Chapter 11 : SPILLWAYS¶

Example 11.1 pg : 538¶

In :
import math

#Given
h = 1.2;         				#head of water
Cd = 2.2;        				#coefficient of discharge
rho = 1;         				#density of water
gamma_w = 9.81;  				#unit weigth of water

q = Cd*h**1.5;

#applying bernaulli's equation at u/s water surface at section A and B
#solving it by error and trial method we get
v1 = 13.7;v2 = 14.7;
d1 = 0.212;d2 = 0.197;

F2 = gamma_w*d2**2/2;
W = gamma_w*60*2*math.pi*3*((d1+d2)/2)/360;
F = (Fx**2+Fy**2)**0.5;
F = round(F*100)/100;

# Results
print "Resultant force = %.2f kN/m."%(F);
Resultant force = 46.68 kN/m.

Example 11.2 pg : 539¶

In :
import math

#Given
C = 2.4;          				#coefficient of discharge
L = 100;          				#length of spillway
wc = 8;           				#heigth of weir crest above bottom
g = 9.81;         				#acceleration due to gravity

# Calculations
h = H+wc;
Q1 = C*L*H**(1.5); 				#neglecting approach velocity and end contractions
va = Q1/(h*L);
ha = va**2/(2*g);
Ha = ha+H;
Q = C*L*Ha**1.5;
Q = round(Q*10)/10;

# Results
print "discharge over oggy weir = %.2f cumecs."%(Q);
discharge over oggy weir = 690.80 cumecs.

Example 11.3 pg : 540¶

In :
import math

#capacity of siphon
#length of oggy weir required

#Given
t = 6;          				#tail water elevation
h = 1;          				#heigth of siphon spillway
w = 4;          				#width of siphon spillway
hw = 1.5;       				#head water elevation
C = 0.6;        				#coefficient of discharge
Co = 2.25;      				#coefficient of discharge of oggy spillway
lo = 4;         				#length of oggy spillway
hc = 1.5;       				#head on weir crest
g = 9.81;       				#acceleration due to gravity

# Calculations and Results
#part (a)
Q = C*h*w*(2*g*(t+hw))**0.5;
Q = round(Q*10)/10;
print "capacity of siphon = %.2f cumecs."%(Q);

#part (b)
h1 = (Q/(Co*lo))**(2./3);
h1 = round(h1*100)/100;
print "head required in oggy spillway = %.2f m"%(h1);

#part (c)
L = Q/(Co*(hc)**1.5);
L = round(L*100)/100;
print "length of oggy weir required = %.2f m."%(L);
capacity of siphon = 29.10 cumecs.
head required in oggy spillway = 2.19 m
length of oggy weir required = 7.04 m.

Example 11.4 pg : 540¶

In :
import math

#Given
rl = 435;         				#full reservior level
cl = 429.6;       				#level of centre of siphon
hfl = 435.85;     				#high flood level
hfd = 600;        				#high flood discharge
w = 4;            				#width of throat
h = 2;            				#heigth of throat
C = 0.65;         				#coefficient of discharge
g = 9.81;       				#acceleration due to gravity

# Calculations
H = hfl-cl;
Q = C*w*h*(2*g*H)**0.5;
n = hfd/Q;
n = round(n*100)/100;

# Results
print " number of siphons units required = %.2f.hence provide 11 siphons units."%(n);
number of siphons units required = 10.42.hence provide 11 siphons units.

Example 11.5 pg : 541¶

In :
import math
from numpy import arange,zeros

#design oggy spillway for concrete gravity dam

#Given
rbl = 250;        				#avarage river bed level
rlc = 350;        				#R.L of spillway crest
s = 0.75;         				#slope on downstream side
Q = 6500;         				#discharge
L = 5*9;          				#length of spillway
Cd = 2.2;         				#coefficient of discharge
t = 2;            				#thickness of each pier

#step 1. computation of design head
H = (Q/(Cd*L))**(2./3);
P = rlc-rbl;

#P/H = 6.15,which is<1.33;it is a high overflow spillway

#H+P/H = 7.15>1.7; hence discharge coefficient is not affected by downstream apron interface

Kp = 0.01;
Ka = 0.1;
N = 4;
He = 17.5;                				#assumed
Le = L-2*(N*Kp+Ka)*He;
He1 = (Q/(Cd*Le))**(2./3);
He1 = round(He1*100)/100;
#He1 is almost equal to He
print "crest profile will be designed for Hd = %.2f m."%(He1);

#step 2. determination of d/s profile

#equating the slope of d/s side and derivative of profile equation suggested by WES
x = 27.03;
y = 0.04372*x**1.85;
print "downstream profile:";
x = arange(1,27)
y = zeros(26)
for i in range(26):
y[i] = 0.04372*x[i]**1.85;
y[i] = round(y[i]*1000)/1000;

print "x          y";
for i in range(26):
print "%i          %.2f"%(x[i],y[i]);

print "27.03          19.48";
#step 3. determination of u/s profile
# math.cosidering equation for vertical u/s face and Hd = 17.58

print "upstream profile:";
x = [-0.5, -0.1, -1.5, -2.0, -3.0, -4.0, -4.75];
y = zeros(7)
for i in range(7):
if i==6:
y[i] = 0.0633*(x[i]+4.75)**1.85+2.2151-1.2643*(x[i]+4.75)**0.625;
y[i] = round(y[i]*1000)/1000;
continue

y[i] = 0.0633*(x[i]+4.7466)**1.85+2.2151-1.2643*(x[i]+4.7466)**0.625;
y[i] = round(y[i]*1000)/1000;

print "x                    y";
for i in range(7):
print "%.2f          %.2f"%(x[i],y[i]);

#step 4.design of d/s bucket

R = P/4;
print "radius of bucket = %i m."%(R);
print "bucket will subtend angle of 60 degree at the centre.";
crest profile will be designed for Hd = 17.58 m.
downstream profile:
x          y
1          0.04
2          0.16
3          0.33
4          0.57
5          0.86
6          1.20
7          1.60
8          2.05
9          2.55
10          3.10
11          3.69
12          4.34
13          5.03
14          5.77
15          6.55
16          7.38
17          8.26
18          9.18
19          10.15
20          11.16
21          12.21
22          13.31
23          14.45
24          15.63
25          16.86
26          18.13
27.03          19.48
upstream profile:
x                    y
-0.50          0.01
-0.10          -0.00
-1.50          0.14
-2.00          0.25
-3.00          0.60
-4.00          1.20
-4.75          2.21
radius of bucket = 25 m.
bucket will subtend angle of 60 degree at the centre.

Example 11.6 pg : 562¶

In :
import math

#design length and depth of stilling bamath.sin

#Given
q = 1;           				#discharge of spillway
Cd = 0.7;        				#coefficient of discharge
h1 = 10;          				#heigth of crest above downstream silting bamath.sin
g = 9.81;           				#acceleration due to gravity
Cv = 0.9;          				#coefficient of velocity

# Calculations
h = (3*q/(2*Cd*(2*g)**0.5))**(2./3);
H = h1+h/2;
vt = (2*g*H)**0.5;
v1 = Cv*vt;
y1 = q/v1;
F1 = v1/(g*y1)**0.5;
#F>1, flow is super-critical
y2 = 1;
v2 = q/y2;
F2 = v2/(g*y2)**0.5;           				#<1
y2 = (y1/2)*((1+8*F1**2)**0.5-1);
de = y2-1;
le = 5*(y2-y1);
de = round(de*1000)/1000;
le = round(le*10)/10;

# Results
print "stilling bamath.sin should be depressed by %.2f m."%(de);
print "length of stilling bamath.sin = %.2f m."%(le);
stilling bamath.sin should be depressed by 0.58 m.
length of stilling bamath.sin = 7.50 m.

Example 11.7 pg : 563¶

In :
import math

#Given
q = 7.83;        				#discharge through spillway
w = 12.5;        				#width of fall
d = 2.;           				#depth of water in downstream
g = 9.8;

y1 = 0.5;
v1 = q/y1;
F1 = v1/(g*y1)**0.5;

#F>1,flow is super-critical

# Calculations
v2 = q/d;
F2 = v2/(g*d)**0.5;
y2 = (y1/2)*((1+8*F1**2)**0.5-1);
de = y2-d;
le = 5*(y2-y1);
de = round(de*100)/100;
le = round(le*10)/10;

# Results
print "stilling bamath.sin should be depressed by %.2f m."%(de);
print "length of stilling bamath.sin = %.2f m."%(le);
stilling bamath.sin should be depressed by 2.76 m.
length of stilling bamath.sin = 21.30 m.

Example 11.8 pg : 564¶

In :
import math

#Given
Ag = 5*2.5;          				#area of gate
miu = 0.25;         				#coefficient of friction
w = 0.5;            				#weigth of gate
h = 2;              				#head of water over crest
g = 9.81;           				#acceleration due to gravity
gamma_w = 1000;     				#unit weigth of water

# Calculations
m = w*g*1000;
F = gamma_w*Ag*h*h*g/10;
ff = miu*F;
tf = (m+ff)/1000;

# Results
print "force to be exerted to lift the gate = %.2f kN."%(tf);
force to be exerted to lift the gate = 17.17 kN.

Example 11.9 pg : 564¶

In :
import math

#Given
q = 19;        				#dischrge through spillway
E = 1;         				#energy loss

#from energy loss equation;E = (y2-y1)**3/4y2y1; and solving it we get
#x = 0.5*(-1+(1+294.39*(x-1)**9/64*x**3))
#by trial and error method x = 2.806
x = 2.806;
y1 = 4*x/(x-1)**3;
y2 = x*y1;
y1 = round(y1*1000)/1000;
y2 = round(y2*1000)/1000;
print "depth of flow at both end of jumps = %.2f m and %.2f m. respectively."%(y1,y2);
depth of flow at both end of jumps = 1.91 m and 5.35 m. respectively.