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Lecture 26: Thermodynamics II

Lecture 26: Thermodynamics II. Heat Engines Refrigerators Entropy 2 nd Law of Thermodynamics Carnot Engines. HEAT ENGINE. REFRIGERATOR. system. T H. T H. Q H. Q H. W. W. Q C. Q C. T C. T C. Engines and Refrigerators. system goes through a closed cycle  U system = 0

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Lecture 26: Thermodynamics II

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  1. Lecture 26:Thermodynamics II • Heat Engines • Refrigerators • Entropy • 2nd Law of Thermodynamics • Carnot Engines

  2. HEAT ENGINE REFRIGERATOR system TH TH QH QH W W QC QC TC TC Engines and Refrigerators • system goes through a closed cycle  Usystem = 0 • therefore, net heat absorbed = work done QH - QC = W(engine) QC - QH = -W(refrigerator) energy into system = energy leaving system

  3. HEAT ENGINE TH QH W QC TC Heat Engine: Efficiency The objective:turn heat from hot reservoir into work The cost:“waste heat” 1st Law:QH -QC = W Efficiencye  W/QH = (QH-QC)/QH = 1-QC/QH

  4. REFRIGERATOR TH QH W QC TC Refrigerator:Coefficient of Performance The objective:remove heat from cold reservoir The cost:work 1st Law:QH = W + QC Coeff. of performance: Kr QC/W = QC/(QH - QC)

  5. Entropy (S) • A measure of “disorder.” • A property of a system (just like P, V, T, U): • related to number of number of different “states” of system • Examples of increasing entropy: • ice cube melts • gases expand into vacuum • Change in entropy: • S = Q/T S > 0 if heat flows into system (Q > 0) S < 0 if heat flows out of system (Q < 0)

  6. Second Law of Thermodynamics • Total entropy NEVER decreases: • S  0 • order to disorder • Consequences: • A “disordered” state cannot spontaneously transform into an “ordered” state. • No engine operating between two reservoirs can be more efficient than one that produces 0 change in entropy.This is called a “Carnot engine.”

  7. Carnot Cycle • Idealized Heat Engine • No Friction • DS = Q/T = 0 • Reversible Process • Isothermal Expansion • Adiabatic Expansion • Isothermal Compression • Adiabatic Compression • Carnot Efficiency • ec = 1 - TC/ TH • e = 1 is forbidden! • e largest if TC << TH

  8. Summary • First Law of Thermodynamics: Energy Conservation • Q = DU + W • Heat Engines: • e = 1-QC/QH • Refrigerators: • Kp = QC/(QH - QC) • Entropy: • DS = Q/T • 2nd Law: • Entropy always increases! • Carnot Cycle: Reversible, Maximum Efficiency • ec = 1 – Tc/Th

  9. Example • An engine operates between Th = 560 K and Tc = 300 K. If the actual efficiency is half its ideal efficiency and the intake heat is Qh = 1285 J every second, at what rate does it do work? First, to find the ideal efficiency we will use: ec = 1 – Tc/Th Second, to find the output work we will use: e = W/Qh

  10. Example • An engine operates between Th = 560 K and Tc = 300 K. If the actual efficiency is half its ideal efficiency and the intake heat is Qh = 1285 J every second, at what rate does it do work? Find the ideal efficiency: ec = 1 – Tc/Th = 1 – (300 K)/(560 K) = 46.4%

  11. Example • An engine operates between Th = 560 K and Tc = 300 K. If the actual efficiency is half its ideal efficiency and the intake heat is Qh = 1285 J every second, at what rate does it do work? Find the work done: e = W/Qh 23.3% = W/(1285 J) W = 298 J every second

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