The Second Law of Thermodynamics

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Milton Marcus Lawson
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1 he Second Law of hermodynamics

2 So far We have studied the second law by looking at its results We don t have a thermodynamic property that can describe it In this chapter we will develop a mathematical definition for entropy Entropy is a quantitative measure of disorder

3 How does the second law relate to disorder? Disorder naturally increases, just like water flows down hill and gases expand Natural processes proceed spontaneously toward disorder

4 Disorder Cleaning your room takes a lot of effort Messing it up is much easier Dissolving sugar in water increases disorder Sugar dissolves spontaneously Sugar and water don t separate spontaneously

5 Energy It takes the same amount of energy to clean your room as to mess it up. It takes the same amount of energy to dissolve sugar as to separate it. We all know that these are not equivalent, reversible, processes.

6 Understanding Entropy What is it? It s a measure of disorder We can define entropy about as well as we can define energy Remember we define energy as the ability to do work

7 Clausius Inequality Consider an energy balance for the entire system E in δq R E exit = E c ( δw ) rev + δwsys = dec δ δ W Q c R = δ δ W W rev c = + δ de W c sys

8 Clausius Inequality If the heat engine is completely reversible then δq R R = δq δ δq Q R = R his is based on Chapter 5, where we said that heat transfer rate in a reversible system is proportional to the temperature

9 Clausius Inequality hat means that we can rewrite the energy balance, and solve for the net work (the work of the combined system. δ δw = Q de c c R c o find the total net work we need to integrate over time W δq = R dec 0

10 Clausius Inequality But we know that it is impossible for a system to exchange heat with only one reservoir, and produce work!! W c = R δ Q hat means W c can t be positive, but it can be zero or negative!! W c Q = 0 R δ

11 Clausius Inequality Since R is always positive δq 0 his is called the Clausius inequality It is equal to 0 for the reversible case, and is less than 0 for the irreversible case

12 Clausius Inequality Can we prove that it is 0 for the reversible case? If the the system is reversible, we can run it backwards. If it produces negative work going one way, it will produce positive work going the other way But we know you can t exchange heat with only one reservoir, and produce work!!

13 Clausius Inequality herefore, for a reversible system, the cyclic integral must be 0!!! δq = rev 0

14 Entropy Entropy (S) is a measure of the disorder or randomness in a system, and is a state variable (like P,V,) that does not depend on the path taken. dq ds = (change in entropy S) where dq is an infinitesimal heat flow

15 Entropy For any reversible process (e.g. Carnot cycle): dq rev = 0 I I b a b a dq dq + II a b = dq II a b = dq 0 = II b a dq he entropy of a system in a given state is independent of the path taken to get there, and is thus a state variable.

16 Entropy he entropy difference between two equilibrium states a and b does not depend on how the system got from a to b. b b dq S = S b S a = = ds a a Entropy is a state variable (like P,V and )

17 Entropy Pressure 2 Reversible Arbitrary 1 Specific Volume Consider a cycle composed of two processes Process 1 (from state 1 to state 2) is reversible Process 2 (from state 2 back to state 1 is arbitrary

18 Entropy Pressure 2 Reversible Arbitrary 1 Specific Volume 2 1 δq net + rev ( S S ) 1 2 Since the process is a cycle, the total change in entropy is 0 he change in entropy from state1 to 2 is = he change in entropy from state 2 to 1 is 0

19 Entropy Pressure 2 Reversible Arbitrary 1 Specific Volume So, the definition of a change in entropy becomes. δ Q 2 S S = net rev ds Or Q = δ net rev

20 Entropy Change he entropy change between two specific states is the same whether the process is reversible or irreversible Entropy is a state function

21 Entropy Change Q 2 S S = net rev Consider an adiabatic reversible process δq = 0 So S=0 δ A reversible, adiabatic process is called isentropic and S is constant. But what if the process isn t reversible? δ Q 2 S S > net irreversible

22 Isolated Systems Consider an isolated system Because it is isolated, there is no heat transfer or work S 2 S δ Q net 0 S 0 he equality holds if all the processes inside the system are reversible

23 Now consider the universe It can be considered to be an isolated system!! S universe = S system + S surroundings 0 his is the second law of thermodynamics, expressed mathematically!!

24 You can consider any system to be isolated, if you draw your system boundaries out far enough S total = S system + S surroundings 0

25 Hopefully you have a better feel for entropy now! How do you calculate entropy changes for real systems? If it s reversible adiabatic, S=0 If it s isothermal heat transfer S 2 Q S = 2 δ 1 1 net = rev Q rev If it s not an ideal gas, just look it up on the property tables

26 Entropy is a property

27 Entropy is a property he entropy of a pure substance is determined from the tables, just as for any other property

28 Some Remarks about Entropy Entropy is a non-conserved property, and there is no such thing as the conservation of entropy principle. he entropy of the universe is continuously increasing. he performance of engineering systems is degraded by the presence of irreversibilities, and entropy generation is a measure of the magnitudes of the irreversibilities present during that process.

29 Engineering Devices Many engineering devices are essentially adiabatic hey perform best when irreversibilities are eliminated Isentropic model serves as an idealization of a real process hese devices work best when they are isentropic

30 he assumption that a process is isentropic, gives us a connection between the inlet and outlet conditions just like assuming constant volume, or constant pressure

31 Property Diagrams Involving Entropy δ S = δ Q int,rev δq = ds int, rev Q int, rev = 2 1 ds his area has no meaning for irreversible processes

32 Consider some special cases of the -s diagram emperature Isothermal Process 1 2 Q emperature Isentropic Process 1 Q=0 2 Entropy Entropy

33 -s Diagram for the Carnot Cycle 1 Isothermal expansion 2 emperature Isentropic compression W=Q Q in -Q out in Isentropic expansion 4 Isothermal compression Q out 3 Entropy

34 Mollier Diagrams (h-s Diagrams) For adiabatic, steady flow devices, h is a measure of work, and s is a measure of irreversibilities

35 Mollier Diagram for Steam

The Kelvin-Planck statement of the second law

The Kelvin-Planck statement of the second law It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work Q W E =ΔE net net net, mass