Numerical methods for engineers 5th e-solution ch 24

1

CHAPTER 24

24.1 Simpson’s rules can be implemented as

The amount of heat required can be computed as

24.2 The first iteration involves computing 1 and 2 segment trapezoidal rules and combining them as

and computing the approximate error as

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 32.410000002 32.58187500 32.410000004 32.45296875

24.3 Change of variable:

Therefore, the transformed function is

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2

Two-point formula:

Three-point formula:

Thus, both give exact results because the errors are

2 point: Et f(4)()3 point: Et f(6)()

which are zero for a second-order polynomial.

24.4 Simpson’s rules can be implemented as

The amount of mass can be computed as

24.5 Newton-Cotes integration formulas can be implemented to evaluate the integral as

The amount of mass in grams can be computed as


3

24.6 Equation (23.9) can be used to compute the derivative as

x0 = 0 f(x0) = 0.06x1 = 1 f(x1) = 0.32x2 = 3 f(x2) = 0.6

The mass flux can be computed as

where the negative sign connotes transport from the sediments into the lake. The amount of mass transported into the lake can be computed as

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

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) = 0.77x1 = 30 f(x1) = 0.88x2 = 45 f(x2) = 1.05

All the results can be summarized as

t V Derivative Method0 0.4 0.0415 Forward

10 0.7 0.0185 Centered20 0.77 0.009 Centered30 0.88 0.011133 Unequal45 1.05 0.009667 Centered60 1.17 0.01 Centered75 1.35 0.014 Backward

The derivatives can be plotted versus time as


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%d and 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 difference formulas from Figs. 23.1 through 23.3. For the first point, we can use

For the intermediate points, we can use centered differences. For example, for the second point

For the last point, we can use a backward difference


5

All the values can be tabulated as

d/d87.8 153 0.519696.6 204 0.8647176 255 1.6314263 306 1.7157351 357 3.0196571 408 4.7353834 459 6.4510

1229 510 7.74511624 561 8.60782107 612 10.33332678 663 12.48043380 714 15.49024258 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

As described in the problem statement, the fit is not very good. We therefore pick a point near the midpoint of the data interval

We can then plot the second model

The result is a much better fit

0

1000

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

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,

All the evaluations are summarized in the following table,

tc,

originalc,

syntheticTrap rule

5 0 07 0.1 0.1 0.19 0.11 0.11 0.21

11 0.4 0.4 0.5113 4.1 4.1 4.515 9.1 9.1 13.217 8 8 17.119 4.2 4.2 12.221 2.3 2.3 6.5


7

23 1.1 1.1 3.425 0.9 0.5902 1.690227 1.75 0.3065 0.896729 2.06 0.1591 0.465631 2.25 0.0826 0.241733 2.32 0.0429 0.125535 2.43 0.0223 0.0652

The cardiac output can then be computed as

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

24.13

The extremes for both cases are at 16 and 92

24.14


9

1 2 3 4n a 11.9724% 0.1646% 0.01412%1 10440.1504 20034.6250 19520.5626 19345.73782 17636.0064 19552.6915 19348.46944 19073.5202 19361.23338 19289.3051

1 2 3 4n a 17.8666% 0.9589% 0.03823%1 348.0050 1219.6400 1440.6846 1476.79732 1001.7312 1426.8693 1476.23304 1320.5848 1473.14788 1435.0070

24.15 Change of variable:



10

Five-point formula:


Five-point formula:

24.16 Trapezoidal rule:


11

Simpson’s 1/3 rule:


24.17 Trapezoidal rule (h = 4):

Trapezoidal rule (h = 2):


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/s0 0.5 0.03 0.0152 1.3 0.06 0.0784 1.25 0.05 0.06255 1.7 0.12 0.2046 1 0.11 0.119 0.25 0.02 0.005

The cross-sectional area can be evaluated using a combination of Simpson’s 1/3 rule and the trapezoidal rule:


12

The flow can be evaluated in a similar fashion:

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 3150200 600 2400 3200400 880 2600 3200600 950 2800 3150800 1100 3000 3000

1000 1600 3200 28001200 1900 3400 27001400 2100 3600 26001600 2300 3800 25251800 2600 4000 25252000 2950 4200 2430

We can use the multi-segment Simpson’s 1/3 rule for the first 18 segments and the Simpson’s 3/8 rule for the last 3 segments:

Therefore, the total area is

24.20 In order to generate the average rate in units of car/min, the integral to be evaluated along with the units is


13

The integral can be evaluated with the trapezoidal and Simpson’s rules as tabulated below:

Time (hr) Rate (car/min) Method Integral0 22 24 0 Simp 1/3 6.6666675 26 6 Simp 1/3 4.6666677 78 239 11 Simp 3/8 40.125

10.5 4 Trap 11.2511.5 1112.5 12 Simp 1/3 2014 8 Trap 1516 7 Trap 1517 2618 20 Simp 1/3 43.6666719 1020 8 Simp 1/3 22.6666721 1022 8 Simp 1/3 18.6666723 724 3 Simp 1/3 13

The summation of the last column yields the integral estimate of 210.7083 carshr/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:

Height l, m

Force, F(l), N/m lF(l)

0 0 030 340 1020060 1200 7200090 1600 144000

120 2700 324000150 3100 465000180 3200 576000210 3500 735000240 3800 912000

Because there are an even number of equally-spaced segments, we can evaluate the integrals with the multi-segment Simpson’s 1/3 rule.


14

The line of action can therefore be computed as

24.22 The values needed to perform the evaluation can be tabulated:

z w(z) gw(z)(60z) zgw(z)(60z)60 200 0 050 190 1.862E+07 9.310E+0840 175 3.430E+07 1.372E+0930 160 4.704E+07 1.411E+0920 135 5.292E+07 1.058E+0910 130 6.370E+07 6.370E+080 122 7.174E+07 0

Because there are an even number of equally-spaced segments, we can evaluate the required integrals with the multi-segment Simpson’s 1/3 rule.

The line of action can therefore be computed as

24.23 One way to determine the volume is to integrate the data from mid-Jan through mid-Jan by duplicating 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 a combination of the trapezoidal and Simpson’s rules.

day Flow Trap/Simp Method15 3045 3875 82

105 125 5793.75 Simp 3/8


15

165 95 6600 Trap255 20 5175 Trap285 22315 24 1320 Simp 1/3345 35375 30 1940 Simp 1/3

The summation of the integral estimates is 20,828.75 m3d/s. We can divide this quantity by the 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.

24.24 The integral to be evaluated is

Simpson’s rules can be used to evaluate the integral,

We can then multiply this result by the area and the efficiency to give

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) = 20x1 = 0.08 f(x1) = 17x2 = 0.16 f(x2) = 15

The coefficient of thermal conductivity can then be estimated as k = –60 W/m2/(–43.75 oC/m) = 1.37143 W/(oCm).

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


16

For the interior points, second-order centered differences can be used. For example, at the second point at (z = 4 m),

The other interior points can be determined in a similar fashion

For the last point, the second-order backward formula yields

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 to determine the average concentration.

z, m V, m3 c, g/m3 As, m2 cAs

0 9817500 10.2 1374450.0 140193904 5105100 8.5 981750.0 83448758 1963500 7.4 589050.0 4358970

12 392700 5.2 245437.5 127627516 0 4.1 0 0

The necessary integrals can then be evaluated with the multi-segment Simpson’s 1/3 rule,

The average concentration can then be computed as


17

24.27 The integral to be determined is

Change of variable:


Five-point formula:

Therefore, the RMS current can be computed as

24.28 The integral to be determined is

Five applications of Simpson’s 1/3 rule can be applied with h = 0.05 as tabulated below:

t i2 Simpsons 1/3 rule0 0

0.05 2.1067740.1 6.726726 0.2525636980.15 11.245920.2 13.7153 1.0904282840.25 13.381540.3 10.68149 1.298715588


18

0.35 6.8209930.4 3.177481 0.6857157160.45 0.7750390.5 0 0.104627262

Sum 3.432050547


24.29

1 2 3 4n a 25.0000% 1.9781% 0.0198933%1 0.00000000 4.46051190 3.38814962 3.431842782 3.34538393 3.45517226 3.431160084 3.42772518 3.432660848 3.43142692


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

For the intermediate equispaced points, we can use centered differences. For example, for the second point

For the last point, we can use a backward difference

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) = 0.56x2 = 0.5 f(x2) = 0.84


19

We can then multiply these derivative estimates by the inductance. All the results are summarized below:

t i di/dt VL

0 0 1.6 6.40.1 0.16 1.6 6.40.2 0.32 2 80.3 0.56 2.066667 8.2666670.5 0.84 3.6 14.40.7 2 8 32

24.31 We can solve Faraday’s law for inductance as

We can evaluate the integral using a combination of the trapezoidal and Simpson’s rules,

The inductance can then be computed as

24.32 The average voltage can be computed as

We can use the formulas to generate values of i(t) and R(i) and their product for various equally-spaced times over the integration interval as summarized in the table below. The last column shows the integral of the product as calculated with Simpson’s 1/3 rule.

t i(t) R(i) i(t) R(i) Simpson's 1/30 3600.000 43669.784 157211223


20

6 2950.461 35816.950 105676498 128714949812 2288.787 27812.788 6365753518 1726.549 21006.431 36268640 45619282924 1260.625 15360.891 1936432130 878.355 10723.694 9419212 12201705436 569.294 6968.908 396735742 327.533 4025.426 1318460 1904834048 151.215 1871.341 28297454 41.250 518.872 21403 73717460 0.000 0.000 0

Sum 1885144894

The average voltage can therefore be computed as

24.33 We can use the trapezoidal rule to integrate the data. For example at t = 0.2, the voltage is computed as

At t = 0.4, we add this value to the integral from t = 0.2 to 0.4,

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 5.6830.4 0.0003819 7.502 13.1850.6 0.0002282 6.101 19.2860.8 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


21

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

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 Simpson’s 1/3 rule.

x F(x) (x) F(x)cos(x) Simpson’s 1/30 0 0.5 05 6.875 1.4 1.168524107 21.81420

10 11.5 0.75 8.41442199215 13.875 0.9 8.62483831 77.7646020 14 1.3 3.74498360125 11.875 1.48 1.076725541 14.3040230 7.5 1.5 0.530529013

Sum 113.88282


For the 4-segment trapezoidal rule, we can compute values of the integrand at equally-spaced values of x 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 8.96256915 13.875 1.325 3.376187 21.623271

22.5 13.21875 1.334375 3.096162 24.27130830 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 summarized below:

Segments W4 58.716788 64.89955


22

16 66.41926


For the 4-segment Simpson’s 1/3 rule, we can compute values of the integrand at equally-spaced values of x with h = 7.5. The results are summarized in the following table,

x F(x) (x) F(x)cos(x) Simpson’s 1/3 Rule0 0 0.8 0

7.5 9.46875 1.315625 2.390018 32.3406515 13.875 1.325 3.376187

22.5 13.21875 1.334375 3.096162 34.2347730 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 W4 66.575428 66.96047

16 66.92583

24.37

1 2 3 4n a 38.5532% 0.9312% 0.0051%1 -31.003903 57.189106 67.201176 66.9827292 35.140854 66.575421 66.9861434 58.716779 66.9604738 64.899549

24.38 The function to be integrated is

Change of variable:


23


Two-point formula:

Three-point formula:

The results for the 2- through 6-point formulas are summarized below:

points W2 73.84963 66.81864 66.77345 66.91456 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),

A table can be set up holding the velocities at evenly spaced times over the integration interval. The Simpson’s 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 rule0 01 4 82 83 12 244 165 17 34.66676 20


24

7 25 50.66678 329 41 82.6667

10 5211 65 130.666712 8013 97 194.666714 116

Sum 525.3333

Thus, the work can be computed as

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

For the intermediate points, we can use centered differences. For example, for the second point

We can analyze the other interior points in a similar fashion. For the last point, we can use a backward difference

All the values can be tabulated as

t T T Ta dT/dt0 80 60 -9.25 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 Newton’s law of cooling holds, we can plot dT/dt versus T – Ta and the points can be fit with a linear regression with zero intercept to estimate the cooling rate. As in the following plot, the result is k = 0.1618/min.


25

y = -0.1618x

R2 = 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

24.42 The function to be integrated is

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 Q2 25.18964 36.16358 40.9621

16 43.070532 44.000964 44.4127

128 44.5955256 44.6767


26

512 44.71281024 44.72892048 44.73614096 44.73928192 44.7407

16384 44.741332768 44.741665536 44.7417

131072 44.7418262144 44.7418

Therefore, the result to 4 significant figures appears to be 44.7418.

24.43 The work is computed as

A table can be set up holding the forces at the various displacements. A combination of Simpson’s 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 Method0 0

0.05 100.1 28 1.133333 Simp 1/3

0.15 460.2 63 4.583333 Simp 1/3

0.25 820.3 110

0.35 130 14.41875 Simp 3/8Sum 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 a0 153 68.26923 -35.14510

0.52 185 54.80769 -16.630051.04 210 50.97398 -13.208421.75 249 35.93855 -26.994782.37 261 16.05181 -23.260463.25 271 6.59273 -10.805613.83 273 0.30382 -10.88031

24.45 The distance traveled is equal to the integral of velocity


27

A table can be set up holding the velocities at evenly spaced times (h = 1) over the integration interval. The Simpson’s 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 rule0 01 6 19.333332 343 84 175.33334 1565 250 507.33336 3667 504 1015.3338 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 2897.33328 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,


28

The total distance traveled is therefore 3416.667 + 10,250 + 13,166.67 = 26,833.33.

24.46 6-segment trapezoidal rule:

6-segment Simpson’s 1/3 rule:

6-point Gauss quadrature:

Romberg integration:

1 2 3 4n a 4.0634% 0.0098% 0.0000291%1 12668.10909 10896.96330 10879.82315 10879.620772 11339.74974 10880.89441 10879.623934 10995.60824 10879.703338 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

For the interior points, second-order centered differences can be used. For example, at the second point at (t = 0.586),

The other interior points can be determined in a similar fashion. For the last point, the second-order backward formula yields

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


29

0.586 0.13 0.05994 6.091 0.39 0.039961.086 0.16 0.04995 6.592 0.41 0.039961.587 0.18 0.03996 7.092 0.43 0.042.087 0.20 0.04995 7.592 0.45 0.039962.588 0.23 0.04995 8.093 0.47 0.049953.088 0.25 0.02997 8.593 0.50 0.049953.589 0.26 0.04995 9.094 0.52 0.039964.089 0.30 0.05994 9.594 0.54 0.039884.590 0.32 0.03996 10.097 0.56 0.039885.090 0.34 0.04995

The derivatives can be plotted:

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 = 0.04328Standard 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


30

(b) The data and the integrand can be tabulated as shown below. Simpson’s 1/3 rule can then be computed and summed in the last column.

r, m v, m/s 2rv Simp 1/30 0.914 0

0.025 0.89 0.139801 0.0068770.05 0.847 0.266093

0.075 0.795 0.374635 0.0184700.1 0.719 0.451761

0.125 0.543 0.426471 0.0213340.15 0.427 0.402438

0.175 0.204 0.22431 0.0108310.2 0 0

Sum 0.057512

(c)

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

This can be added to the core flow to determine the total flow


31

(b) The data and the integrand can be tabulated as shown below. Simpson’s 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 Simpson’s 1/3 and 3/8 rules.

r, ft v, ft/s 2rv Integral Method0.083333 5 2.6179940.166667 4.62 4.8380530.250000 4.01 6.298893 0.785253 Simp 1/3

0.333333 3.42 7.1628310.416667 1.69 4.424410.500000 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

(c)

24.50 The following Excel worksheet solves the problem. Note that the derivative is calculated with a centered difference,

24.51 A single application of the trapezoidal rule yields:

A 2-segment trapezoidal rule gives


32

A 4-segment trapezoidal rule gives

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 4th-order polynomial results in almost a perfect fit. For example, using the Excel trend line gives:

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 Boole’s rule, Romberg integration or Gauss quadrature.

Therefore, we can estimate the errors as

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


33

0

0.2

0.4

0.6

0.8

1

0 20 40 60 80 100

0

0.01

0.02

0.03

(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

0.008

0.012

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 I100 0.073599883200 0.073632607300 0.073632609

Thus, the integral seems to be converging to a value of 0.073633. Sm can be computed as 60.073633 = 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) = 900x1 = 0.01 f(x1) = 480x2 = 0.03 f(x2) = 270

The heat flux can be computed as


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(b) The heat transfer can be computed by multiplying the flux by the area

24.54 (a) The pressure drop can be determined by integrating the pressure gradient

After converting the units to meters, a table can be set up holding the data and the integrand. The trapezoidal and Simpson’s rules can then be used to integrate this data as shown in the last column of the table.

x, m r, m integrand integral method0 0.002 -7957.75

0.02 0.00135 -38333.20.04 0.00134 -39490.3 -1338.5392 Simp 1/30.05 0.0016 -19428.10.06 0.00158 -20430.60.07 0.00142 -31315.3 -713.9319 Simp 3/8

0.1 0.002 -7957.75 -589.0959 TrapSum -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

The numerical evaluations can again be determined by a combination of the trapezoidal and Simpson’s rules.

x, m r, m integral method0 0.00200

0.02 0.001350.04 0.00134 5.83E-05 Simp 1/30.05 0.001600.06 0.001580.07 0.00142 4.61E-05 Simp 3/8

0.1 0.00200 5.13E-05 Trap


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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

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

Then the Reynolds number can be computed as

24.55 After converting the units to meters, a table can be set up holding the data and the integrand. The trapezoidal and Simpson’s 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 method0 10 0

0.016 9.69 1.168974 0.036905 Simp 1/30.032 9.3 2.2438510.048 8.77 3.173964 0.100138 Simp 1/30.064 7.95 3.836262

0.0747 6.79 3.824296 0.040984 Trap0.0787 5.57 3.305149 0.014259 Trap0.0795 4.89 2.931144 0.002495 Trap

0.08 0 0 0.000733 TrapSum 0.195514

Therefore, the mass flow rate is 0.195514 kg/s.


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Numerical methods for engineers 5th e-solution ch 24