西安建筑科技大学：《工程热力学》课程教学资源（电子教材）The Second Law of Thermodynamics_1.pdf

PaRT 1. THE SECOND LAW OF THERMODYNAMICS l.A. Background to the second Law of Thermodynamics AW 23-31(see IAW for detailed VwB&s references); VN Chapters 2, 3, 4 1-A. I Some Properties of engineering Cycles; Work and Efficiency As motivation for the development of the second law, we examine two types of processes that concern interactions between heat and work. The first of these represents the conversion of work into heat. The second which is much more useful. concerns the conversion of heat into work The question we will pose is how efficient can this conversion be in the two cases R Block on rough surface Ⅴ iscous liquid Resistive heating Figure A-1: Examples of the conversion of work into heat Three examples of the first process are given above. The first is the pulling of a block on a rough horizontal surface by a force which moves through some distance Friction resists the pullin After the force has moved through the distance, it is removed. The block then has no kinetic energy and the same potential energy it had when the force started to act. If we measured the temperature of the block and the surface we would find that it was higher than when we started (High temperatures can be reached if the velocities of pulling are high; this is the basis of inertia welding. The work done to move the block has been converted totally to heat The second example concerns the stirring of a viscous liquid. There is work associated with the torque exerted on the shaft turning through an angle. When the stirring stops, the fluid comes to rest and there is(again) no change in kinetic or potential energy from the initial state. The fluid and the paddle wheels will be found to be hotter than when we started, however The final example is the passage of a current through a resistance. This is a case of electrical work being converted to heat, indeed it models operation of an electrical heater. All the examples in Figure A-l have 100% conversion of work into heat. This 100% conversion could go on without limit as long as work were supr plied. Is this true for the conversion of heat into work? To answer the last question, we need to have some basis for judging whether work is done in a given process. One way to do this is to ask whether we can construct a way that the process could result in the raising of a weight in a gravitational field. If so, we can say"Work has been done".It may sometimes be difficult to make the link between a complicated thermodynamic process and the simple raising of a weight, but this is a rigorous test for the existence of work One example of a process in which heat is converted to work is the isothermal(constant temperature)expansion of an ideal gas, as sketched in the figure. The system is the gas inside the internal energy is a function of temperature only, so that if the temperature is constant for some c chamber. As the gas expands, the piston does work on some external device. For an ideal gas process the internal energy change is zero. To keep the temperature constant during the expansion

1A-1 PART 1 - THE SECOND LAW OF THERMODYNAMICS 1.A. Background to the Second Law of Thermodynamics [IAW 23-31 (see IAW for detailed VWB&S references); VN Chapters 2, 3, 4] 1.A.1 Some Properties of Engineering Cycles; Work and Efficiency As motivation for the development of the second law, we examine two types of processes that concern interactions between heat and work. The first of these represents the conversion of work into heat. The second, which is much more useful, concerns the conversion of heat into work. The question we will pose is how efficient can this conversion be in the two cases. Figure A-1: Examples of the conversion of work into heat Three examples of the first process are given above. The first is the pulling of a block on a rough horizontal surface by a force which moves through some distance. Friction resists the pulling. After the force has moved through the distance, it is removed. The block then has no kinetic energy and the same potential energy it had when the force started to act. If we measured the temperature of the block and the surface we would find that it was higher than when we started. (High temperatures can be reached if the velocities of pulling are high; this is the basis of inertia welding.) The work done to move the block has been converted totally to heat. The second example concerns the stirring of a viscous liquid. There is work associated with the torque exerted on the shaft turning through an angle. When the stirring stops, the fluid comes to rest and there is (again) no change in kinetic or potential energy from the initial state. The fluid and the paddle wheels will be found to be hotter than when we started, however. The final example is the passage of a current through a resistance. This is a case of electrical work being converted to heat, indeed it models operation of an electrical heater. All the examples in Figure A-1 have 100% conversion of work into heat. This 100% conversion could go on without limit as long as work were supplied. Is this true for the conversion of heat into work? To answer the last question, we need to have some basis for judging whether work is done in a given process. One way to do this is to ask whether we can construct a way that the process could result in the raising of a weight in a gravitational field. If so, we can say “Work has been done”. It may sometimes be difficult to make the link between a complicated thermodynamic process and the simple raising of a weight, but this is a rigorous test for the existence of work. One example of a process in which heat is converted to work is the isothermal (constant temperature) expansion of an ideal gas, as sketched in the figure. The system is the gas inside the chamber. As the gas expands, the piston does work on some external device. For an ideal gas, the internal energy is a function of temperature only, so that if the temperature is constant for some process the internal energy change is zero. To keep the temperature constant during the expansion, i R + - Viscous liquid Block on rough surface Resistive heating F

heat must be supplied. Because 40=0, the first law takes the form @=w. This is a process that has 100%o conversion of heat into work atm The work exerted by the system is given by Work received. w Work=∫PdV here 2 and i denote the two states at the beginning and end of the process. The equation of state for an ideal gas is P= NRT/ with n the number of moles of the gas contained in the chamber. Using the equation of state, the expression for work can be written as Work during an isothermal expansion= nRTdv/V =NRTIn (A.1.1) For an isothermal process, PV=constant, so that P/P2=v2/V,. The work can be written in terms of the pressures at the beginning and end as Work during an isothermal expansion= NRTin (A.1.2) P The lowest pressure to which we can expand and still receive work from the system is atmospheric pressure. Below this, we would have to do work on the system to pull the piston out further There is thus a bound on the amount of work that can be obtained in the isothermal expansion; we of continuous power, in other words a device that would give power or propulsion as long as fuel was added to it. To do this, we need a series of processes where the system does not progres through a one-way transition from an initial state to a different final state, but rather cycles back to the initial state. What is looked for is in fact a thermodynamic cycle for the system We define several quantities for a cycle Qa is the heat absorbed by the system OR is the heat rejected by the system w is the net work done by the system. The cycle returns to its initial state, so the overall energy change, 40, is zero. The net work done by the system is related to the magnitudes of the heat absorbed and the heat rejected by W=Net work =2A-OR The thermal efficiency of the cycle is the ratio of the work done to the heat absorbed.(Efficiencies are often usefully portrayed as "What you get"versus"What you pay for. Here what we get is work and what we pay for is heat, or rather the fuel that generates the heat. In terms of the heat absorbed and rejected the thermal efficiency is Work done n= thermal efficiency Heat absorbed 4-QR=19k

1A-2 P, T Q Work received, W Patm heat must be supplied. Because ∆U = 0, the first law takes the form Q=W. This is a process that has 100% conversion of heat into work. The work exerted by the system is given by Work = PdV 1 2 ∫ where 2 and 1 denote the two states at the beginning and end of the process. The equation of state for an ideal gas is P = NRT/V, with N the number of moles of the gas contained in the chamber. Using the equation of state, the expression for work can be written as Work during an isothermal expansion = NRT dV V/ 1 2 ∫ = NRTln V V 2 1       . (A.1.1) For an isothermal process, PV = constant, so that PP V V 12 21 / / = . The work can be written in terms of the pressures at the beginning and end as Work during an isothermal expansion = NRTln P P 1 2       . (A.1.2) The lowest pressure to which we can expand and still receive work from the system is atmospheric pressure. Below this, we would have to do work on the system to pull the piston out further. There is thus a bound on the amount of work that can be obtained in the isothermal expansion; we cannot continue indefinitely. For a power or propulsion system, however, we would like a source of continuous power, in other words a device that would give power or propulsion as long as fuel was added to it. To do this, we need a series of processes where the system does not progress through a one-way transition from an initial state to a different final state, but rather cycles back to the initial state. What is looked for is in fact a thermodynamic cycle for the system. We define several quantities for a cycle: QA is the heat absorbed by the system QR is the heat rejected by the system W is the net work done by the system. The cycle returns to its initial state, so the overall energy change, ∆U , is zero. The net work done by the system is related to the magnitudes of the heat absorbed and the heat rejected by W QQ = =− A R Net work . The thermal efficiency of the cycle is the ratio of the work done to the heat absorbed. (Efficiencies are often usefully portrayed as “What you get” versus “What you pay for”. Here what we get is work and what we pay for is heat, or rather the fuel that generates the heat.) In terms of the heat absorbed and rejected, the thermal efficiency is: η = thermal efficiency = Work done Heat absorbed = − = − Q Q Q Q Q A R A R A 1 . (A.1.3)

The thermal efficiency can only be 100%(complete conversion of heat into work) if 2R=0, and a basic question is what is the maximum thermal efficiency for any arbitrary cycle? we examine this for two cases, the Carnot cycle and the Brayton (or Joule) cycle which is a model for the power cycle in a jet engine 1A. 2 Carnot Cycles A Carnot cycle is shown below. It has four processes. There are two adiabatic reversible legs and two isothermal reversible legs. We can construct a Carnot cycle with many different systems, but the concepts can be shown using a familiar working fluid, the ideal gas. The system can be regarded as a chamber filled with this ideal gas and with a piston 易 T Reservoir Insulating stand Figure A-2: Carnot cycle- thermodynamic diagram on left and schematic of the different stages in the cycle for a system composed of an ideal gas on the right The four processes in the Carnot cycle are whi system is at temperature T2 at state(a). It is brought in contact with a heat reservoir which is just a liquid or solid mass of large enough extent such that its temperature does not change appreciably when some amount of heat is transferred to the system. In other words, the heat reservoir is a constant temperature source(or receiver)of heat. The system then undergoes an isothermal expansion from a to b, with heat absorbed 22 2) At state b, the system is thermally insulated (removed from contact with the heat reservoir) and then let expand to c. During this expansion the temperature decreases to T. The heat exchanged during this part of the cycle, Lbc =0 3)At state c the system is brought in contact with a heat reservoir at temperature T. It is then compressed to state d, rejecting heat o, in the process 4)Finally, the system is compressed adiabatically back to the initial state a. The heat exchange Oda =0 The thermal efficiency of the cycle is given by the definition Q (A.2.1) In this equation, there is a sign convention implied. The quantities QA, Qr as defined are the magnitudes of the heat absorbed and rejected. The quantities Q1, Q2 on the other hand are defined with reference to heat received by the system. In this example, the former is negative and the latter is positive. The heat absorbed and rejected by the system takes place during isothermal processes and we already know what their values are from Eq(A 1.1)

1A-3 The thermal efficiency can only be 100% (complete conversion of heat into work) if QR = 0, and a basic question is what is the maximum thermal efficiency for any arbitrary cycle? We examine this for two cases, the Carnot cycle and the Brayton (or Joule) cycle which is a model for the power cycle in a jet engine. 1.A.2 Carnot Cycles A Carnot cycle is shown below. It has four processes. There are two adiabatic reversible legs and two isothermal reversible legs. We can construct a Carnot cycle with many different systems, but the concepts can be shown using a familiar working fluid, the ideal gas. The system can be regarded as a chamber filled with this ideal gas and with a piston. P V Q1 T1 T2 T2 Q2 Q2 T1 Q1 a b d c 1 2 4 3 Reservoir Insulating stand Reservoir Figure A-2: Carnot cycle – thermodynamic diagram on left and schematic of the different stages in the cycle for a system composed of an ideal gas on the right The four processes in the Carnot cycle are: 1) The system is at temperature T2 at state (a). It is brought in contact with a heat reservoir, which is just a liquid or solid mass of large enough extent such that its temperature does not change appreciably when some amount of heat is transferred to the system. In other words, the heat reservoir is a constant temperature source (or receiver) of heat. The system then undergoes an isothermal expansion from a to b, with heat absorbed Q2 . 2) At state b, the system is thermally insulated (removed from contact with the heat reservoir) and then let expand to c. During this expansion the temperature decreases to T1. The heat exchanged during this part of the cycle, Qbc = 0. 3) At state c the system is brought in contact with a heat reservoir at temperature T1. It is then compressed to state d, rejecting heat Q1 in the process. 4) Finally, the system is compressed adiabatically back to the initial state a. The heat exchange Qda = 0. The thermal efficiency of the cycle is given by the definition η =− =+ 1 1 1 2 Q Q Q Q R A . (A.2.1) In this equation, there is a sign convention implied. The quantities Q ,Q A R as defined are the magnitudes of the heat absorbed and rejected. The quantities Q ,Q1 2 on the other hand are defined with reference to heat received by the system. In this example, the former is negative and the latter is positive. The heat absorbed and rejected by the system takes place during isothermal processes and we already know what their values are from Eq. (A.1.1):