The CRC Handbook of Mechanical Engineering. Chapter 2. Engineering Thermodynamics (776125), страница 9

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are evaluated by considering the respective components to be at thepressure p and temperature T of the mixture. The additive volume rule can be expressed alternatively asjZ=∑ y Z ii =1(2.48b)ip,Twhere the compressibility factors Zi are determined assuming that component i exists at the pressure pand temperature T of the mixture.Evaluating ∆h, ∆u, and ∆sUsing appropriate specific heat and p-v-T data, the changes in specific enthalpy, internal energy, andentropy can be determined between states of single-phase regions. Table 2.5 provides expressions forsuch property changes in terms of particular choices of the independent variables: temperature andpressure, and temperature and specific volume.© 1999 by CRC Press LLC2-38Section 2Taking Equation 1 of Table 2.5 as a representative case, the change in specific enthalpy between states1 and 2 can be determined using the three steps shown in the accompanying property diagram.

Thisrequires knowledge of the variation of cp with temperature at a fixed pressure p′, and the variation of [v– T(∂v/∂T)p] with pressure at temperatures T1 and T2:1-a: Since temperature is constant at T1, the first integral of Equation 1 in Table 2.5 vanishes, andha − h1 =∫ [v − T (∂v ∂T ) ]dpp′pp1a-b: Since pressure is constant at p′, the second integral of Equation 1 vanishes, andhb − ha =∫ c (T, p′)dTT2T1pb-2: Since temperature is constant at T2, the first integral of Equation 1 vanishes, andh2 − hb =∫ [v − T (∂v ∂T ) ]dpp2p′pAdding these expressions, the result is h2 – h1. The required integrals may be performed numerically oranalytically.

The analytical approach is expedited when an equation of state explicit in specific volumeis known.Similar considerations apply to Equations 2 to 4 of Table 2.5. To evaluate u2 – u1 with Equation 3,for example, requires the variation of cv with temperature at a fixed specific volume v′, and the variationof [T(∂p/∂T)v – p] with specific volume at temperatures T1 and T2. An analytical approach to performingthe integrals is expedited when an equation of state explicit in pressure is known.As changes in specific enthalpy and internal energy are related through h = u + pv byh2 − h1 = (u2 − u1 ) + ( p2 v2 − p1v1 )(2.49)only one of h2 – h1 and u2 – u1 need be found by integration. The other can be evaluated from Equation2.49. The one found by integration depends on the information available: h2 – h1 would be found whenan equation of state explicit in v and cp as a function of temperature at some fixed pressure is known,u2 – u1 would be found when an equation of state explicit in p and cv as a function of temperature atsome specific volume is known.Example 6Obtain Equation 1 of Table 2.4 and Equations 3 and 4 of Table 2.5.Solution.

With Equation 2.33 and the Maxwell relation corresponding to ψ(T, v) from Table 2.2, Equations3′ and 4′ of Table 2.5 become, respectively, ∂u du = cv dT +   dv ∂v  T ∂s  ∂p ds =   dT +   dv ∂T  v ∂T  vIntroducing these expressions for du and ds in Equation 2.32a, and collecting terms,© 1999 by CRC Press LLCEngineering ThermodynamicsTABLE 2.5 ∆h, ∆u, ∆s Expressions2-39© 1999 by CRC Press LLC2-40Section 2 ∂u   ∂s  ∂p  T   − cv  dT =   + p − T    dv∂∂∂T v TvvTSince T and v are independent, the coefficients of dT and dv must vanish, giving, respectively,c ∂s   = v ∂T  v T ∂u  ∂p   = T  − p ∂v  T ∂T  vThe first of these corresponds to Equation 1 of Table 2.4 and Equation 4 of Table 2.5. The second ofthe above expressions establishes Equation 3 of Table 2.5. With similar considerations, Equation 3 ofTable 2.4 and Equations 1 and 2 of Table 2.5 may be obtained.Fundamental Thermodynamic FunctionsA fundamental thermodynamic function is one that provides a complete description of the thermodynamicstate.

The functions u(s, v) , h(s, p) , ψ(T, v), and g(T, p) listed in Table 2.2 are fundamental thermodynamicfunctions.In principle, all properties of interest can be determined from a fundamental thermodynamic functionby differentiation and combination. Taking the function ψ(T, v) as a representative case, the propertiesv and T, being the independent variables, are specified to fix the state. The pressure p and specific entropys at this state can be determined by differentiation of ψ(T, v), as shown in Table 2.2. By definition, ψ =u – Ts, so specific internal energy is obtained asu = ψ + Tswith u, p, and v known, the specific enthalpy can be found from the definition h = u + pv. Similarly,the specific Gibbs function is found from the definition g = h – Ts.

The specific heat cv can be determinedby further differentiation cv = (∂u/∂T)v .The development of a fundamental function requires the selection of a functional form in terms ofthe appropriate pair of independent properties and a set of adjustable coefficients that may number 50or more. The functional form is specified on the basis of both theoretical and practical considerations.The coefficients of the fundamental function are determined by requiring that a set of selected propertyvalues and/or observed conditions be statisfied in a least-squares sense.

This generally involves propertydata requiring the assumed functional form to be differentiated one or more times, for example p-v-Tand specific heat data. When all coefficients have been evaluated, the function is tested for accuracy byusing it to evaluate properties for which accepted values are known such as velocity of sound and JouleThomson data. Once a suitable fundamental function is established, extreme accuracy in and consistencyamong the thermodynamic properties are possible. The properties of water tabulated by Keenan et al.(1969) and by Haar et al.

(1984) have been calculated from representations of the Helmholtz function.Thermodynamic Data RetrievalTabular presentations of pressure, specific volume, and temperature are available for practically importantgases and liquids. The tables normally include other properties useful for thermodynamic analyses, suchas internal energy, enthalpy, and entropy. The various steam tables included in the references of thischapter provide examples. Computer software for retrieving the properties of a wide range of substancesis also available, as, for example, the ASME Steam Tables (1993) and Bornakke and Sonntag (1996).© 1999 by CRC Press LLC2-41Engineering ThermodynamicsIncreasingly, textbooks come with computer disks providing thermodynamic property data for water,certain refrigerants, and several gases modeled as ideal gases — see, e.g., Moran and Shapiro (1996).The sample steam table data presented in Table 2.6 are representative of data available for substancescommonly encountered in mechanical engineering practice.

Table A.5 and Figures 2.7 to 2.9 providesteam table data for a greater range of states. The form of the tables and figures, and how they are usedare assumed to be familiar. In particular, the use of linear interpolation with such tables is assumedknown.Specific internal energy, enthalpy, and entropy data are determined relative to arbitrary datums andsuch datums vary from substance to substance.

Referring to Table 2.6a, the datum state for the specificinternal energy and specific entropy of water is seen to correspond to saturated liquid water at 0.01°C(32.02°F), the triple point temperature. The value of each of these properties is set to zero at this state.If calculations are performed involving only differences in a particular specific property, the datumcancels.

When there are changes in chemical composition during the process, special care should beexercised. The approach followed when composition changes due to chemical reaction is considered inSection 2.4.Liquid water data (see Table 2.6d) suggests that at fixed temperature the variation of specific volume,internal energy, and entropy with pressure is slight. The variation of specific enthalpy with pressure atfixed temperature is somewhat greater because pressure is explicit in the definition of enthalpy. Thisbehavior for v, u, s, and h is exhibited generally by liquid data and provides the basis for the followingset of equations for estimating property data at liquid states from saturated liquid data:v(T , p) ≈ vf (T )(2.50a)u(T , p) ≈ uf (T )(2.50b)[]h(T , p) ≈ h f (T ) + vf p − psat (T )−−−−−−−−−s(T , p) ≈ sf (T )(2.50c)(2.50d)As before, the subscript f denotes the saturated liquid state at the temperature T, and psat is the corresponding saturation pressure.

The underlined term of Equation 2.50c is often negligible, giving h(T, p)≈ hf (T), which is used in Example 3 to evaluate h1.In the absence of saturated liquid data, or as an alternative to such data, the incompressible modelcan be employed:v = constantIncompressible model: u = u(T )(2.51)This model is also applicable to solids. Since internal energy varies only with temperature, the specificheat cv is also a function of only temperature: cv(T) = du/dT. Although specific volume is constant,enthalpy varies with both temperature and pressure, such thath(T , p) = u(T ) + pv(2.52)Differentiation of Equation 2.52 with respect to temperature at fixed pressure gives cp =cv. The commonspecific heat is often shown simply as c. Specific heat and density data for several liquids and solids are© 1999 by CRC Press LLC2-42TABLE 2.6 Sample Steam Table Data(a)Properties of Saturated Water (Liquid-Vapor): Temperature TableSpecific Volume (m3/kg)Internal Energy (kJ/kg)SaturatedVapor(vg)SaturatedLiquid(uf)SaturatedVapor(ug)SaturatedLiquid(hf)206.136157.232147.120137.734120.9170.0016.7720.9725.1933.592375.32380.92382.32383.62386.40.0116.7820.9825.2033.60Temp(°C)Pressure(bar)SaturatedLiquid(vf × 103).0145680.006110.008130.008720.009350.010721.00021.00011.00011.00011.0002(b)Enthalpy (kJ/kg)Internal Energy (kJ/kg)SaturatedVapor(vg)SaturatedLiquid(uf)SaturatedVapor(ug)SaturatedLiquid(hf)34.80023.73918.10314.6747.649121.45151.53173.87191.82251.382415.22425.02432.22437.92456.7121.46151.53173.88191.83251.40Pressure(bar)Temp(°C)0.040.060.080.100.2028.9636.1641.5145.8160.061.00401.00641.00841.01021.0172Evap.(hfg)SaturatedLiquid(sf)SaturatedVapor(sg)2501.32491.92489.62487.22482.52501.42508.72510.62512.42516.10.00000.06100.07610.09120.12129.15629.05149.02579.00038.9501Properties of Saturated Water (Liquid-Vapor): Pressure TableSpecific Volume (m3/kg)SaturatedLiquid(vf × 103)Entropy (kJ/kg · K)SaturatedVapor(hg)Enthalpy (kJ/kg)Entropy (kJ/kg · K)Evap.(hfg)SaturatedVapor(hg)SaturatedLiquid(sf)SaturatedVapor(sg)2432.92415.92403.12392.82358.32554.42567.42577.02584.72609.70.42260.52100.59260.64930.83208.47468.33048.22878.15027.9085Section 2© 1999 by CRC Press LLC(c)T(°C)Properties of Superheated Water Vaporu(kJ/kg)h(kJ/kg)s(kJ/kg · K)v(m3/kg)p = 0.06 bar = 0.006 MPa (Tsat 36.16°C)Sat.8012016020023.73927.13230.21933.30236.383T(°C)v × 103(m3/kg)2080140200Sat.1.00061.02801.07841.15551.19732425.02487.32544.72602.72661.42567.42650.12726.02802.52879.7(d)8.33048.58048.78408.96939.13984.5264.6255.1635.6966.2282473.02483.72542.42601.22660.42631.42645.62723.12800.62878.47.71587.75647.96448.15198.3237Properties of Compressed Liquid Wateru(kJ/kg)h(kJ/kg)s(kJ/kg · K)p = 25 bar = 2.5 MPa (Tsat 223.99°C)83.80334.29587.82849.9959.1v(m3/kg)u(kJ/kg)h(kJ/kg)s(kJ/kg · K)p = 0.35 bar = 0.035 MPa (Tsat = 72.69°C)Engineering ThermodynamicsTABLE 2.6 Sample Steam Table Data (continued)86.30336.86590.52852.8962.10.29611.07371.73692.32942.5546v × 103(m3/kg)u(kJ/kg)h(kJ/kg)s(kJ/kg · K)p = 50 bar = 5.0 MPa (Tsat = 263.99°C)0.99951.02681.07681.15301.285983.65333.72586.76848.11147.888.65338.85592.15853.91154.20.29561.07201.73432.32552.9202Source: Moran, M.J.

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