SECOND LAW OF THERMODYNAMICS

SECOND LAW OF THERMODYNAMICS
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Dr.NaveenBansal,India,Teacher
Published Date:25-10-2017
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Chapter 5 SECOND LAW OF THERMODYNAMICS 5.1. INTRODUCTION Thus far we have analyzed thermodynamic systems according to the First law of thermodynamics and state properties relationships. Water does not flow up a hill, heat does not flow from a low temperature body to a high temperature body. Our experiences suggest that processes have a definite direction. A proposed thermodynamic system that does not violate the First law of thermodynamics does not ensure that the thermodynamic system will actually occur. The First law of thermodynamics does not give any information to the direction of the process. There is a need to place restrictions on the direction of flow of a process. The limited amount of energy that can be transformed from one form to another form and the direction of flow of heat and work have not been discussed. The Second law of thermodynamics addresses these areas. Homework 5.1. Introduction 1. Give an equivalent statement of First Law of thermodynamics. 2. On what important point is the First Law of thermodynamics entirely silent? 3. Does the First law of thermodynamics give any information to the direction of a process? 4. Why do we need the Second Law of thermodynamics? 5.2. DEFINITIONS 5.2.1. Thermal Reservoirs A thermal reservoir is any object or system which can serve as a heat source or sink for another system. Thermal reservoirs usually have accumulated energy capacities which are very, very large compared with the amounts of heat energy they exchange. Therefore the thermal reservoirs are considered to operate at constant temperatures. Examples of large capacity, constant temperature thermal reservoirs which make convenient heat sources and sinks are: ocean, atmosphere, etc. 158 Chih Wu 5.2.2. Heat Engines A heat engine is a continuous cyclic device which produces positive net work output by adding heat. The energy flow diagram of a heat engine and its thermal reservoirs are shown in Figure 5.2.2.1. Heat (Q ) is added to the heat engine from a high-temperature thermal H reservoir at T , output work (W) is done by the heat engine, and heat (Q ) is removed from H L the heat engine to a low-temperature thermal reservoir at T . L HIGH-TEMPERA TURE THERMAL RESER VOIR T H HEAT Q H ADDED (DESIRABLE) OUTPUT HEAT WORK ENGINE W HEAT Q L REMOVED LOW-TEMPERA TURE THERMAL RESER VOIR T L Figure 5.2.2.1. Heat engine Figure 5.2.2.2. Rankine heat engine For example, a commercial central power station using a heat engine called a Rankine steam power plant is shown in Figure 5.2.2.2. The Rankine heat engine consists of a pump, a boiler, a turbine and a condenser. Heat (Q ) is added to the working fluid (water) in the boiler H from a high-temperature (T ) flue gas by burning coal or oil. Output work (W ) is done by the H o turbine. Heat (Q ) is removed from the the working fluid in the condenser to low-temperature L (T ) lake cooling water. Input work (W) is added to drive the pump. The net work (W) L i produced by the Rankine heat engine is W=W -W . o i Another example of a heat engine is a nuclear helium gas power plant which is made of a nuclear reactor, a gas turbine, a cooler and a compressor as shown in Figure 5.2.1.3. Heat is added to the nuclear gas power plant in the nuclear reactor, work is produced by the turbine, heat is removed from the cooler, and work input is required to operate the compressor. A part of the work produced by the turbine is used to drive the compressor. The net work output of the nuclear helium gas power plant is the difference between the work produced by the turbine and work required to operate the compressor. Second Law of Thermodynamics 159 Figure 5.2.2.3. Nuclear helium gas heat engine The measurement of performance for a heat engine is called the thermal efficiency, η. The thermal efficiency of a heat engine is defined as the ratio of the desirable net output work sought to the heat input of the engine: η=W /Q =Wdot /Qdot (5.2.2.1) net input net input Example 5.2.2.1. Heat is transferred to a Rankine power plant at a rate of 80 MW. If the net power output of the plant is 30 MW. Determine the thermal efficiency of the power plant. To solve this problem by CyclePad, we take the following steps: 1. Build A. Go to LIBRARY, select Rankine cycle unsolved. B. Switch to analysis mode. 2. Analysis A. Go to cycle, then cycle properties. B. Input the given information: (a) rate of heat added is +80 MW, (b) net power output is +30 MW. 3. Display results: The answer is η=0.375. Figure E5.2.2.1. Heat engine efficiency Example 5.2.2.2. A Rankine heat engine with a net power output of 85000 hp has a thermal efficiency of 35%. Determine the rate of heat added to the engine. 160 Chih Wu To solve this problem by CyclePad, we take the following steps: 1. Build A. Go to LIBRARY, select Rankine cycle unsolved. B. Switch to analysis mode. 2. Analysis A. Go to cycle, then cycle properties. B. Input the given information: (a) net power output is +85000 hp, (b) cycle efficiency is 35%. 3. Display results: The answer is the rate of heat added=171657 Btu/s. Figure E5.2.2.2. Heat engine Homework 5.2.2. Heat engines 1. What is a thermal reservoir? 2. Define a heat engine. Define the efficiency of a heat engine. 3. Can heat engine do anything else energy wise other than deliver work? 4. Is a gas turbine a heat engine? 5. Is a steam power plant a heat engine? 6. Is it possible to have a heat engine with efficiency of 100%? 7. Can a heat engine operating with only one thermal reservoir? 8. Is a reversible heat engine more efficient than an irreversible heat engine when operating between the same two thermal reservoirs? 9. Give two expressions to calculate the efficiency of a heat engine. 10. An inventor claims to have developed a heat engine that produces work at 10 kW, while absorbing heat at 9 kW. Evaluate such a claim. 11. An inventor claims to have developed a heat engine that produces work at 10 kW, while absorbing heat at 10 kW. Evaluate such a claim. 12. A closed system cycle has a thermal efficiency of 28%. The heat supplied from the energy source is 1000 Btu/lbm of working substance. Determine (A) heat rejected, and (B) net work of the cycle. ANSWER: (A) -720 Btu/lbm, (B) 280 Btu/lbm. Second Law of Thermodynamics 161 13. A closed system undergoes a cycle in which 600 Btu of heat is transferred to the o o system from a source at 600 F and 350 Btu of heat is rejected to a sink at 250 F. From the start to end of the cycle, A. What is the change in internal energy of the system (Btu)? B. What is the net work of the system (Btu)? C. Is this cycle possible? Justify your answer. ANSWER: (A) 0, (B) 250 Btu, (C) possible. 14. A simple steam power cycle plant receives 100,000 kJ/min as heat transfer from hot combustion gases at 3000 K and rejects 66,000 kJ/min as heat transfer to the environment at 300 K. Determine (A) the thermal efficiency of the cycle, (B) the net power produced by the cycle, and (C) the maximum possible thermal efficiency of any cycle operating between 3000 K and 300 K. ANSWER: (A) 34%, (B) 34000 kJ/min, (C) 90%. 5.2.3. Refrigerators A refrigerator is a continuous cyclic device which removes heat from a low temperature reservoir to a high temperature reservoir at the expense of work input. The energy flow diagram of a refrigerator and its thermal reservoirs are shown in Figure 5.2.3.1. Input work (W) is added to the refrigerator, desirable heat (Q ) is removed from the low-temperature L thermal reservoir at T , and heat (Q ) is added to the high-temperature thermal reservoir at L H T . H An example of a refrigerator is a domestic refrigerator which is made up of a compressor, a condenser, an expansion valve and an evaporator. The domestic refrigerator is illustrated in Figure 5.2.3.2. Work (W) is added to drive the compressor by an electric motor, desirable heat (Q ) is added in the evaporator and removed from the low temperature (T ) refrigerator L L inner space by the working fluid (refrigerant), and heat (Q ) is removed from the condenser H from the working fluid to the high temperature (T ) reservoir (kitchen). H HIGH-TEMPERA TURE THERMAL RESER VOIR T H HEAT Q H REMOVED INPUT WORK REFRIGERA TOR W (DESIRABLE) Q L HEAT ADDED LOW-TEMPERA TURE THERMAL RESER VOIR T L Figure 5.2.3.1. Refrigerator 162 Chih Wu Figure 5.2.3.2. Domestic Refrigerator The measurement of performance for a refrigerator is called the coefficient of performance (COP), and is denoted by β . The coefficient of performance of a refrigerator is R defined as the ratio of the desirable heat removed Q to the work input W of the refrigerator: L β =Q /W=Qdot/Wdot (5.2.3.1) R L L Qdot is called refrigerator capacity and is usually expressed in tons of refrigeration. L One ton of refrigeration is 3.516 kW or 12,000 Btu/h. The term “ton” is derived from the fact that the heat required to melt one ton of ice is about 12,000 Btu/h. Notice that the coefficient of performance of a refrigerator may be larger or smaller than one. Example 5.2.3.1. The inside space of a refrigerator is maintained at low temperature by removing heat (Qdot ) from it at a rate of 6 kW. If the COP of the refrigerator is 1.5, determine the L refrigerator capacity in tons of refrigeration and the required power input to the refrigerator. To solve this problem by CyclePad, we take the following steps: 1. Build A. Go to LIBRARY, select basic refrigerator cycle unsolved. B. Switch to analysis mode. 2. Analysis A. Go to cycle, then cycle properties. B. Input the given information: (a) COP is 1.5, (b) rate of heat removed from the refrigerator but added to the cycle is +6 kW. 3. Display results The answers are: refrig. Capacity is 1.71 ton, and net power input is -4kW. Second Law of Thermodynamics 163 Figure E5.2.3.1. Refrigerator Example 5.2.3.2. The inside space of a refrigerator is maintained at 3ºC by removing heat from it at a rate of 5 kW. If the required power input to the refrigerator is 2 kW. Determine the COP of the refrigerator. To solve this problem by CyclePad, we take the following steps: 1. Build A. Go to LIBRARY, select basic refrigerator cycle unsolved. B. Switch to analysis mode. 2. Analysis A. Go to cycle, then cycle properties. B. Input the given information: (a) power input is -2 kW, (b) heat removed from the refrigerator but added to the cycle is +5 kW. 3. Display results The answer is COP=2.50. Figure E5.2.3.2. Refrigerator 164 Chih Wu Homework 5.2.3. Refrigerator 1. Is air conditioner a refrigerator? 2. Is it possible to have a refrigerator operating with one thermal reservoir? 3. Define the COP of a refrigerator. 4. Is the COP of a refrigerator always larger than 1? 5. What is a ton of refrigeration? 6. It is suggested that the kitchen in your house could be cooled in the summer by closing the kitched from the rest of the house and opening the door to the domestic electric refrigerator. Is it true or false? State the reason for your conclusion. 7. The refrigeration plant of an air conditioning system has a capacity of 12000 Btu/h. The COP of the refrigerator is Determine the power required by the compressor. ANSWER: -3000 Btu/h. 8. The refrigeration plant of an air conditioning system has a cooling capacity of 2400 Btu/h. The compressor power added is 1200 Btu/h. What is the COP of the refrigerator? What is the amount of heat transfer to the atmosphere? ANSWER: 2, 3600 Btu/h. 5.2.4. Heat Pumps A heat pump is a continuous cyclic device which pumps heat to a high temperature reservoir from a low temperature reservoir at the expense of work input. The energy flow diagram of a heat pump and its thermal reservoirs are shown in Figure 5.2.4.1. Input work (W) is added to the heat pump, desirable heat (Q ) is pumped to the high-temperature thermal H reservoir at T , and heat (Q ) is removed from the low-temperature thermal reservoir at T . H L L An example of a heat pump is a house heat pump which is made up of a compressor, a ) is condenser, a throttling valve and an evaporator as illustrated in Figure 5.2.4.2. Heat (Q H removed to the high-temperature house from the working fluid in the condenser, work is added to the compressor by an electric motor, and heat (Q ) is added to the evaporator from L the low-temperature outside atmospheric air in the winter season. HIGH-TEMPERA TURE THERMAL RESER VOIR T H (DESIRABLE) HEAT Q H REMOVED INPUT HEAT WORK PUMP W HEAT ADDED Q L LOW-TEMPERA TURE THERMAL RESER VOIR T L Figure 5.2.4.1. Heat pump Second Law of Thermodynamics 165 Notice that the energy flow diagram and hardware components of a heat pump are exactly the same as those of a refrigerator. The difference between the heat pump and the refrigerator is the function of the cyclic devices. A refrigerator is used to remove heat (Q ) L from a low temperature (T ) thermal reservoir by adding work. A heat pump is used to add L heat (Q ) to a high temperature (T ) thermal reservoir by adding work. H H Figure 5.2.4.2. House Heat pump The measurement of performance for a heat pump is called the coefficient of performance (COP), and is denoted by β . The coefficient of performance of a heat pump is HP defined as the ratio of the desirable heat output Q to the work input W of the heat pump: H β =Q /W=Qdot/Wdot (5.2.4.1) HP H H Notice that the coefficient of performance of a heat pump is always larger than one. Example 5.2.4.1. A heat pump with a COP of 2.8 is selected to meet the heating requirements of a house and maintain it at a comfortable temperature (T ). Heat (Q ) is pumped from the outdoor H L ambient air at low temperature (T ). The house is estimated to lose heat (Q ) at a rate of 1000 L H kW. Determine the electric motor power consumed by the heat pump. To solve this problem by CyclePad, we take the following steps: 1. Build A. Go to LIBRARY, select basic refrigeration cycle unsolved. B. Switch to analysis mode. 2. Analysis A. Go to cycle, then cycle properties. B. Input the given information: (a) COP is 2.8, (b) heat removed is -1000 kW. 3. Display results The answer is net power=357.1 kW. 166 Chih Wu Figure E5.2.4.1. Heat pump Example 5.2.4.2. A heat pump is selected to meet the heating requirements of a house and maintain it at 18ºC. If the outdoor temperature is 2ºC, the house is estimated to lose heat at a rate of 1000 kW and the net power input to the house is 680 kW. Determine the COP of the heat pump. To solve this problem by CyclePad, we take the following steps: 1. Build A. Go to LIBRARY, select basic refrigeration cycle unsolved. B. Switch to analysis mode. 2. Analysis A. Go to cycle, then cycle properties. B. Input the given information: (a) net power input is -680 kW, (b) heat removed is -1000 kW. 3. Display results Figure E5.2.4.2. Heat pump The answer is COP=1.47. Second Law of Thermodynamics 167 Homework 5.2.4. Heat Pump 1. What is the difference between a refrigerator and a heat pump? 2. A heat pump and a refrigerator are operating between the same two thermal reservoirs. Which one has a higher COP? 3. Define the COP of a heat pump. 4. Thermal efficiency of a heat engine is defined to be the ratio of energy output desirable to energy input. Would you think that efficiency is an appropriate definition to be used when a heat pump is being discussed? Why? 5. Is the COP of a heat pump always larger than 1? 6. Indicate whether the following statements are true or false: 7. A process that causes heat to be removed from one reservoir only is feasible. 8. A process that causes heat to be supplied to one reservoir only is feasible. 9. A process that causes heat to be transferred from reservoir A to reservoir B is feasible. 10. For a heat pump, COP1. 11. For a refrigerator, COP1. 12. A heat pump picks up 1000 kJ of heat from well water at 10 C and discharges 3000 kJ of heat to a building to maintain it at 20 C. What is the COP of the heat pump? What is the minimum required cycle net work? ANSWER: 1.5, 2000 kJ. 5.3. SECOND LAW STATEMENTS The principle underlying the directionality of spontaneous change and inefficiency of a heat engine is called the second law of thermodynamics. This law may be stated in various equivalent ways. Among the best known statements are the following two statements. Kelvin-Planck statement: It is impossible to construct a heat engine that produces work with heat interaction only from a single thermal reservoir. This implies that a heat engine requires at least two thermal reservoirs with different temperatures, and that it is impossible to build a heat engine that has a thermal efficiency of 100%. Clausius statement: It is impossible to construct a heat pump or a refrigerator which moves heat from a low temperature thermal reservoir to a high temperature thermal reservoir without adding work. This implies that the coefficient of performance of a heat pump or a refrigerator is always less than infinity. Every relevant experiment that has been conducted verifies the Second law of thermodynamics. Homework 5.3. Second Law Statements 1. Give an equivalent statement of the Kelvin-Planck statement. 2. Give an equivalent statement of the Clausius statement. 168 Chih Wu 5.4. REVERSIBLE AND IRREVERSIBLE PROCESSES A reversible process is an idealized one that is performed in such a way that, at the conclusion of the process, both the system and its surroundings may be restored to their initial states without producing any changes in the rest of the universe. Any process that does not fulfill these stringent requirements is called an irreversible process. The factors that cause a process to be irreversible are called irreversibility factors. The irreversibility factors include friction, heat transfer across a finite temperature difference, free expansion, mixing of two fluids, chemical reactions, etc. A process is called internally reversible if no irreversibilities occur within the boundaries of the system during the process. A process is called externally reversible if no irreversibilities occur outside the boundaries of the system during the process. A process is called totally reversible if no irreversibilities occur neither within nor outside the boundaries of the system during the process. A process is called endo-reversible if no irreversibilities occur within but outside the boundaries of the system during the process. Reversible processes actually do not occur in nature. They serve as idealized models in which actual processes can be compared. It is generally much easier to describe a system’s behavior under reversible conditions rather than irreversible ones. Thus the assumption of reversibility, although not perfect, is often a useful engineering approach. Homework 5.2.4. Reversible and Irreversible Processes 1. Are the following natural phenomena reversible or irreversible process: a waterfall, the weathering of rocks, the rusting of iron, the tearing of a piece of paper, and combustion of a coal pile. 2. List several reversible processes. 3. List several irreversible processes. 5.5. CARNOT CYCLE Considering the concepts of reversible processes, a reversible cycle can be carried out for given thermal reservoirs at temperatures T and T . The reversible cycle was introduced by a H L French engineer N.S. Carnot. The Carnot heat engine cycle on a T-s diagram as shown in Figure 5.5.1 is composed of the following four reversible processes: 1-2 reversible adiabatic (isentropic) compression 2-3 reversible isothermal heating at T H 3-4 reversible adiabatic (isentropic) expansion 4-1 reversible isothermal cooling at T L Referring to Figure 5.5.1, the system undergoes a Carnot heat engine cycle in the following manner: Second Law of Thermodynamics 169 (A) During process 1-2, the system is thermally insulated and the temperature of the working substance is raised from the low temperature T to the high temperature T . L H The process is an isentropic process. The amount of heat transfer during the process is Q =∫Tds=0, because there is no area underneath a constant entropy (vertical) line. 12 (B) During process 2-3, heat is transferred isothermally to the working substance from the high temperature reservoir at T . This process is accomplished reversibly by H bringing the system in contact with the high temperature reservoir whose temperature is equal to or infinitesimally higher than the working substance. The amount of heat transfer during the process is Q =∫Tds= T (S -S ), which can be represented by the 23 H 3 2 area 2-3-5-6-2. Q is the amount of heat added to the Carnot cycle from a high 23 temperature thermal reservoir. p T 3 2 3 T H 4 2 T L 1 1 4 65 S V Figure 5.5.1. Carnot heat engine cycle on p-v and T-s diagram (C) During process 3-4, the system is thermally insulated and the temperature of the working substance is decreased from the high temperature T to the low temperature H T . The process is an isentropic process. The amount of heat transfer during the L process is Q =∫Tds=0, because there is no area underneath a constant entropy 34 (vertical) line. (D) During process 4-1, heat is transferred isothermally from the working substance to the low temperature reservoir at T . This process is accomplished reversibly by L bringing the system in contact with the low temperature reservoir whose temperature is equal to or infinitesimally lower than the working substance. The amount of heat transfer during the process is Q =∫Tds= T (S -S ), which can be represented by the 41 L 1 4 area 1-4-5-6-1. Q is the amount of heat removed from the Carnot cycle to a low 41 temperature thermal reservoir. The net heat added to the cycle is Q =Q +Q +Q +Q =0+T (S -S )+0+T (S -S )= net 12 23 34 41 H 3 2 L 1 4 (T -T )(S -S )=Area 2-3-5-6-2-area 1-4-5-6-1=Area 1-2-3-4-1. Notice that the area 1-2-3- H L 4 1 4-1 is the area enclosed by the cycle. The net work produced to the cycle is W =Q = Area 1-2-3-4-1. net net According to the definition of a heat engine efficiency , the efficiency of the Carnot heat engine is η =W /Q = Area 1-2-3-4-1/Area 2-3-5-6-2= (T -T )(S -S )/T (S - Carnot net output input H L 4 1 H 4 S )=(T -T )/T . 1 H L H Or η =1-T /T (5.5.1) Carnot L H 170 Chih Wu Example 5.5.1. Heat is transferred to a Carnot heat engine at a rate of 500 MW from a high-temperature source at 700 ºC and rejects heat to a low-temperature sink at 40 ºC. Determine the power produced and the efficiency of the Carnot heat engine. To solve this problem by CyclePad, we take the following steps: 1. Build A. Go to LIBRARY, select Rankine cycle unsolved. B. Switch to analysis mode. 2. Analysis A. Go to cycle, then cycle properties. B. Input the given information: (a) Tmax is 700ºC, (b) Tmin is 40ºC, (c) rate of heat added is +500 MW. C. After the Carnot cycle efficiency is posted, input eta-thermal=eta Carnot=67.82%. 3. Display results The answers are η=67.82% and net-power=339,100 kW. Figure E5.5.1. Carnot heat engine Example 5.5.2 A Carnot heat engine operates between 1000ºF and 100ºF develops 5 hp. Determine the thermal efficiency and the rate of heat supplied. If an inventor claims that he has developed a heat engine operating between the same temperature interval producing 5 hp and the rate of heat supplied to his engine is 10,000 Btu/h. Is it possible? To solve this problem by CyclePad, we take the following steps: 1. Build A. Go to LIBRARY, select Rankine cycle unsolved. B. Switch to analysis mode. 2. Analysis A. Go to cycle, then cycle result. Second Law of Thermodynamics 171 B. Input the given information: (a) Tmax is 1000ºF, (b) Tmin is 100ºF, (c) net power output is 5 hp. C. After the Carnot cycle efficiency is posted, input eta-thermal=eta Carnot=61.66%. 3. Display results The answer is rate of heat added for the Carnot cycle is 20,638 Btu/h. Since the Carnot cycle is the most efficient one among heat engines operating between the same two temperature reservoirs, the claim of having the rate of heat supplied to his engine is 10,000 Btu/h is impossible. Figure E5.5.2. Carnot heat engine Homework 5.5. Carnot Cycle 1. Does the Carnot heat engine efficiency depends on the working fluid used in the engine? Which working fluid used in the engine would be more efficient, air or water? 2. There are four distinct events that occur in the Carnot cycle. Name the events and describe each one. 3. Can a real heat engine cycle be more efficient than the Carnot heat engine efficiency? 4. What two parameters determine the limiting efficiency of any cycle? 5. What kind of shape is the Carnot cycle illustrating on a T-s diagram? 6. What is the area enclosed by the cycle area of the Carnot cycle illustrating on a T-s diagram? 7. What is the area enclosed by the cycle area of the Carnot cycle illustrating on a p-V diagram? 8. Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat engine is more efficient than the other. 9. What are the four processes of a Carnot heat engine? 10. A Carnot heat engine operates between a high temperature thermal reservoir at T H and a low temperature thermal reservoir at T . What is the efficiency of the Carnot L heat engine? 172 Chih Wu 11. An inventor claims to have developed a heat engine with better efficiency than the Carnot heat engine efficiency when operating between the same two thermal reservoirs. Is it possible? 12. A heat engine operates between two reservoirs, one at 1273 K and the other at 573 K. For every heat interaction of 1 kJ with the high temperature reservoir it rejects 0.6 kJ heat to the low temperature reservoir. (A) Is such an engine feasible? If yes, is it reversible? (B) If it rejects 0.3 kJ heat to the low temperature reservoir, is it feasible? 13. A Carnot heat engine operating between 300 K and 800 K is modified solely by raising both the high temperature and the low temperature by 100 K. Which of the following statements is false? (A) More work is done during the isothermal expansion process. (B) More work is done during the isentropic expansion process. (C) More work is done during the isothermal compression process. (D) More work is done during the isentropic compression process. (E) More cycle net work is done. (F) Thermal cycle efficiency is increased. 14. A Carnot heat engine operating between 300 K and 900 K is modified solely by raising the high temperature by 100 K. Is the thermal cycle efficiency increased? 15. A Carnot heat engine operates between 300 K and 1000 K. The power produced by the cycle is 100 kW. Calculate the rate of heat transferred from the high temperature reservoir, and the rate of heat transferred from the low temperature reservoir. 16. A Carnot heat engine operates between a low temperature thermal reservoir at T and L 1000 K. The power produced by the cycle is 100 kW, and the rate of heat transferred from the high temperature reservoir is 200 kW. Determine the rate of heat transferred from the low temperature reservoir and T . L 17. Consider the design of a power plant operating between a high-temperature reservoir at 1000ºR and a low-temperature reservoir at 500ºR. (A) What is the maximum possible efficiency of such a power plant? (B) for a production of 1,000,000 kW of power, what is the minimum rate of heat addition? (C)What is the maximum rate of heat rejection? (D) What is the rate of heat addition if the actual efficiency is 10 %? ANSWER: (A) 50%, (B) 2000000 kW, (C) 1000000 kW, (D) 10000000 kW. 18. A midshipman has invented an auto engine cycle that receives heat from combustion gases at 2500 R and rejects heat to ambient air at 500ºR. He claims that for a steady fuel flow of 10 lbm/h, his engine can produce 60000 Btu/h. The heating value of the fuel is 20000 Btu/lbm. How do you evaluate his claim? ANSWER: possible. 19. An inventor claims to have developed a heat engine when receives 800 kJ of heat from a source at 400K and produces 400 kJ of net work while rejecting the waste heat to a sink at 300 K. Is this a reasonable claim? Why? Show your justification in detail. ANSWER: not reasonable claim. 20. An inventor claims to have developed a heat engine that takes in 100 Btu of heat from a source at 1000 R and produces 15 Btu of net work while rejecting the waste heat to a sink at 500 R. Is this a reasonable claim? Would you advise investing money to put this engine on the market? Show your justification in detail. Second Law of Thermodynamics 173 21. Which is the more effective way to increase the efficiency of a Carnot heat engine: (A) to increase T and keeping T constant, or (B) to decrease T and keeping T H L L H constant. 22. Hot water could in theory be used as heat source for the generation of work. If hot water is available at 370 K and if an infinite heat sink is available at 290 K, what is the minimum amount of water that would be required for the production of 1 kJ of work? 5.5.1. Carnot Heat Pump If the Carnot cycle for a heat engine is carried out in the reverse direction, the result will be either a Carnot heat pump or a Carnot refrigerator. Such a cycle is shown in Figure 5.5.2. Using the same graphical explanation that was used in the Carnot heat engine, the heat added from the low temperature reservoir at T is area 1-4-5-6-1. Q is the amount of heat added to L 41 the Carnot cycle from a low temperature thermal reservoir. T p 3 3 2 2 4 1 1 4 S V Figure 5.5.2. Carnot heat pump or Carnot refrigerator cycle on p-v and T-s diagram The net heat added to the cycle is Q =Q +Q +Q +Q =0+T (S -S )+0+T (S -S )= net 12 23 34 41 H 3 2 L 1 4 (T -T )(S -S )=Area 2-3-5-6-2-area 1-4-5-6-1=Area 1-2-3-4-1. Notice that the area 1-2-3- H L 4 1 4-1 is the area enclosed by the cycle. The net work added to the cycle is W =Q = Area 1-2-3-4-1. According to the COP (β) net net definition of a heat pump, the COP of the Carnot refrigerator is β =Q /W = Area 4-1-5-6-4/Area 1-2-3-4-1= (T )(S -S )/(T - Carnot,R desirable output input L 2 3 H T )(S -S )=T /(T -T ). L 2 3 L H L or β =1/(T /T-1) (5.5.2) Carnot,R H L The net work added to the cycle is W =Q = Area 1-2-3-4-1. According to the COP (β) net net definition of a heat pump, the COP of the Carnot heat pump is β =Q /W = Area 2-3-5-6-2/Area 1-2-3-4-1= (T )(S -S )/(T - Carnot,HP desirable output input H 2 3 H T )(S -S )=T /(T -T ). L 2 3 H H L or 174 Chih Wu β =1/(1-T /T) (5.5.3) Carnot,HP L H Referring to Figure 5.5.2, the system undergoes a Carnot heat pump or Carnot refrigerator cycle in the following manner: (A) During process 1-2, the system is thermally insulated and the temperature of the working substance is raised from the low temperature T to the high temperature T . L H (B) During process 2-3, heat is transferred isothermally from the working substance to the high temperature reservoir at T . This process is accomplished reversibly by H bringing the system in contact with the high temperature reservoir whose temperature is equal to or infinitesimally lower than the working substance. (C) During process 3-4, the system is thermally insulated and the temperature of the working substance is decreased from the high temperature T to the low temperature H T . L (D) During process 4-1, heat is transferred isothermally to the working substance from the low temperature reservoir at T . This process is accomplished reversibly by L bringing the system in contact with the low temperature reservoir whose temperature is equal to or infinitesimally higher than the working substance. Example 5.5.3. An inventor claims to have developed a refrigerator that removes heat from a region at 270 K and transfer it to a thermal reservoir at 540 K while maintaining a COP of 8.5. Is this claim reasonable? Solution: To solve this problem, we take the following steps: 1. From Eq. (5.5.2), we have β =1/(T /T -1)=1/(540/270-1)=1. Carnot,R H L 2. Since the Carnot refrigerator has the maximum COP, the claim of 8.5 is not valid. Homework 5.5.1. Carnot Heat Pump and Carnot Refrigerator 1. What is the relationship between the COP of a Carnot refrigerator and the COP of a Carnot heat pump when the two cycles are operating between the same two thermal reservoirs? 2. A reversible heat pump and a Carnot heat pump operate between the same two thermal reservoirs. Which heat pump has higher COP? 3. An irreversible heat pump and a Carnot heat pump operate between the same two thermal reservoirs. Which heat pump has higher COP? Can a real heat pump have higher COP than the COP of the Carnot heat pump? 4. Can a real refrigerator have higher COP than the COP of the Carnot refrigerator? 5. A Carnot refrigerator operates in a room in which the temperature is 27ºC. The refrigerator consumes 500 W of power when operating, and has a COP of 4. Determine the rate of heat removal from the refrigerated space, and the temperature of the refrigerated space. ANSWER: 2000 W, -33ºC. Second Law of Thermodynamics 175 º 6. An air conditioning system is used to maintain a house at 60 F when the outside air º temperature is 100F. If heat enters the house from the outside at a rate of 5,000 Btu/h, Find the COP of the Carnot refrigerator, and the minimum possible power (in Btu/h) required for the air-conditioner. ANSWER: 13, 384.6 Btu/h. º 7. A heat pump is used to maintain a house at 60 F when the outside air temperature is º 40 F. If heat leaves the house at a rate of 10,000 Btu/h, what is the minimum amount of power (in Btu/h) required to run the heat pump? ANSWER: 385 Btu/h. 8. A Carnot power cycle using carbon mono-oxide as a working fluid has a thermal efficiency of 40 percent. At the beginning of the isothermal heating process, the temperature is 500 K. Determine the temperature of the low temperature thermal reservoir, and the COP (coefficient of performance) for the refrigerator obtained by reversing the power cycle. ANSWER: 300 K, 1.5. 9. A Carnot refrigerator operates between the temperatures of 268 K and 283 K. The power consumption is 10 kW. Find the change of the COP if the maximum temperature of the cycle increases by 2 K and the minimum temperature of the cycle decreases by 2 K. ANSWER: -0.137 (0.9333-0.9470). 10. A heat pump is to be used to maintain a building at an average temperature of 295 K. What is the minimum power required to do this when the outside temperature is 276 K and the average total heat loss is 250 kW? ANSWER: -16.10 kW. 11. A refrigerator extracts 291 kW from a cooled space at 253 K, while the ambient temperature is 293 K. (A) Find the maximum COP. (B) Find the minimum power consumption. 12. Find the heat transfer in the air cooler (heat source). ANSWER: (A) 6.325, (B) 46.01 kW, (C) 337.0 kW. 13. A heat pump is used to heat a system and maintain it at 300 K. On a winter day when the outdoor air temperature is 270 K, the system is estimated to lose heat at a rate of 5,000 kW. Determine the minimum power required to operate this heat pump. What would be the COP at this condition. ANSWER: 555.6 kW, 10. 14. Determine the maximum possible COP for a heat pump operating between an ambient temperature of -23ºC and the interior of a house at 27ºC. ANSWER: 6. 15. A refrigerator is to be built to cool water from 310 K to 290 K continuously. Heat is to be discarded to atmosphere at a temperature of 315 K. What is the minimum 3 of water is to be cooled? How much power requirement of the refrigerator if 1 m heat must be discarded to the atmosphere? 16. A ten-ton capacity refrigerator is cooling a space to 263 K by transferring heat to the atmosphere at 300 K. By assuming a Carnot refrigerator, calculate the power required to drive the refrigerator and the COP. 176 Chih Wu 17. A Carnot refrigerator operates between 260 K and 290 K. Find the change in COP if the lower temperature is lowered by 10 K and the high temperature is increased by 5 K. 18. A Carnot air conditioning unit operates between 303 K and 290 K. Find the change in COP if the lower temperature is lowered by 1 K. 19. A Carnot heat pump operates between 310 K and 280 K. Find the change in COP if the high temperature is increased by 1 K. 20. A Carnot heat pump operates between 310 K and 280 K. Find the change in COP if the lower temperature is lowered by 10 K and the high temperature is increased by 5 K. 5.6. CARNOT COROLLARIES Six corollaries deduced from the Carnot cycle are of great use in comparing the performance of cycles. The corollaries are: (1) The efficiency of the Carnot heat engine operating between a fixed high-temperature heat source thermal reservoir at T and a fixed low-temperature heat sink thermal H reservoir at T is independent of the working substance. L (2) No heat engine operating between a fixed high-temperature heat source thermal reservoir and a fixed low-temperature heat sink thermal reservoir can be more efficient than a Carnot heat engine operating between the same two thermal reservoirs. (3) All reversible heat engines operating between a fixed high-temperature heat source thermal reservoir and a fixed low-temperature heat sink thermal reservoir have the same efficiency. (4) The COP (coefficient of performance) of the Carnot heat pump (or refrigerator) operating between a fixed high-temperature thermal reservoir at T and a fixed low- H temperature thermal reservoir at T is independent of the working substance. L (5) No heat pump (or refrigerator) operating between a fixed high-temperature thermal reservoir and a fixed low-temperature thermal reservoir can have higher COP (coefficient of performance) than a Carnot heat pump (or refrigerator) operating between the same two thermal reservoirs. (6) All reversible heat pumps (or refrigerators) operating between a fixed high- temperature thermal reservoir and a fixed low-temperature thermal reservoir have the same COP (coefficient of performance). These corollaries can be proven by demonstrating that the violation of any of the corollary results in the violation of the Second law of thermodynamics.

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