1. 1 Property Relationships Chapter 6

2. 2 Apply the differential form of the first law for a closed stationary system for an internally reversible process The T-ds relations:

3. 3 This equation is known as: First Gibbs equation or First Tds relationship Divide by T,.. Although we get this form for internally reversible process, we still can compute  s for an irreversible process. This is because S is a point function. Divide by the mass, you get

4. 4 Second T-ds (Gibbs) relationship Recall that… Take the differential for both sides Rearrange to find du Substitute in the First Tds relationship Second Tds relationship, or Gibbs equation

5. 5 Divide by T,.. Thus We have two equations for ds To find  s, we have to integrate these equations. Thus we need a relation between du and T (or dh and T). Now we can find entropy change (the LHS of the entropy balance) for liquids and solids

6. 6 2- Entropy Change of Liquids and Solids 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 Thus For solids and liquids Recall also that For solids and liquids,

7. 7 Integrate to give… Only true for solids and liquids!! What if the process is isentropic? The only way this expression can equal 0 is if, Hence, for solids and liquids, isentropic processes are also isothermal.

8. 8 Example(6-7): Effect of Density of a Liquid on Entropy Liquid methane is commonly used in various cryogenic applications. The critical temperature of methane is 191 K (or -82 o C), and thus methane must be maintained below 191 K to keep it in liquid phase. The properties of liquid methane at various temperatures and pressures are given next page. Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa (a)using actual data for methane and (b)approximating liquid methane as an incompressible substance. What is the error involved in the later case?

9. 9

10. 10 Example (6-19): Entropy generated when a hot block is dropped in a lake A 50-kg block of iron casting at 500 K is dropped in a large lake that is at 285 K. The block reaches thermal equilibrium with lake water. Assuming an average specific heat of 0.45 kJ/kg.K for the iron, determine: (a) The entropy change of the iron block, (b) The entropy change of the water lake, (c) the entropy generated during this process.

11. 11 (a) The entropy change of the iron block, (b) The entropy change of the water lake, we need also to find Q coming out of the system. T=500K T surr = 285 K

12. 12 Thus S g =  S tot =  S sys +  S lake T=500K T surr = 285 K System boundary (c) the entropy generated during this process. S g =  S tot = -12.6 + 16.97 = 4.32 Choose the iron block and the lake as the system and treat it is an isolated system.

13. 13 3- The Entropy Change of Ideal Gases, first relation The entropy change of an ideal gas can be obtained by substituting du = C v dT and P /T= R/  into Tds relations: integrating First relation

14. 14 A second relation for the entropy change of an ideal gas for a process can be obtained by substituting dh = C p dT and  /T= R/P into Tds relations: integrating Second relation

15. 15 The integration of the first term on the RHS can be done via two methods: 1. Assume constant Cp and constant Cv (Approximate Analysis) 2. Evaluate these integrals exactly and tabulate the data (Exact Analysis)

16. 16 Method 1: Constant specific heats (Approximate Analysis) First relation Only true for ideal gases, assuming constant heat capacities Second relation Only true for ideal gases, assuming constant heat capacities

17. 17 Sometimes it is more convenient to calculate the change in entropy per mole, instead of per unit mass kJ/kmol. K Ru is the universal gas constant

18. 18 Method 2: Variable specific heats (Exact Analysis) We could substitute in the equations for C v and C p, and perform the integrations C p = a + bT + cT 2 + dT 3 But this is time consuming. Someone already did the integrations and tabulated them for us (table A-17) They assume absolute 0 as the starting point We use the second relation

19. 19 The integral is expressed as: Where is tabulated in Table A-17 Therefore

20. 20 From this equation, It can be seen that the entropy of an ideal gas is not a function only of the temperature ( as was the internal energy) but also of the pressure or the specific volume. The function s° represents only the temperature- dependent part of entropy Temperature dependence Pressure dependence Is s = f (T) only? like u for an ideal gas. Let us see

21. 21 How about the other relation We can develop another relation for the entropy changed based on the above relation but this will require the definition of another function and tabulating it which is not practical.

22. 22 6-4 Isentropic Processes The entropy of a fixed mass can be changed by 1. Heat transfer, 2. Irreversibilities It follows that the entropy of a system will not change if we have 1. Adiabatic process, 2. Internally reversible process. Therefore, we define the following:

23. 23 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 for solids and liquids

24. 24 For the isentropic case,  S=0. Thus Constant specific heats (1 st relation) Recall Recall also from ch 2, the following relations..… Only applies to ideal gases, with constant specific heats

25. 25 Only applies to ideal gases, with constant specific heats Constant specific heats (2nd relation) Recall..…or

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

27. 27 Compact form Full form of Isentropic relations of Ideal Gases Valid for only for 1- Ideal gas 2- Isentropic process 3- Constant specific heats

28. 28 That works if the specific heat constants 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. 29 s 2 0 is only a function of temperature!!! Rename the exponential term as P r, (relative pressure) which is only a function of temperature, and is tabulated on the ideal gas tables

30. 30 You can use either of the following 2 equations This is good if you know the pressure ratio but how about if you know only the volume ratio In this case, we use the ideal gas law where Remember, these relationships only hold for ideal gases and isentropic processes

31. 31 Example (6-10): Isentropic Compression of Air in a Car Engine Air is compressed in a car engine from 22 o C and 95 kPa in a reversible and adiabatic manner. If the compression ratio V 1 /V 2 of this piston-cylinder device is 8, determine the final temperature of the air. Sol: