CHAPTER 24
24.1 Simpson’s rules can be implemented as
32.41 20.484 11.926 8 0.15024 ) 0.14046 0.132 ( 3 0.12486 )) 50 ( 100 ( 6 0.12486 ) 0.11904 ( 4 11454 . 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 32.41 3 33.0975 ) 32.581875 ( 4 = − = I
and computing the approximate error as
0.5303% % 100 32.41 32.581875 32.41 = × − = a ε
The computation can be continues as in the following tableau until εa < 0.01%.
1 2 3 n εa 0.5303% 0.0000% 1 33.09750000 32.41000000 32.41000000 2 32.58187500 32.41000000 4 32.45296875 24.3 Change of variable: d d x x x 25 125 2 ) 150 ( 100 2 150 100 + − = − − + − = d d dx dx dx 125 2 ) 150 ( 100− − = = d d d x dx x I 1 (0.132 1.56 10 ( 25 125 ) 2.64 10 ( 25 125 ) )125 1 2 7 4
∫
− − − − + + × − + × + =) ) 125 25 ( 10 64 2. ) 125 25 ( 10 1.56 (0.132 125 ) ( 4 7 2 d d d x x x f = + × − − + + × − − + Two-point formula: 41 . 32 17.49321 14.91679 3 1 3 1 = + = + − = f f I Three-point formula: 41 . 32 10.03934 14.25167 8.11899 ) 18.07082 ( 5555556 . 0 ) 16.03313 ( 8888889 . 0 ) 14.61418 ( 5555556 . 0 ) 7745967 . 0 ( 5555556 . 0 ) 0 ( 8888889 . 0 ) 7745967 . 0 ( 5555556 . 0 = + + = + + = + + − = f f f I
Thus, both give exact results because the errors are 2 point: Et ∝ f(4)(ξ)
3 point: Et ∝ f(6)(ξ)
which are zero for a second-order polynomial. 24.4 Simpson’s rules can be implemented as
6667 . 1991 6667 . 331 415 1245 6 34 ) 32 ( 4 37 ) 40 50 ( 6 37 ) 40 ( 4 52 ) 30 40 ( 8 52 ) 55 35 ( 3 10 ) 0 30 ( = + + = + + − + + + − + + + + − = I
The amount of mass can be computed as
mg 667 . 966 , 7 m min mg 6667 . 1991 min m 4 3 3 = = M
24.5 Newton-Cotes integration formulas can be implemented to evaluate the integral as
5 . 1089 5 . 811 180 81 17 8 48 ) 70 75 ( 3 58 ) 8 20 ( 6 58 ) 45 ( 4 32 ) 4 8 ( 2 32 22 ) 1 4 ( 2 22 12 ) 0 1 ( = + + + = + + + − + + + − + + − + + − = I
The amount of mass in grams can be computed as
g 611 . 19 mg 1000 g min s 60 m min mg 5 . 1089 s m 3 . 0 3 3 = = M
x0 = 0 f(x0) = 0.06 x1 = 1 f(x1) = 0.32 x2 = 3 f(x2) = 0.6 3 . 0 1 . 0 48 . 0 08 . 0 ) 1 3 )( 0 3 ( 1 0 ) 0 ( 2 6 . 0 ) 3 1 )( 0 1 ( 3 0 ) 0 ( 2 32 . 0 ) 3 0 )( 1 0 ( 3 1 ) 0 ( 2 06 . 0 ) 0 ( ' =− + − = − − − − + − − − − + − − − − = f
The mass flux can be computed as
s cm g 10 56 . 4 cm g 10 3 . 0 s cm 10 52 . 1 flux Mass 13 2 4 6 2 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
kg 695 . 517 g 1000 kg m cm 10,000 d s 86,400 yr d 365 ) m 10 6 . 3 ( s cm g 10 56 . 4 transport Mass 2 2 2 6 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 barrels 0415 . 0 ) 10 ( 2 ) 4 . 0 ( 3 ) 7 . 0 ( 4 77 . 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 ) 30 45 )( 20 45 ( 30 20 ) 30 ( 2 05 . 1 ) 45 30 )( 20 30 ( 45 20 ) 30 ( 2 88 . 0 ) 45 20 )( 30 20 ( 45 30 ) 30 ( 2 77 . 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 Centered 60 1.17 0.01 Centered 75 1.35 0.014 Backward The derivatives can be plotted versus time as
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
5196 . 0 153 255 ) 8 . 87 ( 3 ) 6 . 96 ( 4 176 − = − − + − = ε σ d d
For the intermediate points, we can use centered differences. For example, for the second point 8647 . 0 153 255 8 . 87 176 = − − = ε σ d d
94118 . 18 663 765 2678 ) 3380 ( 4 ) 4258 ( 3 = − + − = ε σ d d
All the values can be tabulated as
σ ε dσ/dε 87.8 153 −0.5196 96.6 204 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
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 (ε = σ =
We can then plot the second model
) 1 ( 5719 . 221 ) 1 ( 1 2107 0.003844 0.003844 ) 612 ( 0.003844 − = − − = ε ε σ e e e
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 t
e c=2132.9 −0.3277
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 . 61 2 0223 . 0 0429 . 0 ) 33 35 ( 2 0429 . 0 0826 . 0 ) 31 33 ( 2 11 . 0 1 . 0 ) 7 9 ( 2 1 . 0 0 ) 5 7 ( = + − + + − + ⋅⋅ ⋅ + + − + + − = I
All the evaluations are summarized in the following table,
t c, original c, synthetic Trap rule 5 0 0 7 0.1 0.1 0.1 9 0.11 0.11 0.21 11 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
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 L 5.48975 60 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:
24.13
The extremes for both cases are at 16 and 92 24.14 dx e I z z 5 z 200z 30 0 15 /
∫
− + =1 2 3 4 n εa 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 dx e I z z 5 z 200 30 0 15 /
∫
− + = 1 2 3 4 n εa 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 . 13 797 . 1476 74 . 19345 = = d 58 . 6448 3 ) 0998 . 13 ( 797 . 1476 = = V 985 . 6480 995 . 0 58 . 6448 = = T 94 . 831 ) 0995 . 0 ( 985 . 6480 797 . 1476 − = = H 24.15 Change of variable: d d x x x 15 15 2 0 30 2 0 30 + = − + + = d d dx dx dx 15 2 0 30 = − = dz e z z z I z 5 200 30 0 15 /∫
− + = d x d d d x e dx x x I d 15 ) 15 15 ( 5 15 15 ) 15 15 200( 1 1 15 / ) 15 15 (∫
− + − + + + + =) 1 ( 2 15 20 ) 15 15 ( 3000 ) ( xd d d d x e x x f − + + + = Five-point formula: 41 . 320 , 19 ) 10852.84 ( 0.236927 2217.47) 0.478629(1 0.236927(844.2582) 0.478629(7601.173) 0.568889(12415.93) ) 0.90618 ( 0.236927 ) 0.53847 ( 0.478629 0.236927 ( 0.90618) 0.478629 ( 0.53847) 0.568889 (0) = + + + + = + + − + + − = f f f f f I dz e z z I z 5 200 30 0 15 /
∫
− + = d x d d e dx x x I d 15 ) 15 15 ( 5 15 15 200 1 1 15 / ) 15 15 (∫
− + − + + + =Therefore, the transformed function is
) 1 ( 15 20 15 15 3000 ) ( xd d d d e x x x f − + + + = Five-point formula: 1,481.865 ) 379.5667 ( 0.236927 29.4212) 0.478629(5 0.236927(599.9124) 0.478629(1097.966) 0.568889(827.7287) ) 0.90618 ( 0.236927 ) 0.53847 ( 0.478629 0.236927 ( 0.90618) 0.478629 ( 0.53847) 0.568889 (0) = + + + + = + + − + + − = f f f f f I 038 . 13 865 . 1481 41 . 19320 = = d 185 . 6440 3 ) 038 . 13 ( 865 . 1481 = = V 548 . 6472 995 . 0 185 . 6440 = = T 846 . 837 ) 0995 . 0 ( 548 . 6472 865 . 1481 − = = H 24.16 Trapezoidal rule: 057 . 1070 ) 6 ( 2 372 . 10 ) 549 . 16 026 . 26 845 . 39 481 . 57 924 . 68 ( 2 0 ) 0 30 ( − + + + + + + = = I
Simpson’s 1/3 rule: 1131.098 ) 6 ( 3 372 . 10 ) 026 . 26 481 . 57 ( 2 ) 549 . 16 845 . 39 924 . 68 ( 4 0 ) 0 30 ( − + + + + + + = = I Simpson’s 3/8 rule: 1119.381 333.6435 785.7375 8 372 . 10 ) 549 . 16 026 . 26 ( 3 845 . 39 ) 15 30 ( 8 845 . 39 ) 481 . 57 924 . 68 ( 3 0 ) 0 15 ( = + = + + + − + + + + − = I 24.17 Trapezoidal rule (h = 4): 2 m 6 . 53 10 0 ) 4 . 3 4 4 2 ( 2 0 ) 0 20 ( − + + + + + = = I Trapezoidal rule (h = 2): 2 m 2 . 63 20 0 ) 8 . 2 4 . 3 6 . 3 4 6 4 4 2 8 . 1 ( 2 0 ) 0 20 ( − + + + + + + + + + + = = I Simpson’s 1/3 rule: 2 m 4 . 66 30 0 ) 4 . 3 4 4 2 ( 2 ) 8 . 2 6 . 3 6 4 8 . 1 ( 4 0 ) 0 20 ( − + + + + + + + + + + = = 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 Simpson’s 1/3 rule and the trapezoidal rule: 2 m 525 . 9 1.875 3.016667 4.633333 2 25 . 0 1 ) 6 9 ( 6 1 ) 7 . 1 ( 4 25 . 1 ) 4 6 ( 6 25 . 1 ) 3 . 1 ( 4 5 . 0 ) 0 4 ( = + + = + − + + + − + + + − = c A
s m 0.761667 1725 . 0 3295 . 0 0.259667 2 005 . 0 11 . 0 ) 6 9 ( 6 11 . 0 ) 204 . 0 ( 4 0625 . 0 ) 4 6 ( 6 0625 . 0 ) 078 . 0 ( 4 015 . 0 ) 0 4 ( 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 Simpson’s 1/3 rule for the first 18 segments and the Simpson’s 3/8 rule for the last 3 segments:
33 7,917,333. 2600]/54 2800) 3150 3200 2950 2300 1900 1100 2(880 ) 2700 3000 3200 3150 2600 2100 1600 950 600 ( 4 0 )[ 0 3600 ( = + + + + + + + + + + + + + + + + + + − = I 500 , 513 , 1 8 2430 ) 2525 2525 ( 3 2600 ) 3600 4200 ( − + + + = = I
Therefore, the total area is
2 ft 9,430,833 500 , 513 , 1 33 . 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
hr 24 min) / car ( rate rate average hr 24 0 dt
∫
=The integral can be evaluated with the trapezoidal and Simpson’s rules as tabulated below:
Time (hr) Rate (car/min) Method Integral
0 2
2 2
4 0 Simp 1/3 6.666667
5 2
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:
Height l, m F(l), N/mForce, 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 Simpson’s 1/3 rule.
600 , 521 24 3800 ) 3200 2700 1200 ( 2 ) 3500 3100 1600 340 ( 4 0 ) 0 240 ( − + + + + + + + + = = F 000 , 728 , 82 24 912000 ) 576000 324000 72000 ( 2 ) 735000 465000 144000 10200 ( 4 0 ) 0 240 ( = + + + + + + + + − = I
158.6043 521,600
82,728,000=
=
d
24.22 The values needed to perform the evaluation can be tabulated: z w(z) ρgw(z)(60−z) zρgw(z)(60−z) 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 Simpson’s 1/3 rule.
9 7 2.54539 10 10 18 0 ) 43 . 3 292 . 5 ( 2 ) 862 . 1 704 . 4 37 . 6 ( 4 174 . 7 ) 0 60 ( − + + + + + + × = × = t f 10 8 5.59253 10 10 18 0 ) 72 . 13 584 . 10 ( 2 ) 31 . 9 112 . 4 371 . 6 ( 4 0 ) 0 60 ( − + + + + + + × = × = t f
The line of action can therefore be computed as
m 21.97 10 54539 . 2 10 59253 . 5 9 10 = × × = d
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 Method
15 30 45 38 75 82 105 125 5793.75 Simp 3/8 165 95 6600 Trap 255 20 5175 Trap 285 22 315 24 1320 Simp 1/3 345 35 375 30 1940 Simp 1/3
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.
3 9 m 10 8246 . 1 yr d 65 3 d hr 4 2 hr min 60 min s 60 57.8576 Flow Annual = × × × × = ×
24.24 The integral to be evaluated is dt q A e h ab t 0
∫
=Simpson’s rules can be used to evaluate the integral,
585 . 79 245 . 28 34 . 51 8 2 . 0 ) 54 . 3 27 . 6 ( 3 03 . 8 ) 8 14 ( 12 03 . 8 ) 8 . 7 ( 2 ) 8 32 . 5 ( 4 1 . 0 ) 0 8 ( = + = + + + − + + + + + − = I
We can then multiply this result by the area and the efficiency to give cal 988 , 371 , 5 cal/cm 585 . 79 ) cm 000 , 100 ( 45 . 0 2 2 = = 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 C 75 . 43 16 . 0 ) 20 ( 3 ) 17 ( 4 15 ) 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 2 m 450 , 374 , 1 8 ) 500 , 817 , 9 ( 3 ) 100 , 105 , 5 ( 4 500 , 963 , 1 ) 0 ( ) 0 ( =− =−− + − = dz dV As
For the interior points, second-order centered differences can be used. For example, at the second point at (z = 4 m),
2 m 750 , 981 8 500 , 817 , 9 500 , 963 , 1 ) 4 ( ) 4 ( =− =− − = dz dV As
The other interior points can be determined in a similar fashion
2 m 050 , 589 8 100 , 105 , 5 700 , 392 ) 8 ( ) 8 ( =− =− − = dz dV As 2 m 5 . 437 , 245 8 500 , 963 , 1 0 ) 12 ( ) 12 ( =− =− − = dz dV As
For the last point, the second-order backward formula yields
2 m 5 . 087 , 49 8 500 , 963 , 1 ) 700 , 392 ( 4 ) 0 ( 3 ) 16 ( ) 16 ( =− =− − + =− dz dV As
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 A s, m2 c×As 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 Simpson’s 1/3 rule,
3 z 0 12 9,948,400m 0 ) 050 , 589 ( 2 ) 5 . 437 , 245 750 , 981 ( 4 450 , 374 , 1 ) 0 16 ( ) ( = − + + + + =
∫
As z dz g 81,629,240 12 0 ) 970 , 358 , 4 ( 2 ) 275 , 276 , 1 875 , 344 , 8 ( 4 390 , 019 , 14 ) 0 16 ( ) ( ) ( z 0 = + + + + − =∫
c z As z dzThe average concentration can then be computed as
3 0 0 m g 8.205263 9,948,400 81,629,240 ) ( ) ( ) ( = = =
∫
∫
dz z A dz z A z c c Z s Z s24.27 The integral to be determined is dt t e I 1/2 (5 tsin2 ) 0 2 25 . 1
∫
− = π Change of variable:d d x x x 0.25 0.25 2 0 5 . 0 2 0 5 . 0 + + − = + = d d dx dx dx 0.25 2 0 5 . 0 = − = d d x x dx e I 1 (5 d sin2 (0.25 0.25 )) 0.25 1 2 ) 25 . 0 25 . 0 ( 25 . 1
∫
− + − + = πTherefore, the transformed function is
2 ) 25 . 0 25 . 0 ( 25 . 1 sin2 (0.25 0.25 )) (5 25 . 0 ) (xd e x xd f = − + d π + Five-point formula: 3.431617 ) 0.040941 ( 0.236927 .050662) 0.478629(1 0.236927(0.127087) 0.478629(2.059589) 0.568889(3.345384) ) 0.90618 ( 0.236927 ) 0.53847 ( 0.478629 0.236927 ( 0.90618) 0.478629 ( 0.53847) 0.568889 (0) = + + + + = + + − + + − = f f f f f I
Therefore, the RMS current can be computed as 1.852463
3.431617
RMS = =
I
24.28 The integral to be determined is dt t e I 1/2 (5 tsin2 ) 0 2 25 . 1
∫
− = πFive applications of Simpson’s 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 6.820993 0.4 3.177481 0.685715716 0.45 0.775039 0.5 0 0.104627262 Sum → 3.432050547
1.852579 3.432051 RMS = = I 24.29 dt t e I 1/2 (5 tsin2 ) 0 2 25 . 1
∫
− = π 1 2 3 4 n εa 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 3.43142692Therefore, the RMS current can be computed as 1.852523
3.431843
RMS = =
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 . 1 0 2 . 0 ) 0 ( 3 ) 16 . 0 ( 4 32 . 0 = − − + − = dt di
For the intermediate equispaced points, we can use centered differences. For example, for the second point 6 . 1 0 2 . 0 0 32 . 0 = − − = dt di
For the last point, we can use a backward difference
8 3 . 0 7 . 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) = 0.32 x1 = 0.3 f(x1) = 0.56 x2 = 0.5 f(x2) = 0.84 2.066667 ) 3 . 0 5 . 0 )( 2 . 0 5 . 0 ( 3 . 0 2 . 0 ) 3 . 0 ( 2 84 . 0 ) 5 . 0 3 . 0 )( 2 . 0 3 . 0 ( 5 . 0 2 . 0 ) 3 . 0 ( 2 56 . 0 ) 5 . 0 2 . 0 )( 3 . 0 2 . 0 ( 5 . 0 3 . 0 ) 3 . 0 ( 2 32 . 0 ) 0 ( ' = − − − − + − − − − + − − − − = f
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 6.4 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 14.4 0.7 2 8 32
24.31 We can solve Faraday’s law for inductance as
i dt V L t L 0
∫
=We can evaluate the integral using a combination of the trapezoidal and Simpson’s rules,
667 . 9811 1320 2050 1830 1620 2655 6667 . 336 2 7 15 ) 280 400 ( 2 15 26 ) 180 280 ( 2 26 35 ) 120 180 ( 2 35 46 ) 80 120 ( 8 6 4 ) 49 44 ( 3 9 2 ) 20 80 ( 6 29 ) 18 ( 4 0 ) 0 20 ( = + + + + + = + − + + − + + − + + − + + + + − + + + − = I
The inductance can then be computed as
H 905833 . 4 ms 1000 s A ms volts 4905.833 2 9811.667 = = = L
24.32 The average voltage can be computed as
60 ) ( ) ( 60 0 it R i dt V =
∫
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 1/3
0 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
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
7 10 3.1419 60 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 . 5 2 0.0003683 0.0002 ) 0 2 . 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 . 13 2 0.0003819 0.0003683 ) 2 . 0 4 . 0 ( 10 1 683 . 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 22.622 1.2 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
dx x . x W 30 (1.6 0045 )cos 0 2
∫
− = θWe 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) θ(x) F(x)cosθ(x) Simpson’s 1/3 0 0 0.5 0 5 6.875 1.4 1.168524107 21.81420 10 11.5 0.75 8.414421992 15 13.875 0.9 8.62483831 77.76460 20 14 1.3 3.744983601 25 11.875 1.48 1.076725541 14.30402 30 7.5 1.5 0.530529013 Sum → 113.88282
24.35 The work can be computed as
dx x x x x . x W 30 (1.6 0045 )cos(0.8 0.125 0.009 0.0002 ) 0 3 2 2
∫
− + − + =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) θ(x) F(x)cosθ(x) Trap Rule
0 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 summarized below:
Segments W
4 58.71678 8 64.89955 16 66.41926
24.36 The work can be computed as
dx x x x x . x W 30 (1.6 0045 )cos(0.8 0.125 0.009 0.0002 ) 0 3 2 2
∫
− + − + =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) θ(x) F(x)cosθ(x) Simpson’s 1/3 Rule
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 dx x x x x . x W 30 (1.6 0045 )cos(0.8 0.125 0.009 0.0002 ) 0 3 2 2
∫
− + − + = 1 2 3 4 n εa 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 64.89954924.38 The function to be integrated is
dx x x x x . x W 30 (1.6 0045 )cos(0.8 0.125 0.009 0.0002 ) 0 3 2 2
∫
− + − + = Change of variable: d d x x x 15 15 2 0 30 2 0 30+ + − = + = d d dx dx dx 15 2 0 30 = − =Therefore, the transformed function is
) ) 15 15 ( 0002 . 0 ) 15 15 ( 009 . 0 ) 15 15 ( 125 . 0 8 . 0 cos( ) ) 15 15 ( 045 0 ) 15 15 (1.6( 15 ) ( 3 2 2 d d d d d d x x x x . x x f + + + − + + × + − + = Two-point formula: 73.84961 38.20683 35.64278 3 1 3 1 = + = + − = f f I
Three-point formula: 66.8186 4.304681 45.01583 17.4981 ) 7.748426 ( 5555556 . 0 ) 50.64281 ( 8888889 . 0 ) 31.49657 ( 5555556 . 0 ) 7745967 . 0 ( 5555556 . 0 ) 0 ( 8888889 . 0 ) 7745967 . 0 ( 5555556 . 0 = + + = + + = + + − = f f f I
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),
dt t v F W t ( ) 0
∫
=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 rule 0 0 1 4 8 2 8 3 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 14 116 Sum → 525.3333
Thus, the work can be computed as m N 105,066.7 ) m 3333 . 525 ( N 200 = ⋅ = W
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 . 9 0 10 ) 80 ( 3 ) 5 . 44 ( 4 30 =− − − + − = dt dT
For the intermediate points, we can use centered differences. For example, for the second point 5 0 10 80 30 − = − − = dt dT
We can analyze the other interior points in a similar fashion. For the last point, we can use a backward difference 06 . 0 15 25 1 . 24 ) 7 . 21 ( 4 ) 7 . 20 ( 3 =− − + − = dt dT
All the values can be tabulated as t T T − Ta dT/dt 0 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 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.
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
12.10417 5.783333 4.358333 1.1625 0.8 12 60 ) 52 ( 2 ) 60 43 ( 4 5 . 37 ) 05 . 0 25 . 0 ( 2 5 . 37 40 ) 02 . 0 05 . 0 ( 2 40 40 ) 0 02 . 0 ( = + + + = + + + + − + + − + + − = I
24.42 The function to be integrated is dr r r r Q 32 1 (2 ) 0 6 / 1 0
∫
− = πA 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 43.0705 32 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 44.7417 131072 44.7418 262144 44.7418
dx F W x 0
∫
=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 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 dt t v y t t ( ) 2 1
∫
=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 rule 0 0 1 6 19.33333 2 34 3 84 175.3333 4 156 5 250 507.3333 6 366 7 504 1015.333
8 664 9 846 1699.333 10 1050 11 1045 2090 12 1040 13 1035 2070 14 1030 15 1025 2050 16 1020 17 1015 2030 18 1010 19 1005 2010 20 1000 21 1052 2105.333 22 1108 23 1168 2337.333 24 1232 25 1300 2601.333 26 1372 27 1448 2897.333 28 1528 29 1612 3225.333 30 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,
[
3.66667 2.5]
3416.667 5 11 3 2 100 10 0 2 − = − = =∫
t t dt t t y[
1100 2.5]
10,250 5 1100 2 1020 20 10 − = − = =∫
t dt t t y 67 . 166 , 3 1 800 15 3 2 ) 20 ( 2 50 0 3 20 2 3 30 20 2 = − + = − + =∫
t t dt t t t yThe total distance traveled is therefore 3416.667 + 10,250 + 13,166.67 = 26,833.33. 24.46 6-segment trapezoidal rule:
10,931.25 2(6) 541 . 44 8 ) 579 . 46 6 448 . 78 4 713 . 33 3 818 . 07 2 422 . 7 9 ( 2 0 ) 0 30 ( − + + + + + + = = y
6-segment Simpson’s 1/3 rule:
10,879.88 3(6) 541 . 44 8 ) 448 . 78 4 818 . 07 2 ( 2 ) 579 . 646 713 . 33 3 422 . 7 9 ( 4 0 ) 0 30 ( − + + + + + + = = y
14 10,879.619 = y Romberg integration: 1 2 3 4 n εa 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 . 0 8 ) 1 . 0 ( 3 ) 13 . 0 ( 4 16 . 0 ) 085 . 0 ( =−− + − = dt dε
For the interior points, second-order centered differences can be used. For example, at the second point at (t = 0.586), 05994 . 0 8 1 . 0 16 . 0 ) 586 . 0 ( =− − = dt dε
The other interior points can be determined in a similar fashion. For the last point, the second-order backward formula yields
03988 . 0 8 52 . 0 ) 54 . 0 ( 4 ) 56 . 0 ( 3 ) 097 . 10 ( =− − + = dt dε
All the results are summarized in the following table.
t ε dε/dt t ε dε/dt 0.085 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
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
[
]
ms 5809161 0 . 0 0.4524275 0.111567 6.09004325 2 ) 904855 . 0 334701 . 0 360173 . 24 ( 2 3 0.2 0 2 3 4 0.2 0 2 3 = + + − = + + − =∫
r r r dr r r r Q π π(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 2πrv Simp 1/3 0 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
0.15 0.427 0.402438 0.175 0.204 0.22431 0.010831 0.2 0 0 Sum→ 0.057512 (c) % 998 . 0 % 100 0.05809161 0.057512 0.05809161 = × − = t ε
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
[
]
fts 2.06379 2.3615 1.585857 7.09072 2 ) 723 . 4 4.75757 28.3629 ( 2 3 0.5 0.08333 2 3 4 0.5 0.08333 2 3 noncore = + + − = + + − =∫
r r r dr r r r Q π πThis can be added to the core flow to determine the total flow
s ft 2.172873 109083 . 0 2.06379 ) ft 083333 . 0 ( s ft 5 s ft 2.170447 3 2 3 core noncore + = + = + = =Q Q π Q
(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 2πrv Integral Method 0.083333 5 2.617994 0.166667 4.62 4.838053 0.250000 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
s ft 2.17748 109083 . 0 2.068397 ) ft 083333 . 0 ( s ft 5 s ft 2.068397 3 2 3 core noncore + = + = + = =Q Q π Q (c) % 21 . 0 % 100 2.172873 2.17748 2.172873− × = = t ε
24.50 The following Excel worksheet solves the problem. Note that the derivative is calculated with a centered difference,
K V V dT dV K K 100 350 450 − =
24.51 A single application of the trapezoidal rule yields: 137 2 2 . 1 5 . 12 ) 3 23 ( − + = = I
A 2-segment trapezoidal rule gives
5 . 86 4 2 . 1 ) 8 . 1 ( 2 5 . 12 ) 3 23 ( − + + = = I
A 4-segment trapezoidal rule gives
75 . 67 8 2 . 1 ) 4 . 1 8 . 1 5 . 3 ( 2 5 . 12 ) 3 23 ( − + + + + = = 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 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
% 76 . 124 % 100 60.955 137 60.955 = × − = I % 91 . 41 % 100 60.955 5 . 86 60.955− × = = I % 15 . 11 % 100 60.955 75 . 67 60.955− × = = 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
(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 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) = 480 x2 = 0.03 f(x2) = 270 m K 500 , 52 ) 01 . 0 03 . 0 )( 0 03 . 0 ( 01 . 0 0 ) 0 ( 2 270 ) 03 . 0 01 . 0 )( 0 01 . 0 ( 03 . 0 0 ) 0 ( 2 480 ) 03 . 0 0 )( 01 . 0 0 ( 03 . 0 01 . 0 ) 0 ( 2 900 ) 0 ( ' =− − − − − + − − − − + − − − − = f
The heat flux can be computed as
2 m W 470 , 1 m K 500 , 52 K m s J 028 . 0 flux Heat = − ⋅ ⋅ − =
(b) The heat transfer can be computed by multiplying the flux by the area W 470 , 1 cm 10,000 m cm) 50 cm 200 ( m W 470 , 1 fer Heat trans = 2 × 2 2 =
dx x r Q p x x ( ) 8 2 1 4
∫
− = π µAfter 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 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
1 2 ) ( 2 1 x x dx x r r x x − =
∫
The numerical evaluations can again be determined by a combination of the trapezoidal and Simpson’s rules. x, m r, m integral method 0 0.00200 0.02 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
2 4 4 m N 2166.95 1 . 0 ) 001557 . 0 ( 00001 . 0 ) 005 . 0 ( 8 8 − = − = ∆ − = ∆ = ∆ π π µ x r Q x dx dp p
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
s m 1.31317 10 7.6152 00001 . 0 ) 001557 . 0 ( 00001 . 0 6 2 = × = = = − π c A Q v
Then the Reynolds number can be computed as
796 . 817 005 . 0 0031138 1.31317)0. ( 10 1 Re= × 3 =
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 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 1/3 0.064 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.
