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Heat Engines ∆=+E QW Thermo Processes Int • Adiabatic – No Heat Exchanged

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Heat Engines ∆=+E QW Thermo Processes Int • Adiabatic – No Heat Exchanged Chapter 19 Heat Engines ∆=+E QW Thermo Processes int • Adiabatic – No heat exchanged – Q = 0 and ∆Eint = W • Isobaric – Constant pressure – W = P (Vf – Vi) and ∆Eint = Q + W • Isochoric – Constant Volume – W = 0 and ∆Eint = Q • Isothermal – Constant temperature V W= nRT ln i ∆Eint = 0 and Q = -W Vf C and C P V Note that for all ideal gases: where R = 8.31 J/mol K is the universal gas constant. Slide 17-80 Heat Engines Refrigerators Important Concepts 2nd Law: Perfect Heat Engine Can NOT exist! No energy is expelled to the cold reservoir It takes in some amount of energy and does an equal amount of work e = 100% It is an impossible engine No Free Lunch! Limit of efficiency is a Carnot Engine Reversible and Irreversible Processes The reversible process is an idealization. All real processes on Earth are irreversible. Example of an approximate reversible process: The gas is compressed isothermally The gas is in contact with an energy reservoir Continually transfer just enough energy to keep the temperature constant The change in entropy is equal to zero for a reversible process and increases for irreversible processes. Section 22.3 The Maximum Efficiency Qc TTcc =andec = 1 − QThh Th Qc Tc Qh Th COPC = = COPH = = W TThc− W TThc− Heat Engines . In a steam turbine of a modern power plant, expanding steam does work by spinning the turbine. The steam is then condensed to liquid water and pumped back to the boiler to start the process again. First heat is transferred to the water in the boiler to create steam, and later heat is transferred out of the water to an external cold reservoir, in the condenser. This steam generator is an example of a heat engine. © 2013 Pearson Education, Inc. Slide 19-33 ∆Eint = 0 for the entire cycle Heat Engine A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work A heat engine carries some working substance through a cyclical process The working substance absorbs energy by heat from a high temperature energy reservoir (Qh) Work is done by the engine (Weng) Energy is expelled as heat to a lower temperature reservoir (Qc) Thermal Efficiency of a Heat Engine ∆Eint = 0 for the entire cycle Weng = QQhc − Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature W QQ− Q e =eng =hc =1 − c QQhh Q h Analyze this engine to determine (a) the net work done per cycle, (b) the engine’s thermal efficiency and (c) the engine’s power output if it runs at 600 rpm. Assume the gas is monatomic and follows the ideal- gas process above. From Last Week….. A 4.00-L sample of a nitrogen gas confined to a cylinder, is carried through a closed cycle. The gas is initially at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. (a) Draw a PV diagram of this cycle. (b) Find the number of moles of the gas. (c) Find the volumes and temperatures at the end of each process (d) Find the Work and heat for each process. (e) What was the net work done on the gas for this cycle? The Brayton Cycle . Many ideal-gas heat engines, such as jet engines in aircraft, use the Brayton Cycle, as shown. The cycle involves adiabatic compression (1-2), isobaric heating during combustion (2-3), adiabatic expansion which does work (3-4), and isobaric cooling (4-1). The efficiency is: © 2013 Pearson Education, Inc. Slide 19-58 Otto Cycle The Otto cycle approximates the processes occurring in an internal combustion engine If the air-fuel mixture is assumed to be an ideal gas, then the efficiency of the Otto cycle is 1 e =1 − γ −1 (VV12) γ is the ratio of the molar specific heats V1 / V2 is called the compression ratio Typical values: Compression ratio of 8 γ = 1.4 e = 56% Efficiencies of real engines are 15% to 20% Mainly due to friction, energy transfer by conduction, incomplete combustion of the air-fuel mixture No Perfect Heat Engines A perfect heat engine connected to a refrigerator would violate the second law of thermodynamics. © 2013 Pearson Education, Inc. Slide 19-46 Rank in order, from largest to smallest, the work Wout performed by these four heat engines. W QQ− Q e =eng =hc =1 − c QQhh Q h A. Wb > Wa > Wc > Wd B. Wb > Wa > Wb > Wc C. Wb > Wa > Wb = Wc D. Wd > Wa = Wb > Wc E. Wd > Wa > Wb > Wc Rank in order, from largest to smallest, the work Wout performed by these four heat engines. W QQ− Q e =eng =hc =1 − c QQhh Q h A. Wb > Wa > Wc > Wd B. Wb > Wa > Wb > Wc C. Wb > Wa > Wb = Wc D. Wd > Wa = Wb > Wc E. Wd > Wa > Wb > Wc QuickCheck 19.8 How much heat is exhausted to the cold reservoir? A. 7000 J. B. 5000 J. C. 3000 J. D. 2000 J. E. 0 J. © 2013 Pearson Education, Inc. Slide 19-54 QuickCheck 19.8 How much heat is exhausted to the cold reservoir? A. 7000 J. B. 5000 J. C. 3000 J. D. 2000 J. E. 0 J. © 2013 Pearson Education, Inc. Slide 19-55 QuickCheck 19.9 Which heat engine has the larger efficiency? A. Engine 1. B. Engine 2. C. They have the same efficiency. D. Can’t tell without knowing the number of moles of gas. © 2013 Pearson Education, Inc. Slide 19-56 QuickCheck 19.9 Which heat engine has the larger efficiency? A. Engine 1. B. Engine 2. C. They have the same efficiency. D. Can’t tell without knowing the number of moles of gas. © 2013 Pearson Education, Inc. Slide 19-57 Heat Pumps and Refrigerators Heat engines can run in reverse This is not a natural direction of energy transfer Must put some energy into a device to do this Devices that do this are called heat pumps or refrigerators energy transferred at high temp Q COP = = h heating work done by heat pump W energy transferred at low temp Q COP = = C cooling work done by heat pump W Refrigerators . In a sense, a refrigerator or air conditioner is the opposite of a heat engine. In a heat engine, heat energy flows from a hot reservoir to a cool reservoir, and work Wout is produced. In a refrigerator, heat energy is somehow forced to flow from a cool reservoir to a hot reservoir, but it requires work Win to make this happen. © 2013 Pearson Education, Inc. Slide 19-41 Refrigerators . Shown is the energy- transfer diagram of a refrigerator. All state variables (pressure, temperature, thermal energy, etc.) return to their initial values once every cycle. The heat exhausted per cycle by a refrigerator is: QH = QC +Win © 2013 Pearson Education, Inc. Slide 19-42 Refrigerators . The purpose of a refrigerator is to remove heat from a cold reservoir, and it requires work input to do this. We define the coefficient of performance K of a refrigerator to be: . If a “perfect refrigerator” could be built in which Win = 0, then heat would move spontaneously from cold to hot. This is expressly forbidden by the second law of thermodynamics: © 2013 Pearson Education, Inc. Slide 19-43 QuickCheck 19.6 The coefficient of performance of this refrigerator is A. 0.40. B. 0.60. C. 1.50. D. 1.67. E. 2.00. © 2013 Pearson Education, Inc. Slide 19-44 QuickCheck 19.6 The coefficient of performance of this refrigerator is A. 0.40. B. 0.60. C. 1.50. D. 1.67. E. 2.00. © 2013 Pearson Education, Inc. Slide 19-45 Refrigerators • Understanding a refrigerator is a little harder than understanding a heat engine. • Heat is always transferred from a hotter object to a colder object. • The gas in a refrigerator can extract heat QC from the cold reservoir only if the gas temperature is lower than the cold-reservoir temperature TC. Heat energy is then transferred from the cold reservoir into the colder gas. • The gas in a refrigerator can exhaust heat QH to the hot reservoir only if the gas temperature is higher than the hot-reservoir temperature TH. Heat energy is then transferred from the warmer gas into the hot reservoir. Refrigerators Heat Pumps Coefficient of Performance The effectiveness of a heat pump is described by a number called the coefficient of performance (COP) In heating mode, the COP is the ratio of the heat transferred in to the work required energy transferred at high temp Q COP = = h work done by heat pump W A heat pump, is essentially an air conditioner installed backward. It extracts energy from colder air outside and deposits it in a warmer room. Suppose that the ratio of the actual energy entering the room to the work done by the device’s motor is 10.0% of the theoretical maximum ratio. Determine the energy entering the room per joule of work done by the motor, given that the inside temperature is 20.0°C and the outside temperature is –5.00°C. energy transferred at high temp Q COP = = h heating work done by heat pump W QQ hh= 0.100 WW Carnot cycle Q T 293 K h = 0.100 h = 0.100 = 1.17 W TThc−−293 K 268 K 1.17 joules of energy enter the room by heat for each joule of work done.
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