S.A. Klein and G.F. Nellis Cambridge University Press, 2011

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1 12.A-1 A mixture of helium and water vapor is flowing through a pipe at T= 90 C and P = 150 kpa. The mole fraction of helium is y He = a.) What is the relative humidity of the mixture? b.) What is the humidity ratio of the mixture? c.) What is the dew point of the mixture?

2 12.A-2 Figure 12.A-2 shows a humidifier that takes in an air-water vapor mixture with 10% relative humidity, 1 = 0.1, at temperature T 1 = 18 C and pressure P 1 = 1 atm and mixes it with a source of steam at P 2 = 50 psia in order to generate an air-water mixture that has 70% relative humidity, = 0.7, with temperature T = 20 C and pressure P = 1 atm. The mass flow rate of the inlet air-water mixture is m 1 = 0.25 kg/s. steam at P2 50psia m 1 025kg/s. T1 18 C P1 1atm humidifier Figure 12.A-2: Humidifier. T =20 C P =1atm =0.7 The humidifier is at steady-state and is externally adiabatic. a.) Determine the humidity ratio of the air-water vapor mixture entering the humidifier at state 1 ( 1 ) and the humidity ratio at state ( ). b.) Determine the mass flow rates of air and the mass flow rate of water vapor entering the humidifier at state 1 and leaving at state. c.) Determine the mass flow rate of steam entering the humidifier at state 2. d.) What is the temperature of the steam entering the humidifier at state 2?

3 12.A- Figure 12.A- shows an air compressor system that is used in a factory. At state 1, the compressor draws in ambient air (which is actually an air/water vapor mixture) with T o = 20ºC and P o = 1 atm. The relative humidity of the ambient air/water mixture is 1 = 0.5. The mass flow rate of the incoming air/water mixture is m 1 = 0.1 kg/s. The rate of heat transfer from the compressor to the atmosphere is Q c = 50 kw. The air/water mixture leaves the compressor at T 2 = 175ºC and P 2 = 90 psig. The hot air/water mixture is cooled in an aftercooler that rejects heat to ambient. The air/water mixture leaving the aftercooler is at T = T o and P = P 2. The aftercooler also serves as a filter and drier. Condensate (i.e., liquid water condensed out of the air) leaves the aftercooler at state 4, T 4 = T o and P 4 = P o. aftercooler Q ac W c compressor T 2 = 175 C P 2 =90psig 1 2 Q 50 kw T1 T 20 C o P1 P 1atm o m 1 01kg/s. Figure 12.A-: Air compressor with aftercooler. c 4 condensate T 4 = T o P 4 = P o T = T o P = P 2 a.) Determine the humidity ratio of the air at state 1. What is the mass flow rate rate of dry air and water vapor entering the compressor ( m a,1 and m v,1 )? b.) Assume that no condensation of water occurs in the compressor. Determine the relative humidity and humidity ratio at state 2. Is the assumption correct? How do you know? c.) Determine the power consumed by the compressor ( W c ) and the rate of entropy generation within the compressor ( S gen, c ). d.) Determine the humidity ratio of the air at state. e.) What is the mass flow rate rate of dry air and water vapor leaving the aftercooler ( m a, and m v, )? What is the mass flow rate of condensate leaving the aftercooler ( m 4 )? What is volumetric flow rate of condensate (in gallons/day)? f.) What is the rate of heat transfer from the aftercooler to the environment ( Q ac )? g.) Use EES to draw a T-s diagram for the states of the water in the problem (i.e., the water vapor at states 1, 2, and and the water leaving at state 4). Label the states.

4 12.A-4 During the winter, indoor air with relative humidity 1 = 0.7, temperature T 1 = 22 C and volumetric flow rate V 1 = 0.04 m /s is used to pre-heat air drawn from the outdoors at = 0.2, temperature T = 10 C, and volumetric flow rate V = 0.06 m /s in a heat exchanger as shown in Figure 12.A-4. The outdoor air leaving the heat exchanger is heated to T 4 = 20 C. The process occurs at atmospheric pressure and the heat exchanger is externally adiabatic. Use the Psychrometric Chart to solve this problem. indoor air V m /s heat exchanger T1 22 C (1) (2) Q (4) () outdoor air T 4 =20 C V 0.06 m /s T 10 C 0.2 Figure 12.A-4: Heat exchanger. a.) What is the relative humidity of the air leaving the heat exchanger at state (4)? b.) What is the rate of heat transfer between the air streams in the heat exchanger, Q? c.) Will there be any condensation as the indoor air flows through the heat exchanger? Justify your answer. d.) Assume that your answer to (c) is yes. At what temperature will the indoor air begin to condense? e.) What is the exit temperature of the indoor air (T 2 )? What is the mass flow rate of the condensate from the indoor air? You may neglect the enthalpy carried by the condensate to answer this question. f.) Locate and label all four states on the Psychrometric chart.

5 12.A-5 Two flows of air/water vapor mixtures enter a steady flow device, as shown in Figure 12.A-5. Flow 2 V 2 00m. /s T2 10 C 2 0. Flow 1 V 1 001m. /s T1 25 C steady flow device Flow Q 100W Figure 12.A-5: Steady flow device. Flow 1 enters the device with temperature T 1 = 25 C and relative humidity 1 = 0.8. The volumetric flow rate at state 1 is V 1 = 0.01 m /s. Flow 2 enters the device with temperature T 2 = 10 C and relative humidity 2 = 0.. The volumetric flow rate at state 2 is V 2 = 0.0 m /s. Heat is added to the device at a rate of Q = 100 W. A single flow of air/water vapor mixture leaves at state and there is no condensation in the device. The pressure of each stream (entering and leaving) is P = 1 atm. Use the psychrometric chart to determine the temperature and relative humidity of the air leaving the device. Indicate each of the states on the chart.

6 12.A-6 A flow of 400 cfm of air at 80 F, 14.7 psia is cooled to a saturated state at 50 F. Calculate and plot the following quantities as a function of the relative humidity of air for relative humidities between 0.5 and a.) humidity ratio of the stream b.) the rate of moisture removal in lb m /hr c.) the minimum energy required to cool the air in Btu/hr

7 12.A-7 An apparatus that can supply air at a controlled dry-bulb temperature and relative humidity has been designed as indicated in the schematic in Figure 12.A-7. Dry air is supplied at locations 1 and 2. One stream is isothermally humidified to saturation (state ) and then mixed with the other stream to produce the supply air at location 4. The entire operation occurs isothermally with a temperature-regulated enclosure. The dry air flow rates at locations 1 and 2 are adjusted in proportion to the desired relative humidity at location 4. In the case at hand, 50% relative humidity is desired so the dry air mass flow rates at location 1 and 2 are equal. Table 12.A-7: Measurements T Constant Temperature Enclosure Dry air Humidifer Dry air Figure 12.A-7: Apparatus for supplying controlled humidity air Relative Humidity C Table 12.A-7 shows the measured conditions at state 4 for several temperatures. For temperatures below 60 C, the apparatus functions well in achieving the desired 50% relative humidity. However, significant errors occur for temperatures above 60 C. The technicians have checked the equipment and calibrated all measuring devices, but cannot find a problem. Can you help?

8 12.A.8 In an HVAC (heating, ventilating, and air conditioning) system, atmospheric air at a volumetric flow rate of 0.5 m /s at 25 C and 60% relative humidity is dehumidified and cooled with a water spray in a counter-current arrangement. The water is supplied to the spray head at 0.5 kg/s and 5 C. The dry-bulb temperature, wet-bulb temperature, and relative humidity of the air exiting the spray are to be determined. Heat and mass transfer theory indicate the process may be described in terms of an effectiveness factor defined by hair, in hair, out air, in air, out h h air, in air, out, sat ' d at Twin, air, in air, out, sat ' d at Twin, where h air, in and h air, out are the specific enthalpies of the entering end exiting air-water vapor mixtures per unit masss of dry air, air, in and air, out are the humidity ratios of entering and exiting air, and hair, out, sat ' d at T and win, air, out, sat ' d at T are the specific enthalpy win, and humidity of saturated air at the water inlet temperature. Prepare a plot of the drybulb and wet-bulb temperatures and the relative humidity as a function of the effectivenss factor for values between 0 and 1.

9 12.A-9 A building space having a volume of 140 m is to be maintained at 25 C, 50% relative humidity on a winter day. Ventilation air is provided at 0.02 m /s, 5 C, and 0% relative humidity. Heat losses from the building space to outdoors occur at a rate of 1.5 kw. A humidifier provides the necessary moisture to the building space in the form of saturated vapor at 100 C. Space heaters within the building space maintain the required room temperature. Assuming the air in the room to be fully mixed, determine: a.) the rate at which moisture must be added b.) the rate at which heating must be provided in the building space c.) the mass of water vapor in the room at any time d.) the temperature of the walls in the room that would produce condensate

10 12.A-10 Ambient air at 85 F and 75 F wetbulb enters a cooling tower at a volumetric flowrate of 54,440 cfm. The air exits the cooling tower at 91 F. The relative humidity of the air exiting the cooling tower is difficult to measure, but it is believed to be somewhere between 90% and saturated. Condenser water from a chiller plant enters the cooling tower at 10 F at a flowrate of 2,848 lb m /min. a.) What is the relative humidity of the ambient air? b.) What is the inlet air flow rate? c.) Calculate and plot the water outlet temperature and make-up water required as a function of the relative humidity exiting the cooling tower for relative humidities between 0.9 and 1.

11 12.A-11 Two tanks, A and B, are connected by piping and a closed valve. Tank A contains 4 kg of butane (C 8 H 18 ) at.5 bar and 175 C. Tank B contains 1.4 kg of water vapor at 0.5 bar and 95 C. The valve is opened and after some time, the contents of the tanks completely mix. A temperature measurement indicates that the mixture is at 126 C. a.) What is the final pressure of the gas mixture? b.) What is the humidity ratio of the gas mixture? c.) Determine the value of the relative humidity of the final gas mixture.

12 12.A-12 A manufacturing plant requires a supply of 100 cfm of air at 70 F, 20% relative humidity and 1 atm for purging equipment. One way to obtain this relatively dry air is to compressed atmospheric air which is available at 70 F dry-bulb, 66 F wet-bulb at 1 atm to a suitable pressure. The compressed air is cooled to 70 F in a heat exchanger with cooling water and then throttled to 1 atm as shown in Figure 12.A-12. The compressor operates adiabatically with an isentropic efficiency of Figure 12.A-12: System for drying air a.) Determine the necessary pressure at the compressor outlet (state 2). b.) Determine the power required to operate the compressor at steady conditions. c.) Determine the rate at which condensate must be removed.

13 12.A-1 A rigid tank with volume V = 1 m initially contains pure refrigerant R14a vapor at P 1 = 100 kpa and T 1 = T amb = 20ºC. The tank is connected to a source of high pressure nitrogen at P s = 500 kpa and T s = 200ºC, as shown in Figure 12.A-1. pure nitrogen P s =500kPa T s =200 C tank initially filled with pure R14a P 1 = 100 kpa T 1 = 20 C V =1m Figure 12.A-1: Tank containing refrigerant vapor connected to a source of nitrogen. The valve is opened, allowing pure nitrogen to flow into the tank. At the conclusion of the flow process, the pressure in the tank is P 2 = P s and the temperature in the tank is T 2 = 100ºC. Model both the nitrogen and the refrigerant as an ideal gas and assume that they form an ideal gas mixture. The properties of the refrigerant should be evaluated using R r = 81.5 J/kg-K, c P,r = 840 J/kg-K, and c v,r = J/kg-K. The properties of nitrogen should be evaluated using R N 2 = 297 J/kg-K, c P, N = 1040 J/kg-K, and c 2 vn, = 74 J/kg- 2 K. a.) Determine the amount of heat transferred from the tank to the surroundings at T amb during the filling process. b.) Determine the entropy generated by the filling process. At the conclusion of the filling process, the valve is closed and the temperature of the tank is gradually reduced. c.) At what temperature will R14a begin to condense? Note that your text includes tables for the properties of saturated R14a.

14 12.A-14 A closed rigid tank having a volume of m contains air at 100 C, 4.4 bar, and 40% relative humidity. The tank contents cool to 20 C as a result of heat transfer with the surroundings. The gas mixture may be assumed to behave as an ideal gas. a.) The mass of water in the tank is kg. b.) The temperature at which condensation begins is C c.). The mass of liquid water in the tank at 20 C is kg.

15 12.B-1 In an air conditioning system that operates at steady-state, return air at V 1 = 5 m /min, T 1 = 22 C, 54% relative humidity is mixed with outdoor ventilation air at V 2 = 12 m /min, T 2 = 0 C, 2 = 55% relative humidity, as shown in Figure 12.B-1. The mixed air at state () first passes over a cooling coil. The air-water mixture exits in a saturated condition at state (4). Some water vapor is condensed in the cooling coil and leaves at state (6). The saturated moist air and condensate streams exit the dehumidifier at the same temperature (T 4 = T 6 ). The moist air at state (4) then passes through a heating coil. The air is exhausted to the building at T 5 = 24 C and 5 = 40% relative humidity. The entire process occurs at atmospheric pressure. building return air V 1 5 m /min T1 22 C Q h (2) outdoor air V 2 12 m /min T2 0 C (1) mixer () cooling coil (6) (4) heating coil condensate Q c Figure 12.B-1: Air conditioning system. (5) air to building T5 24 C a.) Determine the temperature and relative humidity of the air at state (). b.) Determine the temperature of the moist air exiting the cooling coil at state (4). c.) Determine the rate at which cooling must be provided in the cooling coil in kw and in tons. Also determine the mass flow rate of condensate leaving the cooling coil. d.) Plot the cooling required as a function of the outdoor air relative humidity.

16 12.B-2 One type of residential humidifier is shown in Figure 12.B-2. The humidifier is directly connected to the furnace. In a particular case, air from the building at T 1 = 20 C and 0% relative humidity (state 1) is heated in the furnace to T 2 = 95 C. A fraction of furnace outlet air (f) passes through a humidifier; it exits as saturated air at T = 50 C. The humidified air mixes with the air from the furnace and the mixture reenters the building at state (4). The entire process occurs at atmospheric pressure. water T 2 = 95 C T = 50 C, =1 f humidifier (2) () furnace mixer (1-f) (2) (4) air from building T 1 = 20 C, 1 = 0. Figure 12.B-2: Humidifier. a.) What is the humidity ratio of the air in the building at state 1? b.) What is the relative humidity of the air at state (2)? c.) What is the humidity ratio of the air at state? d.) What is the ratio of the mass flow rate of liquid water that must be added to the humidifier to the mass flow rate of the air entering the humidifier? e.) Determine the temperature and relative humidity of the air leaving the mixer that is supplied to the building if f = 0.5 (i.e., 50% of the flow passes through the humidifier). f.) Plot the temperature and relative humidity of the air returning to the building as a function of the fraction of the flow diverted through the humidifier, f.

17 12.B- Your cabin has no air conditioning system. Therefore you have installed a swamp cooler, as shown in Figure 12.B-. makeup water V m /s T1 0 C =0.9 () Q 250W swamp cooler cabin () Figure 12.B-: Swamp cooler. The cooler draws in air at a rate of V 1 = 0.05 m /s from outdoors. Your cabin is located in a hot but dry climate; the outdoor air is at T 1 = 0ºC and 1 = 0.2. The air is forced to flow through a chamber that is filled with a porous structure that is saturated with liquid water. The chamber is adiabatic and the air exits from the chamber at 2 = The air/water mixture passing through the chamber is being adiabatically saturated; therefore, it follows a line of constant adiabatic saturation (or wet bulb) temperature. The air leaving the chamber is directed to your cabin. The cabin has no significant moisture gain but does experience heating at a rate of Q = 250 W. Air leaves the cabin at temperature T, the indoor air temperature of the cabin. a.) Determine the humidity ratio of the outdoor air. b.) Determine the mass flow rate of dry air passing through the system. c.) Determine the humidity ratio and temperature of the air leaving the chamber. d.) Determine the volume of makeup water (in gallon) that must be provided to the chamber each day. e.) Determine the temperature and relative humidity of the air leaving the cabin. f.) Sketch the three states on the psychrometric chart - make sure that you label each of the states. g.) If you wanted to keep your cabin at the temperature that you calculated in (e) and the outdoor air temperature remained at 0ºC, then estimate the outdoor air relative humidity at which the swamp cooler would no longer provide any useful cooling to the cabin (assume the air leaving the swamp cooler will always be at 2 = 0.9).

18 12.B-4 Figure 12.B-4 illustrates a system used to humidify building air. saturated steam T 110 C s m kg/s s V 1 010m. /s 1 Building air T b =20 C T wb,b = 10 C V. steam-spray humidifier 015m /s 2 mixer 4 Figure 12.B-4: Steady flow device. Two streams of air are removed from the building space. The building air has temperature (i.e., dry bulb temperature) T b = 20 C and adiabatic saturation temperature (i.e., wet bulb temperature) T wb,b = 10 C. The stream of air at state 1 has volumetric flow rate V 1 = 0.1 m /s and enters a steam-spray humidifier where it is humidified through the injection of a saturated steam flow. The saturated steam is injected at T s = 110 C with mass flow rate m s = kg/s. A separate flow of air is pulled from the building with volumetric flow rate V = 0.15 m /s and mixes with the air leaving the humidifier. The mixed air flow at state 4 re-enters the building. The pressure of each air stream is at ambient pressure. Use the Psychrometric chart to do this problem. a.) Determine the temperature and relative humidity of the air leaving the steam spray humidifier at state 2. b.) Determine the temperature and relative humidity of the air that is provided to the building at state 4.

19 12.B-5 In an air conditioning system that operates at steady-state, return air at 5 m /min, 22 C, 54% relative humidity (state 1) is mixed with outdoor ventilation air at 12 m /min, 0 C, 55% rh (state 2), as shown in Figure 12.B-5. The mixture (state ) air first passes over a cooling coil such that the air exits in a saturated condition (state 4) and some water vapor is condensed (state 6). The saturated moist air and condensate streams exit the dehumidifier at the same temperature. The moist air then passes through a heating unit, exiting at 24 C, 40% rh (state 5). Building return air 5 m /min 22 C, 54% rh 1 heat 2 mixing 4 5 cooling coil heating coil chamber Outdoor air 24 C, 40% rh 12 m /min 6 0 C, 55% rh heat condenstate Figure 12.B-5: Air conditioning system Determine: a) the temperature and relative humidity of the air state b) the temperature of the moist air exiting the cooling coil (state 4) c) the rate at which cooling must be provided in the cooling coil in kw and in tons d) the rate of condensate in kg/min e) the rate of heat transfer to the air passing through the heating unit in kw. Plot all state points on a psychrometric chart.

20 12.B-6 The total (sensible plus latent) air conditioning load for a building is 7. kw when the outdoor conditions are at 5 C dry-bulb, 25 C wet-bulb and the indoor space is maintained at 24 C dry-bulb, 50% relative humidity. For ventilation purposes, outdoor air is mixed with recirculated air in a 1 to 4 proportion. The mixed air is cooled and dehumidified as it flows through a cooling cool. The air exiting the cooling coil is saturated. The conditioned air is then blown into the space to supply the air conditioning load. The cooling coil heat transfer effectiveness, e, is approximately 0.70, defined as T T air, in air, out air, in T T supply where air, in T is the dry-bulb temperature of the mixed air entering the cooling coil, T, is the dry-bulb temperature of the air leaving the cooling coil and T supply is the temperature of the chilled water entering the coil. a.) Determine the temperature and relative humidity of the air entering the cooling coil. b.) Calculate and plot the sensible heat ratio provided by the conditioned air as a function of T supply for T supply values between 2 C and 12 C. c.) Calculate and plot the required volumetric flowrate of conditioned air needed to meet the total load as a function of T supply for T supply values between 2 C and 12 C. air out

21 12.B-7 Supermarkets have unusual cooling needs during the summer. Because the freezer cases are often left uncovered, frost accumulates on the frozen food products making them unsightly. The frost also reduces the efficiency of the frozen food cases. To reduce this problem, the humidity in the store can be lowered. A cooling system designed for this purpose is shown in Figure 12.B-7. Exhaust Control Valve Bypass Return air 6 Supermarket Ventilation Air Coiling Coil Reheat Coil Figure 12.B-7: Supermarket refrigeration system 4000 cfm of air at 62 F, 55% relative humidity is supplied at state 5 to maintain comfort conditions in the supermarket. Air returns from the supermarket at 74 F, 54% relative humidity at state 6. 15% of the return air (mass basis) is exhausted and replaced with outdoor ventilation air at 82 F, 48% relative humidity (state 1). Some of the recirculated air is mixed with the ventilation air (state 2) and passed through the cooling coil from which it emerges at saturated conditions (state ). The rest of the recirculated air bypasses the cooling coil, mixes with the cooling coil outlet air and then enters the reheat coil at state 4. Calculate and plot the following quantities as a function of the fraction of the recirculation air that bypasses the cooling coil. What is the largest value of the bypass that will allow the air at state to remain above the freezing temperature? a.) the required heat transfer rate in the cooling coil b.) the cooling coil outlet air temperature c.) the rate of heat transfer in the reheat coil

22 12.B-8 A cooling tower operates with a flow of condenser water of 50,000 lb m /hr that enters the tower at 95 F. The ambient air used to cool the water is at a dry bulb temperature of 70 F and a relative humidity of 50 %. The air flow rate is 10,000 cfm. The performance of a cooling tower can be estimated in terms of a cooling tower effectiveness defined as: Q m h h Q m h h a air, out air, in max a sat, Twin, air, in where m a is the dry air mass flowrate, m win, is the inlet water flowrate, h air, in and h air, out are the specific enthalpies of the air entering and leaving the cooling tower and h sat, T win is, the specific enthalpy of air assuming it is saturated at the water inlet temperature. Air exits the tower at a relative humidity of 0.98 at all effectiveness values. a.) Plot the rate of the water outlet temperature and required rate of water replacement as a function of the cooling tower effectiveness for values between 0.2 and 0.8. b.) Plot the cooling tower Range and Approach parameters s a function of the cooling tower effectiveness for values between 0.2 and 0.8. c.). Determine the heat transfer rate that would result for the same effectiveness and air and water flowrates if a dry counterflow heat exchanger were used instead of a cooling tower.

23 12.B-9 Shown in Figure 12.B-9 is a schematic of an indirect/direct evaporative cooling system. Air at 26 C, 40% relative humidity (state 1) entering the equipment is split into two flow streams. One stream passes through an evaporative cooler that lowers the temperature of the air to T 2. This cooled air is then used as the cold stream in a countercurrent heat exchanger to lower the temperature of the other ambient air stream to state 4. The heat exchanger effectiveness is 0.85 based on the minimum capacitance rate. The air exiting the heat exchanger at state 4 then passes through a second evaporative cooler to emerge at the final conditioned state. Evaporative Cooler 2 26 C 40% rel.hum. conditioned air Evaporative Cooler Heat Exchanger exhaust Figure 12.B-9: Indirect/direct evaporative cooling system a.) Assume that the effectivenesses of the evaporative coolers are both 0.75 at any air flow rate. Calculate and plot the energy rate at which conditioned air (state 5) is provided (relative to state 1) as function of the ratio of the mass flow rates at state 1 and. b.) If the effectiveness of one of the evaporative coolers could be increased to 0.85 while the other remained at 0.75, which evaporative cooler would you select to have the higher effectiveness? c.) Indicate advantages and disadvantages of the system in Figure 12.B-9 compared to a traditional evaporative cooling system.

24 12.B-10 The purpose of this problem is to estimate the dry air mass flow rate and cooling capacity of the cooling coil necessary to maintain indoor design conditions of 24 C, 50% relative humidity for a the system shown in Figure 12.B-10. Ventilation air from outdoors is supplied to the building 2 C, 45% relative humidity at a rate of 0.25 kg/s. A portion of the air exiting the building is mixed with the ventilation air and the rest is exhausted. Conditioned air is provided to the building at 12 C. The sensible heat gain occurring in the building due to lights, people and equipment is 10 kw. The moisture gain in the space is not known, so generate your results for sensible heat ratios over the expected range of 0.7 to 0.9. Neglect the fan energy. The result of your analysis should be plots of the mass flow rate at state, the relative humidity at state and the required cooling coil capacity as a function of the sensible heat ratio. 2 C, 45% rh 0.25 kg/s 1 Mixing Section 2 4 Exhaust Conditioned Space 24 C. 50% rel. hum Supply Fan 12 C 10 kw generation SHR 0.7 to 0.9 Figure 12.B-10: Building air-conditioning system

25 12.B-11 An air conditioning system is to be designed to deliver 15,000 ft /min of air at 60 F and a relative humidity than is less that 50% for all outdoor conditions. A schematic of the system is shown in Figure 12.B-11. Hourly values of the outdoor air temperature and relative humidity are provided for a 12 hour period in Table 12.B-11. Write a program that will calculate the total cooling provided and total condensate removed in the cooling coil and the total heating that must be supplied in the reheat coil for the 12 hour period. Outdoor Air cfm 60 F <50% Cooling 1 2 Reheat Coil Coil condensate Figure 12.B-11: Air-conditioning system Table 12.B-11: Outdoor temperature and relative humidity for 12 hours Time Outdoor Temp., F Relative Humidity Time Outdoor Temp., F Relative Humidity 7: : : : : : : : : : : :

26 12.B-12 Figure 12.B-12 shows a novel air conditioning system. It is similar to a conventional air conditioning system, but it includes an additional heat exchanger. Air at 28 C, 45% relative humidity (state 1) is cooled to state 2 without condensation) in the heat exchanger. The air then passes through a conventional cooling coil where cooling and condensation both occur. The pinch point occurs at state where the air leaving the cooling coil is 10 C warmer than the evaporating refrigerant. The air exiting the cooling cool passes through the heat exchanger as the cold stream (without mixing with stream 1-2) and then to the reheat coil where heat is provided so that air exits 18 C, 40% relative humidity. 18 C, 40% rh 5 Reheat Coil 4 Heat Exchanger Cooling Coil Refrigerant out 28 C, 45% rh 1 2 Refrigerant in Condensate Figure 12.B-12: Schematic of a novel air-conditioning system The purpose of this problem is to determine the benefit of the heat exchanger. In your analysis, please do the following: a.) determine the maximum value of the heat exchanger effectiveness that will cool the air at state 2 without condensation occurring. b.) determine the required rates of cooling and reheat per kg of dry air for effectiveness values between 0 and the value you determined in part a). c.) indicate the benefits (if any) and disadvantages of the proposed system.

27 12.B-1 An air-conditioning system for a supermarket uses two cooling coils as shown in Figure 12.B-1. The cooling coil in the outdoor air duct cools and dehumidifies the ventilation air so that air exiting this coil is at 40 F and a humidity ratio of The return air cooling coil operates at higher temperature and does not provide any condensation. The amount of cooling and the air flows for each cooling coil are controlled so as to provide supply air at 55 F, with a wet bulb temperature of 52 F. The return air taken from the supermarket is maintained at 75 F, 50% relative humidity. The total cooling load for the supermarket is 25 tons when the outdoor conditions are 95 F, 50% relative humidity. Exhaust Return Air 5 Supermarket Outdoor Air 95 F, 50% rh 1 Cooling Coil F, 50% rh condensate Figure 12.B-1: Supermarket cooling system with two cooling coils Determine: a.) The supply dry air flow rate b.) The dry air flow rates through each cooling coil. c.) The cooling load for each cooling coil. d.) The operating cost in $/hr if the COP of the outdoor cooling coil is 2.8, the COP of the return air cooling coil is.6 and electricity is purchased at $0.15/kwhr. (Neglect the cost to operate the fans.)

28 12.B-14 In the air-conditioning system shown in Figure 12.B-14, 200 lbm dry air/min at 95 F, 55% relative humidity (state 1) is cooled to 48 F (state 2) in a cooling coil. Water enters the cooling coil at 55 F and exits at 62 F. The cooled air is then mixed with 1000 cfm of outdoor air to produce the conditions at state. The air is then heat exchanged with water that enters at 140 F and exits at 125 F. The water flow rate is controlled so that the mixture exits at 68 F (State 4). Outdoor Air 95 F, =55% 200 lb m /min 1000 cfm 125 F 140 F 68 F =? Cooling 48 F Coil 1 2 Heating 4 Coil 62 F 55 F condensate 48 F Figure 12.B-14: Schematic of an air conditioning system a.) Determine the rate of condensate formation. b.) Determine the tons of cooling that must be provided in the cooling coil and the required chilled water flow rate. c.) What are the temperature and relative humidity at state? d.) What is the rate of rate that hot water is provided to the heating coil? e.) What is the relative humidity of the air delivered at state 4?

29 12.B-15 One method of controlling the amount of cooling required for a coil is to use a bypass method in which only part of the air passes through the coil, and the other bypasses the coil completely as shown in Figure 12.B-15. For a specific situation, the inlet air flow rate is 10,000 lb m /hr at a temperature of 90 F and a relative humidity of 40%. The desired supply state is 65 F with a maximum relative humidity of 60%. The air leaving the cooling coil (state 2) can be assumed to be saturated. cooling coil ,000 lb m /hr bypass damper 4 65 F, =60% 90 F, =40% Figure 12.B-15: Air conditioning unit with bypass damper a.) Plot the required rate of cooling, the required rate of heating, and the required saturation temperature at state 2 as a function of the bypass flow, f, where f=0 indicates that none of the air passes through the cooling coil and f=1 indicates that all of the flow passes through the cooling coil. b.) Based on your plot, what is your recommendation for the bypass damper setting?

30 12.B-16 The air-conditioning system shown in Figure 12.B-16 is intended to provide conditioned air for a greenhouse. Air enters the system at 24 C, 20% relative humidity at 0.2 m /s. This air flow is split into two equal mass flows. One stream passes through an adiabatic evaporative cooler that is supplied with water at 25 C. The effectiveness of the evaporative cooler is The two streams are then combined in a mixing chamber and provided to the greenhouse. The greenhouse is maintained at 27 C, 60% relative humidity. Figure 12.B-16 Air conditioning system for a greenhouse a.) Determine the temperature and relative humidity at state 5. b.) Determine the total cooling load for the greenhouse and the sensible heat ratio of the load.

31 12.B-17 A cooling tower is shown in Figure 12.B-17 with operating conditions indicated. air 4 water from condenser V 6000 gal/hr w T 95 F win, 1 water to condenser 2 5 air T 74 F amb 0.5 amb V 10,000 cfm a makeup water T muw = 80 F Figure 12.B-17: Cooling tower. The cooling tower is designed to cool water with flow rate V w = 6,000 gallons per hour that enters the tower with temperature T w,in = 95 F (5 C) from the condenser of a large air conditioning system. Outdoor air at T amb = 74 F (2. C) and amb = 50% relative humidity is forced through the tower at a rate of V a = 10,000 cfm (4.72 m /s). Makeup water is provided at T muw = 80 F (26.7 C). The temperature of the air leaving the cooling tower at state 4 is T a,out = 89 F (1.7 C) and the relative humidity of the air leaving at state 4 is a,out = Use the Psychrometric Chart to do this problem. If you need properties of pure water use the table in your text. a.) Determine the rate of makeup water (in gal/hr) that must be provided to the cooling tower. b.) Determine the temperature of the water leaving the cooling tower at state 2. c.) Determine the rate of cooling that is provided to the condenser using the cooling tower (in ton). d.) Clearly indicate states and 4 on the Psychrometric chart. Note that one of the reasons that the cooling tower works so well is that the air passing through the tower is not only heated, it is also humidified. Therefore, both its temperature and humidity ratio are increased. What fraction of the total enthalpy change of the air is due to it temperature change only (i.e., what fraction of the enthalpy change of the air would be achieved if its humidity ratio did not change but it left at the same temperature, T a,out )? This is referred to as the sensible heat ratio of the device.

32 12.B-18 Figure 12.B-18 illustrates a building that is being cooled using an earth tube system. outdoor air T out =0 C building Q 5.8 kw out =70% infiltration V 0.08 m /s exhaust at indoor air V m /s 4 conditions 1 2 T g =8 C ground tube Figure 12.B-18: Earth tube system. Outdoor air at out = 70% and T out = 0 C is drawn into tubes at state 1 with volumetric flow rate V 1 = 0.51 m /s. The tubes are buried in the ground. The ground temperature is T g = 8 C and the air leaving the tubes and entering the building at state 2 has been cooled to within T g = C of the ground temperature. There is an infiltration of outdoor air directly into the building at state. The volumetric flow rate of outdoor air that directly enters the building is V = 0.08 m /s. The building experiences a heat transfer of Q = 5.8 kw. Assume that no moisture is added to the air as it passes through the building. Indoor air leaves the building through the exhaust duct at state 4. Assume that the pressure of the air is atmospheric throughout the system and solve this problem using the Psychrometric chart. a.) Is there any condensation within the ground tube? If so then determine the volumetric flow rate of condensation draining into the ground in the ground tube (in gallons/day). b.) Determine the temperature and relative humidity of the indoor air.

33 12.C-1 Consider a summer day with outdoor conditions at 27 C and 40% relative humidity. An air flow is split into two (possibly unequal) streams as shown in Figure 12.C-1. One stream is passed through an evaporative cooler that cools that has an effectiveness of 1. The two streams then enter a device that uses the temperature difference between the streams to produce power. The work producing device may be adiabatic and the streams do not mix as shown in the figure. Assuming that power be produced in this manner: Air 27 C, 40%rh 1 Water, 27 C Evaporative Cooler 2 Work Producing Device Figure 12.C-1 Power production from an evaporative cooler a.) Explain why this power production scheme does not violate the Kelvin-Planck statement of the Second Law which states that It is impossible for any device that operates continuously to receive heat from a single reservoir and produce a net amount of work. b.) How would you define the efficiency of this work producing process? What is the upper bound for this efficiency? c.) Estimate the maximum possible work per kg of processed air that can be produced if the two streams have the same air flow rate. d.) What fraction of the air passing through the evaporative cooler will result in the larges power? e.) What do you see as practical limitations that prevent this work production method from becoming commercially successful? What improvements could be made? 4

34 12.C-2 An auditorium is to be maintained at 25 C dry-bulb temperature and 40% relative humidity when the outdoor conditions are 2 C dry-bulb and 25 C wet bulb. The space will be conditioned by the system shown in the figure. Outdoor air enters the duct at state 5 at 0.25 m /s. The air exiting the cooling coil (state 7) is saturated at 7 C. Air is blown into the space (state 1) at 0.90 m /s. Pressure losses in the ducts and fan power are negligible. As shown in Figure 12.C-2, the system provides a bypass damper that controls how much of the recirculated air passes through the cooling coil. Unless otherwise specified, assume that the bypass damper is positioned such that the 50% of the recirculated air at state passes through the cooling coil. Figure 12.C-2: HVAC system for an auditorium a. Determine the temperature, humidity ratio, relative humidity and mass flow rate at all points. b. Determine the rate of condensate removal in the cooling coil. c. Determine the sensible heat ratio defined as the ratio of the sensible to the total cooling load. d. Determine the cost to condition this space assuming that the cooling coil is served by a cooling system that has a system COP of.5 at these conditions for the conditions of part a. e. Vary the bypass damper position between 0 and 1, where 0 indicates that all of the air passes through the cooling coil and 1 indicates that only the outdoor air that enters at state 5 passes through the cooling coil. Prepare a plot of the total load and operation cost as a function of the bypass damper position. Explain the results you see in the plot.

35

36 12.C- Figure 12.C-(a) illustrates a simple Rankine cycle power plant. The condenser rejects heat from the plant to the ambient. In some cases it is possible to run lake water or river water through the condenser in order to accomplish this heat rejection. Q b T H boiler W p pump condenser turbine W t water from lake Q cond waterreturnedtolake Figure 12.C-(a): Rankine cycle power plant with a water-cooled condenser. However, often a river or lake is not available (or maybe is available, but is protected by environmental regulation). In this case, the power plant must reject heat to the ambient air. Air is not a great heat transfer fluid and therefore it is not usually attractive to replace the water flow with an air flow; the heat exchanger would have to be too large and expensive. Instead, a cooling tower is often used, as shown in Figure 12.C-(b). A flow of water is pumped through the condenser. The water is sprayed into the cooling tower as a mist so that it is in direct contact with air that is pulled through the tower by fans. There is both heat transfer as well as evaporation of water into the air, leading to very high rates of heat transfer. Because the heat transfer takes place directly from the water droplets to the air, there is no need for expensive heat exchanger surfaces. Cooling tower are used extensively for thermal-fluid systems; the very large distinctive towers that you see near nuclear power plants (for example see Figure 12.C-(c)) are actually large, natural draft cooling towers that service the plant (the stuff coming out of the top is actually just clouds of water vapor). You can see cooling towers on top of many buildings, serving as heat rejection devices for the condensers used in the building HVAC systems (for example, see Figure 12.C-(d)).

37 Q b T H W p pump boiler condenser turbine W t 4 air/water mixture T 4 =5 C P 4 =1atm 4 =1.0 Q cond 1 pump water, m w T1 40 C P 4atm 1 cooling tower 2 2 water, m w T 20 C P2 1atm makeup water T 5 =18 C P 5 =1atm Figure 12.C-(b): Rankine cycle power plant with a cooling tower. 5 ambient air/water mixture T =18 C =0. P =1atm (a) (b) Figure 12.C- (c) and (d): Cooling towers used to provide heat rejection for a (c) power plant and (d) building HVAC system. The power plant shown in Figure 12.C-(b) provides and has efficiency = 0.5. a.) What is the rate of heat transfer in the condenser, W net = 1000 MW (1 GW) of power Q cond?

38 Ambient air at temperature T = 18 C and pressure P = 1 atm is drawn into the bottom of the tower by large fans (you may neglect the fan power for this problem). The relative humidity of the ambient air is = 0.. Warm pure liquid water flows from the condenser at T 1 = 40 C and P 1 = 4 atm; the water is sprayed from nozzles at the top of the tower in order to break it into small droplets that mix with the upward stream of air. Therefore, the air leaving the top of the cooling tower is saturated with water and has temperature T 4 = 5 C, relative humidity 4 = 1.0, and pressure P 4 = 1.0 atm. Cooled water collects at the bottom of the tower at temperature T 2 = 20 C and pressure P 2 = 1 atm where it is pumped back to the condenser. Because some water is lost to evaporation, makeup water at temperature T 5 = 18 C and pressure P 5 = 1 atm is added at the bottom of the tower. You can neglect the pump work and assume that the system is at steady state. The cooling tower is adiabatic. Use the internal functions in EES and the substances AirH2O and Water to model air-water mixtures and pure water, respectively. b.) What is the mass flow rate of water pumped through the condenser ( m w in Figure 12.C-(b))? c.) What is the mass flow rate of makeup water required at state 5 ( m muw ) and the total volumetric flow rate of air drawn in at state by the fans ( V )? d.) What is the rate of entropy generation in the cooling tower?

39 12.C-4 In hot dry climates, air conditioning can be achieved simply by evaporative cooling. Water is sprayed into the dry air and subsequently evaporates with a resulting decrease in the temperature of the exit air stream. The evaporative cooler design is characterized by an effectiveness that is defined as Tin Tout Tin Twb where T in the entering air temperature T out is the exiting air temperature T is the wet bulb temperature wb In a particular case, outdoor air at 5 C, 20% relative humidity enters an evaporative cooler at 6.0 m /sec and is evaporatively cooled with liquid water at 5 C. Calculate and plot the following quantities as a function of the evaporator cooler effectiveness. a.) the outlet air temperature b.) the rate of exergy destruction c.) the Second-law efficiency Is the Second-law efficiency of this process equal to one when the effectiveness is one? If not, explain why. What do you see as the advantages and disadvantages of evaporative cooling?

40 12.C-5 Many electric utilities rely on gas turbine generators to help supply their peak electrical demand during the summer months. Gas turbines operate less efficiently than the large Rankine power cycle and this is one reason why electrical rates are higher in summer than in winter. Other disadvantages of gas turbines are that they operate with a nearly constant volumetric flow rate of inlet air. The reduced density of air on hot summer days results in reduced power output at the times in which the power is most needed. One way to increase the capacity and efficiency of a stationary gas turbine power system is to cool the inlet air with an evaporative cooler, as shown in Figure 12.C-5. Cooling the air in this manner lowers its temperature at state 1 which increases the density of the air and thus the mass flow rate of working fluid, assuming a constant volumetric flow rate. Also, the added water in the air at state 1, which acts as an inert component in the air, also increases the mass flow rate, resulting in increased capacity. One purpose of this problem is to compare the efficiency and capacity of the gas turbine power system with and without the evaporative cooler. Liquid water at 25 C Evaporative Cooler Compressor 6 2 Regenerator Combustion 4 chamber Turbine 5 Generator Outdoor air 5 C, 5% rh 0 1 Fuel 46,000 kj/kg Figure 12.C-5: Gas turbine system using evaporative cooler to cool inlet air Assume the isentropic efficiencies of the compressor and turbine to be 0.72 and 0.84, respectively. The evaporative cooler has an effectiveness of The maximum temperature at state 4 is 1150 C. Pressure losses can be neglected in this analysis. The regenerator effectiveness is 0.0. The heating value of the fuel is 46,000 kj/kg. Treat the combustion process as a heat input so the composition of the gas at state 4 is the same as at state. Consider at peak day in which the ambient air (state 0) is at 5 C, 101. [kpa] and 5% relative humidity. Calculate and compare the First-Law efficiencies of the gas turbine system without and with the evaporative cooler unit. In each case, determine the pressure ratio that maximizes the power output. Estimate the increase in power output resulting from the evaporative cooler considering that the mass flow though the turbine is increased. What is your assessment of evaporative cooler based on your calculations?

41 12.C-6 A standard residential dehumidifier consists of a refrigeration unit and a small fan. Air from the space to be dehumidified is blown over the evaporator coil where it is cooled below its dewpoint temperature. Condensate drips off the evaporator coil into a drain pan. This cool dry air is then blown past the condenser coil and returned to the room warmer and dryer than it was initially. Figure 11.C-6 below shows a schematic of a proposed new product which we call the regenerative dehumidifier unit. It is similar to a conventional dehumidifier but in includes an additional cross-flow air to air heat exchanger. The purpose of this problem is to determine the performance of the regenerative dehumidifier as a function of the effectiveness of the heat exchanger which can be approximated to be T1 T2 HX T T Condenser Evaporator C 55% rh 1 2 Heat Exchanger Condensate 6 Compressor Figure 12.C-6: Schematic of a regenerative dehumidifier During typical operation, air at 0.00 kg/sec, 24 C, and 55% relative humidity (state 1) is cooled to state 2 in the heat exchanger. (The heat exchanger is not designed to provide condensation.) The air then passes through a cooling coil that both cools and dehumidifies the air to state. Air leaving the cooling coil at state is saturated at a temperature 10 C above that of the evaporating refrigerant.. Air leaving the evaporator at state reenters the heat exchanger as the cooler fluid and exits at state 4. (Streams 1-2 and -4 do not mix.) The air is then blown past the condenser coil to remove the heat of condensation and returns to the room at state 5. The saturation temperature in the condenser is 10 C higher than the temperature of the air state 4. The fans in the system draw at total of 40 W. The refrigeration cycles uses R14a and it standard in all respects. The compressor is adiabatic with an efficiency of The compressor is sized such that the volumetric flowrate at the compressor inlet is m /s. Assume that saturated vapor exits the evaporator and saturated liquid exits the condenser. Dehumidifier capacity is measured in pints of condensate per day. Calculate and plot the the capacity and pints per kw-hr of the regenerative dehumidifier as a

42 function of the effectiveness of the heat exchanger. How does it compare to a non-regenerative design?

43 12.C-7 The purpose of this problem is to compare the three evaporative cooling systems shown in Figures 12.C-7a, b, and c for a particular case in which the outdoor conditions are 85 F, 0% rel, humidity. (The water supplies to the evaporative coolers are not shown.) The indoor temperature is to be maintained at 75 F requiring a sensible cooling load of tons (6,000 Btu/hr) for these conditions. Assume that there is no water generation in the conditioned space. Since the conditioned air is blown directly into the space, the humidity ratio in the building is approximately equal to that of the conditioned air. The basic design calls for evaporative coolers having an effectiveness of Model the three evaporative cooler systems and use the results of your models to answer the following equations. Figure 12.C-7a, b, and c. Alternative evaporative cooling systems a.) What effect does the regenerator have on the required volumetric air flow rate needed to meet the cooling load. (A plot of volumetric air flow rate versus regenerator effectiveness would be helpful in answering this question. b.) How important is the evaporative cooler on the exhaust side of the building in Figure 12.C-7c? c.) What relative humidity is maintained in the conditioned space with the three systems? Is it acceptable?

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