A chemical constituent flows between three reactors a

PROBLEMS 283 inflow concentration to reactor 2 is halved, what is the concentration of reactor 3? 11.16 As described in Examples 8.2 and 11.2, use the matrix inverse to answer the following: a Determine the change in position of the first jumper, if the mass of the third jumper is increased to 100 kg. b What force must be applied to the third jumper so that the final position of the third jumper is 140 m? 11.17 Determine the matrix inverse for the electric circuit formulated in Sec. 8.3. Use the inverse to determine the new FIGURE P11.18 Q ⫽ 50 m 3 hr 50 m 3 hr 50 m 3 hr 90 m 3 hr Load ⫽ 5000 mghr Room 4 Room 3 Room 2 Room 1 Q ⫽ 50 m 3 hr c ⫽ 40 mgm 3 Q ⫽ 150 m 3 hr c ⫽ 1 mgm 3 Q ⫽ 50 m 3 hr current between nodes 2 and 5 i 52 , if a voltage of 200 V is applied at node 6 and the voltage at node 1 is halved.

11.18 a

Using the same approach as described in Sec. 11.3, develop steady-state mass balances for the room configura- tion depicted in Fig. P11.18. b Determine the matrix inverse and use it to calculate the resulting concentrations in the rooms. c Use the matrix inverse to determine how much the room 4 load must be reduced to maintain a concentration of 20 mgm 3 in room 2. Iterative Methods 284 CHAPTER OBJECTIVES The primary objective of this chapter is to acquaint you with iterative methods for solving simultaneous equations. Specific objectives and topics covered are • Understanding the difference between the Gauss-Seidel and Jacobi methods. • Knowing how to assess diagonal dominance and knowing what it means. • Recognizing how relaxation can be used to improve the convergence of iterative methods. • Understanding how to solve systems of nonlinear equations with successive substitution and Newton-Raphson. I terative or approximate methods provide an alternative to the elimination methods described to this point. Such approaches are similar to the techniques we developed to obtain the roots of a single equation in Chaps. 5 and 6. Those approaches consisted of guessing a value and then using a systematic method to obtain a refined estimate of the root. Because the present part of the book deals with a similar problem—obtaining the val- ues that simultaneously satisfy a set of equations—we might suspect that such approximate methods could be useful in this context. In this chapter, we will present approaches for solving both linear and nonlinear simultaneous equations.

12.1 LINEAR SYSTEMS: GAUSS-SEIDEL

The Gauss-Seidel method is the most commonly used iterative method for solving linear algebraic equations. Assume that we are given a set of n equations: [ A]{x} = {b} Suppose that for conciseness we limit ourselves to a 3 × 3 set of equations. If the diagonal elements are all nonzero, the first equation can be solved for x 1 , the second for x 2 , and the third for x 3 to yield x j 1 = b 1 − a 12 x j − 1 2 − a 13 x j − 1 3 a 11 12.1a x j 2 = b 2 − a 21 x j 1 − a 23 x j − 1 3 a 22 12.1b x j 3 = b 3 − a 31 x j 1 − a 32 x j 2 a 33 12.1c where j and j − 1 are the present and previous iterations. To start the solution process, initial guesses must be made for the x’s. A simple ap- proach is to assume that they are all zero. These zeros can be substituted into Eq. 12.1a, which can be used to calculate a new value for x 1 = b 1 a 11 . Then we substitute this new value of x 1 along with the previous guess of zero for x 3 into Eq. 12.1b to compute a new value for x 2 . The process is repeated for Eq. 12.1c to calculate a new estimate for x 3 . Then we return to the first equation and repeat the entire procedure until our solution converges closely enough to the true values. Convergence can be checked using the criterion that for all i, ε a,i = x j i − x j − 1 i x j i × 100 ≤ ε s 12.2 EXAMPLE 12.1 Gauss-Seidel Method Problem Statement. Use the Gauss-Seidel method to obtain the solution for 3x 1 − 0.1x 2 − 0.2x 3 = 7.85 0.1x 1 + 7x 2 − 0.3x 3 = − 19.3 0.3x 1 − 0.2x 2 + 10x 3 = 71.4 Note that the solution is x 1 = 3, x 2 = − 2.5, and x 3 = 7. Solution. First, solve each of the equations for its unknown on the diagonal: x 1 = 7.85 + 0.1x 2 + 0.2x 3 3 E12.1.1 x 2 = − 19.3 − 0.1x 1 + 0.3x 3 7 E12.1.2 x 3 = 71.4 − 0.3x 1 + 0.2x 2 10 E12.1.3 By assuming that x 2 and x 3 are zero, Eq. E12.1.1 can be used to compute x 1 = 7.85 + 0.10 + 0.20 3 = 2.616667 12.1 LINEAR SYSTEMS: GAUSS-SEIDEL 285