numerical method for engineers-chapter 24
TRANSCRIPT
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1
CHAPTER 24
24.1 Simpsons rules can be implemented as
32.4120.48411.926
8
0.15024)0.140460.132(30.12486))50(100(
6
0.12486)0.11904(411454.0))150(50(
=+=
++++
++=I
The amount of heat required can be computed as
38892)32.41(1200 ==H
24.2 The first iteration involves computing 1 and 2 segment trapezoidal rules and combining
them as
32.413
33.0975)32.581875(4=
=I
and computing the approximate error as
0.5303%%10032.41
32.58187532.41=
=a
The computation can be continues as in the following tableau until a < 0.01%.
1 2 3n a 0.5303% 0.0000%1 33.09750000 32.41000000 32.41000000
2 32.58187500 32.41000000
4 32.45296875
24.3 Change of variable:
dd xxx 125252
)150(100
2
150100+=
+
=
dd dxdxdx 1252
)150(100
=
=
ddd dxxxI 125))12525(10642.)12525(101.56(0.1321
1
274 ++++=
Therefore, the transformed function is
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2
))12525(10642.)12525(101.56(0.132125)( 274 ddd xxxf ++++=
Two-point formula:
41.3217.4932114.916793
1
3
1=+=
+
= ffI
Three-point formula:
41.3210.0393414.251678.11899
)18.07082(5555556.0)16.03313(8888889.0)14.61418(5555556.0
)7745967.0(5555556.0)0(8888889.0)7745967.0(5555556.0
=++=
++=
++= fffI
Thus, both give exact results because the errors are
2 point: Etf(4)
()3 point: Etf(6)()
which are zero for a second-order polynomial.
24.4 Simpsons rules can be implemented as
6667.19916667.3314151245
6
34)32(437)4050(
6
37)40(452)3040(
8
52)5535(310)030(
=++=
+++
+++
+++=I
The amount of mass can be computed as
mg667.966,7m
minmg6667.1991
min
m4
3
3
=
=M
24.5 Newton-Cotes integration formulas can be implemented to evaluate the integral as
5.10895.8111808117
8
48)7075(358)820(
6
58)45(432)48(
2
3222)14(
2
2212)01(
=+++=
++++
+++
++
+=I
The amount of mass in grams can be computed as
g611.19mg1000
g
min
s60
m
minmg5.1089
s
m3.0
3
3
=
=M
24.6 Equation (23.9) can be used to compute the derivative as
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x0 = 0 f(x0) = 0.06x1 = 1 f(x1) = 0.32
x2 = 3 f(x2) = 0.6
3.01.048.008.0)13)(03(
10)0(26.0
)31)(01(
30)0(232.0
)30)(10(
31)0(206.0)0(' =+=
+
+
=f
The mass flux can be computed as
scm
g1056.4
cm
g103.0
s
cm1052.1fluxMass
2
13
4
62
6 ==
where the negative sign connotes transport from the sediments into the lake. The amount of
mass transported into the lake can be computed as
kg695.517
g1000
kg
m
cm10,000
d
s86,400
yr
d365)m106.3(
scm
g1056.4transportMass
2
226
2
13 ==
24.7 For the equally-spaced points, we can use the second-order formulas from Figs. 23.1 through
23.3. For example, for the first point (t= 0), we can use
min
barrels0415.0
)10(2
)4.0(3)7.0(477.0)0(' =
+=f
However, the points around t= 30 are unequally spaced so we must use Eq. (23.9) to
compute the derivative as
x0 = 20 f(x0) = 0.77
x1 = 30 f(x1) = 0.88
x2 = 45 f(x2) = 1.05
011133.0)3045)(2045(
3020)30(205.1
)4530)(2030(
4520)30(288.0
)4520)(3020(
4530)30(277.0)30(' =
+
+
=f
All the results can be summarized as
t V Derivative Method
0 0.4 0.0415 Forward
10 0.7 0.0185 Centered
20 0.77 0.009 Centered
30 0.88 0.011133 Unequal
45 1.05 0.009667 Centered60 1.17 0.01 Centered
75 1.35 0.014 Backward
The derivatives can be plotted versus time as
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0
0.4
0.8
1.2
1.6
0 10 20 30 40 50 60 70 80
0
0.01
0.02
0.03
0.04
0.05
V Derivative
24.8 If we use 2-point Gauss quadrature the depth domain can be mapped onto the [1, 1] domain
as depicted here
0
d
1
1
0.57735
+ 0.57735
This spacing corresponds to 21.1%dand 78.9%d. Thus, if we wanted to get a good estimate,
we would average the two samples taken at about 20% and 80% of the depth. It should be
noted that the U.S. Geological Survey recommends this procedure for making flow
measurements in rivers.
24.9 Because the data is equispaced, we can use the second-order finite divided differenceformulas from Figs. 23.1 through 23.3. For the first point, we can use
5196.0153255
)8.87(3)6.96(4176=
+
=
d
d
For the intermediate points, we can use centered differences. For example, for the second
point
8647.0153255
8.87176=
=
d
d
For the last point, we can use a backward difference
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5
94118.18663765
2678)3380(4)4258(3=
+
=
d
d
All the values can be tabulated as
d/d87.8 153 0.519696.6 204 0.8647
176 255 1.6314
263 306 1.7157
351 357 3.0196
571 408 4.7353
834 459 6.4510
1229 510 7.7451
1624 561 8.6078
2107 612 10.3333
2678 663 12.4804
3380 714 15.4902
4258 765 18.9412
We can plot these results and after discarding the first few points, we can fit a straight line as
shown (note that the discarded points are displayed as open circles).
y = 0.003844x + 2.482254
R2
= 0.991948
-5
0
5
10
15
20
0 1000 2000 3000 4000
Therefore, the parameter estimates are Eo = 2.48225 and a = 0.003844. The data along with
the first equation can be plotted as
0
4000
8000
12000
0 200 400 600 800
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As described in the problem statement, the fit is not very good. We therefore pick a point near
the midpoint of the data interval
)2107,612( ==
We can then plot the second model
)1(5719.221)1(1
2107 0.0038440.003844)612(0.003844
=
= ee
e
The result is a much better fit
01000
2000
3000
4000
5000
0 100 200 300 400 500 600 700 800
24.10 An exponential model can be fit to the four points from t= 17 to 23. The result is
tec 3277.09.2132 =
This equation can then be used to generate data from t= 25 through 35. This synthetic data
along with the previous measured data can then be integrated with the trapezoidal rule,
2049.612
0223.00429.0)3335(
2
0429.00826.0)3133(
211.01.0)79(
21.00)57(
=+
++
++
+++=I
All the evaluations are summarized in the following table,
tc,
originalc,
syntheticTraprule
5 0 0
7 0.1 0.1 0.1
9 0.11 0.11 0.2111 0.4 0.4 0.51
13 4.1 4.1 4.5
15 9.1 9.1 13.2
17 8 8 17.1
19 4.2 4.2 12.2
21 2.3 2.3 6.5
23 1.1 1.1 3.4
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25 0.9 0.5902 1.6902
27 1.75 0.3065 0.8967
29 2.06 0.1591 0.4656
31 2.25 0.0826 0.2417
33 2.32 0.0429 0.1255
35 2.43 0.0223 0.0652
The cardiac output can then be computed as
min
L5.4897560
2049.61
6.5==C
24.11 The following Excel Solver application can be used to estimate: k = 0.09916 and A =
6.98301.
24.12 The following Excel spreadsheet is set up to use (left) a combination of the trapezoidal and
Simpsons rules and (right) just the trapezoidal rule:
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24.13
The extremes for both cases are at 16 and 92
24.14
dxeI z z5
z200z
30
0
15/
+=
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1 2 3 4
n a 11.9724% 0.1646% 0.01412%1 10440.1504 20034.6250 19520.5626 19345.7378
2 17636.0064 19552.6915 19348.4694
4 19073.5202 19361.2333
8 19289.3051
dxeI z z5
z200
30
0
15/
+=
1 2 3 4
n a 17.8666% 0.9589% 0.03823%1 348.0050 1219.6400 1440.6846 1476.7973
2 1001.7312 1426.8693 1476.2330
4 1320.5848 1473.1478
8 1435.0070
0998.13797.1476
74.19345==d
58.64483
)0998.13(797.1476==V
985.6480995.0
58.6448==T
94.831)0995.0(985.6480797.1476 ==H
24.15 Change of variable:
dd xxx 15152
030
2
030+=
+
+=
dd dxdxdx 152
030=
=
dzez
zzI z
5200
30
0
15/
+=
dx
d
dd dxe
x
xxI d 15
)1515(5
1515)1515200(
1
1
15/)1515(+
++
++=
Therefore, the transformed function is
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10
)1(2
1520
)1515(3000)( dx
d
dd e
x
xxf +
++
=
Five-point formula:
41.320,19)10852.84(0.2369272217.47)0.478629(112415.93)(0.568889)7601.173(0.478629844.2582)(0.236927
)0.90618(0.236927)0.53847(0.478629 )0(0.568889)0.53847(0.478629)0.90618(0.236927
=++++=
++ ++= ff fffI
dzez
zI z
5200
30
0
15/
+=
dx
d
d dxex
xI d 15
)1515(5
1515200
1
1
15/)1515(+
++
+=
Therefore, the transformed function is
)1(
1520
15153000)( d
x
d
dd e
x
xxf
+
++
=
Five-point formula:
1,481.865)379.5667(0.23692729.4212)0.478629(5
827.7287)(0.568889)1097.966(0.478629599.9124)(0.236927
)0.90618(0.236927)0.53847(0.478629)0(0.568889)0.53847(0.478629)0.90618(0.236927
=++
++=
++++=
fffffI
038.13865.1481
41.19320==d
185.64403
)038.13(865.1481==V
548.6472995.0
185.6440==T
846.837)0995.0(548.6472865.1481 ==H
24.16 Trapezoidal rule:
057.1070)6(2
372.10)549.16026.26845.39481.57924.68(20)030( =
++++++=I
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Simpsons 1/3 rule:
1131.098)6(3
372.10)026.26481.57(2)549.16845.39924.68(40)030( =
++++++=I
Simpsons 3/8 rule:
1119.381333.6435785.73758
372.10)549.16026.26(3845.39)1530(
8
845.39)481.57924.68(30)015(
=+=+++
+
+++=I
24.17 Trapezoidal rule (h = 4):
2m6.5310
0)4.3442(20)020( =
+++++=I
Trapezoidal rule (h = 2):
2m2.6320
0)8.24.36.3464428.1(20)020( =
++++++++++=I
Simpsons 1/3 rule:
2m4.6630
0)4.3442(2)8.26.3648.1(40)020( =
++++++++++=I
24.18 A table can be set up to hold the values that are to be integrated:
y, m H, m U, m/s UH, m2/s
0 0.5 0.03 0.015
2 1.3 0.06 0.078
4 1.25 0.05 0.0625
5 1.7 0.12 0.204
6 1 0.11 0.11
9 0.25 0.02 0.005
The cross-sectional area can be evaluated using a combination of Simpsons 1/3 rule and the
trapezoidal rule:
2m525.91.8753.0166674.633333
225.01)69(
61)7.1(425.1)46(
625.1)3.1(45.0)04(
=++=
+++++++=cA
The flow can be evaluated in a similar fashion:
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s
m0.7616671725.03295.00.259667
2
005.011.0)69(
6
11.0)204.0(40625.0)46(
6
0625.0)078.0(4015.0)04(
3
=++=
++
+++
++=Q
24.19 A table can be set up to hold the values that are to be integrated:
Abscissa(N to S)
Ordinate(W to E)
Abscissa(N to S)
Ordinate(W to E)
0 0 2200 3150
200 600 2400 3200
400 880 2600 3200
600 950 2800 3150
800 1100 3000 3000
1000 1600 3200 2800
1200 1900 3400 2700
1400 2100 3600 2600
1600 2300 3800 2525
1800 2600 4000 2525
2000 2950 4200 2430
We can use the multi-segment Simpsons 1/3 rule for the first 18 segments and the Simpsons
3/8 rule for the last 3 segments:
337,917,333.2600]/542800)3150320029502300190011002(880
)2700300032003150260021001600950600(40)[03600(
=+++++++++
+++++++++=I
500,513,18
2430)25252525(32600)36004200( =
+++=I
Therefore, the total area is
2ft9,430,833500,513,133.333,917,7 =+=A
24.20 In order to generate the average rate in units of car/min, the integral to be evaluated along
with the units is
hr24
min)/car(raterateaverage
hr24
0dt
=
The integral can be evaluated with the trapezoidal and Simpsons rules as tabulated below:
Time (hr) Rate (car/min) Method Integral
0 2
2 2
4 0 Simp 1/3 6.666667
5 2
6 6 Simp 1/3 4.666667
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7 7
8 23
9 11 Simp 3/8 40.125
10.5 4 Trap 11.25
11.5 11
12.5 12 Simp 1/3 20
14 8 Trap 15
16 7 Trap 15
17 26
18 20 Simp 1/3 43.66667
19 10
20 8 Simp 1/3 22.66667
21 10
22 8 Simp 1/3 18.66667
23 7
24 3 Simp 1/3 13
The summation of the last column yields the integral estimate of 210.7083 cars hr/min.Dividing by 24 hrs yields the average rate of 210.7083/24 = 8.779514 cars/min. This can then
be multiplied by 1,440 min/d to yield 12,642.5 cars/d.
24.21 The values needed to perform the evaluation can be tabulated:
Heightl, m
Force,F(l), N/m l F(l)
0 0 0
30 340 10200
60 1200 72000
90 1600 144000
120 2700 324000
150 3100 465000
180 3200 576000
210 3500 735000
240 3800 912000
Because there are an even number of equally-spaced segments, we can evaluate the integrals
with the multi-segment Simpsons 1/3 rule.
600,52124
3800)320027001200(2)350031001600340(40)0240( =
++++++++=F
000,728,82
24912000)57600032400072000(2)73500046500014400010200(40)0240(
=
++++++++=I
The line of action can therefore be computed as
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14
158.6043521,600
82,728,000==d
24.22 The values needed to perform the evaluation can be tabulated:
z w(z) gw(z)(60z) zgw(z)(60z)60 200 0 0
50 190 1.862E+07 9.310E+08
40 175 3.430E+07 1.372E+09
30 160 4.704E+07 1.411E+09
20 135 5.292E+07 1.058E+09
10 130 6.370E+07 6.370E+08
0 122 7.174E+07 0
Because there are an even number of equally-spaced segments, we can evaluate the required
integrals with the multi-segment Simpsons 1/3 rule.
97 1054539.21018
0)43.3292.5(2)862.1704.437.6(4174.7)060( =++++++=tf
108 1059253.51018
0)72.13584.10(2)31.9112.4371.6(40)060( =
++++++=tf
The line of action can therefore be computed as
m21.971054539.2
1059253.59
10
=
=d
24.23 One way to determine the volume is to integrate the data from mid-Jan through mid-Jan byduplicating the mid-Jan value at the end of the record. Since we are dealing with long-term
averages, this is valid. Further, for simplicity, we can assume that every month is 30 days
long. We can then integrate the data. As shown in the following table, we have used the acombination of the trapezoidal and Simpsons rules.
day Flow Trap/Simp Method
15 30
45 38
75 82
105 125 5793.75 Simp 3/8
165 95 6600 Trap
255 20 5175 Trap285 22
315 24 1320 Simp 1/3
345 35
375 30 1940 Simp 1/3
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The summation of the integral estimates is 20,828.75 m3 d/s. We can divide this quantity bythe integration interval (360 d) to yield an average flow of 57.8576 m3/s. This quantity can
then be multiplied by the number of seconds per year to arrive at the volume of water that
flows down the river over a typical year.
39
m108246.1yr
d
653d
hr
42hr
min
60min
s
6057.8576FlowAnnual ==
24.24 The integral to be evaluated is
dtqAeht
ab 0=
Simpsons rules can be used to evaluate the integral,
585.79245.2834.51
8
2.0)54.327.6(303.8)814(
12
03.8)8.7(2)832.5(41.0)08(
=+=
++++
++++=I
We can then multiply this result by the area and the efficiency to give
cal988,371,5cal/cm585.79)cm000,100(45.0 22 ==h
24.25 Because the values are equally spaced, we can use the second-order forward difference
from Fig. 23.1 to compute the derivative as
x0 = 0 f(x0) = 20
x1 = 0.08 f(x1) = 17
x2 = 0.16 f(x2) = 15
m
C75.43
16.0
)20(3)17(415)0('
=
+=f
The coefficient of thermal conductivity can then be estimated as k= 60 W/m2/(43.75 oC/m)
= 1.37143 W/(oC m).
24.26 First, we can estimate the areas by numerically differentiating the volume data. Because the
values are equally spaced, we can use the second-order difference formulas from Fig. 23.1 to
compute the derivatives at each depth. For example, at the first depth, we can use the forward
difference to compute
2m450,374,18
)500,817,9(3)100,105,5(4500,963,1)0()0( =
+==
dz
dVAs
For the interior points, second-order centered differences can be used. For example, at the
second point at (z= 4 m),
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16
2m750,9818
500,817,9500,963,1)4()4( =
==
dz
dVAs
The other interior points can be determined in a similar fashion
2m050,5898
100,105,5700,392)8()8( === dz
dVAs
2m5.437,2458
500,963,10)12()12( =
==
dz
dVAs
For the last point, the second-order backward formula yields
2m5.087,498
500,963,1)700,392(4)0(3)16()16( =
+==
dz
dVAs
Since this is clearly a physically unrealistic result, we will assume that the bottom area is 0.
The results are summarized in the following table along with the other quantities needed todetermine the average concentration.
z, m V, m3 c, g/m3 As, m2 cAs
0 9817500 10.2 1374450.0 14019390
4 5105100 8.5 981750.0 8344875
8 1963500 7.4 589050.0 4358970
12 392700 5.2 245437.5 1276275
16 0 4.1 0 0
The necessary integrals can then be evaluated with the multi-segment Simpsons 1/3 rule,
3z
0m400,948,9
12
0)050,589(2)5.437,245750,981(4450,374,1)016()( =++++= dzzAs
g81,629,24012
0)970,358,4(2)275,276,1875,344,8(4390,019,14)016()()(
z
0=
++++= dzzAzc s
The average concentration can then be computed as
3
0
0
m
g8.205263
9,948,400
81,629,240
)(
)()(===
dzzA
dzzAzcc
Z
s
Z
s
24.27 The integral to be determined is
dtteI t )sin2(51/2
0
225.1 =
Change of variable:
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17
dd xxx 25.025.02
05.0
2
05.0+=
+
+=
dd dxdxdx 25.0
2
05.0=
=
ddx dxxeI d 0.25))25.025.0(sin2(5
1
1
2)25.025.0(25.1+ +=
Therefore, the transformed function is
2)25.025.0(25.1 ))25.025.0(sin2(525.0)( dx
d xexfd += +
Five-point formula:
3.431617)0.040941(0.236927.050662)0.478629(13.345384)(0.568889)2.059589(0.4786290.127087)(0.236927
)0.90618(0.236927)0.53847(0.478629
)0(0.568889)0.53847(0.478629)0.90618(0.236927
=++++=
++
++=ff
fffI
Therefore, the RMS current can be computed as
1.8524633.431617RMS ==I
24.28 The integral to be determined is
dtteI
t
)sin2(5
1/2
0
225.1
=
Five applications of Simpsons 1/3 rule can be applied with h = 0.05 as tabulated below:
t i2 Simpsons 1/3 rule
0 0
0.05 2.106774
0.1 6.726726 0.252563698
0.15 11.24592
0.2 13.7153 1.090428284
0.25 13.38154
0.3 10.68149 1.298715588
0.35 6.8209930.4 3.177481 0.685715716
0.45 0.775039
0.5 0 0.104627262
Sum 3.432050547
Therefore, the RMS current can be computed as
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18
1.8525793.432051RMS ==I
24.29
dtteI t )sin2(51/2
0
225.1
=
1 2 3 4
n a 25.0000% 1.9781% 0.0198933%1 0.00000000 4.46051190 3.38814962 3.43184278
2 3.34538393 3.45517226 3.43116008
4 3.42772518 3.43266084
8 3.43142692
Therefore, the RMS current can be computed as
1.8525233.431843RMS ==I
24.30 For the equispaced data, we can use the second-order finite divided difference formulas
from Figs. 23.1 through 23.3. For example, for the first point, we can use
6.102.0
)0(3)16.0(432.0=
+
=dt
di
For the intermediate equispaced points, we can use centered differences. For example, for the
second point
6.1
02.0
032.0=
=
dt
di
For the last point, we can use a backward difference
83.07.0
56.0)84.0(4)2(3=
+
=dt
di
One of the points (t= 0.3) has unequally spaced neighbors. For this point, we can use Eq.
(23.9) to compute the derivative as
x0 = 0.2 f(x0) = 0.32x1 = 0.3 f(x1) = 0.56
x2 = 0.5 f(x2) = 0.84
2.066667)3.05.0)(2.05.0(
3.02.0)3.0(284.0
)5.03.0)(2.03.0(
5.02.0)3.0(256.0
)5.02.0)(3.02.0(
5.03.0)3.0(232.0)0(' =
+
+
=f
We can then multiply these derivative estimates by the inductance. All the results are
summarized below:
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19
t i di /dt VL0 0 1.6 6.4
0.1 0.16 1.6 6.4
0.2 0.32 2 8
0.3 0.56 2.066667 8.266667
0.5 0.84 3.6 14.40.7 2 8 32
24.31 We can solve Faradays law for inductance as
i
dtVL
t
L 0=
We can evaluate the integral using a combination of the trapezoidal and Simpsons rules,
667.9811132020501830162026556667.336
2
715)280400(
2
1526)180280(
2
2635)120180(
2
3546
)80120(8
64)4944(392
)2080(6
29)18(40
)020(
=+++++=
++
++
++
++
++++
++=I
The inductance can then be computed as
H905833.4ms1000
s
A
msvolts4905.833
2
9811.667===L
24.32 The average voltage can be computed as
60
)()(60
0dtiRti
V
=
We can use the formulas to generate values ofi(t) and R(i) and their product for various
equally-spaced times over the integration interval as summarized in the table below. The lastcolumn shows the integral of the product as calculated with Simpsons 1/3 rule.
t i(t) R(i) i(t) R(i) Simpson's 1/30 3600.000 43669.784 157211223
6 2950.461 35816.950 105676498 1287149498
12 2288.787 27812.788 63657535
18 1726.549 21006.431 36268640 456192829
24 1260.625 15360.891 19364321
30 878.355 10723.694 9419212 122017054
36 569.294 6968.908 3967357
42 327.533 4025.426 1318460 19048340
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20
48 151.215 1871.341 282974
54 41.250 518.872 21403 737174
60 0.000 0.000 0
Sum 1885144894
The average voltage can therefore be computed as
7103.141960
1885144894==V
24.33 We can use the trapezoidal rule to integrate the data. For example at t= 0.2, the voltage is
computed as
683.52
0.00036830.0002)02.0(
10
1)2.0(
5=
+=
V
At t= 0.4, we add this value to the integral from t= 0.2 to 0.4,
185.132
0.00038190.0003683)2.04.0(
10
1683.5)4.0(
5=
++=
V
The remainder of the values can be computed in a similar fashion,
t, s i, A Trap V(t)
0 0.0002000 0 0
0.2 0.0003683 5.683 5.683
0.4 0.0003819 7.502 13.185
0.6 0.0002282 6.101 19.286
0.8 0.0000486 2.768 22.054
1 0.0000082 0.568 22.6221.2 0.0001441 1.523 24.145
The plot of voltage versus time can be developed as
0
5
10
15
20
25
0 0.2 0.4 0.6 0.8 1 1.2
24.34 The work can be computed as
dxx.xW cos)0450(1.630
0
2 =
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21
We can use the formulas to generate values of the integrand as summarized in the table
below. The last column shows the integral as calculated with Simpsons 1/3 rule.
x F(x) (x) F(x)cos (x) Simpsons 1/30 0 0.5 0
5 6.875 1.4 1.168524107 21.81420
10 11.5 0.75 8.414421992
15 13.875 0.9 8.62483831 77.76460
20 14 1.3 3.744983601
25 11.875 1.48 1.076725541 14.30402
30 7.5 1.5 0.530529013
Sum 113.88282
24.35 The work can be computed as
dxxxxx.xW )0002.0009.0125.08.0cos()0450(1.630
0
322 ++=
For the 4-segment trapezoidal rule, we can compute values of the integrand at equally-spaced
values ofx with h = 7.5. The results are summarized in the following table,
x F(x) (x) F(x)cos (x) Trap Rule0 0 0.8 0
7.5 9.46875 1.315625 2.390018 8.962569
15 13.875 1.325 3.376187 21.623271
22.5 13.21875 1.334375 3.096162 24.271308
30 7.5 1.85 -2.066927 3.859631
Sum 58.716779
The finer-segment versions can be generated in a similar fashion. The results are summarizedbelow:
Segments W
4 58.71678
8 64.89955
16 66.41926
24.36 The work can be computed as
dxxxxx.xW )0002.0009.0125.08.0cos()0450(1.630
0
322 ++=
For the 4-segment Simpsons 1/3 rule, we can compute values of the integrand at equally-
spaced values ofx with h = 7.5. The results are summarized in the following table,
x F(x) (x) F(x)cos (x) Simpsons 1/3 Rule0 0 0.8 0
7.5 9.46875 1.315625 2.390018 32.34065
15 13.875 1.325 3.376187
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22
22.5 13.21875 1.334375 3.096162 34.23477
30 7.5 1.85 -2.066927
Sum 66.57542
The finer-segment versions can be generated in a similar fashion. The results are summarized
below:
Segments W
4 66.57542
8 66.96047
16 66.92583
24.37
dxxxxx.xW )0002.0009.0125.08.0cos()0450(1.630
0
322 ++=
1 2 3 4
n a 38.5532% 0.9312% 0.0051%1 -31.003903 57.189106 67.201176 66.982729
2 35.140854 66.575421 66.986143
4 58.716779 66.960473
8 64.899549
24.38 The function to be integrated is
dxxxxx.xW )0002.0009.0125.08.0cos()0450(1.630
0
322 ++=
Change of variable:
dd xxx 15152
030
2
030+=++=
dd dxdxdx 152
030=
=
Therefore, the transformed function is
))1515(0002.0)1515(009.0)1515(125.08.0cos(
))1515(0450)1515(1.6(15)(
32
2
ddd
ddd
xxx
x.xxf
+++++
++=
Two-point formula:
73.8496138.2068335.642783
1
3
1=+=
+
= ffI
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23
Three-point formula:
66.81864.30468145.0158317.4981
)7.748426(5555556.0)50.64281(8888889.0)31.49657(5555556.0
)7745967.0(5555556.0)0(8888889.0)7745967.0(5555556.0
=++=
++=
++= fffI
The results for the 2- through 6-point formulas are summarized below:
points W
2 73.8496
3 66.8186
4 66.7734
5 66.9145
6 66.9225
24.39 The work is computed as the product of the force times the distance, where the latter can be
determined by integrating the velocity data (recall Eq. PT6.5),
dttvFWt
)(0=
A table can be set up holding the velocities at evenly spaced times over the integration
interval. The Simpsons 1/3 rule can then be used to integrate this data as shown in the last
column of the table
t v Simp 1/3 rule
0 0
1 4 8
2 83 12 24
4 16
5 17 34.6667
6 20
7 25 50.6667
8 32
9 41 82.6667
10 52
11 65 130.6667
12 80
13 97 194.6667
14 116Sum 525.3333
Thus, the work can be computed as
mN105,066.7)m3333.525(N200 ==W
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24
24.40 Because the data is equispaced, we can use the second-order finite divided difference
formulas from Figs. 23.1 through 23.3. For the first point, we can use
2.9010
)80(3)5.44(430=
+
=dt
dT
For the intermediate points, we can use centered differences. For example, for the second
point
5010
8030=
=dt
dT
We can analyze the other interior points in a similar fashion. For the last point, we can use a
backward difference
06.01525
1.24)7.21(4)7.20(3=
+
=dt
dT
All the values can be tabulated as
t T T Ta dT/dt0 80 60 -9.2
5 44.5 24.5 -5
10 30 10 -2.04
15 24.1 4.1 -0.83
20 21.7 1.7 -0.34
25 20.7 0.7 -0.06
If Newtons law of cooling holds, we can plot dT/dtversus TTa and the points can be fit
with a linear regression with zero intercept to estimate the cooling rate. As in the followingplot, the result is k= 0.1618/min.
y = -0.1618x
R
2
= 0.9758-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
24.41 As in the plot, the initial point is assumed to be e = 0, s = 40. We can then use a
combination of the trapezoidal and Simpsons rules to integrate the data as
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25
12.104175.7833334.3583331.16250.8
12
60)52(2)6043(45.37)05.025.0(
2
5.3740)02.005.0(
2
4040)002.0(
=+++
=++++
++
++
=I
24.42 The function to be integrated is
drrr
rQ )2(12
3
0
6/1
0
=
A plot of the integrand can be developed as
0
5
10
15
20
25
0 1 2 3
As can be seen, the shape of the function indicates that we must use fine segmentation to
attain good accuracy. Here are the results of using a variety of segments.
n Q
2 25.1896
4 36.1635
8 40.9621
16 43.070532 44.0009
64 44.4127
128 44.5955
256 44.6767
512 44.7128
1024 44.7289
2048 44.7361
4096 44.7392
8192 44.7407
16384 44.7413
32768 44.7416
65536 44.7417131072 44.7418
262144 44.7418
Therefore, the result to 4 significant figures appears to be 44.7418.
24.43 The work is computed as
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26
dxFWx
0=
A table can be set up holding the forces at the various displacements. A combination of
Simpsons 1/3 and 3/8 rules can be used to determine the integral as shown in the last two
columns,
x, m F, N Integral Method
0 0
0.05 10
0.1 28 1.133333 Simp 1/3
0.15 46
0.2 63 4.583333 Simp 1/3
0.25 82
0.3 110
0.35 130 14.41875 Simp 3/8
Sum 20.13542
24.44 Although the first three points are equally spaced, the remaining values are unequally
spaced. Therefore, a good approach is to use Eq. 23.9 to perform the differentiation for all
points. The results are summarized below:
t x v a
0 153 68.26923 -35.14510
0.52 185 54.80769 -16.63005
1.04 210 50.97398 -13.20842
1.75 249 35.93855 -26.99478
2.37 261 16.05181 -23.26046
3.25 271 6.59273 -10.80561
3.83 273 0.30382 -10.88031
24.45 The distance traveled is equal to the integral of velocity
dttvyt
t)(
2
1
=
A table can be set up holding the velocities at evenly spaced times (h = 1) over the integration
interval. The Simpsons 1/3 rule can then be used to integrate this data as shown in the last
column of the table
t v Simp 1/3 rule
0 01 6 19.333332 343 84 175.33334 1565 250 507.33336 3667 504 1015.333
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27
8 6649 846 1699.333
10 105011 1045 209012 104013 1035 207014 103015 1025 205016 102017 1015 203018 101019 1005 201020 100021 1052 2105.33322 110823 1168 2337.33324 123225 1300 2601.33326 137227 1448 2897.333
28 152829 1612 3225.33330 1700
Sum 26833.33
Since the underlying functions are second order or less, this result should be exact. We can
verify this by evaluating the integrals analytically,
[ ] 3416.6675.266667.3511 1002310
0
2 === ttdttty
[ ] 250,105.211005110002
10
220
10 === ttdtty
67.166,31800153
2)20(250
03
20
2330
20
2 =
+=+= tttdttty
The total distance traveled is therefore 3416.667 + 10,250 + 13,166.67 = 26,833.33.
24.46 6-segment trapezoidal rule:
10,931.252(6)
541.448)579.466448.784713.333818.072422.79(20)030( =
++++++=y
6-segment Simpsons 1/3 rule:
10,879.883(6)
541.448)448.784818.072(2)579.646713.333422.79(40)030( =
++++++=y
6-point Gauss quadrature:
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28
1410,879.619=y
Romberg integration:
1 2 3 4
n a 4.0634% 0.0098% 0.0000291%1 12668.10909 10896.96330 10879.82315 10879.62077
2 11339.74974 10880.89441 10879.62393
4 10995.60824 10879.70333
8 10908.67956
24.47 We can use the second-order difference formulas from Fig. 23.1 to compute the derivatives
at each depth. For example, at the first point, we can use the forward difference to compute
05994.08
)1.0(3)13.0(416.0)085.0( =
+=
dt
d
For the interior points, second-order centered differences can be used. For example, at thesecond point at (t= 0.586),
05994.08
1.016.0)586.0( =
=
dt
d
The other interior points can be determined in a similar fashion. For the last point, the
second-order backward formula yields
03988.08
52.0)54.0(4)56.0(3)097.10( =
+=
dt
d
All the results are summarized in the following table.
t d/dt t d/dt0.085 0.10 0.05994 5.591 0.37 0.04995
0.586 0.13 0.05994 6.091 0.39 0.03996
1.086 0.16 0.04995 6.592 0.41 0.03996
1.587 0.18 0.03996 7.092 0.43 0.04
2.087 0.20 0.04995 7.592 0.45 0.03996
2.588 0.23 0.04995 8.093 0.47 0.04995
3.088 0.25 0.02997 8.593 0.50 0.04995
3.589 0.26 0.04995 9.094 0.52 0.03996
4.089 0.30 0.05994 9.594 0.54 0.03988
4.590 0.32 0.03996 10.097 0.56 0.03988
5.090 0.34 0.04995
The derivatives can be plotted:
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29
0
0.02
0.04
0.06
0.08
0 2 4 6 8 10
After about nine points, the derivatives have settled down and the mean and standard
deviation can be computed as
Mean = 0.04328
Standard deviation = 0.004926
24.48 (a) After converting the radii to units of meters, a polynomial can be fit to the velocity data
y = -24.360173x2
+ 0.334701x + 0.904855
R2
= 0.997577
-0.2
0
0.2
0.4
0.6
0.8
1
0 0.05 0.1 0.15 0.2
The flow can then be evaluated analytically as
[ ]s
m58091610.00.45242750.1115676.090043252
)904855.0334701.0360173.24(2
30.2
0234
0.2
0
23
=++=
++=
rrr
drrrrQ
(b) The data and the integrand can be tabulated as shown below. Simpsons 1/3 rule can then
be computed and summed in the last column.
r, m v, m/s 2 rv Simp 1/30 0.914 0
0.025 0.89 0.139801 0.006877
0.05 0.847 0.266093
0.075 0.795 0.374635 0.018470
0.1 0.719 0.451761
0.125 0.543 0.426471 0.021334
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30
0.15 0.427 0.402438
0.175 0.204 0.22431 0.010831
0.2 0 0
Sum 0.057512
(c)
%998.0%1000.05809161
0.0575120.05809161==t
24.49 (a) After converting the radii to units of feet, a polynomial can be fit to the non-core
velocity data
y = -28.363x2
+ 4.7576x + 4.723
R2
= 0.9939
0
2
4
6
0.0 0.1 0.2 0.3 0.4 0.5 0.6
The noncore flow can then be evaluated analytically as
[ ]s
ft2.063792.36151.5858577.090722
)723.44.7575728.3629(2
30.5
0.08333234
0.5
0.08333
23noncore
=++=
++=
rrr
drrrrQ
This can be added to the core flow to determine the total flow
s
ft2.172873109083.02.06379)ft083333.0(
s
ft5
s
ft2.170447
32
3
corenoncore =+=+=+= QQQ
(b) The data and the integrand can be tabulated as shown below. Simpsons rules can then be
computed and summed in the last column. Note that because we have an odd number of
segments, we must use a combination of Simpsons 1/3 and 3/8 rules.
r, ft v, ft/s 2 rv Integral Method0.083333 5 2.617994
0.166667 4.62 4.8380530.250000 4.01 6.298893 0.785253 Simp 1/3
0.333333 3.42 7.162831
0.416667 1.69 4.42441
0.500000 0 0 1.283144 Simp 3/8
Sum 2.068397
This numerical estimate can be added to the core flow to determine the total flow
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be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, 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.
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31
s
ft2.17748109083.02.068397)ft083333.0(
s
ft5
s
ft2.068397
32
3
corenoncore =+=+=+= QQQ
(c)
%21.0%100
2.172873
2.177482.172873=
=t
24.50 The following Excel worksheet solves the problem. Note that the derivative is calculated
with a centered difference,
K
VV
dT
dV KK
100
350450 =
24.51 A single application of the trapezoidal rule yields:
1372
2.15.12)323(
=
+
=I
A 2-segment trapezoidal rule gives
5.864
2.1)8.1(25.12)323( =
++=I
A 4-segment trapezoidal rule gives
75.678
2.1)4.18.15.3(25.12)323( =
++++=I
Because we do not know the true value, it it would seem impossible to estimate the error.
However, we can try to fit different order polynomials to see if we can get a decent fit to the
data. This yields the surprising result that a 4 th-order polynomial results in almost a perfect fit.
For example, using the Excel trend line gives:
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be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual
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32
y = 3.26667E-04x4
- 2.17200E-02x3
+ 5.33673E-01x2
-
5.82588E+00x + 2.57346E+01
R2
= 1.00000E+00
0
4
8
12
0 6 12 18 24
This can be integrated analytically to give 60.955. Note that the same result would result from
using Booles rule, Romberg integration or Gauss quadrature.
Therefore, we can estimate the errors as
%76.124%10060.955
13760.955 ==I
%91.41%10060.955
5.8660.955=
=I
%15.11%10060.955
75.6760.955=
=I
The ratio of these is 124.76:41.91:11.15 = 11.2:3.8:1. Thus, the result approximates the
quartering of the error that we would expect according to Eq. 21.13.
24.52 (b) This problem can be solved in a number of ways. One approach is to set up Excel with
a series of equally-spaced x values from 0 to 100. Then one of the formulas described in this
Part of the book can be used to numerically compute the derivative. For example, x values
with an interval of 1 and Eq. 23.9 can be used. The resulting plot of the function and its
derivative is
0
0.2
0.4
0.6
0.8
1
0 20 40 60 80 100
0
0.01
0.02
0.03
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be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, 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.
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33
(b) Inspection of this plot indicates that the maximum derivative occurs at about a diameter of
13.3.
(c) The function to be integrated looks like
f(x)/x
0
0.004
0.0080.012
0 20 40 60 80 100
This can be integrated from 1 to a high number using any of the methods provided in this
book. For example, the trapezoidal rule can be used to integrate from 1 to 100, 1 to 200 and 1
to 300 using h = 1. The results are:
h I
100 0.073599883
200 0.073632607
300 0.073632609
Thus, the integral seems to be converging to a value of 0.073633. Sm can be computed as
6 0.073633 = 0.4418.
24.53(a) First, the distance can be converted to meters. Then, Eq. (23.9) can be used to compute
the derivative at the surface as
x0 = 0 f(x0) = 900
x1 = 0.01 f(x1) = 480x2 = 0.03 f(x2) = 270
m
K500,52
)01.003.0)(003.0(
01.00)0(2270
)03.001.0)(001.0(
03.00)0(2480
)03.00)(01.00(
03.001.0)0(2900)0(' =
+
+
=f
The heat flux can be computed as
2m
W470,1
m
K500,52
Kms
J028.0fluxHeat =
=
(b) The heat transfer can be computed by multiplying the flux by the area
W470,1cm10,000
mcm)50cm200(
m
W470,1ferHeat trans
2
2
2==
24.54(a) The pressure drop can be determined by integrating the pressure gradient
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be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, 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.
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34
dxxr
Qp
x
x
)(
8
2
1
4 =
After converting the units to meters, a table can be set up holding the data and the integrand.
The trapezoidal and Simpsons rules can then be used to integrate this data as shown in the
last column of the table.
x, m r, m integrand integral method
0 0.002 -7957.75
0.02 0.00135 -38333.2
0.04 0.00134 -39490.3 -1338.5392 Simp 1/3
0.05 0.0016 -19428.1
0.06 0.00158 -20430.6
0.07 0.00142 -31315.3 -713.9319 Simp 3/8
0.1 0.002 -7957.75 -589.0959 Trap
Sum -2641.5670
Therefore, the pressure drop is computed as 2,641.567 N/m2
.
(b) The average radius can also be computed by integration as
12
)(2
1
xx
dxxr
r
x
x
=
The numerical evaluations can again be determined by a combination of the trapezoidal and
Simpsons rules.
x, m r, m integral method
0 0.002000.02 0.00135
0.04 0.00134 5.83E-05 Simp 1/3
0.05 0.00160
0.06 0.00158
0.07 0.00142 4.61E-05 Simp 3/8
0.1 0.00200 5.13E-05 Trap
Sum 0.0001557
Therefore, the average radius is 0.0001557/0.1 = 0.001557 m. This value can be used to
compute a pressure drop of
244 m
N2166.951.0
)001557.0(
00001.0)005.0(88 ====
x
r
Qx
dx
dpp
Thus, there is less pressure drop if the radius is at the constant mean value.
(c) The average Reynolds number can be computed by first determining the average velocity
as
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be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual
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s
m1.31317
107.6152
00001.0
)001557.0(
00001.062=
===
cA
Qv
Then the Reynolds number can be computed as
796.817005.0
00311381.31317)0.(101Re
3
=
=
24.55 After converting the units to meters, a table can be set up holding the data and the
integrand. The trapezoidal and Simpsons rules can then be used to integrate this data as
shown in the last column of the table.
r, m v, m/s integrand integral method
0 10 0
0.016 9.69 1.168974 0.036905 Simp 1/3
0.032 9.3 2.243851
0.048 8.77 3.173964 0.100138 Simp 1/30.064 7.95 3.836262
0.0747 6.79 3.824296 0.040984 Trap
0.0787 5.57 3.305149 0.014259 Trap
0.0795 4.89 2.931144 0.002495 Trap
0.08 0 0 0.000733 Trap
Sum 0.195514
Therefore, the mass flow rate is 0.195514 kg/s.