Numerical Method for engineers-chapter 24

8/2/2019 Numerical Method for engineers-chapter 24

1/35

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

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 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.


2/35

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





3/35

3

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





4/35

4

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





5/35

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





6/35

6

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





7/35

7

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:





8/35

8

24.13

The extremes for both cases are at 16 and 92

24.14

dxeI z z5

z200z

30

0

15/

+=





9/35

9

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(+

++

++=






10/35

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(+

++

+=


)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





11/35

11

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:





12/35

12

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





13/35

13

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





14/35

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





15/35

15

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),





16/35

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:





17/35

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+ +=


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






18/35

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


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:





19/35

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





20/35

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 =





21/35

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


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


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





22/35

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=

=


))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





23/35

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





24/35

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





25/35

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





26/35

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





27/35

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:





28/35

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:





29/35

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





30/35

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





31/35

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:





32/35

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





33/35

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





34/35

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





35/35

35

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.

Documents

Numerical Method for engineers-chapter 24