VLE from an Equation of State By J.R. Elliott and C.T. Lira FUGACITY IN A MIXTURE BY AN EQUATION OF STATE ∂ G ∑ ∂ dni dG = VdP - SdT + ni i TPn,,ji≠ and noting, ∂ A ∑ ∂ dni dA = -PdV - SdT + ni i TV,, nji≠ we may substitute ∂ G ∑ ∂ dni dA = dG - PdV - VdP = VdP - SdT + ni - PdV - VdP i TPn,,ji≠ ∂ ∂ A G ⇒ ∑ ∂ dni ∑ ∂ dni -PdV - SdT + ni = -PdV - SdT + ni i TV,, nji≠ i TPn,,ji≠ Equating coefficients of dni ∂ A ∂G ∂ ∂ µ ni = ni = i(T,P,V) TV,, nji≠ TPn,,ji≠ Referencing to the ideal gas state: ∂()/AA− ig RT $ ln( f i/yiP) = (µi(T,P)-µiig(T,P))/RT = ∂ - lnZ ni TV,, nji≠ VLE from an Equation of State Slide 1 K-Values from an Equation of State To apply this, consider the PR EOS as an example. AA− ig A ZB++()12 a 11++() 2bρ =−ln()1 −BZ / − ln =−ln()1 −bρ − ln nRT B 8 ZB+−()12 bRT 8 11+−() 2bρ For “random mixing”, the probability of any “i-j interaction” is the same and goes as the product of the “i-j concentrations”. This suggests that we could define Av = ΣΣyiyjAij and Bv = ΣyiBi letting Aij= AAii jj by comparison to the form of the energy equation for mixtures (discussed below). Then differentiation (as detailed below) yields $ f v B v Av ZBvv++()12 2ΣyA B v ln i =−−−−i ()(ZZBvvv1 ln ) ln jij− i v v vv+− v v yPi B B 8 ZB()12 A B Note: AL = ΣΣxixjAij and BL = ΣxiBi but the derivation of the fugacity coefficient would be the same and: $ f L B L A L ZBLL++()12 2ΣxA B L ln i =−−−−i ()(ZZBLLL1 ln ) ln jij− i L L LL+− L L xPi B B 8 ZB()12 A B As we saw in the case of pure fluids, there is no fundamental reason to distinguish between the vapor and liquid phases except by the initial guess for Z. The equation of state approach encompasses this lack of distinction in a very direct way. To obtain an expression for Ki, it is convenient to define the fugacity coefficients of a mixture as $ v $ L L f i v f i L ϕ$ ≡≡ϕϕ$ $ $$v L = i i and i = K v ⇒ Recalling that ffi i at equilibrium, we find that i ϕ$ yPi xPi i VLE from an Equation of State Slide 2 UNIT III. FLUID PHASE EQUILIBRIA FUGACITY IN A MIXTURE BY AN EQUATION OF STATE: Density Dependent Formulas Example. Fugacity coefficient for the virial equation For pressures to 10 bars, a common method is to use the virial equation given by: Z = 1 + Bρ ; where B = ΣΣyiyjBij and Bij= BBii jj . Develop an expression for the fugacity coefficient. ρ ∂()/AA− ig RT AA− ig B dBρ AA− ig Bn2 1 ln()ϕ = − ln Z ==⇒Bρ Bρ ==∑∑nn B k ∂ ; ∫ ρ ijij nk nRT 0 B RT VV TV,, nki≠ 2 BB ∑∑∑∑∑ ==()()() Note: For Bij= ii jj , nni j B ij n i B ii n j B jj n j B jj 2 ig ∂ ∂ ∂()/AA− RT 1 ()∑∑∑nBjjj 2()nBjjj()nB jjj == ∂ ∂ ∂ nVk nk V nk TV,, nki≠ ∂ ()∑ nBjjj 2 = B ()ϕρ=−=−()∑∑() ∂ kk ⇒ lnkkkjjjjjkBnBZ ln2 yBZ ln nk V VLE from an Equation of State Slide 3 UNIT III. FLUID PHASE EQUILIBRIA Example. Fugacity coefficient for the van der Waals EOS The VdW EOS provides a simple but fairly accurate representation of key EOS concepts for mixtures. The main tricks developed for this EOS are the same for other EOS's but the algebra is a little simpler. 1 aρ Z = − 1− bρ RT where a= ΣΣyiyjaij ; aij= aaii jj b= Σyibi Develop an expression for the fugacity coefficient. Solution ∂()/AA− ig RT ln()ϕ = − ln Z k ∂ nk TV,, nki≠ ρρ AA− ig bbdb()ρ bρ a db ()ρ a =−()Z 1 = − bρ =−ln()1 −bρρ − b ∫∫ρ − ρ ρ nRT 00b 1 b bRT b bRT AA− ig an 2 ∑∑nn a =−nbln()11 −ρρ − =−nb ln() − − ijij RT VRT VRT VLE from an Equation of State Slide 4 ∂()/AA− ig RT n ∂ρb 1 ∂()ΣΣnna =−ln()1 −bρ + − ijij ∂ − ρ ∂ ∂ nk 1 b nVRTk nk TV,, nki≠ nyb()∑∑jj()nb jj ∂ρb b bρ ==⇒=k ∂ V V nk V 2 ∑∑∑∑∑nn a==() n a() n a() n a Note: For aij= aaii jj , ijij i ii j jj j jj 2 ∂ ΣΣ ∂ ()∑na ()nnaijij jjj ==2 ana()∑ ∂ ∂ kk j jj nk nk ΣΣρ n b 2 najkj b ρ 2 xajkj ln()ϕρ=− ln()1 −b + k −−=−−+ lnZb ln()1 ρ k −− ln Z k 1 − bρ V VRT 1 − bρ RT B a A a jkA jk b B bρ ≡≡≡;; ; kk ≡ Z bRT B a A b B Σ B 2 x jkjA ln()ϕ =− ln()ZB − + k − k Z − B Z VLE from an Equation of State Slide 5 UNIT III. FLUID PHASE EQUILIBRIA Example. Fugacity coefficient for the PREOS 1 aρ 1 Z = − 1− bρ RT ()12+−bbρρ22 where a= ΣΣyiyjaij ; aij= aaii jj b= Σyibi Develop an expression for the fugacity coefficient. Solution ∂()/AA− ig RT ln()ϕ = − ln Z k ∂ nk TV,, nki≠ From our integration for the pure fluid, AA− ig a 11++() 2bρ =−ln()1 −bρ − ln nRT bRT 8 11+−() 2bρ AA− ig an 2 =−nbln()1 −ρρρ −{} ln[]11 + ( + 2 )bb − ln[] 11 + ( − 2 ) RT nbRT 8 VLE from an Equation of State Slide 6 ∂ρb ∂ρb ()12+ ()12− ∂()/AA− ig RT n ∂ρb an 2 ∂n ∂n =−ln()1 −bρ + − kk− ∂n 1− bρ ∂n ++ ρ +− ρ k TV,, n k nbRT 8 11() 2b 11() 2b ki≠ ∂an 2 ∂nb ∂ 2 ∂ 11++() 2bρ nkkan n − ln − 11+−() 2bρ nbRT 8RT 8()nb 2 nxb()∑∑jj()nb jj ∂ρb b bρ ==⇒=k ∂ V V nk V 2 ∑∑∑∑∑nna==() n a() n a() n a Note: For aij= aaii jj , i j ij i ii j jj j jj 2 2 ∂ ΣΣ ∂ ()∑na ∂ ()an ()nnijij a jjj == =2 ana()∑ ∂ ∂ ∂ kk j jj nk nk nk VLE from an Equation of State Slide 7 b ρ ab ρ ()12+ ()12− ln()ϕρ=− ln()1 −bZ − ln + kk− − k 1− bρ bRT8 11++() 2 bρρ11+−() 2 b Σ a 11++() 2bρ 2 xajjk b − ln − k bRT 8 11+−() 2bρ a b Note: b ρ ab ρ ()12+ ()12− b bρ abρ 1 b kk− [ − ] = k − =− k{}Z 1 1− bρ bRT8 11++() 2 bρρ11+−() 2 b b 1− bρ bRT 8 12+−bbρρ22 b Σ b a 11++() 2bρ 2 xajjk b ln()ϕρ=− ln()11 −bZ − ln +k {}Z − − ln − k k b bRT 8 11+−() 2bρ a b B a A a A b B bρ ≡≡≡≡;; jk jk ; kk Z bRT B a A b B Σ B A ZB++()12 2 xAjjk B ln()ϕ =− ln()ZB − +k {}Z −1 − ln − k k B B 8 ZB+−()12 A B VLE from an Equation of State Slide 8 Example 10.6. Bubble point pressure from PR EOS Use the PREOS(kij = 0) to determine the bubble point pressure of equimolar solution of nitrogen+methane at 100K. Solution: Σ vap is Initial guess P ≈ xPi i ≈ .5*7.851+.5*.340 = 4.1 bars; y1 = .5*7.851/4.1=.96 p=4.1,y1=.963 p=4.5,y1=.944 p=4.24,y1=.945 Comp Tc Pc ω Tb x K y K y K y N2 126.2 33.94 .040 77 .5 1.95 .970 1.796 .898 1.893 .9467 CH4 190.2 46.00 .011 112 .5 .115 .057 .1046 .052 .1097 .0548 1.027 0.950 1.0015 Note: y1 for next guess is computed from estimate of y1 of current guess. For example, 0.944=0.970/1.027, and 0.945=0.898/0.950. Note how quickly the estimate for y1 converges to the final estimate of 0.945. VLE from an Equation of State Slide 9 Example 10.7. Flash of PREOS solutions A distillation is to produce overhead products having the following compositions: Propane Isobutane n-Butane zi 0.23 0.67 0.10 Suppose only a partial condensation at 320 K and 8 bars. What fraction of liquid would be condensed according to the PREOS (kij=0)? Solution: Refer back to the same problem for an ideal solution for guess. L/F = 0.75 ⇒ x = {0.1829,0.7053,0.1117} and y = {0.3713,0.5624,0.0648} L/F= L/F= L/F= 0.75 0.90 0.867 Co Tc Pc ω zi Ki Di Di Di xi yi C3 369.8 42.49 0.152 0.23 1.729 -0.142 -0.1563 -.1528 .2097 .3625 iC4 408.1 36.48 0.177 0.67 0.832 0.118 0.1146 0.1153 .6853 .5700 nC4 425.2 37.97 .193 0.10 0.640 0.040 0.0373 0.0378 .1050 .0673 0.016 -0.004 0.0002 1.000 .9998 For comparison, L/F = 0.87 (PR) vs. 0.75 (IS) Propane Isobutane n-Butane Ki (PR) 1.727 0.833 0.641 Ki (IS) 2.03 0.80 0.58 VLE from an Equation of State Slide 10 Phase Envelope for PREOS Use the PREOS(kij=0) to determine the phase envelope of nitrogen+methane at 150K. Plot P vs. xN2, yN2 and compare the results from PREOS to the results from the "short-cut" result. Solution: use the last guess as the initial guess for the next guess. Above 0.5 mole fraction, increment xN2 by 0.02 each time or you will not converge.