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CHAPTER 2 HYDRAULICS
Review Question Pace References
(16) 49
(17) 49
(18) 50
(19) 50
(20) 51
(1) 27 (6) 31 (11) 37
(2) 28 (7) 33 (12) 39
(3) 28 (8) 34 (13) 41
(4) 28 (9) 35 (14) 45-47
(5) 30 (10) 36 (15) 49
(21) 51 (22) www.iihs.uiowa.edu/projects/index.html
(23) www.usbr.gov/wrrl (24) www.envirosources.com
Solutions to Practice Problems
1. P = 0.43 x h (Equation 2-2b)
P = 0.43 x 50 ft = 22 psi at the bottom of the reservoir
P = 0.43 x (50 - 30) = 0.43 x 20 ft = 8.6 psi above the bottom
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8. Q = A1 x V1 = A2 x V2 (Eq. 2-5)
Since A = πD2/4, we can write
D12 x V1 = D22 x V2 and V2 = V1 x (D12 /D22)
In the constriction, V2 = (2 m/s) x (4) = 8 m/s
11. p1/w + v12/2g = p2/w + v22/2g (Eq. 2-8)
A1 = π(0.300)2/4 = 0.0707 m2 A2 = π(0.100)2/4 = 0.00785 m2
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12. From Figure 2.15, with Q = 200 L/s and D = 600 mm, read S = 0.0013.
13. hL = 2.3 x 20 = 46 ft and S = 46/5280 = 0.0087 (where 1 mi = 5280 ft)
14. S = 10/1000 = 0.01
15. Use (Eq. 2-10): Q = C x A2 x {(2g(p1 – p2)/w)/(1 - (A2/A1)2}1/2
16. Use (Eq. 2-10): Q = C x A2 x {(2g(p1 – p2)/w)/(1 - (A2/ A1 )2)}1/2
17. Use Manning's nomograph (Figure 2.21): With D = 800 mm = 80 cm, and
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20. Q = 7 mgd = 7,000,000 gal/day x 1 day/1440 min ≈ 4900 gpm
21. For full-flow conditions, with D = 300 mm and S = 0.02, read from
23. For full-flow conditions, from Fig. 2.21 read Q = 0.55 m3/s = 550 L/s. From Fig.
25. Q = A x V = 2 x 0.75 x 25/75 = 0.5 m3/s = 500 L/s
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