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The First Law of Thermodynamics & Cyclic Processes Meeting 7 Section 4-1.

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Apresentação em tema: "The First Law of Thermodynamics & Cyclic Processes Meeting 7 Section 4-1."— Transcrição da apresentação:

1 The First Law of Thermodynamics & Cyclic Processes Meeting 7 Section 4-1

2 Thermodynamic Cycle Is a series of processes which form a closed path. The initial and the final states are coincident.

3 For 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

4 Heat Engine Power Cycles Hot body or source Cold body or sink System, or heat engine Q in Q out W cycle

5 Heat Engine Efficiency W cycle Hot body or source Cold body or sink System, or heat engine QHQH QLQL

6 Refrigerators and heat pumps Hot body or source Cold body or sink System Q out Q in W cycle REFRIGERATOR HEAT PUMP

7 Energy analysis of cycles For the cycle, E 1 E 1 = 0, or

8 For cycles, we can write: Q cycle and W cycle represent net amounts which can also be represented as: Q cycle = W cycle

9 TEAMPLAY A 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.

10 TEAMPLAY B A W in = 45 Btu Q in =40 Btu W out =60 Btu 1 2

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12 Carnot Cycle The Carnot cycle is a reversible cycle that is composed of four internally reversible processes. – Two isothermal processes – Two adiabatic processes

13 Carnot cycle for a gas The area represents the net work TLTL

14 P-v Diagram of the Reversed Carnot Cycle TLTL

15 The Carnot cycle for a gas might occur as visualized below. TLTL QLQL T H T L T L T H

16 Execution of the Carnot Cycle in a Closed System (Fig. 5-43)

17 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. TLTL TLTL TLTL

18 Vapor Power Cycles Well 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.

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28 Question ….. 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 10 6 Btu? $467,952.27/day $170,801,979/year

29 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? Lets assume $2.00 per million BTUs fuel charge and 24 hr/day operation. Question …. $12,998.67/day $4,744,499/year

30 Vapor-cycle Power Plants

31 Well simplify the power plant

32 Carnot Vapor Cycle Low thermal efficiency compressor and turbine must handle two phase flows

33 Carnot Vapor Cycle The Carnot cycle is not a suitable model for vapor power cycles because it cannot be approximated in practice.

34 Rankine Cycle The 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.

35 Refrigerator and Heat Pump Objectives The objective of a refrigerator is to remove heat (Q L ) from the cold medium; the objective of a heat pump is to supply heat (Q H ) to a warm medium

36 Inside The Household Refrigerator

37 Ordinary Household Refrigerator

38 Gas Power Cycle A cycle during which a net amount of work is produced is called a power cycle, and a power cycle during which the working fluid remains a gas throughout is called a gas power cycle.

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40 Actual and Ideal Cycles in Spark-Ignition Engines v v

41 Otto Cycle q in q out P-V diagram (Work) T-S diagram (Heat Transfer)

42 Performance of cycle Thermal Efficiency: Need to know Q H and Q L

43 Otto Cycle Heat addition 2-3 Q H = mC V (T 3 -T 2 ) Heat rejection 4-1 Q L = mC V (T 4 -T 1 ) or in terms of the temperature ratios q out q in

44 Otto Cycle 1-2 and 3-4 are adiabatic process, using the adiabatic relations between T and V q out q in

45 T 1 and T 2 are not constant. This looks like the Carnot efficiency, but it is not! T 1 and T 2 are not constant. Cycle performance with cold air cycle assumptions What are the limitations for this expression?

46 Thermal 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 C p /C v.

47 Effect of compression ratio on Otto cycle efficiency k = 1.4

48 Otto Cycle The thermal efficiency of the Otto Cycle increases with the specific heat ratio k of the working fluid

49 Brayton Cycle This is another air standard cycle and it models modern turbojet engines.

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51 Brayton Cycle Proposed by George Brayton in 1870!

52 jet engine with afterburner for military applications.

53 Schematic of A Turbofan Engine

54 Illustration of A Turbofan Engine

55 Turboprop compressorturbine burner

56 Schematic of a Turboprop Engine

57 Other applications of Brayton cycle Power generation - use gas turbines to generate electricity…very efficient Marine applications in large ships Automobile racing - late 1960s Indy 500 STP sponsored cars

58 An Open-Cycle Gas-Turbine Engine

59 A Closed-Cycle Gas-Turbine Engine

60 Brayton Cycle The 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.

61 Turbojet Engine Basic Components and T-s Diagram for Ideal Turbojet Cycle

62 P-v and T-s Diagrams for the Ideal Brayton Cycle

63 Brayton 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 turbine 4 to 1--constant pressure heat rejection to return air to original state

64 Brayton Cycle Because the Brayton cycle operates between two constant pressure lines, or isobars, the pressure ratio is important. The pressure ratio is not a compression ratio.

65 Brayton cycle analysis Lets assume cold air conditions and manipulate the efficiency expression:

66 Using the isentropic relationships, Lets define: Brayton cycle analysis

67 Brayton Cycle Because 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).

68 Brayton Cycle The pressure ratio is Also

69 Brayton cycle analysis Then we can relate the temperature ratios to the pressure ratio: Plug back into the efficiency expression and simplify:

70 Ideal Brayton Cycle What does this expression assume?

71 Thermal Efficiency of Brayton Cycle Under cold-air-standard assumptions, the Brayton cycle thermal efficiency is where r p = P max /P min is the pressure ratio and k is the specific heat ratio. The thermal efficiency of the simple Brayton cycle increases with the pressure ratio.

72 Brayton Cycle k = 1.4

73 Thermal Efficiency of the Ideal Brayton Cycle

74 Um sistema pistão cilindro contêm, inicialmente, três kg de H 2 O 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? Evaporação a pressão constante Representação do processo Fronteira do Sistema

75 processo a pressão const. Primeira Lei: 1 Q 2 – 1 W 2 = U 2 – U 1 mas o trabalho 1 W 2 = P atm *(V 2 -V 1 ), Logo 1 Q 2 = (P 2 V 2 +U 2 )-(P 1 V 1 +U 1 ) = H 2 -H 1 Onde h 2 é a entalpia do vapor saturado e h 1 é a entalpia do líquido. Na tabela 1-2 termodinâmico para 0.6MPa, tem-se que h 2 = 2756,8 KJ/kg e h 1 = 670,56 KJ/kg. Considerando 3kg de H2O, então o calor transferido será de 3*( ) = 6259 KJ.

76 Resfriamento 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. Representação do processo Fronteira do Sistema

77 Resfriamento com Gelo Seco – processo a pressão const. Primeira Lei: 1 Q 2 – 1 W 2 = U 2 – U 1 mas o trabalho 1 W 2 = P atm *(V 2 -V 1 ), Logo 1 Q 2 = (P 2 V 2 +U 2 )-(P 1 V 1 +U 1 ) = H 2 -H 1 Onde h 2 é a entalpia do vapor saturado e h 1 é a entalpia do sólido. No diagrama termodinâmico para Patm, tem-se que h 2 = 340 KJ/kg e h 1 = -220 KJ/kg. Considerando 0.5kg de CO 2, então o calor transferido será de 280 KJ. A temperatura de saturação do CO 2 será de 175K (-98 o C)

78 SOLID LIQUID VAPOR

79 Solução de Exercícios Cap 4

80 Ex4.13) 1000K 300K W 100KJ Ciclo de Carnot W=? Q c =?

81 Ex4.14)

82 Ex4.15) Condensador Turbina Caldeira Bomba W T >0 Q c <0 (3500) W b <0 Q h >0 (5000) Q meio <0 (500)

83 Ex4.16) P2 T s P1>P2 550ºC 30ºC

84 Ex4.17) Q h =+180kW Aquecedor Condensador Turbina Compressor Q L =-110kW W>0 W<0 1000ºC 100ºC

85 Rendimento Máximo -> Rendimento de Carnot Ciclo Brayton Conhecer Pressões


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