ogee design

Design of ogee spillwayDesign of ogee spillway

Offsets and Risers onOffsets and Risers on
Upstream Face :Upstream Face :
13/2/14
Prepared by v.h,khokhani, assistant
professor, DIET
2

PracticalProfileofSpillwaYPracticalProfileofSpillwaY
When the profile for the crest of theWhen the profile for the crest of the
ogee spillway is plotted over theogee spillway is plotted over the
triangular profile the section of atriangular profile the section of a
gravity dam (non-overflowgravity dam (non-overflow
section) ,it is found that it goessection) ,it is found that it goes
beyond vie downstream face of thebeyond vie downstream face of the
dam , thu requiring thickening ofdam , thu requiring thickening of
the section for the spillway .the section for the spillway .

However,this extra concrete can beHowever,this extra concrete can be
saved by shifting the curve of thesaved by shifting the curve of the
nappe in a backward direction untilnappe in a backward direction until
this curve becomes tangential to thethis curve becomes tangential to the
downstream face of the dam .downstream face of the dam .

Offsets and Risers onOffsets and Risers on
Upstream FaceUpstream Face
13/2/14
professor, DIET
7

Factors affecting on spill wayFactors affecting on spill way

Design of spillwayDesign of spillway
Design an ogee spillway forDesign an ogee spillway for
concrete gravity dam, for theconcrete gravity dam, for the
following data :following data :
(1) Average river bed level = 250.0(1) Average river bed level = 250.0
mm
(2) R.L. of spillway crest =350.0 m(2) R.L. of spillway crest =350.0 m
13/2/14
professor, DIET
9

(3) Slope of d/s face of gravity dam(3) Slope of d/s face of gravity dam
= 0.75 H : 1 V= 0.75 H : 1 V
(4) Design discharge = 6500 cumecs(4) Design discharge = 6500 cumecs
(5) Length of spillway = 5 spans(5) Length of spillway = 5 spans
with a clear width of 7 m each.with a clear width of 7 m each.
(6) Thickness of each pier = 2.0 m(6) Thickness of each pier = 2.0 m
13/2/14
professor, DIET
10

Step-1 : Computation ofStep-1 : Computation of
design headdesign head
HHdd = H= Hee + H+ Haa
Where HWhere Haa = V= Vaa
22
/2g/2g
13/2/14
professor, DIET
11

2/3
ee HCLQ =
[ ] eaPe HKKNLL +−= *2
357*5 === LLe
2/3
)7*5(2.26500 eH=
41.842/3
=eH
24.19=eH

50.19=eH
100250350 =−=h
33.112.5
5.19
100
≥==
dH
h
7.112.6
5.19
5.19100
≥=
+
=
+
d
d
H
Hh
VELOCITY APPROACH CAN BE NEGLECTED
EFFECT ON DISCHARGE COEFFICIENT IS NEGLECTED

[ ] eaPe HKKNLL +−= *2
[ ]201.001.0*4235 +−=eL
= 29.4
2/3
eeHCLQ =
2/3
)4.29(2.26500 eH=

6.21=eH
24.27
85.1
X
Y =
75.0
1
=
dx
dy

24.27
85.1 85.0
x
dx
dy
=
85.0
0678.0 x
dx
dy
=
Comparing both dy/dx we get x = 33.26 and Y =24.01
24.27
85.1
X
Y =

X Y
1 0.036
2 0.132
3 0.265
4 0.477
5 0.720
6 1.01
7 1.343
8 1.719
9 2.138
10 2.598
14 4.843
18 7.709
22 11.175
26 15.22
30 19.836
33.6 24.463

u/s profile :u/s profile :
625.085.1
)832.5(365.172.2)832.5(05.0 +−++= xxy

This curve will extend up to,This curve will extend up to,
X=-0.27*21.6
= - 5.832
625.085.1
)832.5(365.172.2)832.5(05.0 +−++= xxy

X Y
-0.5
-1.0
-2.0
-3.0
-4.0
-5.0
-5.832

Design of d/s bucket :Design of d/s bucket :
The radius of the bucket isThe radius of the bucket is
generally kept equal to,generally kept equal to,
The bucket will subtend an angle ofThe bucket will subtend an angle of
60° at the centre.60° at the centre.

Examples - mahajanExamples - mahajan
An ogee spillway has 2.8 m headAn ogee spillway has 2.8 m head
above the crest .Depth of flow at theabove the crest .Depth of flow at the
toe of spillway is measured equal totoe of spillway is measured equal to
0.7 m. The tail watcr depth is 4.0m.0.7 m. The tail watcr depth is 4.0m.
Suggest the type of energySuggest the type of energy
dissipater and give its dimensiondissipater and give its dimension
Assume coefficient C as 2.2 in theAssume coefficient C as 2.2 in the
equation = CHequation = CH3/23/2

DATADATA
Head over spillway H=2.8mHead over spillway H=2.8m
Initial depth y1 = 0.7mInitial depth y1 = 0.7m
Tail water depth y 2 =4.0mTail water depth y 2 =4.0m
Discharge coefficient C=2.2Discharge coefficient C=2.2

STEP 1 Will be to find outSTEP 1 Will be to find out
sequent depth.sequent depth.
q=CHq=CH3/23/2
Q = 2.2 * 2.8Q = 2.2 * 2.83/23/2
= 10.30 M= 10.30 M33
/S/M/S/M
Pre jump DepthPre jump Depth
VV11 = q/y= q/y11
= 10.30/ 0.7= 10.30/ 0.7

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