sm_ch10

Water Resources Engineering

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10 – 1 Chapter 10 10.2.1. T = 50 years. For a return period of 50 years, 02 . 0 50 1 T 1 P P(Q d q 50 ) = 1 – P(Q t q 50 ) = 1 – 0.02 = 0.98 Thus, ± ² ±² » ¼ º « ¬ ª ³ d » » ¼ º « « ¬ ª ³ d 368 . 0 515 . 8 q ln Z P q ln Z P 98 . 0 50 ln ln 50 Q Q V P ± ² Z) ( 368 . 0 515 . 8 q ln Z 98 . 0 50 ) » ¼ º « ¬ ª ³ d ) From Table 10.2.1, Z = 2.055 for ± ² 98 . 0 Z ) Hence, ± ² » ¼ º « ¬ ª ³ ) 368 . 0 515 . 8 q ln 055 . 2 50 q 50 = 17360 m 3 /s. 10.2.2. a) P Q = 8000 m 3 /s, V Q = 3000 m 3 /s. 375 . 0 8000 3000 Q : ± ² 921 . 8 ) 375 . 0 ( 1 8000 ln 2 1 2 2 ln Q » ¼ º « ¬ ª ´ ± ² > @ 132 . 0 1 375 . 0 2 2 ln Q ´ P(Q t 12000) = P[ln Q t ln (12000)] = 1 – P[ln Q d ln (12000)] = » ¼ º « ¬ ª ¸ ¸ ¹ · ¨ ¨ © § d ³ » » ¼ º « « ¬ ª ¸ ¸ ¹ · ¨ ¨ © § d ³ 132 . 0 921 . 8 - (12000) ln Z P 1 - Q ln Z P 1 Q Q ln ln = 1 – P(Z d 1.298) = 1 – 0.903 = 0.097 b) P(Q d q 25 ) = 1 – P(Q t q 25 ) = 1 - 25 1 = 0.96 Thus, Z 132 . 0 921 . 8 - q ln Z P 96 . 0 25 ) » ¼ º « ¬ ª ¸ ¸ ¹ · ¨ ¨ © § d Z = 1.75 for ± ² Z ) = 0.96 Hence, 1.75 = 132 . 0 921 . 8 - q ln 25 q 25 = 14140 m 3 /s Similarly, P(Q d q 100 ) = 1 – P(Q t q 100 ) = 1 - 100 1 = 0.99 Z 132 . 0 921 . 8 - q ln 99 . 0 100 ) » ¼ º « ¬ ª ¸ ¸ ¹ · ¨ ¨ © § )
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10 – 2 Z = 2.33 for ± ² Z ) = 0.99 Hence, 2.33 = 132 . 0 921 . 8 - q ln 100 q 100 = 17460 m 3 /s. 10.2.3. a) P Q = 5000 m 3 /s, V Q = 3000 m 3 /s. 6 . 0 5000 3000 Q : ± ² 363 . 8 ) 6 . 0 ( 1 5000 ln 2 1 2 2 ln Q » ¼ º « ¬ ª ³ P ± ² > @ 307 . 0 1 6 . 0 2 2 ln Q ³ V P(Q t 12000) = P[ln Q t ln (12000)] = 1 – P[ln Q d ln (12000)] = » ¼ º « ¬ ª ¸ ¸ ¹ · ¨ ¨ © § d ´ » » ¼ º « « ¬ ª ¸ ¸ ¹ · ¨ ¨ © § d ´ 307 . 0 363 . 8 - (12000) ln Z P 1 - Q ln Z P 1 Q Q ln ln = 1 – P(Z d 1.858) = 1 – 0.968 = 0.032 b) P(Q d q 25 ) = 1 – P(Q t q 25 ) = 1 - 25 1 = 0.96 Thus, ±² Z 307 . 0 363 . 8 - q ln Z P 96 . 0 25 ) » ¼ º « ¬ ª ¸ ¸ ¹ · ¨ ¨ © § d Z = 1.75 for ± ² Z ) = 0.96 Hence, 1.75 = 307 . 0 363 . 8 - q ln 25 q 25 = 11300 m 3 /s Similarly, P(Q d q 100 ) = 1 – P(Q t q 100 ) = 1 - 100 1 = 0.99 Z 307 . 0 363 . 8 - q ln 99 . 0 100 ) » ¼ º « ¬ ª ¸ ¸ ¹ · ¨ ¨ © § ) Z = 2.33 for ± ² Z ) = 0.99 Hence, 2.33 = 307 . 0 363 . 8 - q ln 100 q 100 = 15590 m 3 /s. 10.3.1. T = 100 years, n = 20 years. 182 . 0 100 1 - 1 - 1 T 1 - 1 - 1 R 20 n ¸ ¹ · ¨ © § ¸ ¹ · ¨ © § The probability that the 25-year flood will occur during the service life of the project is
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10 – 3 558 . 0 25 1 - 1 - 1 T 1 - 1 - 1 R 20 n ¸ ¹ · ¨ © § ¸ ¹ · ¨ © § Thus, the probability that the 25-year flood will not occur during this period is R 1 ± = 1 – 0.558 = 0.442. 10.3.2. T = 25 years, n = 20 years. 558 . 0 25 1 - 1 - 1 T 1 - 1 - 1 R 20 n ¸ ¹ · ¨ © § ¸ ¹ · ¨ © § The probability that the 25-year flood will occur during the service life of the project is the same: 0.558. Thus, the probability that the 25-year flood will not occur during this period is R 1 ± = 1 – 0.558 = 0.442. 10.3.3. T = 2, 5, 10, 25, and 100 years. The corresponding T E ’s are determined using 1 E 1 - T T ln T ± » ¼ º « ¬ ª ¸ ¹ · ¨ © § 44 . 1 1 - 2 2 ) 2 ( T 1 E » ¼ º « ¬ ª ¸ ¹ · ¨ © § ± years 48 . 4 1 - 5 5 ) 5 ( T 1 E » ¼ º « ¬ ª ¸ ¹ · ¨ © § ± years 49 . 9 1 - 10 10 ) 10 ( T 1 E » ¼ º « ¬ ª ¸ ¹ · ¨ © § ± years 50 . 24 1 - 25 25 ) 25 ( T 1 E » ¼ º « ¬ ª ¸ ¹ · ¨ © § ± years 50 . 99 1 - 100 100 ) 100 ( T 1 E » ¼ º « ¬ ª ¸ ¹ · ¨ © § ± years 10.4.1. P = 4.5, S y = 0.6, G s = 0.2.
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sm_ch10 - 10 1 Chapter 10 10.2.1. T = 50 years. For a...

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