Ch 10

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 +1C
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 tothe 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 rateleaving 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 chemicalreaction for the reference
reactant component, k, on Stage j
Then, the formation or disappearance of component i by chemical reaction on Stage j is:
Mj

M
m =1

νi ,mrk ,m, j

The component material balance equations, (10-1), now become:

L j −1

C
i =1

xi , j −1 + V j +1

C
i =1

yi , j +1 + Fj

C
i =1

zi , F j − L j + U j

C
i =1

xi , j − V j + Wj

C
i =1

yi, j +

θ j −1 V j −1 + Wj −1 yi , j −1 − θ j V j + Wj yi , j + φ j +1 L j +1 + U j +1 xi , j +1 − φ j L j + U j xi , j − M j

M
m =1

νi ,mrk ,m, j = 0

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 theelements:

L j −1hL j −1 + V j +1hV j +1 + Fj hFj + θ j +1 V j −1 + Wj −1 hV j −1 + φ j +1 L j +1 + U j +1 hL j +1
− 1 + θ j V j + Wj hV j − 1 + φ j L j + U j hL j = 0

Exercise 10.3
Subject: Revised set of MESH equations.
Given: MESH Eqs. (10-1) to (10-5).
Find: Revised set of MESH equations to account for liquid pumparounds as shown in Fig. 10.2.
Revised M equations like Eq. (10-7).Possible partitioning to tridiagonal sets.
Analysis: As an example, consider the section of equilibrium stages below, which includes a
bypass pumparound and a recycle pumparound. Let Pu,v designate a total molar liquid
pumparound flow rate from Stage u to Stage v.

Exercise 10.3 (continued)
Analysis: (continued)
The revised MESH equations for Stage j are:
M i , j = L j −1 xi , j −1 + V j +1 yi...