sm_ch07

Water Resources Engineering

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7 – 1 Chapter 7 7.2.1. T = 25 years, T d = 30 min. Using equation (7.2.1b), P 30min = 0.51P 15min + 0.49P 60min Also, P 25,60 = aP 2,60 + bP 100,60 . For T = 25 yr., a = 0.293 and b = 0.669 (From Table 7.2.1). Using Figures 7.2.14 (e) and (f), P 2,60 | 1.45 in and P 100,60 | 3.125 in for Chicago, Illinois). Hence, P 25,60 = 0.293(1.45) + 0.669(3.125) = 2.515 in. Similarly, P 25,15 = aP 2,15 + bP 100,15 . Using Figures 7.2.14 (c) and (d), P 2,15 | 0.86 in and P 100,15 | 1.65 in for Chicago, Illinois). Hence, P 25,15 = 0.293(0.86) + 0.669(1.65) = 1.356 in. Thus, P 25,30 = 0.51(1.356) + 0.49(2.515) = 1.924 in 7.2.2. T = 2-, 10-, 25-, 100 year, T d = 15 min From Figures 7.2.14 (c) and (d) (HYDRO-35), P 2,15 | 1.0 in and P 100,15 | 1.78 in (for Memphis, Tennessee). Hence, P 2,15 = 1.0 in, P 100,15 = 1.78 in may be taken. For the remaining return periods, P T,15 = aP 2,15 + bP 100,15 where the values of a and b are given in Table 7.2.1. P 10,15 = 0.496(1.0) + 0.449(1.78) = 1.295 in P 25,15 = 0.293(1.0) + 0.669(1.78) = 1.484 in 7.2.3. T = 2- and 25-years. The intensity-duration-frequency relationship is The intensity-duration-frequency (IDF) curves for Memphis, Tennessee, are determined by using the HYDRO-35 maps. P 2,5 | 0.488 in, P 2,15 | 1.0 in, P 2,60 | 1.76 in. Precipitation values for 10- and 30- minute durations are computed by the interpolation method. P 2,10 = 0.41P 2,5 + 0.59P 2,15 = 0.41(0.488) + 0.59(1.0) = 0.790 in. P 2,30 = 0.41P 2,15 + 0.59P 2,60 = 0.41(1.0) + 0.59(1.76) = 1.448 in. Also, the 25-year return period precipitation depths are computed for the selected durations. P 25yr = aP 2yr + bP 100yr where a = 0.293 and b = 0.669 for T = 25 yr. (Table 7.2.1). P 100,5 | 0.83 in, P 100,15 | 1.775 in, P 100,60 | 3.625 in (from HYDRO-35 maps) P 100,10 = 0.41P 100,5 + 0.59P 100,15 = 0.41(0.83) + 0.59(1.775) = 1.388 in. P 100,30 = 0.41P 100,15 + 0.59P 100,60 = 0.41(1.775) + 0.59(3.625) = 2.867 in. Hence, P 25,5 = 0.293P 2,5 + 0.669P 100,5 = 0.293(0.488) + 0.669(0.83) = 0.698 in P 25,10 = 0.293P 2,10 + 0.669P 100,10 = 0.293(0.790) + 0.669(1.388) = 1.160 in P 25,15 = 0.293P 2,15 + 0.669P 100,15 = 0.293(1.0) + 0.669(1.775) = 1.480 in P 25,30 = 0.293P 2,30 + 0.669P 100,30 = 0.293(1.448) + 0.669(2.867) = 2.342 in P 25,60 = 0.293P 2,60 + 0.669P 100,60 = 0.293(1.76) + 0.669(3.625) = 2.941 in The result is given in tabular form and in Figure as shown below. The rainfall intensities are derived by dividing the rainfall depths by their corresponding duration.
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7 – 2 Derived IDF relations for Memphis, Tennessee. Precipitation (in) Intensity (in/hr) Duration (min) T = 2 yrs T = 25 yrs T = 2 yrs T = 25 yrs 5 0.488 0.698 5.856 8.376 10 0.790 1.160 4.740 6.960 15 1.000 1.480 4.000 5.920 30 1.448 2.342 2.896 4.684 60 1.760 2.941 1.760 2.941 IDF Curves derived for Memphis, Tennessee 1.000 10.000 11 0 1 0 0 Duration (min) Rainfall intensity (in/hr) T = 2 yrs T = 25 yrs 7.2.4. T = 10- and 50-years. The same procedure as in the solution for problem 7.2.3 is followed, in this case for Chicago, Illinois. P 2,5 | 0.433 in, P 2,15 | 0.86 in, P 2,60 | 1.45 in (from HYDRO-35 maps). Precipitation values for 10- and 30-minute durations are computed by the interpolation method. P 2,10 = 0.41P 2,5 + 0.59P 2,15 = 0.41(0.433) + 0.59(0.86) = 0.685 in. P 2,30 = 0.41P 2,15 + 0.59P 2,60 = 0.41(0.86) + 0.59(1.45) = 1.208 in. P 100,5 | 0.78 in, P 100,15 | 1.65 in, P 100,60 | 3.125 in (from HYDRO-35 maps) P 100,10 = 0.41P 100,5 + 0.59P 100,15 = 0.41(0.78) + 0.59(1.65) = 1.293 in. P 100,30 = 0.41P 100,15 + 0.59P 100,60 = 0.41(1.65) + 0.59(3.125) = 2.520 in. The 10- and 50-year return period precipitation depths are computed for the selected durations using equation (7.2.2). P 10yr = aP 2yr + bP 100yr where a = 0.496 and b = 0.449 for T = 10 yr. (Table 7.2.1). Hence, P 10,5 = 0.496P 2,5 + 0.449P 100,5 = 0.496(0.433) + 0.449(0.78) = 0.565 in P 10,10 = 0.496P 2,10 + 0.449P 100,10 = 0.496(0.685) + 0.449(1.293) = 0.920 in P 10,15 = 0.496P 2,15 + 0.449P 100,15 = 0.496(0.86) + 0.449(1.65) = 1.167 in P 10,30 = 0.496P 2,30 + 0.449P 100,30 = 0.496(1.208) + 0.449(2.520) = 1.731 in P 10,60 = 0.496P 2,60 + 0.449P 100,60 = 0.496(1.45) + 0.449(3.125) = 2.122 in Similarly, P 50yr = aP 2yr + bP 100yr where a = 0.146 and b = 0.835 for T = 50 yr. (Table 7.2.1). Hence, P 50,5 = 0.146P 2,5 + 0.835P 100,5 = 0.146(0.433) + 0.835(0.78) = 0.715 in P 50,10 = 0.146P 2,10 + 0.835P 100,10 = 0.146(0.685) + 0.835(1.293) = 1.180 in P 50,15 = 0.146P 2,15 + 0.835P 100,15 = 0.146(0.86) + 0.835(1.65) = 1.503 in P 50,30 = 0.146P 2,30 + 0.835P 100,30 = 0.146(1.208) + 0.835(2.520) = 2.281 in
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7 – 3 P 50,60 = 0.146P 2,60 + 0.835P 100,60 = 0.146(1.45) + 0.835(3.125) = 2.821 in The result is given in tabular form and in Figure as shown below. The rainfall intensities are derived by dividing the rainfall depths by their corresponding duration.
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sm_ch07 - 71 Chapter 7 7.2.1. T = 25 years, Td = 30 min....

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