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

1 v j 1 wj 1 yi j 1 j v j wj yi j j 1 l j 1

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Unformatted text preview: 50), respectively: Variable φAE φSE φΑX φSX EB PX MX 0.160 0.103 0.102 0.0000 0.0000 0.0000 0.102 0.0449 0.0466 0.0001 0.0030 0.00274 OX 0.00175 0.0307 0.0003 0.0823 Now apply Eq. (3) above to obtain the component distribution: Variable f, kmol/h b/d d, kmol/h b, kmol/h EB 100 0.0001 99.99 0.01 PX 100 0.0102 98.99 1.01 MX 200 0.0091 198.20 1.80 OX 100 68.2 0.68 99.32 Comparing these Edmister method results with the FUG method, we see that the Edmister method predicts a somewhat better split of the two key components than does the FUG method for this problem. 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 j −1 C i =1 xi , j −1 + V j +1 C i =1 yi , j +1 + Fj C i =1 zi , Fj − L j + U j C i =1 xi , j − V j + Wj C i =1 yi , j = 0 (1) From Eqs. (10-3) and (10-4), all 5 sums in Eq. (1) are equal to 1. Therefore, Eq. (1) becomes: L j −1 + V j +1 + F j − L j + U j − V j + W j = 0 (2) Writing Eq. (2) for each stage from Stage 1 to Stage j: L0 + V2 + F1 − L1 − U 1 − V1 − W1 = 0 L1 + V3 + F2 − L2 − U 2 − V2 − W2 = 0 L2 + V4 + F3 − L3 − U 3 − V3 − W3 = 0 ....... L j − 2 + V j + Fj −1 − L j −1 − U j −1 − V j −1 − Wj −1 = 0 (3) L j −1 + V j +1 + F j − L j − U j − V j − W j = 0 Summing Eqs. (2), noting that L0 = 0 and that many variables cancel, we obtain: V j +1 + j m =1 o r L j = V j +1 + Fm − U m − Wm − L j − V1 = 0 j m =1 Fm − U m − Wm − V1 But this is Eq. (10-6). Therefore, it is not independent of Eqs. (10-1), (10-3), and (10-4). 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 φj = ratio of entrained liquid (in the exiting vapor) that leaves Stage j to the liquid (Lj + Uj) leaving Stage j. Then, the entrained component liquid flow rate leaving Stage j = φjxi,j (Lj + Uj). Correspondingly, the entrained component liquid flow rate entering Stage j = φj+1xi,j+1 (Lj+1+ Uj+1). Occlusion: Let θj = ratio of occluded vapor (in the exiting liquid) that leaves Stage j to the vapor (Vj + Wj) leaving Stage j. Then the occluded component liquid flow rate leaving Stage j = θj yi,j (Vj + Wj). Correspondingly, the occluded component liquid flow rate entering Stage j = θj-1 yi,j-1(Vj-1 + Wj-1 ). Chemical Reaction: Let: Mj = molar liquid volume holdup on Stage j M = number of independent chemical reactions νi,m = stoichiometric coefficient of component i in chemical reaction m rk,m,j = chemical reaction rate, dck,m,j/dt, of the mth chemical reaction for the reference reactant component, k, on Stage j Then, the forma...
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