# Chapter 7 : Flow Through Open Channels¶

## Example 7.1 Page No : 140¶

In :
# Variables
b = 6.          # m width
i = 1./1000     # slope
d = 2.          # m, depth of water
C = 50          # Constant

# Calculations
A = b*d
m = A/(b+2*d)
Q = A*C*((i*m)**0.5)

# Results
print "flow rate assuming chezys consmath.tant eqaul to 50 in m3/sec",round(Q,4)

flow rate assuming chezys consmath.tant eqaul to 50 in m3/sec 20.7846


## Example 7.2 Page No : 140¶

In :
# Variables
b = 5.          #m wide
d = 3.          #m deep
i = 1./1000     #slope
C = 55.         #constant

# Calculations
A = b*d
m = A/(b+2*d)
Q = A*C*((i*m)**0.5)
v = Q/A

# Results
print "flow rate assuming chezys constant eqaul to 55 in m3/sec & velocity of flow in m/sec : ",v,Q

flow rate assuming chezys constant eqaul to 55 in m3/sec & velocity of flow in m/sec :  2.03100960116 30.4651440174


## Example 7.3 Page No : 141¶

In :
# Variables
b = 2.5       #m wide
d = 2.5       #m depth
C = 56.       #constant
A = b*(7.5+d)*0.5

# Calculations
P = 2.5+((b*b+d*d)**0.5)*2
m = A/P
i = 1./1200
Q = A*C*((m*i)**0.5)

# Results
print "the diacharge through the channel in litres/sec",round((Q*1000),4)

# note : rounding off error.

the diacharge through the channel in litres/sec 23093.0995


## Example 7.4 Page No : 142¶

In :
import math

# Variables
b = 3.5           #m, width
i = 1./1000       # slope
d = 1.5           #m, depth of flow
C = 60.           # degree C
y = 60.           #Constant

# Calculations
x = 1.5/math.tan(math.radians(y))
w = b+x*2
A = (w+b)*0.5*d
P = b+2*((x*x+d*d)**0.5)
m = A/P
Q = A*C*((m*i)**0.5)

# Results
print "discharge carried by the canal in litres/sec : %.2f"%(Q*1000)

# note : rounding off error.

discharge carried by the canal in litres/sec : 12049.94


## Example 7.5 Page No : 142¶

In :
# Variables
b = 9.           #m, width
i = 1./3000      #slope
d = 1.2          #m, water depth

# Calculations
w = b+d
A = (w+b)*0.5*d
P = b+2*((d*d+d*d*0.25)**0.5)
m = A/P
C = 50.
V = C*((m*i)**0.5)
Q = V*A

# Results
print "average velocity of flow, rate of flow",round(V,7),round((Q*1000),3)

average velocity of flow, rate of flow 0.9064695 10442.529


## Example 7.6 Page No : 143¶

In :
# Variables
Q = 0.1
b = 0.6      #m, width
C = 56       # constant
d = 0.3     #m, depth of flow

# Calculations
a = b*d
v = Q/a
p = b+2*d
m = a/p
i = (v*v)/(C*C*m)
k = a*C*(m**0.5)

# Results
print "bottom slope neccessary for uniform flow,conveyance of the channel section",round(i,7),round(k,7)

bottom slope neccessary for uniform flow,conveyance of the channel section 0.0006561 3.9039672


## Example 7.7 Page No : 144¶

In :
# Variables
i = 1./1000     #slope
d = 1.5         #m, depth of water
Cd = 0.55       #co-effient
a = d*d
C = 40.        # constant
g = 9.81

# Calculations
m = d
Q = a*C*((d*i)**0.5)
H = (3*Q/(Cd*2*((2*g)**0.5)))**0.4
height = d+3-H

# Results
print "height of the dam in m",round(height,3)

# note : book answer is wrong. kindly check.

height of the dam in m 3.143


## Example 7.8 Page No : 145¶

In :
# Variables
b = 1.4        #m, width
d = 1.4        #m, depth
n = 1./4       # side slope
i = 1./700     #bad slope

# Calculations
N = 0.025
a = d*(b+(n*d))
p = b+(2*d*((n*n+1)**0.5))
m = a/p
q = (a*(m**0.6666)*(i**0.5))/N

# Results
print "discharge from the trapezoidal channel in litres/sec",round((q*1000),3)

# note : rounding off error.

discharge from the trapezoidal channel in litres/sec 2551.276


## Example 7.9 Page No : 146¶

In :
# Variables
Q = 0.3                        #m**3/s rate
D = 1.5                        #m diameter
N = 0.02                       #N
A = 3.142*D*D/(4*2)
p = 3.142*D/2

# Calculations
m = A/p
i = ((Q*N)/(A*(m**0.6666)))**2

# Results
print "the slope of the sewer",round(i,7)

the slope of the sewer 0.0001705


## Example 7.10 Page No : 147¶

In :
import math

# Variables
D = 2.4         #m diameter
d = 1.5         #m, depth of water
i = 1./1500     #gradient
N = 0.02        # N Manning formula

# Calculations
a = (d-(D/2))/(D/2)
z = math.degrees(math.acos(a))
z1 = math.radians(180 - z)
P = D*z1
A = D*D*0.25*(z1-(math.sin(2*z1)/2))
m = A/P
Q = (A*(m**0.6666)*(i**0.5))/N

# Results
print "the discharge through the sewer in m**3/s :  %.5f"%Q

the discharge through the sewer in m**3/s :  2.96839


## Example 7.11 Page No : 148¶

In :
# Variables
b = 1.5          #m, wide
d = 0.8          #m, depth
Q = 0.75         #m**3/s
i = 1./2500      #slope

# Calculations
A = b*d
P = b+(2*d)
m = A/P
C = Q/(((m*i)**0.5)*A)
z = (157.6/C)-1.81
K = z*(m**0.5)

# Results
print "Chezys constant and coefficient of roughness",round(C,2),round(K,3)

Chezys constant and coefficient of roughness 50.23 0.826


## Example 7.12 Page No : 149¶

In :
# Variables
b = 10.              #m wide
d = 4.               #m, depth of water
i = 1./1000          #slope
N = 0.03             #N Kutter's fprmula

# Calculations
A = b*d
P = b+(2*d)
m = A/P
z1 = 23+(0.00155/i)+(1/N)
z2 = 1+((23+(0.00155/i))*(N/(m**0.5)))
C = z1/z2
Q = A*C*((m*i)**0.5)

# Results
print "discharge through the recmath.tangular channel in litres/sec",round((Q*1000),3)

discharge through the recmath.tangular channel in litres/sec 73053.236


## Example 7.13 Page No : 150¶

In :
# Variables
b = 4.              #m wide
d = 1.5             #m, depth of water
i = 1./1000         #slope
C = 55.             # constant

# Calculations and Results
A = b*d
P = b+(2*d)
m = A/P
Q = A*C*((m*i)**0.5)
d1 = (A/2)**0.5
b1 = d1*2
print "the new dimension of the channel",round(b1,3),round(d1,3)

P1 = b1+(2*d1)
m1 = A/P1
Q1 = A*C*((m1*i)**0.5)
Qf = Q1-Q
print "increase in discharge in m3/sec",round(Qf,3)

the new dimension of the channel 3.464 1.732
increase in discharge in m3/sec 0.05


## Example 7.14 Page No : 151¶

In :
import math

# Variables
i = 1./2500     #slope
N = 0.02        # N, Manning's formula
Q = 14.         #m**3/s

# Calculations
n = 1./(math.radians(math.tan(60)))
a = (3**0.5)
d = ((Q*N*(2**0.6666))/((i**0.5)*a))**(3./8)
b = d*2/(3**0.5)

# Results
print "dimension of the channel",round(b,4),round(d,4)

dimension of the channel 3.0065 2.6037


## Example 7.15 Page No : 152¶

In :
import math

# Variables
Q = 20.2           #m**3/s
i = 1./2500        #slope
C = 60.            #Constant

# Calculations
n = 1./(math.radians(math.tan(60)))
a = (3**0.5)
d = ((Q*(2**0.5))/(C*a*(i**0.5)))**0.4
b = 2*d/(a)

# Results
print "dimension of the cross section in m",round(b,4),round(d,4)

dimension of the cross section in m 3.294 2.8527


## Example 7.16 Page No : 153¶

In :
import math

# Variables
Q = 10.        #m**3/s
V = 2.         #m/s, velocity
A = Q/V
n = 1.         #m length

# Calculations
d = (A/1.828)**0.5
b = 0.828*d
A1 = (b+(2*d*((n*n+1)**0.5)))
print "area in m2 of lining required for 1m canal lenght",round(A1,3)

# note : rounding off error.

area in m2 of lining required for 1m canal lenght 6.047


## Example 7.17 Page No : 154¶

In :
# Variables
n = 1.
Q = 14.            #m**3/s
i = 1./1000        #slope
C = 44.            #constant
a = 1.828

# Calculations
d = ((Q*(2**0.5))/(C*a*(i**0.5)))**0.4
b = d*0.828
cost = (b+n*d)*4
A = 1.828*d*d
C1 = 70.
d1 = ((Q*(2**0.5))/(C1*a*(i**0.5)))**0.4
b1 = 0.828*d1
cost1 = (b1+n*d1)*4
costl = (b1+(2*d1*((n*n+1)**0.5)))
totalcost =  cost1+costl

# Results
print "lined channel is cheaper ,dimension in m",round(b1,4),round(d1,3)

lined channel is cheaper ,dimension in m 1.5626 1.887


## Example 7.18 Page No : 156¶

In :
import math

# Variables
d = 1.2            #m diameter
i = 1./1500        #slope
C = 52.            #constant
z = 1.9-1./1

# Calculations
z1 = math.acos(z)
x = math.pi-z1
A = d*d*0.25*(x-(math.sin(2*x)/2))
P = d*x
m = A/P
Q = A*C*((m*i)**0.5)

# Results
print "the maximium discharge through the channel in litres/sec",round((Q*1000),3)

# note : rounding off error.

the maximium discharge through the channel in litres/sec 873.637

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