Apresentação em tema: "The First Law of Thermodynamics & Cyclic Processes"— Transcrição da apresentação:
1The First Law of Thermodynamics & Cyclic Processes Meeting 7 Section 4-1
2Thermodynamic Cycle Is a series of processes which form a closed path. The initial and the final states are coincident.
3For What Thermodynamics Cycles Are For? Thermal engines work in a cyclic process.A Thermal engines draws heat from a hot source and rejects heat to a cold source producing work
4Heat Engine Power Cycles Hot body or sourceQinSystem, or heat engineWcycleQoutCold body or sink
5Heat Engine Efficiency Hot body or sourceCold body or sinkSystem, or heat engineQHQLWcycle
6Refrigerators and heat pumps Hot body or sourceCold body or sinkSystemQoutQinWcycleREFRIGERATORHEAT PUMP
7Energy analysis of cycles For the cycle, E1 E1 = 0, or4321
8Qcycle = Wcycle For cycles, we can write: Qcycle and Wcycle represent net amountswhich can also be represented as:
9TEAMPLAYA closed system undergoes a cycle consisting of two processes. During the first process, 40 Btu of heat is transferred to the system while the system does 60 Btu of work. During the second process, 45 Btu of work is done on the system.(a) Determine the heat transfer during the second process.(b) Calculate the net work and net heat transfer of the cycle.
15The Carnot cycle for a gas might occur as visualized below. TL THTH TLQLTL
16Execution of the Carnot Cycle in a Closed System (Fig. 5-43)
17It is always reversible -- a Carnot cycle is reversible by definition. This is a Carnot cycle involving two phases it is still two adiabatic processes and two isothermal processes.It is always reversible -- a Carnot cycle is reversible by definition.TLTLTL
18Vapor Power CyclesWe’ll look specifically at the Rankine cycle, which is a vapor power cycle.It is the primary electrical producing cycle in the world.The cycle can use a variety of fuels.
28Question ….. $467,952.27/day $170,801,979/year How much does it cost to operate a gas fired 1000 MW(output) power plant with a 35% efficiency for 24 hours/day for a full year if fuel cost are $2.00 per 106 Btu?$467,952.27/day$170,801,979/year
29Question …. $12,998.67/day $4,744,499/year If you could improve the efficiency of a 1000 MW power plant from 35% to 36%, what would be a reasonable charge for your services? Let’s assume $2.00 per million BTUs fuel charge and 24 hr/day operation.$12,998.67/day$4,744,499/year
32Carnot Vapor Cycle Low thermal efficiency compressor and turbine must handle two phase flows
33Carnot Vapor CycleThe Carnot cycle is not a suitable model for vapor power cycles because it cannot be approximated in practice.
34Rankine CycleThe model cycle for vapor power cycles is the Rankine cycle which is composed of four internally reversible processes: constant-pressure heat addition in a boiler, isentropic expansion in a turbine, constant-pressure heat rejection in a condenser, and isentropic compression in a pump. Steam leaves the condenser as a saturated liquid at the condenser pressure.
35Refrigerator and Heat Pump Objectives The objective of a refrigerator is to remove heat (QL) from the cold medium; the objective of a heat pump is to supply heat (QH) to a warm medium
42Performance of cycleThermal Efficiency:Need to know QH and QL
43Otto Cycle Heat addition 2-3 QH = mCV(T3-T2) Heat rejection 4-1 QL = mCV(T4-T1)or in terms of the temperature ratiosOtto Cycleqoutqin
441-2 and 3-4 are adiabatic process, using the adiabatic relations between T and V Otto Cycleqoutqin
45Cycle performance with cold air cycle assumptions This looks like the Carnot efficiency, but it is not! T1 and T2 are not constant.What are the limitations for this expression?
46Thermal Efficiency of Ideal Otto Cycle Under cold-air-standard assumptions, the thermal efficiency of the ideal Otto cycle is where r is the compression ratio and k is the specific heat ratio Cp /Cv.
47Effect of compression ratio on Otto cycle efficiency k = 1.4
48Otto CycleThe thermal efficiency of the Otto Cycle increases with the specific heat ratio k of the working fluid
49Brayton CycleThis is another air standard cycle and it models modern turbojet engines.
57Other applications of Brayton cycle Power generation - use gas turbines to generate electricity…very efficientMarine applications in large shipsAutomobile racing - late 1960s Indy 500 STP sponsored cars
60Brayton CycleThe ideal cycle for modern gas-turbine engines is the Brayton cycle, which is made up of four internally reversible processes: isentropic compression, constant pressure heat addition, isentropic expansion, and constant pressure heat rejection.
61Turbojet Engine Basic Components and T-s Diagram for Ideal Turbojet Cycle
62P-v and T-s Diagrams for the Ideal Brayton Cycle
63Brayton Cycle 1 to 2--isentropic compression in the compressor 2 to 3--constant pressure heat addition (replaces combustion process)3 to 4--isentropic expansion in the turbine4 to 1--constant pressure heat rejection to return air to original state
64Brayton CycleBecause the Brayton cycle operates between two constant pressure lines, or isobars, the pressure ratio is important.The pressure ratio is not a compression ratio.
65Brayton cycle analysis Let’s assume cold air conditions and manipulate the efficiency expression:
66Brayton cycle analysis Using the isentropic relationships,Let’s define:
67Brayton CycleBecause the Brayton cycle operates between two constant pressure lines, or isobars, the pressure ratio is important.The pressure ratio is just that--a pressure ratio.A compression ratio is a volume ratio (refer to the Otto Cycle).
69Brayton cycle analysis Then we can relate the temperature ratios to the pressure ratio:Plug back into the efficiency expression and simplify:
70What does this expression assume? Ideal Brayton CycleWhat does this expression assume?
71Thermal Efficiency of Brayton Cycle Under cold-air-standard assumptions, the Brayton cycle thermal efficiency is where rp = Pmax/Pmin is the pressure ratio and k is the specific heat ratio. The thermal efficiency of the simple Brayton cycle increases with the pressure ratio.
74Evaporação a pressão constante Um sistema pistão cilindro contêm, inicialmente, três kg de H2O no estado de líquido saturado com 0.6 MPa.Calor é adicionado, vagarosamente, a água fazendo com que o pistão se movimente de tal maneira que a pressão seja constante.Quanto de trabalho é realizado pela água?Quanta energia deve ser transferida para a água de tal maneira que ao final do processo ela esteja no estado de vapor saturado?FronteiradoSistemaRepresentaçãodo processo
75processo a pressão const. Primeira Lei:1Q2 – 1W2 = U2 – U1mas o trabalho 1W2 = Patm*(V2-V1),Logo 1Q2 = (P2V2+U2)-(P1V1+U1) = H2-H1Onde h2 é a entalpia do vapor saturado e h1 é a entalpia do líquido. Na tabela 1-2 termodinâmico para 0.6MPa, tem-se que h2 = 2756,8 KJ/kg e h1 = 670,56 KJ/kg. Considerando 3kg de H2O, então o calor transferido será de 3*( ) = 6259 KJ.
76Resfriamento com Gelo Seco 0.5 kg de gelo seco (CO2) a 1 atm é colocado em cima de uma fatia de picanha. O gelo seco sublima a pressão constante devido ao fluxo de calor transferido pela picanha. Ao final do processo todo CO2 está no estado de vapor (foi completamente sublimado).Determine a temperatura do CO2 e quanto de calor ele recebeu da picanha.Representaçãodo processoFronteiradoSistema
77Resfriamento com Gelo Seco – processo a pressão const. Primeira Lei:1Q2 – 1W2 = U2 – U1mas o trabalho 1W2 = Patm*(V2-V1),Logo 1Q2 = (P2V2+U2)-(P1V1+U1) = H2-H1Onde h2 é a entalpia do vapor saturado e h1 é a entalpia do sólido. No diagrama termodinâmico para Patm, tem-se que h2 = 340 KJ/kg e h1 = -220 KJ/kg. Considerando 0.5kg de CO2, então o calor transferido será de 280 KJ. A temperatura de saturação do CO2 será de 175K (-98 oC)