1. Thermodynamic Properties Property Table w Property Table -- from direct measurement w Equation of State w Equation of State -- any equations that relates P,v, and T of a substance

2. Ideal -Gas Equation of State w Any relation among the pressure, temperature, and specific volume of a substance is called an equation of state. The simplest and best-known equation of state is the ideal-gas equation of state, given as where R is the gas constant. Caution should be exercised in using this relation since an ideal gas is a fictitious substance. Real gases exhibit ideal-gas behavior at relatively low pressures and high temperatures.

3. Universal Gas Constant Universal gas constant is given on R u = 8.31434 kJ/kmol-K = 8.31434 kPa-m 3 /kmol-k = 0.0831434 bar-m 3 /kmol-K = 82.05 L-atm/kmol-K = 1.9858 Btu/lbmol-R = 1545.35 ft-lbf/lbmol-R = 10.73 psia-ft 3 /lbmol-R

4. Example Determine the particular gas constant for air and hydrogen.

5. Ideal Gas “Law” is a simple Equation of State

6. Percent error for applying ideal gas equation of state to steam

7. Question …... Under what conditions is it appropriate to apply the ideal gas equation of state?

8. Ideal Gas Law w Good approximation for P-v-T behaviors of real gases at low densities (low pressure and high temperature). w Air, nitrogen, oxygen, hydrogen, helium, argon, neon, carbon dioxide, …. ( < 1% error).

9. Compressibility Factor w The deviation from ideal-gas behavior can be properly accounted for by using the compressibility factor Z, defined as Z represents the volume ratio or compressibility.

10. Ideal Gas Z=1 Real Gases Z > 1 or Z < 1

11. Real Gases w Pv = ZRT or w Pv = ZR u T, where v is volume per unit mole. w Z is known as the compressibility factor. w Real gases, Z 1.

12. Compressibility factor w What is it really doing? w It accounts mainly for two things Molecular structureMolecular structure Intermolecular attractive forcesIntermolecular attractive forces

13. Principle of corresponding states w The compressibility factor Z is approximately the same for all gases at the same reduced temperature and reduced pressure. Z = Z(P R,T R ) for all gases

14. Reduced Pressure and Temperature where: P R and T R are reduced values. P cr and T cr are critical properties.

15. Compressibility factor for ten substances (applicable for all gases Table A-3)

16. Where do you find critical-point properties? Table A-7 Mol (kg-Mol) R (J/kg.K) Tcrit (K) Pcrit (MPa) Ar28,97287,0(---) O2O2 32,00259,8154,85,08 H2H2 2,0164124,233,31,30 H2OH2O18,016461,5647,122,09 CO 2 44,01188,9304,27,39

17. Reduced Properties w This works great if you are given a gas, a P and a T and asked to find the v. w However, if you are given P and v and asked to find T (or T and v and asked to find P), trouble lies ahead. w Use pseudo-reduced specific volume.

18. Pseudo-Reduced Specific Volume w When either P or T is unknown, Z can be determined from the compressibility chart with the help of the pseudo- reduced specific volume, defined as not v cr !

19. Ideal-Gas Approximation w The compressibility chart shows the conditions for which Z = 1 and the gas behaves as an ideal gas: w (a) P R 1

20. Exercise 3-21 Steam at 600 o C & 1 MPa. Evaluate the specific volume using the steam table and ideal gas law Tsat = 180 o C therefore is superheated steam (Table A-3) Volume = 0,4011 m 3 /kg ToCToCToCToC1MPa v 374 600 180 Tcr

21. Solution - page 1 Part (b) ideal gas law Rvapor = 461 J/kgK v = RT/P, v = 461x873/10 6 = 0.403 m 3 /kg Steam is clearly an ideal gas at this state. Error is less than 0.5% Check: P R = 1/22.09 = 0.05 & T R = 873/647 = 1.35

22. TEAMPLAY Find the compressibility factor to determine the error in treating oxygen gas at 160 K and 3 MPa as an ideal gas.

23. Other Thermodynamic Properties: Isobaric (c. pressure) Coefficient v T P

24. Other Thermodynamic Properties: Isothermal (c. temp) Coefficient v P T

25. Other Thermodynamic Properties: We can think of the volume as being a function of pressure and temperature, v = v(P,T). Hence infinitesimal differences in volume are expressed as infinitesimal differences in P and T, using  and  coefficients If  and  are constant, we can integrate for v:

26. Other Thermodynamic Properties: Internal Energy, Enthalpy and Entropy

27. Other Thermodynamic Properties: Specific Heat at Const. Volume u T v

28. Other Thermodynamic Properties: Specific Heat at Const. Pressure h T P

29. Other Thermodynamic Properties: Ratio of Specific Heat

30. Other Thermodynamic Properties: Temperature s u v T 1

31. Ideal Gases: u = u(T) Therefore, 0

32. We can start with du and integrate to get the change in u: Note that C v does change with temperature and cannot be automatically pulled from the integral.

33. Let’s look at enthalpy for an ideal gas: w h = u + Pv where Pv can be replaced by RT because Pv = RT. w Therefore, h = u + RT => since u is only a function of T, R is a constant, then h is also only a function of T w so h = h(T)

34. Similarly, for a change in enthalpy for ideal gases:

35. Summary: Ideal Gases w For ideal gases u, h, C v, and C p are functions of temperature alone. w For ideal gases, C v and C p are written in terms of ordinary differentials as

36. For an ideal gas, w h = u + Pv = u + RT C p = C v + R

37. Ratio of specific heats is given the symbol, 

38. Other relations with the ratio of specific heats which can be easily developed:

39. For monatomic gases, Argon, Helium, and Neon

40. For all other gases, w C p is a function of temperature and it may be calculated from equations such as those in Table A-5(c) in the Appendice w C v may be calculated from C p =C v +R. w Next figure shows the temperature behavior …. specific heats go up with temperature.

41. Specific Heats for Some Gases w C p = C p (T) a function of temperature

42. Three Ways to Calculate Δu and Δh w Δu = u 2 - u 1 (table) w Δu = w Δu = C v,av ΔT w Δh = h 2 - h 1 (table) w Δh = w Δh = C p,av ΔT

43. Isothermal Process w Ideal gas: PV = mRT

44. For ideal gas, PV = mRT We substitute into the integral Collecting terms and integrating yields:

45. Polytropic Process w PV n = C

46. Ideal Gas Adiabatic Process and Reversible Work w What is the path for process with expand or contract without heat flux? How P,v and T behavior when Q = 0? w To develop an expression to the adiabatic process is necessary employ: 1. Reversible work mode: dW = PdV 2. Adiabatic hypothesis: dQ =0 3. Ideal Gas Law: Pv=RT 4. Specific Heat Relationships 5. First Law Thermodynamics: dQ-dW=dU

47. Ideal Gas Adiabatic Process and Reversible Work (cont) First Law: Using P = MRT/V Integrating from (1) to (2)

48. Ideal Gas Adiabatic Process and Reversible Work (cont) Using the gas law : Pv=RT, other relationship amid T, V and P are developed accordingly:

49. Ideal Gas Adiabatic Process and Reversible Work (cont) An expression for work is developed using PV  = constant. i and f represent the initial and final states

50. Ideal Gas Adiabatic Process and Reversible Work (cont)  The path representation are lines where Pv  = constant.  For most of the gases,   1.4 w The adiabatic lines are always at the righ of the isothermal lines. w The former is Pv = constant (the exponent is unity) P v Q = 0 T=const. i f f

51. Polytropic Process A frequently encountered process for gases is the polytropic process: PV n = c = constant Since this expression relates P & V, we can calculate the work for this path.

52. Polytropic Process w Constant pressure 0 w Constant volume  w Isothermal & ideal gas 1 w Adiabatic & ideal gas k=C p /C v Process Exponent n

53. Boundary work for a gas which obeys the polytropic equation

54. We can simplify it further The constant c = P 1 V 1 n = P 2 V 2 n

55. Polytropic Process

56. Exercise 3-11 A bucket containing 2 liters of water at 100 o C is heated by an electric resistance. a) a)Identify the energy interactions if the system boundary is i) the water, ii) the electric coil b) b) If heat is supplied at 1 KW, then how much time is needed to boil off all the water to steam? (latent heat of vap at 1 atm is 2258 kJ/kg) c) c) If the water is at 25 o C, how long will take to boil off all the water (Cp = 4.18 J/kgºC)

57. Solution - page 1 Part a) If the water is the system then Q > 0 and W = 0. There is a temperature difference between the electric coil and the water. If the electric coil is the system then Q < 0 and W < 0. It converts 100% of the electric work to heat!

58. Solution - page 2 The mass of water is of 2 kg. It comes from 2liters times the specific volume of 0.001 m3/kg To boil off all the water is necessary to supply all the vaporization energy: Evap = 2285*2=4570 KJ The power is the energy rate. The time necessary to supply 4570 KJ at 1KJ/sec is therefore: Time = 4570/1 = 4570 seconds or 1.27 hour Part b)

59. Solution - page 3 The heat to boil off all the water initially at 25 o C is the sum of : (1) the sensible heat to increase the temperature from 25 o C to 100 o C, (2) the vap. heat of part (b) The sensible heat to increase from 25 o C to 100 o C is determined using the specific heat (C P = 4.18 kJ/kg o C) Heat = 4.18*2*(100-25)=627 KJ The time necessary to supply (4570+627)KJ at 1KJ/sec is : Time = 5197/1 = 5197 seconds or 1.44 hour Part c)

60. Exercise 3-12 A bucket containing 2 liters of R-12 is left outside in the atmosphere (0.1 MPa) a) a)What is the R-12 temperature assuming it is in the saturated state. b) b) the surrounding transfer heat at the rate of 1KW to the liquid. How long will take for all R- 12 vaporize?

61. Solution - page 1 Part a) From table A-2, at the saturation pressure of 0.1 MPa one finds: Tsaturation = - 30 o C V liq = 0.000672 m 3 /kg v vap = 0.159375 m 3 /kg h lv = 165KJ/kg (vaporization heat)

62. Solution - page 2 Part b) The mass of R-12 is m = Volume/v L, m=0.002/0.000672 = 2.98 kg The vaporization energy: Evap = vap energy * mass = 165*2.98 = 492 KJ Time = Heat/Power = 492 sec or 8.2 min

63. Exercise 3-17 A rigid tank contains saturated steam (x = 1) at 0.1 MPa. Heat is added to the steam to increase the pressure to 0.3 MPa. What is the final temperature?

64. Solution - page 1 P0.1MPa v 99.63 o C Tcr = 374 o C 0.3MPa 133.55 o C superheated Process at constant volume, search at the superheated steam table for a temperature corresponding to the 0.3 MPa and v = 1.690 m 3 /kg -> nearly 820 o C.

65. Exercise 3-30 Air is compressed reversibly and adiabatically from a pressure of 0.1 MPa and a temperature of 20 o C to a pressure of 1.0 MPa. a) a)Find the air temperature after the compression b) b)What is the density ratio (after to before compression) c) c)How much work is done in compressing 2 kg of air? d) d)How much power is required to compress 2kg per second of air?

66. w In a reversible and adiabatic process P, T and v follows: P v Q = 0 T=const. i f f Solution - page 1

67. Solution - page 2 Part a) The temperature after compression is Part b) The density ratio is

68. Solution - page 3 Part c) The reversible work: Part d) The power is:

69. Exercícios – Capítulo 3 Propriedades das Substâncias Puras Exercícios Propostos: 3.6 / 3.9 / 3.12 / 3.16 / 3.21 / 3.22 / 3.26 / 3.30 / 3.32 / 3.34 Team Play: 3.1 / 3.2 / 3.4

70. Ex. 3.1) Utilizando a Tabela A-1.1 ou A-1.2, determine se os estados da água são de líquido comprimido, líquido-vapor, vapor superaquecido ou se estão nas linhas de líquido saturado ou vapor saturado. w a) P=1,0 MPa; T=207 ºC w b) P=1,0 MPa; T=107,5 ºC w c) P=1,0 MPa; T=179,91 ºC; x=0,0 w d) P=1,0 MPa; T=179,91 ºC; x=0,45 w e) T=340 ºC; P=21,0 MPa w f) T=340 ºC; P=2,1 MPa w g) T=340 ºC; P=14,586 MPa; x=1,0 w h) T=500 ºC; P=25 MPa w i) P=50 MPa; T=25 ºC  a) Vapor superaquecido  b) Líquido comprimido  c) Líquido saturado  d) Líquido-Vapor  e) Líquido comprimido  f) Vapor superaquecido  g) Vapor saturado  h) Vapor superaquecido - Fluido  i) Líquido comprimido - Fluido

71. w Ex. 3.2) Encontre o volume específico dos estados “b”, “d” e “h” do exercício anterior. b) P=1,0 MPa; T=107,5 ºC  v  v l =0,001050 (utilizar referência T=107,5ºC) d) P=1,0 MPa; T=179,91 ºC; x=0,45  v=0,08812 [v=(1-x)v l +x(v v )] h) T=500 ºC; P=25 MPa  v=0,011123 m 3 /kg (Tabela A-1.3 Vapor Superaquecido)

72. w Ex. 3.4) Amônia a P=150 kPa, T=0 ºC se encontra na região de vapor superaquecido e tem um volume específico e entalpia de 0,8697 m 3 /kg e 1469,8 kJ/kg, respectivamente. Determine sua energia interna específica neste estado. u =1339,345 kJ/kg (u = h - P v)