#Design of Ogee Spillway# Problem statement: Design a suitable section for the overflow portion of a concrete gravity dam having the downstream face sloping at a slope of 0.7H:1V . The design discharge for the spillway is 8000 cumecs. The height of the spillway crest is kept at RL 204.0m. The average river bed level at site is 100m. The spillway length consists of 6 spans having a clear width of 10m each. Thickness of each pier may be taken to be 2.5m.
Elementary section of Ogee spillway
Solution: For the given spillway the coefficient of discharge may be assumed Now , Q= C.L.He3/2 Where , Q= Discharge from the spillway le=Effective length of the spillway crest C=Coefficient of discharge which depends upon various factors such as relative depth of approach (i.e d/Hd ratio , relation of actual crest shape to the ideal nappe shape ,slope of upstream face downstream apron interference, and submergence , etc ) He = total head over the crest including the velocity head SD
8,000=2.2 * 60 (He)3/2 He3/2 =8000/(2.2*60) He = 60.62/3 He = 15.5m The height of the spillway above the river bed = h= 204100= 104m Since h/hd ,i.e. 104/15.5>1.33 It is a high spillway, the effect of velocity head can, therefore ,be neglected. Since (hd+ d)/H e =(Hd +h)/He =(15.5+104)/ 15.5>1.7; The discharge coefficient is not affected by tail water conditions, and the spillway remains a high spillway
# Design criteria for effective length: Le = L - 2[N.KP + Ka ] He
Where; Where Le = effective length of the crest L= net length of the crest N= number of piers Kp= pier contraction coefficient Ka= abutments contraction coefficient He= Total design head on the crest including velocity head
# Value of Kp for different cases: S.No.
Pier shape
1.
Square nosed piers without any rounding Square nosed piers with corners rounded on radius equal to 0.1 pier thickness
2.
Contraction coefficient 0.1 0.02
3
Rounded nose piers and 900 cut water nosed piers
0.01
4
Pointed nose piers
0.0
# Value of Ka for different cases: S.No. Shape of abutment 1.
2.
Square abutment with head wall at 900 to the direction of flow Rounded abutment with head wall at 900 to the direction of flow
Contraction coefficient 0.2
0.1
Effective length of spillway (Le ) : Le can be worked out as follows: Le = L-2[N.KP + Ka ] He Assuming that 90⁰ cut water nose piers and rounded abutments shall be provided , we have Ka = 0.01 and KP = 0.1 No. of piers = N = 5. Also assuming that the actual value of He is slightly more than the approximate value worked out (i.e. 15.5 m), say, let it be 16.3 m, we have Therefore , Le = 60 -2[5×0.01 + 0.1]×16.3 =55.1 m.
Hence Q = 2.2 × 55.1 × He3/2 or 8000 = 2.2 × 55.1 × He3/2 He3/2= 8000 2.2 × 55.1 He3/2 = 66.0 He = 16.4 m ≈ 16.3 m(assumed) Hence, the assumed He for calculating Le is all right. The crest profile will be designed for
Hd = 16.4 m (neglecting velocity head).
Note : The velocity head (Ha) can also be calculated as follows : Velocity of approach = Va = 8000 (60 + 5× 2.5)(104 + 16.4) Va = 0.917 m/s Velocity head = Ha = Va2 /2g Ha = (0.917)2 = 0.043 m. 2×9.81 This is very small and was, therefore, neglected.
Design criteria Crest profiles for inclined u/s face: Based on extensive experiments conducted by U.S.B.R, U.S. army corps of engineers have developed several standard crests of overflow spillways at its waterways experiment station (W.E.S) at Vickesburg. These shapes, given in fig for both vertical as well as inclined face, are known as WESstandard spillway crest. IS 6934-1973 has also adopted and recommended these shapes. These shapes can be represented by the following general equation: xn= K Hdn-1 .y Where, (x,y) are co-ordinates of the point on the crest profile with the origin at the highest point Hd is the design head including the velocity head K and n are constants depending upon the slope of upstream face.
Different values of K and n: Slope of u/s face of spillway
K
N
Vertical 1:3 (1H:3V) 1:1 ½ (1H: 1 ½ V)
2.0 1.936 1.939
1.84 1.836 1.810
# Downstream Profile : According to water experimental station (W.E.S.) , for a vertical u/s face the d/s profile is given by: x1.85 = 2. Hd0.85 y (K and n from table)
So, y=
x1.85 = x1.85 2 (Hd )0.85 2 (1.64)0.85
y=
x1.85 21.6
Before we determine the various co-ordinates of the d/s profile, we shall first determine the tangent point. The d/s slope of the dam is given to be 0.7H:1V Hance, dy = 1 dx 0.7 Differentiating the equation of the d/s profile with respect to x, we get dy = dx
1.85x1.85-1 21.6
x0.85 =
21.6 1.85 X 0.7
= 16.7
x= 22.4m y = (22.4)1.85 21.6
= 14.6 m
The co-ordinates from x=0,x=22.4m are worked out in table X(meters)
Y=x1.85 /21.6(meters)
1
0.046
2
0.166
3
0.354
4
0.60
5
0.905
6
1.274
7
1.710
8
2.162
9
2.684
X(meters)
Y=x1.85 /21.6(meters)
10
3.240
12
4.575
14
6.020
16
7.88
18
9.74
20
11.85
22
14.35
22.4
14.60
DESIGN CRITERIA FOR THE U/S CURVE OF THE OGEE SPILLWAY : According to the latest studies of U.S. Army Corps, the u/s curve of the ogee spillway having a vertical u/s face have the following equation: y = 0.724(x + 0.27Hd)1.85/ Hd0.85 + 0.126Hd - 0.4315 Hd0.375 X (x + 0.27Hd)0.625 The u/s profile extends upto
X = -0.27 Hd By the equation we can easily design the u/s side with the help of X & Y co-ordinate.
The u/s profile . The u/s profile may be designed as per equation Y=0.724(x+0.27Hd)1.85 +0.126Hd – 0.4315Hd0.375 (x+0.27Hd)0.625 Hd 0.85 Using Hd = 16.4m ,we get Y= 0.724(x+0.27X16.4)1.85 16.40.85 +0.126(16.4) 0.4315(16.4)0.375(x+0.27X16.4)0.625 Y=0.07(x+4.44)1.85 +2.07 -1.234 (x+4.432)0.625 Or, This curve should go up to x=-0.27Hd Or, x=-0.27X16.4=-4.443m
X in meters
Y in meters
-0.5
0.020
-1.0
0.063
-2.0
0.27
-3.0
0.65
-4.0
1.34
-4.443
2.07
DESIGNED PROFILE
The profile of the spillway has been determined and plotted in previous figure. A reverse curve at the toe with a radius equal to h/4 = 104/4 = 26m can be drawn at angle 60o, as shown in figure.
