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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.
2
Four Mile Dam, Australia –
Ogee Spillway
Upper South Dam, Australia – Ogee
Spillway
3
4
Source:http://www.leanhtuan.com/
5
Hoover Dam – Spillway Crest
6
Hoover Dam – Spillway
New Cronton Dam NY – Stepped
Chute Spillway
7
Sippel Weir, Australia – Drop Spillway
8
Four Mile Dam, Australia – Ogee
Spillway
9
Upper South Dam, Australia – Ogee
Spillway
10
Itaipu Dam, Uruguay – Chute
Spillway
11
Itaipu Dam – Flip Bucket
12
• Common type of spillways:
1) Free over fall/straight drop
spillways
2) Overflow or ogee spillways.
3) Chute spillways
4) Siphon saddle spillway
13
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.
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.
17
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.
18
Overflow Spillway
19
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)
e
a
p
e H
K
K
N
L
L )
*
(
' 

 2
2
3/
e
e
d H
L
C
Q 
a
d
e H
H
H 

Le = effective width of crest
L’ = net width of crest(clear waterway x
no.of spans)
N = number of piers
Kp = pier contraction coefficient
Ka = abutment contraction coefficient
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.
• Design head, Hd
21
2
3/
e
d
d
L
C
Q
H 
•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
8
0
1
.

dx
dy
slope of the d/s face of the overflow section
C. vs. 
24
Cd. vs. (P/Hd)
(P/Hd)>1.33, velocity is neglected
Tailwater Effect on C
26
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.
28
Neglecting the end contraction, so we calculate L= L’
L’ =clear waterway x no.of spans
L=12 x 6 = 72m
2
3/
CLH
Q 
Determine the design head by neglecting the end contraction.
29
2
3 /
CL
Q
H 
m
H 11
7
72
2
2
3000
2
3 .
)
(
.
/ 

a) Determine the design head
b) design head is determined by taking the effect of end
contraction of piers and spans
d
a
p
e H
K
K
N
L
L )
*
( 

 2 N=6
Kp=0.02
Ka=0.2
2
3/
CLH
Q 
m
L
L
e
e
45
67
55
4
72
11
7
2
0
02
0
6
2
72
.
.
.
)
.
.
*
(






m
Hd
42
7
45
67
2
2
3000
2
3 .
)
.
(
.
/ 

c) Determine the tangent point of x ordinate of the downstream profile
from the origin of the crest.
.
y
H
K
n
e
n
X )
(
1


)
(
1

 n
e
n
H
K
X
y
For vertical
upstream
K=2, n=1.85
)
.
( .
.
85
0
85
1
11
7
2
X
y 
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
)
.
( .
.
85
0
85
1
11
7
2
X
y 
Differentiate both sides with respect to x
85
1
85
0
85
1
85
0
85
1
094
0
11
7
2
11
7
2
.
.
.
.
.
.
)
.
(
)
.
(
X
X
X
y 


85
0
1
85
1
85
1
174
0
1
0
174
0
094
0
.
.
.
.
)
.
(
.
.
X
dx
dy
X
dx
dy
X
y




Since slope of the d/s face of the overflow section is 0.8: 1, So...
8
0
1
.

dx
dy
85
0
174
0
8
0
1 .
.
.
X

m
x
x
17
10
174
0
8
0
1 85
0
1
.
.
*
.
.








End

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Ce 3205-lecture-08-spillways

  • 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. 2 Four Mile Dam, Australia – Ogee Spillway
  • 3. Upper South Dam, Australia – Ogee Spillway 3
  • 5. 5 Hoover Dam – Spillway Crest
  • 6. 6 Hoover Dam – Spillway
  • 7. New Cronton Dam NY – Stepped Chute Spillway 7
  • 8. Sippel Weir, Australia – Drop Spillway 8
  • 9. Four Mile Dam, Australia – Ogee Spillway 9
  • 10. Upper South Dam, Australia – Ogee Spillway 10
  • 11. Itaipu Dam, Uruguay – Chute Spillway 11
  • 12. Itaipu Dam – Flip Bucket 12
  • 13. • Common type of spillways: 1) Free over fall/straight drop spillways 2) Overflow or ogee spillways. 3) Chute spillways 4) Siphon saddle spillway 13
  • 14. 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. 17
  • 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. 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. 18
  • 19. Overflow Spillway 19 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) e a p e H K K N L L ) * ( '    2 2 3/ e e d H L C Q  a d e H H H   Le = effective width of crest L’ = net width of crest(clear waterway x no.of spans) N = number of piers Kp = pier contraction coefficient Ka = 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 21 2 3/ e d d L C Q H  •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. 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. 23 8 0 1 .  dx dy slope of the d/s face of the overflow section
  • 25. Cd. vs. (P/Hd) (P/Hd)>1.33, velocity is neglected
  • 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. 28 Neglecting the end contraction, so we calculate L= L’ L’ =clear waterway x no.of spans L=12 x 6 = 72m 2 3/ CLH Q  Determine the design head by neglecting the end contraction.
  • 29. 29 2 3 / CL Q H  m H 11 7 72 2 2 3000 2 3 . ) ( . /   a) Determine the design head b) design head is determined by taking the effect of end contraction of piers and spans d a p e H K K N L L ) * (    2 N=6 Kp=0.02 Ka=0.2 2 3/ CLH Q  m L L e e 45 67 55 4 72 11 7 2 0 02 0 6 2 72 . . . ) . . * (       m Hd 42 7 45 67 2 2 3000 2 3 . ) . ( . /  
  • 30. c) Determine the tangent point of x ordinate of the downstream profile from the origin of the crest. . y H K n e n X ) ( 1   ) ( 1   n e n H K X y For vertical upstream K=2, n=1.85 ) . ( . . 85 0 85 1 11 7 2 X y  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
  • 31. ) . ( . . 85 0 85 1 11 7 2 X y  Differentiate both sides with respect to x 85 1 85 0 85 1 85 0 85 1 094 0 11 7 2 11 7 2 . . . . . . ) . ( ) . ( X X X y    85 0 1 85 1 85 1 174 0 1 0 174 0 094 0 . . . . ) . ( . . X dx dy X dx dy X y     Since slope of the d/s face of the overflow section is 0.8: 1, So... 8 0 1 .  dx dy 85 0 174 0 8 0 1 . . . X  m x x 17 10 174 0 8 0 1 85 0 1 . . * . .        
  • 32. End