Open Channel Weirs

8/4/2019 Open Channel Weirs

1/19

CE 516

Hydraulic Structures

Hydraulic structures play an integral role in the design and analysis of open chan-nel flows. Weirs and dams are used to store water in reservoirs, gates are usedto regulate the flow within structures, and culverts are used to discharge water

under embankments or roads. In this section well discuss:

Flow measuring structures

Sharp-crested weirs Broad-crested weirs Parshall flumes

In the following discussion, well only consider the design and operation of weirs.Weirs have simple designs, but can cause high head losses and have the potential

for upstream sedimentation. For cases where the sedimentation or head losses area concern (i.e. Wastewater treatment plants and irrigation channels), a Parshallflume can be used. Please see your textbook for the design of Parshall flumes.

Regulation structures

Gates

Discharge structures

Culverts

Open Channel Flow 1 of 19 Hydraulic Structures


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CE 516

Weirs and Flumes

Weirs are used to measure flow and/or control outflow elevations from basinsand channels

The vocabulary well be using includes

sharp-crested (or thin plate) weirs - thin plastic or metal plate that is setvertically across a channel

rectangular or v-notch weirs - two types of sharp crested weirs which describetheir geometry

tailwater - flow downstream of the weir

submerged weir - a weir that has its tailwater elevation at least as high as theweir crest

nappe - jet of water flowing over crest



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CE 516

Rectangular Sharp Crested Weirs

A rectangular weir has a rectangular opening and it can be either suppressed orunsuppressed.

A suppressed (uncontracted) weirs has a rectangular opening spans channel

width a vent is often needed to maintain atmospheric pressure. An unsuppressed(contracted) weirs has a rectangular opening which only spans part of thechannel.



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CE 516

To model the flow through a rectangular weir, the energy equation betweenupstream of the weir crest and at the weir crest is defined with

assumptions: no head losses (E1 = E2) (the means no significant turbulence) water surface elevation remains constant

note: this is physically impossible, but actual errors are small pressure at crest (station 2) = patmo throughout depth



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CE 516

Rectangular Sharp Crested Weirs (continued)

Estimate the theoretical flowrate with the conservation of mass

where the velocity is determined with the conservation of energy equation

substituting, we find the flowrate is only a function of the geometry and theelevation of the water surface above the weir

Discrepancies in the estimated flowrate arise from:

1) pressure distribution at crest is not uniformly atmospheric

2) water surface does not remain uniform as it approaches the crest

3) energy losses due to viscous effects are not negligible



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CE 516

To compensate for the discrepancies, we define a discharge coefficient

Cd =Q

Q=

true flow rate

theoretical flow rate

The true flow rate is estimated with

Q =2

3Cd

2gbH3/2

where

Cd = F(Re,We,H

Hw)

W e is the Weber numberHw is the height of the weir crest from the bottom

Empirical estimates of this discharge coefficient were made by Rouse in 1946

Cd = 0.611 + 0.075H

Hwfor

H

Hw< 5 10

A variation of the discharge coefficient is called the weir coefficient that isdesigned to simplify the flowrate equation and is defined with

Cw =2

3Cd

2g

so the flowrate equation becomes

Q = CwbH3/2

Q = 1.83bH3/2 IfH

Hw< .4 (SI units only)

Please note, it is recommended that you measure H at a distance of 4 5 Hupstream of the gate.



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CE 516

Unsuppressed rectangular weirs

Unsuppressed (or contracted) weirs behave similarly to suppressed weirs with twomajor assumptions

1. venting is not needed2. side contractions recude nappe width

where

Unsuppressed trapezoidal sharp-crested weir (Cipolletti Weir)

for this structure, side contractions do not reduce nappe width



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CE 516

Example:An end-contracted weir of total length 286 ft and crest height 5 ft is used todischarge water without exceeding a head of 2.5 ft from a tank 300 ft wide. Theweir carries piers that are 10 ft clear distance apart and 2 ft wide, to suport the

footway. Determine the discharge. Cd = 0.6. Solution:



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CE 516

V-notch weirs

These weirs are typically used in low flow (Q < 0.28 m3/s or 10 cf s) environ-ments in place of rectangular, because they are more accurate.

Recall the weir energy equation

and theoretical flow rate

where

combining

as V-notch weirs are typically applied in low flow environments, we can neglectthe approach velocity



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CE 516

The actual flow rate for a V-notch weir becomes

Q =8

15Cd

2gtan

2

H5/2

where



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CE 516

Example:Water passes over a rectangular weir of 10 ft width at a depth of 1 ft. If theweir is replaced by an 80o V-notch, determine the depth of water over the notch.Disregard end contractions. Cd notch = 0.59 and Cd rectangular = 0.63. Solution:



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CE 516

Broad-Crested weirs

The variety of broad crested weirs have crests that are signigicantly larger thansharp crested weirs and are capable of handling much larger discharges.

Rectangular broad-crested weirs are designed so that the flow above the weiris at critical flow conditions. The theoretical flow rate is given by

Energy equation for a RBC weir

under critical flow conditions



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CE 516

combining this yields an expression for the critical flow depth

where H is the energy of the upstream flow measured relative to the weir location

the theoretical flow rate over a RBC weir become

To account for the non negligible energy losses over the weir, the actual flow

rate, Q, is given by

Q = Cd

gb

2

3H

32

where the discharge coefficient is estimated with (Chow, 1959)

Cd =0.65

1 + HHw

1/2

These equations are valid 0.08 < h1/L < 0.5. Note that for

h1/L < 0.08 head losses can not be neglectedh1/L > 0.5 the streamlines are not horizontal

The weir discharges freely if



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CE 516

Example:Determine the discharge over a broad-crested weir of 100 ft length. The up-stream water level over the crest is 2 ft and the crest has a height of 2.25 ft.The width of approach channel is 150 ft. Cd = 0.95 Solution:



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CE 516

Gates

Gates are used to regulate flow in open channels

Vertical gates are a vertical plate supported by vertical guides on the channelwalls. These gates experience large hydrostatic pressure forces.

Tainter Radial gates more easily resist the hydrostatic forces and are generallymore economical

As with weirs, define the energy equation upstream and downstream of the gate

and the conservation of mass at the same locations



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CE 516

Combining, the flowrate becomes

Weknow the water depth at the gate, yg, but we dont know y2. As the flowmoves under the gate, the streamline becomes contracted. The coefficient ofcontractions is a coefficient which accounts for the contraction of flow down-stream of the gate

where:

vertical sluice gate:

radial gate:

Given that their will likely be energy losses through the gate, the dischargecoefficient is defined with

combining the flow rate becomes

Q = Cdbyg

2gy1

where



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CE 516

Supercritical discharge from the gates

In cases where the gate discharge is supercritical and the depth of flow down-stream exceeds the depth at the gate opening then the outflow may be sub-merged. In this case the previous equations are not valid.

To characterize submerged flow through a gate

1) define the energy between 1 - 2

2) define the momentum between 2 - 3

wherey is they1 is they2 is they3 is the

Solve 1) and 2) simultaneously for Q



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CE 516

Example:Water is ponded behind a vertical gate to a height of 4 m in a rectangular chan-nel of width 7 m. Calculate the gate opening that will release 40 m3/s throughthe gate. How would this discharge be affected by a downstream flow depth of

3.5 m? Solution:



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CE 516

What happens if downstream depth is 3.5 m?


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Open Channel Weirs

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