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Outline o Bubble/Dew Calculations using MRL Making the function for - PowerPoint PPT Presentation

Outline o Bubble/Dew Calculations using MRL Making the function for G E realistic: o 11.4 Van der Waals perspective Van Laar Scatchard/Hildebrand Flory, Flory-Huggins o 11.5 Theory - Skip o 11.6 Local Composition Theory Wilson UNIQUAC


  1. Outline o Bubble/Dew Calculations using MRL Making the function for G E realistic: o 11.4 Van der Waals perspective Van Laar Scatchard/Hildebrand Flory, Flory-Huggins o 11.5 Theory - Skip o 11.6 Local Composition Theory Wilson UNIQUAC UNIFAC 1

  2. 11.4 Van der Waals perspective Regular Solutions (V E = 0, S E = 0) energetics of mixing are described by the same energy equation for mixtures that we previously developed in discussing the simple basis for mixing rules. ∑ ∑ − = − ρ ig U U x x a i j ij Noting that 1/ ρ = V = Σ x i V i according to “regular solution theory,” ∑ ∑ – x i x j a ij U ig ( ) U – = - - - - - - - - - - - - - - - - - - - - - - - - - - - - ∑ x i V i For the pure fluid, taking the limit as x i → 1, 2

  3. a ii x i a ii U ig U is U ig ( ) i ⇒ ( ) ∑ U – = – - - - - - – = – - - - - - - - - - - V i V i For a binary mixture, subtracting the ideal solution result to get the excess energy gives, 2 a 11 2 a 22   a 11 a 22 x 1 + 2 x 1 x 2 a 12 + x 2 U E   = x 1 - - - - - - - + x 2 - - - - - - - – - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -   V 1 V 2 x 1 V 1 + x 2 V 2 3

  4. van Laar 2 F a a I x x V V ≡ − ⇒ = 11 22 E 1 2 1 2 Q U x V Q G J + V V x V H K 1 2 1 1 2 2 A A x x = = E E 12 21 1 2 G U + ( x A x A ) 1 12 2 21 A 12 γ 1 ln = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 A 12 x 1 1 + - - - - - - - - - - - - - A 21 x 2 4

  5. Fitting Van Laar to a single experiment: γ 2 2 x 2 ln γ 1 1 A 12 = ln + - - - - - - - - - - - - - - - γ 1 x 1 ln γ 1 x 1 ln 2 γ 2 1 A 21 = ln + - - - - - - - - - - - - - - - γ 2 x 2 ln Azeotrope point can be used. See example 11.6 5

  6. Infinite dilution can be used. See example 11.7 6

  7. Scatchard Hildebrand a 12 = a a 11 22   x 1 x 2 V 1 V 2 - a 11 a 22 2 a 11 - a 22 U E   = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + - - - - - - - – - - - - - - - - - - - - -   2 2 2 2 x 1 V 1 + x 2 V 2 V 1 V 2 V 1 V 2 2   x 1 x 2 V 1 V 2 a 11 a 22 U E   = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - – - - - - - - - - - - -   x 1 V 1 + x 2 V 2 V 1 V 2 = = Φ Φ δ − δ + 2 E E G U f ( x V x V ) a 1 2 1 2 1 1 2 2 = = ΦΦ δ − δ + 2 E E f ( ) G U xV x V a 1 2 1 2 1 1 2 2 Â F i ∫ / is known as the "volume fraction" x V x V i i i i 7

  8. d i ∫ a / V is known as the "solubility parameter" ii i ∆ ∆ − vap vap U H RT δ i ≡ = V V i i This is a predictive technique valid for nonpolar substances. See table 11.1 for parameters. Also can make adjustable. 8

  9. Van Laar and Scatchard-Hildbrand 355 350 345 T(K) Scatchard-Hildebrand 340 k ij =-0.038 335 van Laar 330 0 0.2 0.4 0.6 0.8 1 MeOH x,y methanol + benzene 9

  10. Flory’s Equation Recall Ideal Solution result (pg 95) V T V T ∆ S 1 = n 1 R ln - - - - - - = n tot x 1 R ln - - - - - - i i V 1 V 1 V T V T ∆ = ln - - - - - - = ln - - - - - - S 2 n 2 R n tot x 2 R i i V 2 V 2 10

  11.   V T V T ∆ ∆ ∆   = + = ln - - - - - - + ln - - - - - - S S 1 S 2 n tot R x 1 x 2  i i  V 1 V 2 11

  12. Φ Φ = − = + E E E 1 2 G H TS RT x ( ln x ln ) 1 2 x x 1 2 12

  13. Flory-Huggins Model Φ 1 Φ 2   G E Φ 1 Φ 2 x 1 ( )χ RT = RT x 1 ln - - - - - - + x 2 ln - - - - - - + + x 2 R   x 1 x 2 13

  14. 11.6 Local composition models (nonrandom) Common Features o Lattice Model - Fixed number of neighbors = 10 o Uses G E = A E approximation (good) o Local Composition 14

  15. x 21 x 2 x 12 x 1 - Ω 21 - Ω 12 - - - - - - - = - - - - - - - - - - - = - - - - x 11 x 1 x 22 x 2 o Two-fluid Theory ( ) ( ) 1 2 U ig U ig U ig ( ) ( ) ( ) U – = x 1 U – + x 2 U – - x 1 x 2 Ω 21 ε 21 ( ε 11 ) x 2 x 1 Ω 12 ε 12 ( ε 22 ) N A z – – U E = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - x 2 Ω 21 x 1 Ω 12 2 x 1 + + x 2 A E U E – - T d ∫ - - - - - - - = - - - - - - - - - - - - - - - + C RT RT T 15

  16. Wilson’s equation V 1 – A   Ω 12 Λ 21 21 = = - - - - - - exp - - - - - - - - - - -   V 2 RT Λ 12 Λ 21   γ 1 ( x 2 Λ 12 ) ln = – ln x 1 + + x 2 - - - - - - - - - - - - - - - - - - - - - - - - - – - - - - - - - - - - - - - - - - - - - - - - - - -   x 2 Λ 12 x 1 Λ 21 x 1 + + x 2 16

  17. UNIQUAC A z ε ij ( ε jj ) – N – q i q i – a q i     Ω ij - τ ij ij = - - - - exp - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = - - - - exp - - - - - - - - - = - - -     q j 2 RT q j T q j r i q i θ i τ ij 17

  18. E G = Φ Φ ln / + x ln / x x x a f a f 1 1 1 2 2 2 RT − Φ θ + Φ θ 5 ln / ln / q x q x a f a f 1 1 1 1 2 2 2 2 − θ + θ τ − θ τ + θ ln( ) ln( ) q x q x 1 1 1 2 21 2 2 1 12 2 18

  19. Φ 1 Φ 1 Φ 1 Φ 1     γ 1 ln = ln - - - - - - + 1 – - - - - - - – 5 q 1 ln - - - - - - + 1 – - - - - - -     θ 1 θ 1 x 1 x 1 θ 2 τ 12 θ 1 ( θ 1 θ 2 τ 21 ) + q 1 1 – ln + – - - - - - - - - - - - - - - - - - - - - - - - - – - - - - - - - - - - - - - - - - - - - - - - - - θ 1 θ 2 τ 21 θ 1 τ 12 θ 2 + + Φ 2 Φ 2 Φ 2 Φ 2     γ 2 ln = ln - - - - - - + 1 – - - - - - - – 5 q 2 ln - - - - - - + 1 – - - - - - -     θ 2 θ 2 x 2 x 2 θ 1 τ 21 θ 2 ( θ 1 τ 12 θ 2 ) + q 2 1 – ln + – - - - - - - - - - - - - - - - - - - - - - - - - – - - - - - - - - - - - - - - - - - - - - - - - - θ 1 θ 2 τ 21 θ 1 τ 12 θ 2 + + 19

  20. Group parameters for the UNIFAC and UNIQUAC equations. AC in the table means aromatic carbon. (DIFFERS SLIGHTLY FROM TEXT) Main Sub- R(rel.vol.) Q(rel.area) Example Group group CH2 CH3 0.9011 0.8480 CH2 0.6744 0.5400 n-hexane: 4 CH2+2 CH3 CH 0.4469 0.2280 isobutane: 1CH+3 CH3 C 0.2195 0 neopentane: 1C+ 4 CH3 C=C CH2=CH 1.2454 1.1760 1-hexene: 1 CH2=CH+3 CH2+1 CH3 CH=CH 1.1167 0.8670 2-hexene: 1 CH=CH+2 CH2+2 CH3 CH2=C 1.1173 0.9880 CH=C 0.8886 0.6760 C=C 0.6605 0.4850 ACH ACH 0.5313 0.4000 benzene: 6ACH AC 0.3652 0.1200 benzoic acid: 5ACH+1AC+1COOH ACCH2 ACCH3 1.2663 0.9680 toluene: 5ACH+1ACCH3 ACCH2 1.0396 0.6600 ethylbenzene: 5ACH+1ACCH2+1CH2 ACCH 0.8121 0.3480 OH OH 1.0000 1.2000 n-propanol: 1OH+1 CH3+2 CH2 20

  21. Group parameters for the UNIFAC and UNIQUAC equations. AC in the table means aromatic carbon. (DIFFERS SLIGHTLY FROM TEXT) CH3O CH3OH 1.4311 1.4320 methanol is an independent group H water H2O 0.9200 1.4000 water is an independent group furfural furfural 3.1680 2.4810 furfural is an independent group DOH (CH2OH) 2.4088 2.2480 ethylene glycol is an independent 2 group ACOH ACOH 0.8952 0.6800 phenol: 1ACOH+5ACH CH2CO CH3CO 1.6724 1.4880 dimethylketone: 1 CH3CO+CH3 CH2CO 1.4457 1.1800 diethylketone=1 CH2CO+2 CH3+1 CH2 CHO CHO 0.9980 0.9480 acetaldehyde: 1CHO+1 CH3 CCOO CH2COO 1.9031 1.7280 methyl acetate: 1 CH3COO+1 CH3 CH2COO 1.6764 1.4200 methyl propanate: 1 CH2COO+2 CH3 benzoic acid: COOH COOH 1.31013 1.2240 5ACH+1AC+1COOH 21

  22. UNIFAC Pure 2-propanol CHOHCH 3 CH 3 CH 3 CHOHCH 3 ( ) 1 o ( ) µ CH 3 µ CH 3 SOG 1 µ CH 3 µ CH 3 – – CH 3 CHOHCH 3 O H 2 CH 3 CH 3 CHOHCH 3 CH 3 H 2 O CH CH 3 CHOHCH 3 3 CH 3 Hypothetical H O 2 solution of SOG o µ CH 3 µ CH 3 “pure” CH 3 – Real mixture (SOG) 22

  23. g = g + g COMB RES ln ln ln k k k ( ) ( ) SOG 1 SOG o 1 o µ CH 3 µ CH 3 µ CH 3 µ CH 3 µ CH 3 µ CH 3 – – – ( ) 1 Γ CH 3 Γ CH 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - – - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = ln – ln RT RT RT o µ 1 µ 1 – ( ) ( ) resid 1 1 γ 1 ν m [ Γ m Γ m ] ∑ ln = - - - - - - - - - - - - - - - - - = ln – ln RT m 23

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