001 02 l the chief effect is due to turbulent

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Unformatted text preview: .33 H Y 0.08 0.56 (10.58) Surface roughness is not a significant factor here. For H/L 0.08 there is large scatter ( 10 percent) in the data. For H/L 0.33 and H/Y 0.56, Cd increases up to 10 percent due to each parameter, and complex charts are needed for the discharge coefficient [19, chap. 6]. EXAMPLE 10.9 A weir in a horizontal channel is 1 m high and 4 m wide. The water depth upstream is 1.6 m. Estimate the discharge if the weir is (a) sharp-crested and (b) round-nosed with an unfinished cement broad crest 1.2 m long. Neglect V2/(2g). 1 Solution Part (a) We are given Y 1 m and H Y 1.6 m, hence H 0.6 m. Since H weir is “wide.” For a sharp crest, Eq. (10.56) applies: Cd 0.564 0.0846 0.6 m 1m b, we assume that the 0.615 Then the discharge is given by the basic correlation, Eq. (10.55): Q Cd b gH3/2 (0.615)(4 m) (9.81 m/s2)(0.6 m)3/2 3.58 m3/s We check that H/Y 0.6 2.0 for Eq. (10.56) to be valid. From continuity, V1 3.58/[(4.0)(1.6)] 0.56 m/s, giving a Reynolds number V1H/ 3.4 E5. Part (b) 0.001 0.2 /L 0.001 0.2 0.0024 m 1.2 m 1/2 Then Eq. (10.57) predicts the discharge...
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