AUC AprMay 2011 Given Data b 45 m h 2 h 1 5 3 m 2 m N 0025 1 10000 e S i

Auc aprmay 2011 given data b 45 m h 2 h 1 5 3 m 2 m n

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(AUC Apr/May 2011) Given Data b = 45 m h 2 h 1 = 5-3 m = 2 m N = 0.025 1 10000 e S i Solution: Step-1: To Determine Approximate Length of the Curve: 2 1 1 1 1 1 1 1 2 1 3 2 1 1 2 2 3 45 3 135 2 45 (2 3) 51 135 2.647 51 1 1 1 (2.64) 0.025 10000 0.765 / sec A b h m P b h m A m m P V m S N m Specific Energy E 1 2 1 1 1 2 V E h g
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IV Semester CE2253 / Applied Hydraulic Engineering by P.Dhanabal AP / Civil Page 15 2 1 1 2 2 1 1 2 2 2 2 2 0.765 3 2 9.81 3.029 0.765 135 0.765 135 0.459 / sec 45 5 2 45 (2 5) 55 m V A V A V A V m A b h P b h m Specific Energy E 2 2 2 2 2 2 1 2 1 1 1 1 1 2 2 1 3 2 1 2 2 3 2 0.459 5 5.011 2 9.81 3 5 4 2 2 0.765 3 0.574 / sec 2 2.647 4.091 3.369 2 2 1 ( ) ( ) 1 (3.369) 0.899 / sec 0.025 10000 avg avg avg avg avg avg avg V E h g m h h h m V A V b h V m A b h m m m m V m i n i m
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IV Semester CE2253 / Applied Hydraulic Engineering by P.Dhanabal AP / Civil Page 16 1 2 ( ) 0.0064 0.000064 i i Substitute the above values in the given formulae 2 1 e E E L i i 5.011 3.029 0.0001 0.00004 33.03 L Km 5. The bed width of a rectangular channel is 24 m and the depth of flow is 6m. The discharge in the canal is 86 cumecs. The bed slope of the channel is 1 in 4000. Assume Chezy’s constant C = 60. Determine the slope of the free water surface. (AUC Apr/May 2012) Given Data b = 24 m y = 6 m Q = 86 m 3 /sec 1 4000 S C = 60 Solution: Step-1:To Determine the Slope of the Free water Surface : 0 2 3 1 : f f S S dy dx Q T A g Q A V note V C RS
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IV Semester CE2253 / Applied Hydraulic Engineering by P.Dhanabal AP / Civil Page 17 3 5 0 2 3 5 2 3 86 60 24 6 86 60 24 12.6 4.98 10 2.48 10 1 1 2.48 10 4000 86 1 (24 6) 9.81 f f f f f A S A P S A S S S S dy dx Q T A g 0 2 3 4 4 4 1 2.252 10 2.252 10 177504 0.993940292 1 29292503.04 2.266 10 f S S dy dx Q T A g This the required Slope value for this problem
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IV Semester CE2253 / Applied Hydraulic Engineering by P.Dhanabal AP / Civil Page 18 6. In a rectangular channel of bed width 0.5 m, a hydraulic jump occurs at a point where depth of flow is 0.15 m and Fro ude’s number is 2.5. Determine (1) The specific energy (2) The critical depth (3) The subsequent depths (4) Loss of head (5) Energy dissipated. (AUC Apr/May 2012) Given Data b = 0.5 m 3 /sec y = 0.15 m F = 2.5 Solution: Step-1: To Determine the Specific Energy: 2 2 2 2 : 2 V E y g Q Q y note V V g A : 2.5 0.15 3.033 / sec V A b y F note D y T b gD V F gy V g V m From Equation 1 , we can Get 2 2 2 3.033 0.15 0.619 2 9.81 V E y g m
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IV Semester CE2253 / Applied Hydraulic Engineering by P.Dhanabal AP / Civil Page 19 Step-2: To Determine the Critical Depth: 1 2 3 ( ) c q h g Q A V b d V q V d b b b 2 1 2 3 1 2 3 3.033 0.15 0.455 / sec 0.455 9.81 0.276 c q m q h g m Step-3: To Determine the Subsequent Depth: 1 2 1 1 8 1.0 2 0.15 0.15 1 8(2.5) 1.0 6.14 2 2 0.46 y y F m Step-4: To Determine the loss of Head: 3 2 1 1 2 ( ) 4 0.104 loss y y H y y m
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IV Semester CE2253 / Applied Hydraulic Engineering by P.Dhanabal AP / Civil Page 20 Step-5: To Determine the Energy Dissipated: 1000 9.81 (3.033 0.5 0.15) 0.104 232.08 / sec L loss E g Q H Nm
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IV Semester CE2253 / Applied Hydraulic Engineering by P.Dhanabal AP / Civil
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