PART 1 THE SECOND LAW OF THERMODYNAMICS
PART 1 THE SECOND LAW OF THERMODYNAMICS
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 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):
e2=Wab-NRT[In(v/Va e=Wad=NRT [In(va/V)=[In(v/va).(@, is negative. The efficiency can now be written in terms of the volumes at the different states as 们=T(vV (A.2.2) T2[In(vb/a)l The path from states b to c and from a to d are both adiabatic and reversible. For a reversible adiabatic process we know that Pv:= constant. Using the ideal gas equation of state, we have TVY-=constant. Along curve b-c, therefore T, Vr-=TV- Along the curve d-a, ,V-=TVI (2/T)V which means that d==a, or V/V=v/V (T2/7)(v Comparing the expression for thermal efficiency Eq(A 2.1)with Eq(A 2.2)shows two consequences. First, the heats received and rejected are related to the temperatures of the isothermal parts of the cycle by Q1,Q2 (A.2.3) T72 Second, the efficiency of a Carnot cycle is given compactly by Carnot cycle efficiency (A24) The efficiency can be 100%o only if the temperature at which the heat is rejected is zero. The heat and work transfers to and from the system are shown schematically in Figure A-3 T2 w(net work Figure A-3: Work and heat transfers in a Carnot cycle between two heat reservoirs
1A-4 Q W 2 = = ab NRT2 ln V V b a [ ] ( ) / Q W 1 = = cd NRT1 ln V V d c [ ] ( ) / = - ln V V c d [ ] ( ) / . (Q1 is negative.) The efficiency can now be written in terms of the volumes at the different states as: η= + [ ] ( ) [ ] ( ) 1 1 2 T VV T VV d c b a ln ln / / . (A.2.2) The path from states b to c and from a to d are both adiabatic and reversible. For a reversible adiabatic process we know that PVγ = constant. Using the ideal gas equation of state, we have TVγ −1 = constant. Along curve b-c, therefore TV TV 2 b c 1 1 γ γ − −1 = . Along the curve d-a, TV TV 2 a d 1 1 γ γ − −1 = . Thus, V V T T T T V V d c a b = ( ) ( ) γ −1 γ − 2 1 2 1 1 / / , which means that V V V V d c a b = , orVV VV d c ab / / = . Comparing the expression for thermal efficiency Eq. (A.2.1) with Eq. (A.2.2) shows two consequences. First, the heats received and rejected are related to the temperatures of the isothermal parts of the cycle by Q T Q T 1 1 2 2 + = 0 . (A.2.3)) Second, the efficiency of a Carnot cycle is given compactly by ηc T T = −1 1 2 . Carnot cycle efficiency (A.2.4) The efficiency can be 100% only if the temperature at which the heat is rejected is zero. The heat and work transfers to and from the system are shown schematically in Figure A-3. Q2 Q1 T1 W (net work) T2 System Figure A-3: Work and heat transfers in a Carnot cycle between two heat reservoirs
Muddy points Since n=1-=_looking at the P-V graph, does that mean the farther apart the T1,T2 isotherms are, the greater efficiency? And that if they were very close, it would be very nefficient?(MP 1A.1 In the Carnot cycle, why are we only dealing with volume changes and not pressure therms?(MP 1A.2) Is there a physical application for the carnot cycle? can we design a carnot engine for a propulsion device?(MP 1A.3 How do we know which cycles to use as models for real processes?(MP 1A. 4 1.A Brayton Cycles (or oule cycles): The power Cycle for a Gas Turbine et engine engine and the corresponding cycle are given in Figure A-g stant pressure legs. Sketches of an For a Brayton cycle there are two adiabatic legs and two cor Combustor Combustor citb Inlet Nozzle Turbine and nozzle P Compressor Turbine compressor a Heat rejectio Figure A-4: Sketch of the jet engine components and corresponding thermodynamic states Gas turbines are also used for power generation and for closed cycle operation(for example for space power generation). a depiction of the cycle in this case is shown in Figure A-5
1A-5 Muddy points Since η = −1 T T 1 2 ,looking at the P-V graph, does that mean the farther apart the T1, T2 isotherms are, the greater efficiency? And that if they were very close, it would be very inefficient? (MP 1A.1) In the Carnot cycle, why are we only dealing with volume changes and not pressure changes on the adiabats and isotherms? (MP 1A.2) Is there a physical application for the Carnot cycle? Can we design a Carnot engine for a propulsion device? (MP 1A.3) How do we know which cycles to use as models for real processes? (MP 1A.4) 1.A.3 Brayton Cycles (or Joule Cycles): The Power Cycle for a Gas Turbine Jet Engine For a Brayton cycle there are two adiabatic legs and two constant pressure legs. Sketches of an engine and the corresponding cycle are given in Figure A-4. Turbine and nozzle Heat rejection to atmosphere Inlet and compressor Combustor q2 PCompressor exit Patm q1 a d P V b c Combustor Compressor Turbine Inlet Nozzle Figure A-4: Sketch of the jet engine components and corresponding thermodynamic states Gas turbines are also used for power generation and for closed cycle operation (for example for space power generation). A depiction of the cycle in this case is shown in Figure A-5
Equivalent heat transfer at constant pressure Compressor Turbine Equivalent heat transfer at constant pressure Figure A-5: Thermodynamic model of gas turbine engine cycle for power generation The objective now is to find the work done, the heat absorbed, and the thermal efficiency of the cycle. Tracing the path shown around the cycle from a-b-c-d and back to a, the first law gives (writing the equation in terms of a unit mass) Aua-b-c-d-a =0=92+91-w The net work done is where qu q2 are defined as heat received by the system( q, is negative). We thus need to evaluate the heat transferred in processes b-c and d-a. For a constant pressure process the heat exchange per unit mass is dh=CpdT=dq, or [dq]onstant p=dh The heat exchange can be expressed in terms of enthalpy differences between the relevant states Treating the working fluid as an ideal gas, for the heat addition from the combustor, he-hb=Cp(T-T The heat rejected is, similarly, q1=ha-ha=Cp(Ta-Ta) The net work per unit mass is given by Net work per unit mass=q1+q2=CpI(T-b)+(Ta-Ta)] The thermal efficiency of the Brayton cycle can now be expressed in terms of the temperatures 1A-6
1A-6 Wnet 2 3 1 4 Equivalent heat transfer at constant pressure Equivalent heat transfer at constant pressure Turbine Compressor ⋅ Wcomp ⋅ Q ⋅ Q ⋅ Figure A-5: Thermodynamic model of gas turbine engine cycle for power generation The objective now is to find the work done, the heat absorbed, and the thermal efficiency of the cycle. Tracing the path shown around the cycle from a-b-c-d and back to a, the first law gives (writing the equation in terms of a unit mass), ∆u qqw abcda −−− − == + − 0 2 1 . The net work done is wq q = +2 1, where q q 1 2 , are defined as heat received by the system ( q1 is negative). We thus need to evaluate the heat transferred in processes b-c and d-a. For a constant pressure process the heat exchange per unit mass is dh c dT dq = = p , or dq dh cons t P [ ] = tan . The heat exchange can be expressed in terms of enthalpy differences between the relevant states. Treating the working fluid as an ideal gas, for the heat addition from the combustor, q h h cT T 2 =−= − c b pc b ( ). The heat rejected is, similarly, q h h cT T 1 =−= − a d p a ( ) d . The net work per unit mass is given by Net work per unit mass = qq cTT T T 1 2 += − pc b a [ ] ( ) + − ( ) d . The thermal efficiency of the Brayton cycle can now be expressed in terms of the temperatures:
Net work_cp(T-TD)-(Ta-Ta). Heat in (7- (-0)1、(T4-1) T(T/7-1) (A3.1) To proceed further, we need to examine the relationships between the different temperatures. We know that points a and d are on a constant pressure process as are points b and c, and Pa=Pd; P=P. The other two legs of the cycle are adiabatic and reversible, so Pa P /)- y P T T Therefore -=-,or, finally,=. Using this relation in the expression for thermal T efficiency, Eq (A1.3) yields an expression for the thermal efficiency of a Brayton cycle Ideal Brayton cycle efficiency: nB=1-A (A3.2) Th pressor exit The temperature ratio across the compressor, Tb/Ta=TR. In terms of compressor temperature ratio, and using the relation for an adiabatic reversible process we can write the efficiency in terms of the compressor(and cycle) pressure ratio, which is the parameter commonly used y-1)/y (A.33) R Figure A-6 shows pressures and temperatures through a gas turbine engine(the afterburning J57 which powers the F-8 and the F-101) AFTERBURNING MILITARY TURBOJET TYPICAL SEA LEVEL STATIC INTERNAL PRESSURES AND TEMPERATURES DATA FOR PRATT WHITNEY J57"B" SERIES AXIMUM AFTERBURNERJ ITITTTT STATION 2 pt(psia)14.7 540167.0158036.0 °F)59 330660 15701013 2540 Figure A-6: Gas turbine engine pressures and temperatures
1A-7 η = = [ ] ( ) − − − ( ) [ ] − Net work Heat in cTT T T cT T pc b d a pc b = − ( ) − ( ) − = − ( ) − ( ) − 1 1 1 1 T T T T TT T TT T d a c b a d a bc b / / . (A.3.1) To proceed further, we need to examine the relationships between the different temperatures. We know that points a and d are on a constant pressure process as are points b and c, and PPPP a = = d b c ; . The other two legs of the cycle are adiabatic and reversible, so P P P P T T T T d c a b d c a b = == = ( ) − ( ) − > γ γ / 1 γ γ / 1 . Therefore T T T T d c a b = , or, finally, T T T T d a c b = . Using this relation in the expression for thermal efficiency, Eq. (A.1.3) yields an expression for the thermal efficiency of a Brayton cycle: Ideal Brayton cycle efficiency:η B a b T T = −1 (A.3.2) = −1 T T atmospheric compressor exit . The temperature ratio across the compressor, T T TR b a / = . In terms of compressor temperature ratio, and using the relation for an adiabatic reversible process we can write the efficiency in terms of the compressor (and cycle) pressure ratio, which is the parameter commonly used: η B γ γ TR PR =− =− ( )( ) − 1 1 1 1 1 / . (A.3.3) Figure A-6 shows pressures and temperatures through a gas turbine engine (the afterburning J57, which powers the F-8 and the F-101). Figure A-6: Gas turbine engine pressures and temperatures
Equation(A3.3)says that for a high cycle efficiency, the pressure ratio of the cycle should be increased. Figure A-7 shows the history of aircraft engine pressure ratio versus entry into service and it can be seen that there has been a large increase in cycle pressure ratio. The thermodynamic concepts apply to the behavior of real aerospace devices Trent 890 Trent 77 cF6-80c2A8 CF6-80E1A4 CF6-80C2A8 M565c RB211524D4 CF6-50A RB211-22 CFM56-5B TF39-1 CF6-6d JT9D-70 CFM56-3C JT9D-3A Spey 512Spey 512-14 JT8D-219 BD- Spey 505- JT8D-17 Tay 611 Tay 651 占10 0 2000 Year of certification Figure A-7: Gas turbine engine pressure ratio trends (Janes Aeroengines, 1998) Muddy points When flow is accelerated in a nozzle, doesn t that reduce the internal energy of the flow and therefore the enthalpy? (MP 1A.5 Why do we say the combustion in a gas turbine engine is constant pressure?(MP 1A.6 Why is the Brayton cycle less efficient than the Carnot cycle? (MP 1A.7) still within the system boundary? (MP 1A(( g in the exhaust outside tsg-that Does it matter what labels we put on the corners of the cycle or not?(MP 1A.9) u the work done in the compressor always equal to the work done in the turbine plus Is vork out(for a Brayton cyle)?(MP 1A.10) L.A. 4 Gas Turbine Technology and Thermodynamics The turbine entry temperature, T, is fixed by materials technology and cost. (If the temperature is too high, the blades fail. )Figures A-8 and A-9 show the progression of the turbine entry temperatures in aeroengines. Figure A-8 is from Rolls royce and Figure A-9 is from Pratt whitney. Note the relation between the gas temperature coming into the turbine blades and the blade melting temperature
1A-8 Equation (A.3.3) says that for a high cycle efficiency, the pressure ratio of the cycle should be increased. Figure A-7 shows the history of aircraft engine pressure ratio versus entry into service, and it can be seen that there has been a large increase in cycle pressure ratio. The thermodynamic concepts apply to the behavior of real aerospace devices! Figure A-7: Gas turbine engine pressure ratio trends (Jane’s Aeroengines, 1998) Muddy points When flow is accelerated in a nozzle, doesn’t that reduce the internal energy of the flow and therefore the enthalpy? (MP 1A.5) Why do we say the combustion in a gas turbine engine is constant pressure? (MP 1A.6) Why is the Brayton cycle less efficient than the Carnot cycle? (MP 1A.7) If the gas undergoes constant pressure cooling in the exhaust outside the engine, is that still within the system boundary ? (MP 1A.8) Does it matter what labels we put on the corners of the cycle or not? (MP 1A.9) Is the work done in the compressor always equal to the work done in the turbine plus work out (for a Brayton cyle) ? (MP 1A.10) 1.A.4 Gas Turbine Technology and Thermodynamics The turbine entry temperature, Tc , is fixed by materials technology and cost. (If the temperature is too high, the blades fail.) Figures A-8 and A-9 show the progression of the turbine entry temperatures in aeroengines. Figure A-8 is from Rolls Royce and Figure A-9 is from Pratt&Whitney. Note the relation between the gas temperature coming into the turbine blades and the blade melting temperature. 1960 0 10 20 JT3D Conway 508 Conway 550 JT8D-17 JT8D-219 CFM56-3C PW4084 PW4168 CFM56-5B CFM56-5C4 CFM56-2 JT8D-1 Tay 611 Spey 555 Spey 512 JT9D-3A JT9D-70 TF39-1 RB211-22 RB211-524D4 CF6-50E PW4052 CF6-80C2A8 CF6-50A CF6-6 JT9D-7R4G Spey 512-14 Spey 505 Tay 651 GE90 Trent 890 Trent 775 30 40 1970 1980 Year of Certification Overall Pressure Ratio (OPR), Sea Level, T-O 1990 2000 CF6-80C2A8 CF6-80E1A4
十端 Demonstration Take off turbine Engine TET temperatur wooL D Materid capablity cast alloys Convent ly cast alloys 19401950196019601980199020002010 A gure A-8: Rolls-Royce high temperature Fi Figure A-9: Turbine blade coolin technology [Pratt WhitneyI For a given level of turbine technology (in other words given maximum temperature)a design question is what should the compressor tr be? What criterion should be used to decide this? Maximum thermal efficiency? Maximum work? We examine this issue below. Cycle with Th→T。 P Cycle with lower T T=T Figure A-10: Efficiency and work of two Brayton cycle engines The problem is posed in Figure A-10, which shows two Brayton cycles. For maximum efficiency we would like TR as high as possible. This means that the compressor exit temperature approaches the turbine entry temperature. The net work will be less than the heat received; as Tb T the heat received approaches zero and so does the net work The net work in the cycle can also be expressed as fPdv, evaluated in traversing the cycle. This is the area enclosed by the curves, which is seen to approach zero as T-T
1A-9 Figure A-8: Rolls-Royce high temperature Figure A-9: Turbine blade cooling technology technology [Pratt & Whitney] For a given level of turbine technology (in other words given maximum temperature) a design question is what should the compressor TR be? What criterion should be used to decide this? Maximum thermal efficiency? Maximum work? We examine this issue below. Figure A-10: Efficiency and work of two Brayton cycle engines The problem is posed in Figure A-10, which shows two Brayton cycles. For maximum efficiency we would like TR as high as possible. This means that the compressor exit temperature approaches the turbine entry temperature. The net work will be less than the heat received; as T T b c → the heat received approaches zero and so does the net work. The net work in the cycle can also be expressed as ∫ Pdv, evaluated in traversing the cycle. This is the area enclosed by the curves, which is seen to approach zero as T T b c → . V T = Tatm = Ta T = Tc Cycle with Tb → Tc Tb2 Tb1 Patm P Cycle with lower Tb
