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Ch 6a 2nd law

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The document discusses the Second Law of Thermodynamics. It states that heat always flows spontaneously from hotter to colder bodies and it is impossible for any device operating in a cycle to receive heat from a single reservoir and produce work without rejecting some heat to the reservoir. The document also discusses heat engines, refrigerators, heat pumps, and how they are characterized. It introduces the concepts of thermal efficiency, coefficient of performance, and explains how refrigerators and heat pumps differ in their intended use but work on the same principles.

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1 Second Law of Thermodynamics 6CHAPTER
2 Heat always flows from high temperature to low temperature. So, a cup of hot coffee does not get hotter in a cooler room. Yet, doing so does not violate the first law as long as the energy lost by air is the same as the energy gained by the coffee. Room at 25属 C !? Example 1
3 The amount of EE is equal to the amount of energy transferred to the room. Example 2
4 It is clear from the previous examples that.. Processes proceed in certain direction and not in the reverse direction. The first law places no restriction on the direction of a process. Therefore we need another law (the second law of thermodynamics) to determine the direction of a process.
5 Thermal Energy Reservoir If it supplies heat then it is called a source. It is defined as a body to which and from which heat can be transferred without a change in its temperature. If it absorbs heat then it is called a sink.
6 Some obvious examples are solar energy, oil furnace, atmosphere, lakes, and oceans Another example is two- phase systems, and even the air in a room if the heat added or absorbed is small compared to the air thermal capacity (e.g. TV heat in a room). Air Thermal Pollution- disrupt marine life!
7 We all know that doing work on the water will generate heat. However transferring heat to the liquid will not generate work. Yet, doing so does not violate the first law as long as the heat added to the water is the same as the work gained by the shaft. Heat Engines
8 Previous example leads to the concept of Heat Engine!. We have seen that work always converts directly and completely to heat, but converting heat to work requires the use of some special devices. These devices are called Heat Engines and can be characterized by the following:
9 Characteristics of Heat Engines.. They receive heat from high-temperature source. They convert part of this heat to work. They reject the remaining waste heat to a low- temperature sink. They operate on (a thermodynamic) cycle. High-temperature Reservoir at TH Low-temperature Reservoir at TL QH QL WHE
10 Difference between Thermodynamic and Mechanical cycles A heat engine is a device that operates in a thermodynamic cycle and does a certain amount of net positive work through the transfer of heat from a high-temperature body to a low- temperature body. A thermodynamic cycle involves a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid. Internal combustion engines operate on a mechanical cycle (the piston returns to its starting position at the end of each revolution) but not on a thermodynamic cycle. However, they are still called heat engines
11 Steam power plant is another example of a heat engine..
12 Thermal efficiency Thermal Efficiency < 100 % in out Q Q 1 inputRequired outputDesired ePerformanc 緒 in out,net th Q W in outin Q QQ
13 Thermal efficiency Thermal Efficiency < 100 %1 L H Q Q ,net out th H W Q H L H Q Q Q QH= magnitude of heat transfer between the cycle device and the H-T medium at temperature TH QL= magnitude of heat transfer between the cycle device and the L-T medium at temperature TL
14 thermal efficiency can not reach 100% Even the Most Efficient Heat Engines Reject Most Heat as Waste Heat 40 0.4 100 th Automobile Engine 20% Diesel Engine 30% Gas Turbine 30% Steam Power Plant 40%
15 Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine. <Answers: 30 MW, 0.375> Example 6-1: Net Power Production of a Heat Engine
6.2. Statements of the Second Law of Thermodynamics 16
17 The Second Law of Thermodynamics: Kelvin-Plank Statement (The first) The Kelvin-Plank statement: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.
18 It can also be expressed as: No heat engine can have a thermal efficiency of 100%, or as for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace. Note that the impossibility of having a 100% efficient heat engine is not due to friction or other dissipative effects. It is a limitation that applies to both idealized and the actual heat engines.
19 Example 1 at the beginning of the notes leads to the concept of Refrigerator and Heat Pump.. Heat can not be transferred from low temperature body to high temperature one except with special devices. These devices are called Refrigerators and Heat Pumps Heat pumps and refrigerators differ in their intended use. They work the same. They are characterized by the following:
20 High-temperature Reservoir at TH Low-temperature Reservoir at TL QH QL W RefQL = QH - W Objective Refrigerators
21 An example of a Refrigerator and a Heat pump ..
22 Coefficient of Performance of a Refrigerator The efficiency of a refrigerator is expressed in term of the coefficient of performance (COPR). Desired output Required input RCOP , 1 1 L L Hnet in H L L Q Q QW Q Q Q
23 Heat Pumps High-temperature Reservoir at TH Low-temperature Reservoir at TL QH QL W HP QH = W + QL Objective
24 Heat Pump
25 Coefficient of Performance of a Heat Pump The efficiency of a heat pump is expressed in term of the coefficient of performance (COPHP). Desired output Required input HPCOP , 1 1 H H Lnet in H L H Q Q QW Q Q Q
26 Relationship between Coefficient of Performance of a Refrigerator (COPR) and a Heat Pump (COPHP). , , net in LH H HP net in H L H L W QQ Q COP W Q Q Q Q , 1net in L HP R H L H L W Q COP COP Q Q Q Q 1HP RCOP COP
27 The second Law of Thermodynamics: Clausius Statement The Clausius statement is expressed as follows: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. Both statements are negative statements!
28 High-temperature Reservoir at TH Low-temperature Reservoir at TL QH + QL QL W = QH RefHE QH Net QIN = QL Net QOUT = QL HE + Ref Equivalence of the Two Statements Consider the HE-RF combination shown below
29 Example (6-2): Heating a House by a Heat Pump A heat pump is used to meet the heating requirements of a house and maintain it at 20oC. On a day when the outdoor air temperature drops to -2oC, the house is estimated to lose heat at rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from the cold outdoor air. Sol:
30 6.3. Perpetual Motion Machines Any device that violates the first or second law is called a perpetual motion machine If it violates the first law, it is a perpetual motion machine of the first type (PMM1) If it violates the second law, it is a perpetual motion machine of the second type (PMM2) Perpetual Motion Machines are not possible
31 The second law of thermodynamics state that no heat engine can have an efficiency of 100%. Then one may ask, what is the highest efficiency that a heat engine can possibly have. Before we answer this question, we need to define an idealized process first, which is called the reversible process.

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Ch 6a 2nd law

  • 2. 2 Heat always flows from high temperature to low temperature. So, a cup of hot coffee does not get hotter in a cooler room. Yet, doing so does not violate the first law as long as the energy lost by air is the same as the energy gained by the coffee. Room at 25属 C !? Example 1
  • 3. 3 The amount of EE is equal to the amount of energy transferred to the room. Example 2
  • 4. 4 It is clear from the previous examples that.. Processes proceed in certain direction and not in the reverse direction. The first law places no restriction on the direction of a process. Therefore we need another law (the second law of thermodynamics) to determine the direction of a process.
  • 5. 5 Thermal Energy Reservoir If it supplies heat then it is called a source. It is defined as a body to which and from which heat can be transferred without a change in its temperature. If it absorbs heat then it is called a sink.
  • 6. 6 Some obvious examples are solar energy, oil furnace, atmosphere, lakes, and oceans Another example is two- phase systems, and even the air in a room if the heat added or absorbed is small compared to the air thermal capacity (e.g. TV heat in a room). Air Thermal Pollution- disrupt marine life!
  • 7. 7 We all know that doing work on the water will generate heat. However transferring heat to the liquid will not generate work. Yet, doing so does not violate the first law as long as the heat added to the water is the same as the work gained by the shaft. Heat Engines
  • 8. 8 Previous example leads to the concept of Heat Engine!. We have seen that work always converts directly and completely to heat, but converting heat to work requires the use of some special devices. These devices are called Heat Engines and can be characterized by the following:
  • 9. 9 Characteristics of Heat Engines.. They receive heat from high-temperature source. They convert part of this heat to work. They reject the remaining waste heat to a low- temperature sink. They operate on (a thermodynamic) cycle. High-temperature Reservoir at TH Low-temperature Reservoir at TL QH QL WHE
  • 10. 10 Difference between Thermodynamic and Mechanical cycles A heat engine is a device that operates in a thermodynamic cycle and does a certain amount of net positive work through the transfer of heat from a high-temperature body to a low- temperature body. A thermodynamic cycle involves a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid. Internal combustion engines operate on a mechanical cycle (the piston returns to its starting position at the end of each revolution) but not on a thermodynamic cycle. However, they are still called heat engines
  • 11. 11 Steam power plant is another example of a heat engine..
  • 12. 12 Thermal efficiency Thermal Efficiency < 100 % in out Q Q 1 inputRequired outputDesired ePerformanc 緒 in out,net th Q W in outin Q QQ
  • 13. 13 Thermal efficiency Thermal Efficiency < 100 %1 L H Q Q ,net out th H W Q H L H Q Q Q QH= magnitude of heat transfer between the cycle device and the H-T medium at temperature TH QL= magnitude of heat transfer between the cycle device and the L-T medium at temperature TL
  • 14. 14 thermal efficiency can not reach 100% Even the Most Efficient Heat Engines Reject Most Heat as Waste Heat 40 0.4 100 th Automobile Engine 20% Diesel Engine 30% Gas Turbine 30% Steam Power Plant 40%
  • 15. 15 Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine. <Answers: 30 MW, 0.375> Example 6-1: Net Power Production of a Heat Engine
  • 16. 6.2. Statements of the Second Law of Thermodynamics 16
  • 17. 17 The Second Law of Thermodynamics: Kelvin-Plank Statement (The first) The Kelvin-Plank statement: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.
  • 18. 18 It can also be expressed as: No heat engine can have a thermal efficiency of 100%, or as for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace. Note that the impossibility of having a 100% efficient heat engine is not due to friction or other dissipative effects. It is a limitation that applies to both idealized and the actual heat engines.
  • 19. 19 Example 1 at the beginning of the notes leads to the concept of Refrigerator and Heat Pump.. Heat can not be transferred from low temperature body to high temperature one except with special devices. These devices are called Refrigerators and Heat Pumps Heat pumps and refrigerators differ in their intended use. They work the same. They are characterized by the following:
  • 20. 20 High-temperature Reservoir at TH Low-temperature Reservoir at TL QH QL W RefQL = QH - W Objective Refrigerators
  • 21. 21 An example of a Refrigerator and a Heat pump ..
  • 22. 22 Coefficient of Performance of a Refrigerator The efficiency of a refrigerator is expressed in term of the coefficient of performance (COPR). Desired output Required input RCOP , 1 1 L L Hnet in H L L Q Q QW Q Q Q
  • 23. 23 Heat Pumps High-temperature Reservoir at TH Low-temperature Reservoir at TL QH QL W HP QH = W + QL Objective
  • 25. 25 Coefficient of Performance of a Heat Pump The efficiency of a heat pump is expressed in term of the coefficient of performance (COPHP). Desired output Required input HPCOP , 1 1 H H Lnet in H L H Q Q QW Q Q Q
  • 26. 26 Relationship between Coefficient of Performance of a Refrigerator (COPR) and a Heat Pump (COPHP). , , net in LH H HP net in H L H L W QQ Q COP W Q Q Q Q , 1net in L HP R H L H L W Q COP COP Q Q Q Q 1HP RCOP COP
  • 27. 27 The second Law of Thermodynamics: Clausius Statement The Clausius statement is expressed as follows: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. Both statements are negative statements!
  • 28. 28 High-temperature Reservoir at TH Low-temperature Reservoir at TL QH + QL QL W = QH RefHE QH Net QIN = QL Net QOUT = QL HE + Ref Equivalence of the Two Statements Consider the HE-RF combination shown below
  • 29. 29 Example (6-2): Heating a House by a Heat Pump A heat pump is used to meet the heating requirements of a house and maintain it at 20oC. On a day when the outdoor air temperature drops to -2oC, the house is estimated to lose heat at rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from the cold outdoor air. Sol:
  • 30. 30 6.3. Perpetual Motion Machines Any device that violates the first or second law is called a perpetual motion machine If it violates the first law, it is a perpetual motion machine of the first type (PMM1) If it violates the second law, it is a perpetual motion machine of the second type (PMM2) Perpetual Motion Machines are not possible
  • 31. 31 The second law of thermodynamics state that no heat engine can have an efficiency of 100%. Then one may ask, what is the highest efficiency that a heat engine can possibly have. Before we answer this question, we need to define an idealized process first, which is called the reversible process.