1 =22C 1 = 50% V& V&1 & ma = 1 = v 1 0.8585 m 3 /kg The mass flow rate of condensate water is expressed as & & m w = ma (1 - 2 ) = V&1 0.8585 (0.008337 - 0.006065) = 0.002646V&1 The enthalpy of condensate water is hw 2 = h f@ 8C = 33.63 kJ/kg (Table A - 4) An energy balance on the control volume gives & & & & ma h1 = Qout + ma h2 + mw hw2 V&1 V&1 & (43.05) = Qout + (23.11) + 0.002646V&1 (33.63) 0.8585 0.8585 (1) The properties of the R-134a at the inlet of the compressor and the enthalpy at the exit for the isentropic process are (R-134a tables) PR1 = 375 kPa h R1 = 254.48 kJ/kg x R1 = 1 s R1 = 0.9278 kJ/kg.K PR 2 = 1800 kPa h R 2, s = 286.90 kJ/kg s R 2 = s R1 The enthalpies of R-134a at the condenser exit and the throttle exit are h R3 = h f@ 1800 kPa = 144.07 kJ/kg h R 4 = h R 3 = 144.07 kJ/kg The mass flow rate of the refrigerant can be determined from the expression for the compressor power: PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-55 & & WC = m R & 6 kW = m R h R 2, s - h R1 C (286.90 - 254.48) kJ/kg 0.85 & R = 0.1573 kg/s = 9.439 kg/min m The rate of heat absorbed by the R-134a in the evaporator is & & Q R,in = m R (h R1 - h R 4 ) = (9.439 kg/min)(254.48 - 144.07) kJ/kg = 1042.1 kJ/min The rate of heat lost from the air in the evaporator is absorbed by the refrigerant-134a. That is, & & Q R ,in = Qout . Then, the volume flow rate of the air at the inlet of the evaporator can be determined from Eq. (1) to be V&1 0.8474 (43.05) = 1042.1 + V&1 0.8474 (23.11) + 0.002646V1 (33.63) V&1 = 44.87 m 3 /min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-56 14-95 Air flows through an air conditioner unit. The inlet and exit states are specified. The rate of heat transfer and the mass flow rate of condensate water are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet state of the air is completely specified, and the total pressure is 98 kPa. The properties of the air at the inlet state may be determined from (Fig. A-31) or using EES psychrometric functions to be (we used EES) h1 = 77.88 kJ/kg dry air Cooling coils Tdb2 = 25C Tdp2 = 6.5C 2 25C Condensate 1 Condensate removal Tdb1 =30C Twb1 =25C P = 98 kPa 1 = 0.01866 kg H 2 O/kg dry air 1 = 0.6721 The partial pressure of water vapor at the exit state is Pv 2 = Psat@ 6.5C = 0.9682 kPa (Table A - 4) The saturation pressure at the exit state is Pg 2 = Psat@ 25C = 3.17 kPa (Table A - 4) Then, the relative humidity at the exit state becomes 2 = Pv 2 0.9682 = = 0.3054 3.17 Pg 2 Now, the exit state is also fixed. The properties are obtained from EES to be h2 = 40.97 kJ/kg dry air 2 = 0.006206 kg H 2 O/kg dry air v 2 = 0.8820 m 3 /kg The mass flow rate of dry air is & ma = V&2 1000 m 3 /min = = 1133.8 kg/min v 2 0.8820 m 3 /kg The mass flow rate of condensate water is & & m w = m a (1 - 2 ) = (1133.8 kg/min)(0.01866 - 0.006206) = 14.12 kg/min = 847.2 kg/h The enthalpy of condensate water is h w 2 = h f@ 25C = 104.83 kJ/kg (Table A - 4) An energy balance on the control volume gives & & & & m a h1 = Qout + m a h2 + m w h w2 & + (1133.8 kg/min)(40.97 kJ/kg) + (14.12 kg/min)(104.83 kJ/kg) (1133.8 kg/min)(77.88 kJ/kg) = Qout & Q = 40,377 kJ/min = 672.9 kW out PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-57 Evaporative Cooling 14-96C In steady operation, the mass transfer process does not have to involve heat transfer. However, a mass transfer process that involves phase change (evaporation, sublimation, condensation, melting etc.) must involve heat transfer. For example, the evaporation of water from a lake into air (mass transfer) requires the transfer of latent heat of water at a specified temperature to the liquid water at the surface (heat transfer). 14-97C During evaporation from a water body to air, the latent heat of vaporization will be equal to convection heat transfer from the air when conduction from the lower parts of the water body to the surface is negligible, and temperature of the surrounding surfaces is at about the temperature of the water surface so that the radiation heat transfer is negligible. 14-98C Evaporative cooling is the cooling achieved when water evaporates in dry air. It will not work on humid climates. 14-99 Air is cooled by an evaporative cooler. The exit temperature of the air and the required rate of water supply are to be determined. Analysis (a) From the psychrometric chart (Fig. A-31) at 36C and 20% relative humidity we read Twb1 = 19.5C 1 = 0.0074 kg H 2 O/kg dry air & Water, m v 1 = 0.887 m /kg dry air 3 Humidifier 1 atm 36C 20% AIR 90% Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature. That is, Twb2 Twb1 = 19.5C At this wet-bulb temperature and 90% relative humidity we read T2 = 20.5 C 2 = 0.0137 kg H 2 O / kg dry air Thus air will be cooled to 20.5C in this evaporative cooler. (b) The mass flow rate of dry air is & ma = V&1 4 m 3 / min = = 4.51 kg/min v 1 0.887 m 3 / kg dry air Then the required rate of water supply to the evaporative cooler is determined from & & & & msupply = m w2 - m w1 = m a ( 2 - 1 ) = (4.51 kg/min)(0.0137 - 0.0074) = 0.028 kg/min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-58 14-100E Air is cooled by an evaporative cooler. The exit temperature of the air and the required rate of water supply are to be determined. Analysis (a) From the psychrometric chart (Fig. A-31E) at 90F and 20% relative humidity we read Twb1 = 62.8F 1 = 0.0060 lbm H 2O/lbm dry air v1 = 14.0 ft 3/lbm dry air Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature. That is, Twb2 Twb1 = 62.8F & Water, m Humidifier 1 atm 90F 20% AIR 90% At this wet-bulb temperature and 90% relative humidity we read T2 = 65F 2 = 0.0116 lbm H 2 O/lbm dry air Thus air will be cooled to 64F in this evaporative cooler. (b) The mass flow of rate dry air is & ma = V&1 150 ft 3 / min = = 10.7 lbm/min v 1 14.0 ft 3 / lbm dry air Then the required rate of water supply to the evaporative cooler is determined from & & & & msupply = mw2 - mw1 = ma (2 - 1 ) = (10.7 lbm/min)(0.0116 - 0.0060) = 0.06 lbm/min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-59 14-101 Air is cooled by an evaporative cooler. The final relative humidity and the amount of water added are to be determined. Analysis (a) From the psychrometric chart (Fig. A-31) at 32C and 30% relative humidity we read Twb1 = 19.4C 1 = 0.0089 kg H 2O/kg dry air v1 = 0.877 m3/kg dry air Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wetbulb temperature. That is, Twb2 Twb1 = 19.4C Water Humidifier 32C 30% 2 m3/min AIR 22C At this wet-bulb temperature and 22C temperature we read 2 = 79% 2 = 0.0130 kg H 2 O/kg dry air (b) The mass flow rate of dry air is & ma = 5 m3 / min V&1 = = 5.70 kg/min v1 0.877 m3 / kg dry air Then the required rate of water supply to the evaporative cooler is determined from & & & & m supply = m w2 - m w1 = m a ( 2 - 1 ) = (5.70 kg/min)(0.0130 - 0.0089) = 0.0234 kg/min 14-102 Air enters an evaporative cooler at a specified state and relative humidity. The lowest temperature that air can attain is to be determined. Analysis From the psychrometric chart (Fig. A-31) at 29C and 40% relative humidity we read Twb1 = 19.3C Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature, which is the lowest temperature that can be obtained in an evaporative cooler. That is, Tmin = Twb1 = 19.3C Water Humidifier 1 atm 29C 40% AIR 100% PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-60 14-103 Air is first heated in a heating section and then passed through an evaporative cooler. The exit relative humidity and the amount of water added are to be determined. Analysis (a) From the psychrometric chart (Fig. A-31) at 15C and 60% relative humidity we read 1 = 0.00635 kg H 2 O / kg dry air The specific humidity remains constant during the heating process. Therefore, 2 = 1 = 0.00635 kg H2O / kg dry air. At this value and 30C we read Twb2 = 16.7C. Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature. That is, Twb3 Twb2 = 16.7C. At this Twb value and 25C we read 15C 60% 1 Water Heating coils Humidifier 30C 25C AIR 1 atm 2 3 3 = 42.6% 3 = 0.00840 kg H 2 O/kg dry air (b) The amount of water added to the air per unit mass of air is 23 = 3 - 2 = 0.00840 - 0.00635 = 0.00205 kg H 2 O/kg dry air 14-104E Desert dwellers often wrap their heads with a water-soaked porous cloth. The temperature of this cloth on a desert with specified temperature and relative humidity is to be determined. Analysis Since the cloth behaves as the wick on a wet bulb thermometer, the temperature of the cloth will become the wet-bulb temperature. According to the pshchrometric chart, this temperature is T2 = Twb1 = 73.7F Water Humidifier 1 atm 120F 10% AIR 100% This process can be represented by an evaporative cooling process as shown in the figure. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-61 Adiabatic Mixing of Airstreams 14-105C This will occur when the straight line connecting the states of the two streams on the psychrometric chart crosses the saturation line. 14-106C Yes. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-62 14-107 Two airstreams are mixed steadily. The specific humidity, the relative humidity, the dry-bulb temperature, and the volume 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 Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31) to be h1 = 62.7 kJ/kg dry air 1 = 0.0119 kg H 2 O/kg dry air v 1 = 0.882 m 3 /kg dry air and h2 = 31.9 kJ/kg dry air 1 32C 40% 20 m3/min P = 1 atm AIR 25 m3/min 12C 90% 2 = 0.0079 kg H 2 O/kg dry air 3 3 T3 3 v 2 = 0.819 m 3 /kg dry air Analysis The mass flow rate of dry air in each stream is & ma1 = & ma 2 = 2 V&1 20 m 3 / min = = 22.7 kg/min v 1 0.882 m 3 / kg dry air V&2 25 m 3 / min = = 30.5 kg/min v 2 0.819 m 3 / kg dry air From the conservation of mass, & & & ma 3 = ma1 + ma 2 = ( 22.7 + 30.5) kg / min = 53.2 kg / min The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: & ma1 2 - 3 h2 - h3 = = & a 2 3 - 1 h3 - h1 m 22.7 0.0079 - 3 319 - h3 . = = 30.5 3 - 0.0119 h3 - 62.7 which yields, 3 = 0.0096 kg H 2O / kg dry air h3 = 45.0 kJ / kg dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: T3 = 20.6C 3 = 63.4% v 3 = 0.845 m 3 /kg dry air Finally, the volume flow rate of the mixture is determined from & V&3 = m a 3v 3 = (53.2 kg/min)(0.845 m 3 / kg) = 45.0 m 3 /min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-63 14-108 Two airstreams are mixed steadily. The specific humidity, the relative humidity, the dry-bulb temperature, and the volume 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. Analysis The properties of each inlet stream are determined to be Pv1 = 1 Pg1 = 1 Psat @ 32C = (0.40)(4.760 kPa) = 1.90 kPa Pa1 = P1 - Pv1 = 90 - 1.90 = 88.10 kPa R a T1 (0.287 kPa m 3 / kg K)(305 K) = 88.10 kPa Pa1 1 v1 = = 0.994 m 3 / kg dry air 0.622 Pv1 0.622(1.90 kPa) 1 = = P1 - Pv1 (90 - 1.90) kPa = 0.0134 kg H 2 O/kg dry air h1 = c p T1 + 1 h g1 = (1.005 kJ/kg C)(32C) + (0.0134)(2559.2 kJ/kg) = 66.45 kJ/kg dry air 32C 40% 20 m3/min P = 90 kPa AIR 25 m /min 12C 90% 3 3 3 3 T3 2 and Pv 2 = 2 Pg 2 = 2 Psat@12C = (0.90)(1.403 kPa) = 1.26 kPa Pa 2 = P2 - Pv 2 = 90 - 1.26 = 88.74 kPa R a T2 (0.287 kPa m 3 / kg K)(285 K) = = 0.922 m 3 / kg dry air 88.74 kPa Pa 2 0.622 Pv 2 0.622(1.26 kPa) = = 0.00883 kg H 2 O/kg dry air (90 - 1.26) kPa P2 - Pv 2 v2 = 2 = h2 = c p T2 + 2 h g 2 = (1.005 kJ/kg C)(12C) + (0.00883)(2522.9 kJ/kg) = 34.34 kJ/kg dry air Then the mass flow rate of dry air in each stream is & m a1 = & ma2 = V&1 20 m 3 / min = = 20.12 kg/min v 1 0.994 m 3 / kg dry air V&2 25 m 3 / min = = 27.11 kg/min v 2 0.922 m 3 / kg dry air From the conservation of mass, & & & m a 3 = m a1 + m a 2 = (20.12 + 27.11) kg/min = 47.23 kg/min The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: & ma1 2 - 3 h2 - h3 20.12 0.00883 - 3 34.34 - h3 = = = = & ma 2 3 - 1 h3 - h1 27.11 3 - 0.0134 h3 - 66.45 which yields 3 = 0.0108 kg H 2O/kg dry air h3 = 48.02 kJ/kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-64 These two properties fix the state of the mixture. Other properties are determined from h3 = c pT3 + 3hg 3 c pT3 + 3 (2501.3 + 1.82T3 ) 48.02 kJ/kg = (1.005 kJ/kg C)T3 + (0.0108)(2500.9 + 1.82T3 ) kJ/kg T3 = 20.5C 3 = 0.0108 = 0.622 Pv 3 P3 - Pv 3 0.622 Pv 3 Pv 3 = 1.54 kPa 90 - Pv 3 Pv 3 Pv 3 1.54 kPa = = = 0.639 or 63.9% Pg 3 Psat @ T3 2.41 kPa 3 = Finally, Pa 3 = P3 - Pv3 = 90 - 1.54 = 88.46 kPa v3 = Ra T3 (0.287 kPa m 3 / kg K)(293.5 K) = = 0.952 m 3 /kg dry air Pa 3 88.46 kPa & V&3 = m a 3v 3 = (47.23 kg/min)(0.952 m 3 / kg) = 45.0 m 3 /min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-65 14-109E Two airstreams are mixed steadily. The temperature and the relative humidity 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 Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31E or from EES) to be h1 = 66.7 Btu/lbm dry air 1 = 0.0386 lbm H 2 O/lbm dry air 1 v 1 = 14.98 ft 3 /lbm dry air and h2 = 14.5 Btu/lbm dry air 100F 90% 3 ft3/s 2 = 0.0023 lbm H 2 O/lbm dry air P = 1 atm AIR 1 ft3/s 50F 30% 3 3 3 T3 v 2 = 12.90 ft 3 /lbm dry air Analysis The mass flow rate of dry air in each stream is & m a1 = & ma2 = 2 V&1 3 ft 3 / s = = 0.2002 lbm/s v 1 14.98 ft 3 / lbmdry air V&2 1 ft 3 / s = = 0.07755 lbm/s v 2 12.90 ft 3 / lbm dry air From the conservation of mass, & & & m a 3 = m a1 + m a 2 = (0.2002 + 0.07755) lbm/s = 0.2778 lbm/s The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: & m a1 2 - 3 h2 - h3 = = & m a 2 3 - 1 h3 - h1 0.2002 0.0023 - 3 14.5 - h3 = = 0.07755 3 - 0.0386 h3 - 66.7 which yields 3 = 0.0284 lbm H 2 O/lbm dry air h3 = 52.1 Btu/lbm dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: T3 = 86.7F 3 = 1.0 = 100% PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-66 14-110E Two airstreams are mixed steadily. The rate of entropy generation is 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 Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31 or from EES) to be h1 = 66.7 Btu/lbm dry air 1 = 0.0386 lbm H 2 O/lbm dry air v 1 = 14.98 ft 3 /lbm dry air and h2 = 14.5 Btu/lbm dry air 1 100F 90% 3 ft3/s 2 = 0.0023 lbm H 2 O/lbm dry air P = 1 atm AIR 1 ft /s 50F 30% 3 3 3 3 T3 v 2 = 12.90 ft /lbm dry air 3 The entropies of water vapor in the air streams are s g1 = s g @ 100 F = 1.9819 Btu/lbm R s g 2 = s g @ 50 F = 2.1256 Btu/lbm R 2 Analysis The mass flow rate of dry air in each stream is & m a1 = & ma2 = V&1 3 ft 3 / s = = 0.2002 lbm/s v 1 14.98 ft 3 / lbmdry air V&2 1 ft 3 / s = = 0.07755 lbm/s v 2 12.90 ft 3 / lbm dry air From the conservation of mass, & & & m a 3 = m a1 + m a 2 = (0.2002 + 0.07755) lbm/s = 0.2778 lbm/s The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: & m a1 2 - 3 h2 - h3 = = & m a 2 3 - 1 h3 - h1 0.2002 0.0023 - 3 14.5 - h3 = = 0.07755 3 - 0.0386 h3 - 66.7 which yields 3 = 0.0284 lbm H 2 O/lbm dry air h3 = 52.1 Btu/lbm dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: T3 = 86.7F 3 = 1.0 = 100% The entropy of water vapor in the mixture is PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-67 s g 3 = s g @ 86.7 F = 2.0168 Btu/lbm R An entropy balance on the mixing chamber for the water gives & & & & S w = m a 3 3 s 3 - m a11 s1 - m a 2 2 s 2 = 0.2778 0.0284 2.0168 - 0.2002 0.0386 1.9819 - 0.07755 0.0023 2.1256 = 2.169 10 - 4 Btu/s R The partial pressures of water vapor and dry air for all three air streams are Pv1 = 1 Pg1 = 1 Psat @ 100F = (0.90)(0.95052 psia) = 0.8555 psia Pa1 = P1 - Pv1 = 14.696 - 0.8555 = 13.84 psia Pv 2 = 2 Pg 2 = 2 Psat @ 50 F = (0.30)(0.17812 psia) = 0.0534 psia Pa 2 = P2 - Pv 2 = 14.696 - 0.0534 = 14.64 psia Pv 3 = 3 Pg 3 = 3 Psat @ 86.7C = (1.0)(0.6298 psia) = 0.6298 psia Pa 3 = P3 - Pv 3 = 14.696 - 0.6298 = 14.07 psia An entropy balance on the mixing chamber for the dry air gives & & & Sa = ma1( s3 - s1) + ma 2 ( s3 - s2 ) T P T P & & = ma1 c p ln 3 - R ln a3 + ma 2 c p ln 3 - R ln a3 T1 Pa1 T2 Pa 2 14.07 546.7 14.07 546.7 = 0.2002(0.240) ln - (0.06855) ln + 0.07755(0.240) ln - (0.06855) ln 14.64 510 13.84 560 = (0.2002)(-0.006899) + (0.07755)(0.01940) = 1.233 10- 4 Btu/s R The rate of entropy generation is then & & & S gen = S a + S w = 1.233 10 -4 + 2.169 10 -4 = 3.402 10 -4 Btu/s R PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-68 14-111 Two airstreams are mixed steadily. The mass flow ratio of the two streams for a specified mixture relative humidity and the temperature 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 Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31 or from EES) to be h1 = 29.4 kJ/kg dry air 1 = 0.0077 kg H 2 O/kg dry air and h2 = 94.6 kJ/kg dry air 1 10C 100% P = 1 atm AIR 32C 80% 2 = 0.0244 kg H 2 O/kg dry air Analysis An application of Eq. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams gives & m a1 2 - 3 h2 - h3 = = & a 2 3 - 1 h3 - h1 m & m a1 0.0244 - 3 94.6 - h3 = = & m a 2 3 - 0.0077 h3 - 29.4 3 3 T3 70% 2 This equation cannot be solved directly. An iterative solution is needed. A mixture temperature T3 is selected. At this temperature and given relative humidity (70%), specific humidity and enthalpy are read from the psychrometric chart. These values are substituted into the above equation. If the equation is not satisfied, a new value of T3 is selected. This procedure is repeated until the equation is satisfied. Alternatively, EES software can be used. We used the following EES program to get the results: "Given" P=101.325 [kPa] T_1=10 [C] phi_1=1.0 T_2=32 [C] phi_2=0.80 phi_3=0.70 "Analysis" Fluid$='AirH2O' "1st stream properties" h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) "2nd stream properties" h_2=enthalpy(Fluid$, T=T_2, P=P, R=phi_2) w_2=humrat(Fluid$, T=T_2, P=P, R=phi_2) (w_2-w_3)/(w_3-w_1)=(h_2-h_3)/(h_3-h_1) Ratio=(w_2-w_3)/(w_3-w_1) "mixture properties" T_3=temperature(Fluid$, h=h_3, P=P, R=phi_3) h_3=enthalpy(Fluid$, T=T_3, P=P, R=phi_3) The solution of this EES program is T3 = 24.0C, 3 = 0.0149 kg H 2 O/kg dry air & m a1 = 1.31 h3 = 57.6 kJ/kg dry air, & ma2 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-69 14-112 A stream of warm air is mixed with a stream of saturated cool air. The temperature, the specific humidity, and the relative humidity 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 (Fig. A-31) to be h1 = 110.2 kJ/kg dry air 1 = 0.0272 kg H 2 O/kg dry air and 1 h2 = 50.9 kJ/kg dry air 40C 8 kg/s Twb1 = 32C 2 = 0.0129 kg H 2 O/kg dry air Analysis The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: & m a1 2 - 3 h2 - h3 = = & m a 2 3 - 1 h3 - h1 50.9 - h3 8.0 0.0129 - 3 = = 6.0 3 - 0.0272 h3 - 110.2 P = 1 atm AIR 6 kg/s 18C 100% 3 3 3 T3 2 which yields, (b) 3 = 0.0211 kg H 2O / kg dry air h3 = 84.8 kJ / kg dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: (a) (c) T3 = 30.7 C 3 = 75.1% PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-70 14-113 EES Problem 14-112 is reconsidered. The effect of the mass flow rate of saturated cool air stream on the mixture temperature, specific humidity, and relative humidity is to be investigated. Analysis The problem is solved using EES, and the solution is given below. P=101.325 [kPa] Tdb[1] =40 [C] Twb[1] =32 [C] m_dot[1] = 8 [kg/s] Tdb[2] =18 [C] Rh[2] = 1.0 m_dot[2] = 6 [kg/s] P[1]=P P[2]=P[1] P[3]=P[1] "Energy balance for the steady-flow mixing process:" "We neglect the PE of the flow. Since we don't know the cross sectional area of the flow streams, we also neglect theKE of the flow." E_dot_in - E_dot_out = DELTAE_dot_sys DELTAE_dot_sys = 0 [kW] E_dot_in = m_dot[1]*h[1]+m_dot[2]*h[2] E_dot_out = m_dot[3]*h[3] "Conservation of mass of dry air during mixing:" m_dot[1]+m_dot[2] = m_dot[3] "Conservation of mass of water vapor during mixing:" m_dot[1]*w[1]+m_dot[2]*w[2] = m_dot[3]*w[3] m_dot[1]=V_dot[1]/v[1]*convert(1/min,1/s) m_dot[2]=V_dot[2]/v[2]*convert(1/min,1/s) h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],B=Twb[1]) Rh[1]=RELHUM(AirH2O,T=Tdb[1],P=P[1],B=Twb[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) Tdb[3]=TEMPERATURE(AirH2O,h=h[3],P=P[3],w=w[3]) Rh[3]=RELHUM(AirH2O,T=Tdb[3],P=P[3],w=w[3]) v[3]=VOLUME(AirH2O,T=Tdb[3],P=P[3],w=w[3]) Twb[2]=WETBULB(AirH2O,T=Tdb[2],P=P[2],R=RH[2]) Twb[3]=WETBULB(AirH2O,T=Tdb[3],P=P[3],R=RH[3]) m_dot[3]=V_dot[3]/v[3]*convert(1/min,1/s) m2 [kga/s] 0 2 4 6 8 10 12 14 16 Tdb3 [C] 40 35.69 32.79 30.7 29.13 27.91 26.93 26.13 25.45 Rh3 0.5743 0.6524 0.7088 0.751 0.7834 0.8089 0.8294 0.8462 0.8601 w3 [kgw/kga] 0.02717 0.02433 0.02243 0.02107 0.02005 0.01926 0.01863 0.01811 0.01768 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-71 40 38 36 Tdb[3] [C] 34 32 30 28 26 24 0 2 4 6 8 10 12 14 16 m[2] [kga/s] 0.9 0.85 0.8 Rh[3] 0.75 0.7 0.65 0.6 0.55 0 2 4 6 8 10 12 14 16 m[2] [kga/s] 0.028 0.026 w [3] [kgw /kga] 0.024 0.022 0.02 0.018 0.016 0 2 4 6 8 10 12 14 16 m [2] [kga/s] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-72 Wet Cooling Towers 14-114C The working principle of a natural draft cooling tower is based on buoyancy. The air in the tower has a high moisture content, and thus is lighter than the outside air. This light moist air rises under the influence of buoyancy, inducing flow through the tower. 14-115C A spray pond cools the warm water by spraying it into the open atmosphere. They require 25 to 50 times the area of a wet cooling tower for the same cooling load. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-73 14-116 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus 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. 4 The cooling tower is adiabatic. & & & Analysis (a) The mass flow rate of dry air through the tower remains constant (ma1 = ma 2 = ma ) , but 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. Applying the mass and energy balances yields Dry Air Mass Balance: & & ma ,i = ma ,e & & & ma1 = ma 2 = ma AIR 34C 2 EXIT 90% Water Mass Balance: & & & & & & m w,i = m w,e m3 + m a11 = m 4 + m a 2 2 & & & & m3 - m 4 = m a ( 2 - 1 ) = m makeup Energy Balance: & & & E - E = E in out 0 (steady) system =0 WARM WATER 40C 60 kg/s 3 & & Ein = Eout & & & & mi hi = me he (since Q = W = 0) & & 0 = me he - mi hi & & & & 0 = ma 2 h2 + m4 h4 - ma1h1 - m3h3 & & & & 0 = ma ( h2 - h1 ) + ( m3 - mmakeup )h4 - m3h3 & Solving for ma , & m3 (h3 - h4 ) & ma = (h2 - h1 ) - ( 2 - 1 )h4 COOL WATER 4 26C Makeup water 1 AIR INLET 1 atm Tdb = 22C Twb = 16C From the psychrometric chart (Fig. A-31), h1 = 44.7 kJ/kg dry air 1 = 0.0089 kg H 2 O/kg dry air v 1 = 0.849 m 3 /kg dry air and h2 = 1135 kJ / kg dry air . 2 = 0.0309 kg H 2 O / kg dry air From Table A-4, h3 h f @ 40C = 167.53 kJ/kg H 2 O h4 h f @ 26C = 109.01 kJ/kg H 2 O Substituting, & ma = (60 kg/s)(167.53 - 109.01)kJ/kg = 52.9 kg/s (113.5 - 44.7) kJ/kg - (0.0309 - 0.0089)(109.01) kJ/kg Then the volume flow rate of air into the cooling tower becomes & V& = m v = (52.9 kg/s)(0.849 m 3 / kg ) = 44.9 m 3 /s 1 a 1 (b) The mass flow rate of the required makeup water is determined from & & mmakeup = ma ( 2 - 1 ) = (52.9 kg / s)(0.0309 - 0.0089) = 1.16 kg / s PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-74 14-117 Water is cooled by air in a cooling tower. The mass flow rate of dry air is to be determined. Assumptions 1 Steady operating conditions exist and thus 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. 4 The cooling tower is adiabatic. & & & Analysis The mass flow rate of dry air through the tower remains constant (m a1 = m a 2 = m a ) , but 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. Applying the mass and energy balances yields Dry Air Mass Balance: & & & & & ma ,i = m a ,e ma1 = m a 2 = m a Water Mass Balance: & & & & & & m w,i = m w,e m3 + m a11 = m 4 + m a 2 2 & & & & m3 - m 4 = m a ( 2 - 1 ) = m makeup AIR EXIT 2 18C 95% Energy Balance: & & & Ein - Eout = Esystem & & Ein = Eout & & & & mi hi = me he (since Q = W = 0) & & 0 = me he - mi hi 0 (steady) =0 WARM WATER 30C 5 kg/s 3 & & & & 0 = ma 2h2 + m4h4 - ma1h1 - m3h3 & & & & 0 = ma (h2 - h1 ) + (m3 - mmakeup )h4 - m3h3 4 1 AIR & Solving for m a , & ma = & m 3 ( h3 - h 4 ) (h2 - h1 ) - ( 2 - 1 )h4 h1 = 21.8 kJ/kg dry air COOL WATER 22C Makeup water INLET 1 atm 15C 25% From the psychrometric chart (Fig. A-31), 1 = 0.00264 kg H 2 O/kg dry air v 1 = 0.820 m 3 /kg dry air and h2 = 49.3 kJ/kg dry air 2 = 0.0123 kg H 2 O/kg dry air From Table A-4, h3 h f @ 30C = 125.74 kJ/kg H 2 O h4 h f @ 22C = 92.28 kJ/kg H 2 O Substituting, & ma = (5 kg/s)(125.74 - 92.28)kJ/kg = 6.29 kg/s (49.3 - 21.8) kJ/kg - (0.0123 - 0.00264)(92.28) kJ/kg PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-75 14-118 Water is cooled by air in a cooling tower. The exergy lost in the cooling tower is to be determined. Assumptions 1 Steady operating conditions exist and thus 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. 4 The cooling tower is adiabatic. & & & Analysis The mass flow rate of dry air through the tower remains constant (m a1 = m a 2 = m a ) , but 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. Applying the mass and energy balances yields Dry Air Mass Balance: & & & & & ma ,i = m a ,e ma1 = m a 2 = m a Water Mass Balance: & & & & & & m w,i = m w,e m3 + m a11 = m 4 + m a 2 2 & & & & m3 - m 4 = m a ( 2 - 1 ) = m makeup AIR EXIT 2 18C 95% Energy Balance: & & & Ein - Eout = Esystem & & Ein = Eout & & & & mi hi = me he (since Q = W = 0) & & 0 = me he - mi hi 0 (steady) =0 WARM WATER 30C 5 kg/s 3 & & & & 0 = ma 2h2 + m4h4 - ma1h1 - m3h3 & & & & 0 = ma (h2 - h1 ) + (m3 - mmakeup )h4 - m3h3 & Solving for m a , & ma = & m 3 ( h3 - h 4 ) (h2 - h1 ) - ( 2 - 1 )h4 h1 = 21.8 kJ/kg dry air COOL WATER 4 22C Makeup water 1 AIR INLET 1 atm 15C 25% From the psychrometric chart (Fig. A-31), 1 = 0.00264 kg H 2 O/kg dry air v 1 = 0.820 m 3 /kg dry air and h2 = 49.3 kJ/kg dry air 2 = 0.0123 kg H 2 O/kg dry air From Table A-4, h3 h f @ 30C = 125.74 kJ/kg H 2 O h4 h f @ 22C = 92.28 kJ/kg H 2 O Substituting, & ma = (5 kg/s)(125.74 - 92.28)kJ/kg = 6.29 kg/s (49.3 - 21.8) kJ/kg - (0.0123 - 0.00264)(92.28) kJ/kg The mass flow rate of water stream at state 3 per unit mass of dry air is PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-76 m3 = & m3 5 kg water/s = = 0.7949 kg water/kg dry air & m a 6.29 kg dry air/s The mass flow rate of water stream at state 4 per unit mass of dry air is m 4 = m3 - ( 2 - 1 ) = 0.7949 - (0.0123 - 0.00264) = 0.7852 kg water/kg dry air The entropies of water streams are s 3 = s f @ 30C = 0.4368 kJ/kg K s 4 = s f @ 22C = 0.3249 kJ/kg K The entropy change of water stream is s water = m 4 s 4 - m3 s 3 = 0.7852 0.3249 - 0.7949 0.4368 = -0.09210 kJ/K kg dry air The entropies of water vapor in the air stream are s g1 = s g @ 15C = 8.7803 kJ/kg K s g 2 = s g @ 18C = 8.7112 kJ/kg K The entropy change of water vapor in the air stream is s vapor = 2 s g 2 - 1 s g1 = 0.0123 8.7112 - 0.00264 8.7803 = 0.08397 kJ/K kg dry air The partial pressures of water vapor and dry air for air streams are Pv1 = 1 Pg1 = 1 Psat @ 15C = (0.25)(1.7057 kPa) = 0.4264 kPa Pa1 = P1 - Pv1 = 101.325 - 0.4264 = 100.90 kPa Pv 2 = 2 Pg 2 = 2 Psat @ 18C = (0.95)(2.065 kPa) = 1.962 kPa Pa 2 = P2 - Pv 2 = 101.325 - 1.962 = 99.36 kPa The entropy change of dry air is s a = s 2 - s1 = c p ln = (1.005) ln P T2 - R ln a 2 Pa1 T1 291 99.36 - (0.287) ln = 0.01483 kJ/kg dry air 288 100.90 The entropy generation in the cooling tower is the total entropy change: s gen = s water + s vapor + s a = -0.09210 + 0.08397 + 0.01483 = 0.00670 kJ/K kg dry air Finally, the exergy destruction per unit mass of dry air is x dest = T0 s gen = (288 K)(0.00670 kJ/K kg dry air) = 1.93 kJ/kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-77 14-119E Water is cooled by air in a cooling tower. The relative humidity of the air at the exit and the water's exit temperature are to be determined. Assumptions 1 Steady operating conditions exist and thus 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. 4 The cooling tower is adiabatic. & & & Analysis The mass flow rate of dry air through the tower remains constant (m a1 = m a 2 = m a ) , but 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. Applying the mass and energy balances yields Dry Air Mass Balance: & & & & & ma ,i = m a ,e ma1 = m a 2 = m a Water Mass Balance: & & & & & & m w,i = m w,e m3 + m a11 = m 4 + m a 2 2 & & & & m3 - m 4 = m a ( 2 - 1 ) = m makeup AIR EXIT 2 =0.018 75F Energy Balance: & & & E - E = E in out 0 (steady) system =0 WARM WATER 100F 10,000 lbm/h 3 & & E in = E out & & & & mi hi = m e he (since Q = W = 0) & & 0 = m e he - mi hi & a 2 h2 + m 4 h4 - m a1 h1 - m 3 h3 & & & 0=m & & & & 0 = m a (h2 - h1 ) + (m3 - m makeup )h4 - m3 h3 1 AIR 4 Solving for h4, & & m h - m a (h2 - h1 ) h4 = 3 3 & & m3 - m makeup From the psychrometric chart (Fig. A-31E), h1 = 16.8 Btu/lbm dry air COOL WATER Makeup water INLET 1 atm 60F 20% 7000 lbm/h 1 = 0.00219 kg H 2 O/kg dry air v 1 = 13.15 ft 3 /lbm dry air and h2 = 37.7 Btu/lbm dry air 2 = 0.957 = 95.7% From Table A-4, h3 h f @ 100F = 68.03 Btu/lbm H 2 O Also, & & m makeup = m a ( 2 - 1 ) = (7000 / 3600 lbm/s)(0.018 - 0.00219) = 0.03075 lbm/s Substituting, & & m h - m a (h2 - h1 ) (10,000 / 3600)(68.03) - (7000 / 3600)(37.7 - 16.8) h4 = 3 3 = = 53.99 Btu/lbm & & (10,000 / 3600) - 0.03075 m3 - m makeup The exit temperature of the water is then (Table A-4E) T4 = Tsat @ h f =53.99 Btu/lbm = 85.9F PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-78 14-120 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus 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. 4 The cooling tower is adiabatic. & & & Analysis (a) The mass flow rate of dry air through the tower remains constant (ma1 = ma 2 = ma ) , but 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. Applying the mass and energy balances yields Dry Air Mass Balance: & & m a ,i = m a , e & & & m a1 = m a 2 = m a 2 35C 100% Water Mass Balance: & & m w , i = m w, e & & & & m3 + m a11 = m 4 + m a 2 2 & 3 - m 4 = m a ( 2 - 1 ) = m makeup & & & m WATER 40C 25 kg/s System boundary 1 4 30C Makeup AIR 3 Energy Balance: & & & Ein - Eout = Esystem 0 (steady) & & =0 Ein = Eout & & & & mi hi = me he (since Q = W = 0) & & 0 = me he - mi hi & & & & 0 = ma 2 h2 + m4 h4 - ma1h1 - m3h3 & & & & 0 = ma (h2 - h1 ) + (m3 - mmakeup )h4 - m3h3 & ma = & m3 (h3 - h4 ) (h2 - h1 ) - ( 2 - 1 )h4 96 kPa 20C 70% The properties of air at the inlet and the exit are Pv1 = 1Pg1 = 1Psat @ 20C = (0.70)(2.3392 kPa) = 1.637 kPa Pa1 = P - Pv1 = 96 - 1.637 = 94.363 kPa 1 v1 = 1 = RaT1 (0.287 kPa m3 / kg K)(293 K) = = 0.891 m3 / kg dry air Pa1 94.363 kPa 0.622 Pv1 0.622(1.637 kPa) = = 0.0108 kg H 2O/kg dry air P - Pv1 (96 - 1.637) kPa 1 h1 = c pT1 + 1hg1 = (1.005 kJ/kg C)(20C) + (0.0108)(2537.4 kJ/kg) = 47.5 kJ/kg dry air and Pv 2 = 2 Pg 2 = 2 Psat @ 35C = (1.00)(5.6291 kPa) = 5.6291 kPa 2 = 0.622 Pv 2 0.622(5.6291 kPa) = = 0.0387 kg H 2O/kg dry air P2 - Pv 2 (96 - 5.6291) kPa h2 = c pT2 + 2 hg 2 = (1.005 kJ/kg C)(35C) + (0.0387)(2564.6 kJ/kg) = 134.4 kJ/kg dry air From Table A-4, PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-79 h3 h f @ 40C = 167.53 kJ/kg H 2 O h4 h f @ 30C = 125.74 kJ/kg H 2 O Substituting, & ma = (25 kg/s)(167.53 - 125.74)kJ/kg = 12.53 kg/s (134.4 - 47.5) kJ/kg - (0.0387 - 0.0108)(125.74) kJ/kg Then the volume flow rate of air into the cooling tower becomes & V&1 = m av 1 = (12.53 kg/s)(0.891 m 3 / kg ) = 11.2 m 3 /s (b) The mass flow rate of the required makeup water is determined from & & m makeup = m a ( 2 - 1 ) = (12.53 kg/s)(0.0387 - 0.0108) = 0.35 kg/s PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-80 14-121 A natural-draft cooling tower is used to remove waste heat from the cooling water flowing through the condenser of a steam power plant. The mass flow rate of the cooling water, the volume flow rate of air into the cooling tower, and the mass flow rate of the required makeup water are to be determined. Assumptions 1 All processes are steady-flow and the mass flow rate of dry air remains constant during the & & & entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and exit states of the moist air for the tower are completely specified. The properties may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES) h1 = 50.74 kJ/kg dry air 1 = 0.01085 kg H 2 O/kg dry air v 1 = 0.8536 m 3 /kg dry air h2 = 142.83 kJ/kg dry air T2 = 37C 2 =100% 2 AIR 1 Makeup water T1 = 23C Twb1 = 18C 2 = 0.04112 kg H 2 O/kg dry air The enthalpies of cooling water at the inlet and exit of the condenser are (Table A-4) hw3 = h f@ 40C = 167.53 kJ/kg hw 4 = h f@ 26C = 109.01 kJ/kg The steam properties for the condenser are (Steam tables) Ps1 = 200 kPa h s1 = 504.71 kJ/kg x s1 = 0 hs 2 = 2524.3 kJ/kg s s 2 = 7.962 kJ/kg.K Ps 3 = 10 kPa h s 3 = 191.81 kJ/kg x s1 = 0 Ps 2 = 10 kPa The mass flow rate of dry air is given by V& V&1 & ma = 1 = v 1 0.8536 m 3 /kg The mass flow rates of vapor at the inlet and exit of the cooling tower are V&1 & & V mv1 = 1ma = (0.01085) = 0.01271 &1 0.8536 V&1 & & mv 2 = 2 ma = (0.04112) = 0.04817V&1 0.8536 Mass and energy balances on the cooling tower give & & & & m v1 + m cw3 = m v 2 + m cw4 & & & & m a h1 + m cw3 hw3 = m a h2 + m cw4 h w4 The mass flow rate of the makeup water is determined from & & & & & m makeup = m v 2 - m v1 = m cw3 - m cw4 An energy balance on the condenser gives & & & & & & 0.18m s h s1 + 0.82m s h s 2 + m cw4 h w4 + m makeup h w 4 = m s h s 3 + m cw3 h w3 Solving all the above equations simultaneously with known and determined values using EES, we obtain & m cw3 = 1413 kg/s V& = 47,700 m 3 /min 1 & m makeup = 28.19 kg/s PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-81 Review Problems 14-122 Air is compressed by a compressor and then cooled to the ambient temperature at high pressure. It is to be determined if there will be any condensation in the compressed air lines. Assumptions The air and the water vapor are ideal gases. Properties The saturation pressure of water at 20C is 2.3392 kPa (Table A-4).. Analysis The vapor pressure of air before compression is Pv1 = 1 Pg = 1 Psat @ 25C = (0.50)(2.3392 kPa) = 1.17 kPa The pressure ratio during the compression process is (800 kPa)/(92 kPa) = 8.70. That is, the pressure of air and any of its components increases by 8.70 times. Then the vapor pressure of air after compression becomes Pv 2 = Pv1 (Pressure ratio) = (1.17 kPa)(8.70) = 10.2 kPa The dew-point temperature of the air at this vapor pressure is Tdp = Tsat @ Pv 2 = Tsat @ 10.2 kPa = 46.1C which is greater than 20C. Therefore, part of the moisture in the compressed air will condense when air is cooled to 20C. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-82 14-123E The mole fraction of the water vapor at the surface of a lake and the mole fraction of water in the lake are to be determined and compared. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly soluble in water and thus Henry's law is applicable. Properties The saturation pressure of water at 60F is 0.2564 psia (Table A-4E). Henry's constant for air dissolved in water at 60F (289 K) is given in Table 16-2 to be H = 62,000 bar. Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 60F, Pvapor = Psat @60 F = 0.2564 psia Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air at the surface of the lake is determined to be y vapor = Pvapor P = 0.2564 psia = 0.0186 (or 1.86 percent) 13.8 psia Air 13.8 psi 60F The partial pressure of dry air just above the lake surface is Pdry air = P - Pvapor = 13.8 - 0.2564 = 13.54 psia Lake Then the mole fraction of air in the water becomes ydry air, liquid side = Pdry air, gasside H = 1354 psia(1 atm / 14.696 psia ) . = 151 10 -5 . 62,000 bar (1 atm / 1.01325 bar) which is very small, as expected. Therefore, the mole fraction of water in the lake near the surface is y water,liquid side = 1 - y dry air, liquid side = 1 - 1.5110 -5 1.0 Discussion The concentration of air in water just below the air-water interface is 1.51 moles per 100,000 moles. The amount of air dissolved in water will decrease with increasing depth. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-83 14-124 The mole fraction of the water vapor at the surface of a lake at a specified temperature is to be determined. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air at the lake surface is saturated. Properties The saturation pressure of water at 18C is 2.065 kPa (Table A-4). Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 18C, Pvapor = Psat @18C = 2.065 kPa Assuming both the air and vapor to be ideal gases, the partial pressure and mole fraction of dry air in the air at the surface of the lake are determined to be Pdry air = P - Pvapor = 100 - 2.065 = 97.94 kPa y dry air = Pdry air P = 97.94 kPa = 0.979 (or 97.9%) 100 kPa Air 100 kPa 18C Lake Therefore, the mole fraction of dry air is 97.9 percent just above the air-water interface. 14-125E A room is cooled adequately by a 7500 Btu/h air-conditioning unit. If the room is to be cooled by an evaporative cooler, the amount of water that needs to be supplied to the cooler is to be determined. Assumptions 1 The evaporative cooler removes heat at the same rate as the air conditioning unit. 2 Water evaporates at an average temperature of 70F. Properties The enthalpy of vaporization of water at 70F is 1053.7 Btu/lbm (Table A-4E). Analysis Noting that 1 lbm of water removes 1053.7 Btu of heat as it evaporates, the amount of water that & & needs to evaporate to remove heat at a rate of 7500 Btu/h is determined from Q = mwater h fg to be & m water = & Q 7500 Btu/h = = 7.12 lbm/h h fg 1053.7 Btu/lbm PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-84 14-126E The required size of an evaporative cooler in cfm (ft3/min) for an 8-ft high house is determined by multiplying the floor area of the house by 4. An equivalent rule is to be obtained in SI units. Analysis Noting that 1 ft = 0.3048 m and thus 1 ft2 = 0.0929 m2 and 1 ft3 = 0.0283 m3, and noting that a flow rate of 4 ft3/min is required per ft2 of floor area, the required flow rate in SI units per m2 of floor area is determined to 1 ft 2 4 ft 3 / min 0.0929 m 2 4 0.0283 m 3 / min 1 m 2 1.22 m 3 / min Therefore, a flow rate of 1.22 m3/min is required per m2 of floor area. 14-127 A cooling tower with a cooling capacity of 440 kW is claimed to evaporate 15,800 kg of water per day. It is to be determined if this is a reasonable claim. Assumptions 1 Water evaporates at an average temperature of 30C. 2 The coefficient of performance of the air-conditioning unit is COP = 3. Properties The enthalpy of vaporization of water at 30C is 2429.8 kJ/kg (Table A-4). Analysis Using the definition of COP, the electric power consumed by the air conditioning unit when running is & Win = & Qcooling COP = 440 kW = 146.7 kW 3 Then the rate of heat rejected at the cooling tower becomes & & & Qrejected = Qcooling + Win = 440 + 146.7 = 586.7 kW Noting that 1 kg of water removes 2429.8 kJ of heat as it evaporates, the amount of water that needs to & & evaporate to remove heat at a rate of 586.7 kW is determined from Qrejected = mwater h fg to be & m water = & Qrejected = 586.7 kJ/s = 0.2415 kg/s = 869.3 kg/h = 20,860 kg/day 2429.8 kJ/kg h fg In practice, the air-conditioner will run intermittently rather than continuously at the rated power, and thus the water use will be less. Therefore, the claim amount of 15,800 kg per day is reasonable. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-85 14-128 Air is cooled by evaporating water into this air. The amount of water required and the cooling produced are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 64.0 kJ/kg dry air 1 = 0.0092 kg H 2 O/kg dry air and h2 = 66.0 kJ/kg dry air Water 20C 1 atm 40C 20% (Table A-4) AIR 25C 80% 2 = 0.0160 kg H 2 O/kg dry air Also, hw h f @ 20C = 83.92 kJ/kg Analysis The amount of moisture in the air increases due to humidification ( 2 > 1). Applying the water mass balance and energy balance equations to the combined cooling and humidification section, Water Mass Balance: & & & & & m w,i = m w, e m a11 = m a 2 2 + m w = 2 - 1 = 0.0160 - 0.0092 = 0.0068 kg H 2 O/kg dry air Energy Balance: & & & E in - E out = E system 0 (steady) =0 & & E in = E out & & & m i hi = Qout + m e he & & & & & & Qout = m a1 h1 + m w h w - m a 2 h2 = m a (h1 - h2 ) + m w hw q out = h1 - h2 + ( 2 - 1 )h w = (64.0 - 66.0)kJ/kg + (0.0068)(83.92) = -1.43 kJ/kg dry air The negative sign shows that the heat is actually transferred to the system. PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-86 14-129 Air is humidified adiabatically by evaporating water into this air. The temperature of the air at the exit is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31) to be h1 = 64.0 kJ/kg dry air 1 = 0.0092 kg H 2 O/kg dry air and h w h f @ 20C = 83.92 kJ/kg (Table A-4) Analysis The amount of moisture in the air increases due to humidification ( 2 > 1). Applying the water mass balance and energy balance equations to the combined cooling and humidification section, Water Mass Balance: & & & & & m w, i = m w, e m a11 = m a 2 2 + m w & & m w = m a ( 2 - 1 ) Energy Balance: & & & E in - E out = E system 0 (steady) Water 20C =0 & & E in = E out & & m i hi = m e h e & a1 h1 + m w hw = m a 2 h2 & & m & & m w hw = m a (h2 - h1 ) ( 2 - 1 )hw = h2 - h1 1 atm 40C 20% AIR T2=? 80% Substituting, ( 2 - 0.0092)(83.92) = h2 - 64.0 The solution of this equation requires a trial-error method. An air exit temperature is assumed. At this temperature and given relative humidity, the enthalpy and specific humidity values are obtained from psychrometric chart and substituted into this equation. If the equation is not satisfied, a new value of exit temperature is assumed and this continues until the equation is satisfied. Alternatively, an equation solver such as EES may be used for the direct results. We used the following EES program. "Given" P=101.325 "[kPa]" T_1=40 "[C]" phi_1=0.20 phi_2=0.80 "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, R=phi_2, P=P) PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-87 w_2=humrat(Fluid1$, T=T_2, R=phi_2, P=P) h_w=enthalpy(Fluid2$, T=20, x=0) q=0 q=h_1-h_2+(w_2-w_1)*h_w The results of these equations are T2 = 24.6C h2 = 64.56 kJ/kg dry air 2 = 0.01564 kg H 2 O/kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-88 14-130E Air is cooled and dehumidified at constant pressure. The rate of cooling and the minimum humid air temperature required to meet this cooling requirement are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 50.6 Btu/lbm dry air 1 = 0.0263 lbm H 2 O/lbm dry air v 1 = 14.44 ft 3 /lbm dry air and h2 = 28.2 Btu/lbm dry air Cooling coils T2 = 75F 2 =50% 2 82.5F Condensate removal T1 = 90F 1 =85% 2 = 0.0093 lbm H 2 O/lbm dry air 1 atm Condensate 1 We assume that the condensate leaves this system at the average temperature of the air inlet and exit. Then, hw h f @ 82.5F = 50.56 Btu/lbm (Table A-4) Analysis The amount of moisture in the air decreases due to dehumidification ( 2 < 1). The mass of air is ma = V1 1000 ft 3 = = 69.25 lbm v 1 14.44 ft 3 / lbm dry air Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: & & & & mw,i = mw ,e ma1 1 = ma 2 2 + mw & m w = m a (1 - 2 ) = (69.25 kg)(0.0263 - 0.0093) = 1.177 lbm Energy Balance: & & & E in - E out = E system 0 (steady) =0 & & E in = E out & & & mi hi = Qout + m e he & & & & & & Qout = m a1 h1 - (m a 2 h2 + m w h w ) = m a (h1 - h2 ) - m w hw Qout = m a (h1 - h2 ) - m w hw Qout = (69.25 kg)(50.6 - 28.2)Btu/lbm - (1.177 lbm)(50.56 Btu/lbm) = 1492 Btu For the desired dehumidification, the air at the exit should be saturated with a specific humidity of 0.0093 lbm water/lbm dry air. That is, 2 = 1.0 2 = 0.0093 lbm H 2 O/lbm dry air The temperature of the air at this state is the minimum air temperature required during this process: T2 = 55.2F PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-89 14-131E Air is cooled and dehumidified at constant pressure by a simple ideal vapor-compression refrigeration system. The system's COP is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 50.6 Btu/lbm dry air 1 = 0.0263 lbm H 2 O/lbm dry air v 1 = 14.44 ft 3 /lbm dry air and h2 = 28.2 Btu/lbm dry air T2 = 75F 2 =50% 2 Cooling coils T1 = 90F 1 =85% 1 atm Condensate 1 82.5F Condensate removal 2 = 0.0093 lbm H 2 O/lbm dry air For the desired dehumidification, the air at the exit should be saturated with a specific humidity of 0.0093 lbm water/lbm dry air. That is, 2 = 1.0 2 = 0.0093 lbm H 2 O/lbm dry air The temperature of the air at this state is the minimum air temperature required during this process: T2, min = 55.2F From the problem statement, the properties of R-134a at various states are (Tables A-11E through A-13E or from EES): T1 = 55.2 - 10 = 45.2F h1 = h g @ 55 psia = 109.49 Btu/lbm P2 = Psat @ 45.2 F = 55 psia s1 = s g @ 55 psia = 0.22156 Btu/lbm R sat. vapor Tsat = 90 + 19.5 = 109.5F P2 = Psat @ 109.5F = 160 psia h2 = 119.01 kJ/kg s 2 = s1 @ 160 psia T QH 3 109.5F 2 Win P3 = 160 psia h3 = h f sat. liquid = 48.52 Btu/lbm h4 h3 = 48.52 Btu/lbm ( throttling) 45.2F 4s 4 QL 1 The COP of this system is then COP = qL h - h4 109.49 - 48.52 = 6.40 = = 1 win h2 - h1 119.01 - 109.49 s PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-90 14-132E Air at a specified state is heated to a specified temperature. The relative humidity after the heating is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis There is no correspondence of inlet state from the psychrometric chart. Therefore, we have to use EES psychrometric functions to obtain the specific humidity: 1 = 0.0023 lbm H 2 O/lbm dry air As the outside air infiltrates into the dacha, it does not gain or lose any water. Therefore the humidity ratio inside the dacha is the same as that outside, 70F 1 32F 60% RH 2 1 atm AIR 2 = 1 = 0.0023 lbm H 2 O/lbm dry air From EES or Fig. A-31E, at this humidity ratio and the temperature inside the dacha gives 2 = 0.146 = 14.6% PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-91 14-133E Air is humidified by evaporating water into this air. The amount of heating is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (m a1 = m a 2 = m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 19.3 Btu/lbm dry air 1 = 0.0023 kg H 2 O/kg dry air v 1 = 13.40 m 3 /kg dry air and h2 = 27.1 Btu/lbm dry air Water 60F 1 atm 70F 14.6% AIR 70F 60% 2 = 0.0094 lbm H 2 O/lbm dry air Also, hw h f @ 60F = 28.08 Btu/lbm (Table A-4E) Analysis The amount of moisture in the air increases due to humidification ( 2 > 1). Applying the water mass balance and energy balance equations to the combined cooling and humidification section, Water Mass Balance: & & & & & m w, i = m w, e m a11 = m a 2 2 + m w Energy Balance: & & & E in - E out = E system 0 (steady) =0 & & E in = E out & & & mi hi + Qin = m e he & = m h - m h - m h = m (h - h ) - m h &w w Qin & a 2 2 & a1 1 & w w & a 2 1 q in = h2 - h1 - ( 2 - 1 )h w = (27.1 - 19.3)Btu/lbm - (0.0094 - 0.0023)(28.08) = 7.59 Btu/lbm dry air The mass of air that has to be humidified is ma = V 16,000 ft 3 = = 1194 lbm dry air v 1 13.40 ft 3 /lbm dry air The total heat requirement is then Qin = m a q in = (1194 lbm dry air )(7.59 Btu/lbm dry air ) = 9062 Btu PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-92 14-134E It is estimated that 190,000 barrels of oil would be saved per day if the thermostat setting in residences in summer were raised by 6F (3.3C). The amount of money that would be saved per year is to be determined. Assumptions The average cooling season is given to be 120 days, and the cost of oil to be $20/barrel. Analysis The amount of money that would be saved per year is determined directly from (190,000 barrel/day)(120 days/year)($70/barrel) = $1,596,000 ,000 Therefore, the proposed measure will save more than one and half billion dollars a year. 14-135 Shading the condenser can reduce the air-conditioning costs by up to 10 percent. The amount of money shading can save a homeowner per year during its lifetime is to be determined. Assumptions It is given that the annual air-conditioning cost is $500 a year, and the life of the airconditioning system is 20 years. Analysis The amount of money that would be saved per year is determined directly from ($500 / year)(20 years)(0.10) = $1000 Therefore, the proposed measure will save about $1000 during the lifetime of the system. 14-136 A tank contains saturated air at a specified state. The mass of the dry air, the specific humidity, and the enthalpy of the air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The air is saturated, thus the partial pressure of water vapor is equal to the saturation pressure at the given temperature, Pv = Pg = Psat @ 25C = 3.1698 kPa Pa = P - Pv = 97 - 3.1698 = 93.83 kPa Treating air as an ideal gas, ma = PaV (93.83 kPa)(3 m3 ) = = 3.29 kg RaT (0.287 kPa m3 / kg K)(298 K) 3 m3 25C 97 kPa (b) The specific humidity of air is determined from = 0.622 Pv (0.622)(3.1698 kPa) = = 0.0210 kg H 2 O/kg dry air P - Pv (97 - 3.1698) kPa (c) The enthalpy of air per unit mass of dry air is determined from h = ha + hv c p T + h g = (1.005 kJ/kg C)(25C) + (0.0210)(2546.5 kJ/kg) = 78.6 kJ/kg dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-93 14-137 EES Problem 14-136 is reconsidered. The properties of the air at the initial state are to be determined and the effects of heating the air at constant volume until the pressure is 110 kPa is to be studied. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" Tdb[1] = 25 [C] P[1]=97 [kPa] Rh[1]=1.0 P[2]=110 [kPa] Vol = 3 [m^3] w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) m_a=Vol/v[1] h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],w=w[1]) "Energy Balance for the constant volume tank:" E_in - E_out = DELTAE_tank DELTAE_tank=m_a*(u[2] -u[1]) E_in = Q_in E_out = 0 [kJ] u[1]=INTENERGY(AirH2O,T=Tdb[1],P=P[1],w=w[1]) u[2]=INTENERGY(AirH2O,T=Tdb[2],P=P[2],w=w[2]) "The ideal gas mixture assumption applied to the constant volume process yields:" P[1]/(Tdb[1]+273)=P[2]/(Tdb[2]+273) "The mass of the water vapor and dry air are constant, thus:" w[2]=w[1] Rh[2]=RELHUM(AirH2O,T=Tdb[2],P=P[2],w=w[2]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],w=w[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) PROPERTIES AT THE INITIAL STATE h[1]=78.67 [kJ/kga] m_a=3.289 [kga] v[1]=0.9121 [m^3/kga] w[1]=0.02101 [kgw/kga] 100 P2 [kPa] 97 99 101 103 105 107 109 110 Qin [kJ] 0 15.12 30.23 45.34 60.45 75.55 90.65 98.2 75 Qin [kJ] 50 25 0 96 98 100 102 104 106 108 110 P[2] [kPa] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-94 14-138E Air at a specified state and relative humidity flows through a circular duct. The dew-point temperature, the volume flow rate of air, and the mass flow rate of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The vapor pressure of air is Pv = Pg = Psat @ 60F = (0.50)(0.2564 psia) = 0.128 psia Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 0.128 psia = 41.3F (from EES) AIR 15 psia 50 f/s 60F, 50% (b) The volume flow rate is determined from V& = VA = V D 2 4 (8 / 12 ft ) 2 = (50 ft/s) 4 = 17.45 ft 3 /s (c) To determine the mass flow rate of dry air, we first need to calculate its specific volume, Pa = P - Pv = 15 - 0.128 = 14.872 psia RaT1 (0.3704 psia ft 3 / lbm R)(520 R) = = 12.95 ft 3 / lbm dry air Pa1 14.872 psia v1 = Thus, & ma1 = V&1 17.45 ft 3 / s = = 1.35 lbm/s v1 12.95 ft 3 / lbm dry air PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-95 14-139 Air enters a cooling section at a specified pressure, temperature, and relative humidity. The temperature of the air at the exit and the rate of heat transfer are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air also remains constant ( 1 = 2 ) as it flows through the cooling section since the process involves no humidification or dehumidification. The total pressure is 97 kPa. The properties of the air at the inlet state are Pv1 = 1Pg1 = 1Psat @ 35C = (0.3)(5.629 kPa) = 1.69 kPa Pa1 = P - Pv1 = 97 - 1.69 = 95.31 kPa 1 Cooling coils 1 35C 30% 6 m3/min 2 97 kPa AIR v1 = RaT1 (0.287 kPa m3 / kg K)(308 K) = Pa1 95.31 kPa = 0.927 m3 / kg dry air 1 = 0.622 Pv1 0.622(1.69 kPa) = = 0.0110 kg H 2O/kg dry air (= 2 ) P - Pv1 (97 - 1.69) kPa 1 h1 = c pT1 + 1hg1 = (1.005 kJ/kgC)(35C) + (0.0110)(2564.6 kJ/kg) = 63.44 kJ/kg dry air The air at the final state is saturated and the vapor pressure during this process remains constant. Therefore, the exit temperature of the air must be the dew-point temperature, Tdp = Tsat @ Pv = Tsat @ 1.69 kPa = 14.8C (b) The enthalpy of the air at the exit is h2 = c pT2 + 2 hg 2 = (1.005 kJ/kg C)(14.8C) + (0.0110)(2528.1 kJ/kg) = 42.78 kJ/kg dry air Also & ma = V&1 6 m3 / s = = 6.47 kg/min v 1 0.927 m 3 / kg dry air Then the rate of heat transfer from the air in the cooling section becomes & & Qout = m a (h1 - h2 ) = (6.47 kg/min)(63.44 - 42.78)kJ/kg = 134 kJ/min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-96 14-140 The outdoor air is first heated and then humidified by hot steam in an air-conditioning system. The rate of heat supply in the heating section and the mass flow rate of the steam required in the humidifying section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The amount of moisture in the air also remains constants it flows through the heating section ( 1 = 2 ) , but increases in the humidifying section ( 3 > 2 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Fig. A-31) to be h1 = 17.7 kJ/kg dry air 1 = 0.0030 kg H 2 O/kg dry air (= 2 ) v 1 = 0.807 m 3 /kg dry air h2 = 29.8 kJ / kg dry air Heating coils 10C 40% 22 m3/min 1 1 atm AIR 2 22C 3 25C 55% 2 = 1 = 0.0030 kg H 2 O / kg dry air h3 = 52.9 kJ / kg dry air 3 = 0.0109 kg H 2 O / kg dry air Analysis (a) The mass flow rate of dry air is & ma = V&1 22 m3 / min = = 27.3 kg/min v1 0.807 m3 / kg Then the rate of heat transfer to the air in the heating section becomes & & Qin = ma ( h2 - h1 ) = ( 27.3 kg / min)(29.8 - 17.7 )kJ / kg = 330.3 kJ / min (b) The conservation of mass equation for water in the humidifying section can be expressed as & & & ma 2 2 + mw = ma 3 3 & & or mw = ma ( 3 - 2 ) Thus, & mw = ( 27.3 kg / min)(0.0109 - 0.0030) = 0.216 kg / min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-97 14-141 Air is cooled and dehumidified in an air-conditioning system with refrigerant-134a as the working fluid. The rate of dehumidification, the rate of heat transfer, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The saturation pressure of water at 30C is 4.2469 kPa. Then the dew point temperature of the incoming air stream at 30C becomes Tdp = Tsat @ Pv = Tsat @ 0.74.2469 kPa = 24C Since air is cooled to 20C, which is below its dew point temperature, some of the moisture in the air will condense. The mass flow rate of dry air remains constant during the entire process, but the amount of moisture in the air decreases due to dehumidification ( 2 < 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. Then the properties of the air at both states are determined from the psychrometric chart (Fig. A-31) to be h1 = 78.3 kJ/kg dry air 1 = 0.0188 kg H 2 O/kg dry air 3 R-134a 700 kPa x3 = 20% 4 700 kPa sat. vapor AIR 1 atm 20C 2 v 1 = 0.885 m 3 /kg dry air and h2 = 57.5 kJ / kg dry air 2 = 0.0147 kg H 2 O / kg dry air Also, Then, hw h f @ 20C = 83.915 kJ/kg (Table A-4) 1 30C 70% 4 m3/min & ma1 = V&1 4 m 3 / min = = 4.52 kg/min v 1 0.885 m 3 / kg dry air Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the refrigerant), Water Mass Balance: & & mw,i = mw,e & & & ma1 1 = ma 2 2 + mw & & mw = ma ( 1 - 2 ) = (4.52 kg / min)(0.0188 - 0.0147) = 0.0185 kg / min (b) Energy Balance: & & & E - E = E in out 0 (steady) system =0 & & Ein = Eout & & & mi hi = Qout + me he & & & & & & Qout = ma1h1 - ( ma 2 h2 + mw hw ) = ma ( h1 - h2 ) - mw hw & Qout = (4.52 kg/min)(78.3 - 57.5)kJ/kg - (0.0185 kg/min)(83.915 kJ/kg) = 92.5 kJ/min (c) The inlet and exit enthalpies of the refrigerant are h3 = h g + x 3 h fg = 88.82 + 0.2 176.21 = 124.06 kJ/kg h4 = h g @ 700 kPa = 265.03 kJ/kg Noting that the heat lost by the air is gained by the refrigerant, the mass flow rate of the refrigerant becomes & QR 92.5 kJ/min & & & = = 0.66 kg/min Q R = m R (h4 - h3 ) m R = h4 - h3 (265.03 - 124.06) kJ/kg PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-98 14-142 Air is cooled and dehumidified in an air-conditioning system with refrigerant-134a as the working fluid. The rate of dehumidification, the rate of heat transfer, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow 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. Analysis (a) The dew point temperature of the incoming air stream at 30C is Pv1 = 1Pg1 = 1Psat @ 30C = (0.7)(4.247 kPa) = 2.973 kPa Tdp = Tsat @ Pv = Tsat @ 2.973 kPa = 24C Since air is cooled to 20C, which is below its dew point temperature, some of the moisture in the air will condense. The amount of moisture in the air decreases due to dehumidification ( 2 < 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 95 kPa. The properties of the air at both states are determined to be Pa1 = P1 - Pv1 = 95 - 2.97 = 92.03 kPa 3 R-134a 30C 70% 4 m3/min 700 kPa x3 = 20% 4 700 kPa sat. vapor AIR 95 kPa 20C 1 2 v1 = 1 = Ra T1 (0.287 kPa m 3 / kg K)(303 K) = = 0.945 m 3 / kg dry air Pa1 92.03 kPa 0.622 Pv1 0.622(2.97 kPa) = = 0.0201 kg H 2 O/kg dry air P1 - Pv1 (95 - 2.97) kPa h1 = c p T1 + 1hg1 = (1.005 kJ/kg C)(30C) + (0.0201)(2555.6 kJ/kg) = 81.50 kJ/kg dry air and Pv 2 = 2 Pg 2 = (1.00) Psat @ 20C = 2.3392 kPa 2 = 0.622 Pv 2 0.622(2.3392 kPa) = = 0.0157 kg H 2 O/kg dry air P2 - Pv 2 (95 - 2.3392) kPa h2 = c p T2 + 2 hg 2 = (1.005 kJ/kg C)(20C) + (0.0157)(2537.4 kJ/kg) = 59.94 kJ/kg dry air hw h f @ 20C = 83.915 kJ/kg Also, Then, (Table A-4) & ma1 = 4 m3 / min V&1 = = 4.23 kg/min v1 0.945 m3 / kg dry air Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the refrigerant), Water Mass Balance: & & mw,i = mw,e & & & ma1 1 = ma 2 2 + mw & & mw = ma ( 1 - 2 ) = (4.23 kg / min)(0.0201 - 0.0157) = 0.0186 kg / min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-99 (b) Energy Balance: & & & Ein - Eout = Esystem 0 (steady) =0 & & & & & & Qout = ma1h1 - ( ma 2 h2 + mw hw ) = ma ( h1 - h2 ) - mw hw & & Ein = Eout & & & mi hi = Qout + me he & Qout = (4.23 kg/min)(81.50 - 59.94)kJ/kg - (0.0186 kg/min)(83.915 kJ/kg) = 89.7 kJ/min (c) The inlet and exit enthalpies of the refrigerant are h3 = h g + x 3 h fg = 88.82 + 0.2 176.21 = 124.06 kJ/kg h4 = h g @ 700 kPa = 265.03 kJ/kg Noting that the heat lost by the air is gained by the refrigerant, the mass flow rate of the refrigerant is determined from & & Q R = m R (h4 - h3 ) & QR 89.7 kJ/min & = = 0.636 kg/min mR = h4 - h3 (265.03 - 124.06) kJ/kg PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-100 14-143 Air is heated and dehumidified in an air-conditioning system consisting of a heating section and an evaporative cooler. The temperature and relative humidity of the air when it leaves the heating section, the rate of heat transfer in the heating section, and the rate of water added to the air in the evaporative cooler are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) Assuming the wet-bulb temperature of the air remains constant during the evaporative cooling process, the properties of air at various states are determined from the psychrometric chart (Fig. A-31) to be h1 = 23.5 kJ/kg dry air T1 = 10C = 0.00532 kg/ H 2 O/kg dry air 1 = 70% 1 v 1 = 0.810 m 3 / kg Water Heating coils 10C 70% 30 m3/min 1 1 atm AIR 2 T2 3 20C 60% 2 = 1 Twb2 = Twb3 T2 = 28.3C 2 = 22.3% h h = 42.3 kJ / kg dry air 2 3 h = 42.3 kJ/kg dry air T3 = 20C 3 = 0.00875 kg/ H 2O/kg dry air 3 = 60% 3 Twb3 = 15.1C (b) The mass flow rate of dry air is & ma = V&1 30 m 3 / min = = 37.0 kg/min v 1 0.810 m 3 / kg dry air Then the rate of heat transfer to air in the heating section becomes & & Qin = m a (h2 - h1 ) = (37.0 kg/min)(42.3 - 23.5)kJ/kg = 696 kJ/min (c) The rate of water added to the air in evaporative cooler is & & & & mw, added = mw3 - mw2 = ma (3 - 2 ) = (37.0 kg/min)(0.00875 - 0.00532) = 0.127 kg/min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-101 14-144 EES Problem 14-143 is reconsidered. The effect of total pressure in the range 94 to 104 kPa on the results required in the problem is to be studied. Analysis The problem is solved using EES, and the solution is given below. P=101.325 [kPa] Tdb[1] =10 [C] Rh[1] = 0.70 Vol_dot[1]= 50 [m^3/min] Tdb[3] = 20 [C] Rh[3] = 0.60 P[1]=P P[2]=P[1] P[3]=P[1] "Energy balance for the steady-flow heating process 1 to 2:" "We neglect the PE of the flow. Since we don't know the cross sectional area of the flow streams, we also neglect theKE of the flow." E_dot_in - E_dot_out = DELTAE_dot_sys DELTAE_dot_sys = 0 [kJ/min] E_dot_in = m_dot_a*h[1]+Q_dot_in E_dot_out = m_dot_a*h[2] "Conservation of mass of dry air during mixing: m_dot_a = constant" m_dot_a = Vol_dot[1]/v[1] "Conservation of mass of water vapor during the heating process:" m_dot_a*w[1] = m_dot_a*w[2] "Conservation of mass of water vapor during the evaporative cooler process:" m_dot_a*w[2]+m_dot_w = m_dot_a*w[3] "During the evaporative cooler process:" Twb[2] = Twb[3] Twb[3] =WETBULB(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) {h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],B=Twb[2])} h[2]=h[3] Tdb[2]=TEMPERATURE(AirH2O,h=h[2],P=P[2],w=w[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) h[3]=ENTHALPY(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) w[3]=HUMRAT(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) mw [kg/min] 0.2112 0.2112 0.2111 0.2111 0.211 0.2109 Qin [kJ/min] 1119 1131 1143 1155 1168 1180 Rh2 0.212 0.2144 0.2167 0.219 0.2212 0.2233 Tdb2 [C] 29.2 29 28.82 28.64 28.47 28.3 P [kPa] 94 96 98 100 102 104 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-102 0.224 0.222 0.22 Rh[2] 0.218 0.216 0.214 0.212 94 96 98 100 102 104 P [kPa] 29.2 29.1 29 Tdb[2] [C] 28.9 28.8 28.7 28.6 28.5 28.4 28.3 94 96 98 100 102 104 P [kPa] 1180 1170 0.2113 Qin [kJ/min] 1150 0.2111 1140 1130 1120 1110 94 0.2109 104 0.211 96 98 100 102 P [kPa] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. mw [kg/min] 1160 0.2112 14-103 14-145 Air is heated and dehumidified in an air-conditioning system consisting of a heating section and an evaporative cooler. The temperature and relative humidity of the air when it leaves the heating section, the rate of heat transfer in the heating section, and the rate at which water is added to the air in the evaporative cooler are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) Assuming the wet-bulb temperature of the air remains constant during the evaporative cooling process, the properties of air at various states are determined to be Water Pv1 = 1Pg1 = 1Psat @ 10C = (0.70)(1.2281 kPa) = 0.86 kPa Heating coils Pa1 = P - Pv1 = 96 - 0.86 = 95.14 kPa 1 v1 = RaT1 (0.287 kPa m 3 / kg K)(283 K) = Pa1 95.14 kPa = 0.854 m 3 / kg dry air 10C 70% 30 m3/min 96 kPa AIR 2 T2 3 20C 60% 1 0.622 Pv1 0.622(0.86 kPa) 1 = = = 0.00562 kg H 2O/kg dry air P - Pv1 (96 - 0.86) kPa 1 and Pv 3 = 3 Pg 3 = 3 Psat @ 20C = (0.60)(2.3392 kPa) = 1.40 kPa Pa 3 = P3 - Pv3 = 96 - 1.40 = 94.60 kPa h1 = c pT1 + 1hg1 = (1.005 kJ/kg C)(10C) + (0.00562)(2519.2 kJ/kg) = 24.21 kJ/kg dry air 3 = 0.622 Pv3 0.622(1.40 kPa) = = 0.00923 kg H 2 O/kg dry air (96 - 1.40) kPa P3 - Pv 3 h3 = c p T3 + 3 hg 3 = (1.005 kJ/kg C)(20C) + (0.00921)(2537.4 kJ/kg) = 43.52 kJ/kg dry air Also, h2 h3 = 43.52 kJ/kg 2 = 1 = 0.00562 kg H 2 O/kg dry air Thus, h2 = c pT2 + 2 hg 2 c pT2 + 2 (2500.9 + 1.82T2 ) = (1.005 kJ/kg C)T2 + (0.00562)(2500.9 + 1.82T2 ) Solving for T2, T2 = 29.0C Pg 2 = Psat@29C = 4.013 kPa Thus, 2 = 2 P2 (0.00562)(96) = = 0.214 or 21.4% (0.622 + 2 ) Pg 2 (0.622 + 0.00562)(4.013) (b) The mass flow rate of dry air is V& 30 m 3 / min & ma = 1 = = 35.1 kg/min v 1 0.854 m 3 / kg dry air Then the rate of heat transfer to air in the heating section becomes & & Q = m (h - h ) = (35.1 kg/min)(43.52 - 24.21)kJ/kg = 679 kJ/min in a 2 1 (c) The rate of water addition to the air in evaporative cooler is & & & & mw, added = mw3 - mw 2 = ma (3 - 2 ) = (35.1 kg/min)(0.00923 - 0.00562) = 0.127 kg/min PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-104 14-146 [Also solved by EES on enclosed CD] Waste heat from the cooling water is rejected to air in a natural-draft cooling tower. The mass flow rate of the cooling water, the volume flow rate of air, and the mass flow rate of the required makeup water 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 cooling tower is adiabatic. & & & Analysis (a) The mass flow rate of dry air through the tower remains constant (ma1 = ma 2 = ma ) , but 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 is made up later in the cycle using water at 27C. Applying the mass balance and the energy balance equations yields Dry Air Mass Balance: & & ma ,i = ma ,e & & & ma1 = ma 2 = ma Water Mass Balance: & & m w , i = m w, e & & & & m3 + m a11 = m 4 + m a 2 2 & & & & m3 - m 4 = m a ( 2 - 1 ) = m makeup Energy Balance: & & & Ein - Eout = Esystem 0 (steady) & & =0 Ein = Eout & & & & mi hi = me he (since Q = W = 0) & & 0 = me he - mi hi & a 2 h2 + m4 h4 - ma1h1 - m3h3 & & & 0=m & & & & 0 = ma (h2 - h1 ) + (m3 - mmakeup )h4 - m3h3 & Solving for ma , & ma = & m3 (h3 - h4 ) (h2 - h1 ) - ( 2 - 1 )h4 WARM WATER 27C 4 COOL WATER 42C AIR EXIT 2 37C saturated 3 From the psychrometric chart (Fig. A-31), h1 = 50.8 kJ/kg dry air 1 AIR INLET Tdb = 23C Twb = 18C 1 = 0.0109 kg H 2O/kg dry air v1 = 0.854 m3/kg dry air and h2 = 143.0 kJ / kg dry air 2 = 0.0412 kg H 2 O / kg dry air From Table A-4, h3 h f @ 42C = 175.90 kJ/kg H 2 O h4 h f @ 27C = 113.19 kJ/kg H 2 O Substituting & ma = & m3 (175.90 - 113.19)kJ/kg & = 0.706 m 3 (143.0 - 50.8) kJ/kg - (0.0412 - 0.0109)(113.25) kJ/kg PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-105 The mass flow rate of the cooling water is determined by applying the steady flow energy balance equation on the cooling water, & & & & & & & Qwaste = m3h3 - ( m3 - mmakeup )h4 = m3h3 - [m3 - ma ( 2 - 1 )]h4 & & & = m3h3 - m3[1 - 0.706( 0.0412 - 0.0109 )]h4 = m3 ( h3 - 0.9786h4 ) & & 50,000 kJ/s = m3 (175.90 - 0.9786 113.19) kJ/kg m3 = 768.1 kg/s and & & ma = 0.706m3 = (0.706)(7681 kg / s) = 542.3 kg / s . (b) Then the volume flow rate of air into the cooling tower becomes & V&1 = m av 1 = (542.3 kg/s)(0.854 m 3 / kg ) = 463.1 m 3 /s (c) The mass flow rate of the required makeup water is determined from & & mmakeup = ma ( 2 - 1 ) = (542.3 kg / s)(0.0412 - 0.0109) = 16.4 kg / s PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-106 14-147 EES Problem 14-146 is reconsidered. The effect of air inlet wet-bulb temperature on the required air volume flow rate and the makeup water flow rate is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P_atm =101.325 [kPa] T_db_1 = 23 [C] T_wb_1 = 18 [C] T_db_2 = 37 [C] RH_2 = 100/100 "%. relative humidity at state 2, saturated condition" Q_dot_waste = 50 [MW]*Convert(MW, kW) T_cw_3 = 42 [C] "Cooling water temperature at state 3" T_cw_4 = 27 [C] "Cooling water temperature at state 4" "Dry air mass flow rates:" "RH_1 is the relative humidity at state 1 on a decimal basis" v_1=VOLUME(AirH2O,T=T_db_1,P=P_atm,R=RH_1) T_wb_1 = WETBULB(AirH2O,T=T_db_1,P=P_atm,R=RH_1) m_dot_a_1 = Vol_dot_1/v_1 "Conservaton of mass for the dry air (ma) in the SSSF mixing device:" m_dot_a_in - m_dot_a_out = DELTAm_dot_a_cv m_dot_a_in = m_dot_a_1 m_dot_a_out = m_dot_a_2 DELTAm_dot_a_cv = 0 "Steady flow requirement" "Conservation of mass for the water vapor (mv) and cooling water for the SSSF process:" m_dot_w_in - m_dot_w_out = DELTAm_dot_w_cv m_dot_w_in = m_dot_v_1 + m_dot_cw_3 m_dot_w_out = m_dot_v_2+m_dot_cw_4 DELTAm_dot_w_cv = 0 "Steady flow requirement" w_1=HUMRAT(AirH2O,T=T_db_1,P=P_atm,R=RH_1) m_dot_v_1 = m_dot_a_1*w_1 w_2=HUMRAT(AirH2O,T=T_db_2,P=P_atm,R=RH_2) m_dot_v_2 = m_dot_a_2*w_2 "Conservation of energy for the SSSF cooling tower process:" "The process is adiabatic and has no work done, ngelect ke and pe" E_dot_in_tower - E_dot_out_tower = DELTAE_dot_tower_cv E_dot_in_tower= m_dot_a_1 *h[1] + m_dot_cw_3*h_w[3] E_dot_out_tower = m_dot_a_2*h[2] + m_dot_cw_4*h_w[4] DELTAE_dot_tower_cv = 0 "Steady flow requirement" h[1]=ENTHALPY(AirH2O,T=T_db_1,P=P_atm,w=w_1) h[2]=ENTHALPY(AirH2O,T=T_db_2,P=P_atm,w=w_2) h_w[3]=ENTHALPY(steam,T=T_cw_3,x=0) h_w[4]=ENTHALPY(steam,T=T_cw_4,x=0) "Energy balance on the external heater determines the cooling water flow rate:" E_dot_in_heater - E_dot_out_heater = DELTAE_dot_heater_cv E_dot_in_heater = Q_dot_waste + m_dot_cw_4*h_w[4] E_dot_out_heater = m_dot_cw_3 * h_w[3] DELTAE_dot_heater_cv = 0 "Steady flow requirement" PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-107 "Conservation of mass on the external heater gives the makeup water flow rate." "Note: The makeup water flow rate equals the amount of water vaporized in the cooling tower." m_dot_cw_in - m_dot_cw_out = DELTAm_dot_cw_cv m_dot_cw_in = m_dot_cw_4 + m_dot_makeup m_dot_cw_out = m_dot_cw_3 DELTAm_dot_cw_cv = 0 "Steady flow requirement" Vol1 [m3/s] 408.3 420.1 433.2 447.5 463.4 481.2 501.1 523.7 549.3 578.7 mmakeup [kgw/s] 16.8 16.72 16.63 16.54 16.43 16.31 16.18 16.03 15.87 15.67 mcw3 [kgw/s] 766.6 766.7 766.8 767 767.2 767.4 767.7 767.9 768.2 768.6 ma1 [kga/s] 481.9 495 509.4 525.3 542.9 562.6 584.7 609.7 638.1 670.7 Twb1 [C] 14 15 16 17 18 19 20 21 22 23 16.8 560 520 480 440 400 14.0 16.4 16.2 16.0 15.8 15.6 16.0 18.0 20.0 22.0 Twb,1 [C] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. mmakeup [kgw/s] 16.6 Vol1 [m^3/s] 14-108 14-148 Atmospheric air enters an air-conditioning system at a specified pressure, temperature, and relative humidity. The heat transfer, the rate of condensation of water, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during & & & the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at the inlet and exit states may be determined from the psychrometric chart (Figure A31) or using EES psychrometric functions to be (we used EES) R-134a h1 = 78.24 kJ/kg dry air 350 kPa 350 kPa 1 = 0.01880 kg H 2O/kg dry air x = 1.0 x = 0.20 v1 = 0.8847 m3 / kg dry air h2 = 27.45 kJ/kg dry air Cooling coils T2 =20C 2 = 20% 2 20C Condensate removal T1 =30C 1 = 70% 4 m3/min 1 2 = 0.002885 kg H 2O/kg dry air The mass flow rate of dry air is & ma = 1 atm AIR Condensate V&1 4 m3/min = = 4.521 kg/min v1 0.8847 m3 The mass flow rates of vapor at the inlet and exit are & & m v1 = 1 m a = (0.01880)(4.521 kg/min) = 0.0850 kg/min & & m v 2 = 2 m a = (0.002885)(4.521 kg/min) = 0.01304 kg/min An energy balance on the control volume gives & & & & m a h1 = Qout + m a h2 + m w h w2 where the the enthalpy of condensate water is h w 2 = h f@ 20C = 83.91 kJ/kg (Table A - 4) and the rate of condensation of water vapor is & & & m w = m v1 - m v 2 = 0.0850 - 0.01304 = 0.07196 kg/min Substituting, & & & & m a h1 = Qout + m a h2 + m w hw 2 & (4.521 kg/min)(78.24 kJ/kg) = Qout + (4.521 kg/min)(27.45 kJ/kg) + (0.07196 kg/min)(83.91 kJ/kg) & Q = 223.6 kJ/min = 3.727 kW out The properties of the R-134a at the inlet and exit of the cooling section are PR1 = 350 kPa h R1 = 97.56 kJ/kg x R1 = 0.20 PR 2 = 350 kPa h R 2 = 253.34 kJ/kg x R 2 = 1.0 Noting that the rate of heat lost from the air is received by the refrigerant, the mass flow rate of the refrigerant is determined from PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-109 & & & m R h R1 + Qin = m R h R 2 & Qin 223.6 kJ/min & = = 1.435 kg/min mR = h R 2 - h R1 (253.34 - 97.56) kJ/kg 0,050 Pressure = 101.3 [kPa] 0,045 0,040 AirH2O 35C 0,035 0.8 Humidity Ratio 0,030 30C 0,025 0,020 0,015 0,010 10C 15C 20C 25C 0.6 1 0.4 0.2 0,005 0,000 0 2 5 10 15 20 25 30 35 40 T [C] PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-110 14-149 An uninsulated tank contains moist air at a specified state. Water is sprayed into the tank until the relative humidity in the tank reaches a certain value. The amount of water supplied to the tank, the final pressure in the tank, and the heat transfer during the process are to be determined. Assumptions 1 Dry air and water vapor are ideal gases. 2 The kinetic and potential energy changes are negligible. Analysis The initial state of the moist air is completely specified. The properties of the air at the inlet state may be determined from the psychrometric chart (Figure A-31) or using EES psychrometric functions to be (we used EES) 1 = 0.005433 kg H 2 O/kg dry air h1 = 49.16 kJ/kg dry air, v 1 = 0.6863 m 3 / kg dry air The initial mass in the tank is V 0.5 m 3 ma = 1 = = 0.7285 kg v 1 0.6863 m 3 The partial pressure of dry air in the tank is = 128.8 kPa (0.5 m 3 ) Then, the pressure of moist air in the tank is determined from P2 = Pa 2 1 + 2 = (128.8 kPa)1 + 2 0.622 0.622 We cannot fix the final state explicitly by a hand-solution. However, using EES which has built-in functions for moist air properties, the final state properties are determined to be 2 = 0.02446 kg H 2O/kg dry air P2 = 133.87 kPa Pa 2 = ma Ra T2 V = (0.7285 kg)(0.287 kJ/kg.K)(35 + 273 K) h2 = 97.97 kJ/kg dry air v 2 = 0.6867 m3 / kg dry air The partial pressures at the initial and final states are Pv1 = 1 Psat@35C = 0.20(5.6291 kPa) = 1.126 kPa Pa1 = P 1 - Pv1 = 130 - 1.126 = 128.87 kPa Pv 2 = P 2 - Pa 2 = 133.87 - 128.81 = 5.07 kPa The specific volume of water at 35C is v w1 = v w2 = v g @35C = 25.205 m 3 /kg The internal energies per unit mass of dry air in the tank are u1 = h1 - Pa1v 1 - w1 Pv1v w1 = 49.16 - 128.87 0.6863 - 0.005433 1.126 25.205 = -39.44 kJ/kg u 2 = h2 - Pa 2v 2 - w 2 Pv 2v w 2 = 97.97 - 128.81 0.6867 - 0.02446 5.07 25.205 = 6.396 kJ/kg The enthalpy of water entering the tank from the supply line is h w1 = hf @50C = 209.34 kJ/kg The internal energy of water vapor at the final state is u w 2 = u g @35C = 2422.7 kJ/kg The amount of water supplied to the tank is m w = m a ( 2 - 1 ) = (0.7285 kg)(0.02446 - 0.005433) = 0.01386 kg An energy balance on the system gives E in = E tank Qin + (0.01386 kg )(209.34 kJ/kg) = (0.7285 kg)[6.396 - (-39.44)kJ/kg ] + (0.01386 kg)(2422.7 kJ/kg) & Q = 64.1 kJ in Qin + m w h w1 = m a (u 2 - u1 ) + m w u w2 PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-111 Fundamentals of Engineering (FE) Exam Problems 14-150 A room is filled with saturated moist air at 25C 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 Answer (d) 2.04 kg (b) 1.97 kg (c) 2.96 kg (d) 2.04 kg (e) 3.17 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" P=100 "kPa" m_air=100 "kg" RH=1 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v w=0.622*P_v/(P-P_v) w=m_v/m_air "Some Wrong Solutions with Common Mistakes:" W1_vmass=m_air*w1; w1=0.622*P_v/P "Using P instead of P-Pv in w relation" W2_vmass=m_air "Taking m_vapor = m_air" W3_vmass=P_v/P*m_air "Using wrong relation" 14-151 A room contains 50 kg of dry air and 0.6 kg of water vapor at 25C and 95 kPa total pressure. The relative humidity of air in the room is (a) 1.2% Answer (c) 56.7% (b) 18.4% (c) 56.7% (d) 65.2% (e) 78.0% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" P=95 "kPa" m_air=50 "kg" m_v=0.6 "kg" w=0.622*P_v/(P-P_v) w=m_v/m_air P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g "Some Wrong Solutions with Common Mistakes:" W1_RH=m_v/(m_air+m_v) "Using wrong relation" W2_RH=P_g/P "Using wrong relation" PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-112 14-152 A 40-m3 room contains air at 30C 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 Answer (c) 39.9 kg (b) 29.9 kg (c) 39.9 kg (d) 41.4 kg (e) 52.3 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=40 "m^3" T1=30 "C" P=90 "kPa" RH=0.75 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v R_air=0.287 "kJ/kg.K" m_air=P_air*V/(R_air*(T1+273)) "Some Wrong Solutions with Common Mistakes:" W1_mass=P_air*V/(R_air*T1) "Using C instead of K" W2_mass=P*V/(R_air*(T1+273)) "Using P instead of P_air" W3_mass=m_air*RH "Using wrong relation" 14-153 A room contains air at 30C 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 Answer (c) 92.8 kPa (b) 85.8 kPa (c) 92.8 kPa (d) 90.6 kPa (e) 72.0 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=30 "C" P=96 "kPa" RH=0.75 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v "Some Wrong Solutions with Common Mistakes:" W1_Pair=P_v "Using Pv as P_air" W2_Pair=P-P_g "Using wrong relation" W3_Pair=RH*P "Using wrong relation" PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-113 14-154 The air in a house is at 20C 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) 8.7C Answer (d) 9.3C (b) 11.3C (c) 13.8C (d) 9.3C (e) 10.0C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=20 "C" RH1=0.50 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH1=P_v/P_g T_dp=TEMPERATURE(Steam_IAPWS,x=0,P=P_v) "Some Wrong Solutions with Common Mistakes:" W1_Tdp=T1*RH1 "Using wrong relation" W2_Tdp=(T1+273)*RH1-273 "Using wrong relation" W3_Tdp=WETBULB(AirH2O,T=T1,P=P1,R=RH1); P1=100 "Using wet-bulb temperature" 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) Answer (e) diagonal downwards to the left (SW direction) 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) Answer (c) diagonal upwards to the right (NE direction) PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-114 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 Answer (a) the dry bulb temperature at the given state 14-158 Air is cooled and dehumidified as it flows over the coils of a refrigeration system at 85 kPa from 30C and a humidity ratio of 0.023 kg/kg dry air to 15C and a humidity 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 Answer (e) 25 kJ/s (b) 10 kJ/s (c) 15 kJ/s (d) 20 kJ/s (e) 25 kJ/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P=85 "kPa" T1=30 "C" w1=0.023 T2=15 "C" w2=0.015 m_air=0.7 "kg/s" m_water=m_air*(w1-w2) h1=ENTHALPY(AirH2O,T=T1,P=P,w=w1) h2=ENTHALPY(AirH2O,T=T2,P=P,w=w2) h_w=ENTHALPY(Steam_IAPWS,T=T2,x=0) Q=m_air*(h1-h2)-m_water*h_w "Some Wrong Solutions with Common Mistakes:" W1_Q=m_air*(h1-h2) "Ignoring condensed water" W2_Q=m_air*Cp_air*(T1-T2)-m_water*h_w; Cp_air = 1.005 "Using dry air enthalpies" W3_Q=m_air*(h1-h2)+m_water*h_w "Using wrong sign" PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 14-115 14-159 Air at a total pressure of 90 kPa, 15C, and 75 percent relative humidity is heated and humidified to 25C 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 Answer (a) 0.032 kg/s (b) 0.013 kg/s (c) 0.019 kg/s (d) 0.0079 kg/s (e) 0 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P=90 "kPa" T1=15 "C" RH1=0.75 T2=25 "C" RH2=0.75 m_air=4 "kg/s" w1=HUMRAT(AirH2O,T=T1,P=P,R=RH1) w2=HUMRAT(AirH2O,T=T2,P=P,R=RH2) m_water=m_air*(w2-w1) "Some Wrong Solutions with Common Mistakes:" W1_mv=0 "sine RH = constant" W2_mv=w2-w1 "Ignoring mass flow rate of air" W3_mv=RH1*m_air "Using wrong relation" 14-160 14-164 Design and Essay Problems PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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