Introductory Chemical Engineering Thermodynamics

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Introductory Chemcal Engneerng Thermodynamcs Chapter 9 - Introducton to Multcomponent Systems By J.R. Ellott and C.T. Lra

INTRODUCTION TO MULTICOMPONENT SYSTEMS The prmary dfference between pure and multcomponent systems s that we must now consder the mpacts of changng the composton on the Gbbs energy. Beyond that, the Gbbs energy must stll be mnmzed, the calculus of classcal thermodynamcs must be appled, the fugactes of the components n the phases must be equal, and, n general, the problem s pedagogcally the same as the phase equlbrum problem for pure components. The computatonal methodology gets more complcated because there s more to keep track of. And the best choce of "equaton of state" can change qute a bt dependng on the nature of the soluton. In ths context, "equaton of state" refers to any model equaton for the thermodynamc propertes of mxtures. But computatons and equaton of state development should not seem unfamlar at ths stage n the course. eg. du(t,p,n) = ( U/ P)T,ndP + ( U/ T)P,n dt + Σ ( U/ n )P,T,n j dn eg. dg(t,p,n) = ( G/ P)T,ndP + ( G/ T)P,n dt + Σ ( G/ n )P,T,n j dn At constant moles of materal, the mxture must follow the same constrants as any other flud. That s, we only have two state varables left f we keep the composton constant. ( G/ P)T,n = V and ( G/ T)p,n = -S ; Fundamental propertes therefore dg = VdP - SdT + Σ ( G/ n )T,P,n j dn µ ( G/ n )T,P,n j "chemcal potental" Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 1

Equlbrum Constrant: equalty of chemcal potentals dg = 0 = Σ µ dn equlbrum, constant T and P Defne system of two components (eg. EtOH+H2O) between two phases (eg. vapor and lqud). µ L 1 dn L 1 + µ L 2 dn L 2 + µ V 1 dn V 1 + µ V V 2 dn2 = 0 but f component 1 leaves the lqud phase then t must enter the vapor phase (and smlarly for component 2). dn L = dn V and dn L = dn V 1 1 2 2 V V V V L V µ 1 µ 1 dn 1 + µ 2 µ 2 dn2 = ( ) ( ) 0 Therefore, V L V L µ 1 = µ 1 and µ 2 = µ 2. Chemcal potental of a pure flud ( ) ( ) ( ) ( ) ( ) µ G / n = ng / n = G n / n + n G / n j n n T, P, n j n n n n n n / = / and G =, / = 0. n n Therefore, µ = j n n G Note: Ths means G L = G V for pure flud, as before. Thus, µ s a "generalzed G" for components n mxtures. but n( G n ) n( G n) G( T P) f ( n) n( G n) Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 2

Equlbrum Constrant: equalty of fugactes Fugacty - "escapng tendency" For a pure flud: ln(f/p) (G-G g )/RT For mxtures, generalze: ln(f/p) (G-G g )/RT to RTln( $ f /P) (µ - µ g ) where $ f fugacty of component n a mxture. Equalty of fugactes as a crteron of phase equlbrum V eq eq µ T, P = µ L T eq, P eq ( ) ( ) V eq eq By defnton, µ ( T, P ) µ L ( T eq, P eq ) = RT ln f $ V ( T eq, P eq ) / f$ L ( T eq, P eq ) [ ] V eq eq but, µ ( T, P ) µ L ( T eq, P eq ) = 0 ln ( f $ V L / f$ ) = 0 $ V f f$ = at equlbrum. L Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 3

Actvty, Actvty Coeffcent, and Fugacty Coeffcent "Actvty" a = $ f /f Ionc solutons, reactons "Actvty coeffcent" = γ = $ f /(x f ) Low pressure phase equlbra "Fugacty coeffcent" = φ = $ f /(x P) Hgh pressure phase equlbra where f s the value of the fugacty at standard state. The most common standard state s the pure component at the same temperature and pressure as the system of nterest. In that case f = f. Note that the defnng equaton for actvty coeffcent can be rearranged to become an equaton for calculatng the fugacty of a component n a mxture. Ths s the general procedure used for lquds. L o o $f = γ x f = γ x f γ x φ P Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 4

IDEAL SOLUTIONS The smplest soluton model s to assume that nteracton energess n the mxture are lke nteracton energes n the pure flud. e.g. 35 Cl 2 mxed wth 37 Cl 2 s s then, H mx = 0 and H Tot = Σ n H As for the entropy change of mxng, the loss of order due to mxng s unavodable s Smx / R = Σ x ln ( x ) Combnng G s s H s mx mx Smx = = + x ln( x ) RT RT R The defnton of fugacty n terms of a change n free energes gves: G n ( µ G ) f mx = = x ln $ RT n RT f G ( µ G ) f mx Thus, for an deal soluton, = x = x ln $ RT RT f $ s f / f = x f$ s = x f Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 5

Raoult's Law V = Substtutng the deal soluton result y f = x f V L f $ f $ V ( f / P) ( Tr,Pr, ) = ϕ ω for the vapor We correct the fugacty of the urated lqud for the effect of pressure on the lqud. f = P L L ( f ) RT ln( f f ) = G G dg = RTd ln / ( G / P) = ( H / P) T( S / P) = V T ( V / T ) + T( V / T ) T ( Tr, Pr ω ) ϕ, T L L L L ( G G ) = V dp V ( P P ) = RT ln( f / f ) L L f / f = exp[ V ( P P )/ RT ] Poyntng Correcton L Substtutng, y P ( Tr, Pr ) xp ϕ ( Tr,Pr ) exp[ V ( P P )/ RT ] (, P ) [ V ( P ) / ] K Note: at reasonably low pressures, = x P L [ V ( P P )/ RT ] 1 T ϕ p L ( Tr, r ) P ϕ 1and exp K = P p = V Raoult s Law Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 6

Classes of VLE calculatons Type Informaton_requred Informaton_computed Convergence BP T,x P,y Easest DP T,y P,x Not bad BT P,x T,y dffcult DT P,y T,x dffcult FL P,T x,y,l/f Most dffcult Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 7

Short-cut Estmaton of VLE K-ratos and Intal Guesses for VLE [ 7 / 3( 1+ ω )( 1 1 / Tr 10 )] K = Pr 7 3 ( 1+ )( 1 1 ) /Tr Bubble Pont Pressure: P = P = P = Pc10 y y P 1 y Dew pont pressure,: Σ x = 1= Σ = Σ = 7 K ( + )( ) 3 Pc 10 1 ω 1 1 / Tr P 7 ( + )( ) 3 10 1 w Pc 1 1 / Tr For bubble or dew pont temperatures, t s not so straghtforward. One guess would be: T = Σx T T = ΣyTc T / ΣyTc But these are somewhat cruder guesses than for the pressures. Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 8

Example 9.1 Bubble temperature, dew temperature, and flash A dstllaton column s to produce overhead and bottoms products havng the followng compostons: z(overhead) Propane 0.23 Isobutane 0.67 n-butane 0.10 Total 1.00 a) Calculate the temperature at whch the condenser must operate n order to condense the overhead product completely at 8 bars. b) Assumng the overhead product vapors are n equlbrum wth the lqud on the top plate of the column, calculate the temperature of the overhead vapors and the composton of the lqud on the top plate when operatng at the pressure of part (a). c) Suppose only a partal condenon at 320 K. What fracton of lqud would be condensed? Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 9

Example 9.1 Bubble temperature Soluton: Use the short-cut estmates of the K-ratos a) The maxmum temperature s the bubble pont temperature. Guess T=310K Guess T=320K K y K y C3 1.61 0.370 2.03 0.466 C4 0.616 0.413 0.80 0.536 nc4 0.438 0.044 0.58 0.058 0.827 1.061 1.000-0.827 T = 310 + ( 320 310) = 317 1. 061 0. 827 Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 10

Example 9.1 Dew temperature b) The urated vapor s at ts dew pont temperature. Guess T=315K Guess T=320K y K x K x C3 0.23 1.81 0.127 2.03 0.113 C4 0.67 0.70 0.957 0.80 0.838 nc4 0.10 0.50 0.200 0.58 0.172 1.284 1.123 1. 00 1123. T = 320 + ( 320 315) = 324 1. 281 1123. Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 11

Example 9.1 Flash c) Ths s an sothermal flash problem. M-Bal: zc3 = xc3 (L/F) + yc3 (V/F) zic4 = xc4 (L/F) + yc4 (V/F) znc4 = xnc4 (L/F) + ync4 (V/F) Where z s the feed composton and L/F s the lqud to feed rato. Overall, V/F = 1-L/F, z [( ) = x L / F + K ( 1 L / F) ] z x = K ( + L / F)( 1 K ) z K y = K + L / F 1 K ( )( ) Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 12

Solve (Σx -Σy )=0. That s z ( 1 K ) ( / )( 1 ) Σx Σy = Σ ΣD = K + L F K 0 Guess L/F=0.5 L/F=0.6 L/F=0.77 z K D K D K D C3 0.23 2.03-0.1564 2.03-0.1678 2.03-0.1915 C4 0.67 0.80 0.1489 0.80 0.1457 0.80 0.1405 nc4 0.10 0.58 0.0532 0.58 0.0505 0.58 0.0465 0.0457 0.0284-0.0045 0 + 0.0045 L / F = 0.77 + = 0.0284 + 0.0045 ( 0.6 0.77) 0. 7467 Ellott and Lra: Chapter 9 - Introducton to Multcomponent Systems Slde 13