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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 (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 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. (103) and (104), 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. (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 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 = θj1 yi,j1(Vj1 + Wj1
).
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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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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