Module 03 Lecture Slides on Evaluating Properties PDF

Chapter 3 Evaluating Properties (Adapted from: Moran, M. J., Shapiro, H. N., Boettner, D. D., Bailey, M. B. (2018). Fundamentals of Engineering Thermodynamics, 9th Edition.)  Learning Outcomes ►Explain key concepts, including phase and pure substance, state principle for sim...

Chapter 3 Evaluating Properties (Adapted from: Moran, M. J., Shapiro, H. N., Boettner, D. D., Bailey, M. B. (2018). Fundamentals of Engineering Thermodynamics, 9th Edition.)  Learning Outcomes ►Explain key concepts, including phase and pure substance, state principle for simple compressible systems, p-v-T surface, saturation temperature and saturation pressure, two-phase liquid- vapor mixture, quality, enthalpy, and specific heats. ►Analyze closed systems, including applying the energy balance with property data.  Learning Outcomes, cont. ►Sketch T-v, p-v, and phase diagrams, and locate states on these diagrams. ►Retrieve property data from Tables A-1 through A-23. ►Apply the ideal gas model for thermodynamic analysis, including determining when use of the model is warranted.  Phase ►A quantity of matter that is homogeneous throughout in both chemical composition and physical structure. ►Homogeneity in physical structure means that the matter is all solid, or all liquid, or all vapor (gas). ►Examples: ►The air we breathe is a gas phase consisting of a mixture of different gases. ►Drinking water with ice cubes contains two phases of water: liquid and solid. ►Vinegar and olive oil salad dressing contains two different liquid phases.  Pure Substance ►A substance that is uniform and invariable in chemical composition. ►A pure substance can exist in more than one phase, but its chemical composition must be the same in each phase. ►Examples: ►Drinking water with ice cubes can be regarded as a pure substance because each phase has the same composition. ►A fuel-air mixture in the cylinder of an automobile engine can be regarded as a pure substance until ignition occurs.  State Principle for Simple Compressible Systems (1 of 3) ►Systems of commonly encountered pure substances are called simple compressible systems. These substances include those in appendix tables A-2 through A-18, A-22, and A-23. ►The intensive state of a simple compressible system at equilibrium is described by its intensive properties, including temperature, pressure, specific volume, density, specific internal energy, and specific enthalpy. ►Properties such as velocity and elevation are excluded because their values depend on arbitrary datum choices, such as zero values at the surface of the earth. For the state principle, these properties are not relevant.  State Principle for Simple Compressible Systems (2 of 3) ►Not all of the relevant intensive properties are independent. ►Some are related by definitions – for example, density is 1/v and specific enthalpy, h = u + pv (Eq. 3.4). ►Others are related through expressions developed from experimental data. ►Some intensive properties may be independent in a single phase, but become dependent when there is more than one phase present.  State Principle for Simple Compressible Systems (3 of 3) ►For a simple compressible system, values for any two independent intensive properties determine the values of all other intensive properties. This is the state principle or state postulate for simple compressible systems. ►Among alternative sets of two independent intensive properties, (T, v) and (p, v) are frequently convenient. We soon show that temperature and pressure (p, T) are not always an independent set.  p-v-T Surface (1 of 2) ► For pure, simple compressible systems, pressure can be determined as a function of temperature and specific volume: p = p(T, v) The graph of this relation for water is indicated by the p-v-T surface shown. ► Single-phase regions on the surface include solid, liquid, and vapor. ► Two-phase regions are located between single-phase regions, where two phases exist in equilibrium: liquid- vapor, solid-vapor, solid-liquid.  p-v-T Surface (2 of 2) ► The dome-shaped region composed of the two-phase liquid-vapor states is called the vapor dome. ► A state at which a phase change begins or ends is called a saturation state. Lines bordering the vapor dome are called the saturated liquid and saturated vapor lines. ► At the top of the dome, where saturated liquid and saturated vapor lines meet, is the critical point. ►Critical temperature (Tc) is the maximum temperature at which liquid and vapor phases can coexist in equilibrium. ►Critical pressure (pc) is pressure at the critical point.  Projections of the p-v-T Surface (1 of 2) ►Projection of the p-v-T surface onto the pressure-temperature plane is a phase diagram. ►Saturation temperature designates the temperature at which a phase change takes place at a given pressure. ►Saturation pressure designates the pressure at which a phase change takes place at a given temperature. ►Within two-phase regions pressure and temperature are not independent.  Projections of the p-v-T Surface (2 of 2) ► Projection of the p-v-T surface onto the pressure-specific volume plane results in a p-v diagram. Critical Isobar Critical Isotherm ► Projection of the p-v-T surface onto the temperature-specific volume plane results in a T-v diagram.  Phase Change ► Consider a closed system consisting of a unit mass of liquid water at 20oC contained within a piston- cylinder assembly. ► This state is represented by l (highlighted by the blue dot). ► Liquid states such as this, where temperature is lower than the saturation temperature corresponding to the pressure at the state, are called compressed liquid states. l  Saturated Liquid ► As the system is heated at constant pressure, the temperature increases considerably while the specific volume increases slightly. ► Eventually, the system is brought to the state represented by f (highlighted by the blue dot). ► This is the saturated liquid state corresponding to the specified pressure. f  Two-Phase Liquid-Vapor Mixture (1 of 2) ► When the system is at the saturated liquid state, additional heat transfer at fixed pressure results in the formation of vapor without change in temperature but with a considerable increase in specific volume as shown by movement of the blue dot. ► With additional heating at fixed pressure, more vapor is formed and specific volume increases further as shown by additional movement of the blue dot. ► At these states, the system now consists of a two-phase liquid-vapor mixture. f  Two-Phase Liquid-Vapor Mixture (2 of 2) ► When a mixture of liquid and vapor exists in equilibrium, the liquid phase is a saturated liquid and the vapor phase is a saturated vapor. ► For a two-phase liquid-vapor mixture, the ratio of the mass of vapor present to the total mass of the mixture is its quality, x. ► The value of quality ranges from 0 to 1. ► At saturated liquid states, x = 0.  Saturated Vapor ► If the system is heated further until the last bit of liquid has vaporized it is brought to the saturated vapor state. ► This state is represented by g (highlighted by the blue dot). ► At saturated vapor states, x = 1. g  Superheated Vapor ► When the system is at the saturated vapor state, further heating at fixed pressure results in increases in both temperature and specific volume. ► This state is represented by s (highlighted by the blue dot). ► Vapor states such as this, where temperature is higher than the saturation temperature corresponding to the pressure at the state, are called superheated vapor states. s  Animations Liquid to Vapor A.13 Vapor to Liquid A.14 from Moran and from Moran and Shapiro, 9th Ed. Book. Shapiro, 9th Ed. Book. Boiling Point of Water? Boiling point of water is 100 deg C. True / False Can you boil water at room temperature? Yes, if the pressure is low enough (in this case, we pull a vacuum to accomplish “boiling”) The correct statement is that the boiling point of water is 100 deg C only at atmospheric pressure. Saturation temperature is a better way to define boiling temperature since it will always be accompanied by a pressure! Thermocouple reader measuring water temperature in deg C  Note: In this document, y represents General rules for determining the properties of pure substances any of the following: u, v, h or s Pure substance – need 2 independent intensive properties to fix thermodynamic state u: Specific internal energy (kJ/kg) Pure substance = Water, Ammonia, R-12, R-22, R-134a, etc. Tables to be used – A-2 – A-18, A22-A23, Fund. Eng Thermodynamics, Moran et al., 9 th Ed. h: Specific Enthalpy (kJ/kg) s: Specific Entropy (kJ/kg/K) START with v: Specific volume (m 3/kg) Saturated liquid tables NO YES Superheated vapor use superheat tables YES NO Sat liq. 2 Phase Mixture Sat vap. Compressed liquid T = Tsat P = Psat Find NO Are quality, x compressed liquid tables available? NOTE: y(T,P) above means YES the value of the property ‘y’ use compressed liquid at temperature T and tables pressure P. Similarly, yf (T) means the value of yf at the Adapted from Thermodynamics, An integrated learning system – thermonet 2.0, temperature T. The same is Schmidt et al., 2006 true for h(T,P)  Property Approximations for Liquids or Compressed Liquid Approximation ► Approximate values for v, u, and h at liquid states can be obtained using saturated liquid data. ► Since the values of v and u for liquids change very little with pressure at a fixed Saturated temperature, Eqs. 3.11 and 3.12 can be used liquid to approximate their values. v(T, p) ≈ vf(T) (Eq. 3.11) u(T, p) ≈ uf(T) (Eq. 3.12) ► An approximate value for h at liquid states can be obtained using Eqs. 3.11 and 3.12 in the definition h = u + pv: h(T, p) ≈ uf(T) + pvf(T) or alternatively h(T, p) ≈ hf(T) + vf(T)[p – psat(T)] (Eq. 3.13) where psat denotes the saturation pressure at the given temperature ► When the underlined term in Eq. 3.13 is small h(T, p) ≈ hf(T) (Eq. 3.14)  Steam Tables ►Tables of properties for different substances are frequently set up in the same general format. The tables for water, called the steam tables, provide an example of this format. The steam tables are in appendix tables A-2 through A-5. ►Table A-4 applies to water as a superheated vapor. ►Table A-5 applies to compressed liquid water. ►Tables A-2 and A-3 apply to the two-phase, liquid-vapor mixture of water.  Single-Phase Regions (1 of 3) ► Since pressure and temperature are independent properties in the single-phase liquid and Table A-5/A-5E vapor regions, they can be used to fix the state in these regions. ► Tables A-4/A-4E (Superheated Water Vapor) and A-5/A-5E (Compressed Liquid Water) provide several properties as Table A-4/A-4E functions of pressure and temperature, as considered next.  Single-Phase Regions (2 of 3) Properties tabulated in Tables A-4 and A-5 include ► Temperature (T) Table A-5/A-5E ► Pressure (p) ► Specific volume (v) ► Specific internal energy (u) ► Specific enthalpy (h), which is a sum of terms that often appears in thermodynamic analysis: Table A-4/A-4E h = u + pv Enthalpy is a property because it is defined in terms of properties; physical significance is associated with it in Chapter 4. ► Specific entropy (s), an intensive property developed in Chapter 6  Single-Phase Regions (3 of 3) ► Example: Properties associated with superheated water vapor at 10 MPa and 400oC are found in Table A-4. ►v = 0.02641 m3/kg ►h = 3096.5 kJ/kg ►u = 2832.4 kJ/kg ►s = 6.2120 kJ/kg∙K Table A-4  Linear Interpolation ► When a state does not fall exactly on the grid of values provided by property tables, linear interpolation between adjacent entries is used. ► Example: Specific volume (v) associated with superheated water vapor at 10 bar and 215oC is found by linear interpolation between adjacent entries in Table A-4. (0.2275 – 0.2060) m3/kg (v – 0.2060) m3/kg slope = (240 – 200)oC = (215 – 200)oC → v = 0.2141 m3/kg Table A-4  Two-Phase Liquid-Vapor Region (1 of 4) ► Tables A-2/A-2E (Temperature Table) and A- 3/A-3E (Pressure Table) provide ►saturated liquid (f) data ►saturated vapor (g) data Table note: For saturated liquid specific volume, the table heading is vf×103. At 8oC, vf × 103 = 1.002 → vf = 1.002/103 = 1.002 × 10–3. Table A-2  Two-Phase Liquid-Vapor Region (2 of 4) ► The specific volume of a two-phase liquid- vapor mixture can be determined by using the saturation tables and quality, x. ► The total volume of the mixture is the sum of the volumes of the liquid and vapor phases: V = V + V liq vap ► Dividing by the total mass of the mixture, m, an average specific volume for the mixture is: ► With Vliq = mliqvf , Vvap = mvapvg , mvap/m = x , and mliq/m = 1 – x : v = (1 – x)vf + xvg = vf + x(vg – vf) (Eq. 3.2)  Two-Phase Liquid-Vapor Region (3 of 4) ►Since pressure and temperature are NOT independent properties in the two-phase liquid- vapor region, they cannot be used to fix the state in this region. ►The property, quality (x), defined only in the two- phase liquid-vapor region, and either temperature or pressure can be used to fix the state in this region. v = (1 – x)vf + xvg = vf + x(vg – vf) (Eq. 3.2) u = (1 – x)uf + xug = uf + x(ug – uf) (Eq. 3.6) h = (1 – x)hf + xhg = hf + x(hg – hf) (Eq. 3.7)  Two-Phase Liquid-Vapor Region (4 of 4) ► Example: A system consists of a two-phase liquid-vapor mixture of water at 6oC and a quality of 0.4. Determine the specific volume, in m3/kg, of the mixture. ► Solution: Apply Eq. 3.2, v = vf + x(vg – vf) Substituting values from Table 2: vf = 1.001×10–3 m3/kg and vg = 137.734 m3/kg: v = 1.001×10–3 m3/kg + 0.4(137.734 – 1.001×10–3) m3/kg v = 55.094 m3/kg Table A-2 For Water, Critical Temperature = 374.14 deg C Critical Pressure = 22.09 MPa Critical Sp. Volume = 0.003155 m3/kg Isobars are “constant pressure” lines (iso = same, bar = pressure) Isotherms are “constant temperature” lines (iso = same, therm = temperature)  Property Data Use in the Closed System Energy Balance (1 of 4) Example: A piston-cylinder assembly contains 2 kg of water at 100oC and 1 bar. The water is compressed to a saturated vapor state where the pressure is 2.5 bar. During compression, there is a heat transfer of energy from the water to its surroundings having a magnitude of 250 kJ. Neglecting changes in kinetic energy and potential energy, determine the work, in kJ, for the process of the water. T State 1 State 2 p2 = 2.5 bar T1 = 100oC 2 kg Saturated vapor 2 p1 = 1 bar p1 = 1 bar of water p2 = 2.5 bar 1 T1 = 100oC Q = –250 kJ v  Property Data Use in the Closed System Energy Balance (2 of 4) Solution: An energy balance for the closed system is 0 0 DKE + DPE +DU = Q – W where the kinetic and potential energy changes are neglected. Thus W = Q – m(u2 – u1) State 1 is in the superheated vapor region and is fixed by p1 = 1 bar and T1 = 100oC. From Table A-4, u1 = 2506.7 kJ/kg. State 2 is saturated vapor at p2 = 2.5 bar. From Table A-3, u2 = ug = 2537.2 kJ/kg. W = –250 kJ – (2 kg)(2537.2 – 2506.7) kJ/kg = –311 kJ The negative sign indicates work is done on the system as expected for a compression process.  Lectures 12 and 13 Compressed Liquid Approximation – Review Specific Heats Incompressible substance model Generalized compressibility charts Ideal gas model Using ideal gas tables Using algebraic approximations for temperature dependent ideal gasproperties  Specific Heats ► Three properties related to specific internal energy and specific enthalpy having important applications are the specific heats cv and cp and the specific heat ratio k. (Eq. 3.8) (Eq. 3.9) (Eq. 3.10) ► In general, cv is a function of v and T (or p and T), and cp depends on both p and T (or v and T). ► Specific heat data are provided in Fig 3.9 and Tables A-19 through A-21. ► Although cv and cp are referred to as specific heats, there is no general relationship between them and the heat transfer term of the energy balance denoted by Q.  Incompressible Substance Model ►For a substance modeled as incompressible ►v = constant ►u = u(T) ►For a substance modeled as incompressible, cp = cv; the common specific heat value is represented by c. ►For a substance modeled as incompressible with constant c: u2 – u1 = c(T2 – T1) (Eq. 3.20a) h2 – h1 = c(T2 – T1) + v(p2 – p1) (Eq. 3.20b) ►In Eq. 3.20b, the contribution of the underlined term is often small enough to be ignored. Generalized Compressibility Chart (1 of 3) ►The p- -T relation for 10 common gases is shown in the generalized compressibility chart.  Generalized Compressibility Chart (2 of 3) ► In this chart, the compressibility factor, Z, is plotted versus the reduced pressure, pR, and reduced temperature TR, where pR = p/pc TR = T/Tc (Eq. 3.23) (Eq. 3.27) (Eq. 3.28) is the universal gas constant 8.314 kJ/kmol∙K 1.986 Btu/lbmol∙oR (Eq. 3.22) 1545 ft∙lbf/lbmol∙oR The symbols pc and Tc denote the temperature and pressure at the critical point for the particular gas under consideration. These values are obtained from Tables A-1 and A-1E.  Generalized Compressibility Chart (3 of 3) ► When p, pc, T, Tc, v , and are used in consistent units, Z, pR, and TR are numerical values without units. ► Example: For air at 200 K, 132 bar, TR = 200 K/133 K = 1.5, pR = 132 bar/37.7 bar = 3.5 where Tc and pc for air are from Table A-1. With these TR, pR values, the generalized compressibility chart gives Z = 0.8. ► With v = Mv, an alternative form of the compressibility factor is pv Z= (Eq. 3.24) RT where R R= (Eq. 3.25) M Studying the Generalized Compressibility Chart (1 of 3) ► The solid lines labeled with TR values represent best fits to experimental data. For the 10 different gases represented there is little scatter in data about these lines. ► At the lowest reduced temperature value shown, TR = 1.0, the compressibility factor varies between 0.2 and 1.0. Less variation is observed as TR takes higher values. Studying the Generalized Compressibility Chart (2 of 3) ► For each specified value of TR, the compressibility factor approaches a value of 1.0 as pR approaches zero. Studying the Generalized Compressibility Chart (3 of 3) ► Low values of pR, where Z ≈ 1, do not necessarily correspond to a range of low absolute pressures. ► For instance, if pR = 0.05, then p = 0.05pc. With pc values from Table A-1 Water vapor pc = 220.9 bar → p = 11 bar Ammonia pc = 112.8 bar → p = 5.6 bar Carbon dioxide pc = 73.9 bar → p = 3.7 bar Air pc = 37.7 bar → p = 1.9 bar ► These pressure values range from 1.9 to 11 bar, which in engineering practice are not normally considered as low pressures.  Introducing the Ideal Gas Model (1 of 4) ► To recap, the generalized compressibility chart shows that at states where the pressure p is small relative to the critical pressure pc (where pR is small), the compressibility factor Z is approximately 1. ► At such states, it can be assumed with reasonable accuracy that Z = 1. Then pv = RT (Eq. 3.32)  Introducing the Ideal Gas Model (2 of 4) ►Three alternative forms of Eq. 3.32 can be derived as follows: ► With v = V/m, Eq. 3.32 gives pV = mRT (Eq. 3.33) ► With v = v/M and R = R/M, Eq. 3.32 gives is the universal gas constant 8.314 kJ/kmol∙K 1.986 Btu/lbmol∙oR (Eq. 3.34) 1545 ft∙lbf/lbmol∙oR ► Finally, with v = V/n, Eq. 3.34 gives pV = nRT (Eq. 3.35)  Introducing the Ideal Gas Model (3 of 4) ► Investigation of gas behavior at states where Eqs. 3.32- 3.35 are applicable indicates that the specific internal energy depends primarily on temperature. Accordingly, at such states, it can be assumed with reasonable accuracy that it depends on temperature alone: u = u(T) (Eq. 3.36) ► With Eqs. 3.32 and 3.36, the specific enthalpy also depends on temperature alone at such states: h = u + pv = u(T) + RT (Eq. 3.37) ► Collecting results, a gas modeled as an ideal gas adheres to Eqs. 3.32-3.35 and Eqs. 3.36 and 3.37.  Introducing the Ideal Gas Model (4 of 4) ►While the ideal gas model does not provide an acceptable approximation generally, in the limiting case considered in the discussion of the compressibility chart, it is justified for use, and indeed commonly applied in engineering thermodynamics at such states. ►Appropriateness of the ideal gas model can be checked by locating states under consideration on one of the generalized compressibility charts provided by appendix figures Figs. A-1 through A-3.  Internal Energy and Enthalpy of Ideal Gases (1 of 4) ►For a gas obeying the ideal gas model, specific internal energy depends only on temperature. Hence, the specific heat cv, defined by Eq. 3.8, is also a function of temperature alone. That is, (ideal gas) (Eq. 3.38) ►On integration, T2 (ideal gas) (Eq. 3.40) T1  Internal Energy and Enthalpy of Ideal Gases (2 of 4) ►Similarly, for a gas obeying the ideal gas model, specific enthalpy depends only on temperature. Hence, the specific heat cp, defined by Eq. 3.9, is also a function of temperature alone. That is, (ideal gas) (Eq. 3.41) ►On integration, T2 (ideal gas) (Eq. 3.43) T1  Internal Energy and Enthalpy of Ideal Gases (3 of 4) ►In applications where the specific heats are modeled as constant, u(T2) – u(T1) = cv[T2 – T1] (Eq. 3.50) h(T2) – h(T1) = cp[T2 – T1] (Eq. 3.51) ►For several common gases, evaluation of changes in specific internal energy and enthalpy is facilitated by use of the ideal gas tables: Tables A-22 and A-23. ►Table A-22 applies to air modeled as an ideal gas.  Internal Energy and Enthalpy of Ideal Gases (4 of 4) ► Example: Using Table A-22, determine the change in specific enthalpy, in kJ/kg, for a process of air from an initial state where T1 = 300 K, p1 = 1 bar to a final state where T2 = 1500 K, p2 = 10 bar. ► Solution: h1 = 300.19 kJ/kg; h2 = 1635.97 kJ/kg Over such a wide temperature interval, h2 – h1 = 1335.78 kJ/kg use of h2 – h1 = cp[T2 – T1], Eq. 3.51, would not be appropriate. TABLE A-22  Property Data Use in the Closed System Energy Balance (3 of 4) Example: A closed, rigid tank consists of 1 kg of air at 300 K. The air is heated until its temperature becomes 1500 K. Neglecting changes in kinetic energy and potential energy and modeling air as an ideal gas, determine the heat transfer, in kJ, during the process of the air. p2 T 2 T2 = 1500 K p1 State 1 State 2 T1 = 300 K 1 kg of air T2 = 1500 K T1 = 300 K 1 Q v  Property Data Use in the Closed System Energy Balance (4 of 4) Solution: An energy balance for the closed system is 0 0 0 DKE + DPE +DU = Q – W where the kinetic and potential energy changes are neglected and W = 0 because there is no work mode. Thus Q = m(u2 – u1) Substituting values for specific internal energy from Table A-22 TABLE A-22 Q = (1 kg)(1205.41 – 214.07) kJ/kg = 991.34 kJ  Polytropic Process ►A polytropic process is a quasiequilibrium process described by pV n = constant (Eq. 3.52) ►The exponent, n, may take on any value from –∞ to +∞ depending on the particular process. ►For any gas (or liquid), when n = 0, the process is a constant-pressure (isobaric) process. ►For any gas (or liquid), when n = ±∞, the process is a constant-volume (isometric) process. ►For a gas modeled as an ideal gas, when n = 1, the process is a constant-temperature (isothermal) process.