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EGR 334 Thermodynamics Chapter 3: Section 1-5 Lecture 06: Evaluating Properties Quiz Today? Today’s main concepts: Number of properties needed to set simple compressible system. State properties: Temperature, Pressure, Specific Volume Phase and Quality. Given two properties of a pure compressible substance, find other state properties. • Using the Thermodynamic Property tables. • • • • Reading Assignment: • Read Chap 3: Sections 6-8 Homework Assignment: From Chap 3: 5, 7, 10, 29 Sec 3.1.1: Phase and Pure Substance 3 Pure Substance: a substance that is uniform and invariable in chemical composition. May each of these be considered a pure substance? Helium gas in a tank? Air Ice Yes Oil Water Honey A contents of a glass containing water and ice? Yes Atmospheric air? Yes…maybe A jar of honey, water, oil, and ice topped with air? No Sec 3.1.1: Phase and Pure Substance 4 Phase: Matter which is homogeneous throughout in both its chemical composition and physical structure. Bottle Contents : Air (N2, O2, ect.) Homogeneous chemical composition? Homogeneous physical structure? Bottle : Glass Homogeneous chemical composition? Homogeneous physical structure? Cap : Aluminum? Homogeneous chemical composition? Homogeneous physical structure? Sec 3.1.1: Phase and Pure Substance 5 Phase: Matter which is homogeneous throughout in both its chemical composition and physical structure. - Gaseous Phase - Liquid Phase - Solid Phase Pure substances can exist in multiple phases. Pure substances can undergo changes in Phase. liquid gas: vaporization solid liquid: melting gas liquid: condensation liquid solid: freezing solid gas: sublimation gas solid: deposition 6 The intensive state of a closed system at equilibrium is its condition as described by the values of its intensive thermodynamic properties. In this class we will be dealing with simple compressible systems. Examples of Simple Compressible Systems: --Standard Air (Oxygen-Nitrogren mix) --Ideal Gases --Superheated Water Vapor For a simple compressible system, specifying any ___2_____ independent intensive thermodynamic properties will fix the other intensive thermodynamics properties of the system. Intensive Properties include: Pressure Temperature Specific Volume Enthalpy Density Internal Energy Entropy 7 Most important skill for today’s lecture: • Given any two intensive properties of H20, be able to specify other intensive properties. Given any two of Temperature Specific Volume and Pressure ...find the missing intensive property value. p p (T , v) or T T ( p, v ) or v v(T , p ) Sec 3.2: P-v-T relation 3D p-v-T surface model: A representation of how p, v, and T take on specific values depending upon their intensive properties and phase changes. The different surfaces of the model represent different phases of a pure substance (like H20) which depend only on the pressure, temperature, and specific volume of the state. 8 Sec 3.2: P-v-T relation 9 Phase Diagram from the 3D p-v-T surface model: If you pull off the Pressure-Temperature projection you create a plot given as the Phase Diagram. Critical Point Pressure • • Triple point Sec 3.2: P-v-T relation 10 Pressure p-V Diagram from the 3D p-v-T surface model: The Pressure-Specific Volume projection of the surface model is a useful tool for showing processes involving ice-water-water vapor system. Sec 3.2: P-v-T relation 11 Projections of p-v-T surface: Triple Point--the state at which solid, liquid, and gas coexist Critical Point-- the point where saturated liquid and saturated vapor lines meet. Isotherms--Lines of constant temperature Isotherms Phase diagram p-V Diagram Sec 3.2: P-v-T relation 12 T-v diagram from the 3D p-v-T surface model: If you pull off the Temperature-Specific Volume projection you create a plot that can be useful for showing lines of constant pressure. • isobar: line of constant pressure 13 You should be able to recognize: Temperature i) Saturated Liquid Line (identified values of Tf , vf, and pf ) ii) Saturated Vapor Line (identified values of Tg , vg, and pg iii) Critical Point (inflection point at top of dome) iv) On p-v diagram, lines of constant temperature run from high left to low right. v) On T-v diagram, lines of constant pressure run from low left to high right. vi) While traversing the liquid-vapor phase (the area under the dome), both temperature and pressure remain constant for changes in spec. volume. Sec 3.3: Phase Change Consider a Constant Pressure Process •j •i water sub-cooled liquid (compressed liquid) water saturated liquid Run Animation vapor water vapor water two phase liquid vapor two phase liquid-vapor sat. vapor saturated vapor super vapor super heated vapor (steam) Sec 3.3: Phase Change Two Phase, Liquid-Vapor Mix Can have different mixtures of liquid and vapor (10% liquid, 90% vapor) Distinguish mixture of liquid/vapor using quality. x mvapor mliquid mvapor x = 1, saturated vapor (g) x = 0, saturated liquid (f) Can also be expressed as a percentage (%) Sec 3.5: Saturation Two Phase, Liquid-Vapor Mix Quality (x) is also a property. It is a way to express the relative amount of a substance that contains two different phases of material. For example, in a given vessel, we know that the two volumes must add to the total volume VT VL VV ml vT mT mv v f v g mT with 1 x ml mT ; x mv mT v 1 x v f xvg v v f x vg v f 17 Quantitative Thermodynamic Properties: Given any two intensive properties, can you a) determine the phase (liquid, saturated, mixture, solid, or gas?) b) determine the other intensive property values at that state. Methods: 1) Read off a p-v or a T-v diagram 2) Look up on Steam Tables: (Appendix and Handout) i) Table A.2: Prop. of Saturated H20: Temperature Table ii) Table A.3: Prop. of Saturated H20: Pressure Table iii) Table A.4: Prop. of Superheated Water Vapor iv) Table A.5: Prop. of Compressed Water v) Table A.6: Prop. of Saturated H20: Solid-vapor table 3) Use a Computer Steam Application i) IT (download from www.wiley.com/college/moran) ii) http://www.dofmaster.com/steam.html Link to IT dofmaster * pressure Link to IT download (p, v, T) spec. vol. 18 Examples of Property Diagrams Given two properties, locate and read other values. Molliere chart Pyschrometric chart 19 Saturated Steam Table Given T= 25 deg C and saturated vapor Read p and v. Given two property values, look up other intensive values. 20 Example 1: Determine property values Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the following Identify the specific volume, v, if possible: a) p = 5 bar T = 151.9 deg C. b) p = 5 bar T = 200 deg. C. c) p = 2.5 MPa T = 200 deg C. d) p = 2.8 bar T = 160 deg C. e) p = 1 bar T = -12 deg C. 21 Example 1: Determine property values Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the following: Identify the specific volume, v, if possible: a) p = 5 bar T = 151.9 deg C. from table A.3: at 5 bar the saturated temperature is 151.9 which means it could be anywhere on the liq-vapor line as a two phase material. 22 Example 1: Determine property values Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the following Identify the specific volume, v, if possible: a) p = 5 bar T = 151.9 deg C. b) p = 5 bar T = 200 deg. C. c) p = 2.5 MPa T = 200 deg C. d) p = 2.8 bar T = 160 deg C. e) p = 1 bar T = -12 deg C. 23 Example 1: Determine property values Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the following: b) p = 5 bar T = 200 deg. C. from table A.3: at 5 bar the temp. of 200 > 151.9 which means the substance is vapor or superheated vapor…need to consult table A.4 from table A.4 at 5 bar and T = 200 deg C….v = 0.4249 m3/kg 24 Example 1: Determine property values Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the following Identify the specific volume, v, if possible: a) p = 5 bar T = 151.9 deg C. b) p = 5 bar T = 200 deg. C. c) p = 2.5 MPa T = 200 deg C. d) p = 2.8 bar T = 160 deg C. e) p = 1 bar T = -12 deg C. 25 Example 1: Determine property values c) p = 2.5 MPa T = 200 deg C. from table A.3 at 2.5 MPa = 25 bar the temp of 200 is 200< 224 which means that the substance is compressed liquid. For compressed liquid we assume that the spec. vol, will be similar to vf = 1.1973x10-3. This can be checked on Table A.5 for compressed liquids which gives 1.1555x10-3 m3/kg 26 Example 1: Determine property values Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the following Identify the specific volume, v, if possible: a) p = 5 bar T = 151.9 deg C. b) p = 5 bar T = 200 deg. C. c) p = 2.5 MPa T = 200 deg C. d) p = 2.8 bar T = 160 deg C. e) p = 1 bar T = -12 deg C. 27 Example 1: Determine property values d) p = 2.8 bar T = 160 deg C. from Table A.2 at Tsat = 160, psat = 6.178 > 2.8 which means the substance will be vapor or superheated…so refer to table A.4. from Table A.4 at p = 1.5 and T = 160….v = 1.317 at p = 3 and T = 160….v = 0.651 Interpolation is required: v 1.317 2.8 1.5 0.651 1.317 3.0 1.5 v 0.7398 28 Example 1: Determine property values Determine phase or phases of H20 at the following conditions and sketch p-v and T-v diagrams showing the positions of each of the following Identify the specific volume, v, if possible: a) p = 5 bar T = 151.9 deg C. b) p = 5 bar T = 200 deg. C. c) p = 2.5 MPa T = 200 deg C. d) p = 2.8 bar T = 160 deg C. e) p = 1 bar T = -12 deg C. 29 Example 1: Determine property values e) p = 1 bar T = -12 deg C. from table A.3 at psat of 1 bar, Tsat = 99.63 C > -12 which means the substance will be compressed liquid or solid. Noticing that Table A-5 doesn’t handle pressures this low, if you assumed that the material was liquid, you would use the saturated spec. vol. at the given temperature of -12 deg. However, you recognize that -12 degrees is below the freezing point of water and therefore you expect that this is in solid phase (ice). Table A-6 also has information on saturated solid-vapor data. Assume v = vi for a T of -12 deg. C. v = 1.0888 x 10-3 = 0.0010888 m3/kg 30 Linear Interpolation: yunknown y1 xknown x1 y2 y1 x2 x1 yunknown y2 y1 y1 ( xknown x1 ) x2 x1 * y2 yunknown y1 * x1 xknown x2 Sec 3.5: Evaluating P, v, T You try it: What is the specific volume of a water at 40 bar and 140 oC? Linear Interpolation will be needed. Given the following: @ 140°C & 25 bar, v1 = 1.0784 m3/kg @ 140°C & 50 bar, v2 = 1.0768 m3/kg Find the spec. vol at p = 40 bar. v v1 v2 v1 ( p p1 ) p2 p1 v 1.0784 (1.0768 1.0784) (40 25) 1.0774 (50 25) Sec 3.5: Evaluating P, v, T Example 2: This time try it for the same pressure with different temperatures. Given: 40 bar and 180 oC, the specific volume of water is 1.1248 m3/kg. 40 bar and 140 oC, the specific volume of water is 1.0774 m3/kg. What is the specific volume of water at 40 3bar and 150 oC? @ 180°C & 40 bar, v1 = 1.1248 m /kg v2 v1 v v1 (T T1 ) T2 T1 v 1.0774 (1.1248 1.0774) (150 140) 1.0893 (180 140) m3 / kg Sec 3.5.2 : Saturation Tables Example: A cylinder-piston assembly initially contains water at 3 MPa and 300oC. The water is cooled at constant volume to 200 oC, then compressed isothermally to a final pressure of 2.5 MPa. Sketch the process on a T-v diagram and find the specific volume at the 3 states. Sec 3.5.2 : Saturation Tables Example: TV diagram 35 State 1: p1 = 3 MPa= 30 bar and T1 = 300 deg C. i) Start with superheated vapor table A-4 ii) Interpolate between 280 and 320 deg C. vb va v1 va (T1 Ta ) Tb Ta v1 0.0771 (0.0850 0.0771) (300 280) (320 280) v1 0.0811 m3 / kg 36 State 2: v2 = v1 = 0.0811 m3/kg and T2 = 200 deg C. Recognize that this is in the liquid-vapor mixture range. Refer to Table A-2 i) Locate vg and vf at T = 200 deg C. v f 1.1565 103 vg 0.1274 v2 0.0811 ii) Determine quality, x2 v v f x (v g v f ) x v vf vg v f 0.0811 0.0011565 0.633 63.3% 0.1274 0.0011565 37 State 3: p3 = 2.5 MPa = 25 bar and T2 = T3= 200 deg C. Recognize that this is in the compressed liquid. Refer to Table A-5 i) Locate p3=25 bar and T3=200 deg C ii) Read the value of v directly. v3 1.1555 103 m3 / kg 38 End of Slides for Lecture 06