Ciclo Brayton - Sistemas de Potência a Gás e Termodinâmica - Exercícios Resolvidos

UNIVERSIDADE ESTADUAL PAULISTA Departamento de Engenharia Mecânica Termodinâmica II Prof Dr Délson Luiz Módolo Capítulo 9 Sistemas de Potência à Gás Ciclo Brayton Energia solar Unidade de armazen amento solar pressurizada Compressor Turbina Ar T1310 K p11 bar nt08 T2520 K p25 bar T3760 K T41140 K Trocador de calor Escape 1 bar Fig P968 972 O diagrama esquemático Ts de uma turbina de dois estágios operando sob regime estacionário com reaquecimento a uma pressão p1 entre os dois estágios encontrase representada na Fig P972 Valores de p1 T1 e p2 são conhecidos A temperatura na entrada de cada turbina é a mesma as expansões nas turbinas são isentrópicas e efeitos de energia cinética e potencial são desprezíveis Assumindo o modelo de gás ideal com k constante para o ar mostre que a o trabalho máximo por unidade de massa de ar fluindo é alcançado quando a razão de pressão é a mesma em ambos os estágios b a temperatura na saída de cada estágio de turbina é o mesmo c o trabalho por unidade de massa de ar fluindo é o mesmo para cada estágio de turbina d o calor transferido por unidade de massa fluindo é igual ao trabalho determinado em c Qentra Reaquecedor combustor Turbina estágio 1 Turbina estágio 2 Wt p1 pi variável Fig P972 974 Ar é admitido sob 100 kPa 300 K e 6 kgs em um compressor de um ciclo arpadrão Brayton com regeneração e reaquecimento A razão de pressão do compressor é 10 e a temperatura de entrada em cada estágio de turbina é 1400 K As razões de pressão para os estágios de turbina são iguais Os estágios de turbina e compressor têm eficiências isentrópicas de 80 e o regenerador tem uma efetividade de 80 Para k 14 calcule a a eficiência térmica do ciclo b a razão de trabalho reverso c a potência líquida em kW d as razões de destruição de exergia no compressor e em cada estágio de turbina assim como no regenerador em kW para T0 300 K Outras Aplicações para Sistemas de Potência a Gás 985 Ar a 26 kPa 230 K e 220 ms entra no motor de um turbojato em voo A vazão mássica de ar é 25 kgs A razão de pressão ao longo do compressor é 11 a temperatura na entrada da turbina é 1400 K e a pressão de saída no bocal é 26 kPa Os processos no difusor e no bocal são isentrópicos as eficiências isentrópicas do compressor e da turbina valem respectivamente 85 e 90 e não há perda de carga no escoamento ao longo do combustor Os efeitos de energia cinética são desprezíveis exceto na entrada do difusor e na saída do bocal Tomando como base uma análise de arpadrão determine a as pressões e temperaturas em cada estado principal em kPa e K respectivamente b a taxa de adição de calor para o ar que passa através do combustor em kJs c a velocidade na saída do bocal em ms SOLUÇÃO DOS EXERCÍCIOS Capítulo 9 Sistemas de Potência à Gás Ciclo Brayton Problem 942 Continued Page 2 4 There are no pressure drops for flow through the heat exchangers 5 Kinetic and potential energy effects are negligible 6 The working fluid is air modeled as an ideal gas ANALYSIS The Ts diagram for the cycle is shown below a The mass flow rate of air is found as follows Mass and energy rate balances for control volumes enclosing the turbine and compressor give Wiṁh3h4 and Wcṁh2h1 The net power of the cycle is ẆcycleẆiẆcṁh3h4 h2h1 Solving for ṁ ṁẆcycleh3h4 h2h1 Inserting values ṁ10000 kW157557 kJkg80078 kJkg61065 kJkg30019 kJkgkJs kW 2154 kgs b The rate of heat transfer to the working fluid passing through the heat exchanger can be determined by applying mass and energy balances to a control volume around the heat exchanger to give Qinṁh3h42154 kgs157557 kJkg61065 kJkgkW kJs20784 kW c The thermal efficiency is ηẆcycle Qin10000 kW207844 kW0481 481 PROBLEM 945 945 For an ideal Brayton cycle on a cold airstandard basis show that a the back work ratio is given by bwr T1T4 where T1 is the temperature at the compressor inlet and T4 is the temperature at the turbine exit b the temperature at the compressor exit that maximizes the net work developed per unit of mass flowing is given by T2 T1T312 where T1 is the temperature at the compressor inlet and T3 is the temperature at the turbine inlet KNOWN An ideal cold airstandard Brayton cycle is under consideration FIND a Show the back work ratio takes a specific form b Show that the compressor exit temperature that maximizes the net work developed per unit of mass takes a specific form SCHEMATIC GIVEN DATA ENGINEERING MODEL See Example 95 a bwr h2h1h3h4 cpT2T1cpT3T4 T1T2T1 1 T4T3T4 1 For the isentropic processes T2T1 P2P1k1k T3T4 P3P4k1k Then since P3 P4 and P4 P1 we get T2T1 T3T4 Finally bwr T1T4 as required b Using Eq a of the solution to Example 95 the compressor pressure ratio for maximum net work developed per unit of mass flow is P2P1 T2T1kk1 Further as noted in part a for the isentropic compression P2P1 T2T1kk1 Collecting results T2T1k1 T2T1kk1 T2T1 T3T112 Finally T2 T1T312 as required PROBLEM 946 KNOWN Air enters a cold airstandard ideal Brayton cycle with a given flow rate and at a specified state The compressor pressure ratio and maximum cycle temperature are known FIND Determine a the thermal efficiency b the back work ratio and c the net power SCHEMATIC GIVEN DATA T T31400K P2P110 P1 100 kPa T1 300 K S ENGINEERING MODEL See Example 94 Also assume k14 and cp 1005 kJkgK ANALYSIS First determine T2 and T4 as follows Eg 923 T2 T1P2P1k1k 3001041414 5792 K Eg 924 T4 T3P1P4k1k 140011041414 72513 K a To evaluate thermal efficiency use ηt 1 QoutQin Qin Qin ṁh3h2 ṁ cpT3T2 6 kgs1005 kJkgK14005792k 49494 kJs Qout Qout ṁh4h1 ṁ cpT4T1 6100572513300 25635 kJs Thus ηt 1 QoutQin 1 2563549494 0482 482 b The back work ratio is bwr WcWt Wc ṁh2h1 ṁ cpT2T1 610055792 300 16836 kJs Wt ṁh3h4 ṁ cpT3T4 61005140072513 40695 kJs 1 and bwr WcWt 1683640695 04137 c The net power developed is Wcycle Wt Wc 40695 16836 kJs 23859 kW 1 Using the result of Problem 945a bwr T1T4 300K72513K 04137 which agrees with the value determined as expected PROBLEM 950 KNOWN Air enters the compressor of an ideal Brayton cycle with known conditions The turbine inlet temperature is specified FIND Determine for various compressor pressure ratios a the thermal efficiency b the back work ratio and c the net power developed SCHEMATIC GIVEN DATA T T31400K P2P1 6 8 12 T1 300K S ENG MODEL See Example 94 ANALYSIS First fix each of the principal states Table A22 State 1 T1300K h130019 kJkg Pr113860 State 2 For isentropic compression Pr2 P2P1Pr1 T2 h2 Thus P2P1 6 Pr2 8316 T24984K h250136 kJkg 8 Pr2 11088 T25298 K h2 54418 kJkg 12 Pr3 16632 T3 6034 K h3 61065 kJkg State 3 T31400K h3151542 kJkg Pr3 4505 State 4 For the isentropic expansion Pr4 P4P3Pr3 T4 h4 Thus Pr4 75083 T48993 K h493220 kJkg 56312 T48349 K h486040 kJkg 37542 T4751 K h4 76838 kJkg a The thermal efficiency is ηt 1 h4 h1h3 h2 b The back work ratio is bwr h2 h1h3 h4 c The net power developed is Wnet ṁ h3 h4 h2 h1 AV P1RT1 h3 h4 h2 h1 4 The results are summarized in the following table P2P1 η bwr Wnet kW 6 3767 345 22186 8 4232 372 23868 12 4875 416 25552 1 As the compressor ratio increases ηt increases bwr increases and net power developed increases 953 The cycle of Problem 942 is modified to include the effects of irreversibilities in the adiabatic expansion and compression processes If the states at the compressor and turbine inlets remain unchanged the cycle produces 10 MW of power and the compressor and turbine isentropic efficiencies are both 80 determine a the pressure kPa temperature K and enthalpy kJkg at each principal state of the cycle and sketch the Ts diagram b the mass flow rate of air in kgs c the rate of heat transfer in kW to the working fluid passing through the heat exchanger d the thermal efficiency KNOWN An airstandard Brayton cycle operates with known states at the turbine and compressor inlets and known compressor and turbine isentropic efficiencies The net power output of the cycle is given FIND Determine the mass flow rate of air the rate of heat transfer to the working fluid passing through the heat exchanger and the thermal efficiency SCHEMATIC AND GIVEN DATA p2 1200 kPa T3 1450 K p3 p4 1200 kPa ηc 80 ηt 80 T1 300 K p1 p4 100 kPa p4 100 kPa ENGINEERING MODEL 1 Each component is analyzed as a control volume at steady state The control volumes are shown on the accompanying sketch by dashed lines 2 The turbine and compressor operate adiabatically 3 There are no pressure drops for flow through the heat exchangers 4 Kinetic and potential energy effects are negligible 5 The working fluid is air modeled as an ideal gas Problem 953 Continued Page 3 In summary State p kPa T K h kJkg 1 100 300 30019 2s 1200 6035 61065 2 1200 6767 68827 3 1200 1450 157557 4s 100 7807 80078 4 100 9203 95574 b The mass flow rate of air is found as follows Mass and energy rate balances for control volumes enclosing the turbine and compressor give Wt ṁh3 h4 and Wo ṁh2 h1 The net power of the cycle is Wcycle Wt Wo ṁh3 h4 h2 h1 Solving for ṁ ṁ Wcycle h3 h4 h2 h1 Inserting values ṁ 10000 kW 157557 kJkg 95574 kJkg 68827 kJkg 30019 kJkg 4315 kgs c The rate of heat transfer to the working fluid passing through the heat exchanger can be determined by applying mass and energy balances to a control volume around the heat exchanger to give Qin ṁh3 h2 4315 kgs x 157557 kJkg 68827 kJkg 38287 kW d The thermal efficiency is η Wcycle Qin 10000 kW382870 kW 02612 2612 PROBLEM 954 KNOWN Air enters the compressor of an airstandard Brayton cycle at a specified state and a given volumetric flow rate The compressor pressure ratio and maximum cycle temperature are known The compressor and turbine isentropic efficiencies are known FIND Determine a the net power b the rate of heat addition and c the thermal efficiency SCHEMATIC GIVEN DATA T3 2100 K p2p1 20 T1 280 K p1 08 bar ηc 092 ηt 095 ṁ Wcycle ENGINEERING MODEL See Example 97 Also ηc 092 ηt 095 ANALYSIS First fix each of the principal states State 1 T1 280 K h1 28013 kJkg Pr 10089 State 2 Pr2 P2P1 Pr1 20100889 21778 T2s 6493 K h2s 65913 kJkg Using the isentropic compressor efficiency ηc h2s h1 h2 h1 h2 h1 h2s h1 ηc 28013 65913 28013 092 69209 kJkg State 3 T3 2100 K h3 23774 kJkg p3 2559 State 4 Pr4 P4P3 Pr3 12559 0000391 T4s 10292 K h4s 10794 kJkg Using the isentropic turbine efficiency ηt h3 h4 h3 h4s h4 h3 ηt h3 h4s 23774 09523774 10794 11443 kJkg Now determine the mass flow rate ṁ AV vi AV p1 R T1 60 m3s 08 bar 8314 kJkgk 280 K 5973 kgs a Wc ṁ h2 h1 597369209 28013 kJs 2461 x 104 kW Wt ṁ h3 h4 597323774 11443 kJs 7365 x 104 kW Wcycle Wt Wc 4904 x 104 kW b Qin ṁ h3 h2 597323774 69209 10066 x 105 kW c η Wcycle Qin 4904 x 104 10066 x 105 0487 487 Problem 959 Continued Page 2 ENGINEERING MODEL 1 Each component is analyzed as a control volume at steady state The control volumes are shown on the accompanying sketch by dashed lines 2 All processes of the working fluid are internally reversible 3 The turbine and compressor operate adiabatically 4 There are no pressure drops for flow through the regenerator and combustor 5 Kinetic and potential energy effects are negligible 6 The working fluid is air modeled as an ideal gas ANALYSIS The Ts diagram for the cycle is shown below a The mass flow rate of air is found as follows Mass and energy rate balances for control volumes enclosing the turbine and compressor give The net power of the cycle is Solving for m Inserting values b The rate of heat transfer to the working fluid passing through the combustor can be determined by applying mass and energy balances to a control volume around the combustor to give c The thermal efficiency is PROBLEM 968 KNOWN Data are known for a regenerative gas turbine power plant using solar energy as the source of heat addition FIND Determine a the thermal efficiency and b the mass flow rate for a net power output of 500 kW SCHEMATIC GIVEN DATA ENGINEERING MODEL 1 Each component is modeled as a control volume at steady state 2 The compressor and turbine operate adiabatically 3 There are no stray heat loses or pressure drops in the heat exchangers or interconnecting piping 4 Kinetic and potential energy effects are negligible 5 The working fluid is air modeled as an ideal gas ANALYSIS For states 1 4 the temperatures are known Thus T1 310 K h1 31024 kJkg T2 520 K h2 52363 kJkg T3 760 K h3 77818 kJkg T4 1140 K h4 120757 kJkg State 5 Pr5 P5P4 Pr4 154931 3862 h55 77479 kJkg Using nt h4 h5 ch4 hss hs h4 nth4 hss 86111 kJkg a The thermal efficiency is b For Wcycle 500 kW PROBLEM 970 KNOWN Air expands in two stages through a turbine with reheat between the stages The states are specified at the inlet and exit of each component FIND Determine per unit mass of air flowing a the work developed by each stage b the heat transfer for reheat and c the increase in net work compared to a single stage of expansion with no reheat SCHEMATIC GIVEN DATA ENGINEERING MODEL 1 Each control volume is at steady state 2 The turbines operate isentropically 3 Kinetic and potential energy effects are negligible 4 The working fluid is air modeled as an ideal gas ANALYSIS First fix each of the principal states Table A22 State 1 T1 1200 K h1 127779 kJkg P1 2380 State 2 Pr2 P2P1 Pr1 69417 h2 91211 kJkg State 3 T3 1200 K h3 h1 127779 Pr3 Pr1 2380 State 4 Pr4 P4P3 Pr3 68 h4 90685 kJkg a The work developed by each stage is b For the reheater c To determine the work for a single stage of expansion determine ha as follows Pra PaPr Pr1 19833 ha 63858 kJkg Thus and increase x100 152 PROBLEM 972 The schematic and Ts diagram of a twostage turbine operating at steady state with reheat at pressure p1 between the two stages is shown in Fig P972 Values for p1 T1 and p2 are known The temperature at the inlet to each turbine stage is the same the turbine expansions are isentropic and kinetic and potential energy effects are negligible Assuming the ideal gas model with constant k for the air show that a the maximum total work per unit of mass flowing is developed when the pressure ratio is the same across each stage b the temperature at the exit of each turbine stage is the same c the work developed per unit mass of air flowing is the same for each turbine stage d the heat transfer per unit of mass flowing equals the work value determined in part c KNOWN A twostage turbine with reheat operates at steady state under specified conditions FIND On the basis of a cold airstandard analysis demonstrate four distinct performance features ENGINEERING MODEL 1 Each component is analyzed as a control volume at steady state 2 The turbine expansions are isentropic 3 There is no pressure change through the reheater 4 The temperatures at the turbine inlets are equal 5 Kinetic and potential effects are negligible 6 The working fluid is air modelled as an ideal gas with constant k and cp ANALYSIS The turbine work developed per unit of mass flowing is Wtm h1 ha hb h2 cpT1 Ta Tb T1 With T1 Tb Wtm cp 2T1 Ta T2 cpT1 2 TaT1 T2T1 For the isentropic expansions Eq 643 gives TgT1 PiP1k1k and T2T1 P2Pik1k 1 Thus Wtm cpT12 PiP1k1k P2Pik1k To determine the pressure pi that maximizes the total form the derivative dWtmdpi cpT1k1kPiP1k1k 11P1 P2Pik1kP2pi2 cpT1k1k1piPiP1k1k P2P1k1k For dWtmdpi 0 PiP1 P2Pi a By checking the second derivative it can be verified that the total turbine work is a maximum PROBLEM 974 KNOWN Air enters a cold airstandard regenerative Brayton cycle with reheat at a specified state with a given mass flow rate The compressor pressure ratio maximum cycle temperature and reheat temperature are known The compressor and turbines each have isentropic efficiencies of 08 and the regenerator effectiveness is 08 FIND Determine a the thermal efficiency b the back work ratio c the net power and d the rates of exergy destruction in the compressor each turbine stage and the regenerator for T0 300K SCHEMATIC GIVEN DATA Engineering Model See Example 911 Also ηcηt1ηt208 and ηreg08 The specific heats are constant with k14 and cp1005 kJkgK Let T0300K ANALYSIS From the solution to Problem 955 T2649K To find T3 first find Tas as follows Tas PaP3k1k T3 31623100004141400 10076K Using the firststage turbine efficiency ηt1 h3 ha h3 h4s cpT3 Ta cpT3 Tas Thus Ta T3 ηt1T3 Tas 1400 08140010076 10861K Similarly for the secondstage turbine T4s P4Pbk1k T0 1003162304141400 10076K T4 Tb ηt2CTb T4s 10861K The regenerator effectiveness is ηreg hx hz h4 hz cpTx Tz cpT4 Tz or Tx Tz ηregT4 Tz 9987K For the control volume enclosing the regenerator 0 hz hx h4 hy 0 cpΤz Tx cpT4 Ty or Ty T4 Tx Tz 7364K a The rate of heat addition is Qin Q1 Q2 ṁcpT3 Tx cpTb Ta 6 kgs1005 kJkgK14009987 140010861 43127 kJs And Qout ṁ cpTy T0 610057364 300 26315 kJs PROBLEM 974 Contd Page 2 The thermal efficiency is η 1 Qout Qin 1 26315 43127 0390 390 b Wc ṁ cp T2 T1 6 kgs1005 kJkgK649 300K 21045 kJs For turbine 1 Wt1 ṁ cp T3 Ta 610051400 10861 18928 kJs Since Tb T3 and Ta T4 We Wt1 18928 kJs Thus Wt 218928 37856 kJs and bwr Wc Wt 21045 37856 0556 bwr c The net power is Wcycle Wt Wc 37856 21045 kW 16811 kW Wcycle d For the compressor O ΣQTj0 ṁs1 s2 δwcomp Ėdcomp T0σcvcomp T0ṁcp lnT2T1 R lnp2p1 300K6 kgs1005 ln649300 8314 kJkmolK 2897 kgkmol ln1000100 Ėdcomp 2065 kW Turbine stage 1 Ėdturb1 T0σwturb1 T0ṁsa s3 Ėdturb1 T0ṁcp lnTaT3 R lnPaP3 30061005 ln108611400 83142897 ln316231000 Ėdturb1 1355 kW Since Tb T3 T4 Ta and PaP3 PbPb Ėdturb2 Ėdturb1 1355 kW Finally for the regenerator O ΣQTj0 ṁs2 s1 σcvreg and Ėdreg T0σcvreg T0ṁcp lnTxT2 R lnPbP2 cp lnTyT4 R lnPaPb 30061005ln9987649 ln736410861 768 kW Ėdreg PROBLEM 985 KNOWN A turbojet engine is analyzed on an airstandard basis Data are known at various locations and the mass flow rate is specified FIND Determine a the pressures and temperatures at each principal state b the rate of heat addition and c the velocity at the nozzle exit SCHEMATIC GIVEN DATA ENGINEERING MODEL Same as in Example 913 except ηc085 and ηt 090 ANALYSIS a Fix each of the principal states State a Ta 230 K Pa 26 kPa ha 23002 kJkg Pra 05477 State 1 From an energy balance for the diffuser 0 ha V2a2 h1 Thus h1 ha V2a2 23002 2202 m2s22 25422 kJkg Ti 25415 K T See A22 Interpolating in TSee A22 Pr1 077759 For this isentropic process from a to 1 P1 Pr1Pra Pa 07775905477 26 kPa 3691 kPa State 2 Pr2s Pr1 P3P2 07775911 85535 Than h2s 50539 kJkg With the compressor efficiency h2 h1 h2s h1ηc 25422 50539 25422085 54971 kJkg T2 5452 K P2 113691 kPa 40601 kPa State 3 T3 1400 K h3 151542 kJkg P3 P2 40601 kPa State 4 The turbine is assumed to drive the compressor only Wc Wt h2 h1 h3 h4 h4 h3 h2 h1 151542 54971 25422 121993 kJkg From Table A22 T4 11506 K Pr4 200564 With the turbine efficiency h4s h3 h3 h4ηt 151542 151542 12199309 118710 kJkg Interpolating Pr4s 18132 Then P5 P3 Pr5Pr3 40601 kPa 181324505 1634 kPa Pas PROBLEM 985 Continued State 5 Pr5 Pr4 P5P4 200564 261634 31913 Interpolating h5 73412 kJkg Ts 7193 K a Summary State p kPa T K h kJkg 1 3691 25415 25422 2 40601 5452 54971 3 40601 1400 151542 4 1634 11506 121993 5 26 7193 73412 b 𝑄𝑖𝑛 ṁ h3 h2 25 kgs151542 54971 kJkg 2414 x 104 kJs c For the nozzle 0 h4 h5 V52 V252 Thus V5 2h4 h5 2121993 73412 kJkg 103 Nm 1 kJ kgms2 1 N 9857 ms