The CRC Handbook of Mechanical Engineering. Chapter 2. Engineering Thermodynamics (776125), страница 10

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and Shapiro, H.N. 1995. Fundamentals of Engineering Thermodynamics, 3rd ed. Wiley, New York, as extracted fromKeenan, J. H., Keyes, F.G., Hill, P.G., and Moore, J.G. 1969. Steam Tables. Wiley, New York.2-43© 1999 by CRC Press LLC2-44Section 2provided in Tables B.2, C.1, and C.2. As the variation of c with temperature is slight, c is frequentlytaken as constant.When the incompressible model is applied. Equation 2.49 takes the formh2 − h1 =∫T2T1c(T ) dT + v( p2 − p1 )(2.53)= cave (T2 − T1 ) + v( p2 − p1 )Also, as Equation 2.32a reduces to du = Tds, and du = c(T)dT, the change in specific entropy is∆s =∫T2T1c(T )dTT(2.54)T= cave ln 2T1Ideal Gas ModelInspection of the generalized compressibility chart, Figure 2.10, shows that when pR is small, and formany states when TR is large, the value of the compressibility factor Z is close to 1.

In other words, forpressures that are low relative to pc, and for many states with temperatures high relative to Tc, thecompressibility factor approaches a value of 1. Within the indicated limits, it may be assumed withreasonable accuracy that Z = 1 — that is,pv = RT or pv = RT(2.55a)where R = R /M is the specific gas constant.

Other forms of this expression in common use arepV = nRT ,pV = mRT(2.55b)Referring to Equation 3′ of Table 2.5, it can be concluded that (∂u/∂v)T vanishes identically for a gaswhose equation of state is exactly given by Equation 2.55, and thus the specific internal energy dependsonly on temperature. This conclusion is supported by experimental observations beginning with the workof Joule, who showed that the internal energy of air at low density depends primarily on temperature.These considerations allow for an ideal gas model of each real gas: (1) the equation of state is givenby Equation 2.55 and (2) the internal energy and enthalpy are functions of temperature alone.

The realgas approaches the model in the limit of low reduced pressure. At other states the actual behavior maydepart substantially from the predictions of the model. Accordingly, caution should be exercised wheninvoking the ideal gas model lest significant error is introduced.Specific heat data for gases can be obtained by direct measurement. When extrapolated to zeropressure, ideal gas-specific heats result. Ideal gas-specific heats also can be calculated using molecularmodels of matter together with data from spectroscopic measurements. Table A.9 provides ideal gasspecific heat data for a number of substances. The following ideal gas-specific heat relations are frequentlyuseful:c p (T ) = cv (T ) + Rcp =© 1999 by CRC Press LLCkR,k −1cv =Rk −1(2.56a)(2.56b)2-45Engineering Thermodynamicswhere k = cp /cv .With the ideal gas model, Equations 1 to 4 of Table 2.5 give Equations 1 to 4 of Table 2.7, respectively.Equation 2 of Table 2.7 can be expressed alternatively using s°(T) defined bys°(T ) ≡∫T0c p (T )T(2.57)dTass(T2 , p2 ) − s(T1 , p1 ) = s°(T2 ) − s°(T1 ) − R lnp2p1(2.58)Expressions similar in form to Equations 2.56 to 2.68 can be written on a molar basis.TABLE 2.7 Ideal Gas Expressions for ∆h, ∆u, and ∆sVariable Specific Heatsh(T2 ) − h(T1 ) =∫ c (T )dTT2T1ps (T2 , p2 ) − s (T1 , p1 ) =u (T2 ) − u (T1 ) =Constant Specific Heats∫T2T1∫T2c p (T )TT1dT − R lnp2p1cv (T ) dTs (T2 , v 2 ) − s(T1 , v1 ) =∫T2T1c v (T )vdT + R ln 2Tv1(1)h(T2 ) − h(T1 ) = c p (T2 − T1 )(2)s (T2 , p2 ) − s (T1 , p1 ) = c p ln(3)u (T2 ) − u (T1 ) = cv (T2 − T1 )(3′)(4)s (T2 , v2 ) − s (T1 , v1 ) = cv ln(4′)s2 = s1pr (T2 )pr (T1 )v r (T2 )v r (T1 )(1′)T2p− R ln 2T1p1T2v+ R ln 2T1v1(2′)s2 = s1=p2p1(5)T2  p2 =T1  p1 =v2v1(6)T2  v1 =T1  v2 ( k −1)k(5′)k −1(6′)For processes of an ideal gas between states having the same specific entropy, s2 = s1, Equation 2.58gives[[]]p2 exp s°(T2 ) R=p1 exp s°(T1 ) Ror with pr = exp[s°(T)/R]p (T )p2= r 2p1pr (T1 )(s2 = s1 )(2.59a)A relation between the specific volume and temperatures for two states of an ideal gas having the samespecific entropy can also be developed:v2 vr (T2 )=v1 vr (T1 )© 1999 by CRC Press LLC(s2 = s1 )(2.59b)2-46Section 2Equations 2.59 are listed in Table 2.7 as Equations 5 and 6, respectively.Table A.8 provides a tabular display of h, u, s°, pr , and vr vs.

temperature for air as an ideal gas.Tabulations of h , u , and s° for several other common gases are provided in Table A.2. Property retrievalsoftware also provides such data; see, e.g., Moran and Shapiro (1996). The use of data from Table A.8for the nozzle of Example 2 is illustrated in Example 7.When the ideal gas-specific heats are assumed constant, Equations 1 to 6 of Table 2.7 becomeEquations 1′ tο 6′, respectively. The specific heat cp is taken as constant in Example 2.Example 7Using data from Table A.8, evaluate the exit velocity for the nozzle of Example 2 and compare with theexit velocity for an isentropic expansion to 15 lbf/in.2.Solution.

The exit velocity is given by Equation 2.27fv e = v i2 + 2(hi − he )At 960 and 520°R, Table A.8 gives, respectively, hi = 231.06 Btu/lb and he = 124.27 Btu/lb. Then210 ft Btu   778.17 ft ⋅ lbf   32.174 lb ⋅ ft sec 2 ve = + 2(231.06 − 124.27)  lb    s 1 Btu1 lbf= 2312.5 ft secUsing Equation 2.59a and pr data from Table A.8, the specific enthalpy at the exit for an isentropicexpansion is found as follows:pr (Te ) = pr (Ti )pe15 = 10.61= 1.061pi150 Interpolating with pr data, he = 119.54 Btu/lb. With this, the exit velocity is 2363.1 ft/sec. The actualexit velocity is about 2% less than the velocity for an isentropic expansion, the maximum theoreticalvalue.

In this particular application, there is good agreement in each case between velocities calculatedusing Table A.8 data and, as in Example 2, assuming cp constant. Such agreement cannot be expectedgenerally, however. See, for example, the Brayton cycle data of Table 2.15.Polytropic ProcessesAn internally reversible process described by the expression pvn = constant is called a polytropic processand n is the polytropic exponent. Although this expression can be applied with real gas data, it mostgenerally appears in practice together with the use of the ideal gas model.

Table 2.8 provides severalexpressions applicable to polytropic processes and the special forms they take when the ideal gas modelis assumed. The expressions for ∫pdv and ∫vdp have application to work evaluations with Equations 2.10and 2.30, respectively. In some applications it may be appropriate to determine n by fitting pressurespecific volume data.Example 8 illustrates both the polytropic process and the reduction in the compressor work achievableby cooling a gas as it is compressed.Example 8A compressor operates at steady state with air entering at 1 bar, 20°C and exiting at 5 bar.

(a) If the airundergoes a polytropic process with n = 1.3, determine the work and heat transfer, each in kJ/kg of airflowing. Repeat for (b) an isothermal compression and (c) an isentropic compression.© 1999 by CRC Press LLC2-47Engineering ThermodynamicsTABLE 2.8 Polytropic Processes: pvn = ConstantaIdeal GasbGeneralp2  v1 =p1  v2 nn = 0: constant pressuren = ±∞: constant specific volumennnnn=1n=1∫ pdv = p v ln v21∫2∫ pdv = RT ln v(2)v21p2p1∫(2′)12− vdp = − RT ln(3)1n≠1(1′)0: constant pressure±∞: constant specific volume1: constant temperaturek: constant specific entropy when k is constant1− vdp = − p1 v1 lnp2p1(3′)n≠1∫ pdv =21p2 v2 − p1 v11− npv  p = 1 1 1 −  2 n − 1   p1 ∫2− vdp =1=a====2v21 11n ( n −1)nT p2  v1 =  = 2p1  v2  T1 (1)∫ pdv =21( n −1)n(4)( n −1) n np1 v1   p2 1 −n − 1   p1 1− n( n −1) n RT1   p2 1 −=n − 1   p1 ∫n( p v − p1v1 )1− n 2 2R(T2 − T1 )2− vdp =1(5)=(4′)nR(T − T1 )1− n 2nRT1   p2 1 −n − 1   p1 ( n −1)n(5′)For polytropic processes of closed systems where volume change is the only work mode, Equations 2, 4, and 2′, 4′are applicable with Equation 2.10 to evaluate the work.

When each unit of mass passing through a one-inlet, oneexit control volume at steady state undergoes a polytropic process, Equations 3, 5, and 3′, 5′ are applicable withEquations 2.30a and 2.30b to evaluate the power. Also note that generally, −∫ vdp = n∫21b© 1999 by CRC Press LLC21pdv.2-48Section 2Solution. Using Equation 5′ of Table 2.8 together with Equation 2.30b,( n −1) n W˙ cv nRT1   p2 1 −  =m˙n − 1   p1 [1.3   8.314 kJ =(293 K) 1 − (5)0.3 1.3 0.3   28.97 kg ⋅ K = −163.9]kJkg(The area behind process 1-2 of Figure 2.11, area 1-2-a-b, represents the magnitude of the work required,per unit mass of air flowing.) Also, Equation 1′ of Table 2.8 gives T2 = 425 K.FIGURE 2.11 Internally reversible compression processes.An energy rate balance at steady state and enthalpy data from Table A.8 givesQ˙ cv W˙ cv=+ h2 − h1m˙m˙= −163.9 + (426.3 − 293.2) = −30.8(b) Using Equation 3′ of Table 2.8 together with Equation 2.30b,© 1999 by CRC Press LLCkJkg2-49Engineering ThermodynamicsW˙ cvp= − RT ln 2m˙p18.314 = −(293) ln 5 28.97 = −135.3kJkgArea 1-2′-a-b on Figure 2.11 represents the magnitude of the work required, per unit of mass of airflowing.

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