Notes_18_Entropy - Meeting 17 Sections 7-1-7-6 This is...

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1 This is going to seem pretty abstract..so Hang on for the ride! Meeting 17 Sections 7-1—7-6
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What we covered Monday Carnot Corollaries Maximum (Carnot) Efficiency Absolute Temperature Scale Carnot Cycle Maximum (Carnot) COP 2
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Topics for Today Clausius Inequality Entropy Generation Obtain Entropy Values Entropy Change Constant Entropy (___________) Processes Ts Diagram 3
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4 Clausius Inequality Another corollary of the 2nd Law. Now we will deal with increments of heat and work, Q and W, rather than Q and W. We will employ the symbol , which means to integrate over all the parts of the cycle.
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5 Look at a reversible power cycle Hot reservoir Cold reservoir System H Q L Q outR W
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6 Look at a reversible cycle: 29 L H cycle Q Q Q - = δ We know: And: 29 0 W Q outR cycle = δ
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7 For the reversible cycle cycle T Q δ - = L L H H T Q T Q δ δ Look at Q/T: Since the heat transfer occurs at ______________ temperature, we can pull T out of integrals: cycle T Q δ L L H H T Q T Q - =
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8 For the reversible cycle H L rev H L T T Q Q = or L L H H T Q T Q = This allows us to write: cycle T Q δ 0 T Q T Q H H H H = - =
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9 For an irreversible cycle Hot reservoir Cold reservoir System H Q LI Q outI W
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10 For an irreversible cycle outR outI W W < For the same heat input: For both cycles we can write: L H outR Q Q W - = LI H outI Q Q W - = and Apply inequality: L H LI H Q Q Q Q - < - or L LI Q Q - < - L LI Q Q or
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11 Apply cyclic integral cycle T Q δ L LI H H T Q T Q - = 0 T Q T Q L L H H = - < For the irreversible cycle: 0 T Q cycle < δ
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12 Clausius Inequality 0 T Q cycle δ
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