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- 1. Engineering Thermodynamics: Properties of Pure Substances To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 2. Outline 2 • 2−1 Pure Substances • 2−2 Phases of a Pure Substance • 2−3 Phase Change processes of a Pure Substance • 2−4 Property Diagrams for Phase Change Processes • 2−5 Property Tables • 2-6 The Ideal gas Equation of State • 2−7 Compressibility Factor • 2-9 Specific Heats • 2-10&11 Internal Energy and Enthalpy Formulation For Ideal Gases and Condensed Phases (Solid and Liquids) To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 3. Pure Substances 3 A pure substance does not change chemical composition but can change phase. Is the working fluid a two-phase substance? H2O R-134a An “ideal” gas? Air To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 4. Phases of a Pure Substance liquid vapor 2 Pure H O vapor liquid 4 Water Air Not pure: different condensation temperatures for different components To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 5. Phase Change Processes of Pure Substances 5 Terminology • Compressed liquid -- not about to evaporate • Saturated liquid -- about to evaporate • Saturated liquid-vapor mixture --two phase • Saturated Vapor -- about to condense • Superheated Vapor -- not about to condense To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 6. Heating Water- Phases 6 Compressed Liquid Saturated Liquid Saturated Liquid-Vapor Mixture Liquid Liquid Liquid/Vapor
- 7. Heating Water- Phases Vapor 7 Vapor Saturated Vapor Superheated Vapor To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 8. T-v Diagram for Heating Water Isobaric process 8 To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 9. Saturation 9 • At a given pressure, a substance changes phase at a fixed temperature, called the saturation temperature. • At a given temperature, the pressure at which a substance changes phase is called the saturation pressure. • T and P are dependant during the phase change, i.e., Tsat = f(Psat) To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 10. Latent Heat 10 • Latent heat is the amount of energy absorbed or released during phase change • Latent heat of fusion: melting/freezing. Equals 333.7 kJ/kg for 1 atm H2O • Latent heat of vaporization: boiling/condensation. Equals 2257.1 kJ/kg for 1 atm H2O To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 11. Property Diagrams for Phase Change Processes T-v diagram 11 A point beyond which T Tcr , a liquid-vapor transition is no longer possible at constant pressure. If T Tcr , the substance cannot be liquefied, no matter how great the pressure. To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 12. 12 Scenarios for Phase Change Processes • Compressed liquid or subcooled liquid: - If heat is added, the temperature will rise but there will be no phase change. - “Not about to vaporize” - Example: liquid water at 20oC/1 atm. • Saturated liquid: - If heat is added, the temperature does not change but some of the liquid vaporizes. - Example: liquid water at 100oC/1 atm. • Saturated liquid/vapor mixture: - As heat is added, liquid turns to vapor. - Example: vapor water at 100oC/l atm. • Saturated vapor: - If heat is removed, the temperature does not change, but some of the vapor condenses. - Example: vapor water at 100oC/1 atm. • Superheated vapor: - If heat is added, the temperature will rise but there will be no phase change. - Example: vapor vapor at 120oC/1 atm. QH
- 13. Phase-Change: P-T Diagram 13 To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 14. P-T Diagram: Isothermal Process 14 GAS @ g Weight GAS State d LIQUID GAS Weight LIQUID @ a P Compressed Liquid a d Superheated Vapor g T To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 15. P-T Diagram: Isobaric Process 15 Gas @ b Q GAS STATE f LIQUID Q GAS LIQUID Q P T b f Superheated Vapor Subcooled Liquid a To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 16. P-V Diagram: Substance that Contracts on Freezing 16 For water, the triple point is at 273.16 K (32.018 F) and 0.6113 kPa (0.0887 psia)
- 17. P-V Diagram: Substance that Expands on Freezing 17
- 18. Using T-v Diagrams 18 Sat. Vapor line Sat. Liquid line T v 1. Draw saturation lines 2. Note given information on state properties T P v U H To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 19. Using T-v Diagrams 19 v P const. T v T const. P 3. Draw constant pressure or constant temperature lines To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 20. Using T-v Diagrams 20 T const. P v P sat. vf; uf; hf vg; ug; hg critical point v P const. T vf; uf; hf vg; ug; hg T sat. critical point 4. Note saturation temperature or saturation pressure, sat. liquid and sat. vap. states To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 21. T v vf; uf; hf vg; ug; hg P > P const. T sat. P v P sat. T> T const. vf; uf; hf vg; ug; hg Using T-v Diagrams 5. Add lines showing higher or lower temperature or pressure To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 22. Property Tables 22 Enthalpy – A Combination Property H = U + PV or h = u + Pv To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 23. Using Property Tables 23 Saturated Liquid and saturated vapor water Tables A-4 and A-5 Example: A rigid tank contains 50 kg of saturated liquid water at 90oC. Determine the pressure in the tank and the volume of the tank. (Answers: 70.14 kPa, 0.0518 m) To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 24. Dependence of Tsat and Psat 24 Elevation (m) Atmospheric Pressure (kPa) Boiling Temp. ( oC) 0 101.33 100.0 1000 89.50 96.3 5000 54.05 83.0 10,000 26.50 66.2 Substance Tsat (oC) H20 100 R-134a -26 N2 -196 @ Patm To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 25. Quality- practical application to finding properties 25 v – vf x = vg – vf u – uf = ug – uf h – hf hg – hf v = vf + x vfg , (vfg = vg – vf) Same procedure applies in finding specific internal energy or enthalpy values. = To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 26. Quality- An Alternative Approach 26 P or T vf vf<v<vg vg v Sat. vapor vg Sat. liquid vf • Since mtotal = mliquid + mvapor , f or m = m + mg Also, V = Vf + Vg & V = mv mv = mf vf + mg vg, or mv = (m-mg) vf + mgvg v = (1-x) vf + xvg = vf + x vfg Since Quality is a mass fraction of the vapor content in mixture, we define x = mg / m To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 27. Example: Determine the specific volume of 2 kg of water at 100oC with a quality of 70%, i.e., moisture content of 30%? Also, obtain mass fraction of the vapor? Ans. From Table (A-4): vf=0.0010435 m3/kg & vg=1.673 m3/kg v = vf + x.vfg = 0.0010435 + 0.7(1.673 -0.0010435) v = 1.7141 m3/kg Since, x = mg/m then mg = 0.7x2 = 1.4 kg Quality: Example Problem 27 To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 28. Fixing the State 28 Is P > Psat. or T < Tsat. ? Yes: Sub-cooled liquid (vf @T; uf @T; hf @T) No: Liquid-vapor mix. or Super-heated vapor P = Psat. T = Tsat. Liquid-vapor mixture Quality calculation required Is P < Psat. or T > Tsat. ? Super-heated vapor (super-heated vapor tables) To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 29. Using Superheated Vapor Table 29 Table (A-6) In the region to the right of the saturated vapor line, a substance exists as superheated vapor. Example: Determine the temperature of water at a state of P = 0.5 MPa and h = 2890 kJ/kg. (Answers: 216.4 oC) To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 30. - In the absence of compressed liquid tables, a general rule of thumb for properties v & u is to treat compressed liquid as saturated liquid at the given temperature: y = yf @T i.e., v = vf @T & u = uf @T - For ‘h’ values in the compressed region: h = hf @T + vf (P - Psat) Compressed Liquid 30 To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 31. Determine the specific enthalpy of liquid water at 25oC and 300 kPa? Compressed Liquid: Example Problem 31 Ans. The state is compressed liquid as Psat@25C = 3.169 kPa However, the pressure of 300 kPa is below the minimum pressure (0.5 MPa) listed in Table (A-7): So we will use the approximation! v = vf @T = 0.001003 m3/kg h = hf @T + vf (P - Psat) = 104.89 + 0.001003(300 – 3.169) = 105.19 kJ/kg To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 32. 32 Reference State and Reference Values • Values of u and h are arbitrary since we are only interested in changes of these properties. • For water, the reference state is u = 0 for saturated liquid @ 0.01°C. • For R-134a and many other substances, the reference state is h = 0 for the saturated liquid at -40 °C. To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 33. 33 Basics of Interpolation Assuming a linear variation in the examined property over a relatively short range. Temp. T (oF) Sp. Volume v (ft3/lbm) Enthalpy h (Btu/lbm) 80 0.5408 113.47 93 v=? h=? 100 0.5751 118.39 (TableA-13E) Superheated R-134a To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 34. 34 Steam Table: Example Problem Complete the following steam table. Locate each point on a PV diagram. # T,oC P , kPa v, m3/kg u, kJ/kg h, kJ/kg x Phase 1 160 1300 2 150 1.0 3 200 1000 4 200 4000 To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 35. 35 The Ideal Gas Equation of State M = mass of one mol of the gas (TableA-1) = m / N • Equation of State: Property relation that involves other properties of a substance at equilibrium states. Ideal (imaginary) Gas Equation is the simplest among other equation of states • History Background: - Boyle (1662): P α 1/v =➔ for constant temperature (P1v1=P2v2) - Charles (1802): for const pressure =➔ v/T=const. or combined: Pv = RT at low pressures - Ideal gas is an imaginary substance that could be resembled by gases at low densities (low P & high T) M Molar mass (molecular weight) Ru = 8.314 kJ/(kmol.K) = 1.986 Btu/(lbmol.R) (kJ/kg.K) • Gas Constant R = Ru Universal gas constant = To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 36. 36 Property Diagrams for Ideal Gases P const. T v P > T const. P v T > Boyle P α 1/v =➔ for constant temperature Charles for const pressure =➔ v/T=const. To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 37. Special processes: - Constant Pressure (Isobaric) - Constant Volume (Isochoric) - Constant Temperature (Isothermal) - Polytropic (PV n = C) - Ideal Gas moves along lines of constant pressure Moves along lines of constant specific volume (if mass constant) Moves along lines of constant temperature As pressure increases, volume decreases (non-linear relationship) P2 v2 = P1 v1 P v = RT , R = T1 T2 Drawing Process Paths 37 To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 38. 38 Drawing Process Path (cont.) Read the problem statement and relate to the property diagram ! Using other information: “ Pressure increases to three times its present …..” “ Volume decreases to ¼ its present value …..” “ Heat losses cause temperature to decrease ….” To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 39. 39 Water Vapor and 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. Q. Is water vapor an Ideal gas? To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 40. ▪ Z represents the volume ratio or compressibility. ▪ Z < 1 or Z > 1 for real gases. RT 40 ideal Compressibility Factor The deviation from the ideal-gas behavior can be corrected by compressibility factor Z. Z P = actual To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 41. 41 Principle of corresponding states • The compressibility factor Z is approximately the same for all gases at the same reduced temperature and reduced pressure. Z = Z(PR,TR) for all gases PR = P/Pcr & TR = T/Tcr where, PR and TR are reduced values. Pcr and Tcr are critical properties (Table A-1). To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 42. 42 Reduced Properties cr not v ! • This works great if you are given a gas with known P &T and asked to find the v. • However, if you were given P and v and asked to find T (or T and v and asked to find P), you need to use: pseudo-reduced specific volume, which is defined as: • Ideal gas approximation (Z=1), if (see Fig. A-30): ✓ PR < 10 and TR > 2 ✓ PR << 1 R v = vactual RTcr / Pcr To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 43. 43 Ideal Gas Example Calculate the specific volume of nitrogen at (a) 300 K and 8.0 MPa (b) 200 K and 20.0 MPa. Tcr = 126.2 K and Pcr = 3.39 MPa for nitrogen (T able A-1) Part (a): TR = T/Tcr = 300 K / 126.2 K =2.38 PR = P/Pcr = 8.0 MPa / 3.39 MPa = 2.36 Since, TR > 2 and PR < 10, we can use ideal gas equation of state. ZRT P m3 kg m3 kPa kJ kJ )(300K) kg.K 8,000 kPa (0.2968 = 0.01113 = v = To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 44. ZRT P m3 kg m3 kPa kJ kJ )(200K) kg.K 20,000 kPa (0.91)(0.2968 = 0.002701 = v = Part (b): TR = T/Tcr = 200 K / 126.2 K =1.58 PR = P/Pcr = 20.0 MPa / 3.39 MPa = 5.90 From Figure (A-30)b: Z=0.91 Ideal Gas Example 44 To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 45. Specific Heats 45 m = 1 kg T = 1o C Specific heat = 5 kJ / kg.o C Definition: The amount of energy required to raise the temperature of a unit mass of a substance by one degree. 5 kJ To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 46. Internal Energy & Enthalpy Formulation Using Specific Heats 46 v v T C = u p C p T = h Cv & Cp are properties J/kg.K OR Btu/lbm.R du = Cv dT Constant volume process dh = Cp dT Constant pressure process To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 47. Specific Heat of Ideal Gases For real gases, specific heats vary with pressure but for most practical purposes a suitable average value may be used. h = u + Pv = u + RT dh = du + RdT Cp = Cv + R (kJ/kg.K) Cp = Cv + Ru (kJ/kmol.K) Introducing another ideal gas property called specific heat ratio: k = Cp / Cv > 1 In general, k is about 1.4, 1.6 and 1.3 for diatomic gases (CO, N2, O2), monatomic gases (Ar, He) and triatomic gases (CO2, SO2), respectively. 47 To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 48. Special Cases for a Polytropic Process (PVn = C) 48 Exponent ‘n’ Value Process Type 0 Constant Pressure 1 Constant Temperature k = Cp / Cv Adiabatic Constant Volume To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 49. Specific Heat of Ideal Gases 49 du = Cv dT 2 Δ u = Cv(T) dT = Cv,average (T2 – T1) 1 dh = Cp dT 2 Δ h = Cp(T) dT = Cp,average (T2 – T1) 1 To get academic assistance in mechanical engineering Click below: MECHOLOGIST
- 50. 50 Specific Heats of Incompressible Solids & Liquids Δ u = Cavg (T2 – T1) Constant pressure process: Constant temperature process (for liquids): Incompressible means or v is essentially constant. Similar to ideal gases, specific heat depends only on temperature. Δh = Δ u + v ΔP Δ h = Cavg ΔT + v ΔP Δ P = 0 Δh = Cavg ΔT Δ T = 0 Δh = v ΔP Cv = Cp = Cavg h = u + Pv dh = du + d(Pv) To get academic assistance in mechanical engineering Click below: MECHOLOGIST