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Change of the flow state

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Presentation on theme: "Change of the flow state"β€” Presentation transcript:

1 Change of the flow state
Hydraulic Drop: A rapid change in the depth of flow from high stage to a low stage will result in a steep depression in the water surface. Such a phenomenon is generally caused by a abrupt change in the channel slope or cross section. Free overfall: is a special case of the hydraulic drop. It occurs where the bottom of a flat channel is discontinued. Hydraulic Jump: when the rapid change in the depth of flow is from a low stage to a high stage the result is usually an abrupt rise of water surface. It occurs frequently in a canal below a regulating sluice, at the foot of a spillway, or at the place where a steep channel slope suddenly turns flat.

2 Hydraulic drop Example
A broad-crested weir is placed in a channel of width 𝑏. If the upstream depth of flow is 𝑦 1 and the upstream velocity head of the flow and frictional losses can be neglected, develop a theoretical equation for the discharge in terms of the upstream depth of flow Weirs are commonly used to alter the flow of rivers to prevent flooding, measure discharge, and help render rivers navigable

3 Using specific energy at critical depth, then 𝐸 𝑐 = 3 2 𝑦 𝑐 = 𝑦 1 or
Bernoulli energy equation between points 1 and 2: 𝑣 𝑔 + 𝑦 1 = 𝑣 𝑐 2 2𝑔 + 𝑦 𝑐 = 𝐸 𝑐 For a deep relatively slowly moving flow in the upstream section 𝑣 𝑔 β‰ͺ 𝑦 1 and hence 𝐸 𝑐 = 𝑦 1 Using specific energy at critical depth, then 𝐸 𝑐 = 3 2 𝑦 𝑐 = 𝑦 1 or 𝑦 𝑐 = 2 3 𝑦 1 Using 𝑦 𝑐 = π‘ž 2 𝑔 then 𝑦 𝑐 3 = 𝑦 = π‘ž 2 𝑔 Or π‘ž 2 𝑔 = ( 2 3 ) 3 𝑦 1 3 Therefore 𝑄=𝑏 𝑔 𝑦 =3.09𝑏 𝑦 The foregoing example demonstrates an important practical application of the concepts of specific energy and critical depth; i.e., many structural means of flow measurements are based on the flow passing through critical depth since for this depth of flow there is an explicit relationship between the depth of flow and the flow rate.

4 Hydraulic Jump Hydraulic jump is formed in a channel whenever supercritical flow changes to subcritical flow Jump location, sharp discontinuity in the water surface and considerable amount of energy is dissipated due to turbulence y1 and y2 are called sequent depths (Chaudhry, 2008)

5 Hydraulic Jump The sluice gate is often used in hydroelectric facilities to control the flow rate of water from the dam onto a spillway Specific force at section 1 is equal to specific force at section 2 𝑄 2 𝑔 𝐴 𝑧 1 𝐴 1 = 𝑄 2 𝑔 𝐴 𝑧 2 𝐴 2 𝑦 2 𝑦 1 = 1 2 βˆ’ 𝐹 π‘Ÿ1 2 If the flow depth and flow velocity on one side of the jump are known then their values on the other side can be determined (continuity equation) (Chaudhry, 2008)

6 Example on Hydraulic Jump
The discharge in a 20-m wide, rectangular, horizontal channel is 80 m3/s at a flow depth of 0.5 m upstream of a hydraulic jump. Determine the flow depth downstream of the jump and the head losses in the jump.

7 Example solution 𝑉 1 = 80 0.5Γ—20 =8 π‘š 𝑠𝑒𝑐 𝐹 π‘Ÿ1 = 8 9.81Γ—0.5 =3.612
𝐹 π‘Ÿ1 = Γ—0.5 =3.612 𝑦 2 𝑦 1 = 1 2 βˆ’ 𝐹 π‘Ÿ1 2 𝑦 2 = Γ— βˆ’1 =2.316 π‘š Head losses = βˆ†πΈ=𝐸 1 βˆ’ 𝐸 2 = 𝑦 𝑉 𝑔 βˆ’ 𝑦 2 βˆ’ 𝑉 𝑔 βˆ†πΈ= Γ—9.81 βˆ’2.316βˆ’ 1 2Γ— Γ— =1.294 π‘š

8 Hydraulic Jump at a Sluice Gate Outlet
Hydraulic jump is formed just downstream of the gate Combined used of the specific energy and specific force The thrust on the gate 𝐹 𝑔 =𝛾 𝐹 𝑠1 βˆ’ 𝐹 𝑠2 and the energy losses in the jump = 𝐸 2 βˆ’ 𝐸 3 (Chaudhry, 2008)

9 Hydraulic Jump at a Sluice Gate Outlet Example
A hydraulic jump is formed just downstream of a sluice gate located at the entrance of a channel. There is a constant-level lake upstream of the sluice gate. The flow depth and velocity in the channel downstream of the jump are 5.2 ft and 4.3 ft/sec, respectively. Determine the water level in the lake. Assume the losses for flow through the gate are negligible. Assume rectangular cross section and horizontal channel bed.

10 Solution 𝐹 π‘Ÿ3 = 𝑉 3 𝑔 𝑦 3 = Γ—5.2 =0.3323 𝑦 2 𝑦 3 = 1 2 βˆ’ 𝐹 π‘Ÿ3 2 𝑦 2 = Γ— βˆ’1 =0.968 𝑓𝑑 π‘ž=𝑉×𝑦=4.3Γ—5.2= 𝑓𝑑 3 𝑠𝑒𝑐 /𝑓𝑑 𝑉 2 = π‘ž 𝑦 2 = = 𝑓𝑑 𝑠𝑒𝑐 Since losses through gate are negligible then: 𝐸 1 = 𝐸 2 𝑦 1 + ( π‘ž 𝑦 1 ) 2 2𝑔 = 𝑔 𝑦 1 3 βˆ’ 𝑦 =0, with trial and error solution 𝑦 1 =9.158 𝑓𝑑

11 Force on a Sluice Gate Example
A sluice gate is used to control the water flow rate over a dam. The gate is 20 ft wide, and the depth of the water above the bottom of the sluice gate is 16 ft. The depth of the water upstream of the gate is 20 ft, and the depth downstream is 3 ft. Estimate the flow rate under the gate and the force on the gate. The water density is 62.4 lbm/ft3

12 Force on a sluice gate Example
Assumptions: Velocity profiles are uniformly distributed. Streamlines are straight at station 1 and 2. Viscous effects are negligible. 𝜌=62.4 π‘™π‘π‘š 𝑓𝑑 3 =1.94 𝑠𝑙𝑒𝑔𝑠 𝑓𝑑 3 , π‘Žπ‘›π‘‘ 𝛾=62.4 𝑙𝑏𝑓 𝑓𝑑 3

13 The force diagram shows forces due to pressure and the force on the gate.
The momentum diagram shows an influx and outflux of momentum. The Bernoulli equation between points a and b along streamline is: 𝑝 π‘Ž 𝛾 + 𝑧 π‘Ž + 𝑣 𝑔 = 𝑝 𝑏 𝛾 + 𝑧 𝑏 + 𝑣 𝑔 The piezometric pressure is constant across sections 1 and 2, so 𝑝 π‘Ž 𝛾 + 𝑧 π‘Ž = 𝑑 1 and 𝑝 𝑏 𝛾 + 𝑧 𝑏 = 𝑑 2 Combine Bernoulli and continuity equations: 2𝑔( 𝑑 1 βˆ’ 𝑑 2 )= 𝑣 2 2 βˆ’ 𝑣 1 2 = 𝑣 βˆ’ 𝑑 𝑑 1 2 𝑣 2 = 1 1βˆ’ 𝑑 𝑑 𝑔( 𝑑 1 βˆ’ 𝑑 2 )

14 𝑣 2 = 1 1βˆ’ Γ— βˆ’ 3 =33.5 𝑓𝑑/𝑠 𝑣 1 = 𝑑 2 𝑑 1 𝑣 2 =33.5Γ— 3 20 =5.02 𝑓𝑑 𝑠 𝑄= 𝑣 2 Γ— 𝑑 2 Γ—π‘€π‘–π‘‘π‘‘β„Ž=33.5Γ—3Γ—20=2010 𝑓𝑑 3 𝑠 Sum of the forces from force diagram 𝐹 π‘₯ = 𝐹 1 βˆ’ 𝐹 2 βˆ’ 𝐹 𝐺 From equation for force on planar surface (F=pA) 𝐹 1 = 𝛾 𝑑 𝑑 1 𝑀 , 𝐹 2 = 𝛾 𝑑 𝑑 2 𝑀 (w is width) Force on sluice gate: 𝐹 𝐺 = 𝛾 2 𝑀 𝑑 1 2 βˆ’ 𝑑 πœŒπ‘„ 𝑣 1 βˆ’ 𝑣 2 𝐹 𝐺 = 62.4 𝑙𝑏𝑓/ 𝑓𝑑 3 2 Γ—20 𝑓𝑑× βˆ’ 𝑓𝑑 𝑠𝑙𝑒𝑔 𝑓𝑑 3 Γ—2010 𝑓𝑑 3 𝑠 Γ— 5.02βˆ’33.5 𝑓𝑑 𝑠 =1.33Γ— 10 5 𝑙𝑏𝑓× 1 π‘‘π‘œπ‘› 2000 𝑙𝑏𝑓 =66.5 π‘‘π‘œπ‘›π‘ 


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