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Thermodynamic Properties

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

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Thermodynamic Properties

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

  2. Ideal -Gas Equation of State • 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 ofstate, given aswhere 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 hightemperatures.

  3. Universal Gas Constant Universal gas constant is given on Ru = 8.31434 kJ/kmol-K = 8.31434 kPa-m3/kmol-k = 0.0831434 bar-m3/kmol-K = 82.05 L-atm/kmol-K = 1.9858 Btu/lbmol-R = 1545.35 ft-lbf/lbmol-R = 10.73 psia-ft3/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 • Good approximation for P-v-T behaviors of real gases at low densities (low pressure and high temperature). • Air, nitrogen, oxygen, hydrogen, helium, argon, neon, carbon dioxide, …. ( < 1% error).

  9. Compressibility Factor • 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. Compressibility Factor Real Gases Ideal Gas Z > 1 or Z < 1 Z=1

  11. Real Gases • Pv = ZRT or • Pv = ZRuT, where v is volume per unit mole. • Z is known as the compressibility factor. • Real gases, Z < 1 or Z > 1.

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

  13. Principle of corresponding states • The compressibility factor Z is approximately the same for all gases at the same reduced temperatureand reduced pressure. Z = Z(PR,TR) for all gases

  14. Reduced Pressure and Temperature where: PR and TR are reduced values. Pcr and Tcr 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

  17. Reduced Properties • This works great if you are given a gas, a P and a T and asked to find the v. • 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. • Use pseudo-reduced specific volume.

  18. Pseudo-Reduced Specific Volume • 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 vcr !

  19. Ideal-Gas Approximation • The compressibility chart shows the conditions for which Z = 1 and the gas behaves as an ideal gas: • (a) PR < 0.1 and TR > 1

  20. ToC 1MPa 600 Tcr 374 180 v Exercise 3-21 Steam at 600 oC & 1 MPa. Evaluate the specific volume using the steam table and ideal gas law Tsat = 180oC therefore is superheated steam (Table A-3) Volume = 0,4011 m3/kg

  21. Solution - page 1 Part (b) ideal gas law Rvapor = 461 J/kgK v = RT/P , v = 461x873/106 = 0.403 m3/kg Steam is clearly an ideal gas at this state. Error is less than 0.5% Check: PR = 1/22.09 = 0.05 & TR = 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. P Other Thermodynamic Properties:Isobaric (c. pressure) Coefficient v T

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

  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 k and b coefficients If k and b are constant, we can integrate for v:

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

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

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

  29. Other Thermodynamic Properties:Ratio of Specific Heat

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

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

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

  33. Let’s look at enthalpy for an ideal gas: • h = u + Pv where Pv can be replaced by RT because Pv = RT. • 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 • so h = h(T)

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

  35. Summary: Ideal Gases • For ideal gases u, h, Cv, and Cp are functions of temperature alone. • For ideal gases, Cv and Cp are written in terms of ordinary differentials as

  36. For an ideal gas, • h = u + Pv = u + RT Cp = Cv + R

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

  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, • Cpis a function of temperature and it may be calculated from equations such as those in Table A-5(c) in the Appendice • Cv may be calculated from Cp=Cv+R. • Next figure shows the temperature behavior …. specific heats go up with temperature.

  41. Specific Heats for Some Gases • Cp = Cp(T)a function of temperature

  42. Δu = u2 - u1 (table) Δu = Δu = Cv,av ΔT Δh = h2 - h1 (table) Δh = Δh = Cp,av ΔT Three Ways to Calculate Δu and Δh

  43. Isothermal Process • Ideal gas: PV = mRT

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

  45. Polytropic Process • PVn = C

  46. Ideal Gas Adiabatic Process and Reversible Work • What is the path for process with expand or contract without heat flux? How P,v and T behavior when Q = 0? • To develop an expression to the adiabatic process is necessary employ: • Reversible work mode: dW = PdV • Adiabatic hypothesis: dQ =0 • Ideal Gas Law: Pv=RT • Specific Heat Relationships • 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 PVg = 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 Pvg = constant. • For most of the gases, g1.4 • The adiabatic lines are always at the righ of the isothermal lines. • The former is Pv = constant (the exponent is unity) P f f Q = 0 T=const. i v

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