# Chapter 5 : Flow in Channels¶

## Example 5.1 Page No : 67¶

In [1]:
import math

#initialisation of variables
h= 2.5 	#ft  depth of water
a= 45. 	#degrees  side slope
x= 5. 	#ft
Q= 45. 	#cuses
v= 2.6 	#ft/sec velocity
w= 6.92 	#ft
C= 120.

#CALCULATIONS
b= (Q/(v*h))-h
p= b+2*(h+math.sqrt(2))
A= h*w
m= A/p
i= (v/(C*math.sqrt(m)))**2

#RESULTS
print  'Width = %.2f ft'%(b)
print  ' Slope = %.6f '%(i)

Width = 4.42 ft
Slope = 0.000332


## Example 5.2 Page No : 69¶

In [2]:
import math

#initialisation of variables
a= 60.   	#degrees sides inclined
i= 1./1600
Q= 8.*10**6 	#gal/hr discharge
M= 110.
w= 6.24 	#lb/ft**3

#CALCULATIOS
d= ((Q*2**(2./3)*math.sqrt(1./i))/(w*3600*math.sqrt(3)*M))**(3./8)
b=6.93   	#ft

#RESULTS
print  'Diameter = %.f ft'%(d)
print ' breadth = %.2f ft'%(b)

Diameter = 6 ft


## Example 5.3 Page No : 71¶

In [3]:
import math

#initialisation of variables
g= 32.2 	#ft/swc**2
Q= 40. 	#cuses rate
w= 5.5 	#ft
h= 9. 	#in  depth
d= 0.75 	#ft
V= 3. 	#ft/sec

#CALCULATIONS
D= ((Q*2)**2/(g*(w*2)**2))**(1./3)
v= Q*d/w
D1= math.sqrt((2*v**2*d/g)+h/64)-(d/2)
dD= D1-d
El= -dD+((v**2*(1-(V/v)**2))/(2*g))
Els= Q*El*62.4/550

#RESULTS
print 'Critical depth = %.2f ft'%(D)
print ' Rise in level = %.f ft'%(D1)
print  ' Horse-power lost = %.3f hp'%(Els)

#The answer is a bit different due to rounding off error in textbook

Critical depth = 1.18 ft
Rise in level = 1 ft
Horse-power lost = 0.961 hp


## Example 5.6 Page No : 77¶

In [4]:
import math

#initialisation of variables
b= 3.5 	#ft
H= 2.5 	#ft
w= 3. 	#ft depth
h= 6. 	#ft  wide
g= 32.2 	#ft/sec**2

#CALCULATIONS
Q= 3.09*b*H**1.5
v= Q/(w*h)
H1= H+(v**2/(2*g))
Q1= 3.09*b*H1**1.5
hc= (Q1**2/(b**2*g))**(1./3)
h2= 0.5*(math.sqrt(hc**2+8*hc**2)-hc)
dh= h2+b-w

#RESULTS
print "Flow rate  = %.1f cusecs"%(Q)
print " Flow rate  = %d cusecs"%(Q1)
print  ' maximum depth of water downstream = %.3f ft'%(dh)
print  ' Shooting flow depth at hump = %.3f ft'%(h2)

Flow rate  = 42.8 cusecs
Flow rate  = 45 cusecs
maximum depth of water downstream = 2.226 ft
Shooting flow depth at hump = 1.726 ft


## Example 5.7 Page No : 79¶

In [5]:
import math

#initialisation of variables
m= 60./26
i= 1./2000
h1= 3. 	#ft depth
h2= 5. 	#ft depth
m1= 10./3
C= 90.      # constant
l= 500. 	#ft depth
H1= 29.62 	#ft
g= 32.2 	#ft/s**2

#CALCULATIONS
v= 90*math.sqrt(m*i)
v1= v*h1/h2
dh= (i-(v1**2/(C**2*m1)))*l/(1-v1**2/(g*h2))
h3= h2-dh
V= h1*v/h3

#RESULTS
print  'Height of water 1000 ft upstream = %.3f ft'%(h3)
print  ' Height of water upstream = %.3f ft'%(h3)

#The answer is a bit different due to rounding off error in textbook

Height of water 1000 ft upstream = 4.808 ft
Height of water upstream = 4.808 ft


## Example 5.8 Page No : 80¶

In [1]:
import math

#initialisation of variables
v= 5. 	    #ft/sec
m= 60./26
i= 1./2000
h= 5.5   	#ft
m1= 110./31
d= 3. 	    #ft
g= 32.2 	#ft/sec**2

#CALCULATIONS
C= v/(math.sqrt(m*i))
v1= v*d/h
r= (i-(v1**2/(C**2*m1)))/(1-(v1**2/(g*h)))
x= 1/r

#RESULTS
print  'Distance upstream = %.f ft'%(round(x,-1))

Distance upstream = 2380 ft


## Example 5.9 Page No : 81¶

In [12]:
import math
from numpy import *
from numpy.linalg import *

#initialisation of variables
g= 32.2 	#ft/sec**2
Q= 12 	#cuses

#CALCULATIONS
hc= (Q/(3*math.sqrt(g)))**(2./3)
vec=roots([1,6,12,8,0,-8.95,-8.95])
H=vec[2]

#RESULTS
print  'Critical depth = %.2f ft'%(hc)
print  ' Critical depth = %.2f ft'%(H)

Critical depth = 0.79 ft
Critical depth = 0.89 ft

-c:17: ComplexWarning: Casting complex values to real discards the imaginary part


## Example 5.11 Page No : 85¶

In [3]:
import math

#initialisation of variables
Cd= 0.64    # coefficient
g= 32.2 	#ft/sec**2
A= 12.5 	#ft**2
H= 24.8 	#ft
Q= 3200. 	#cuses
b= 150. 	#ft  wide
A1= 5.*10**6  # avg surface area
h= 9.   	#ft
h1= 6. 	    #in

#CALCULATIONS
N= Q/(Cd*A*math.sqrt(2*g*H))
H1= (Q/(3.2*b))**(2./3)
ES= (H1-(h1/12))*A1*h

#RESULTS
print  'number of siphons = %.f '%(N)
print  ' Extra Storage = %.2e ft**3'%(ES)

number of siphons = 10
Extra Storage = 1.37e+08 ft**3