Thermodynamics Notes 18

Thermodynamics Notes 18 - ME 311 FALL 2007 CHAPTER 6...

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ME 311 THERMODYNAMICS S. Masutani FALL 2007 CHAPTER 6 (continued) Example (limiting efficiencies) Combustion power system: T H = 2000 K; T C = 290 K Material limitations typical result in T H ~ 1000 K. Thus, for an ideal cycle, the limiting efficiency is 1000 290 1 1 = = H L T T η = 0.71 or 71%. Actual combustion power system operate in the range 20 η 40%. Thermal energy conversion efficiencies of state-of-the-art heat engines do not exceed 50%. Note that for an OTEC power plant: T H = 298 K; T C = 278 K 298 278 1 1 = = H C T T = 0.067 or 6.7% (actual efficiencies typically ~ 1-2%) Consider a 2T heat engine: 1 T H T L W Q H Q L For a cycle: Production = outflow – inflow + storage or Storage = 0 = production + inflow – outflow (there is no storage for a cycle) H H L L s L L H H s s T Q T Q P T Q T Q P T Q P dS = = + = + = ∫∫ 0 0 δ Consider the case when Q L = 0: P s 0 Q H 0 for any cyclically operating device. Since W 0 Q H must be 0; hence a 1T (continuous; cyclic) heat engine is impossible (a 1T engine can operate for a short time but not continuously). This is the Kelvin- Planck statement.
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ME 311 THERMODYNAMICS S. Masutani FALL 2007 CLAUSIUS STATEMENT T A > T B T A T B Q A Q B 1 st Law: for a cycle: Q A = Q B = Q = = = + = ∫∫ B A s B B A A s s T T Q P T Q T Q P T Q P dS 1 1 0 δ But T A > T B 0 1 1 < B A T T Since , a zero-work heat pump is impossible. This is the 0 s P Clausius Statement. Note, however, that short-term (non-cyclic) operation is possible. We have observed that for an isolated system such as the one shown below T TER δ Q Isolated system A T Q P dS s + = 2
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ME 311 THERMODYNAMICS S. Masutani FALL 2007 Consider now only the subsystem A which exchanges energy (and entropy) with its environment.
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Thermodynamics Notes 18 - ME 311 FALL 2007 CHAPTER 6...

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