Solution Manual For System Dynamics, 3rd Edition

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Solutions Manual
to accompany
System Dynamics, Third Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter One

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 ω = 3000(2π)/60 = 314.16 rad/sec. Period P = 2π/ω = 60/3000 = 1/50 sec.
1.12 ω = 5 rad/sec. Period P = 2π/ω = 2π/5 = 1.257 sec. Frequency f = 1/P = 5/2π =
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/ .

1.16 The script file is
x = [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′),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.

1.17 For θ near π/4,
f (θ) ≈ sin π
4 +
(
cos π
4
) (
θ − π
4
)
For θ near 3π/4,
f (θ) ≈ sin 3π
4 +
(
cos 3π
4
) (
θ − 3π
4
)
1.18 For θ near π/3,
f (θ) ≈ cos π
3 −
(
sin π
3
) (
θ − π
3
)
For θ near 2π/3,
f (θ) ≈ cos 2π
3 −
(
sin 2π
3
) (
θ − 2π
3
)
1.19 For h near 25,
f (h) ≈ √25 + 1
2√25(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) ≈ √16 + 1
2√16 (h − 16) = 4 + 1
8 (h − 16)
f (h) ≥ 0 if h > −16.

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) = 0.002√900 = 0.06. This straight line is
f (p) = 0.06
900 p = 1
15, 000p
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.

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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.

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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.

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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.

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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.

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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.

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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.

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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) gives more precise results: f = 4.1595h0.8745
and f = 3.2028h.

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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) gives more precise results: f = 4.1796h0.9381
and f = 3.6735h.

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Solutions Manual c©
to accompany
System Dynamics, Third Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Ten
c©Solutions Manual Copyright 2013 The McGraw-Hill Companies.
ibuted
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.

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10.1 a) A traffic light may be either closed or open loop, depending on whether or not it
uses 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 the
amount of dirt in the wash water and adjust their cycle accordingly. These would be
closed-loop devices.
c) If the toaster uses a timer, it is open loop. If it uses a sensor to tell when the desired
temperature has been reached, it is closed loop.
d) Cruise control is closed loop because it uses a measurement of the vehicle speed to
control 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 temperature
near 98.6◦.

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