PROBLEMS

1. Develop a spread sheet to calculate the mean, standard deviation, and skew coefficient of a series of annual maximum flows. Test your work using the data of Example 6-1 in the text.

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   TBD

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2. The annual maximum flows of a certain stream have been found to be normally distributed with mean 22,500 ft³/s and standard deviation 7500 ft³/s. Calculate the probability that a flow larger than 39,000 ft³/s will occur.

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   The standard unit corresponding to the value of 39,000 ft3 i /s is: z = (39,000 - 22,500)/ (7500) = 2.2. Since the standard unit is positive,the value of 39,000 is located 2.2 standard deviations to the right of the mean. From Table A-5 (Appendix A), for z = 2.2: F (z) = 0.4861. Therefore, the cumulative probability is: 0.5 + 0.4861 = 0.9861. The probability that a flow larger than 39,000 ft3/s will occur is the complementary cumulative probability: 1.0 - 0.9861 = 0.0139; or 1.39 percent.  ANSWER.

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3. The 10-y and 25-y floods of a certain stream are 73 and 84 m³/s, respectively. Assuming a normal distribution, calculate the 50-y and 100-y floods.

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   For return period T = 10 y, the complementary cumulative probability is: 1/T = 0.1. Therefore, the cumulative probability to the right of the mean is: F (z) = 0.5 - 0.1 = 0.4. The corresponding standard unit (Table A-5, Appendix A) is z = 1.281. Likewise, for return period T = 25 y, the corresponding values are: 1/T = 0.04; F (z) = 0.46; and z = 1.751. Therefore: 73 = x̄ + 1.281 s; and 84 = x̄ + 1.751 s. Solving simultaneously: x̄ = 43.02; s = 23.4. For return period T = 50 y: 1/T = 0.02, F (z) = 0.48, z = 2.054. Therefore, the 50-y flood is: Q50 = 43.02 + (2.054 × 23.4) = 91.1 m3/s.  ANSWER. For return period T = 100 y: 1/T = 0.01, F(z) = 0.49, z = 2.327. Therefore, the 100-y flood is: Q100 = 43.02 + (2.327 × 23.4) = 97.5 m/s. ANSWER.

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4. The low flows of a certain stream have been shown to follow a normal distribution. The flows expected to be exceeded 95% and 90% of the time are 15 and 21 m³/s, respectively. What flow can be expected to be exceeded 80% of the time?

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   For 95 percent exceedence probability, the cumulative probability to the left of the mean F(z) is: F(z) = 0.95 - 0.50 = 0.45. The corresponding standard unit (Table A-5, Appendix A) is z = 1.645. Likewise, for 90 percent exceedence probability, F(z) = 0.40, and z = 1.281. Therefore: 15 = x̄ - 1.645 s; and 21 = x̄ - 1.281 s. Solving simultaneously: x̄ = 42.11; s = 16.48. For 80 percent exceedence probability, F(z) = 0.30, and z = 0.842. Therefore, the flow expected to be exceeded 80% of the time is: 42.11 - (0.842 × 16.48) = 28.2 m3/s. ANSWER.

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5. A temporary cofferdam for a 5-y dam construction period is designed to pass the 25-y flood. What is the risk that the cofferdam may fail before the end of the construction period? What design return period is needed to reduce the risk to less than 10%?

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   For a return period T = 25 y, and construction period n = 5 y, the risk of failure of the cofferdam during the construction period (Eq. 6-24) is: R = 1 - [1 - (1/T)]n = 1 - [1 - (1/25)]5 = 0.185 R = 18.5 percent. ANSWER. The design return period needed to reduce the risk to less than 10% (from Eq. 6-24) is: 10/100 = 1 - [1 - (1/T )]5. Solving for T: T = 48 y. ANSWER.

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6. Use the Weibull formula (Eq. 6-26) to calculate the plotting positions for the following series of annual maxima, in cubic feet per second: 1305, 3250, 4735, 5210, 4210, 2120, 2830, 3585, 7205, 1930, 2520, 3250, 5105, 4830, 2020, 2530, 3825, 3500, 2970, 1215.

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   The record length is: n = 20. The values are ranked in descending order, with rank m. Using the Weibull formula, the probability of exceedence (percent) is: P = 100 [m / (n + 1)]. The return period is: T = (n + 1)/m. The calculations are shown in the following table. Year Annual Flood (ft3/s) Ranked values (in descending order) Rank m Probability of exceedence P (percent) Return period T (y) 1 1305 7205 1 4.76 21.00 2 3250 5210 2 9.52 10.50 3 4735 5105 3 14.29 7.00 4 5210 4830 4 19.05 5.25 5 4210 4735 5 23.81 4.20 6 2120 4210 6 28.57 3.50 7 2830 3825 7 33.33 3.00 8 3585 3585 8 38.09 2.63 9 7205 3500 9 42.86 2.33 10 1930 3250 10 47.62 2.10 11 2520 3250 11 52.38 1.91 12 3250 2970 12 57.14 1.75 13 5105 2830 13 61.90 1.62 14 4830 2530 14 66.66 1.50 15 2020 2520 15 71.43 1.40 16 2530 2120 16 76.19 1.31 17 3825 2020 17 80.95 1.24 18 3500 1930 18 85.71 1.17 19 2970 1305 19 90.48 1.11 20 1215 1215 20 95.24 1.05

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7. Use the Gringorten formula to calculate the plotting positions for the following series of annual maxima, in cubic meters per second: 160, 350, 275, 482, 530, 390, 283, 195, 408, 307, 625, 513.

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   The record length is: n = 12. The values are ranked in descending order, with rank m. Using the Gringorten formula (Eq . 6-27 with a = 0.44), the probability of exceedence (percent) is: P = 100 [(m - 0.44) / (n + 0.12)]. The return period is: T = (n + 0.12) / (m - 0.44). The calculations are shown in the following table. Year Annual Flood (ft3/s) Ranked values (in descending order) Rank m Probability of exceedence P (percent) Return period T (y) 1 160 625 1 4.62 21.64 2 350 530 2 12.87 7.77 3 275 513 3 21.12 4.73 4 482 482 4 29.37 3.40 5 530 408 5 37.62 2.66 6 390 390 6 45.87 2.18 7 283 350 7 54.13 1.85 8 195 307 8 62.38 1.60 9 408 283 9 70.63 1.42 10 307 275 10 78.88 1.27 11 625 195 11 87.13 1.15 12 513 160 12 95.38 1.05

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8. Modify the spread sheet of Problem 6-1 to calculate the mean, standard deviation, and skew coefficients of the logarithms of a series of annual maximum flows. Test your work using the results of Example 6-4 in the text.

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   TBD

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9. Fit a log Pearson III curve to the data of Problem 6-6. Plot the calculated distribution on log probability paper, along with the Weibull plotting positions calculated in Problem 6-6.

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   The statistics of the logarithms can be calculated with the computer program developed in Problem 8. The results are: mean: x̄ = 3.491; standard deviation: sy = 0.201; skew coefficient: Csy = -0.382. The calculations are shown in the following table. (1) (2) (3) (4) (5) Return period T (y) Probability of exceedence P (percent) Frequency factor K (Csy = -0.382) y Flood Discharge Q = x (ft3/s) 1.05 95 -1.746 3.140 1380 1.11 90 -1.316 3.226 1683 1.25 80 -0.818 3.327 2123 2 50 0.064 3.504 3192 5 20 0.855 3.663 4603 10 10 1.234 3.739 5483 25 4 1.612 3.815 6531 50 2 1.844 3.862 7278 100 1 2.043 3.902 7980 200 0.5 2.218 3.937 8650 The log Pearson fit to the data is given in Cols. 1 (or 2) and 5. These results are plotted on log probability paper, as shown by the solid line of Fig. M-6-9. Also shown in this figure are the plotting positions calculated inProblem 6. ANSWER. Fig. M-6-9  Flood frequency analysis by Log Pearson III method.

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10. Fit a Gumbel curve to the data of Problem 6-6. Plot the calculated distribution on Gumbel paper, along with the Weibull plotting positions calculated in Problem 6-6.

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    The statistics can be calculated with the computer program developed in Problem 1. The results are: mean x̄ = 3407 ft3/s; standard deviation s = 1492 ft3/s. From Table A-8 (Appendix A), for n = 20, the mean and standard deviation of the Gumbel variate are 0.5236 and 1.0628, respectively. The calculations are shown in the following table. (1) (2) (3) (4) (5) Return period T (y) Probability of exceedence P (percent) Gumbel Variate y Frequency factor K Flood Discharge Q = x (ft3/s) 1.05 95 -1.113 -1.540 1109 1.11 90 -0.838 -1.281 1496 1.25 80 -0.476 -0.941 2003 2 50 0.367 -0.147 3188 5 20 1.500 0.919 4778 10 10 2.250 1.624 5830 25 4 3.199 2.517 7162 50 2 3.902 3.179 8150 100 1 4.600 3.836 9130 200 0.5 5.296 4.490 10,106 The Gumbel fit to the data is given in Cols. 1 (or 2) and 5. These results are plotted in Gumbel (extreme value) paper, as shown by the solid line of Fig. M-6-10. Also shown in this figure are the plotting positions calculated in Problem 6-6. ANSWER. Fig. M-6-10  Flood frequency analysis by GumbelI method.

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11. Develop a spread sheet to read a series of annual maxima, sort the data in descending order, and compute the corresponding plotting positions (percent chance and retum period) by the Weibull and Gringorten formulas.

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    TBD

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12. Given the following statistics of annual maxima for stream X: number of years n = 35; mean = 3545 ft³/s; standard deviation = 1870 ft³/s. Compute the 100-y flood by the Gumbel method.

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    With return period T = 100 y, use Eq. 6-39 to calculate the Gumbel variate: y = 4.6. From Table A-8 (Appendix A), for record length n = 35 y, the mean and standard deviation of the Gumbel variate are: ȳn = 0.5403; and σn = 1.1285. Therefore, the frequency factor (from Eq. 6-40) is: K = (y - ȳn)/ σn = 3.5974. The 100-year flood is: Q100 = x̄ + Ks = 3545 + (3.5974 × 1870) = Q100 = 10,270 ft3/s. ANSWER.

13. Given the following statistics of annual maxima for river Y: number of years n = 45; mean = 2700 m³/s; standard deviation 1300 m³/s; mean of the logarithms = 3.1; standard deviation of the logarithms = 0.4; skew coefficient of the logarithms = - 0.35. Compute the 100-y flood using the following probability distributions: (a) normal, (b) Gumbel, and (c) log Pearson III.

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    (a) For return period T = 100 y, the complementary cumulative probability is: 1/T = 0.01. Therefore, the cumulative probability to the right of the mean is: F(z) = 0.50 - 0.01 = 0.49. The corresponding standard unit (Table A-5, Appendix A) is: z = 2.327. The 100-y flood is: Q100 = 2700 + (2.327 × 1300) = 5725 m3/s. ANSWER. (b) For return period T = 100 y, the Gumbel variate (Eq. 6-39) is: y = 4.6. For record length n = 45, the mean and standard deviation of the Gumbel variate (Table A-8) are 0.5463 and 1.1519, respectively. Therefore, the frequency factor (from Eq. 6-40) is: K = 3.519. The 100-y flood is: Q100 = 2700 + (3.519 × 1300) = 7275 m3/s. ANSWER. (c) For return period T = 100 y and skew coefficient of the logarithms -0.35, the frequency factor is obtained from Table A-6 by linear interpolation: K = 2.066. Therefore: y = 3.1 + (2.066 × 0.4) = 3.9264. The 100-y flood is:Q100 = log-1 (3.9264) = 8441 m3/s. ANSWER.

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14. A station near Denver, Colorado, has flood records for 48 y, with station skew C_sy = - 0.18. Calculate a weighted skew coefficient.

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    The station skew is Csy = -0.18. The quantity G is the absolute value of the station skew: G = 0.18. Using Eqs. 6-34a and 6-34c: A = -0.3156; B = 0.8932. With record length n = 48, the mean square error of the station skew is (Eq. 6-34): (MSE)sy = 0.119. The regional skew near Denver, Colorado, is obtained from Fig. 6-5: Csr = -0.1. When Fig. 6-5 is used to obtain the regional skew, the mean square error of the regional skew is: (MSE)sr = 0.302. Therefore, the weighted skew (Eq. 6-33) is: Csw = [(0.302 × -0.18) + (0.119 × -0.1)] / (0.302 + 0.119) = -0.157. Use C = -0.16. ANSWER.

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15. Determine if the value Q = 13,800 ft³/s is a high outlier in a 45-y flood series with the following statistics: mean of the logarithms = 3.572; standard deviation of the logarithms = 0.215.

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    From Table A-8, the outlier frequency factor for record length n = 45 y is: Kn = 2.727. Using Eq. 6-35, the high outlier threshold (in log units) is: yH = 3.572 + (2.727 × 0.215) = 4.158. Therefore, the high outlier threshold is: QH = log-1 yH = 14,398 ft3/s. Since the discharge Q = 13,800 ft3/s is less than the high outlier threshold, it is not considered to be a high outlier. ANSWER.

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16. Using the Lettenmaier and Burges modification to the Gumbel method, fit a Gumbel curve to the data of Example 6-6 in the text. Plot the calculated distribution on Gumbel paper, along with plotting positions calculated by the Gringorten formula.

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    The statistics of the flood series are: mean x̄ = 1704 m3/s; and standard deviation s = 795 m3/s. Using Eq. 6-45: x = 1704 - [0.78 ln ln (T / (T - 1)) + 0.45] × 795. With record length n = 16 and m = rank (in descending order), the Gringorten formula (Eq. 6-27 with a = 0.44) leads to: P (percent) = 100 [(m - 0.44) / 16.12]; and T = 16.12 / (m - 0.44). The calculations are summarized in the following table. Return period T (y) Flood Discharge Q = x (m3/s) 1.05 656 1.11 827 1.25 1051 2 1574 5 2276 10 2742 25 3330 50 3766 100 4199 200 4630 Year Annual Flood (m3/s) Ranked values (m3/s) Rank m Probability of exceedence P (percent) Return period T (y) 1972 2520 3320 1 3.47 28.79 1973 1850 3170 2 9.68 10.33 1974 750 2520 3 15.88 6.30 1975 1100 2160 4 22.08 4.53 1976 1380 3320 5 28.29 3.54 1977 1910 1910 6 34.49 2.90 1978 3170 1850 7 40.69 2.46 1979 1200 1730 8 46.90 2.13 1980 820 1480 9 53.10 1.88 1981 690 1380 10 59.31 1.69 1982 1240 1240 11 65.51 1.53 1983 1730 1200 12 71.71 1.39 1984 1950 1100 13 77.92 1.28 1985 2160 820 14 84.12 1.19 1986 3320 750 15 90.32 1.11 1987 1480 690 16 96.53 1.04 Using the first table, return period is plotted against flood discharge on Gumbel paper, as shown by the solid line of Fig. M-6-16. ANSWER. Using the second table, return period is plotted against ranked values on Gumbel paper, as shown by the data points of Fig. M-6-16. Fig. M-6-10  Gumbel method: Lettenmeir anf Burges modification.

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