y i x P vap 10 A T SOLUTION TO HOMEWORK #7 #Problem

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1 SOLUTION TO HOMEWORK #7 #roblem a. In order to solve ths problem, we need to know what happens at the bubble pont; at ths pont, the frst bubble s formed, so we can assume that all of the number of moles for the components n the gas phase are close to zero (yet there s an equlbrum). Therefore, we use the compostons gven n the problem statement and we assume they are the compostons of the lqud. (x=z, z beng the composton gven). x ET 0.05 x 0.10 x NB 0.40 x M 0.45 Now, we know, from Raoult s law that: y x vap and from the problem statement we know that the pressure s 5 bar. Therefore, we proceed to calculate the bubble pont temperature. The way the bubble pont temperature s done s by dong the followng. We also know that: y 1 and therefore, substtutng y x vap 1 Therefore: 1 x vap ET ET x vap x NB vap vap NB x M M Now, we do the followng: 1 x ET 10 A T B x 10 A T ET B x NB 10 A T B x M 10 A T NB B M

2 From ths equaton, we can fnd the bubble pont temperature. Ths equaton can be solved for the temperature, whch comes out to be 294K. You can use bsecton, guessng (better be clever guessng), or Goal Seek n excel. Now, we need to calculate the compostons. Ths s done by usng Raoult s law yeldng: y ET y y NB y M One way of checkng f the answer s correct s to use then somethng was done wrong. y 1. If ths equalty s not correct, b. The dew pont calculatons are smlar, but n ths case the equatons change slghtly. x y vap also, snce we are at the dew pont (the pont n whch the frst drop of lqud s formed), we can say that almost no lqud s present. x y vap 1 whch, when the correct terms are substtuted n, we obtan: 1 y ET 10 A T B ET y 10 A T B y NB 10 A T B NB y M 10 A T B M where we use the composton gven (z) as the composton of the vapor. Solvng for the dew pont temperature yelds that the temperature s 314K. Now, by usng Raoult s law, we get that the compostons of the lqud phase are: x x x x ET NB M

3 c. In order to solve the flash problem, we need the help of other equatons. Frst, we begn by calculatng the K-factors for each component. Ths s done by usng the followng equaton: vap vap y K x Where we set the actvty and fugacty coeffcents to one because (Raoult law assumpton). Then, by calculatng each K-factor, we obtan: K ET K K NB K M Now, n a flash, the feed stream s separated nto the lqud and vapor streams. Ths s shown below: Vapor Feed V Lqud Therefore, we need to know what are the compostons of the vapor and lqud streams, so these would be part of our equatons used for solvng ths problem. We have: Vapor Stream y 1 Lqud Stream x 1

4 Now, by dong a mass balance around the flash, for each component, we can obtan four other equatons that could help us n solvng ths problem. These are (z s represent the component s compostons of the feed): z ET F x ET L y ET V Smlarly for propane z F x L y V for n-butane z NB F x NB L y NB V and for 2-methyl propane z M F x M L y M V Now, we can use a bass of 1 mol for the feed (F=1), whch wll gve us: 1 L V Now, we can replace the V n all equatons: ET ET ET ET z F z x L y L z F z x L y 1 L z F z x L y L NB NB NB NB z F z x L y L M M M M Now, we can solve ths complex problem by dong the followng: frst, guess L (the lqud flow). Once ths value s guessed, we can proceed to solve the compostons by usng the equatons of the K-factors and the mass balances. As stated before, from the K-factors we obtan: K y x so, for each component, we get: y ET x ET y 2.238x y NB 0.546x NB y M 0.743x M

5 Now, these relatonshps can be substtuted nto the mass balances as stated above as follows: z F x L x 1 L ET ET ET z F x L x 1 L z F x L x 1 L NB NB NB z F x L x 1 L M M M whch wll allow us to calculate the compostons of each stream. x z F /[ L L ] ET ET x z F /[ L L ] x z F /[ L L ] NB NB x z F /[ L L ] M M We are now left wth only two equatons, namely y 1 and x 1 Thus we equate them y x Whch can be rewrtten as: K x x 0 or K x K x K x K x ( x x x x ) 0 ET ET NB NB M M ET NB M or ( KET 1) xet ( K 1) x ( KNB 1) xnb ( KM 1) xm 0 Thus substtutng the x s:

6 ( ) z F /[ L L ] ( ) z F /[ L L ] ET ( ) z F /[ L L ] (0.7431) z F /[ L L ] 0 NB By solvng ths equaton for L (usng F=1), we obtan that: M Therefore, we have solved the flash problem.

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10 #roblem 5 roblem 6: a) Obtan an expresson relatng the mnmum amount of work needed to separate a mxture nto ts pure components (at constant T and ) as a functon of the fugactes of the components n the mxtures and the fugactes of the pure components. b) Show that the expresson can be wrtten only n terms of temperature and the molar fractons, when deal mxture s assumed and the Lews and Randall rule s used. a) We start by wrtng (Chapter 4) G W rev T, rev Thus, f Gm s the gbbs free energy of the mxture of two components, G1 s the Gbbs free energy of the stream of pure component 1 and G2 s the Gbbs free energy of the stream of pure component 2, we wrte W G G G rev m 1 2 But G x G x G m m,1 1 m,2 2 However, o f m, Gm, G (, T ) RT ln o f o o and G G (, T), G G (, T) Substtutng, we get: W G G G n G n G n G (, T ) n G (, T ) o o rev m 1 2 m,1 m,1 m,2 m,

11 After realzng that nm,1 n1 and nm,2 n2, we get f m,1 f m,2 Wrev RT n1ln n2ln o o f 1 f 2 art b) We can wrte f, x f (Lews and Randall) and therefore: o m Wrev RT n m,1 ln x1 nm,2 ln x2 or Wrev / n RT x m,1 ln x1 xm,2 ln x2 Recognze ths? It s somehow smlar to the gbbs free energy of mxng of deal mxtures, rght?

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