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EGR 334 Thermodynamics Chapter 3: Section 1-5 Lecture 06: Evaluating Properties Quiz Today?

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Todays 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: Homework Assignment: Read Chap 3: Sections 6-8 From Chap 3: 5, 7, 10, 29

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Pure Substance: a substance that is uniform and invariable in chemical composition. 3 Sec 3.1.1: Phase and Pure Substance Oil Ice Water Honey Air May each of these be considered a pure substance? Helium gas in a tank? A contents of a glass containing water and ice? A jar of honey, water, oil, and ice topped with air? Atmospheric air? Yes Yes…maybe Yes No

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Phase: Matter which is homogeneous throughout in both its chemical composition and physical structure. 4 Sec 3.1.1: Phase and Pure Substance 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?

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Phase: Matter which is homogeneous throughout in both its chemical composition and physical structure. 5 Sec 3.1.1: Phase and Pure Substance Pure substances can exist in multiple phases. - Gaseous Phase - Liquid Phase - Solid Phase Pure substances can undergo changes in Phase. liquid gas: vaporization gas liquid: condensation solid gas: sublimation gas solid: deposition solid liquid: melting liquid solid: freezing

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The intensive state of a closed system at equilibrium is its condition as described by the values of its intensive thermodynamic properties. 6 For a simple compressible system, specifying any ___2_____ independent intensive thermodynamic properties will fix the other intensive thermodynamics properties of the system. 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 Intensive Properties include: Pressure Temperature Density Specific Volume Internal Energy Enthalpy Entropy

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Most important skill for todays lecture: Given any two intensive properties of H 2 0, be able to specify other intensive properties. 7 Given any two of Temperature Specific Volume and Pressure...find the missing intensive property value. or

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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. 8 Sec 3.2: P-v-T relation The different surfaces of the model represent different phases of a pure substance (like H 2 0) which depend only on the pressure, temperature, and specific volume of the state.

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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. 9 Sec 3.2: P-v-T relation Triple point Critical Point Temperature Pressure

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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. 10 Sec 3.2: P-v-T relation Specific Vol. Pressure

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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. 11 Sec 3.2: P-v-T relation Phase diagram p-V Diagram Isotherms--Lines of constant temperature Isotherms

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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. 12 Sec 3.2: P-v-T relation Temperature Specific Volume isobar: line of constant pressure

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13 i) Saturated Liquid Line (identified values of T f, v f, and p f ) ii) Saturated Vapor Line (identified values of T g, v g, and p g 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. Temperature You should be able to recognize:

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Consider a Constant Pressure Process Sec 3.3: Phase Change C steam water sub-cooled liquid (compressed liquid) vapor water super vapor sat. vapor water two phase liquid-vapor two phase liquid vapor saturated liquid saturated vapor super heated vapor (steam) i j Run Animation

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Sec 3.3: Phase Change Distinguish mixture of liquid/vapor using quality. Two Phase, Liquid-Vapor Mix Can have different mixtures of liquid and vapor (10% liquid, 90% vapor) x = 1, saturated vapor (g) x = 0, saturated liquid (f) Can also be expressed as a percentage (%)

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Sec 3.5: Saturation 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. Two Phase, Liquid-Vapor Mix For example, in a given vessel, we know that the two volumes must add to the total volume with

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pressure spec. vol. * (p, v, T) Quantitative Thermodynamic Properties: 17 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 H 2 0: Temperature Table ii) Table A.3: Prop. of Saturated H 2 0: 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 H 2 0: 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 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. Link to IT download Link to IT dofmaster

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Examples of Property Diagrams 18 Molliere chart Pyschrometric chart Given two properties, locate and read other values.

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Saturated Steam Table 19 Given two property values, look up other intensive values. Given T= 25 deg C and saturated vapor Read p and v.

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Example 1: Determine property values 20 Determine phase or phases of H 2 0 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.

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Example 1: Determine property values 21 Determine phase or phases of H 2 0 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.

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Example 1: Determine property values 22 Determine phase or phases of H 2 0 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.

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Example 1: Determine property values 23 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 m 3 /kg

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Example 1: Determine property values 24 Determine phase or phases of H 2 0 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.

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Example 1: Determine property values 25 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 v f = 1.1973x10 -3. This can be checked on Table A.5 for compressed liquids which gives 1.1555x10 -3 m 3 /kg

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Example 1: Determine property values 26 Determine phase or phases of H 2 0 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.

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Example 1: Determine property values 27 d) p = 2.8 bar T = 160 deg C. from Table A.2 at T sat = 160, p sat = 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:

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Example 1: Determine property values 28 Determine phase or phases of H 2 0 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.

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Example 1: Determine property values 29 e) p = 1 bar T = -12 deg C. from table A.3 at p sat of 1 bar, T sat = 99.63 C > -12 which means the substance will be compressed liquid or solid. Noticing that Table A-5 doesnt 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 = v i for a T of -12 deg. C. v = 1.0888 x 10 -3 = 0.0010888 m 3 /kg

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Linear Interpolation: 30 * * y2y2 y unknown x known x1x1 y1y1 x2x2

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You try it: What is the specific volume of a water at 40 bar and 140 o C? Linear Interpolation will be needed. Given the following: Sec 3.5: Evaluating P, v, T @ 140°C & 25 bar, v 1 = 1.0784 m 3 /kg @ 140°C & 50 bar, v 2 = 1.0768 m 3 /kg Find the spec. vol at p = 40 bar.

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Sec 3.5: Evaluating P, v, T @ 180°C & 40 bar, v 1 = 1.1248 m 3 /kg This time try it for the same pressure with different temperatures. Given: 40 bar and 180 o C, the specific volume of water is 1.1248 m 3 /kg. 40 bar and 140 o C, the specific volume of water is 1.0774 m 3 /kg. What is the specific volume of water at 40 bar and 150 o C? Example 2:

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A cylinder-piston assembly initially contains water at 3 MPa and 300 o C. The water is cooled at constant volume to 200 o C, 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:

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TV diagram Sec 3.5.2 : Saturation Tables Example:

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State 1: p 1 = 3 MPa= 30 bar and T 1 = 300 deg C. i) Start with superheated vapor table A-4 ii) Interpolate between 280 and 320 deg C. 35

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State 2: v 2 = v 1 = 0.0811 m 3 /kg and T 2 = 200 deg C. Recognize that this is in the liquid-vapor mixture range. Refer to Table A-2 i) Locate v g and v f at T = 200 deg C. 36 ii) Determine quality, x 2

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State 3: p 3 = 2.5 MPa = 25 bar and T 2 = T 3 = 200 deg C. Recognize that this is in the compressed liquid. Refer to Table A-5 i) Locate p 3 =25 bar and T 3 =200 deg C 37 ii) Read the value of v directly.

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End of Slides for Lecture 06 38

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