Solution Manual For System Dynamics, 3rd Edition

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Solutions Manualto accompanySystem Dynamics, Third EditionbyWilliam J. Palm IIIUniversity of Rhode IslandSolutions to Problems in Chapter One

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1.1W=mg= 3(32.2) = 96.6 lb.1.2m=W/g= 100/9.81 = 10.19 kg.W= 100(0.2248) = 22.48 lb.m= 10.19(0.06852) =0.698 slug.1.3d= (50 + 5/12)(0.3048) = 15.37 m.1.4d= 3(100)(0.3048) = 91.44 m1.5d= 100(3.281) = 328.1 ft1.6d= 50(3600)/5280 = 34.0909 mph1.7v= 100(0.6214) = 62.14 mph1.8n= 1/[60(1.341×10−3)] = 12.43, or approximately 12 bulbs.1.95(70−32)/9 = 21.1◦C1.109(30)/5 + 32 = 86◦F1.11ω= 3000(2π)/60 = 314.16 rad/sec. PeriodP= 2π/ω= 60/3000 = 1/50 sec.1.12ω= 5 rad/sec. PeriodP= 2π/ω= 2π/5 = 1.257 sec. Frequencyf= 1/P= 5/2π=0.796 Hz.1.13Speed = 40(5280)/3600 = 58.6667 ft/sec.Frequency = 58.6667/30 = 1.9556 timesper second.1.14x= 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.1.15Physical considerations require the model to pass through the origin, so we seek amodel of the formf=kx. A plot of the data shows that a good line drawn by eye is givenbyf= 0.2x. So we estimatekto be 0.2 lb/.

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1.16The script file isx = [0:0.01:1];subplot(2,2,1)plot(x,sin(x),x,x),xlabel(′x (radians)′),ylabel(′x and sin(x)′),...gtext(′x′),gtext(′sin(x)′)subplot(2,2,2)plot(x,sin(x)-x),xlabel(′x (radians)′),ylabel(′Error:sin(x) - x′)subplot(2,2,3)plot(x,100*(sin(x)-x)./sin(x)),xlabel(′x (radians)′),...ylabel(′Percent Error′),gridThe plots are shown in the figure.00.5100.20.40.60.81x (radians)x and sin(x)xsin(x)00.51−0.2−0.15−0.1−0.050x (radians)Error: sin(x) − x00.51−20−15−10−50x (radians)Percent ErrorFigure : for Problem 1.16.From the third plot we can see that the approximation sinx≈xis accurate to within5% if|x| ≤0.5 radians.

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1.17Forθnearπ/4,f(θ)≈sinπ4 +(cosπ4) (θ−π4)Forθnear 3π/4,f(θ)≈sin 3π4+(cos 3π4) (θ−3π4)1.18Forθnearπ/3,f(θ)≈cosπ3−(sinπ3) (θ−π3)Forθnear 2π/3,f(θ)≈cos 2π3−(sin 2π3) (θ−2π3)1.19Forhnear 25,f(h)≈ √25 +12√25(h−25) = 5 + 110(h−25)1.20Forrnear 5,f(r)≈52+ 2(5)(r−5) = 25 + 10(r−5)Forrnear 10,f(r)≈102+ 2(10)(r−10) = 100 + 20(r−10)1.21Forhnear 16,f(h)≈ √16 +12√16 (h−16) = 4 + 18 (h−16)f(h)≥0 ifh >−16.

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1.22Construct a straight line the passes through the two endpoints atp= 0 andp= 900.Atp= 0,f(0) = 0. Atp= 900,f(900) = 0.002√900 = 0.06. This straight line isf(p) = 0.06900p=115,000p1.23(a) The data is described approximately by the linear functiony= 54x−1360. Theprecise values given by the least squares method (Appendix C) arey= 53.5x−1354.5.(b) Only the loglog plot of the data gives something close to a straight line, so the data isbest described by a power functiony=bxmwhere the approximate values arem=−0.98andb= 3600.The precise values given by the least squares method (Appendix C) arey= 3582.1x−0.9764.(c) Both the loglog and semilog plot (with theyaxis logarithmic) give something closeto a straight line, but the semilog plot gives the straightest line, so the data is best describedby a exponential functiony=b(10)mxwhere the approximate values arem=−0.007 andb= 2.1×105.The precise values given by the least squares method (Appendix C) arey= 2.0622×105(10)−0.0067x.

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1.24With this problem, it is best to scale the data by lettingx=year−2005, to avoidraising large numbers like 2005 to a power.Both the loglog and semilog plot (with theyaxis logarithmic) give something close to a straight line, but the semilog plot gives thestraightest line, so the data is best described by a exponential functiony=b(10)mx. Theapproximate values arem= 0.035 andb= 9.98.Sety= 20 to determine how long it will take for the population to increase from 10 to20 million. This gives 20 = 9.98(10)0.03x. Solve it forx:x= (log(20)−log(9.98))/0.035.The answer is 8.63 years, which corresponds to 8.63 years after 2005.

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1.25(a) IfC(t)/C(0) = 0.5 whent= 500 years, then 0.5 =e−5500b, which givesb=−ln(0.5)/5500 = 1.2603×10−4.(b) Solve fortto obtaint=−ln[C(t)/C(0)]/busingC(t)/C(0) = 0.9 andb= 1.2603×10−4. The answer is 836 years. Thus the organism died 836 years ago.(c) Usingb=1.1(1.2603×10−4) int=−ln(0.9)/bgives 760 years.Usingb=0.9(1.2603×10−4) int=−ln(0.9)/bgives 928 years.

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1.26Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential functiony=b(10)mxwhereyis the temperaturein degrees C andxis the time in seconds.The approximate values arem=−3.67 andb= 356.The alternate exponential form isy=be(mln 10)x= 356e−8.451x.The timeconstant is 1/8.451 = 0.1183 s.The precise values given by the least squares method (Appendix C) arey= 356.0199(10)−3.6709x.

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1.27Only the semilog plot of the data gives something close to a straight line, so the data isbest described by an exponential functiony=b(10)mxwhereyis the bearing life thousandsof hours andxis the temperature in degrees F. The approximate values arem=−0.007andb= 142. The bearing life at 150◦F is estimated to bey= 142(10)−0.007(150)= 12.66,or 12,600 hours.The alternate exponential form isy=be(mln 10)x= 142e−0.0161x.Thetime constant is 1/0.0161 = 62.1 or 6.21×104hr.The precise values given by the least squares method (Appendix C) arey= 141.8603(10)−0.0070x.

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1.28Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential functiony=b(10)mxwhereyis the voltage andxis the time in seconds.The first data point does not lie close to the straight line onthe semilog plot, but a measurement error of±1 volt would account for the discrepancy.The approximate values arem=−0.43 andb= 96.The alternate exponential form isy=be(mln 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) arey= 95.8063(10)−0.4333x.

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1.29A semilog plot generated by the following script file shows that the exponential functionT−70 =bemtfits 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 valuesm=−4×10−4andb= 125.Thecorresponding function isT(t) = 70 + 125e−4×10−4t.The precise values given by the least squares method (Appendix C) arem=−4.0317×10−4andb= 125.1276.

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1.30Plots of the data on a log-log plot and rectilinear scales both give something close toa straight line, so we try both functions. (Note that the flow should be 0 when the heightis 0, so we do not consider the exponential function and we must force the linear functionto pass through the origin by settingb= 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 heighthis approximatelyf= 4h0.9.Note that the exponent is not close to 0.5, as it is for orifice flow.This is because theflow through the outlet is pipe flow. For the linear functionf=mh, the best fit by eye isapproximatelyf= 3.2h.Using the least squares method (Appendix C) gives more precise results:f= 4.1595h0.8745andf= 3.2028h.

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1.31Plots of the data on a log-log plot and rectilinear scales both give something close toa straight line, so we try both functions. (Note that the flow should be 0 when the heightis 0, so we do not consider the exponential function and we must force the linear functionto pass through the origin by settingb= 0.) The variablexis the height and the variableyis the flow rate. The three lowest heights give the same time, so we discard the heightsof 1 and 2 cm.The power function fitted by eye in terms of the heighthis approximatelyf= 4h0.9.Note that the exponent is not close to 0.5, as it is for orifice flow.This is because theflow through the outlet is pipe flow. For the linear functionf=mh, the best fit by eye isapproximatelyf= 3.7h.Using the least squares method (Appendix C) gives more precise results:f= 4.1796h0.9381andf= 3.6735h.

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Solutions Manual c©to accompanySystem Dynamics, Third EditionbyWilliam J. Palm IIIUniversity of Rhode IslandSolutions to Problems in Chapter Tenc©Solutions Manual Copyright 2013 The McGraw-Hill Companies.ibutedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation.Any other reproductionor translation of this work is unlawful.

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10.1a) A traffic light may be either closed or open loop, depending on whether or not ituses a sensor to detect the presence of vehicles in the roadway.b) Most washing machines are open loop, but some now have sensors that detect theamount of dirt in the wash water and adjust their cycle accordingly.These would beclosed-loop devices.c) If the toaster uses a timer, it is open loop. If it uses a sensor to tell when the desiredtemperature has been reached, it is closed loop.d) Cruise control is closed loop because it uses a measurement of the vehicle speed tocontrol the engine.e) A aircraft autopilot is closed loop because it uses measurements of a variety of vari-ables (speed, altitude, angle, etc.) to adjust the control surfaces (ailerons, rudder, elevators,etc.).f) Closed loop because it reacts to changes in the environment to keep the temperaturenear 98.6◦.

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