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Unformatted text preview: Exercise 10.1 Subject: Independency of MESH equations Given: Equations (101), (103), (104), and (106). Prove: Equation (106) is not independent of the other 3 equations Analysis: Eq. (106) can be derived from the other 3 equations as follows, as outlined in the text between Eqs. (105) and (106).. Summing Eq. (101) 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 , , , , , & ¡ & ¡ (1) From Eqs. (103) and (104), 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 & ¡ & ¡ (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 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 + + = + + = + + = + + = + + = + ....... (3) Summing Eqs. (2), noting that L = 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 ¢ £ ¢ £ But this is Eq. (106). Therefore, it is not independent of Eqs. (101), (103), and (104). Exercise 10.2 Subject: Revision of MESH equations to account for entrainment, occlusion, and chemical reaction (in the liquid phase). Given: MESH Eqs. (101) to (105). 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 j1 y i,j1 ( V j1 + W j1 ). 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, (101), 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 , , , , , , , , , , , , & ¡ & ¡ & ¡ & ¡ & ¡ & ¡ θ θ φ φ ν The equilibrium equations, (12) and the summation equations (103) and (104) do not change. The equilibrium equations, (12) and the summation equations (103) and (104) do not change....
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This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .
 Spring '06
 selebi
 Mass Transfer

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