Chapter 9, Problem 9.2.2P

Find the flowrate (in cubicmeter/ second). Discharge y1 =1.6 m Pw=1 m Broad-crested weir

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Note:hand written solution should be avoided.

Subject: Hydraulics 2. Please show your complete solution.

: Water flows over a smooth submerged object in a channel of width 1.2 m. The water depth upstream of the object is 2.2 m and the water depth downstream is 0.5 m. i) Use the Bernoulli and Continuity equations to calculate the discharge within the channel, and sketch the energy grade line and hydraulic grade line for this problem. ii) The measured discharge in the channel is 2.4 m³/s. Calculate the discharge coefficient, and describe what this physically represents. iii) A pitot tube (or stagnation tube) is placed in the downstream part of the channel, where the water depth is 0.5 m. Calculate the height to which water will rise within this pitot tube (above the depth of water in the channel).

Example 12.3. A small reservoir has a spillway crest elevation 200.0 m. Above this elevation, the storage and outflow from the reservoir can be expressed as: Storage: S=36000 + 18000y m³, Outflow Q=10y m³/s, where y is height of water above spillway crest in m. Route an inflow hydrograph which can be approximated by a triangle as: 1=0 at t=0, 1=30 m³/s at t=6 h and I = 0 at t = 26 h Assume at t= 0.y=0 and use At = 2 h

The three-reservoir system is shown below: A The elevation of Reservoir at A is 130m and at C is 30m. The flow rate at Pipe 1 from Reservoir A is 2.1 m³/s. All pipes have a length of 500 meters, friction factor f = 0.02, and sizes namely: Pipe Diameter 1 (mm) = 600mm Pipe Diameter 2 (mm) = 500mm Pipe Diameter 3 (mm) = 400mm 2 B Show a solution sheet containing the following: 1. Headloss at Pipe 1 2. Headloss at Pipe 3 3. Flow rate at Pipe 3 4. Flow rate at Pipe 2 5. Headloss at Pipe 2 6. Elevation of Reservoir B

Calculate the discharge through the channel and the floodway (shown in the Figure below) for steady uniform flow, with S = 0.0009 and y = 8 ft. n=0.025 12 m 5 m n=0.040 POPSPAREN [Σ, Pn] + 120 m Side slopes 1:1 Use the following equivalent channel roughness formula due to Pavlovskii (1931)* ne = For 25,000 cfs flow through the channel above, find the depth of flow in the floodway when the slope of the energy grade line is 0.0004.