University:
Kentucky State University
Course:
COS 232| Programming I
Academic year:

2024
Views:

349

Pages:

26
Author:

Hundebuchesn

24.1 I = 300[0.11854 + 4(0.11943 + 0.15024) + 2(0.12486 + 0.14046) + 0.16134] / 18 = 40.194 Note that using only 3 points also gives exact result because C(x) is 2nd order 2-point formula gives: I = (17.61576 + 23.05343) x 0.5555556 + 19.5 x 0.8888889 = 40.19395 ΔH = 40.194 x 1500 = 60,291 24.2 |E_T| ∝ h^2 |E_S| ∝ h^4 24.3 x = (150 + (-150))/2 + (150 - (-150))/2 * d = 150d dx = 150d*d 2-point formula gives: I = 1996.667 M = 4 * 1996.667 = 9986.667 24.5 Use combination of 2 and 3 point Simpson's Rule I = (9-0)[(10+4(20+40)+2(10+60))/12] + (20-9)[(60+3(72+70)+50)/8] = 246.6667 + 804 = 1050.667 24.6 Use Equation 23.9 x_i-1 | x_i | f(x_i) -----|-----|----- 0 | 0 | 0.1 1 | 1 | 0.4 3 | 3 | 0.9 At x = 0 f'(0) = 0.1[(-1-3)/(0-1)(0-3)] + 0.4[(-3)/(1-0)(1-3)] + 0.9[(-1)/(3-0)(3-1)] = -0.1333 + 0.6 - 0.15 = 0.3167 Mass flux = 0.3167 x 10^-6 x 2 x 10^4 = 0.6334 x 10^-2 Mass = 0.6334 x 10^-2 x (3 x 10^6) x 10^4 x 60 x 60 x 24 x 365 = 5.9428 x 10^12 kg 24.15 Use 5 point Gauss Quadrature I = 200 * ∫₀³ (x³e^(-x/5)) dx I = 0.2369269(694.2578 + 10052.84) + 0.4786287(1401.72 + 12317.47) + 0.5688889(12415.93) = 19320.41 24.16 Trapezoidal Rule gives: I = (30-0)[0 + 2(68.92 + 57.48 + 39.84 + 26.03 + 16.55) + 10.37] / 12 = 1070.057 Simpson's 1/3 Rule I = (30-0)[0 + 4(68.92 + 39.84 + 16.55) + 2(57.48 + 26.03)] / 18 = 1131.098 Simpson's 3/8 Rule I = (30-0)[0 + 3(68.92 + 57.48 + 26.03 + 16.55) + 2(39.84 + 10.37)] / 24 = 1119.381 24.17 With h = 4, (Trapezoidal Rule): I = 20[0 + 2(4 + 4 + 5 + 5) + 10] / 10 = 53.6 With h = 2, (Trapezoidal Rule): I = 20[0 + 2(2 + 4 + 4 + 4 + 3.6 + 3.4 + 2.8) + 10] / 20 = 63.2 With h = 2, (Simpson's 1/3 Rule): I = 20[0 + 4(1.8 + 4 + 6 + 3.6 + 2.9) + 2(2 + 4 + 3.4) + 0] / 18 = 66.4 24.18 Distance (Meters) Ordinate (w/sec) 0 0 200 600 400 880 600 950 800 1100 1000 1200 1200 1400 1400 1900 1600 2200 1800 2400 Use Simpson's 1/3 Rule on first 3 segments and last 3 segments I = (600-0)[0 + 4(600 + 950) + 2(880 + 1100)] / 18 + (1800-1200)[1400 + 4(1900 + 2200) + 2(1600 + 2400)] / 18 = 791733.33 Total = 7,430,833 m³ 24.19 Use combination of Trapezoidal and Simpson's Rule because data are unequally spaced Time after midnight Rate 0 2 2 5 4 8 5 10 6 12 7 12 8.5 10 10 5 12 2 14 0 16 7 17 8 18 11 19 10 20 9 21 9 22 5 24 0 I_1 = (4-0)[(2+4(2)+0)/2] = 6.67 I_2 = (7-4)[(8+2(5+2)+9)/6] = 11 ... I_8 = (24-21)[(9+4(5+2)+0)/2] = 21.75 I = 6.67 + 11 + ... + 21.75 = 107.33 24.20 t F(t) ∆F(t) 0 0 10,500 30 350 35,000 60 1500 450,000 90 2600 312,000 120 3000 450,000 150 3500 525,000 180 3600 964,100 210 3500 735,000 240 3600 864,100

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