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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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CE 3205 Water and Environmental Engineering

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  1. CE 3205 Water and Environmental Engineering Spillways

  2. 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. Four Mile Dam, Australia – Ogee Spillway

  3. Upper South Dam, Australia – Ogee Spillway

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

  5. Hoover Dam – Spillway Crest

  6. Hoover Dam – Spillway

  7. New Cronton Dam NY – Stepped Chute Spillway

  8. Sippel Weir, Australia – Drop Spillway

  9. Four Mile Dam, Australia – Ogee Spillway

  10. Upper South Dam, Australia – Ogee Spillway

  11. Itaipu Dam, Uruguay – Chute Spillway

  12. Itaipu Dam – Flip Bucket

  13. Common type of spillways: • Free over fall/straight drop spillways • Overflow or ogee spillways. • Chute spillways • Siphon saddle spillway

  14. 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.

  15. 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.

  16. 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

  17. 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.

  18. 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. • Inflow flood hydrograph-Indicates rate of inflow respect to time. • Reservoir capacity curve-indicates the reservoir storage at different reservoir elevations. • Outflow discharge curve-indicates the rate of outflow through spillways at different reservoir elevations.

  19. Overflow Spillway Basic equation flow over weirs, Where Q=discharge m3/s Cd=coefficient of discharge Le=effective length He=actual effective head Hd=design head Ha=head due to velocity of approach (sometimes neglected) 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

  20. Contraction Coefficients Table 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.

  21. Design head, Hd • Downstream profile • d/s profile of spillway can be represented by x,y= coordinates of the point on the spillway surface Hd=design head K,n= constant, depend on inclination of the upstream face of spillway

  22. 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

  23. slope of the d/s face of the overflow section

  24. C. vs. 

  25. Cd. vs. (P/Hd) (P/Hd)>1.33, velocity is neglected

  26. Tailwater Effect on C

  27. 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

  28. 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. Determine the design head by neglecting the end contraction. Neglecting the end contraction, so we calculate L= L’ L’ =clear waterway x no.of spans L=12 x 6 = 72m

  29. a) Determine the design head b) design head is determined by taking the effect of end contraction of piers and spans N=6 Kp=0.02 Ka=0.2

  30. 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 1.33 So effect of velocity can be neglected He=Hd+Ha(due to velocity~0) He = Hd c) Determine the tangent point of x ordinate of the downstream profile from the origin of the crest. . For vertical upstream K=2, n=1.85

  31. Differentiate both sides with respect to x Since slope of the d/s face of the overflow section is 0.8: 1, So...

  32. End

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