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2300_les15 - V. The Second Law of Thermodynamics E. The...

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Unformatted text preview: V. The Second Law of Thermodynamics E. The Carnot Cycle 1. The reversible heat engine (a piston cylinder device) that operates on a cycle between heat reservoirs at TH and TL, as shown below on a P-v diagram, is the Carnot cycle. 2. Because the cycle is reversible, the Carnot cycle, operating between TH and TL, is the most efficient cycle possible. We want to know how much heat, QL, must be rejected by the engine. 3. If the Carnot heat engine is operated in reverse, we call it a Carnot refrigeration cycle. P 1 Q Q QH ηth = 1 − out = 1 − L Qin QH Wnet,out = shaded area Q41=0 2 TH = constant 4 QL Q23=0 3 TL = constant v •1-2 •2-3 •3-4 •4-1 rev., isothermal expansion rev., adiabatic expansion rev., isothermal compression rev., adiabatic compression Lesson 15, Geof Silcox, Chemical Engineering, University of Utah V. The Second Law of Thermodynamics F. The Carnot Principles 1. “The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.” (p. 302) 2. “The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.” (p. 302) Both principles are based on the Kelvin-Plank and Clausius statements of the second law. From the second principle we conclude that the thermal efficiency is a function only of the temperatures of the reservoirs. ηth = 1 − QL = 1 − f (TH ,TL ) QH (6-13) Lesson 15, Geof Silcox, Chemical Engineering, University of Utah 1 V. The Second Law of Thermodynamics 3. Proof of the Second Carnot Principle Suppose we have a “super engine” that is more efficient than a reversible High-T reservoir at TH engine. Let both engines receive the same amount of thermal energy, QH, from a high temperature reservoir. QH QH Then Wrev < Wsrev and QLS < QL. Rev. Super Wrev engine engine If we run the “normal” engine in reverse, and power it with Wsrev, then the combined engine takes heat from a single reservoir and produces a net QL QLS amount of work. This violates the Kelvin-Plank statement of the 2nd law. Low-T reservoir at TL We conclude that both reversible engines have the same efficiency. Lesson 15, Geof Silcox, Chemical Engineering, University of Utah Wsrev V. The Second Law of Thermodynamics G. The Thermodynamic Temperature Scale 1. Define such that the T scale is independent of the properties of the substance being used to measure the temperature. 2. Derivation High-T reservoir at T1 Consider three reversible engines. By the 2nd Carnot principle, Q1 QL ηth = 1 − = 1 − f (TH ,TL ) (6-13) Q1 Rev. QH WA HE A Apply (6-13) to each engine: Q2 Rev. Q3 Q3 Q2 T2 WC = f (T1 ,T2 ), = f (T2 ,T3 ), = f (T1 ,T3 ) HE C Q2 Q1 Q2 Q1 Rev. Q3 Q2 Q3 WB = = f (T1 ,T3 ) = f (T1 ,T2 )f (T2 ,T3 ) HE B Q3 Q1 Q1 Q2 Q3 φ (T3 ) Q φ (TL ) Q or L = ∴ 3= Low-T reservoir at T3 Q1 φ (T1 ) QH φ (TH ) Lesson 15, Geof Silcox, Chemical Engineering, University of Utah 2 V. The Second Law of Thermodynamics G. The Thermodynamic Temperature Scale 2. Derivation Because it is the simplest possible choice, Kelvin proposed taking φ(T) = T to give the Kelvin scale: QL TL = QH TH (6-16) The magnitude of a kelvin is 1/273.16 of the temperature interval between absolute zero and the triple-point of water (0.01ºC). The magnitude of the Kelvin and Celsius scales are the same (1 K = 1ºC) and T(ºC) = T(K) - 273.15. The smallest possible value of QL is QL = TL QH TH Lesson 15, Geof Silcox, Chemical Engineering, University of Utah V. The Second Law of Thermodynamics H. The Carnot Heat Engine For any heat engine, Q Q ηth = 1 − L = 1 − L QH QH For any reversible heat engine, Carnot or otherwise, TH QH,in Carnot Heat Engine Wnet,out QL TL = QH TH and (6-16) QL,out TL TH TL ηth,rev = 1 − (6-18) Lesson 15, Geof Silcox, Chemical Engineering, University of Utah 3 V. The Second Law of Thermodynamics I. The Carnot Refrigerator and Heat Pump COPR ,rev = 1 1 = (6-20) QH / QL − 1 TH / TL − 1 High-T sink, TH QH Carnot heat pump QL Low-T source, TL COPHP ,rev = 1 1 = (6-21) 1 − QL / QH 1 − TL / TH Wnet,in Lesson 15, Geof Silcox, Chemical Engineering, University of Utah V. The Second Law of Thermodynamics Example - An inventor claims to have a heat engine that will use the temperature difference between the top and bottom layers of a lake to produce Wnet,out = 1 MW of work for centuries. Is this person telling the truth? The lake and conditions are shown below. Average solar flux = 240 W/m2 • 25 C • 15 C Area of lake = 104 m2 Lesson 15, Geof Silcox, Chemical Engineering, University of Utah 4 V. The Second Law of Thermodynamics 7. Example a. First law analysis: not violated since 2.4 MW > 1 MW QH = 240 W ⋅10 4 m 2 = 2.4 MW 2 m b. Second law analysis (what is the best we can hope for?) ηmax = Wout T 15 + 273 =1− L =1− = 0.0336 TH 25 + 273 QH Wout , max = ηQH = 0.0336( 2.4 MW ) = 0.0805 MW c. Conclusion: inventor is not telling the truth because claimed work output is too high. Lesson 15, Geof Silcox, Chemical Engineering, University of Utah 5 ...
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