# Chapter 17 : CANAL OUTLETS¶

## Example 17.1 pg : 788¶

In [2]:
import math

#Given
D = 100.0;         				#F.S.L of distributory
wc = 99.90;      				#F.S.L of water course
L = 9.;           				#length of pipe
d = 20.;          				#diameter of pipe
f = 0.005;     				#coefficient of friction
g = 9.81;      				#acceleration due to gravity

# Calculations
C = (d/((1.5*d/(400*f)+L)*f))**0.5/20;
A = math.pi*d**2/(4*10000);
q = C*A*(2*g*H)**0.5;
q = round(q*10000)/10000;

# Results
print "discharge through the outlet = %.5f cumec."%(q);

discharge through the outlet = 0.02840 cumec.


## Example 17.2 pg : 788¶

In [3]:
import math

#design a submerged pipe

#Given
q = 0.04;         				#discharge through outlet
D = 100.0;         				#F.S.L of distributing canal
wc = 99.90;      				#F.S.L of water course
dep = 1.1;       				#full supply depth distributing canal
C = 0.7;         				#average value of coefficient of discharge
g = 9.81;      				#acceleration due to gravity

# Calculations

A = q/(C*(2*g*H)**0.5);
d = (4*A/math.pi)**0.5*100;
d = round(d*10)/10;

# Results
print "diameter of pipe required = %.2f cm."%(d);
print "use pipe of diameter 25 cm.";

diameter of pipe required = 22.80 cm.
use pipe of diameter 25 cm.


## Example 17.3 pg : 788¶

In [4]:
import math

#design submerged pipe

#Given
q = 0.04;         				#discharge through outlet
D = 100.0;         				#F.S.L of distributing canal
wc = 99.90;      				#F.S.L of water course
dep = 1.1;       				#full supply depth distributing canal
f = 0.01;         				#coefficient of friction
g = 9.81;      				#acceleration due to gravity
L = 9.;         				#Length of pipe

# Calculations
#first trial
#taking d = 22.8 cm
d = 22.8;
C = (d/((1.5*d/(400*f)+L)*f))**0.5/20;
A = q/(C*(2*g*H)**0.5);
d = (4*A/math.pi)**0.5*100;
#second trial
C = (d/((1.5*d/(400*f)+L)*f))**0.5/20;
A = q/(C*(2*g*H)**0.5);
d = (4*A/math.pi)**0.5*100;
d = round(d*100)/100;

# Results
print "diameter of pipe required = %.2f cm."%(d);
print "provide diameter of pipe as 25 cm.";

diameter of pipe required = 24.94 cm.
provide diameter of pipe as 25 cm.


## Example 17.4 pg : 795¶

In [5]:
import math
#design an open flume outlet

#Given
Q = 0.06;           				#discharge
D = 0.85;           				#full supply depth
Hw = 15.;             				#available working head
Bt = 7.;
C = 1.6;      				#let us choose

# Calculations and Results
H = (Q*100/(C*Bt))**(2./3);
mh = 0.2*H;        				#minimum modular head
mh = round(mh*1000)/1000;
print "minimum modular head = %.2f m. < available working head.hemce,design is safe."%(mh);
o = H/D;
o = round(o*1000)/1000;
print "setting of outlet = %.2f. <0.9.hence,outlet will work as hyper propotional outlet."%(o);

minimum modular head = 0.13 m. < available working head.hemce,design is safe.
setting of outlet = 0.78. <0.9.hence,outlet will work as hyper propotional outlet.