sm_ch05

# Water Resources Engineering

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5 – 1 Chapter 5 5.1.1. The following Figure depicts the cross-section of the given channel. 1.5 m 2 m Using Manning’s equation, 1/2 o 2/3 S AR n 1 Q A = 2(1.5) = 3 m 2 P = 2 + 2(1.5) = 5 m R = A/P = 3/5 = 0.6 m Hence, 2 / 1 2/3 ) 0005 . 0 ( (3)(0.6) 0.015 1 Q = 3.181 m 3 /s. 5.1.2. B w = 2.5 m, Q = 3 m 3 /s, S o = 0.0004, n = 0.015. A tolerance of 0.01 is assumed. Using Manning’s equation, 1/2 o 2/3 S AR n 1 Q 2/3 5/3 1/2 y) 2 5 . 2 ( y 14 . 6 (0.00004) 2y 2.5 2.5y (2.5y) 0.015 1 3 ± ¸ ¸ ¹ · ¨ ¨ © § ± This equation is solved using Newton’s iteration method (see appendix A) Let y 1 = 1.5 m 3 / 2 3 / 5 1 )) 5 . 1 ( 2 5 . 2 ( ) 5 . 1 ( 14 . 6 Q ± = 3.873 m 3 /s (> 3.0 m 3 /s) >@ (1.5) 2 2.5 (1.5) 3 (1.5) 6 5(2.5) 3/3.873 1 5 . 1 ) y 2 B ( y 3 y 6 5B Q/Q 1 y y 1 w 1 1 w 1 1 2 ± ± ² ² ± ± ² ² = 1.24. 3 / 2 3 / 5 2 )) 24 . 1 ( 2 5 . 2 ( ) 24 . 1 ( 14 . 6 Q ± = 3.01 m 3 /s (close enough to 3.0 m 3 /s). Thus, the normal depth, y | 1.24 m. 5.1.3. B w = 8 ft, z = 2, Q = 100 ft 3 /s, n = 0.015, S o = 0.0004. The normal depth of flow, y, is determined using Newton iteration method. From Table 5.1.2,

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5 – 2 ±² ± ² 2 w w 2 2 2 w w z 1 2y B zy) y(B 3 z 1 zy 4 z 1 y 6 B 5 2 B y A A 1 y R R 3 2 ³ ³ ³ ³ ³ ³ ³ ³ ³ d d d d ± ² ± ² ± ² 2 2 2 2 2 1 8 2y) y(8 3 2 1 (2)y 4 2 1 y 6 (8) 5 (2)y 2 8 y A A 1 y R R 3 2 ³ ³ ³ ³ ³ ³ ³ ³ ³ d d d d ± ²± ² 4.472y 8 ) 6y (24y 17.889y y 416 . 13 40 4y 8 y A A 1 y R R 3 2 2 2 ³ ³ ³ ³ ³ ³ d d d d (i) Trial 1: Let y 1 = 1.0 ft Substituting the known values in the right hand side of (i), y A A 1 y R R 3 2 d d d d ³ = 1.761/ft A 1 = [8 + 2(1)](1) = 10 ft 2 >@ 2 1 2 1 ) 1 ( 2 8 ) 1 ( ) 1 ( 2 8 R ³ ³ ³ = 0.802 ft Using Manning’s equation: s / ft 149 . 17 ) 802 . 0 )( 10 ( 987 . 1 AR 987 . 1 ) 0004 . 0 ( ) R )( A ( 015 . 0 49 . 1 S AR n 49 . 1 Q 3 3 / 2 2/3 2 / 1 3 / 2 1/2 o 2/3 1 y A A 1 y R R 3 2 Q/Q 1 y y j j 1 j d d d d ³ ´ ´ ³ Hence, 761 . 1 100/17.149 1 1.0 y 2 ´ ´ = 3.743 ft. Trial 2: y 2 = 3.743 ft. Substituting this value of y in the right hand side of (i), y A A 1 y R R 3 2 d d d d ³ = 0.540/ft A 2 = [8 + 2(3.743)](3.743) = 57.964 ft 2 2 2 2 1 ) 743 . 3 ( 2 8 ) 743 . 3 ( ) 743 . 3 ( 2 8 R ³ ³ ³ = 2.343 ft 3 / 2 2 ) 343 . 2 )( 964 . 57 ( 987 . 1 Q = 203.175 ft 3 /s. 540 . 0 5 100/203.17 1 3.743 y 3 ´ ´ = 2.803 ft. Trial 3: y 3 = 2.803 ft y A A 1 y R R 3 2 d d d d ³ = 0.694/ft A 3 = [8 + 2(2.803)](2.803) = 38.138 ft 2 2 3 2 1 ) 803 . 2 ( 2 8 ) 803 . 2 ( ) 803 . 2 ( 2 8 R ³ ³ ³ = 1.857 ft
5 – 3 3 / 2 3 ) 857 . 1 )( 138 . 38 ( 987 . 1 Q = 114.489 ft 3 /s. 694 . 0 9 100/114.48 1 2.803 y 4 ± ± = 2.621 ft. Trial 4: y 4 = 2.621 ft y A A 1 y R R 3 2 d d d d ² = 0.736/ft A 4 = [8 + 2(2.621)](2.621) = 34.707 ft 2 >@ 2 4 2 1 ) 621 . 2 ( 2 8 ) 621 . 2 ( ) 621 . 2 ( 2 8 R ² ² ² = 1.760 ft 3 / 2 4 ) 760 . 1 )( 707 . 34 ( 987 . 1 Q = 100.529 ft 3 /s. 736 . 0 9 100/100.52 1 2.621 y 5 ± ± = 2.614 ft. Trial 5: y 5 = 2.614 ft y A A 1 y R R 3 2 d d d d ² = 0.738/ft A 5 = [8 + 2(2.614)](2.614) = 34.578 ft 2 2 5 2 1 ) 614 . 2 ( 2 8 ) 614 . 2 ( ) 614 . 2 ( 2 8 R ² ² ² = 1.756 ft 3 / 2 5 ) 756 . 1 )( 578 . 34 ( 987 . 1 Q = 100.003 ft 3 /s (which is very close to 100 ft 3 /s) Hence, the normal flow depth can be taken as y = 2.614 ft. 5.1.4. B w = 2.5 m, z = 2, Q = 3 m 3 /s, n = 0.015, S o = 0.0004. As in the solution to problem 5.1.3, the normal depth of flow, y, is determined using Newton iteration method. From Table 5.1.2, ³´ ³ ´ 2 w w 2 2 2 w w z 1 2y B zy) y(B 3 z 1 zy 4 z 1 y 6 B 5 2 B y A A 1 y R R 3 2 ² ² ² ² ² ² ² ² ² d d d d ³ ´ ³ ´ ³ ´ 2 2 2 2 2 1 2.5 2y) y(2.5 3 2 1 (2)y 4 2 1 y 6 (2.5) 5 (2)y 2 2.5 y A A 1 y R R 3 2 ² ² ² ² ² ² ² ² ² d d d d ³ ´³ ´ 4.472y 2.5 ) 6y (7.5y 17.889y y 416 . 13 12.5 4y 2.5 y A A 1 y R R 3 2 2 2 ² ² ² ² ² ² d d d d 3 2 2 26.832y 48.54y 18.75y 71.553y y 54 . 83 25 . 31 y A A 1 y R R 3 2 ² ² ² ² ² d d d d (i) Trial 1: Let y 1 = 1.0 ft Substituting the known values in the right hand side of (i), y

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sm_ch05 - 51 Chapter 5 5.1.1 The following Figure depicts...

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