Question

11.14 The peak-flow data on an annual basis from Cedar River near Austin, MN, are listed in the following table. Plot the flood-frequency curve on lognormal probability paper. Determine the (a) magnitude of a flood having a return period of 100 years (probability of 1%), and (b) probability of a flow of 100 cfs. Peak- flow cfs 7750 5440 Peak- flow cfs Peak- flow cfs Peak- flow cfs Year Year Year 1991 979 3880 2003 8690 1992 2004 3580 1993 Year 1965 1966 1967 1968 1969 1970 1971 2005 2110 8270 5740 1978 1979 1980 1981 1982 1983 1984 1985 4940 4260 9400 9530 1994 1995 2006 2007 4140 2330 1996 3910 2008 3410 2190 4440 4070 1400 3290 7580 4640 3190 10800 990 1997 1300 2009 1972 9400 5260 4000 8800 7070 7520 6990 5570 2710 2190 2250 2010 1973 3740 3250 2011 12400 4720 3250 1974 1998 1999 2000 2001 2002 2012 1986 1987 1988 1989 1990 2920 4810 1975 1976 1977 2013 3830 3060 2014 2500 1430

Answer 1

solution Given that * The Peak flow data is given in the table. + Normal distribution | Y= logx & Magnitude of an event with retuan period | Y = y + kr55 Re: W- 2.5155A + 0-80285361+ 0.010 32W? 171-43278 W + 0-189 26902 +0.00368w3. wa [km]* Deros OLP Lois

C D E F Mean Std. dev 3.607049 0.267114 A B logx 7750 3.889302 5440 3.735599 3580 3.553883 5260 3.720986 4000 3.60206 8800 3.944483 7070 3.849419 7520 3.876218 6990 3.844477 5570 3.745855 2710 3.432969 2190 3.340444 2250 3.352183 979 2.990783 4940 3.693727 4260 3.62941 9400 3.973128 9530 3.979093 2330 3.367356 990 2.995635 9400 3.973128 3740 3.572872 3250 3.511883 2920 3.465383 3830 3.583199 1430 3.155336 3880 3.588832 2110 3.324282 8270 3.917506 5740 3.758912 4140 3.617 3910 3.592177 1300 3.113943 12400 4.093422 4720 3.673942 3250 3.511883 4810 3.682145 3060 3.485721 8690 3.93902 3410 3.532754 2190 3.340444 4440 3.647383 4070 3.609594 1400 3.146128 3290 3.517196 7580 3.879669 4640 3.666518 3190 3.503791 10800 4.033424 2500 3.39794

(a). Magnitude of a flood haring a return period of loo for loo years P= /100 P = 0.01 Lu = 3.035 ki = 2.35 Y = 3.607049+ 2.52% 0.267114 X = 104-28 * 19054.6685 (6) Probability of a flow of loo cts : tog (1000) =3 3 = 360749 4k,* 0.267114 1 * when Pjos, P is replaced by lop and k., has negative Sign * By hit trail Method we get 4=2.87; W.2.185 and so l.p = 0.0116 Pa 0.988 IP- 98.1.

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