Numerical Method for engineers-chapter 24

1

CHAPTER 24 24.1 Simpson’s rules can be implemented as I = ( −50 − (−150))

0.11454 + 4(0.11904) + 0.12486 6

+ (100 − (−50))

0.12486 + 3(0.132 + 0.14046) + 0.15024 8 = 11.926 + 20.484 = 32.41

The amount of heat required can be computed as ∆H = 32.41(1200) = 38892 24.2 The first iteration involves computing 1 and 2 segment trapezoidal rules and combining them as I=

4(32.581875) − 33.0975 = 32.41 3

and computing the approximate error as

εa =

32.41 − 32.581875 × 100% = 0.5303% 32.41

The computation can be continues as in the following tableau until εa < 0.01%. 1 εa  33.09750000 32.58187500 32.45296875

n 1 2 4

2 0.5303% 32.41000000 32.41000000

3 0.0000% 32.41000000

24.3 Change of variable: x=

100 − 150 100 − ( −150) + x d = −25 + 125 x d 2 2

dx = I=

∫

100 − ( −150) dx d = 125dx d 2 1 −1

(0.132 + 1.56 × 10 − 4 ( −25 + 125 x d ) + 2.64 × 10 −7 ( −25 + 125 x d ) 2 ) 125 dx d

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 the publisher, 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 f ( x d ) = 125 (0.132 + 1.56 × 10 −4 ( −25 + 125 x d ) + 2.64 × 10 −7 ( −25 + 125 x d ) 2 ) Two-point formula:  −1   + I = f   3

 1   = 14.91679 + 17.49321 = 32.41 f   3

Three-point formula: I = 0.5555556 f ( −0.7745967) + 0.8888889 f (0) + 0.5555556 f (0.7745967) = 0.5555556(14.61418) + 0.8888889(16.03313) + 0.5555556(18.07082) = 8.11899 + 14.25167 + 10.03934 = 32.41 Thus, both give exact results because the errors are Et ∝ f(4)(ξ) Et ∝ f(6)(ξ)

2 point: 3 point:

which are zero for a second-order polynomial. 24.4 Simpson’s rules can be implemented as I = (30 − 0)

10 + 3(35 + 55) + 52 52 + 4( 40) + 37 37 + 4(32) + 34 + ( 40 − 30) + (50 − 40) 8 6 6 = 1245 + 415 + 331.6667 = 1991.6667

The amount of mass can be computed as M =4

m3  mg min  1991.6667  = 7,966.667 mg min  m3 

24.5 Newton-Cotes integration formulas can be implemented to evaluate the integral as I = (1 − 0)

12 + 22 22 + 32 32 + 4( 45) + 58 58 + 3(75 + 70) + 48 + ( 4 − 1) + (8 − 4) + (20 − 8) 2 2 6 8 = 17 + 81 + 180 + 811.5 = 1089.5

The amount of mass in grams can be computed as M = 0.3

m3 s

mg min  60 s g  = 19.611 g 1089.5  3 m   min 1000 mg

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

3 x0 = 0 x1 = 1 x2 = 3

f(x0) = 0.06 f(x1) = 0.32 f(x2) = 0.6

f ' (0) = 0.06

2(0) − 1 − 3 2(0) − 0 − 3 2(0) − 0 − 1 + 0.32 + 0.6 = −0.08 + 0.48 − 0.1 = 0.3 (0 − 1)(0 − 3) (1 − 0)(1 − 3) (3 − 0)(3 − 1)

The mass flux can be computed as Mass flux = −1.52 × 10 −6

cm 2 g g 0.3 × 10 −6 = −4.56 × 10 −13 4 s cm cm 2 s

where the negative sign connotes transport from the sediments into the lake. The amount of mass transported into the lake can be computed as 2 g kg 6 2 365 d 86,400 s 10,000 cm ( 3 . 6 × 10 m ) = 517.695 kg 2 2 yr d 1000 g cm s m 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

Mass transport = 4.56 × 10 −13

f ' (0) =

− 0.77 + 4(0.7) − 3(0.4) barrels = 0.0415 2(10) min

However, the points around t = 30 are unequally spaced so we must use Eq. (23.9) to compute the derivative as x0 = 20 x1 = 30 x2 = 45

f(x0) = 0.77 f(x1) = 0.88 f(x2) = 1.05

2(30) − 30 − 45 2(30) − 20 − 45 2(30) − 20 − 30 + 0.88 + 1.05 = 0.011133 (20 − 30)(20 − 45) (30 − 20)(30 − 45) (45 − 20)(45 − 30) All the results can be summarized as f ' (30) = 0.77

t 0 10 20 30 45 60 75

V 0.4 0.7 0.77 0.88 1.05 1.17 1.35

Derivative 0.0415 0.0185 0.009 0.011133 0.009667 0.01 0.014

Method Forward Centered Centered Unequal Centered Centered Backward

The derivatives can be plotted versus time as

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 the publisher, 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.

4

1.6

0.05 0.04

1.2

0.03

0.8

0.02

0.4

0.01

0

0 0

10

20

30

40 V

50

60

70

80

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

−1 − 0.57735

+ 0.57735 d

1

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 dσ − 176 + 4(96.6) − 3(87.8) = = −0.5196 dε 255 − 153 For the intermediate points, we can use centered differences. For example, for the second point dσ 176 − 87.8 = = 0.8647 dε 255 − 153 For the last point, we can use a backward difference


5 dσ 3( 4258) − 4(3380) + 2678 = = 18.94118 dε 765 − 663 All the values can be tabulated as σ 87.8 96.6 176 263 351 571 834 1229 1624 2107 2678 3380 4258

ε 153 204 255 306 357 408 459 510 561 612 663 714 765

dσ /dε −0.5196 0.8647 1.6314 1.7157 3.0196 4.7353 6.4510 7.7451 8.6078 10.3333 12.4804 15.4902 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

20

R2 = 0.991948

15 10 5 0 -5 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 12000 8000 4000 0 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 (ε = 612, σ = 2107) We can then plot the second model 2107   0.003844 ε σ =  0.003844 ( 612 ) − 1) = 221.5719(e 0.003844 ε − 1)  (e −1 e The result is a much better fit 5000 4000 3000 2000 1000 0 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 c = 2132.9e −0.3277 t 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, I = (7 − 5)

0 + 0.1 0.1 + 0.11 + (9 − 7) 2 2 + ⋅ ⋅ ⋅ + (33 − 31)

0.0826 + 0.0429 0.0429 + 0.0223 + (35 − 33) = 61.2049 2 2

All the evaluations are summarized in the following table, t 5 7 9 11 13 15 17 19 21 23

c, original 0 0.1 0.11 0.4 4.1 9.1 8 4.2 2.3 1.1

c, synthetic 0 0.1 0.11 0.4 4.1 9.1 8 4.2 2.3 1.1

Trap rule 0.1 0.21 0.51 4.5 13.2 17.1 12.2 6.5 3.4


7 25 27 29 31 33 35

0.9 1.75 2.06 2.25 2.32 2.43

0.5902 0.3065 0.1591 0.0826 0.0429 0.0223

1.6902 0.8967 0.4656 0.2417 0.1255 0.0652

The cardiac output can then be computed as C=

5.6 L 60 = 5.48975 61.2049 min

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 I=

∫

30 0

 z  − z / 15 200z  dx e 5 + z 


9

1 εa  10440.1504 17636.0064 19073.5202 19289.3051

n 1 2 4 8

I=

∫

30 0

3 0.1646% 19520.5626 19348.4694

4 0.01412% 19345.7378

2 17.8666% 1219.6400 1426.8693 1473.1478

3 0.9589% 1440.6846 1476.2330

4 0.03823% 1476.7973

 z  − z / 15 200 dx e 5 + z  1 εa  348.0050 1001.7312 1320.5848 1435.0070

n 1 2 4 8

2 11.9724% 20034.6250 19552.6915 19361.2333

d=

19345.74 = 13.0998 1476.797

V=

1476.797(13.0998) = 6448.58 3

T=

6448.58 = 6480.985 0.995

H = 1476.797 − 6480.985(0.0995) = 831.94 24.15 Change of variable: x=

30 + 0 30 − 0 + x d = 15 + 15 x d 2 2

dx =

I=

I=

∫

∫

30 − 0 dx d = 15dx d 2 30 0

1 −1

 z  − z / 15 200 z  dz e 5 + z   15 + 15 x d  −(15+15 xd ) / 15 200(15 + 15 x d )  15 dx d e  5 + (15 + 15 x d ) 

Therefore, the transformed function is


10 (15 + 15 x d ) 2 − (1+ xd ) f ( x d ) = 3000 e 20 + 15 x d Five-point formula: I = 0.236927 f ( −0.90618) + 0.478629 f ( −0.53847) + 0.568889 f (0) + 0.478629 f (0.53847) + 0.236927 f (0.90618) = 0.236927(844.2582) + 0.478629(7601.173) + 0.568889(12415.93) + 0.478629(12217.47) + 0.236927(10852.84) = 19,320.41 I=

I=

∫ ∫

30 0

1 −1

 z  − z / 15 200 dz e 5 + z   15 + 15 x d  −(15+15 xd ) / 15 200  15 dx d e  5 + (15 + 15 x d ) 

Therefore, the transformed function is f ( x d ) = 3000

15 + 15 x d −(1+ xd ) e 20 + 15 x d

Five-point formula: I = 0.236927 f ( −0.90618) + 0.478629 f ( −0.53847) + 0.568889 f (0) + 0.478629 f (0.53847) + 0.236927 f (0.90618) = 0.236927(599.9124) + 0.478629(1097.966) + 0.568889(827.7287) + 0.478629(529.4212) + 0.236927(379.5667) = 1,481.865 d=

19320.41 = 13.038 1481.865

V=

1481.865(13.038) = 6440.185 3

T=

6440.185 = 6472.548 0.995

H = 1481.865 − 6472.548(0.0995) = 837.846 24.16 Trapezoidal rule: I = (30 − 0)

0 + 2(68.924 + 57.481 + 39.845 + 26.026 + 16.549) + 10.372 = 1070.057 2(6)


11 Simpson’s 1/3 rule: 0 + 4(68.924 + 39.845 + 16.549) + 2(57.481 + 26.026) + 10.372 = 1131.098 3(6)

I = (30 − 0)

Simpson’s 3/8 rule: 0 + 3(68.924 + 57.481) + 39.845 8

I = (15 − 0)

39.845 + 3( 26.026 + 16.549) + 10.372 = 785.7375 + 333.6435 = 1119.381 8 24.17 Trapezoidal rule (h = 4): + (30 − 15)

I = ( 20 − 0)

0 + 2( 2 + 4 + 4 + 3.4) + 0 = 53.6 m 2 10

Trapezoidal rule (h = 2): I = ( 20 − 0)

0 + 2(1.8 + 2 + 4 + 4 + 6 + 4 + 3.6 + 3.4 + 2.8) + 0 = 63.2 m 2 20

Simpson’s 1/3 rule: I = ( 20 − 0)

0 + 4(1.8 + 4 + 6 + 3.6 + 2.8) + 2(2 + 4 + 4 + 3.4) + 0 = 66.4 m 2 30

24.18 A table can be set up to hold the values that are to be integrated: y, m 0 2 4 5 6 9

H, m 0.5 1.3 1.25 1.7 1 0.25

U, m/s 0.03 0.06 0.05 0.12 0.11 0.02

UH, m2/s 0.015 0.078 0.0625 0.204 0.11 0.005

The cross-sectional area can be evaluated using a combination of Simpson’s 1/3 rule and the trapezoidal rule: Ac = ( 4 − 0)

0.5 + 4(1.3) + 1.25 1.25 + 4(1.7) + 1 1 + 0.25 + (6 − 4) + (9 − 6) 6 6 2 = 4.633333 + 3.016667 + 1.875 = 9.525 m 2

The flow can be evaluated in a similar fashion:


12

Q = (4 − 0)

0.015 + 4(0.078) + 0.0625 0.0625 + 4(0.204) + 0.11 0.11 + 0.005 + (6 − 4) + (9 − 6) 6 6 2 = 0.259667 + 0.3295 + 0.1725 = 0.761667

m3 s

24.19 A table can be set up to hold the values that are to be integrated: Abscissa (N to S) 0 200 400 600 800 1000 1200 1400 1600 1800 2000

Ordinate (W to E) 0 600 880 950 1100 1600 1900 2100 2300 2600 2950

Abscissa (N to S) 2200 2400 2600 2800 3000 3200 3400 3600 3800 4000 4200

Ordinate (W to E) 3150 3200 3200 3150 3000 2800 2700 2600 2525 2525 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: I = (3600 − 0)[0 + 4(600 + 950 + 1600 + 2100 + 2600 + 3150 + 3200 + 3000 + 2700) + 2(880 + 1100 + 1900 + 2300 + 2950 + 3200 + 3150 + 2800) + 2600]/54 = 7,917,333.33 2600 + 3( 2525 + 2525) + 2430 I = ( 4200 − 3600) = 1,513,500 8 Therefore, the total area is A = 7,917,333.33 + 1,513,500 = 9,430,833 ft 2 24.20 In order to generate the average rate in units of car/min, the integral to be evaluated along with the units is

∫ average rate =

24 hr

0

rate(car / min) dt 24 hr

The integral can be evaluated with the trapezoidal and Simpson’s rules as tabulated below: Time (hr) 0 2 4 5 6

Rate (car/min) 2 2 0 2 6

Method

Integral

Simp 1/3

6.666667

Simp 1/3

4.666667


13 7 8 9 10.5 11.5 12.5 14 16 17 18 19 20 21 22 23 24

7 23 11 4 11 12 8 7 26 20 10 8 10 8 7 3

Simp 3/8 Trap

40.125 11.25

Simp 1/3 Trap Trap

20 15 15

Simp 1/3

43.66667

Simp 1/3

22.66667

Simp 1/3

18.66667

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: Height l, m 0 30 60 90 120 150 180 210 240

Force, F(l), N/m 0 340 1200 1600 2700 3100 3200 3500 3800

l× F(l) 0 10200 72000 144000 324000 465000 576000 735000 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. F = ( 240 − 0) I = (240 − 0)

0 + 4(340 + 1600 + 3100 + 3500) + 2(1200 + 2700 + 3200) + 3800 = 521,600 24

0 + 4(10200 + 144000 + 465000 + 735000) + 2(72000 + 324000 + 576000) + 912000 24 = 82,728,000

The line of action can therefore be computed as


14

d=

82,728,000 = 158.6043 521,600

24.22 The values needed to perform the evaluation can be tabulated: z 60 50 40 30 20 10 0

w(z) 200 190 175 160 135 130 122

ρ gw(z)(60−z) 0 1.862E+07 3.430E+07 4.704E+07 5.292E+07 6.370E+07 7.174E+07

zρ gw(z)(60−z) 0 9.310E+08 1.372E+09 1.411E+09 1.058E+09 6.370E+08 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. f t = (60 − 0)

7.174 + 4(6.37 + 4.704 + 1.862) + 2(5.292 + 3.43) + 0 × 10 7 = 2.54539 × 10 9 18

f t = (60 − 0)

0 + 4(6.371 + 4.112 + 9.31) + 2(10.584 + 13.72) + 0 × 10 8 = 5.59253 × 1010 18

The line of action can therefore be computed as d=

5.59253 × 1010 = 21.97 m 2.54539 × 10 9

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 15 45 75 105 165 255 285 315 345 375

Flow 30 38 82 125 95 20 22 24 35 30

Trap/Simp

Method

5793.75 6600 5175

Simp 3/8 Trap Trap

1320

Simp 1/3

1940

Simp 1/3


15 The summation of the integral estimates is 20,828.75 m3⋅ d/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. Annual Flow = 57.8576 × 60

s min hr d × 60 × 24 × 365 = 1.8246 × 10 9 m 3 min hr d yr

24.24 The integral to be evaluated is h = eab A

∫

t 0

q dt

Simpson’s rules can be used to evaluate the integral, I = (8 − 0)

0.1 + 4(5.32 + 8) + 2(7.8) + 8.03 8.03 + 3(6.27 + 3.54) + 0.2 + (14 − 8) 12 8 = 51.34 + 28.245 = 79.585

We can then multiply this result by the area and the efficiency to give h = 0.45(100,000 cm 2 )79.585 cal/cm 2 = 5,371,988 cal 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 x1 = 0.08 x2 = 0.16 f ' (0) =

f(x0) = 20 f(x1) = 17 f(x2) = 15

− 15 + 4(17) − 3(20) °C = −43.75 0.16 m

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 As (0) = −

dV − 1,963,500 + 4(5,105,100) − 3(9,817,500) (0) = − = 1,374,450 m 2 dz 8

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


16

As ( 4) = −

dV 1,963,500 − 9,817,500 ( 4) = − = 981,750 m 2 dz 8

The other interior points can be determined in a similar fashion As (8) = −

dV 392,700 − 5,105,100 (8) = − = 589,050 m 2 dz 8

As (12) = −

dV 0 − 1,963,500 (12) = − = 245,437.5 m 2 dz 8

For the last point, the second-order backward formula yields As (16) = −

dV 3(0) − 4(392,700) + 1,963,500 (16) = − = −49,087.5 m 2 dz 8

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. V, m3 9817500 5105100 1963500 392700 0

z, m 0 4 8 12 16

c, g/m3 10.2 8.5 7.4 5.2 4.1

As, m2 1374450.0 981750.0 589050.0 245437.5 0

c× As 14019390 8344875 4358970 1276275 0

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

∫

z 0

As ( z ) dz = (16 − 0)

1,374,450 + 4(981,750 + 245,437.5) + 2(589,050) + 0 = 9,948,400 m 3 12

14,019,390 + 4(8,344,875 + 1,276,275) + 2( 4,358,970) + 0 = 81,629,240 g 12 The average concentration can then be computed as

∫

z

0

c=

c( z ) As ( z ) dz = (16 − 0)

∫

Z 0

∫

c( z ) As ( z ) dz Z 0

As ( z ) dz

=

81,629,240 g = 8.205263 3 9,948,400 m

24.27 The integral to be determined is I=

∫

1/2 0

(5e −1.25t sin2πt ) 2 dt

Change of variable: 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 the publisher, 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.

17

x=

0.5 + 0 0.5 − 0 + x d = 0.25 + 0.25 x d 2 2

dx = I=

∫

0.5 − 0 dx d = 0.25dx d 2 1 −1

(5e −1.25( 0.25+ 0.25 xd ) sin2π (0.25 + 0.25 x d )) 2 0.25 dx d

Therefore, the transformed function is f ( x d ) = 0.25 (5e −1.25( 0.25+ 0.25 xd ) sin2π (0.25 + 0.25 x d )) 2 Five-point formula: I = 0.236927 f ( −0.90618) + 0.478629 f ( −0.53847) + 0.568889 f (0) + 0.478629 f (0.53847) + 0.236927 f (0.90618) = 0.236927(0.127087) + 0.478629(2.059589) + 0.568889(3.345384) + 0.478629(1.050662) + 0.236927(0.040941) = 3.431617 Therefore, the RMS current can be computed as I RMS = 3.431617 = 1.852463 24.28 The integral to be determined is I=

∫

1/2 0

(5e −1.25t sin2πt ) 2 dt

Five applications of Simpson’s 1/3 rule can be applied with h = 0.05 as tabulated below: t 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

i2 0 2.106774 6.726726 11.24592 13.7153 13.38154 10.68149 6.820993 3.177481 0.775039 0 Sum →

Simpsons 1/3 rule

0.252563698 1.090428284 1.298715588 0.685715716 0.104627262 3.432050547

Therefore, the RMS current can be computed as 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 the publisher, 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.

18

I RMS = 3.432051 = 1.852579 24.29 I=

∫

1/2 0

n 1 2 4 8

(5e −1.25t sin2πt ) 2 dt 1 εa  0.00000000 3.34538393 3.42772518 3.43142692

2 25.0000% 4.46051190 3.45517226 3.43266084

3 1.9781% 3.38814962 3.43116008

4 0.0198933% 3.43184278

Therefore, the RMS current can be computed as I RMS = 3.431843 = 1.852523 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 di − 0.32 + 4(0.16) − 3(0) = = 1.6 dt 0.2 − 0 For the intermediate equispaced points, we can use centered differences. For example, for the second point di 0.32 − 0 = = 1.6 dt 0.2 − 0 For the last point, we can use a backward difference di 3( 2) − 4(0.84) + 0.56 = =8 dt 0.7 − 0.3 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 x1 = 0.3 x2 = 0.5

f(x0) = 0.32 f(x1) = 0.56 f(x2) = 0.84

f ' (0) = 0.32

2(0.3) − 0.3 − 0.5 2(0.3) − 0.2 − 0.5 2(0.3) − 0.2 − 0.3 + 0.56 + 0.84 = 2.066667 (0.2 − 0.3)(0.2 − 0.5) (0.3 − 0.2)(0.3 − 0.5) (0.5 − 0.2)(0.5 − 0.3)

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

19

t

i

0 0.1 0.2 0.3 0.5 0.7

0 0.16 0.32 0.56 0.84 2

di/dt 1.6 1.6 2 2.066667 3.6 8

VL 6.4 6.4 8 8.266667 14.4 32

24.31 We can solve Faraday’s law for inductance as

∫ L=

t 0

V L dt i

We can evaluate the integral using a combination of the trapezoidal and Simpson’s rules, I = ( 20 − 0)

0 + 4(18) + 29 29 + 3( 44 + 49) + 46 46 + 35 + (80 − 20) + (120 − 80) 6 8 2

+ (180 − 120)

35 + 26 26 + 15 15 + 7 + ( 280 − 180) + ( 400 − 280) 2 2 2 = 336.6667 + 2655 + 1620 + 1830 + 2050 + 1320 = 9811.667

The inductance can then be computed as L=

9811.667 volts ms s = 4905.833 = 4.905833 H 2 A 1000 ms

24.32 The average voltage can be computed as

∫ V =

60 0

i (t ) R (i ) dt 60

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 0 6 12 18 24 30 36 42

i(t) 3600.000 2950.461 2288.787 1726.549 1260.625 878.355 569.294 327.533

R(i) 43669.784 35816.950 27812.788 21006.431 15360.891 10723.694 6968.908 4025.426

i(t) × R(i) 157211223 105676498 63657535 36268640 19364321 9419212 3967357 1318460

Simpson's 1/3 1287149498 456192829 122017054 19048340


20 48 54 60

151.215 41.250 0.000

1871.341 518.872 0.000

282974 21403 0 Sum →

737174 1885144894

The average voltage can therefore be computed as V =

1885144894 = 3.1419 × 10 7 60

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

1 0.0002 + 0.0003683 (0.2 − 0) = 5.683 −5 2 10

At t = 0.4, we add this value to the integral from t = 0.2 to 0.4, V (0.4) = 5.683 +

1 0.0003683 + 0.0003819 (0.4 − 0.2) = 13.185 −5 2 10

The remainder of the values can be computed in a similar fashion, t, s 0 0.2 0.4 0.6 0.8 1 1.2

i, A 0.0002000 0.0003683 0.0003819 0.0002282 0.0000486 0.0000082 0.0001441

Trap

V(t)

0 5.683 7.502 6.101 2.768 0.568 1.523

0 5.683 13.185 19.286 22.054 22.622 24.145

The plot of voltage versus time can be developed as 25 20 15 10 5 0 0

0.2

0.4

0.6

0.8

1

1.2

24.34 The work can be computed as W=

∫

30 0

(1.6 x − 0.045 x 2 ) cos θ dx


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 Simpson’s 1/3 rule. x 0 5 10 15 20 25 30

F(x)

θ (x)

0 6.875 11.5 13.875 14 11.875 7.5

0.5 1.4 0.75 0.9 1.3 1.48 1.5

F(x)cosθ (x) 0 1.168524107 8.414421992 8.62483831 3.744983601 1.076725541 0.530529013 Sum →

Simpson’s 1/3 21.81420 77.76460 14.30402 113.88282


∫

30 0

(1.6 x − 0.045 x 2 ) cos(0.8 + 0.125 x − 0.009 x 2 + 0.0002 x 3 ) dx

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)

0 7.5 15 22.5 30

0 9.46875 13.875 13.21875 7.5

θ (x) 0.8 1.315625 1.325 1.334375 1.85

F(x)cosθ (x)

Trap Rule

0 2.390018 3.376187 3.096162 -2.066927 Sum →

8.962569 21.623271 24.271308 3.859631 58.716779

The finer-segment versions can be generated in a similar fashion. The results are summarized below: Segments 4 8 16

W 58.71678 64.89955 66.41926


∫

30 0

(1.6 x − 0.045 x 2 ) cos(0.8 + 0.125 x − 0.009 x 2 + 0.0002 x 3 ) dx

For the 4-segment Simpson’s 1/3 rule, we can compute values of the integrand at equallyspaced values of x with h = 7.5. The results are summarized in the following table, x 0 7.5 15

F(x) 0 9.46875 13.875

θ (x) 0.8 1.315625 1.325

F(x)cosθ (x) 0 2.390018 3.376187

Simpson’s 1/3 Rule 32.34065


22 22.5 30

13.21875 7.5

1.334375 1.85

3.096162 -2.066927 Sum →

34.23477 66.57542

The finer-segment versions can be generated in a similar fashion. The results are summarized below: Segments 4 8 16

W 66.57542 66.96047 66.92583

24.37 W=

∫

30 0

(1.6 x − 0.045 x 2 ) cos(0.8 + 0.125 x − 0.009 x 2 + 0.0002 x 3 ) dx 1 εa  -31.003903 35.140854 58.716779 64.899549

n 1 2 4 8

2 38.5532% 57.189106 66.575421 66.960473

3 0.9312% 67.201176 66.986143

4 0.0051% 66.982729

24.38 The function to be integrated is W=

∫

30 0

(1.6 x − 0.045 x 2 ) cos(0.8 + 0.125 x − 0.009 x 2 + 0.0002 x 3 ) dx

Change of variable: x=

30 + 0 30 − 0 + x d = 15 + 15 x d 2 2

dx =

30 − 0 dx d = 15dx d 2

Therefore, the transformed function is f ( x d ) = 15(1.6(15 + 15 x d ) − 0.045(15 + 15 x d ) 2 ) × cos(0.8 + 0.125(15 + 15 x d ) − 0.009(15 + 15 x d ) 2 + 0.0002(15 + 15 x d ) 3 ) Two-point formula:  −1   + I = f   3

 1   = 35.64278 + 38.20683 = 73.84961 f   3


23 Three-point formula: I = 0.5555556 f ( −0.7745967) + 0.8888889 f (0) + 0.5555556 f (0.7745967) = 0.5555556(31.49657) + 0.8888889(50.64281) + 0.5555556(7.748426) = 17.4981 + 45.01583 + 4.304681 = 66.8186 The results for the 2- through 6-point formulas are summarized below: points 2 3 4 5 6

W 73.8496 66.8186 66.7734 66.9145 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), W =F

∫

t 0

v(t ) dt

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 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

v 0 4 8 12 16 17 20 25 32 41 52 65 80 97 116 Sum →

Simp 1/3 rule 8 24 34.6667 50.6667 82.6667 130.6667 194.6667 525.3333

Thus, the work can be computed as W = 200 N (525.3333 m) = 105,066.7 N ⋅ m


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 dT − 30 + 4( 44.5) − 3(80) = = −9.2 dt 10 − 0 For the intermediate points, we can use centered differences. For example, for the second point dT 30 − 80 = = −5 dt 10 − 0 We can analyze the other interior points in a similar fashion. For the last point, we can use a backward difference dT 3( 20.7) − 4( 21.7) + 24.1 = = −0.06 dt 25 − 15 All the values can be tabulated as t 0 5 10 15 20 25

T 80 44.5 30 24.1 21.7 20.7

T − Ta 60 24.5 10 4.1 1.7 0.7

dT/dt -9.2 -5 -2.04 -0.83 -0.34 -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. 0

10

20

30

40

50

60

0 -2 -4 -6 -8

y = -0.1618x

-10

R2 = 0.9758

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

I = (0.02 − 0)

37.5 + 4(43 + 60) + 2(52) + 60 40 + 40 40 + 37.5 + (0.05 − 0.02) + (0.25 − 0.05) = 2 2 12 0.8 + 1.1625 + 4.358333 + 5.783333 = 12.10417

24.42 The function to be integrated is  r  Q = 21 −  0  r0 

∫

3

1/ 6

( 2πr ) dr

A plot of the integrand can be developed as 25 20 15 10 5 0 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 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144

Q 25.1896 36.1635 40.9621 43.0705 44.0009 44.4127 44.5955 44.6767 44.7128 44.7289 44.7361 44.7392 44.7407 44.7413 44.7416 44.7417 44.7418 44.7418

Therefore, the result to 4 significant figures appears to be 44.7418. 24.43 The work is computed as 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 the publisher, 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.

26

W=

∫

x 0

F dx

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

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

0 10 28 46 63 82 110 130 Sum→

Integral

Method

1.133333

Simp 1/3

4.583333

Simp 1/3

14.41875 20.13542

Simp 3/8

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 153 185 210 249 261 271 273

0 0.52 1.04 1.75 2.37 3.25 3.83

v 68.26923 54.80769 50.97398 35.93855 16.05181 6.59273 0.30382

a -35.14510 -16.63005 -13.20842 -26.99478 -23.26046 -10.80561 -10.88031

24.45 The distance traveled is equal to the integral of velocity y=

∫

t2 t1

v(t ) dt

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 0 1 2 3 4 5 6 7

Simp 1/3 rule 0 6 34 84 156 250 366 504

19.33333 175.3333 507.3333 1015.333


27 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

664 846 1050 1045 1040 1035 1030 1025 1020 1015 1010 1005 1000 1052 1108 1168 1232 1300 1372 1448 1528 1612 1700 Sum →

1699.333 2090 2070 2050 2030 2010 2105.333 2337.333 2601.333 2897.333 3225.333 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, y= y=

y=

∫

10 0

∫

20

∫

30

10

20

[

11t 2 − 5t dt = 3.66667t 3 − 2.5t 2

[

1100 − 5t dt = 1100t − 2.5t 2

]

20 10

]

10 0

= 3416.667

= 10,250 30

2  50t + 2(t − 20) 2 dt =  t 3 − 15t 2 + 800t  = 13,166.67 3   20

The total distance traveled is therefore 3416.667 + 10,250 + 13,166.67 = 26,833.33. 24.46 6-segment trapezoidal rule: y = (30 − 0)

0 + 2(97.422 + 207.818 + 333.713 + 478.448 + 646.579) + 844.541 = 10,931.25 2(6)

6-segment Simpson’s 1/3 rule: 0 + 4(97.422 + 333.713 + 646.579) + 2(207.818 + 478.448) + 844.541 = 10,879.88 3(6) 6-point Gauss quadrature: y = (30 − 0)


28 y = 10,879.61914 Romberg integration: n 1 2 4 8

1 εa  12668.10909 11339.74974 10995.60824 10908.67956

2 4.0634% 10896.96330 10880.89441 10879.70333

3 0.0098% 10879.82315 10879.62393

4 0.0000291% 10879.62077

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 dε − 0.16 + 4(0.13) − 3(0.1) (0.085) = − = 0.05994 dt 8 For the interior points, second-order centered differences can be used. For example, at the second point at (t = 0.586), dε 0.16 − 0.1 (0.586) = − = 0.05994 dt 8 The other interior points can be determined in a similar fashion. For the last point, the second-order backward formula yields dε 3(0.56) − 4(0.54) + 0.52 (10.097) = − = 0.03988 dt 8 All the results are summarized in the following table. t 0.085 0.586 1.086 1.587 2.087 2.588 3.088 3.589 4.089 4.590 5.090

ε 0.10 0.13 0.16 0.18 0.20 0.23 0.25 0.26 0.30 0.32 0.34

dε /dt t 0.05994 5.591 0.05994 6.091 0.04995 6.592 0.03996 7.092 0.04995 7.592 0.04995 8.093 0.02997 8.593 0.04995 9.094 0.05994 9.594 0.03996 10.097 0.04995

ε 0.37 0.39 0.41 0.43 0.45 0.47 0.50 0.52 0.54 0.56

dε /dt 0.04995 0.03996 0.03996 0.04 0.03996 0.04995 0.04995 0.03996 0.03988 0.03988

The derivatives can be plotted:


29

0.08 0.06 0.04 0.02 0 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

1

R2 = 0.997577

0.8 0.6 0.4 0.2 0 -0.2 0

0.05

0.1

0.15

0.2

The flow can then be evaluated analytically as Q=

∫

0.2 0

2π ( −24.360173r 3 + 0.334701r 2 + 0.904855r ) dr

[

= 2π − 6.09004325r 4 + 0.111567 r 3 + 0.4524275r 2

]

0.2 0

= 0.05809161

m3 s

(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 0 0.025 0.05 0.075 0.1 0.125

v, m/s 0.914 0.89 0.847 0.795 0.719 0.543

2π rv 0 0.139801 0.266093 0.374635 0.451761 0.426471

Simp 1/3 0.006877 0.018470 0.021334


30 0.15 0.175 0.2

0.427 0.204 0

0.402438 0.22431 0 Sum→

0.010831 0.057512

(c)

εt =

0.05809161 − 0.057512 × 100% = 0.998% 0.05809161

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

6

R2 = 0.9939

4 2 0 0.0

0.1

0.2

0.3

0.4

0.5

0.6

The noncore flow can then be evaluated analytically as Qnoncore =

∫

0.5 0.08333

2π (−28.3629r 3 + 4.75757r 2 + 4.723r ) dr

[

= 2π − 7.09072r 4 + 1.585857 r 3 + 2.3615r 2

]

0.5 0.08333

= 2.06379

ft 3 s

This can be added to the core flow to determine the total flow ft 3 ft ft 3 + 5 π (0.083333 ft ) 2 = 2.06379 + 0.109083 = 2.172873 s s s (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. Q = Qnoncore + Qcore = 2.170447

r, ft 0.083333 0.166667 0.250000 0.333333 0.416667 0.500000

v, ft/s 5 4.62 4.01 3.42 1.69 0

2π rv 2.617994 4.838053 6.298893 7.162831 4.42441 0 Sum→

Integral

Method

0.785253

Simp 1/3

1.283144 2.068397

Simp 3/8

This numerical estimate can be added to the core flow to determine the total flow 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 the publisher, 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.

31

Q = Qnoncore + Qcore = 2.068397

ft 3 ft ft 3 + 5 π (0.083333 ft ) 2 = 2.068397 + 0.109083 = 2.17748 s s s

(c)

εt =

2.172873 − 2.17748 × 100% = 0.21% 2.172873

24.50 The following Excel worksheet solves the problem. Note that the derivative is calculated with a centered difference, dV V450 K − V350 K = dT 100 K

24.51 A single application of the trapezoidal rule yields: I = ( 23 − 3)

12.5 + 1.2 = 137 2

A 2-segment trapezoidal rule gives I = ( 23 − 3)

12.5 + 2(1.8) + 1.2 = 86.5 4

A 4-segment trapezoidal rule gives I = ( 23 − 3)

12.5 + 2(3.5 + 1.8 + 1.4) + 1.2 = 67.75 8

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:


32

4

3

2

y = 3.26667E-04x - 2.17200E-02x + 5.33673E-01x 5.82588E+00x + 2.57346E+01

12

2

R = 1.00000E+00

8 4 0 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 I=

60.955 − 137 × 100% = 124.76% 60.955

I=

60.955 − 86.5 × 100% = 41.91% 60.955

I=

60.955 − 67.75 × 100% = 11.15% 60.955

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 1

0.03

0.8 0.02

0.6 0.4

0.01

0.2 0

0 0

20

40

60

80

100


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

0.012 0.008 0.004 0 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 100 200 300

I 0.073599883 0.073632607 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 x1 = 0.01 x2 = 0.03

f(x0) = 900 f(x1) = 480 f(x2) = 270

2(0) − 0.01 − 0.03 2(0) − 0 − 0.03 2(0) − 0 − 0.01 K + 480 + 270 = −52,500 (0 − 0.01)(0 − 0.03) (0.01 − 0)(0.01 − 0.03) (0.03 − 0)(0.03 − 0.01) m The heat flux can be computed as f ' (0) = 900

Heat flux = −0.028

J K W   − 52,500  = 1,470 2 s⋅m⋅K  m m

(b) The heat transfer can be computed by multiplying the flux by the area Heat transfer = 1,470

W m2 ( 200 cm × 50 cm) = 1,470 W m2 10,000 cm 2

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


34

p=

∫

x2 x1

−

8µQ dx πr ( x) 4

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 0 0.02 0.04 0.05 0.06 0.07 0.1

r, m 0.002 0.00135 0.00134 0.0016 0.00158 0.00142 0.002

integrand -7957.75 -38333.2 -39490.3 -19428.1 -20430.6 -31315.3 -7957.75 Sum→

integral

method

-1338.5392

Simp 1/3

-713.9319 -589.0959 -2641.5670

Simp 3/8 Trap

Therefore, the pressure drop is computed as –2,641.567 N/m2. (b) The average radius can also be computed by integration as

∫ r=

x2 x1

r ( x) dx

x 2 − x1

The numerical evaluations can again be determined by a combination of the trapezoidal and Simpson’s rules. x, m 0 0.02 0.04 0.05 0.06 0.07 0.1

r, m integral 0.00200 0.00135 0.00134 5.83E-05 0.00160 0.00158 0.00142 4.61E-05 0.00200 5.13E-05 Sum→ 0.0001557

method

Simp 1/3

Simp 3/8 Trap

Therefore, the average radius is 0.0001557/0.1 = 0.001557 m. This value can be used to compute a pressure drop of ∆p =

dp 8µQ 8(0.005)0.00001 N ∆x = − 4 ∆x = − 0.1 = −2166.95 2 4 dx πr π (0.001557) m

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 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 the publisher, 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.

35

v=

Q 0.00001 0.00001 m = = = 1.31317 2 −6 Ac π (0.001557 ) s 7.6152 × 10

Then the Reynolds number can be computed as Re =

1 × 10 3 (1.31317)0.0031138 = 817.796 0.005

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 0 0.016 0.032 0.048 0.064 0.0747 0.0787 0.0795 0.08

v, m/s 10 9.69 9.3 8.77 7.95 6.79 5.57 4.89 0

integrand 0 1.168974 2.243851 3.173964 3.836262 3.824296 3.305149 2.931144 0 Sum→

integral

method

0.036905

Simp 1/3

0.100138

Simp 1/3

0.040984 0.014259 0.002495 0.000733 0.195514

Trap Trap Trap Trap

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


Numerical Method for engineers-chapter 24

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