1. AGUS HARYANTO 20-02-2010 http://www.energimasadepan.wordpress.com PROPERTY TABLES + EQUATION OF STATE

2. WHY TABLES? The relationships among thermodynamic properties are too complex to be expressed by simple equations. Therefore, properties are frequently presented in the form of tables. Some thermodynamic properties can be measured easily, but others cannot and are calculated by using the relations between them and measurable properties. The results of these measurements and calculations are presented in tables in a convenient format. Property tables of other substances are used in the same manner. For each substance, the thermodynamic properties are listed in more than one table.

3. Enthalpy—A Combination Property Two new properties: enthalpy h and entropy s. Entropy is a property associated with the second law of thermodynamics. In the analysis of certain processes, such in power generation & refrigeration, we encounter the combination of properties u + Pv. This combination is defined as enthalpy, h: h = u + Pv (kJ/kg)

4. Use of Enthalpy The widespread use of the property enthalpy is due to Professor Richard Mollier (Germany), who recognized the importance of the group u + Pv in the analysis of steam turbines. The properties of the of steam are represented in tabular and graphical form (as in the famous Mollier chart). The term enthalpy (from Greek enthalpien, which means to heat) was used in the 1930s.

5. Saturated Liquid and Saturated Vapor Table A–4  based on T Table A–5  based on P Subscript used: f = for saturated liquid g = for saturated vapor. fg = difference of saturated vapor and saturated liquid v f = v of saturated liquid v g = v of saturated vapor v fg = difference of v g and v f v fg = v g - v f Table A–4

6. EXAMPLE 3–1 A rigid tank contains 50 kg of saturated liquid water at 90°C. Determine the pressure in the tank and the volume of the tank. Schematic and T-v diagram

7. EXAMPLE 3–1 : Solution Analysis Since saturation conditions exist in the tank, the pressure must be the saturation pressure at 90°C: P = P sat @ 90°C = 70.183 kPa (Table A–4) The specific volume of the saturated liquid at 90°C is v = v f @ 90°C = 0.001036 m 3 /kg (Table A–4) Then the total volume of the tank becomes V = mv = 50 kg (0.001036 m 3 /kg) = 0.0518 m 3

8. EXAMPLE 3–3 A mass of 200 g of saturated liquid water is completely vaporized at a constant pressure of 100 kPa. Determine (a) the volume change and (b) the amount of energy transferred to the water. Schematic and P-v diagram

9. EXAMPLE 3–3 : Solution Analysis (a) The volume change per unit mass during a vaporization process is v fg, which is the difference between v g and v f. Using Table A–5 at 100 kPa and substituting yield: v fg = v g - v f = 1.6941 - 0.001043 = 1.6931 m 3 /kg  V = m.v fg = (0.2 kg)(11.6931 m 3 /kg) = 0.3386 m 3 (b) The amount of energy needed to vaporize a unit mass of water at a given pressure is the enthalpy of vaporization at that pressure, which is h fg = 2257.5 kJ/kg for water at 100 kPa. Thus, the amount of energy transferred is: H fg = m  h fg = (0.2 kg)(2257.5 kJ/kg) = 451.5 kJ

10. Saturated Liquid–Vapor Mixture To analyze this mixture properly, we need to know the proportions of the liquid and vapor phases in the mixture. A new property called the quality x as the ratio of the mass of vapor to the total mass of the mixture: where:

11. Quality, x Quality has significance for saturated mixtures only. Its value is between 0 and 1. The quality of a system that consists of saturated liquid is 0 (or 0 percent), and the quality of a system consisting of saturated vapor is 1 (or 100 percent). In saturated mixtures, quality can serve as one of the two independent intensive properties needed to describe a state. Note that the properties of the saturated liquid are the same whether it exists alone or in a mixture with saturated vapor. During the vaporization process, only the amount of saturated liquid changes, not its properties. The same can be said about a saturated vapor during condensation.

12. The Use of Quality, x Consider a tank that contains a saturated liquid–vapor mixture. The volume of saturated liquid is V f, and the volume of saturated vapor is V g. The total volume V is : V = V f + V g V = mv  m t.v avg = m f.v f + m g.v g m f = m t – m g  m t.v avg = (m t – m g ) v f + m g.v g Dividing by mt gives: v avg = (1 – x) v f + x.v g since x = m g /m t. This relation can also be expressed as v avg = v f + x.v fg (m 3 /kg) where v fg = v g – v f. Solving for quality, we obtain

13. The Use of Quality, x Analog: u avg = u f + x.u fg (kJ/kg) h avg = h f + x.h fg (kJ/kg) y avg = y f + x.y fg where y is v, u, or h. y f  y avg  y g The subscript “avg” (average) is usually dropped for simplicity. y f  y  y g

14. EXAMPLE 3–4 Pressure and Volume A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in the liquid form and the rest is in the vapor form, determine (a) the pressure in the tank and (b) the volume of the tank. Schematic and T-v diagram

15. EXAMPLE 3–4 : Solution Analysis (a) Since the two phases coexist in equilibrium, we have a saturated mixture, and the pressure must be the saturation pressure at the given temperature: P = P sat @ 90°C = 70.183 kPa (Table A–4) (b) At 90°C, we have v f = 0.001036 m 3 /kg and v g = 2.3593 m 3 /kg (Table A–4). To find the volume of the tank is to determine the volumes of each phase and then add them: V = V f + V g = m f.v f + m g.v g = (8 kg)(0.001036 m 3 /kg) + (2 kg)(2.3593 m 3 /kg) = 4.73 m 3

16. EXAMPLE 3–4 : Solution (Alt) (b) Another way is to first determine the quality x, then the average specific volume v, and finally the total volume: and

17. EXAMPLE 3–5 Saturated Mixture An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160 kPa. Determine (a) the temperature, (b) the quality, (c) the enthalpy of the refrigerant, and (d) the volume occupied by the vapor phase. Schematic and P-v,h diagram

18. EXAMPLE 3–5 : Solution Analysis (a) At this point we do not know whether the refrigerant is in the compressed liquid, superheated vapor, or saturated mixture region. From the information in the problem: v = V/m = (0.080 m 3 )/4 kg = 0.02 m 3 /kg From Table A–12 @ 160 kPa  v f = 0.0007437 m 3 /kg  v g = 0.12348 m 3 /kg Obviously, vf < v < vg, and, the refrigerant is in the saturated mixture region. Thus, the temperature must be the saturation temperature: T = T sat @ 160 kPa = 15.60°C

19. EXAMPLE 3–5 : Solution (Cont’d) (b) Quality can be determined from (c) At 160 kPa, we also read from Table A–12 that h f = 31.21 kJ/kg and h fg = 209.90 kJ/kg. Then,

20. EXAMPLE 3–5 : Solution (Cont’d) (d) The mass of the vapor is (remember definition of x): m g = x.m t = (0.157)(4 kg) = 0.628 kg and the volume occupied by the vapor phase is V g = m g.v g = (0.628 kg)(0.12348 m 3 /kg) V g = 0.0775 m 3 = 77.5 L The rest of the volume (2.5 L) is occupied by the liquid.

21. Superheated Vapor (Table A-6) Superheated vapor is characterized by: Lower pressure (P < P sat at a given T) Higher temperature (T > T sat at a given P) Higher specific volume (v > v g at a given P or T) Higher internal energy (u > u g at a given P or T) Higher enthalpy (h > h g at a given P or T) Table A-6

22. EXAMPLE 3–6 Internal Energy Determine the internal energy of water at 20 psia and 400°F. Analysis At 20 psia, the saturation temperature is 227.92°F. Since T > T sat, the water is in the superheated vapor region. Then the internal energy at the given temperature and pressure is determined from the superheated vapor table (Table A–6E) to be u = 1145.1 Btu/lbm

23. EXAMPLE 3–7 T of Superheated Vapor Determine the temperature of water at a state of P = 0.5 MPa and h = 2890 kJ/kg. Analysis At 0.5 MPa, the enthalpy of saturated water vapor is h g = 2748.1 kJ/kg. Since h > h g, then we have superheated vapor. Under 0.5 MPa in Table A–6 we read: T is between 200 and 250°C. By linear interpolation : T = 216.3°C

24. Compressed Liquid (Table A-7) Compressed liquid tables are not commonly available. Variation of properties due to P is insignificant. Increasing the P 100 times, often causes properties to change < 1 %. If no data for compressed liquid, compressed liquid is treated as saturated liquid at the given T. Thus, To reduce error in h at low to moderate P and T :

25. Compressed Liquid In general, a compressed liquid is characterized by Higher pressures (P > P sat at a given T) Lower temperatures (T < T sat at a given P) Lower specific volumes (v < v f at a given P or T) Lower internal energies (u < u f at a given P or T) Lower enthalpies (h < h f at a given P or T)

26. EXAMPLE 3–8 Compressed Liquid Determine the internal energy of compressed liquid water at 80°C and 5 MPa, using (a) data from the compressed liquid table and (b) saturated liquid data. What is the error involved in the second case? Schematic and T-u diagram 0.047 MPa

27. EXAMPLE 3–8 : Solution Analysis At 80°C, the saturation pressure of water is 47.416 kPa, and since 5 MPa > P sat, we obviously have compressed liquid. (a) From the compressed liquid table (Table A–7) (b) From the saturation table (Table A–4), we read u = u f @ 80°C = 334.97 kJ/kg The error is: < 1 percent.

28. Example 9 : Missing Properties

29. Example 9 : Solution (a) The quality is given to be x = 0.6, which implies that 60 percent of the mass is in the vapor phase and the rest is liquid. Therefore, we have saturated liquid–vapor mixture at a P of 200 kPa. The T must be the T sat at the given pressure: T = T sat @ 200 kPa = 120.21°C (Table A–5) At 200 kPa, from Table A–5 : u f = 504.50 kJ/kg and u fg = 2024.6 kJ/kg. The average internal energy :

30. Example 9 : Solution (Cont’d) (b) The temperature and the internal energy are given, but we do not know which table to be used. First, we go to Table A–4 and find u f and u g at 125°C, where u f = 524.83 kJ/kg and u g = 2534.3 kJ/kg. Next we compare the given u to u f and u g. Mind that: If u < u f  we have compressed liquid If u f < u < u g  we have saturated mixture If u > u g  we have superheated vapor In our case u = 1600 is in between the u f and u g at 125°C. Therefore, we have saturated liquid–vapor mixture. The pressure must be P sat at the given temperature: P = P sat @ 125°C = 232.23 kPa (Table A–4) The quality is determined from:

31. Example 9 : Solution (Cont’d) (c) This is similar to case (b), except pressure is given instead of temperature. Following the argument given above, we read the u f and u g values at the specified pressure. At 1 MPa, we have u f = 761.39 kJ/kg and u g = 2582.8 kJ/kg. The specified u value is 2950 kJ/kg, which is greater than the u g value at 1 MPa. Therefore, we have superheated vapor, and the temperature at this state is determined from the superheated vapor table by interpolation to be: T = 395.2°C (Table A–6) We would leave the quality column blank in this case since quality has no meaning for a superheated vapor.

32. Example 9 : Solution (Cont’d) (d) In this case the temperature and pressure are given, but again we cannot tell which table to use. We go to Table A–5 to find T sat at 500 kPa, where we have T sat = 151.83°C. Compare T to T sat, mind that: T < T sat @ given P  compressed liquid T = T sat @ given P  saturated mixture T > T sat @ given P  superheated vapor In our case, T = 75°C < T sat so we have compressed liquid. But no data available, that we treat as saturated liquid at 75°C, and Again, we would leave the quality column blank

33. Example 9 : Solution (Cont’d) (e) The quality is given to be x = 0, and thus we have saturated liquid at the specified pressure of 850 kPa. Then the temperature must be the saturation temperature at the given pressure, and the internal energy must have the saturated liquid value:

34. IDEAL-GAS EQUATION OF STATE Property tables provide very accurate information about the properties, but they are bulky and prone to typo errors. A practical approach is to have some simple relations among the properties that are sufficiently general and accurate. Any equation that relates the pressure, temperature, and specific volume of a substance is called an equation of state (EOS). Property relations that involve other properties of a substance at equilibrium states are also referred to as EOS. The simplest and best-known equation of state for substances in the gas phase is the ideal-gas EOS. This equation predicts the P-v-T behavior of a gas quite accurately within some properly selected region.

35. Ideal-gas EOS Also known as Boyle-Gay Lussac Law Pv = RTor A gas that obeys this relation is called an ideal gas R is called the gas constant, which is different for each gas and is determined from: R u = universal gas constant. M = molar mass (molecular weight). Eg. N 2 = 28.

36. R u = universal gas constant The constant R u is the same for all substances, and its value is:

37. Forms of Ideal-gas EOS The ideal-gas equation of state can be written in several different forms: Pv = RT is the molar specific volume, that is, the volume per unit mole (m 3 /kmol or ft 3 /lbmol).

38. Forms of Ideal-gas EOS By writing ideal gas EOS twice for a fixed mass and simplifying, the properties of an ideal gas at two different states are related to each other by An ideal gas is an imaginary substance that obeys the relation Pv = RT

39. EXAMPLE 3–10 Mass of Air in a Room Determine the mass of the air in a room whose dimensions are 4 m x 5 m x 6 m at 100 kPa and 25°C. Analysis Assume air at that condition can be treated as an ideal gas. From Table A–1, gas constant of air is R air = 0.287 kPa·m 3 /kg·K, and absolute temperature is T = 25°C + 273 = 298 K. The volume of the room, V = 4 x 5 x 6 = 120 m 3. From the ideal-gas relation:

40. Is Water Vapor an Ideal Gas? Percentage of error involved in assuming steam to be an ideal gas, and the region where steam can be treated as an ideal gas with less than 1 percent error.

41. COMPRESSIBILITY FACTOR, Z The ideal-gas equation is very simple and thus very convenient to use. Real gases deviate from ideal-gas behavior significantly at states near the saturation region and the critical point. This deviation from ideal-gas behavior at a given temperature and pressure can accurately be accounted for by the introduction of a correction factor called the compressibility factor Z: or

42. Reduced Pressure and Temperature Gases behave differently at a given temperature and pressure, but they behave very much the same at temperatures and pressures normalized with respect to their critical temperatures and pressures. The normalization is done as P R = reduced pressure ; T R = reduced temp. The Z factor for all gases is same at the same P R and T R.

46. generalized compressibility chart

47. Ggeneralized compressibility chart The following observations can be made from the generalized compressibility chart: 1. At very low pressures (P R << 1), gases behave as an ideal gas (Z  1) regardless of temperature (Fig. 3– 52). 2. At high temperatures (T R > 2), ideal-gas behavior can be assumed with good accuracy regardless of pressure (except when P R >> 1). 3. The deviation of a gas from ideal- gas behavior is greatest in the vicinity of the critical point (Fig. 3–53).

48. EXAMPLE 3–11 Use of Generalized Charts Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the generalized compressibility chart. Compare the values obtained to the actual value of 0.021796 m3/kg and determine the error involved in each case.

49. EXAMPLE 3–11 : Solution Analysis From Table A–1 we have: R = 0.0815 kPa.m 3 /kg.K P cr = 4.059 Mpa T cr = 374.2 K (a) Using ideal-gas assumption : Error = (0.026325 - 0.021796)/0.021796 = 0.208, or 20.8 percent.

51. EXAMPLE 3–11 : Solution (Cont’d) (b) To determine the correction factor Z from the compressibility chart, we first need to calculate the reduced pressure and temperature: Thus:

52. Pseudo-reduced specific volume, v R When P and v, or T and v, are given instead of P and T, the generalized compressibility chart can still be used to determine the third property by defining pseudo-reduced specific volume v R : Go to Fig. A–15

53. EXAMPLE 3–12 Generalized Charts Determine the pressure of water vapor at 600°F and 0.51431 ft3/lbm, using (a) the steam tables, (b) the ideal-gas equation, and (c) the generalized compressibility chart.

54. EXAMPLE 3–12 : Solution From Table A–1E we have: R = 0.5956 psia.ft 3 /lbm.R P cr = 3200 psiaand T cr = 1164.8 R (a) The pressure at the specified state is determined from Table A–6E: (b) Using ideal-gas assumption:

55. EXAMPLE 3–12 : Solution (Cont’d) error = (1228 - 1000)/1000 = 0.228, or 22.8 % (c) To determine the correction factor Z from the compressibility chart : Using Fig. A–15, at v R = 2.372 and T R = 0.91, we have P R = 0.33. Thus, P = P R.Pcr = (0.33)(3200 psia) = 1056 psia Error = 5.6 percent

56. OTHER EQUATIONS OF STATE The ideal-gas equation of state is very simple, but its range of applicability is limited. Van der Waals EOS The constants a and b are provided in Table A–1.

57. OTHER EQUATIONS OF STATE Beattie-Bridgeman The constants for Beattie-Bridgeman EOS are given in Table 3–4 for various substances. The Beattie-Bridgeman EOS is reasonably accurate for densities up to about 0.8  cr (  at critical point).

58. OTHER EQUATIONS OF STATE Benedict-Webb-Rubin:

60. OTHER EQUATIONS OF STATE Virial Equation of State

62. EXAMPLE 3–13 Different EOS Predict the pressure of nitrogen gas at T = 175 K and v = 0.00375 m 3 /kg on the basis of (a) the ideal-gas equation of state, (b) the van der Waals equation of state, (c) the Beattie-Bridgeman equation of state, and (d) the Benedict-Webb-Rubin equation of state. Compare the values obtained to the experimentally determined value of 10,000 kPa.

63. EXAMPLE 3–13 : Solution Analysis Gas constant for N 2 (Table A–1) is R = 0.2968 kPa.m 3 /kg.K. (a) Using the ideal-gas EOS: Error = 38.5 percent (b) Using van der Waals EOS. For nitrogen (Table A-1), T cr = 126.2 K and P cr = 3.39 Mpa = 3390 kPa. = a = 0.175 m 6.kPa/kg 2

64. EXAMPLE 3–13 : Solution (Cont’d) = b = 0.00138 m 3 /kg Error = 5.3 percent (c) The constants in the Beattie-Bridgeman EOS (Table 3-4) : A = 102.29;B = 0.05378; C = 4.2  10 4.

65. EXAMPLE 3–13 : Solution (Cont’d) Substituting the values into Equation, we obtain: error = 1.1 percent. (d) The constants in the Benedict-Webb-Rubin equation are (Table 3-4): a = 2.54 A 0 = 106.73 b = 0.002328 B 0 = 0.04074 c = 7.379  10 4 C 0 = 8.164  10 5  = 1.272  10 4  = 0.0053

66. EXAMPLE 3–13 : Solution (Cont’d) Substituting these values into Equation gives: Error = 0.09 percent

67. HOMEWORK