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Chapter 6 - Engineering Equations of State II

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Chapter 6 - Engineering Equations of State II Introductory Chemical Engineering Thermodynamics By J.R. Elliott and C.T. Lira Chapter 6 - Engineering Equations of State II. Generalized Fluid Properties The principle of two-parameter corresponding states 50 50 40 40 30 30 T=705K 20 20 ρ T=286K L 10 10 T=470 T=191K T=329K 0 0 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 -10 T=133K -10 ρV Density (g/cc) Density (g/cc) VdW Pressure (bars) in Methane VdW Pressure (bars) in Pentane Critical Definitions: Tc - critical temperature - the temperature above which no liquid can exist. Pc - critical pressure - the pressure above which no vapor can exist. ω - acentric factor - a third parameter which helps to specify the vapor pressure curve which, in turn, affects the rest of the thermodynamic variables. ∂ ∂ 2 P = P = 0 and 0 at Tc and Pc Note: at the critical point, ∂ρ ∂ρ 2 T T Chapter 6 - Engineering Equations of State Slide 1 II. Generalized Fluid Properties The van der Waals (1873) Equation Of State (vdW-EOS) Based on some semi-empirical reasoning about the ways that temperature and density affect the pressure, van der Waals (1873) developed the equation below, which he considered to be fairly crude. We will discuss the reasoning at the end of the chapter, but it is useful to see what the equations are and how we use them before deriving the details. The vdW-EOS is: bρ aρ 1 aρ Z =+1 −= − ()1− bρ RT()1− bρ RT van der Waals’ trick for characterizing the difference between subcritical and supercritical fluids was to recognize that, at the critical point, ∂P ∂ 2 P = 0 and = 0 at T , P ∂ρ ∂ρ 2 c c T T Since there are only two “undetermined parameters” in his EOS (a and b), he has reduced the problem to one of two equations and two unknowns. Running the calculus gives: 2 2 a = 0.475 R Tc /Pc; b = 0.125 RTc/Pc The capability of this simple approach to represent all of the properties and processes that we will discuss below is a tribute to the genius of van der Waals. Chapter 6 - Engineering Equations of State Slide 2 II. Generalized Fluid Properties The principle of three-parameter corresponding states Reduced vapor pressure behavior: 1 1.2 1.4 1.6 1.8 2 2.2 2.4 1/Tr To improve our accuracy over the VdW EOS, we can generate a different set of PVT curves for each family of compounds. We specify the family of compounds via the "third parameter" i.e. ω. Note: The specification of Tc , pc , and ω provides two points on the vapor pressure curve. The key to accurate characterization of the vapor liquid behavior of mixtures of fluids is the accurate characterization of the vapor pressure of pure fluids. VLE was central to the development of distillation technology for the petrochemical industry and provided the basis for most of today’s process simulation technology. Chapter 6 - Engineering Equations of State Slide 3 II. Generalized Fluid Properties The Peng-Robinson (1976) Equation of State RTρ aρ 2 1 a bρ p = − or Z = − ⋅ ()1− bρ 1+ 2bρ − b 2 ρ 2 ()1-b bRT 1+ 2bρ − b 2 ρ 2 where ρ = molar density = n/V By fitting the critical point, where ∂P/∂V = 0 and ∂2P/∂V2 = 0, 2 2 a = 0.457235528 αR Tc /Pc; b = 0.0777960739 RTc/Pc α = [1+ κ (1-√Tr)]2 ; κ = 0.37464 + 1.54226ω - 0.26992ω2 sat ω ≡ -1 - log10(P /Pc)Tr =0.7 ≡ “acentric factor” ω Tc , Pc , and are reducing constants according to the principle of corresponding states. By applying Maxwell's relations, we can calculate the rest of the thermodynamic properties (H,U,S) based on this single equation. Chapter 6 - Engineering Equations of State Slide 4 II. Generalized Fluid Properties Solving the Equation of State for Z 1 A B / Z Z= −⋅ ()1-B / Z B 12+−BZ/(/) BZ2 bρ ≡ B/Z where B ≡ bP/RT Z ≡ P/ρRT 2 2 A ≡ aP/R T Rearranging yields a cubic function in Z Z3 -(1-B)Z2 + (A-3B2-2B)Z - (AB-B2-B3) = 0 Naming this function F(Z), we can plot F(Z) vs. Z to gain some understanding about its roots Chapter 6 - Engineering Equations of State Slide 5 II. Generalized Fluid Properties Solving the Equation of State for Z F F 0.1 P= Pvap 0.1 P> > Pvap T< Tc T< Tc 0.08 0.08 0.06 0.06 0.04 0.04 0.02 0.02 0 0 -0.02 00.51 -0.02 00.51 Z -0.04 Z -0.04 F 0.1 P<<Pvap F 0.1 T> Tc 0.08 T< Tc 0.08 0.06 0.06 0.04 0.04 0.02 0.02 0 0 -0.02 00.51 -0.02 00.51 Z -0.04 -0.04 -0.06 Z Chapter 6 - Engineering Equations of State Slide 6 II. Generalized Fluid Properties Solving the Equation of State for Z (cont.) 1. Guess Zold=1 or Zold=0 and compute Fold(Zold). 2. Compute Fnow(Z) at Z=1.0001 or Z=0.0001 3. "Interpolate" between these guesses to estimate where F(Z)=0. ∆Z = (0 - Fnow)*(Znow-Zold)/(Fnow - Fold). 4. Set Fold=Fnow, Zold=Znow, Znow=Zold+∆Z 5. If |∆Z/Znow| < 1.E-5, print the value of Znow and stop. 6. Compute Fnow(Znow) and return to step 3 until step 5 terminates. Chapter 6 - Engineering Equations of State Slide 7 II. Generalized Fluid Properties An Introduction to the Radial Distribution Function The body centered cubic unit cell R o Nc= ρπ∫ g() r 4 r2 dr; o where g(r) is our "weighting factor" henceforth referred to as the radial distribution function (rdf). g 1 The radial distribution function for the bcc hard sphere fluid 1 2 r / s Chapter 6 - Engineering Equations of State Slide 8 An Introduction to the Radial Distribution Function Ro Nc = ρ ∫ g(r) 4πr 2dr o 4 g g 3 2 1 1 0 01234r/σ r / s The low density hard-sphere fluid The high density hard sphere fluid 4 g g βε 3 η=0 =1.0 2 1 1 η=0.4 0 r / s 01234 r/σ The low density square-well fluid The high density square-well fluid Chapter 6 - Engineering Equations of State Slide 9 II. Generalized Fluid Properties The connection from the molecular scale to the macroscopic scale The Energy Equation ∞ UU− id N ρ = A Nu g 4π r2 dr ∫ A nRT 2RT 0 The Pressure Equation ρ ∞ P NA du 2 =−1 ∫rNA g 4π r dr ρ RT 6RT 0 dr ρ σ ρ ∞ NA du 22NA du PRT/ ρ =−1 ∫∫rNA g 4π r dr - rNA g 4π r dr 6RT 0 dr 6RT σ dr Chapter 6 - Engineering Equations of State Slide 10 II. Generalized Fluid Properties The Van der Waals Equation of State − ρ σ ρ NA du 2 b ∫rNA g 4π r dr ≈ ()ρ where b~close-packed volume 6RT O dr 1-b ρ ∞∞ρσ3 ε NA du 2 NNAA du 2 ∫∫rNA g 4π r dr = x g 4πσ x dx where x≡ r / 66RT σ dr RT 1 dx 3 ∞ NNAAσε du / ε a ≡ ∫ x g 4π x2 dx 6 σ dx The resulting equation of state is: bρ aρ 1 aρ Z =+1 −= − ()1− bρ RT()1− bρ RT By fitting the critical point, where ∂P/∂V = 0 and ∂2P/∂V2 = 0, 2 2 a = 0.475 R Tc /Pc; b = 0.125 RTc/Pc Chapter 6 - Engineering Equations of State Slide 11 II. Generalized Fluid Properties Example. From the molecular scale to the continuum Suppose that the radial distribution function can be reasonably represented by: bρε 1 g ≈+1 ≡ σ ≡ π σ3 kT x 4 where x r/ and b /6 NA Derive expressions for the compressibility factors of fluids that can be accurately represented by the square-well potential. Evaluate this expression at bρ = 0.2 and ε/kT = 1. Solution: First read Appendix C then note that, for the square-well potential, exp(-u/kT) = exp(ε/kT) H(r-σ) + [exp(ε/kT)-1] [1-H(r-1.5σ)] Taking the derivative of the Heaviside function gives the Dirac delta in two places: N ρ Z = 1 + A ∫[]δ(r − σ)y(r)exp()ε / kT − δ(r −1.5σ)y(r)[]exp ()ε / kT −1 4πr3dr 6 4πN ρσ 3 Z = 1 + A {}y(σ)exp()ε / kT −1.53 y(1.5σ)[]exp ()ε / kT −1 6 Noting that y(r) ≡ g(r) exp(u/kT) and that exp(u/kT) is best evaluated inside the well: Z = 1+ 4bρ { g(σ+) - 1.53[1-exp(-ε/kT)] g(1.5σ-)} This is true for SW with any g(r). For the above expression: g(σ+) = 1 + bρ ε/kT and g(1.5σ-) = 1 + 0.198 bρ ε/kT Z = 1+ 4bρ {1 + bρ ε/kT - 1.53[1-exp(-ε/kT)]( 1 + 0.198 bρ ε/kT)} Z = 1+4(0.2){1+0.2-2.1333*1.0396} = 0.1858 Chapter 6 - Engineering Equations of State Slide 12.
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