## Chapter 10 : Open Channel Flow¶

### Example 10.1 Page no 363¶

In [1]:
# Top width, area of lfow, hydraulic radius

from math import *

from __future__ import division

# Given

b = 3                            # base of the channel

z = 0.5                          # slope of the channel

y = 2                            # depth of the channel

# Solution

T = b + 2*z*y

print "Top width =",round(T,0),"m"

A = (b+z*y)*y

print "Area of flow =",round(A,0),"m**2"

P = b + 2*y*sqrt(1+z**2)

print "Wetted perimeter =",round(P,3),"m"

R = A/P

D = A/T

print "Hydraulic depth =",round(D,2),"m"

Z = A*sqrt(D)

print "Secton Factor =",round(Z,2),"m**2"

Top width = 5.0 m
Area of flow = 8.0 m**2
Wetted perimeter = 7.472 m
Hydraulic depth = 1.6 m
Secton Factor = 10.12 m**2


### Example 10.2 Page no 366¶

In [2]:
# Discharge for the trapezoidal channel

from math import *

from __future__ import division

# Given

z = 1.0                # slide slope

b = 3.0                # base width

y = 1.5                # depth

S = 0.0009

n = 0.012             # for concrete

# Solution

A = (b+z*y)*y

P = P = b + 2*y*sqrt(1+z**2)

R = A/P

# from mannings eqquation

Q = A*(1/n)*(R**(2/3)*S**(1/2))

print "Discharge for the channel =",round(Q,2),"m**3/s"

Discharge for the channel = 16.1 m**3/s


### Example 10.4 Page no 373¶

In [3]:
# Determine cross sectional area

from __future__ import division

from math import *

# Given

z = 1

Q = 10000/60                     # discharge of water in ft**#/s

# Solution

y = (Q/(1.828*2.25*sqrt(0.5)))**(2/5)

print "depth(y) =",round(y,2),"ft"

b = 0.828*y

print "base width(b) =",round(b,2),"ft"

depth(y) = 5.05 ft
base width(b) = 4.18 ft


### Example 10.5 Page no 378¶

In [4]:
# Calculate criticcal depth

from math import *

from __future__ import division

# Given

y = 2.5                 # depth

V = 8                  # velocity in m/s

g = 9.81               # acceleration due to gravity in m/s**2

# Solution

Yc = (20**2/g)**(1/3)

print "Critical depth =",round(Yc,2),"m"

Critical depth = 3.44 m


### Example 10.6 Page no 380¶

In [7]:
# determine normal depth, flow regine, critical depth

from math import *

from __future__ import division

# Given

Q = 15 # flow rate in m**3/s

w = 4.0  # bottom width

S = 0.0008 # bed slope

n = 0.025 # manning constant

z = 0.5 # slope

# Solution

# We use a trial and error method here to find the value of y i.e. normal depth

y = 2.22 # we take the value of y as 2.2 m

Q = ((4+0.5*y)*(y/(n))*(((4+0.5*y)*y)/(4+2.236*y))**(0.667)*(S)**(0.5))

print "a )Normal Depth =",round(y,2),"m"

A = (4+0.5*y)*y

T = (w+2*z*y)

D = A/T

V = (Q/A)

F =V/(sqrt(9.81*D))

print "b )F = ",round(F,2)," Since the Froude number is less than 1, the flow is subcritical"

# we use trail and error to find the value of yc for critical depth

yc = 1.08

Q1 = (4+z*yc)*yc*sqrt((9.81*(4+0.5*yc)*yc)/(4+2*z*yc))

print "c )Critical depth is = ",round(yc,2),"m"

a )Normal Depth = 2.22 m
b )F =  0.31  Since the Froude number is less than 1, the flow is subcritical
c )Critical depth is =  1.08 m


### Example 10.8 Page no 390¶

In [8]:
# Height, type and length of jump and loss of energy

from math import *

from __future__ import division

# Given

b = 60                   # base width in ft

y1 = 2.5                 # base depth in ft

Q = 2500                 # discharge in ft**3/s

g = 32.2

# Solution

V1 = Q/(b*y1)

F1 = V1/sqrt(g*y1)

y2 = y1*0.5*(sqrt(1+8*F1**2)-1)

V2 = Q/(b*y2)

print "Since F1 =",round(F1,2)," It is a weak jump"

L = y2*4.25

print "Length of the jump =",round(L,0),"ft"

E1 = y1+(V1**2/(2*g))

E2 = y2+(V2**2/(2*g))

El = E1-E2

Te = El*62.4*Q/543

print "Total energy loss =",round(Te,2),"HP"

Since F1 = 1.86  It is a weak jump
Length of the jump = 23.0 ft
Total energy loss = 133.7 HP


### Example 10.12 Page no 409¶

In [9]:
# Determine flow rate

from math import *

from __future__ import division

# Given

d = 6                      # depth of the channel

w= 12                      # width of the channel

h  = 1.0                   # height of the channel

p = 9                      # pressure drop in m

g = 32.2

# Solution

y2 = y1 - h - 0.75

V1 = sqrt(2*g*0.75/((1.41)**2-1))

Q = w*b*V1/10

print "Discharge =",round(Q,0),"cfs"

Discharge = 503.0 cfs

In [ ]: