S.A. Klein and G.F. Nellis Cambridge University Press, 2011 = 90 F. compressor. condenser. 5 evaporator 1. evap

Transcription

1 9.A-1 Dedicated subcooling is a novel modification for frozen food refrigeration in supermarkets. With the dedicated subcooling modification, liquid refrigerant leaving the condenser is further cooled at constant pressure to an intermediate temperature, T 4, as shown in Figure 9.A-1. The cooling needed for this purpose is provided by another, smaller refrigeration cycle. The overall coefficient of performance is defined as the ratio of the heat removal from the food cases to the total work input for both refrigeration cycles. throttle valve 8 condenser T H = 90 F 7 compressor 9 sub-cooler 6 T H = 90 F throttle valve 4 3 condenser 2 compressor 5 evaporator 1 T C = -20 F Q evap Figure 9.A-1: Dedicated subcooling system for a supermarket application. Consider a refrigeration cycle designed to maintain food products at T C = 20 F. Air-cooled condensers in the refrigeration cycles reject heat to outdoors at a design temperature of T H = 90 F. At these design conditions, all heat exchange equipment is sized such that there is a ΔT hx = 12 F difference between the entering cold stream temperature and the exiting hot stream temperature (e.g., the saturation temperature in the evaporator of the refrigerated cases is -32 F and the temperature at state 6 is 12 F lower than the temperature at state 4). The isentropic efficiency of the compressors is η c = 0.7. Assume that the refrigerant leaving the evaporator is saturated vapor and the refrigerant leaving the condensers is saturated liquid. Neglect pressure losses in this analysis. The refrigerant is R22. a.) Determine the value of T 4 that maximizes the COP of the combined cycle. b.) Compare the COP of the optimized dedicated subcooling cycle to that of a simple vapor compression cycle that does not use subcooling. c.) Calculate the Second-Law efficiencies of the optimized dedicated subcooling cycle and a cycle that do not use subcooling. d.) What do you see as the advantages and disadvantages of the subcooling modification?

2 9.A-2 Refrigeration equipment is often installed with more cooling capacity than is needed. The best way to operate the equipment when the cooling load is lower than the capacity is to reduce the compressor speed with a variable speed motor. However, variable speed motors and controllers are expensive. Shown in Figures 9.A-2(a) and (b) are two alternative methods for reducing cooling capacity. Figure (a) and (b): Alternative methods for reducing cooling capacity Both systems are based on a standard vapor compression refrigeration cycle with a bypass modification for capacity control. System (a) recycles a fraction (y) of the outlet of the compressor back to the compressor inlet whereas System (b) directs a fraction (y) of the compressor outlet to the evaporator inlet. Both systems use ammonia as the refrigerant with a compressor that has an isentropic efficiency of The volumetric flow rate of the refrigerant entering the compressor is given by v 1 V 1 V d 1R 1 v 2 where V d is the compressor displacement rate = 0.20 m 3 /s R is the clearance volume ratio = v1 and v 2 are the specific volumes of the refrigerant at the compressor inlet and outlet. Assume state 5 to be saturated vapor at -15 C and state 3 to be saturated liquid at 40 C for all cases. Pressures losses in the heat exchangers and piping may be neglected. a.) Calculate the all state points, the cooling capacity, and the COP when the bypass is closed (i.e. y=0). Systems (a) and (b) should be identical in this case. b.) Determine the value of y at which the cooling capacity is 80% of the full-load value found in part a) for both System (a) and System (b). Calculate the quantities indicated in part a) for this operating state for both systems. c.) Which capacity control system do you recommend and why? d.) Determine the Second-Law efficiency of System (a) and System (b) when operating at a capacity that is 80% of full load.

3 9.A-3 Carbon dioxide has received a lot of attention as a refrigerant because it is non-toxic, non-flammable, inexpensive and not harmful to the environment. Unfortunately, carbon dioxide has a critical temperature of about 31C and its vapor pressure at 25 C is 6.4 MPa. One alternative is to employ carbon dioxide in a cascade refrigeration cycle, as shown in Figure 9.A-3, using ammonia (another natural refrigerant) in the high temperature cycle and carbon dioxide in the low temperature cycle. 7 T L =320 K 6 8 ammonia cycle 5 W H 3 2 carbon dioxide cycle W L kw T L =225 K Figure 9.A-3: Cascade refrigeration cycle The purpose of this problem is to provide a thermodynamic analysis of a two-stage cascade ammonia - carbon dioxide refrigeration cycle designed to provide 50 kw of cooling at 225 K with heat rejection to the surroundings at 320 K. The isentropic compressor efficiency can be assumed to be 0.76 for both compressors. The heat transfer rates in the three heat exchangers can each be described by relations of the form: Q UA T In the low temperature evaporator, the T is the difference between the evaporation temperature (T 4 ) and T L (225 K). In the condenser the T is difference between T 7 and 320 K. In the heat exchanger between the two cycles, T is difference between saturation temperatures T 8 and T 3. The UA values for all three heat exchangers are 25 kw/k. Make the usual assumptions associated with vapor compression system analysis, i.e., saturated vapor entering the compressors, saturated liquid exiting the condensers, adiabatic compressor operation and negligible pressure losses in the heat exchangers and lines. a.) Assume that T 3 is 250 K. Calculate the thermodynamic properties at all 8 points in the cycle. Determine the COP at these conditions. b.) Investigate the effect of T 3 by calculating the COP for values between 230 K and 280 K. Is there an optimum value for T 3? If so, what is it and why does it occur? c.) Determine the Second-Law efficiency of the cycle and the rate of exergy destruction in each system component for the value of T 3 determined in part b. Which component results in the largest rate of exergy destruction? What, if anything, could be done to reduce this irreversibility rate in this component? d.) Although this cycles using natural refrigerants, it still poses several major safety hazards. What do you see as potential safety problems with this refrigeration system?

4 9.A-4 A refrigeration cycle is being designed for a small freezer room that is to be maintained at 10 C in a 35 C environment. The refrigeration load is estimated to be 10 kw. A refrigeration cycle employing a liquid-to-suction heat exchanger is being considered with the configuration shown in Figure 9.A-4. At design conditions, the refrigerant exits the evaporator as saturated vapor and enters the cool end of the heat exchanger emerging in a superheated state. The isentropic efficiency of the compressor is The condenser heat transfer rate is (approximately) described by Q C UACTC TH where UA C = 3.50 kw/k T C is the saturation temperature in the condenser T H is the design temperature for the surroundings = 35 C Similarly, the evaporator heat transfer rate is Q UA T T where UA T F T E E E E F E = 1.50 kw/k is the desired freezer temperature = -10 C is the saturation temperature in the evaporator REFRIGERATION CYCLE WITH HX 5 HTEX 4 VALVE 1 2 COND 6 EVAP COMP 3 Figure 9.A-4: Refrigeration cycle with a liquid-to-suction heat exchanger a.) Plot the COP of this cycle for suction to discharge heat exchanger effectivenesses between 0 and 1 for refrigerant R134a. b.) Calculate and plot the COP for refrigerants propane, and ammonia on the same plot used for R134a. What is your conclusion regarding the benefits of the liquid-tosuction heat exchanger? c.) Calculate the exergy destruction per unit mass in each of system component for the conditions of part (a). Can the trends evident in your plots for part (a) be explained in terms of the exergy destruction terms? d.) List three modifications you would suggest to improve the cycle performance in order of priority.

5 9.A-5 Many homes rely on electric water heaters. Consider a typical situation in which water is available at 10 C and heated to 55 C in a 300 liter tank. The tank is equipped with two 4500 W heaters. The overall heat loss coefficient from the heated water to the surrounding 20 C environment is 15 W/K. Please answer the following questions concerning conventional electric water heating. a.) If the tank is initially filled with water at 10C, how much time is required to heat the water to 55 C with the electric resistance heaters? b.) What is the First Law efficiency of the process described in part a? c.) What is the Second Law efficiency of the process described in part a? Heat pump water heaters have been proposed as an energy-saving alternative to conventional electric water heaters. A heat pump water heater is a vapor compression refrigeration cycle in which the condensing unit is located within the water storage tank. The refrigerant for this specific unit considered here is R134a. The isentropic efficiency of the compressor and electric motor is The rate of heat transfer between the R134a and the water in the condensing unit can be described by: Q UA ( T T ), c sat c where UA c = 1000 W/K T sat, c is the condenser saturation temperature T w is the average temperature of the water in the tank at the current time. The volumetric flow rate of the refrigerant entering the compressor is given by v 2 V 2 V d 1R 1 v 3 where V d is the compressor displacement rate = m 3 /s R is the clearance volume ratio = v2 and v 3 are the specific volumes of the refrigerant at the compressor inlet and outlet, respectively. Assume the compressor to operate adiabatically, with saturated vapor at the compressor inlet and saturated liquid at the condenser outlet, and negligible pressure losses in the evaporator and condenser. d.) If the tank is initially filled with water at 10C, how much time is required to heat the water to 55 C with the heat pump water heater? e.) Plot the temperature of water in the tank, the saturation temperature in the condenser and the refrigerant flow rate as a function of time for the time period identified in part (d). f.) What is the First Law efficiency of the process described in part d? g.) What is the Second Law efficiency of the process described in part d? w

6 h.) Compare the exergy destroyed in the throttle with the work needed to power the compressor for the process in part (d).

7 9.A-6 A Q evap = 3-ton single-state vapor compression refrigeration system is being designed for an application in which the condensing temperature is T cond = 90 F and the evaporator temperature is T evap = 0 F. Refrigerants R134a, R22, R717 (ammonia), R290 (propane) and R410A are being considered for the system. The compressor efficiency may be assumed to be η c = 0.8 for all refrigerants. Assume that the pressure losses in the heat exchangers and piping are negligible. Assume steady-state operation. a.) Calculate and compare the refrigerant mass flow rate, the compressor power, the cycle COP, and the required volumetric flow rate at the compressor inlet for each refrigerant. Which refrigerant would you recommend based on these comparisons? b.) If a reversible expansion device were used in place of a throttle valve, some power could be produced which would offset the power needed to operate the compressor. In addition, a greater refrigeration effect would occur. Calculate the lost power and the reduction in the refrigeration effect resulting from throttling rather than isentropic expansion for each of the refrigerants.

8 9.A-7 You are considering installation of the conventional vapor compression refrigeration system shown in Figure 9.A-7(a). The system must provide cooling to a space at T C = -10ºF and reject heat to ambient air at T H = 85ºF. Q cond expansion valve 4 condenser T H 3 compressor W c 1 evaporator 2 T C Q evap Figure 9.A-7(a): Conventional vapor compression system. The efficiency of the compressor is η c = 0.60, the volumetric efficiency of the compressor is η vol = 0.65, and the displacement rate of the compressor is V disp = 50 cfm. The condenser has an approach temperature difference ΔT cond = 5 K and the pressure drop associated with the flow of refrigerant through the condenser is ΔP cond = 50 kpa. The refrigerant leaves the condenser with a subcool of ΔT sc = 2 K. The evaporator has an approach temperature difference of ΔT evap = 6 K and the pressure drop associated with the flow of refrigerant through the evaporator is ΔP evap = 15 kpa. The refrigerant leaves the evaporator with a superheat of ΔT sh = 5 K. The refrigerant is R134a. a.) Analyze the cycle. Print out the Arrays table containing the state point information and generate a temperature-entropy diagram that shows all states. What is the rate of refrigeration provided and rate of power consumed by the cycle? b.) Check your answer by carrying out an overall energy balance. c.) Determine the rate of entropy generation in each of the components. Check your answer by carrying out an overall entropy balance. d.) Determine the minimum power required to provide the same amount of refrigeration at T C while rejecting heat to T H (i.e., the power required by a reversible refrigeration system operating between the same two temperatures). The difference between the actual compressor power and this minimum power is the lost work associated with entropy generation in the cycle and therefore should be equal to the product of the rate of entropy generation and the ambient temperature. Show that this is so. e.) What is the Coefficient of Performance of the cycle? The liquid overfed system shown in Figure 9.A-7(b) may provide some advantages relative to the conventional system shown in Figure 9.A-7(a). The refrigerant leaving the condenser is expanded into a tank. Saturated liquid is pumped from the tank into the evaporator at state 1. By controlling the flow rate provided by the pump (which is independent of the flow rate provided by the compressor) it is possible to keep the quality of the refrigerant everywhere in the evaporator low; this dramatically improves the thermal performance of the evaporator. Saturated vapor is pulled from the tank by the compressor at state 4.

9 Q cond expansion valve 6 condenser T H 5 compressor W c 7 4 tank 1 2 evaporator 3 T C Q evap W p Figure 9.A-7(b): Liquid overfed vapor compression system. The characteristics of the compressor are the same. The efficiency of the compressor η c = 0.60, the volumetric efficiency of the compressor is η vol = 0.65, and the displacement of the compressor is V disp = 50 cfm. The condenser characteristics are also unchanged. The condenser has an approach temperature difference ΔT cond = 5 K and the pressure drop associated with the flow of refrigerant through the condenser is ΔP cond = 50 kpa. The refrigerant leaves the condenser with a subcool temperature difference of ΔT sc = 2 K. The pump has an efficiency of η p = 0.5 and the pump is controlled so that the quality of the refrigerant leaving the evaporator is x evap,out = 0.5. The performance of the evaporator improves with the liquid overfed modification; therefore, the evaporator approach temperature difference is reduced to ΔT evap = 3 K. The pressure drop associated with the flow of refrigerant through the evaporator is ΔP evap = 15 kpa. The refrigerant is R134a. The system must provide cooling at T C = -10ºF and reject heat to ambient air at T H = 85ºF. There is no pressure loss associated with the tank and the tank is insulated. f.) Analyze the liquid overfed system. Print out the Arrays table containing the state point information and generate a temperature-entropy diagram that shows all states. What is the rate of refrigeration provided and rate of power consumed by the cycle? g.) What is the COP of the liquid overfed cycle? Compare this value to the COP of the conventional system calculated in part (e).

10 9.A-8 The compressor in an ammonia refrigeration system must be liquid-cooled because the refrigerant would otherwise exit at very high temperature, which would decompose the oil. In a particular case, ammonia steadily enters a compressor at -20F and 10 psia with a mass flow rate of 126 lb m /hr. The ammonia exits the compressor at 355F, 215 psia. Cooling water is supplied to the compressor at 70F and 850 lb m /hr. The cooling water exits at 78F. The ammonia is condensed to a liquid at 105 F. Neglecting pressure losses in the heat exchangers and other heat losses, estimate the COP and cooling capacity of this refrigeration cycle.

11 9.A-9 A heat-powered refrigeration cycle is shown in Figure 9.A-9. In this cycle a Rankine cycle is used to produce the power that is used to drive the compressor in the vapor compression refrigeration cycle. The system operates at steady-state with Refrigerant R134a as the working fluid in both cycles. The isentropic efficiencies of the turbine, compressor, and pump are 0.78, 0.72, and 0.48, respectively. At the design condition, the evaporator must provide 54 kw of cooling at -10 C with saturated vapor at the evaporator exit and 40 C saturated liquid at the condenser exit. The maximum temperature for the saturated vapor at the boiler exit (state 1) is 95 C. The turbine, compressor, pump, and valve operate adiabatically. Neglect pressure losses in the heat exchangers and piping. 1 Turbine Compressor Q B Boiler 2 Q C Condenser 7 Evaporator 6 Q E Pump a.) Determine the mass flow rate of R134a through the evaporator. b.) Determine the mass flow rate of R134a through the boiler. c.) The coefficient of performance (COP) for this cycle is the ratio of the refrigeration heat input to boiler heat input. What is the COP at design conditions? d.).compare the COP determined in part c with the maximum possible COP for a heat-powered refrigeration system operating within the same temperature limits. Valve

12 9.A-10 An ice-making system uses R134a as the working fluid. The cooling load at design conditions is 200 kw. The saturation temperature in the evaporator is -20 C. The compressor exit pressure is 12 bar. Pressure losses in the lines and heat exchangers are negligible. At design conditions, saturated vapor enters the compressor and the saturated liquid exits the compressor. The compressor isentropic efficiency is The external fluid circulated through the evaporator is a 40% propylene glycol-water solution that enters the evaporator at -6 C and exits at -14 C. After leaving the evaporator, the glycol solution circulates through many small tubes in an ice storage unit and ice forms on the outside surface of these tubes. Initially, the ice storage tank contains water at 0 C. After 12 hours of operation, all of the water has frozen to become ice at 0 C. a.) What is the required mass flow rate of the glycol solution? (Note that property data for propylene glycol and other brines are available in the BrineProp2 external routine. Information on BrineProp2 is available from the Function Info menu item in the Options menu. Click the EES library routines button and then click BrineProp2.lib in the list. Clicking the Function Info button will provide documentation. b.) Determine the heat exchanger effectiveness of the evaporator. c.) What is the COP of this refrigeration cycle? d.) Determine the mass of ice that is produced in the 12 hour operating period.

13 9.A-11 Cooling at two different temperatures is needed in many situations, such as in household refrigerators and in supermarket refrigeration systems. Figure 9.A-11 shows a schematic diagram for a vapor compression refrigeration system in a supermarket that has a single compressor, a single condenser and two evaporators providing cooling capacity at two different temperatures and operating at steady conditions. The compressor isentropic efficiency is Saturated liquid exits the condenser at 105 F. The low temperature evaporator provides a cooling capacity of 3 tons at a saturation temperature -5F which is used to maintain a freezer at 5 F. The higher temperature evaporator provides a cooling capacity of 2 tons at 25F, which is used to maintain a dairy case at 35 F. The refrigerant is R-134a. Assume saturated vapor exists at the evaporator exits and neglect pressure losses in the piping and heat exchangers. 90 F expansion valve 3 condenser 2 compressor W c 4 high temperature evaporator 1 expansion valve 35 F 7 8 2tons low temperature evaporator F 3tons Figure 9.A-11: Refrigeration cycle with two evaporators a.) Determine the properties at all points in the cycle. b.) Determine the power required to operate the cycle at steady-state. c.) Indicate how you would define a COP for this cycle and determine its value. d.) Determine the total rate of entropy generation for steady-state operation. e.) Define a Second-Law efficiency and determine its value.

14 9.A-12 A vapor compression refrigeration system is to be installed in a dairy for the purpose of cooling 0.05 liters/sec of milk from 30 C to 5 C. (Assume milk to have the same thermodynamic properties as water.) In addition, 1300 W must be removed from the storage unit to maintain the stored milk at 5 C. While the milking is taking place, 0.08 liter/sec of water at 60 C are needed for cleaning purposes. The water supply is available at 10 C. It has been suggested that the heat rejected form the condenser of the cooling unit could be used to supply some or all of the energy needed to heat the water. A proposed system is shown in Figure 9.A-12. In this system, the water that is needed for cleaning provides the sink to which heat is rejected. The condensing unit consists of two heat exchangers. The first, called the condenser, is where the refrigerant changes from saturated vapor (state 4) to saturated liquid (state 5) at the condenser saturation temperature. The second heat exchanger is called the desuperheater. The refrigerant is cooled from state 3 to saturated vapor in this heat exchanger. The isentropic efficiency of the condenser is The saturation temperature in the evaporator is -2 C. Assume that saturated liquid is produced at the condenser exit and saturated vapor exits the evaporator at design conditions. Water 10 C 0.08 l/s x=1 condenser 4 desuperheater 3 compressor 5 Valve 1 evaporator 2 x=1-2 C Milk, 0.05 l/s, 30 C Milk, 5 C Figure 9.A-12: Refrigeration system for a dairy application a.) Determine the steady-state cooling load for this system b.) Determine the required refrigerant flow rate c.) Assume the heat exchanger effectiveness is 1 in order to calculate the lowest possible saturation temperature in the condenser. What is this temperature? d.) Determine the saturation temperature in the condenser if the effectiveness of the condenser is 0.9 e.) Determine the COP for this system with the heat exchanger as indicated in d.) f.) What temperature is the water at state 8 heated to? What fraction of the energy needed to heat the cleaning water is supplied by this system?

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16 9.A-13 It has been suggested that the refrigeration equipment in an old household refrigerator, which is no longer needed for refrigeration, could be modified to serve as a heat pump for heating water in residence. The purpose of this problem is to evaluate this suggestion. The refrigerator uses R134a. The compressor is driven by an electric motor that spins at 1750 revolutions per minute. The compressor has a displacement of 4.5 in 3, an adiabatic efficiency of 0.78 and a volumetric efficiency of 0.70 when operated at condensing temperature of 110 F and an evaporating temperature of 0 F. a.) Determine the COP and cooling capacity (in tons) of the refrigerator when it is operated as designed. The air-cooled condenser is replaced with a water-cooled condenser in order to heat water for domestic purposes. The evaporator temperature increases to 60 F since energy is supplied to the evaporator from indoors. The condenser must operate at 140 C to ensure that the water is heated to 130 F. The volumetric efficiency is reduced to 0.60 at these conditions. Assume that the isentropic efficiency is unaffected by the change in conditions. b.) Determine the number of gallons per hour of water that could be heated from 50 F to 130 F with this system and the COP. c.) What is your evaluation of the suggestion to use the refrigerator as a water heater?

17 9.B-1 The single-stage LiBr-H 2 O absorption system shown in Figure 9-B-1 will be used to cool water that enters at state 14 at 0.5 kg/s from 12 C to 8 C. The evaporator heat exchanger is designed such that the saturated refrigerant vapor exiting the evaporator at state 10 is at 4.5 C. Cooling water at atmospheric pressure is provided to the absorber (state 11) at 0.8 kg/s and 25 C to maintain the temperature of the dilute absorbent at state 1 at 32 C. The generator uses low pressure steam to concentrate the absorbent solution. The temperatures of the released water vapor (state 7) and the concentrated absorbent solution (state 4) and 7 are both 95 C. Saturated liquid refrigerant (water) exits the condenser at state 8 at a temperature that is 5 C higher than the cooling water exit temperature at state 13. Assume the pump to operate with a 50% isentropic efficiency. Neglect pressure losses and assume that all of the equipment operates adiabatically. Figure 9.B-1: Schematic diagram of the basic lithium bromide-water absorption cooling cycle a.) Determine the COP, and all energy flow rates for this absorption cooling cycle assuming that a solution heat exchanger is not employed. b) Include a solution heat exchanger as shown in Figure 9.B-1 in the analysis and prepare a plot of the COP as a function of the solution heat exchanger effectiveness. c) Prepare a plot that shows how the COP varies with generator temperature between 80 C and 120 C for system that has a solution heat exchanger effectiveness of 40%.

18 9.B-2 A solar air-conditioning system uses a lithium bromide-water absorption chiller having the configuration shown in Figure 9.B-2. The solar collectors provide thermal energy at a rate sufficient to maintain the solution in the generator at 200 F (states 4 and 7). The temperature of the strong solution entering the generator at state 3 is 160 F, while the temperature of the solution leaving the absorber (state 1) is 100 F, the evaporation temperature (states 9 and 10) is 40 F, and the temperature of the saturated liquid water at state 8 is 100 F. The evaporator provides 2 tons of cooling capacity. Assume state 10 is saturated vapor at 40 F and neglect the pressure losses in the piping. The pump power should be small so it is sufficient to assume that it operates isentropically. Heat Rejection to solar collectors from solar collectors condenser 7 generator throttle evaporator 10 absorber 1 pump Cooling at 40 F Heat Rejection Figure 9.B-2: Solar air conditioning system a.) Determine the system coefficient of performance at these operating conditions. b.) Determine the required collector area if the solar radiation is 950 W/m 2 and the collector efficiency at these conditions is 0.30.

19 9.B-3 Figure 9.B-3 shows a schematic diagram of a gas-based low temperature refrigeration cycle. High pressure argon gas enters a concentric-tube counter-current heat exchanger (the recuperator) at 8.5 MPa and 300 K with mass flow rate 0.25 kg/s (state 1) where it is cooled by argon gas at state 4 returning from the load heat exchanger. The gas returning at state 4 has the same flow rate and is at pressure 0.1 MPa. The high pressure argon gas leaving the recuperator at state 2 is throttled from 8.5 MPa to 0.1 MPa (state 3) in an expansion valve. The argon at state 3 enters the load heat exchanger where the argon picks up heat from a refrigerant that is evaporating at a saturation temperature of 200 K. Pressure losses in both heat exchangers can be neglected. During a test the temperatures at states 4 and 5 were measured to be 190 K and 280 K, respectively. m 025kg/s. P =8.5MPa high P 0.1 MPa T =300K low in recuperator K K load heat exchanger Figure 9.B-3: Schematic of a gas-based cooling system a.) Determine the temperatures at all points in the cycles. b.) Determine the effectiveness of both heat exchangers. c.) The argon provided to cycle at state 1 must be compressed from 0.1 MPa. What is the minimum power required to compress the argon from the low to the high pressure assuming that is it done isothermally at 300 K. d.) Determine the COP for this cycle using the power determined in part c. 3 expansion valve

20 9.B-4 Figure 9.B-4 illustrates a cryogenic refrigerator that is used to maintain a detector at 70 K. The refrigerator utilizes a reverse-brayton cycle with neon as the working fluid. Neon enters the compressor at state 1 with pressure P low = 152 kpa and mass flow rate m = kg/s. The isentropic efficiency of the compressor is The neon leaves the compressor at state 2 with pressure P high = 532 kpa. The neon leaving the compressor is quite hot and is therefore cooled in an aftercooler. The aftercooler rejects heat at rate Q H to the ambient temperature at T H = 20ºC with an approach temperature difference is T ac = 10 K. Neon flows through the recuperative heat exchanger where it is cooled by heat transfer with the cold neon returning from the cold end of the device. The neon exits the turbine at state 5 with pressure P low. The turbine operates with an isentropic efficiency of The neon leaving the turbine enters the load heat exchanger where it is warmed by a heat transfer from the detector which is at temperature T C = 70 K. The approach temperature difference for the load heat exchanger is T LHX = 2 K. Finally, the neon leaving the load heat exchanger at state 6 passes through the recuperator where it is warmed by heat transfer with the high pressure neon flowing in the opposite direction. The recuperator has an effectiveness of Neglect pressure losses in the heat exchangers and assume neon behaves as a real fluid for the conditions in this cycle. Figure 9.B-4: Reverse-Brayton refrigeration cycle. a.) Determine the temperature and pressure at each of the six numbered states in Figure 9.B-4. b.) Determine the rate of work transfer required by the compressor ( W ), the rate of work transfer from the turbine ( W ), and the rate of heat transfer to the neon in the load t heat exchanger ( Q C ). c.) Determine the Coefficient of Performance (COP) of the cryocooler. c

21 d.) Conduct numerical experiments with your model to determine the importance of the isentropic efficiencies of the compressor and turbine on the COP of this cycle.

22 9.B-5. A reverse Brayton cycle using air as the working fluid has been proposed as an alternative to the conventional automobile air-conditioning system. A major advantage of this refrigeration cycle is that it does not involve ozone depletion or global warming issues that are associated with leakage of conventional refrigerants. The necessary equipment consists of a belt-driven compressor and turbine mounted on the same shaft. (Similar equipment is commonly used in modern cars for turbocharging). In a particular case shown in Figure 9.B-5, outdoor air enters the compressor at 35 C, 100 kpa, kg/s and is adiabatically compressed with an isentropic efficiency of The air is then in a heat exchanger by air that is blown across the outside of the heat exchanger. Because the external flow rate is high, the effectiveness of the heat exchanger can be estimated to be NTU hx 1 exp The overall heat transfer coefficient-area product (UA) for the heat exchanger is 225 W/K. (See Section for a discussion of heat exchangers.) After leaving the heat exchanger, the air expands through a turbine to 100 kpa. The turbine operates adiabatically with an isentropic efficiency of The cold air exiting at state 4 is blown into the cabin of the automobile to provide the air-conditioning. Figure 9.B-5: Air-based cooling system for a vehicle a.) Calculate and plot the temperature at state 4, the cooling capacity, the required power and the COP of this cooling system as a function of compressor pressure ratio for ratios between 1.5 and 10. b.) What pressure ratio would you recommend for this application?

23 9.B-6. A simplified ammonia-water absorption refrigeration cycle is shown in Figure 9.B-6. Solution having an ammonia mass fraction of 0.48 steadily enters the generator at point 1 at 80 C, 13.5 bar and 0.05 kg/s. Heat is supplied to the flash generator at a rate sufficient to maintain its contents at 115 C. Assume that the vapor and liquid streams (2 and 3) leave the generator in equilibrium at 13.5 bar, 115 C. Cooling water is used to maintain the temperatures at points 4 and 7 at 27 C. The evaporator pressure is 3.0 bar. The refrigerant exits the evaporator at 5 C. Pressure drops in lines and components (except the valves) are assumed to be negligible. Assume the pump to operate ideally. Note: Properties for ammonia-water mixtures are provided in EES with the NH3H2O external procedure. Information relating to the use of these properties is available by selecting Function Info from the Options menu. Click the External Routines button and scroll to the NH3H2O entry in the list of routines. Select the NH3H2O and click the Function Info button to view the documentation. 2 1 Condenser 4 Flash Generator 3 Solution heat exchanger Evaporator 6 10 Absorber 7 Pump Figure 9.B-6: Ammonia-water refrigeration system a. What is the temperature of the refrigerant mixture entering the evaporator? b. What is the effectiveness of the solution heat exchanger? c. What would the temperature at state have to be to allow all of the refrigerant to evaporate? d. What is the coefficient of performance for this cycle?

24 9.B-7 Figure 1 illustrates a system that is used to provide short term cooling for a cryogenic detector on a spacecraft. Q ac T amb 1 aftercooler P reg P bottle P exh 5 recuperator bottle of argon throttle valve 2 3 load heat exchanger 4 T C Q load Figure 1: Joule-Thomson system for detector cooling. A bottle of argon is stored onboard the spacecraft. The initial pressure of the bottle is P bottle = 3500 psi and the volume of the bottle is V bottle = 10 liter. Heat transfer from the ambient environment onboard the spacecraft maintains temperature of the contents of the bottle at T amb = 20ºC. When cooling is desired, a valve on the bottle is opened and argon passes through a pressure regulator that maintains an exit pressure of P reg = 2000 psi. The temperature of the argon may change as it passes through the regulator; therefore an aftercooler is placed downstream of the regulator so that the argon is returned to ambient temperature before it enters the recuperative heat exchanger at state 1. The recuperator has an approach temperature difference of T rec = 5 K and the pinch point is at the warm end. The argon leaving the high pressure side of the recuperator at state 2 is throttled to the exhaust pressure, P exh = 0.25 atm at state 3 before being heated in a load heat exchanger to state 4 and passing through the low pressure side of the recuperator where it is exhausted to space at state 5. Pressure losses in the recuperator are negligible for both the low and high pressure streams. This problem will ignore the initial cool down transient for the detector and consider only its steady state operation at T C = 140 K. The approach temperature difference associated with the load heat exchanger is T LHX = 1 K. The total refrigeration load required to maintain the temperature of the detector at T C is Q = 15 W. The system is deactivated when the bottle pressure reaches the regulator pressure. load a.) Determine the time that the system can maintain the detector at its operating temperature. b.) Plot the operating time as a function of the regulator pressure and determine the optimal regulator pressure that maximizes the operating time of the system.

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26 9.C-1 Carbon dioxide is of interest as a refrigerant because it is inexpensive, non-toxic, nonflammable, and it has excellent thermodynamic and transport properties. However, carbon dioxide has a critical temperature of about 31C. Since the ambient temperature is often this high or higher, and a temperature difference is needed to drive heat transfer rates, the carbon-dioxide refrigeration cycle must operate with the high temperature cooler at supercritical pressures. The purpose of this problem is to design a supercritical carbondioxide air conditioning cycle to provide 10 kw of cooling at 5C with heat rejection to ambient air at 35C. We will assume that the high temperature heat exchanger can be designed to cool the high pressure refrigerant to 40C. The mass flow rate of refrigerant can be represented in terms of a volumetric efficiency with the following relation. 1. p n disch e V RPM m arg 1 C C p suction vb where C is the clearance volume ratio, n is the polytropic index, V is the compressor cylinder displacement, RPM is the compressor speed and v b is the specific volume of the refrigerant at the compressor inlet. A typical value for C is The compressor speed is 3500 rpm. The polytropic index can be assumed to be the isentropic index (ratio of c p /c v ) at the compressor inlet conditions. The compressor power can be represented by defining a combined compressor and motor efficiency defined as the ratio of the isentropic to actual power. The combined efficiency is Compressors are often assumed to be adiabatic. However experience has shown that small compressors have significant heat loss. In this case, the ratio of the heat transfer rate to the compressor power input is It will be necessary to determine the optimum compressor discharge pressure, the associated discharge temperature, the refrigerant mass flow rate, the compressor displacement, and the compressor power. A schematic of the proposed refrigeration cycle is shown in Figure 9.C- 1. Note that this figure includes a liquid-to-suction heat exchanger. At design conditions, the refrigerant exits the evaporator as saturated vapor and enters the cool end of the heat exchanger emerging in a superheated state. The advantage of the liquid-to-suction heat exchanger should be investigated in your analysis. Neglect pressure losses in the heat exchangers and piping.

27 Evaporator Suction-line HX 2 Compressor Heat exchanger 3 4 Air at 35 C Figure 9.C-1: Carbon dioxide refrigeration cycle with suction-line heat exchanger a.) Assume that the liquid-to-suction heat exchanger is not present (or alternatively, it has a small heat exchanger effectiveness) and that the discharge pressure is 95 bar. Determine the COP of this cycle and plot the state point information on temperature-entropy and pressure-enthalpy diagrams. b.) Determine optimum discharge pressure (at state 4) between 75 bar and 125 bar that maximizes the COP of this cycle for the conditions or part a. If you find that there is an optimum discharge pressure, please explain why it exists. c.) Prepare a plot of the optimum COP as a function of the liquid-to-suction heat exchanger effectiveness. (You will need to determine the optimum discharge pressure at each heat exchanger effectiveness value.) What is your conclusion regarding the benefits of the liquid-to-suction heat exchanger? d.) How does carbon dioxide compare to other refrigerants under similar operating conditions? What is your overall assessment of carbon dioxide as a refrigerant?

28 9.C-2 The purpose of this problem is to determine the performance of residential vapor compression heat pump operating in the heating mode using refrigerant R134a. The heat pump system has a fixed speed reciprocating compressor that provides a constant displacement rate ( V d ) when it is operating. The volumetric flow rate of the refrigerant entering the compressor is given by: v 2 V 2 V d 1R 1 v 3 where V d is the compressor displacement rate = m 3 /s R is the clearance volume ratio = v2 and v 3 are the specific volumes of the refrigerant at the compressor inlet and outlet, respectively. The compressor isentropic efficiency is a function of temperatures and it can be expressed approximately as: c Tcond Tevap where T cond and T evap are the saturation temperatures ( C) in the condenser and evaporator, respectively. The heat exchange rates in the condenser and evaporator can both be modeled with effectiveness relations of the form: Q = C min T max where is the heat exchanger effectiveness which can be determined as (1-exp(-NTU)) C min is the minimum capacitance rate T max is the maximum temperature difference between the entering air and the saturation temperature of the refrigerant Air is supplied to the condenser at 25 C and 1 kg/s. The outdoor evaporator coil sees an air supply at the outdoor temperature and 1 kg/s. The overall heat transfer coefficients for the condenser and evaporator are both 40 W/m 2 -K. Assume that the evaporator outlet is saturated vapor and the condenser outlet is saturated liquid. Pressure drops in the heat exchangers can be neglected. a. Assume the outdoor air temperature is -5 C and the heat transfer surface area of condenser and evaporator are each 35 m 2. Calculate the heating capacity (kw) and COP at this condition and plot the cycle on a pressure-enthalpy diagram. b. Calculate and plot the heating capacity and COP as a function of the outdoor temperature between -30 C and 20 C. Explain the behavior you observe in the plot.

29 9.C-3 Determine the heating season performance factor (HSPF) for operation of the heat pump described in 9.C-2 for a building located in Madison, WI. The HSPF is defined as a heat pump s estimated seasonal heating output in BTU divided by the amount of electrical energy that it consumes in watt-hours. The heating season is defined as the hours during the year that the ambient temperature is less than 15 C. Include the fan power and auxiliary energy in your calculated HSPF value. When operating, the condenser and evaporator fans consume a combined power of 68 W. Compare your calculated value with recommendation of SHPF>7.7 provided by the U.S. DOE. Table 9.C-3 provides bin temperature information for Madison WI. The first column shows the building load, i.e., the rate of heat transfer required to maintain the building at a comfortable temperature when the outdoor temperature is at the value indicated in the second column in C. The third columns provides the number of hours during the heating season that the temperature is between -2.5 C and 2.5 C of the temperature shown in the second column. For example, there are 136 hours for which the ambient temperature is between C and C. If the heat pump capacity is less than the building load, then auxiliary electrical heating is provided with a COP=1. If the heat pump capacity exceeds the building load, the heat pump is cycled off and on as necessary to supply the load. In this case, the heat pump operates for a period equal to hours*load/capacity. Table 9.C-3: Heating load and number of hours at each bin temperature in Madison, WI Load [W] T_avg [C] Hours [hr]

30 9.C-4 You own and operate a natural gas-fired gas turbine engine that provides electrical power to the grid. The gas turbine engine is shown in Figure 9.C-4(a). fuel, m f combustor compressor 2 3 net power 1 turbine 4 ambient air m,t,p amb amb Figure 9.C-4(a): Gas turbine engine. Both the ambient air temperature and the price that you can sell your electricity vary dramatically. During the day, the ambient air temperature is high and the resulting airconditioning load causes a high demand for power and therefore leads to high electricity prices. During the night, the ambient air temperature is low and the electricity demand is also low. As a result, you cannot sell your electricity for very much money. For this problem, we will divide the day into two periods. The daytime period has duration time day = 10 hr/day and ambient temperature T day = 96 F. During the daytime period you can sell the electricity that you produce at a rate ev day = 0.25 $/kw-hr. The nighttime period has duration time night = 14 hr/day and ambient temperature T night = 72 F. During the nighttime period you can sell the electricity that you produce at a rate ev night = 0.05 $/kw-hr. The compressor takes in air with ambient conditions, T amb (equal to T day or T night, depending on the time of day) and P amb = 1 atm. The compressor pressure ratio is PR = 5 and the compressor efficiency is c = The compressor is a fixed volume device - it is designed to process a specific volumetric flow rate of air. The volumetric flow rate at the inlet to the compressor is V c,in = cfm. The air leaving the compressor at state 2 enters the combustor where it is mixed with natural gas and combusted. The heat of combustion for natural gas is HC = 50 MJ/kg and the cost of natural gas is ngc = 0.25 $/kg. The air-fuel ratio is adjusted so that the combustion products leaving the combustor are at T t,in = 900 C. The combustion products leaving the turbine are exhausted to atmosphere. The efficiency of the turbine is t = Net power produced by the turbine is provided to a generator and converted to electricity (assume that the generator is 100% efficient). The combustion products can be modeled as air and air can be modeled as an ideal gas. Neglect pressure drop through the combustor. a.) Model the cycle for the daytime period. Prepare a T-s diagram and overlay the states on it. Print out an arrays table that includes each of the states and their entropy, enthalpy, temperature, and pressure. Determine the efficiency, power output, and the air-fuel ratio.

31 b.) Determine the profit that you make each day when you run the power plant during the daytime period. c.) Determine the profit that you make each day when you run the power plant during the nighttime period. Your analysis should have shown that you lose money when you operate at night. The value of the electricity that you produce is not sufficient to pay for the fuel required because you get paid so little for electricity produced during off-peak times. You are considering installing the system shown in Figure 9.C-4(b). Q cond high pressure expansion valve 10 T night 9 condenser 2 nd stage compressor W rc, 2 flash chamber 11 low pressure 12 expansion valve P int 14 mixer st stage compressor W rc, 1 fuel, m f combustor 13 evaporator net power cold storage system T ics =-5 C ambient air m,t,p amb amb Q evap 1 Q pc 2 compressor precooler Figure 9.C-4(b): Gas turbine engine with precooling. turbine The air entering gas turbine engine is precooled using a cold storage system. The cold storage system is an ice system, except that the temperature of the ice can be adjusted by changing the concentration of antifreeze in the frozen solution. The gas turbine engine is operated only during the daytime period. Therefore, each day the ice in the cold storage system is melted as it cools the air entering the compressor in the precooler. Because the air entering the compressor is colder, it has higher density and therefore the mass flow rate processed by the gas turbine is larger and its power output higher. The result is more daytime profits. The cold storage system is recharged (i.e., the ice is re-solidified) during the nighttime period when the gas turbine is off. The flash intercooled refrigeration system shown in Figure 9.C-4(b) is operated at night in order to re-solidify the ice. This is advantageous because the cost of electricity at night is less than during the day and 5

32 because the temperature for heat rejection is much lower at night, increasing the COP of the refrigeration system. During the daytime period, ambient air at T day = 96ºF and P amb = 1 atm is drawn into the precooler at state 1. The temperature of the ice in the cold storage system is T ics = -5ºC and the approach temperature difference for the precooler is T pc = 5 K. Air at T 2 = T ics + T pc enters the compressor. The remaining parameters for the gas turbine are the same ( t = 0.84, c = 0.80, T t,in = 900ºC, PR = 5, HC = 50 MJ/kg, V cin, = cfm). There is no pressure loss in the precooler. d.) Model the gas turbine portion of the cycle shown in Figure 9.C-4(b) for the daytime period. Prepare a T-s diagram and overlay the states on it. Print out an arrays table that includes each of the states and their entropy, enthalpy, temperature, and pressure. Determine the efficiency, power output, and the air-fuel ratio of the system. Determine the profit that you make each day when you run the power plant during the daytime period, neglecting for now the cost associated with purchasing and running the cold storage system. The flash intercooled refrigeration system is used to resolidify the ice in the cold storage system during the night. The condenser rejects heat to the ambient air at T night = 72ºF. The condenser has an approach temperature difference of T cond = 4 K and the refrigerant leaving the condenser is subcooled by T sc = 0.5 K. The evaporator accepts heat from the ice at T ics and has an approach temperature difference of T evap = 5 K. The refrigerant leaving the evaporator is superheated by T sh = 5 K. The isentropic efficiency of the first stage compressor is rc,1 = 0.65 and the isentropic efficiency of the second stage compressor is rc,2 = The volumetric efficiency of both compressors is vol = The working fluid in the refrigeration system is R134a. Neglect pressure loss in all of the heat exchangers. The intercooling pressure (P int ) is the average of the evaporating and condensing pressures. The flash chamber is a large insulated tank. Saturated liquid is pulled from the bottom of the tank at state 12 and sent to the low pressure expansion valve and evaporator. Saturated vapor is pulled from the top at state 14 and used to provide intercooling by mixing with the superheated vapor leaving the first stage compressor at state 7. e.) Model the refrigeration cycle. Prepare a T-s diagram and overlay the states on it. Print out an arrays table that includes each of the states and their entropy, enthalpy, temperature, and pressure. Determine the refrigeration provided by the system (in tons). Determine the coefficient of performance of the cycle and the power required by the refrigerant compressors. The cost of electricity purchased during nighttime hours is ec night = 0.12 $/kw-hr (yes, the cost of electricity to you is higher than what you could sell it for because the power company wants to strongly discourage you from producing power during the night). You have determined that the cost of the refrigeration system can be broken down into the cost of the compressors, condenser, and cold storage system. The cold storage system cost is Cost ics = $50,000. The cost of the condenser is given by:

33 Cost cond $-K Q cond 1.5 W T T where T cond is the saturation temperature at the condenser pressure. The cost of the compressors is given by: Costcomp cond night $-s Displacement m 3 where Displacement is the total displacement of the installed compressors. The displacement of the compressor is the volumetric flow rate at the suction port of the compressor divided by the volumetric efficiency of the compressor. f.) Determine the net profit associated with purchasing the refrigeration system and operating the precooled power plant for Time ec = 1 year. As a system designer you have a few free parameters that you can use to optimize your system. The temperature of the cold storage system, T ics, the approach temperature difference in the condenser, T cond, and the intercooling pressure, P int. g.) Plot the net profit as a function of T ics. Explain why an optimal cold storage temperature exists. h.) Plot the net profit as a function of T cond. Explain why an optimal condenser approach temperature difference exits. i.) Optimize your system - determine the optimal values of T ics, T cond, and P int and the associated maximum value of the net profit.

34 9.C-5 Shown in the figure below is a schematic diagram of a two-stage refrigeration cycle that is currently in the design phase. The cycle is designed to provide 50 kw of refrigeration capacity at 225 K with a heat sink of 320 K. As with all cycles of this nature, the heat transfer processes are responsible for most of the irreversible processes. In this problem, assume that all of the irreversibilities are a result of heat transfer in the heat exchangers i.e., we will assume that the two compressors operate reversibly between constant temperature source and sink streams. The heat transfer rate across each heat exchanger can be represented in the usual manner as follows: Q UA T T L L L z Q UA T T W L y w Q UA T T H H x H Q H W H Q W T H =320 K T x T w where UA is the heat transfer area product and temperatures are identified in Figure 9.C-5. W L Q L =50 kw T y T z T L =225 K Figure 9.C-5 2-stage refrigeration cycle The U values (heat transfer coefficient) for all three exchangers are approximately equal. However, the heat transfer areas can be adjusted to maximize the system performance. In the present case, the sum of the three UA values is constrained to 40 kw/k that fixes the total heat exchanger area for the cycle. a.) Identify how to best distribute the area, i.e., what are the UA values for the three heat exchangers that maximize system performance when T y = 250 K The intermediate cycle temperatures, T w and T y, have not yet determined. How does your result depend on the value on your choice for T y? b.) How does the performance of the optimum cycle compare to that for a single stage cycle with the same total UA value distributed over two heat exchangers? What advantages does a two-stage cycle offer over a single stage cycle?

35 9.C-6 You have been hired as a consultant by a food processing company. The company employs a blast freezing process in which very cold air (T f = -40ºC) is blown over product in order to cause it to freeze very quickly. The rapid freezing process prevents the growth of large ice crystals and therefore improves the quality of the frozen food. You have been asked to design the refrigeration system required by the blast freezer. The typical refrigeration system used for this application is a compound ammonia cycle, shown in Figure 9.C-6(a). Q cond T amb =26 C expansion valve 1 7 condenser 6 high pressure compressor W htc flash chamber 8 m h 4 5 low pressure compressor W ltc 1 3 expansion valve 2 2 evaporator m l Q air evap 9 10 V 12,000 cfm a air at T f = -40 C T Tamb to blast freezer P P 1atm amb Figure 9.C-6(a): Compound ammonia cycle. The condenser rejects heat to the ambient air at T amb = 26ºC. We will begin by assuming that the condenser has an approach temperature difference of T cond = 5 K although the size of the condenser, and therefore the value of T cond will eventually be adjusted during the design process. The refrigerant leaving the condenser at state 7 is not subcooled. The evaporator cools air from state 9 where T = T amb and P = P amb = 1 atm to state 10 where T = T f. The cold air leaving the evaporator is directed to the blast freezer. The volumetric flow rate of air entering the evaporator is V a = 12,000 cfm. We will begin by assuming that the evaporator has an approach temperature difference of T evap = 5 K although the size of the evaporator and therefore the value of T evap will also be optimized. The refrigerant leaving the evaporator is not superheated. The isentropic efficiencies of both compressors depend on the pressure ratio according to: P out c Pin where P in and P out are the pressures entering and leaving the compressor, respectively. The working fluid in the refrigeration system is ammonia. Neglect pressure loss in all of

36 the heat exchangers. The refrigerant leaving the condenser is throttled to an intermediate pressure (P int ) and enters the flash chamber. The intermediate pressure is initially assumed to be half-way between the evaporating and condensing pressures (P evap - the pressure of the ammonia in the evaporator and P cond - the pressure of the ammonia in the condenser): P P f P P int evap int cond evap where f int = 0.5 (for now, but we can adjust the value of f int and therefore change the intermediate pressure during the design in order to optimize performance). The flash chamber is a large insulated tank. Saturated liquid is pulled from the bottom of the tank at state 1 and sent to the low pressure expansion valve and then the evaporator. The superheated vapor leaving the low pressure compressor is directed back to the flash chamber at state 4. Saturated vapor at state 5 is pulled from the top of the flash chamber at state 5 and sent to the high pressure compressor. Note that the mass flow rate passing through the high pressure compressor ( m ) will not be equal to the mass flow rate h passing through the low pressure compressor ( m l ). There is no pressure drop associated with the flash chamber (i.e., states 4, 5, 8, and 1 are all at the intermediate pressure, P int ). a.) Model the refrigeration cycle. Prepare a T-s diagram and overlay the states on it. Print out an arrays table that includes each of the states (numbered as in the figure) and their entropy, enthalpy, temperature, and pressure. Determine the refrigeration provided by the system (in tons). Determine the coefficient of performance of the cycle and the power required by the refrigeration compressors. You are going to use your model developed in part (a) to design a system. The company has specified that the design should be based on minimizing the total cost associated with purchasing and operating the system for 5 years (neglecting the time value of money). Therefore, you need to determine the capital cost of the system as well as the operating cost. The volumetric efficiency of the compressors is estimated according to: v in vol vout where v in and v out are the specific volumes at the inlet and exit of the compressor, respectively. The cost of the compressor can be estimated according to the displacement rate: $-s Costcomp V m 3 disp The conductance of the evaporator (a measure of its size) is computed from: T evap UAevap m a cp, a ln Tamb T evap where m a is the mass flow rate of air in the evaporator, c P,a is the constant pressure specific heat capacity of the air in the evaporator, and T evap is the temperature of the ammonia in the evaporator. The cost of the evaporator is estimated according to: $-K Costevap 0.75 UAevap W

37 The conductance of the condenser is computed from: Q cond UAcond T where Q cond is the rate of heat transfer in the condenser. The cost of the condenser is estimated according to: $-K Costcond 0.75 UAcond W Assume that the cost of the system is entirely related to the cost of the compressors and the heat exchangers. The plant runs 16 hours per and day 340 days per year. The cost of the electricity used to power the compressors is ec = 0.10 $/kw-hr. b.) Estimate the capital cost of the system and the operating cost of the system over five years. Determine the total cost of purchasing and operating the system for 5 years. As a system designer you have a few free parameters that you can use to optimize your system. The intermediate pressure (set by the parameter f int ), the approach temperature difference in the condenser (T cond ), and the approach temperature difference in the evaporator (T evap ) can each be adjusted in order to determine an optimal design. c.) Plot the total cost as a function of f int. Your plot should demonstrate that there is an optimal value of the intermediate pressure. d.) Plot the total cost as a function of the condenser approach temperature difference. Your plot should demonstrate that there is an optimal value of T cond. Explain why this is so. e.) Simultaneously optimize the values of f int, T cond, and T evap in order to come up with the optimal design of the compound refrigeration cycle for this application. Fill in the table below with the salient features of the design; this table can be submitted to the company as the final design specification. Table 1: Characteristics of the optimal compound refrigeration cycle. Parameter Optimal value for compound cycle Optimal value for cascade cycle Refrigeration rate Air exit temperature Evaporator approach temp. difference Evaporator conductance Evaporator cost Condenser approach temp. difference Condenser conductance Condenser cost Cascade HX approach temp. difference cond

38 Cascade HX conductance Cascade HX cost Intermediate pressure Intermediate temperature Low P/T compressor displacement Low P/T compressor cost High P/T compressor displacement High P/T compressor cost Capital cost of system Total compressor power Coefficient of performance Operating cost Total cost of ownership for 5 years The company has asked you to also evaluate an alternative refrigeration system for their application. They have heard that the cascade refrigeration cycle, shown in Figure 9.C- 6(b), is more attractive for low temperature systems. The cascade cycle consists of a low temperature cycle that is interfaced to a high temperature cycle by the cascade heat exchanger. The advantage of the cascade cycle over the compound cycle is that the low temperature circuit can utilize a high density working fluid, in this case carbon dioxide. The high temperature circuit utilizes ammonia. The high density of carbon dioxide results in a much smaller low temperature compressor which is therefore much less expensive than the low pressure compressor required by the compound cycle. (Your analysis from the previous part of the project should show that the cost of the low pressure compressor is pretty large.) The disadvantage of the cascade cycle is the requirement of a cascade heat exchanger which adds cost to the system.

39 Q cond upper stage throttle valve 7 T amb = 26 C 6 condenser high temperature compressor W htc ammonia 8 m h 5 carbon dioxide cascade heat exchanger 4 low temperature compressor low stage throttle valve 1 W ltc 2 evaporator 3 m l Q evap air 9 10 V 12,000 cfm a air at T f = -40 C T Tamb to blast freezer PP 1atm amb Figure 9.C-6(b): Cascade ammonia cycle. The condenser rejects heat to the ambient air at T amb = 26ºC. Begin by assuming that the condenser has an approach temperature difference of T cond = 5 K, although the size of the condenser will eventually be adjusted during the design process. The refrigerant leaving the condenser at state 7 is not subcooled. The cascade system must provide the same refrigeration load; therefore, the evaporator cools air from state 9 where T = T amb and P = P amb = 1 atm to state 10 where T = T f and the volumetric flow rate of air entering the evaporator is V a = 12,000 cfm. Begin by assuming that the evaporator has an approach temperature difference of T evap = 5 K; the optimization of this component will follow. The refrigerant leaving the evaporator is not superheated. The isentropic efficiencies of both compressors depend on the pressure ratio as indicated above. Neglect pressure loss in all of the heat exchangers. The ammonia leaving the condenser is throttled to state 8 where it enters the cascade heat exchanger. The ammonia evaporates in the cascade heat exchanger as it absorbs heat from the condensing carbon dioxide in the other side of the cascade heat exchanger. The ammonia leaves the cascade heat exchanger as saturated vapor at state 5. The ammonia is compressed to state 6 and then enteres the condenser. The carbon dioxide leaves the cascade heat exchanger as saturated liquid at state 1. The temperature at state 1 is the intermediate temperature in the cycle (T int ). The intermediate temperature is initially assumed to be half-way between the evaporating and condensing temperatures (T evap - the temperature of the

40 carbondioxide in the evaporator and T cond - the temperature of the ammonia in the condenser): T T f T T int evap int cond evap where f int = 0.5 initially (but we can adjust the value of f int and therefore change the intermediate temperature during the design in order to optimize performance). The flash chamber is a large insulated tank. Initially assume that the cascade heat exchanger approach temperature difference is T cascade = 5 K (we will optimize this eventually). The pinch point for the cascade heat exchanger is at the end where ammonia enters and carbon dioxide leaves (the left side of the heat exchanger in the figure). There is no pressure loss associated with either the flow of ammonia or carbon dioxide in the cascade heat exchanger. f.) Model the cascade refrigeration cycle. Print out an arrays table that includes each of the states (numbered as in the figure) and their entropy, enthalpy, temperature, and pressure. Determine the refrigeration provided by the system (in tons). Determine the coefficient of performance of the cycle and the power required by the refrigeration compressors. You are going to use your model to design a cascade system. Again, the company has specified that the design should be based on minimizing the total cost associated with purchasing and operating the system for 5 years (neglecting the time value of money). The capital cost of the compressors, condenser, and evaporator can be estimated as in part (b). The cost of the cascade heat exchanger must also be considered. The conductance of the cascade heat exchanger is computed from: Q cascade UAcascade Tcascade where Q cascade is the rate of heat transfer in the condenser. The cost of the cascade heat exchanger is estimated according to: $-K Costcascade 0.75 UAcascade W Assume that the cost of the system is entirely related to the cost of the compressors and the heat exchangers. The plant runs 16 hours per and day 340 days per year. The cost of the electricity used to power the compressors is ec = 0.10 $/kw-hr. g.) Estimate the capital cost of the system and the operating cost of the system over five years. Determine the total cost of purchasing and operating the system for 5 years. As a system designer you have a few free parameters that you can use to optimize the cascade system. The intermediate temperature (set by the parameter f int ), the approach temperature difference in the condenser (T cond ), the approach temperature difference in the evaporator (T evap ), and the approach temperature difference in the cascade heat exchanger (T cascade ) can each be adjusted in order to determine an optimal design. h.) Simultaneously optimize the values of f int, T cond, T evap, and T cascade in order to come up with the optimal design of the cascade refrigeration cycle for this application. Fill in the table with the salient features of the design; this table can be

41 submitted to the company as the final design specification. Based on your analysis, which system would you recommend?

42 9.C-7 The absorption refrigeration cycle shown in Figure 9.C-7 is an example of a heat-driven heat pump. Heat-driven heat pumps operate between three thermal reservoirs T L <T M <T H. In an absorption refrigeration cycle, T L is the temperature of the cooled space maintained by the evaporator heat transfer process. T M is the ambient temperature to which thermal energy is rejected from the absorber and condenser. T H is the temperature of high temperature energy (generally C) supplied to the generator. In the following analysis, T L = 273 K, T M = 310 K and T H = 400 K. Heat transfer across a finite temperature difference is a major source of thermodynamic irreversibility in the absorption cycle. In the following analysis, assume that irreversible heat transfer is the only source of irreversibility. The heat transfer rates to or from the components of the absorption cycle can be expressed as linear functions of a temperature difference as indicated in the following diagram. All four UA values are equal to 2 kw/k. Q UA T T Q UA T T C C C M G G H G Condenser Generator Evaporator Absorber Q UA T T Q UA T T E E L E A A A M Figure 9.C-7: Schematic of an absorption heat pump a. What is the maximum refrigeration coefficient of performance (COP) for this cycle (defined as the ratio of the evaporator heat transfer rate to the generator heat transfer rate ( Q E / Q G), the corresponding evaporator heat transfer rate ( Q E ) and temperatures operating temperatures of the four components, T E, T A, T G, and T C? b. If the cycle is operated so as to maximize the evaporator heat transfer rate, what are the corresponding COP and values of the temperatures T E, T A, T G, and T C? How different are T A and T G at the optimum heat transfer rate? c. Vary the UA values. Can you draw any general conclusions from your numerical experiments?

43 9.C-8 The performance of a heat pump cycle in a heating application can be expressed in terms of its coefficient of performance (COP H ), which is defined as the rate of energy transfer to the heated space, Q H, to the compressor power input. The maximum possible COP for a machine operating between an outdoor temperature T L and a heated space temperature T H is given by the Carnot limit T H /(T H -T L ). The point of this problem is to investigate the operation of a heat-transfer limited Carnot heat pump cycle. Consider a heat pump cycle having no internal irreversibilities that is designed to provide a heating capacity of 10 kw for a building. (The ideal heat pump cycle is a Carnot cycle working in reverse.) Air from the building is blown past the condenser at 25 C and 2.0 kg/s. The condenser heat transfer rate can be expressed as: Q H Hm HCpTc TH, in where H is the effectiveness of the condenser given which is equal to (1-exp(-NTU H )) m H is the condenser air flow rate (2.0 kg/s) C p is the specific heat of air T H, in is the entering temperature of the air, (25 C) T c is the refrigerant condensation temperature NTU H = (U H A H )/( m HCp) A H is the effective heat transfer for the condenser Heat is supplied to the evaporator from outdoors with a fan-forced air stream at 2.0 kg/s at -5 C. The rate of heat supplied from outdoors can be represented as Q m C T T L L L p L, in - e where is the effectiveness of the evaporator given which is equal to (1-exp(-NTU L )) L m is the evaporator air flow rate (2 kg/s) L T is the entering temperature of the air, -5 C Lin, T e is the refrigerant evaporation temperature NTU L = (U L A L )/( mc L p ) A L is the effective heat transfer for the evaporator The overall heat transfer coefficients for the condenser, U H, and evaporator, U L, are both about 40 W/m 2 -K, since the air flow rates through both heat exchangers are equal and they have similar geometry. However, the total heat exchanger area for the evaporator and condenser is fixed for cost reasons to be 35 m 2. One concern is how to optimally distribute this heat exchange area. a.) What fraction of the total (condenser plus evaporator) heat exchanger area should be allocated to the condenser in order maximize the performance of this heat pump cycle?

44 b.) How is your answer changed if the total heat exchanger area is increased to 70 m 2 or decreased to 25 m 2? Can you draw a general conclusion? c.) Using the optimum heat exchanger area allocation determined in part a for 35 m 2 of total heat exchanger surface area, plot the COP of this refrigeration cycle and the evaporator and condenser temperatures as a function of heating capacity for capacities ranging between 0 and 20 kw. Explain the trends that are demonstrated by the plot.

45 9.C-9 A large central chilled water plant uses a vapor compression cycle with an adiabatic flash chamber and two-stage compression, as shown in Figure 9.C-9. The refrigerant is R134a. The isentropic efficiencies of the high and low pressure compressors are 0.82 and 0.78, respectively. Cooling water passing through the condenser maintains 35 C saturated liquid at state 5. The refrigerant is saturated vapor at 1 C at the evaporate exit. Pressure losses in the piping and heat exchangers are negligible, except of course, across the throttle valves. The plant chills 1625 kg/s of water from 15 C to 5 C. Figure 9.C-9: Two-stage refrigeration cycle with flash chamber a.) Determine the temperature and pressure at all points in the cycle assuming that the pressure at state 7 is the algebraic average of the pressures at states 5 and 1. b.) Determine the pressure at state 7 that results in the optimum COP for this cycle. c.) Plot the optimum cycle on a pressure-enthalpy diagram. d.) Indicate in a short paragraph the advantages and disadvantages of the two-stage system in Figure 9.C-9. e.) Determine the Second-Law efficiency of the optimum cycle.

46 9.C-10 Temperatures and pressures at various places in a simple Linde-Hampson gas liquefaction cycle are indicated in Figure 9.C-10. The uncondensed vapor is recycled through a heat exchanger from which it emerges at 250 K, 1 bar. Gas from the intercooled compressor enters the heat exchanger at 15.0 MPa and 275 K and exits at 155 K and essentially the same pressure. Assume an environment at 298 K, 1 bar. Make-up N K, 1 bar 7 1bar 6 heat exchanger 5 intercooled compressor 1 2 valve K 163 K 15 MPa 15 MPa vapor liquid 4 Sat d liquid at 1 bar Figure 9.C-10: Simple nitrogen liquefaction system a.) Determine the mass of liquid per kg of nitrogen gas passing through the compressor during steady operation. b.) What is the effectiveness of the heat exchanger effectiveness for this system? c.) What is the Second-law efficiency of the heat exchanger? d.) If the compressor work per kg of nitrogen compressed is 475 kj/kg, what is the Second Law efficiency of this liquefaction cycle?

47 9.C-11 A schematic of a heat-powered refrigeration system is shown in Figure 9.C-11. This system uses the exergy of thermal energy at 160C to move heat from a 5C source while rejecting heat to the ambient at 25C. The heat transfer rates may be expressed: Q ( T T) Q ( T T ) Q ( T T ) H H 1 L L 2 A 3 amb Figure 9.C-11: Schematic of a heat-powered refrigeration system where are 100, 150, and 300 W/K, respectively and temperatures are as shown in Figure 9.C-11. a.) What is the maximum possible refrigeration COP (coefficient of performance) for this heat-powered refrigeration system based on the heat input? b.) What is the maximum possible refrigeration capacity ( Q )? L