ch10 - Exercise 10.1 Subject Independency of MESH equations...

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Exercise 10.1 Subject: Independency of MESH equations Given: Equations (10-1), (10-3), (10-4), and (10-6). Prove: Equation (10-6) is not independent of the other 3 equations Analysis: Eq. (10-6) can be derived from the other 3 equations as follows, as outlined in the text between Eqs. (10-5) and (10-6).. Summing Eq. (10-1) over all C components: L x V y F z L U x V W y j i j j i j i C i C j i F j j i C i j i C j j i j i C j - - + + = = = = = + + - + - + = ° ° ° ° ° 1 1 1 1 1 1 1 1 1 0 , , , , , ° ± ° ± (1) From Eqs. (10-3) and (10-4), all 5 sums in Eq. (1) are equal to 1. Therefore, Eq. (1) becomes: L V F L U V W j j j j j j j - + + + - + - + = 1 1 0 ° ± ° ± (2) Writing Eq. (2) for each stage from Stage 1 to Stage j: L V F L U V W L V F L U V W L V F L U V W L V F L U V W L V F L U V W j j j j j j j j j j j j j j 0 2 1 1 1 1 1 1 3 2 2 2 2 2 2 4 3 3 3 3 3 2 1 1 1 1 1 1 1 0 0 0 0 0 + + - - - - = + + - - - - = + + - - - - = + + - - - - = + + - - - - = - - - - - - - + ....... (3) Summing Eqs. (2), noting that L 0 = 0 and that many variables cancel, we obtain: or V F U W L V L V F U W V j m m m j m j j j m m m m j + = + = + - - - - = = + - - - ° ° 1 1 1 1 1 1 0 ² ³ ² ³ But this is Eq. (10-6). Therefore, it is not independent of Eqs. (10-1), (10-3), and (10-4).

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Exercise 10.2 Subject: Revision of MESH equations to account for entrainment, occlusion, and chemical reaction (in the liquid phase). Given: MESH Eqs. (10-1) to (10-5). Find: Revised set of MESH equations. Analysis: Entrainment: Let φ j = ratio of entrained liquid (in the exiting vapor) that leaves Stage j to the liquid ( L j + U j ) leaving Stage j . Then, the entrained component liquid flow rate leaving Stage j = φ j x i,j ( L j + U j ). Correspondingly, the entrained component liquid flow rate entering Stage j = φ j+1 x i,j +1 ( L j +1 + U j +1 ). Occlusion: Let θ j = ratio of occluded vapor (in the exiting liquid) that leaves Stage j to the vapor ( V j + W j ) leaving Stage j . Then the occluded component liquid flow rate leaving Stage j = θ j y i,j ( V j + W j ). Correspondingly, the occluded component liquid flow rate entering Stage j = θ j -1 y i,j -1 ( V j -1 + W j -1 ). Chemical Reaction: Let: M j = molar liquid volume holdup on Stage j M = number of independent chemical reactions ν i,m = stoichiometric coefficient of component i in chemical reaction m r k,m,j = chemical reaction rate, dc k,m,j /dt , of the m th chemical reaction for the reference reactant component, k , on Stage j Then, the formation or disappearance of component i by chemical reaction on Stage j is: M r j i m k m j m M ν , , , = ° 1 The component material balance equations, (10-1), now become: L x V y F z L U x V W y V W y V W y L U x L U x M r j i j j i j i C i C j i F j j i C i j i C j j i j i C j j j i j j j j i j j j j i j j j j i j j i m k m j m M j - - + + = = = = = - - - - + + + + = + + - + - + + + - + + + - + - = ° ° ° ° ° ° 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 , , , , , , , , , , , , ° ± ° ± ° ± ° ± ° ± ° ± θ θ φ φ ν The equilibrium equations, (1-2) and the summation equations (10-3) and (10-4) do not change. The energy balance is as follows, where the heat of reaction does not appear because it is assumed that enthalpies are referred to the elements:
L h V h F h V W h L U h V W h L U h j L j V j F j j j V j j j L j j j V j j j L j j j j j j j - + + - - + + + - + - + + + + + + + - + + - + + = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 θ φ θ φ ° ± ° ± ° ±° ± ° ±° ±

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Exercise 10.3 Subject: Revised set of MESH equations.
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