VI. Entropy. VI. Entropy

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1 A. Introduction. he first law: energy cannot be created or destroyed. he second law: certain processes do occur and certain processes don t 3. he magic vortex tube. Will it work or won t it? Cold air kg at atm 73 K Magic Vortex ube Compressed air kg at 4 atm, 3 K ot air kg at atm, 333 K We need a quantitative answer. his suggests that we are looking for a property (like, P, u, or V). ow can we find such a property? lesson 6 B. he Clausius Inequality For a ersible heat engine, (6-6), is Q Q L = L and Q Q QL = or = L Rev. eat Engine W out,net For an irersible heat engine, with Q constant, Q Q L ir QL,ir > QL, or < or < L L Q L In eral, we have the Clausius inequality, (7-) lesson 6

2 C. A New Property Called Entropy he equality in (7-) holds for a ersible process, the inequality for an irersible one. (7-) We know that the cyclic integral of a property is zero. Clausius recognized that the equality in (7-) implies the existence of a new property, entropy, which we will give the symbol. lesson 6 kj = or d= = (7-4) and d K d Q kj Equation 7-4 can also be written as = dt s K Equation 7-4 can be used to calculate changes in entropy for an internally ersible, isothermal heat transfer process. As piously noted, isothermal heat transfer processes are internally ersible and (7-4) can be integrated to give Q = = = = (7-6) Example: kg of sat. liquid water at ºC and atm is heated at constant temperature and pressure until it completely vaporizes. Find the change in entropy of the water. s = Q 57. kj / kg = ( ) K = 6.4 kj kg K Note that is positive because heat is added to the system. lesson 6

3 D. he Increase of Entropy Principle. We know that energy is conserved. What can we say about entropy? Consider the cycle sketched below. It consists of a ersible and an irersible process. ir We apply the Clausius inequality (7-) to this cycle. δ Q + ir Because / is a perfect differential or property, the first integral is the entropy change,, and the inequality becomes lesson 6 D. Entropy balance for a closed system Reminder: is the absolute. he equality only holds for a ersible process. In differential form we have d Q dt or d (7-8) Conclusion: the entropy change for a closed system, undergoing an irersible change, is always greater than the entropy transfer due to heat transfer. In other words, entropy is erated during an irersible process. We will denote the amount of entropy erated by (kj/k) and the rate of entropy eration by. lesson 6 3

4 D. Entropy balance for a closed system We can use the entropy eration term,, to eliminate the inequality where sys = = + (7-9) > irersible process = ersible process < impossible process lesson 6 Equation 7-9 can also be written in rate form as dsys Qj = + dt j j ee (7-83) E. he Increase of Entropy Principle If a system is isolated, =, and = isolated (7-) his known as the increase of entropy principle. Because a system and its surroundings can be considered an isolated system, we can also write = = + (7-) total sys surr lesson 6 4

5 F. Reversible and Actual Work and the Generation of Entropy Consider two closed systems, each undergoing a process, one ersible and the other irersible (actual), with the same initial and final conditions. Q or Q in, act in, u, s u, s lesson 6 W or W out, act out, Energy balances: mu ( u ) = Wout, act Qin, act = Wout, Qin, Entropy balances: Qin, act = ms ( s ) andqin, = ms ( s) Conclusion: W, = W, + out out act represents the work that is lost due to irersibilities. G. Example Maximum work obtainable from a finite, hot reservoir kg sat. liquid water Q Engine Q L Large reservoir atm i = C = 373 K f = 3 K We want to find the maximum work that we can extract from the hot water. Its initial and W final temperatures are i and out f. he 3 K reservoir is so large that its temperature is constant. L = 3 K lesson 6 5

6 lesson 6 G. Example Maximum work obtainable from a finite, hot reservoir t he total work produced is Wout = W outdt where t is time. A Carnot or ersible heat engine will produce the maximum amount of work: W out L η = = and W out = ηq Q An energy balance on the hot water gives du dt d = Q = mc dt Our equation for maximum work produced is t t t f L d L Wout = W outdt = ηq dt = mc dt = mc d dt i G. Example Maximum work obtainable from a finite, hot reservoir Performing the integration and inserting numerical values gives f Wout = mc ( i f ) + L ln i kj 3 Wout = kg ln K = 33 kj kgk 373 lesson 6 6

7 . ummary. Entropy is not conserved. > for all real processes.. Isolated systems will only change in a direction that results in an increase in entropy. 3. Irersibilities result in the eration of entropy. ighly irersible processes have high rates of entropy eration. igh rates of entropy eration 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. Cold air kg at atm 73 K lesson 6 Magic Vortex ube Compressed air kg at 4 atm, 3 K ot air kg at atm, 333 K 7