1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 8 8.1 Ideal gas law: 0.082054(375)15.385132RTvp van der Waals equation: RTbvvapvf )()(2 212.02( ) 2 ( 0.08407) 0.082054(375)fv vv Any of the techniques in Chaps 5 or 6 can be used to determine the root as v = 15.0727 L/mol. The Newton-Raphson method would be a good choice because (a) the equation is relatively simple to differentiate and (b) the ideal gas law provides a good initial guess. The Newton-Raphson method can be formulated as 32212)()(iiiiiiivabvvapRTbvvapvv Using the ideal gas law for the initial guess results in an accurate root determination in a few iterations: i xi f(xi) f'(xi) a 0 15.38513 0.608865 1.949774 1 15.07285 0.000323 1.947683 2.0718% 2 15.07268 9.51011 1.947682 0.0011% 8.2 The function to be solved is 0)]1(1[1)1()1(1ln)( AfAfAfXRRRXRXRRf or substituting XAf = 0.9, 10.1 1() ln 0(0.1) (1 0.1 )RRfRRRR A plot of the function indicates a root at about R = 0.43.2 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. -6-4-2020 0.2 0.4 0.6 0.8 1 Bisection with initial guesses of 0.01 and 1 can be used to determine a root of 0.4299373 after 16 iterations with a = 0.004%. 8.3 The function to be solved is 6( ) 0.05 012xfxxx A plot of the function indicates a root at about x = 0.028. -0.100.10 0.02 0.04 0.06 Because the function is so linear, false position is a good choice. Using initial guesses of 0.02 and 0.04, the first iteration is 0.021458(0.02 0.04)0.04 0.0281730.01483 0.021458rx After 3 iterations, the result is 0.028249 with a = 0.003%. 8.4 The function to be solved is 0.04 0.04( ) 10 1 4 9.3 0ttft e e A plot of the function indicates a root at about t = 55. -6-4-2020 20406080100 Bisection with initial guesses of 0 and 60 can be used to determine a root of 53.711 after 16 iterations with a = 0.002%. 8.5 The function to be solved is 2(4 )( ) 0.015 0(42 2 ) (28 )xfxxx3 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. (a) A plot of the function indicates a root at about x = 16. -0.200.20.40.60.80 5 10 15 20 (b) The shape of the function indicates that false position would be a poor choice (recall Fig. 5.14). Bisection with initial guesses of 0 and 20 can be used to determine a root of 16.01563 after 8 iterations with a = 0.488%. Note that false position would have required many more iterations to attain comparable accuracy. i xl xu xr f(xl) f(xr)f(xl)f(xr)a 1 0 20 10 -0.01492 -0.01355 0.000202 100.000%2 10 20 15 -0.01355 -0.00620 8.41E-05 33.333%3 15 20 17.5 -0.0062 0.02010 -0.00012 14.286%4 15 17.5 16.25 -0.0062 0.001318 -8.2E-06 7.692%5 15 16.25 15.625 -0.0062 -0.00319 1.98E-05 4.000%6 15.625 16.25 15.9375 -0.00319 -0.00117 3.73E-06 1.961%7 15.9375 16.25 16.09375 -0.00117 6.92E-06 -8.1E-09 0.971%8 15.9375 16.09375 16.01563 -0.00117 -0.0006 6.99E-07 0.488% 8.6 The functions to be solved are )()2()(10,2210,210,1xcxxcxxcKbac ))(2()(20,210,210,2xcxxcxxcKdac or 4122121211104)20()250(5),(xxxxxxxf 222121212107.3)10)(250()5(),(xxxxxxxf Graphs can be generated by specifying values of x1 and solving for x2 using a numerical method like bisection. first equation second equationx1 x2 x1x20 8.6672 0 4.4167 1 6.8618 1 3.9187 2 5.0649 2 3.4010 3 3.2769 3 2.8630 4 1.4984 4 2.3038 5 -0.2700 5 1.7227 These values can then be plotted to yield4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. -4048120123451st eq2nd eq Therefore, the root seems to be at about x1 = 3.3 and x2 = 2.7. Employing these values as the initial guesses for the two-variable Newton-Raphson method gives f1(3.3, 2.7) = –2.3610–6 f2(3.3, 2.7) = 2.3310–5 31251110185.5 109.9xfxf 3225211035.9 1057.5xfxf 71037.6J 3367.31037.6)1057.5(1033.2)1035.9(1036.23.3755361x 677.21037.6)10185.5)(1036.2()109.9(1033.27.2736552x The second iteration yields x1 = 3.3366 and x2 = 2.677, with a maximum approximate error of 0.003%. 8.7 Using the given values, a = 12.5578 and b = 0.0018626. Therefore, the roots problem to be solved is 000,65233)0018626.0(5578.12)0018626.0()233(518.0)( vvvvf A plot indicates a root at about 0.0028. -1200000-800000-40000004000008000000.001 0.002 0.003 0.004 Using initial guesses of 0.002 and 0.004, bisection can be employed to determine the root as 0.002807 after 12 iterations with a = 0.017%. The mass of methane contained in the tank can be computed as 3/0.002807 =
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