Unformatted text preview: B2 − A2 p1 −350 − 50( −125) q2 = D2 − A2 q1
−50 − 50(0)
.
=
= 01739
B2 − A2 p1 −350 − 50( −1.25) q1 = D1
0
=
=0
B1 −160 p3 = 0
q3 = .
D3 − A3q2
0 − 150(01739)
=
= 01917
.
B3 − A3 p2 −230 − 150( −0.626) Using Eqs. (1017) and (1018):
x3 = q3 = 0.1917
x2 = q2  p2x3 = 0.1739  (0.626)(0.1917) = 0.2939
x1 = q1  p1x2 = 0  (1.25)(0.2939) = 0.3674 x1 0 180 • x2 = −50 150  230 Applying the algorithm Eqs. (1013) to (1015): p1 = 0 x3 0 Exercise 10.5
Subject: Application of tridiagonal matrix algorithm.
Given: Thomas algorithm equations (1013) to (1018) and a 3x3 matrix equation below. Find: Solution to the matrix equation.
Analysis: The matrix equation is:
−6 B1 C1 0 0 0 x1 D1 3 A2 B2 C2 0 0 x2 D2 0 A3 B3 C3 0 • x3 = D3 = 0 1.5  7.5 0 0 A4 B4 C4 x4 D4 0 0 4.5  7.5 0 0 0 A5 B5 x5 D5 0 0 0 3  4.5 0 0 0 x1 0 3 0 0 x2 0 3 0 4.5 Applying the algorithm Eqs. (1013) to (1015): p1 = C1
3
=
= −0.5
B1 −6 p2 = C2
3
=
= −1.0
B2 − A2 p1 −4.5 − 3( −0.5) p3 = C3
3
=
= −0.5
B3 − A3 p2 −7.5 − 15( −1.0)
. q3 = 100 − 15(0)
D3 − A3q2
.
=
= −16.67
B3 − A3 p2 −7.5 − 15( −10)
. p4 = 3
C4
=
= −0.57143
B4 − A4 p3 −7.5 − 4.5( −0.5) q4 = D4 − A4 q3 0 − 4.5( −16.67)
=
= −14.28571
B4 − A4 p3 −7.5 − 4.5( −0.5) q5 = D5 − A5q4
0 − 4.5( −14.28571)
=
= −33.333
B5 − A5 p4 −4.5 − 4.5( −0.57143) q1 = D1
0
=
=0
B1 −6
q2 = 0 Using Eqs. (1017) and (1018):
x5 = q5 = 33.333
x4 = q4  p4x5 = 14.28571  (0.57143)(33.333) = 33.333
x3 = q3  p3x4 = 16.6667  (0.5)(33.333) = 33.333
x2 = q2  p2x3 = 0.0  (1.0)(33.333) = 33.333
x1 = q1 – p1x2 = 0.0  (0.5)(33.333) = 16.667 • x3 = 100 3 x4 0  4.5 x5 0 Exercise 10.6
Subject: Solution of tridiagonal matrix equation for liquidphase mole fractions to avoid
subtraction of nearly equal quantities as claimed by Wang and Henke.
Given: Equations (101) to (1018)
Prove: Claim of Wang and Henke.
Analysis: First investigate the values of pj. From Eq. (1014), a subtraction occurs in the
denominator:
Cj
pj =
B j − A j p j −1
Starting at the top of the column, p1 = C1/B1, therefore, the denominator of Eq. (1014) becomes:
C
B2 − A2 p1 = B2 − A2 1
Clearing the fraction, we obtain B1 B2 − A2 C1
B1
Examine this difference by substituting for each of the four quantities. From Eq. (108),
A2 = V2 + F1 − W1 − U1 − V1 However, by material balance around Stage 1 in Fig. 10.3,
V2 + F1 − W1 − U 1 − V1 = L1
Therefore, A2 = L1
From Eq. (109),
B1 = −[V2 + F1 − W1 − U 1 − V1 + U 1 + V1 + W1 Ki ,1 ]
However, by material balance around Stage 1 in Fig. 10.3,
V2 + F1 − W1 − U 1 − V1 = L1 Therefore, B1 = − V1 + W1 Ki ,1 + L1 + U 1
Similarly, B2 = − V2 + W2 Ki ,2 + L2 + U 2
Thus, it is clear that B1 B2 = a summation of all positive quantities, that is no subtractions.
From Eq. (1010), C1 = V2 Ki ,2 and
A2 C1 = L1V2 Ki ,2
When B1B2 is expanded, it contains a term L1V2Ki,2 .
Thus, this potential...
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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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