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Separation Process Principles- 2n - Seader & Henley - Solutions Manual

5102 0000055 h lbmol 1750 2154 k12 k13 v r21 z2 1 1 1

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Unformatted text preview: due to reaction be: νi rk , j (1) νk where: Mj = volumetric holdup of liquid on stage j. rk,j = chemical reaction rate, dck,j /dt , for the limiting reactant, k, on stage j. Let the rate law be a power-law expression, in terms of the reference limiting reactant, k, ∆ni , j = − M j C − where: k f , j = A f e − dck , j dt = kf ,j C ∏c i =1 ni i, j − ∏c i =1 mi i, j KC , j Ef RT j ci , j = xi , j cT , j cT , j = total concentration of all components in the liquid on stage j K C , j = chemical equilibrium constant in terms of concentrations on stage j superscripts n, m = reaction orders for forward and backward reactions, respectively, which are related to the stoichiometric coefficients by ν i = mi − ni Exercise 12-2 (continued) Analysis (a): (continued) (a) Chemical equilibrium Eqs. (12-5), (12-7) to (12-15), and (12-17) to (12-18) apply . In Eq. (12-4), liquid mole fractions xi,j are in chemical equilibrium as governed by the chemical equilibrium constant and the stoichiometry, such that: M iLj ≡ (1 + rjL ) L j xi , j − L j −1 xi , j −1 − ∆ni , j − f i ,Lj − N iLj = 0, i = 1, 2, ..., C , , In energy balance, Eq. (12-6), liquid enthalpies include heats of formation. Eq. (12-16) becomes: C C νi M TL, j ≡ (1 + rjL ) L j − L j −1 − ∆nk , j − f i ,Lj − N T , j = 0 i =1 ν k i =1 (b) Kinetic rate law Eqs. (12-5), (12-7) to (12-15), and (12-17) to (12-18) apply. Eq. (12-4) becomes: M iLj ≡ (1 + rjL ) L j xi , j − L j −1 xi , j −1 − ∆ni , j − f i ,Lj − N iLj = 0, i = 1, 2, ..., C , , where ∆ni , j is computed from the reaction rate equation, Eq. (1), above. In Eq. (12-6), enthalpies must include heat of formation. Eq. (12-16) becomes: C C νi M TL, j ≡ (1 + rjL ) L j − L j −1 − ∆nk , j − f i ,Lj − N T , j = 0 i =1 ν k i =1 Exercise 12.3 Subject: Reduction of the number of equations in the rate-based model. Given:. Rate-based model based on Eqs. (12-4) to (12-18). Find: Method to reduce the number of equations. Analysis: either: In Chapter 10, equilibrium-stage models are written for each equilibrium stage in terms of Case 1: xi , j , yi , j , L j ,V j , and Tj (2C + 3) variables Case 2: li , j , υ i , j , and Tj (2C + 1) variables In Section 12.1, the rate-based model is written in terms of : xi , j , yi , j , xiI, j , yiI, j , N i , j , TjL , TjV , TjI , L j , and V j (5C + 5) variables The equations could be rewritten to replace: xi , j , yi , j , L j , and V j by li , j and υ i , j to give (5C + 3) variables. For example, Eq. (12-4) would become: M iLj ≡ (1 + rjL )li , j − li , j −1 − f i ,Lj − N iLj = 0, i = 1, 2, ..., C , , This would eliminate Eqs. (12-11) and (12-12). In Ref. 16, Taylor and his colleagues use component flow rates. This has the advantage that the component material balances are linear and fewer equations are needed because there are two fewer variables per stage. In Ref. 17, Taylor and his colleagues use component mole fractions and total flow rates. This choice is preferable when total flow rates are useful, such...
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