ce 3205 water and environmental engineering

CE 3205 Water and Environmental

Engineering

Spillways

SPILLWAY

• A spillway is a structure used to provide for the controlled release of flows from a dam or levee into a downstream area, typically being the river that was dammed.

• to prevent overtopping and possible failure of the dam.

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Four Mile Dam, Australia – Ogee Spillway

Upper South Dam, Australia – Ogee Spillway

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4Source:http://www.leanhtuan.com/

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Hoover Dam – Spillway Crest

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Hoover Dam – Spillway

New Cronton Dam NY – Stepped Chute Spillway

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Sippel Weir, Australia – Drop Spillway

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Four Mile Dam, Australia – Ogee Spillway

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Upper South Dam, Australia – Ogee Spillway

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Itaipu Dam, Uruguay – Chute Spillway

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Itaipu Dam – Flip Bucket

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• Common type of spillways:

1) Free over fall/straight drop spillways

2) Overflow or ogee spillways.3) Chute spillways4) Siphon saddle spillway

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Free overfall or straight drop spillway

•In this type, water drops freely from the crest.

•Occasionally the crest is extended in the form of overhanging lip to direct small discharges away from the face of overfall section.

Ogee or overflow spillway

•The Ogee spillway is generally provided in rigid dams and forms a part of the main dam itself if sufficient length is available. •The overflow type spillway has a crest shaped in the form of an ogee or S-shape.•The upper curve at the crest may be made either larger or sharper than the nappe.

Chute spillway

•chute spillways are used in flow ways where water is to be lowered from one level to another and where it is desirable to avoid a stilling basin. •These are mostly used with earth dams and have the following merit.

It can be provided on any type of foundations.Simplicity of design.However this type of spillway should not be provided where too many bends are to be given as per topography.

Baffle apron or chute spillway

Saddle spillways

•A siphon spillway is a closed conduit system formed in the shape of an inverted U.

•This type of siphon is also called a Saddle siphon spillway.

•Siphonic action takes place after the air in the bend over the crest has been exhausted.

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Required spillway capacity

• Spillway capacity should be equal to the max. outflow rate determined by flood routing. The following data are required for the flood routing.

I. Inflow flood hydrograph-Indicates rate of inflow respect to time.

II. Reservoir capacity curve-indicates the reservoir storage at different reservoir elevations.

III.Outflow discharge curve-indicates the rate of outflow through spillways at different reservoir elevations.

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Overflow Spillway

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Basic equation flow over weirs,WhereQ=discharge m3/sCd=coefficient of dischargeLe=effective lengthHe=actual effective head

Hd=design headHa=head due to velocity of approach (sometimes neglected)

eape HKKNLL )*(' 2

23 /eed HLCQ

ade HHH

Le = effective width of crestL’ = net width of crest(clear waterway x no.of spans)N = number of piersKp = pier contraction coefficientKa = abutment contraction coefficient

Contraction CoefficientsTable 1: Pier Contraction Coefficient (Kp)

Table 2: Abutment Contraction Coefficient (Ka)

*Pier contraction coefficient depends on several factors such as shape and location of pier nose, thickness of piers and velocity of approach.*Abutment contraction coefficient depends on factors such as shape of abutment and velocity of approach.

• Design head, Hd

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23 /

edd LC

QH

•Downstream profile

•d/s profile of spillway can be represented by

x,y= coordinates of the point on the spillway surfaceHd=design headK,n= constant, depend on inclination of the upstream face of spillway

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Different inclination of upstream face of spillway

*For overspillway/ogee, the upstream face is vertical

•The slope of the d/s face of the overflow dam usually varies in the range of 0.7:1 to 0.8:1

•Z is total fall from the upstream water level to the floor level

•P is height of spillway crest above the bed.•Y depth of flow at toe•R is radius•V is velocity of flow at toe

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80

1

.

dx

dy

slope of the d/s face of the overflow section

C. vs.

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Cd. vs. (P/Hd)

(P/Hd)>1.33, velocity is neglected

Tailwater Effect on C

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Problem 01

Problem 01: An overflow spillway with the upstream face vertical is to be designed for a flood peak of 3000 m3/s. The height of the spillway crest is kept at RL 130.50 m. The average river bed level at the site is 102.50 m. The number of spans is 6, clear waterway between piers is 12 m, thickness of the pier is 2 m, pier contraction coefficient, Kp = 0.02 and abutment contraction coefficient, Ka = 0.20 for the effect of end contraction. Assume the coefficient of discharge is 2.20 and the slope of the d/s face of the overflow section is 0.8: 1.

• Determine the design head by neglecting the end contraction.

• What will happen if the design head is determined by taking the effect of end contraction of piers and spans?

• Determine the tangent point of x ordinate of the downstream

profile from the origin of the crest.

*R.L is reservoir level

Solution• Peak flow, Q=3000 m3/s. • The no. of spans is 6, • clear waterway between piers is 12 m,• thickness of the pier is 2 m, • pier contraction coefficient, Kp = 0.02• abutment contraction coefficient, Ka = 0.20• Coefficient of discharge, C is 2.20• Slope of the d/s face of the overflow section is 0.8: 1.

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Neglecting the end contraction, so we calculate L= L’L’ =clear waterway x no.of spansL=12 x 6 = 72m

23 /CLHQ Determine the design head by neglecting the end contraction.

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23 /

CL

QH

mH 1177222

300023 .

)(./

a) Determine the design head

b) design head is determined by taking the effect of end contraction of piers and spans

dape HKKNLL )*( 2 N=6Kp=0.02Ka=0.2

23 /CLHQ

mL

L

e

e

456755472

117200206272

..

.)..*(

mH d427

456722

300023 .

).(./

c) Determine the tangent point of x ordinate of the downstream profile from the origin of the crest.. yHK n

e

n

X )( 1

)( 1ne

n

HK

Xy

For vertical upstream

K=2, n=1.85

).( .

.

850

851

1172

Xy

P=Height of spillway crest at R.L- average river bed level at the site =130.5-102.5 =28 m

Check,P/Hd = 28/7.11 = 3.94 ~~greater than 1.33So effect of velocity can be neglected

He=Hd+Ha(due to velocity~0)He = Hd

).( .

.

850

851

1172

Xy

Differentiate both sides with respect to x

851850

851

850

851

094011721172

..

.

.

.

.).().(

XXX

y

850

1851

851

1740

101740

0940

.

.

.

.

).(.

.

Xdx

dy

Xdx

dy

Xy

Since slope of the d/s face of the overflow section is 0.8: 1, So...

80

1

.

dx

dy

850174080

1 ...

Xmx

x

1710

174080

1 850

1

.

.*.

.

End