Exam (elaborations) TEST BANK FOR System Dynamics 3rd Edition By William J. Palm III (Solution Manual)

Exam (elaborations) TEST BANK FOR System Dynamics 3rd Edition By William J. Palm III (Solution Manual) Solutions Manual c to accompany System Dynamics, Third Edition by William J. Palm III University of Rhode Island Solutions to Problems in Chapter One c Solutions Manual Copyright 2013 The McGraw-Hill Companies. All rights reserved. No part of this manual may be displayed, reproduced, or distributed in any form or by any means without the written permission of the publisher or used beyond the limited distribution to teachers or educators permitted by McGraw-Hill for their individual course preparation. Any other reproduction or translation of this work is unlawful. 1.1 W = mg = 3(32.2) = 96.6 lb. 1.2 m = W/g = 100/9.81 = 10.19 kg. W = 100(0.2248) = 22.48 lb. m = 10.19(0.06852) = 0.698 slug. 1.3 d = (50 + 5/12)(0.3048) = 15.37 m. 1.4 d = 3(100)(0.3048) = 91.44 m 1.5 d = 100(3.281) = 328.1 ft 1.6 d = 50(3600)/5280 = 34.0909 mph 1.7 v = 100(0.6214) = 62.14 mph 1.8 n = 1/[60(1.341× 10−3)] = 12.43, or approximately 12 bulbs. 1.9 5(70 − 32)/9 = 21.1 C 1.10 9(30)/5+ 32 = 86 F 1.11 ! = 3000(2)/60 = 314.16 rad/sec. Period P = 2/! = 60/3000 = 1/50 sec. 1.12 ! = 5 rad/sec. Period P = 2/! = 2/5 = 1.257 sec. Frequency f = 1/P = 5/2 = 0.796 Hz. 1.13 Speed = 40(5280)/3600 = 58.6667 ft/sec. Frequency = 58.6667/30 = 1.9556 times per second. 1.14 x = 0.005 sin 6t, x˙ = 0.005(6) cos 6t = 0.03 cos 6t. Velocity amplitude is 0.03 m/s. ¨x = −6(0.03) sin 6t = −0.18 sin 6t. Acceleration amplitude is 0.18 m/s2. Displacement, velocity and acceleration all have the same frequency. 1.15 Physical considerations require the model to pass through the origin, so we seek a model of the form f = kx. A plot of the data shows that a good line drawn by eye is given by f = 0.2x. So we estimate k to be 0.2 lb/in. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.16 The script file is x = [0:0.01:1]; subplot(2,2,1) plot(x,sin(x),x,x),xlabel(0x (radians)0),ylabel(0x and sin(x)0),... gtext(0x0),gtext(0sin(x)0) subplot(2,2,2) plot(x,sin(x)-x),xlabel(0x (radians)0),ylabel(0Error: sin(x) - x0) subplot(2,2,3) plot(x,100*(sin(x)-x)./sin(x)),xlabel(0x (radians)0),... ylabel(0Percent Error0),grid The plots are shown in the figure. 0 0.5 1 0 0.2 0.4 0.6 0.8 1 x (radians) x and sin(x) x sin(x) 0 0.5 1 −0.2 −0.15 −0.1 −0.05 0 x (radians) Error: sin(x) − x 0 0.5 1 −20 −15 −10 −5 0 x (radians) Percent Error Figure : for Problem 1.16. From the third plot we can see that the approximation sin x  x is accurate to within 5% if |x|  0.5 radians. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.17 For  near /4, f()  sin  4 + cos  4  −  4  For  near 3/4, f()  sin 3 4 + cos 3 4  − 3 4  1.18 For  near /3, f()  cos  3 − sin  3  −  3  For  near 2/3, f()  cos 2 3 − sin 2 3  − 2 3  1.19 For h near 25, f(h)  p25 + 1 2p25 (h − 25) = 5 + 1 10 (h − 25) 1.20 For r near 5, f(r)  52 + 2(5)(r − 5) = 25 + 10(r − 5) For r near 10, f(r)  102 + 2(10)(r − 10) = 100 + 20(r − 10) 1.21 For h near 16, f(h)  p16 + 1 2p16 (h − 16) = 4 + 1 8 (h − 16) f(h)  0 if h −16. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.22 Construct a straight line the passes through the two endpoints at p = 0 and p = 900. At p = 0, f(0) = 0. At p = 900, f(900) = 0.002p900 = 0.06. This straight line is f(p) = 0.06 900 p = 1 15, 000 p 1.23 (a) The data is described approximately by the linear function y = 54x − 1360. The precise values given by the least squares method (Appendix C) are y = 53.5x− 1354.5. (b) Only the loglog plot of the data gives something close to a straight line, so the data is best described by a power function y = bxm where the approximate values are m = −0.98 and b = 3600. The precise values given by the least squares method (Appendix C) are y = 3582.1x−0.9764. (c) Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx where the approximate values are m = −0.007 and b = 2.1 × 105. The precise values given by the least squares method (Appendix C) are y = 2.0622× 105(10)−0.0067x. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.24 With this problem, it is best to scale the data by letting x = year − 2005, to avoid raising large numbers like 2005 to a power. Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx. The approximate values are m = 0.035 and b = 9.98. Set y = 20 to determine how long it will take for the population to increase from 10 to 20 million. This gives 20 = 9.98(10)0.03x. Solve it for x: x = (log(20) − log(9.98))/0.035. The answer is 8.63 years, which corresponds to 8.63 years after 2005. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.25 (a) If C(t)/C(0) = 0.5 when t = 500 years, then 0.5 = e−5500b, which gives b = −ln(0.5)/5500 = 1.2603 × 10−4. (b) Solve for t to obtain t = −ln[C(t)/C(0)]/b using C(t)/C(0) = 0.9 and b = 1.2603× 10−4. The answer is 836 years. Thus the organism died 836 years ago. (c) Using b = 1.1(1.2603 × 10−4) in t = −ln(0.9)/b gives 760 years. Using b = 0.9(1.2603× 10−4) in t = −ln(0.9)/b gives 928 years. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.26 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx where y is the temperature in degrees C and x is the time in seconds. The approximate values are m = −3.67 and b = 356. The alternate exponential form is y = be(m ln 10)x = 356e−8.451x. The time constant is 1/8.451 = 0.1183 s. The precise values given by the least squares method (Appendix C) are y = 356.0199(10)−3.6709x. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.27 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx where y is the bearing life thousands of hours and x is the temperature in degrees F. The approximate values are m = −0.007 and b = 142. The bearing life at 150  F is estimated to be y = 142(10)−0.007(150) = 12.66, or 12,600 hours. The alternate exponential form is y = be(m ln 10)x = 142e−0.0161x. The time constant is 1/0.0161 = 62.1 or 6.21× 104 hr. The precise values given by the least squares method (Appendix C) are y = 141.8603(10)−0.0070x. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.28 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx where y is the voltage and x is the time in seconds. The first data point does not lie close to the straight line on the semilog plot, but a measurement error of ±1 volt would account for the discrepancy. The approximate values are m = −0.43 and b = 96. The alternate exponential form is y = be(m ln 10)x = 96e−0.99x. The time constant is 1/0.99 = 1.01 s. The precise values given by the least squares method (Appendix C) are y = 95.8063(10)−0.4333x. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.29 A semilog plot generated by the following script file shows that the exponential function T − 70 = bemt fits the data well. t = [0:300:3000]; temp = [207,182,167,155,143,135,128,123,118,114,109]; DT = temp-70; semilogy(t,DT,t,DT,’o’) Fitting a line by eye gives the approximate values m = −4 × 10−4 and b = 125. The corresponding function is T(t) = 70 + 125e−4×10−4t. The precise values given by the least squares method (Appendix C) are m = −4.0317× 10−4 and b = 125.1276. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.30 Plots of the data on a log-log plot and rectilinear scales both give something close to a straight line, so we try both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. The power function fitted by eye in terms of the height h is approximately f = 4h0.9. Note that the exponent is not close to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye is approximately f = 3.2h. Using the least squares method (Appendix C) givesmore precise results: f = 4.1595h0.8745 and f = 3.2028h. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 1.31 Plots of the data on a log-log plot and rectilinear scales both give something close to a straight line, so we try both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The variable x is the height and the variable y is the flow rate. The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. The power function fitted by eye in terms of the height h is approximately f = 4h0.9. Note that the exponent is not close to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye is approximately f = 3.7h. Using the least squares method (Appendix C) givesmore precise results: f = 4.1796h0.9381 and f = 3.6735h. c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. Solutions Manual c to accompany System Dynamics, Third Edition by William J. Palm III University of Rhode Island Solutions to Problems in Chapter Two c Solutions Manual Copyright 2014 The McGraw-Hill Companies. All rights reserved. No part of this manual may be displayed, reproduced, or distributed in any form or by any means without the written permission of the publisher or used beyond the limited distribution to teachers or educators permitted by McGraw-Hill for their individual course preparation. Any other reproduction or translation of this work is unlawful. 2.1 a) Nonlinear because of the y¨y term. b) Nonlinear because of the sin y term. c) Nonlinear because of the p y term. d) Variable coefficient, but Linear. e) Nonlinear because of the sin y term. f) Variable coefficient, but linear. c 2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 2.2 a) 4 Z x 2 dx = 3 Z t 0 t dt x(t) = 2 + 3 8t2 b) 5 Z x 3 dx = 2 Z t 0 e−4t dt x(t) = 3.1 − 0.1e−4t c) Let v = x˙ . 3 Z v 7 dv = 5 Z t 0 t dt v(t) = dx dt = 7 + 5 6t2 Z x 2 dx = Z t 0  7 + 5 6t2  dt x(t) = 2 + 7t + 5 18t3 d) Let v = x˙ . 4 Z v 2 dv = 7 Z t 0 e−2t dt v(t) = 23 8 − 7 8e−2t Z x 4 dx = Z t 0  23 8 − 7 8e−2t  dt x(t) = 57 16 + 23 8 t + 7 16e−2t e) x˙ = C1, but x¨(0) = 5, so C1 = 5. x = 5t + C2, but x(0) = 2, so C2 = 2. Thus x = 5t + 2. c 2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful. 2.3 a) Z x 3 dx 25 − 5x2 = Z t 0 dt = t Z x 3 dx 25 − 5x2 = p 5 25 " arctanh p 5x 5 ! − arctanh 3 p 5 5 !# = t Let C = arctanh 3 p 5 5 ! Solve for x to obtain x = p 5 tanh(5 p 5t + C) b) Z x 10 dx 36 + 4x2 = Z t 0 dt = t 1 12 tan−1 x 3  x 10 = t x(t) = 3 tan(12t + C) C = tan−1 10 3 c) Z x 4 x dx 5x + 25 = Z t 0 dt x 5 − ln(x + 5)  x 4 = x 5 − ln(x + 5) − 4 5 + ln 9 = t x − 5 ln(x + 5) = 5t + 4 − 5 ln 9 So a closed form solution does not exist. (continued on the next page) c 2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher’s consent is unlawful.