2300_les16 - VI. Entropy A. Introduction 1. The first law:...

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Unformatted text preview: VI. Entropy A. Introduction 1. The first law: energy cannot be created or destroyed 2. The second law: certain processes do occur and certain processes don’t 3. The magic vortex tube. Will it work or won’t it? Compressed air 2 kg at 4 atm, 300 K Cold air 1 kg at 1 atm 273 K Hot air 1 kg at 1 atm, 333 K Magic Vortex Tube We need a quantitative answer. This suggests that we are looking for a property (like T, P, u, or V). How can we find such a property? Lesson 16, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy B. The Clausius Inequality For a reversible heat engine, (6-16), is TH QH =0 Rev. Heat Engine QL TL = QH TH and QH TH = QL TL or ∫ δ Qrev T Wout,net For an irreversible heat engine, with QH constant, QL TL QL,irrev > QL,rev or QH TH < QL TL or ∫ δ Qirrev T <0 In general, we have the Clausius inequality, ∫ δQ T ≤0 (7-1) Lesson 16, Geof Silcox, Chemical Engineering, University of Utah 1 VI. Entropy C. A New Property Called Entropy The equality in (7-1) holds for a reversible process, the inequality for an irreversible one. ≤0 (7-1) T We know that the cyclic integral of a property is zero. Clausius recognized that the equality in (7-1) implies the existence of a new property, entropy, which we will give the symbol S. ∫ δQ ∫ δ Qrev T = 0 or ∫ dS = 0 and dS = δQrev T ⎛ kJ ⎞ ⎜ ⎟ (7-4) ⎝K⎠ ⎛ kJ ⎞ ⎜ ⎟ ⎝s K ⎠ Equation 7-4 can also be written as dS Qrev = dt T Lesson 16, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy Equation 7-4 can be used to calculate changes in entropy for an internally reversible, isothermal heat transfer process. As previously noted, isothermal heat transfer processes are internally reversible and (7-4) can be integrated to give ∆S = S2 − S1 = ∫ 1 12 Q δQrev = δQrev = 1T T0 ∫1 T0 2 (7-6) Example: 1 kg of sat. liquid water at 100ºC and 1 atm is heated at constant temperature and pressure until it completely vaporizes. Find the change in entropy of the water. ∆s = Q 2257 .0 kJ / kg kJ = = 6.04 T0 (100 + 273.15 ) K kg K Note that ∆S is positive because heat is added to the system. Lesson 16, Geof Silcox, Chemical Engineering, University of Utah 2 VI. Entropy D. The Increase of Entropy Principle 1. We know that energy is conserved. What can we say about entropy? Consider the cycle sketched below. It consists of a reversible and an irreversible process. 2 irrev We apply the Clausius inequality (7-1) to this cycle. rev 1 ∫ 1 δQrev T 2 +∫ 2 δQirrev T 1 ≤0 Because δQrev/T is a perfect differential or property, the first integral is the entropy change, S1 – S2, and the inequality becomes S2 − S1 ≥ ∫ 2 δQ T 1 Lesson 16, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy D. Entropy balance for a closed system Reminder: T is the absolute T. S2 − S1 ≥ ∫ 2 δQ T 1 The equality only holds for a reversible process. In differential form we have dS Q ≥ dt T or dS ≥ δQ T (7-8) Conclusion: the entropy change for a closed system, undergoing an irreversible change, is always greater than the entropy transfer due to heat transfer. In other words, entropy is generated during an irreversible process. We will denote the amount of entropy generated by Sgen (kJ/K) and the rate of entropy generation by Sgen. Lesson 16, Geof Silcox, Chemical Engineering, University of Utah 3 VI. Entropy D. Entropy balance for a closed system We can use the entropy generation term, Sgen, to eliminate the inequality S2 − S1 ≥ ∫ where 2 δQ T 1 ∆Ssys = S2 − S1 = ∫ 2 δQ T 1 + Sgen (7-9) Sgen > 0 irreversible process Sgen = 0 reversible process Sgen < 0 impossible process Equation 7-9 can also be written in rate form as dSsys dt =∑ j Qj Tj + Sgen See (7-83) Lesson 16, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy E. The Increase of Entropy Principle If a system is isolated, δQ = 0, and ∆S = S2 − S1 ≥ ∫ 2 δQ T 1 ∆Sisolated ≥ 0 (7-10) This known as the increase of entropy principle. Because a system and its surroundings can be considered an isolated system, we can also write Sgen = ∆Stotal = ∆Ssys + ∆Ssurr ≥ 0 (7-11) Lesson 16, Geof Silcox, Chemical Engineering, University of Utah 4 VI. Entropy F. Reversible and Actual Work and the Generation of Entropy Consider two closed systems, each undergoing a process, one reversible and the other irreversible (actual), with the same initial and final conditions. Qin,act or Qin,rev u1, s1 u2, s2 Wout ,act or Wout ,rev Energy balances: m(u1 − u2 ) = Wout ,act − Qin,act = Wout ,rev − Qin,rev Entropy balances: Qin,act = mT (s2 − s1 ) − TSgen and Qin,rev = mT (s2 − s1 ) Conclusion: ∴Wout ,rev = Wout ,act + TSgen TSgen represents the work that is “lost” due to irreversibilities. Lesson 16, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy G. Example – Maximum work obtainable from a finite, hot reservoir 100 kg sat. liquid water 1 atm Ti = 100°C = 373 K Tf = 300 K We want to find the maximum work that we can extract from the hot water. Its initial and final temperatures are Ti and Wout Tf. The 300 K reservoir is so large that its temperature is constant. TL = 300 K QH Engine QL Large reservoir Lesson 16, Geof Silcox, Chemical Engineering, University of Utah 5 VI. Entropy G. Example – Maximum work obtainable from a finite, hot reservoir The total work produced is Wout = ∫W 0 t out dt where t is time. A Carnot or reversible heat engine will produce the maximum amount of work: η= Wout T = 1 − L and Wout = ηQH T QH An energy balance on the hot water gives dU dT = −QH = mc dt dt Our equation for maximum work produced is t t t⎛ Tf ⎛ T ⎞ dT T Wout = ∫ Wout dt = ∫ ηQH dt = −mc ∫ ⎜ 1 − L ⎟ dt = −mc ∫ ⎜ 1 − L 0 0 0 Ti ⎝ T ⎠ dt ⎝T ⎞ ⎟dT ⎠ Lesson 16, Geof Silcox, Chemical Engineering, University of Utah VI. Entropy G. Example – Maximum work obtainable from a finite, hot reservoir Performing the integration and inserting numerical values gives ⎡ T⎤ Wout = mc ⎢(Ti − Tf ) + TL ln f ⎥ Ti ⎦ ⎣ ⎛ 300 ⎤ kJ ⎞ ⎡ Wout = 100kg ⎜ 4.2 K = 3230 kJ ⎟ ⎢73.15 + 300ln 373 ⎥ kgK ⎠ ⎣ ⎦ ⎝ Lesson 16, Geof Silcox, Chemical Engineering, University of Utah 6 VI. Entropy H. Summary 1. Entropy is not conserved. Sgen > 0 for all real processes. 2. Isolated systems will only change in a direction that results in an increase in entropy. 3. Irreversibilities result in the generation of entropy. Highly irreversible processes have high rates of entropy generation. High rates of entropy generation usually degrade the performance of a process. 4. Is it possible or not? We need to be able to calculate the entropy changes of substances to answer this question. Compressed air 2 kg at 4 atm, 300 K Cold air 1 kg at 1 atm 273 K Hot air 1 kg at 1 atm, 333 K Magic Vortex Tube Lesson 16, Geof Silcox, Chemical Engineering, University of Utah 7 ...
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This note was uploaded on 09/02/2010 for the course PHYS 2300 taught by Professor Silcox during the Fall '09 term at University of Utah.

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