Chapter3纯流体热力学性质(PartB)

1、1Chap er 3Part B22. Residual properties by EOS1）Virial EOS If P 1.5 MPa，the compressibility factor is given by the two-term Virial EOS，(3-27)(3-28)And recall(2-7)(3-26)3Since B and dB/dT are only functions of temperature, independent of pressure.4(3-52)(3-53)(3-54)5方程中以1/V，即为变量。用该公式计算剩余性质时，必须对剩余性质计算

2、公式（3-2628）进行积分变量的替换，将压力P 替换成密度 。If 1.5 MPa P Pc，the compressibility factor is given by the three-term Virial EOS，6From the definition of compressibility, Differentiation at constant T gives:7(3-60) Note that 0 and Z1, when P0. GR(3-26)8Since (3-22)(3-19)Division by dT and restriction to constant yie

3、lds: HR9SinceAnd since(3-60)(3-61)Substitution leads to:10前面，将积分变量为 dP 的Eqs(3-2628)变换成了积分变量为 d 的Eqs(3-6062)。通过积分变量的替换后，即可用三项Virial EOS计算GR、HR 和SR 。Since (3-25)(3-62) SR11Substitution of (A) and (B) into Eqs. (3-6062) and integration lead to:From the three-term virial EOS,(A)(B)(3-57)(3-55)(3-56)122）

4、Residual properties by Cubic EOS 与三项Virial EOS类似，Cubic EOS是以V()和T 为隐函数的方程，用它们计算剩余性质时，同样也需对剩余性质计算公式中的积分变量进行替换，将积分变量P 替换成摩尔体积V 或密度 由于具体的Cubic EOS形式众多，结合Eqs.(3-6062)（p75）会得出不同的剩余性质表达式，不便查阅，因此，下面以通用的Cubic EOS 来介绍剩余性质的计算方法13The General Cubic EOS is given bySince(3-63a)Let(3-65)(3-64) Note: q is the funct

5、ion of T only14计算剩余性质公式(3-6062)中需要的两个积分项：And You can prove this.15LetThen the two integrals simplify to :The two integrals needed ：16LetThen GR(3-66)(3-69)(3-60)17 HR and SR(3-68)(3-67)(3-61)(3-62)18For integral 1#: Since2#: =(3-70)(3-71)19Example 3-4试用RK EOS 估算25 MPa、410 K时CO2 的HR、SR 和GR。Solution:比

6、较RK EOS与General Cubic EOS：RK EOS 中的参数a(T) 和 b 的计算公式:(2-16a)(2-16b)20查附表1，CO2 的 Pc = 7.375 MPa，Tc = 304.2 K，1) 求25 MPa、410 K 时CO2 的压缩因子Z。将Cubic EOS 重写成下面的迭代形式用理想气体摩尔体积作为初值，迭代求出CO2的摩尔体积，然后计算出压缩因子Z。也可直接写成求气体压缩因子的迭代方程，再用迭代法求CO2的压缩因子。21将General Cubic EOS左右两边同乘以 P/RT，然后用 Z 表示 V，得到引用定义用General Cubic EOS 求气

7、体压缩因子的迭代方程为 迭代初始值为理想气体的压缩因子(A), p7722求CO2气体压缩因子的迭代RK方程为迭代结果为计算结果为：232) 计算剩余性质， Case 1: (3-70)24将各参数代入Eqs(3-6668)25Calculate the residual properties普遍化方法计算剩余性质1） Pitzer 普遍化压缩因子法各项数值通过查图3-13-8 获得与压缩因子的关系：与PVT的关系262) Pitzer 普遍化第二维里系数法 普遍化方法计算 HR 和 SR 的适用范围与用普遍化方法计算流体 P-V-T 关系的适用范围相同27Residual propertie

8、s by EOS1）Virial EOS 两项维里EOS 三项维里EOS 282）Residual properties by Cubic EOS 29上述将剩余性质计算公式中积分变量由压力P 替换成密度，再用Cubic EOS计算流体剩余性质的方法通用性强，但过程比较复杂。另一种较为简单的方法：从剩余性质与PVT关系式出发，得到直接计算MR的公式。缺点是MR公式随不同的Cubic EOS 而有不同的形式。Residual properties by an alternative method30 HR 的计算 曾记否？ SR 的计算 31Problem用RK EOS 估算25 MPa、410

9、 K 时CO2 的HR、SR。SolutionR-K EOS： 剩余焓公式：32积分项：33剩余熵公式：积分项：34RK EOS 中的参数 a 和 b 的计算公式为对于CO2 气体，Pc = 7.375 MPa，Tc = 304.2 K，用迭代法求CO2 气体的摩尔体积。35V 的迭代初始值为迭代方程为经5次迭代，36 与例 3-4 结果相符 剩余焓：37 与例 3-4 结果相符 剩余熵：383. 剩余性质计算流体热力学性质变化量真实流体由 state 1 (T1，P1)变化到 state 2 (T2，P2)， 其热力学性质的变化量M 的计算，可设计下面路线：Step 2 Step 1 Ste

10、p 3 State 1 real fluidT1, P1State 2 real fluidT2, P2 (ideal gas) T1, P1 (ideal gas) T2, P2三步解题法：39 Step 1：Step2：是真实流体在状态1时的剩余性质； 是理想气体的性质变化量：(3-12)(3-13)40Step 3： 是真实流体在状态2时的剩余性质。41ProblemEstimate H and S for n-butene vapor at 473.15 K and 7 MPa if H and S are set equal to zero for saturated liquid

11、at 273.15 K. Assume that the only data available are:Tn= 267 K (normal boiling point) Tc= 420.0K Pc= 4.02MPa = 0.191 42SolutionAnalysis: Since the properties of n-butene at the state of saturated liquid at 273.15 K are zero (reference state), we must design a path, which is the change of n-butene fr

12、om the state of saturated liquid at 273.15 K to the state of vapor at 473.15 K and 7 MPa, and the properties of n-butene vapor at 473.15 K and 7 MPa are the same as the property changes of the process. Here we use four-step calculational path, which is similar to three-step calculational path, to do

13、 this job. The four-step calculational path is illustrated with the following figure. 43 Step 1 Saturated VaporT1, P1Step 2 Ideal Gas T1, P1 Step 3 Ideal Gas T2, P2Step 4 State 1 Saturated LiquidState 2 Vapor 273.15K, P1T2, P244P1:The saturated pressure of n-butene liquid at 273.15K is determined by

14、 the following method:At normal boiling point, Tn = 267 K, Psat =1.0133 bar; at critical point, Tc = 419.6 K, Pc = 40.2 bar. Therefore we can obtain the solution of A and B as A = 10.126 B = 2699.11 Then we calculate the saturated pressure of n-butene liquid at 273.15 K as Psat = 1.2771 bar. (P1=1.2

15、7 bar)45Step 1: Vaporization of saturated n-butene liquid at 273.15KH lv is the latent heat of vaporization. The latent heat of vaporization at normal point, , is calculated by the following Riedel equation With Tn = 267 K, Tr,n= 267/419.6 = 0.636, and Pc = 4.02 MPa, the calculation result is.The va

16、porization of saturated liquid is an reversible process.46The latent heat of vaporization at 273.15 K is estimated using the following Watson equationWith Tr =273.15/420=0.65, the calculation result is 47Step 2: The transformation of saturated n-butene vapor into an ideal gas at the conditions of T1

17、 = 273.15 K and P1 = 1.2771 bar.Due to the EOS of saturated n-butene vapor at T1 = 273.15K and P1 = 1.277 bar is not known, we should use the Pitzer generalized methods to calculate the residual properties.Tr,1 = 0.651, Pr,1 = 0.036148Since the pressure is relatively low, we can use the generalized

18、Pitzer Correlations for the second virial coefficient to evaluate 4950Step 3: Property changes in the ideal-gas state from (273.15 K, 1.2771 bar) to (473.15 K, 70 bar). 5152Step 4: Transformation of n-butene from the ideal gas into vapor at the state of T2 = 473.15 K and P2 = 70 bar. Also due to the

19、 EOS of n-butene vapor at T2 = 473.15 K and P2 = 70 bar is not known, we should use the Pitzer generalized methods to calculate the residual properties.53Tr,2 = 1.13, Pr,2 = 1.74we should use the generalized Pitzer Correlations for the compressibility factor to evaluate Since,From Figs 3-2, 3-4, 3-6

20、 and 3-8, the following results are obtained :54Finally,553.5 液体的热力学性质 Section 3.4 建立的用剩余性质计算流体热力学性质的各种方法，对纯气体和纯液体，或者组成不变的气体和液体混合物都适用。对于液体，剩余性质计算时涉及到相态的变化，即液体的汽化，必须考虑相变过程中热力学性质的变化量，如焓变和熵变，因此，液体剩余性质的计算相比气体而言要复杂一些，需要的基础数据也相对较多。56对于温度为T 压力为 P 的液体，其剩余性质的计算过程可设计下面的路线进行：Step 1 Saturated LiquidTs, PsStep 2 Ts, PsStep 3 Unsaturated LiquidT, PT, P Saturated VaporIdeal Gas Unsaturated VaporT, PStep 4 57Step 1：不饱和液体温度、压力发生变化，变成饱和液体，M(L)；Step2：饱和液体发生可逆汽化，变成饱和蒸汽，MLV；Step 3：饱和蒸汽温度、压力发生变