Unformatted text preview: subtraction in B1 B2 − A2 C1 is cancelled.
There are no subtractions in p2 . The same is found for all pj .
Now investigate values of qj. From Eq. (10-15), subtractions occur in the numerator and
denominator: Exercise 10.6 (continued)
qj = D j − A j q j −1
B j − A j p j −1 The denominator is identical to that already examined in pj. So we only have to investigate the
Using Eqs. (10-9), as above, and (10-11),
q1 = − F1zi ,1
B1 − V1 + W1 Ki ,1 + L1 + U 1 which contains no subtraction problems Therefore, the numerator in Eq. (10-15) becomes:
D2 − A2 q1 = − F2 zi ,2 − L1 − F1zi ,2
− V1 + W1 Ki ,1 + L1 + U 1 = F2 zi ,2 V1 + W1 Ki ,1 + L1 + U 1 + L1 F1zi ,1
− V1 + W1 Ki ,1 + L1 + U 1 Thus, no subtractions appear in this numerator. The result is obtained for all qj.
Therefore, the claim of Wang and Henke is verified. Exercise 10.7
Subject: Derivation of stagewise component material balance equations in terms of component
vapor flow rates rather than liquid-phase mole fractions.
Given: Eq. (10-7).
Find: Component flow rates form of Eq. (10-7). Whether equations can still be partitioned
into C tridiagonal matrix equations.
Analysis: Start with Eqs. (10-1), (10-2), and (10-6) for Lj and Lj-1:
M i , j = L j −1 xi , j −1 + V j +1 yi , j +1 + Fj zi , j − L j + U j xi , j − V j + Wj yi , j = 0 (1) Ei , j = yi , j − Ki , j xi , j = 0 (2) j L j = V j +1 + Fm − U m − Wm − V1 (3) Fm − U m − Wm − V1 (4) m=1
j −1 L j −1 = V j + m=1 Substituting Eqs. (2) to (4) into (1), to eliminate variables in L and x,
M =V +
−V j −1 υ m i, j − 1
F − U − W −V
i, j + 1
j i, j
i, j − 1 i, j − 1 j υ υ i, j
F − U − W −V + U
j i, j
+ Collecting terms in the component vapor flow rates,
− V j −1 j +1 υ
i, j − 1
i, j − 1 i, j − 1 ( Fm − U m − Wm ) − V1 V
m + j ( Fm − U m − Wm ) − V1 + U j
+F z =0
i, j + 1
j i, j
j i, j
j Thus, we will have a set of equations that can be partitioned into C tridiagonal sets. Exercise 10.8
Subject: Computer storage locations for tridiagonal matrix equations
Given: Eqs. (10-7) to (10-11) Find: Minimum storage locations needed.
Analysis: For a 100-stage column, only need:
99 storage locations for pj
100 storage locations for qj
4 storage locations for current values of Aj, Bj, Cj, and Dj
This is total of 203 storage locations.
If Aj, Bj, Cj, and Dj expressions are substituted into the pj and qj equations, then only need
199 storage locations. Exercise 10.9
Subject: Solving a set of two nonlinear equations by the Newton-Raphson method.
Given: Two nonlinear equations in x1 and x2:
x12 + x2 = 17 8 x1 1/ 3 1
+ x2 / 2 = 4 Find: Solutions for four different starting guesses
(a) Starting guess is x1 = 2, x2 = 5
The Newton-Raphson method converges in 5 iterations:
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- Spring '11
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