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10.0 Spillway conduitdesign

Hydraulic Analyses for Spillways Association of State Dam Safety Officials

Regional Technical Seminar

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Drop inlet spillway components

Crest

Transition

Conduit

USBR Design of Small Dams

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Possible flow conditions

1. Inlet spillway crest control weir equation for circular

spillway.2. Orifice control in conduit throat orifice equation.3. Full conduit flow energy equation for pipe flow.

4. For cases I and 2 the flow in the conduit is openchannel flow use energy equation with standard stepmethod.

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Rating curve for a drop inlet

spillway

Note point of change from weirflow to orifice flow and fromorifice flow to pipe flow

USBR Design of Small Dams

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Case 1 - Crest control

USBR Design of Small Dams

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Case 2 Tube or orifice control

USBR Design of Small Dams

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Case 3 Full pipe control

USBR Design of Small Dams

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Full flow in conduit

The head at which the

conduit just flows full isrepresented by point h At heads above point h,the conduit flows fullunder pressure

At heads less than h theconduit flows partly full

with controllingconditions dictated bythe transition design

Point h

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What happens when the conduit is sealed?

If the conduit flows at such a stage that the downstream

end flows full, both the inlet and outlet will be sealedTo forestall siphon action by the withdrawal of air fromthe conduit would require an adequate venting system

Unless venting is effected over the entire length ofconduit, it may prove inadequate to preventsubatmospheric pressures along some portion of the

lengthThis is due to the possibility of sealing at any point bysurging, wave action, or turbulence

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Problems with sealing at inlet

Thus, if no venting is provided or if the venting is

inadequate, a make and- break siphon action will attendthe flow in the range of discharges approaching full-flowconditions.

This action is accompanied by erratic discharges, bythumping and vibration, and by surges at the entranceand outlet of the spillway

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Avoiding subatmospheric conditions

Limit the flow area to 75 percent full at the downstream

end at maximum discharge.Under this limitation, air will be able to pass up theconduit from the downstream portal and thus prevent

the formation of subatmospheric pressure along theconduit length.Care must be taken, however, in selecting the vertical

and horizontal curvatures of the conduit profile andalignment to prevent sealing along some portion bysurging or wave action

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Conduit design example

Drop inlet design example

USBR Design of Small Dams (p. 418)

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Conduit design

Need to determine the minimum uniform conduit diameter

to pass the flow from the transition to the outlet with theconduit flowing less than 75 percent full.Procedure:1. Select a trial conduit and throat diameter and find the

corresponding throat location.2. Compute the length from transition throat to outlet.3. Approximate the friction losses assuming the conduit

flows three-fourths full.4. Check the elevation of the invert at the outlet portal

required to pass the design discharge through the

selected size conduit.

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Final check of hydraulics

After the approximate conduit size has been

determined in this manner, it should be checked bycomputing the water surface profile through the conduitby open channel flow computations.

This can be done using HEC-RAS.

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Assumptions

For this problem assume a conduit diameter of 9.0 feet.

From figure above, a radius of 4.5 feet is found to be at6.9 feet below the crest; therefore, the elevation of the9.0-foot-diameter throat is 93.1.

The tunnel length may be scaled or calculated byapproximate methods. In this example the approximatetunnel length is 270 feet.

Design discharge = 2000 cfs.

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Hydraulic calculations (1)

Assuming that the conduit flows 75 percent full,

area = 0.75 (4.5 2) = 47.7 ft 2velocity = 2,000/47.7 = 41.9 ft/shv = 41.9 2/64.4 = 27.3 feet.

From hydraulic tables for partially full pipes, whenconduit flows 75% full, d/D = 0.702,the hydraulic radius R = 0.2964(9.0) = 2.67.

Using a value of n = 0.014 to maximize the losses, byMannings equation.

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Hydraulic calculations (2)

Using a value of n = 0.014 to maximize the losses, by

Mannings equation

h f = 0.04(270) = 10.8 ft

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Invert elevation at downstream portal

The invert elevation at the downstream portal of theconduit will then be equal to:1. The elevation of the throat, plus2. The velocity head at the throat, minus3. The velocity head in the conduit flowing 75 percent

full, minus4. The friction losses in the conduit, minus5. The depth of flow at the downstream portal.The required portal invert elevation for this trial conduitdiameter is approximately93.1 + (1/1.1)(110.0 - 93.1) - 27.3 - 10.8 - 0.702(9.0) =

64.1.

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Conduit diameter check

This elevation is somewhat higher than the established

portal invert elevation, 60.0, actual losses through theconduit will be larger than those estimated because theconduit will flow 75 percent full throughout its length.

Therefore, the 9.0-foot-diameter conduit appears to be,for all practical purposes, the minimum uniformdiameter conduit that will meet the requirements of theproblem.This should be checked using standard step openchannel flow calculations (HEC-RAS?).