1. Entropy Property Relationships Chapter 7b

2. The T-ds relations Consider an internally reversible process occurring in a closed system

3. Or… To find  s all you have to do is integrate!!! First Gibbs equation – also called First Tds relationship

4. 2 nd Gibbs relationship Recall that… Find the derivative, dh Rearrange to find du

5. Second Tds relationship, or Gibbs equation To find  s all you have to do is integrate First Tds relationship

6. We have two equations for ds To find  s, integrate the equation that is the easiest, or for which you have the data

7. First lets look at solids and liquids Solids and liquids do not change specific volume appreciably with pressure That means that dv=0, so the first equation is the easiest to use. 0

8. For solids and liquids…

9. Recall that… For solids and liquids, so… Integrate to give… Only true for solids and liquids!!

10. What if the process is isentropic? What happens to s? The only way this expression can equal 0 is if T 2 = T 1 For solids and liquids, isentropic processes are also isothermal, if they are truly incompressible

11. Entropy change of ideal gases Some equations we know for ideal gases

12. Let’s use these relationships with the Gibbs equations

13. We can integrate these equations if we assume constant C p and constant C v Only true for ideal gases, assuming constant heat capacities First Gibbs equation

14. We can integrate these equations if we assume constant C p and constant C v Only true for ideal gases, assuming constant heat capacities Second Gibbs equation

15. Use which ever equation is easiest!! Which should you use?

16. Sometimes it is more convenient to calculate the change in entropy per mole, instead of per unit mass

18. What if it’s not appropriate to assume constant specific heats?  We could substitute in the equations for C v and C p, and perform the integrations C p = a + bT + cT 2 + dT 3 That would be time consuming and error prone There must be a better way!!

19. What if it’s not appropriate to assume constant specific heats?  Someone already did the integrations and tabulated them for us They assume absolute 0 as the starting point See Table A-17, pg 910

20. So…. These two equations are good for ideal gases, and consider variable specific heats Remember

21. Entropy of an Ideal Gas 6-12 The entropy of an ideal gas depends on both T and P. The function s° represents only the temperature-dependent part of entropy

22. Isentropic Processes of Ideal Gases  Many real processes can be modeled as isentropic  Isentropic processes are the standard against which we should measure efficiency  We need to develop isentropic relationships for ideal gases, just like we developed them for solids and liquids

23. For the isentropic case,  S=0 Constant specific heats

24. First isentropic relation for ideal gases andso Recall that…

25. Similarly Second isentropic relationship Only applies to ideal gases, with constant specific heats

26. Since… and Which can be simplified to… Third isentropic relationship

27. Compact form

28. That works if the heat capacities can be approximated as constant, but what if that’s not a good assumption? We need to use the exact treatment 0 This equation is a good way to evaluate property changes, but it can be tedious if you know the volume ratio instead of the pressure ratio

29. Relative Pressure and Relative Specific Volume s 2 0 is a function only of temperature!!!

30. Rename the exponential P r, (relative pressure) which is only a function of temperature, and is tabulated on the ideal gas tables You can use this equation or

31. What if you know the volume ratio? Ideal gas law Rename this v r2 Rename this 1/v r1 Relative specific volumes are also tabulated in the ideal gas tables Remember, these relationships only hold for ideal gases and isentropic processes

32. Summary  We developed the first and second Gibbs relationships

33. Which can also be expressed as

34. Summary For solids and liquids Solids and liquids do not change specific volume appreciably with pressure, so dv=0 0 C can be approximated as a constant in solids and liquids

35. Summary For ideal gases if we assume constant heat capacities…

36. Which can be integrated to give True for ideal gas with constant heat capacities

37. Summary For ideal gases with variable heat capacity

38. Summary What if its not an ideal gas?  You’ll need to use the tables

39. Summary Isentropic Processes – Ideal Gas and Constant Heat Capacity

40. Summary Isentropic processes for Ideal gases – Variable Heat Capacities

41. Summary Isentropic processes if the gas is not ideal and the heat capacities are variable  Use the tables!!