Chapter 8 : Uniform Open Channel Flow

Example 8.1 Page No : 306

In [1]:
import math 
			
#Initialization of variables
b = 4. 			    #m
y = 1.2 			#m
sf = 0.001
n = 0.012
gam = 9.81*1000
			
#calculations
A = b*y
R = A/(b+ 2*y)
Q = 1/n *A*R**(2./3) *sf**(1./2)
T = gam*R*sf
			
#results
print "Discharge  =  %.3f m**3/s"%(Q)
print " bed shear  =  %.2f N/m**2"%(T)
#The answer in textbook is wrong for discharge. Please use a calculator.

Discharge  =  10.442 m**3/s
 bed shear  =  7.36 N/m**2

Example 8.2 Page No : 306

In [2]:
import math 
			
#Initialization of variables
b = 6. 		    	#m
y = 2. 			    #m
sf = 0.005
slope  =  2.
gam = 9.81*1000
Q = 65. 			#m**3/s
			
#calculations
A = (b+ 2*y)*slope
P = b+ 2*y*math.sqrt(slope**2 +1)
R = A/P
V = Q/A
n = R**(2./3) *sf**(1./2) /V
			
#results
print "Value of mannings coefficient  =  %.3f"%(n)

Value of mannings coefficient  =  0.026

Example 8.3 Page No : 307

In [3]:
import math 
			
#Initialization of variables
b = 3. 			#m
y = 1. 			#m
sf = 0.005      #slope
n = 0.028       
gam = 9.81*1000
Q = 0.25 			#discharge - m**3/s
slope = 1.5
			
#calculations
A =  0.5 *b*y
P = 2*math.sqrt(1 + (slope)**2)
R = A/P
yx =  Q*n/(slope * R**(2./3) *sf**(1./2))
y =  yx**(3./8)
			
#results
print "depth  =  %.2f m"%(y)

depth  =  0.45 m

Example 8.4 Page No : 307

In [4]:
import math 
			
#Initialization of variables
sf = 0.0064         #slope
n = 0.015
Q = 6.   			#discharge - m**3/s
gam = 9.81*1000
			
#calculations
AR =  n*Q/math.sqrt(sf)
print ("On trial and error, ")
y = 0.385 			#m
print "normal depth  =  %.3f m"%(y)

On trial and error, 
normal depth  =  0.385 m

Example 8.5 Page No : 308

In [1]:
import math 
from numpy import *
			
#Initialization of variables\
sf = 0.00007
n = 0.013
gam = 9.81*1000
V = 0.45 			#velocity - m/s
Q = 1.4 			#m**3/s
			
#calculations
by = Q/V
#x = poly(0,"x")
#y = roots(x**2 -2.66*x +1.55)
y = roots([1,-2.66,1.55])
b = by/y
			
#results
print "y  =  ", y 
print "corresponding b = " ,b

y  =   [ 1.7978675  0.8621325]
corresponding b =  [ 1.73044516  3.60862294]

Example 8.6 Page No : 310

In [2]:
import math 
			
#Initialization of variables
sf = 0.0016         #slope
n = 0.02
Q = 0.84 			#m**3/s
gam = 9.81*1000
			
#calculations
y53 =  Q*n/math.sqrt(sf)
y = y53**(3./5)
			
#results
print "depth of flow  =  %.2f m"%(y)

depth of flow  =  0.59 m

Example 8.7 Page No : 313

In [7]:
import math 
			
#Initialization of variables
n = 0.015
Q = 1.3 			#m**3/s
V = 0.6 			#m/s
gam = 9.81*1000
			
#calculations
alpha = 60. 			#degrees
A = 0.5 *(1./2)**2 *(180-alpha)/180 *math.pi -(1./4)**2 *math.radians(math.tan(alpha))
A = 0.206
P = 0.5*(180-alpha)/180 *math.pi
R = A/P
d2 = V*n/(R**(2./3))
d8 =  Q*n*4*4**(2./3) /math.pi
d = math.sqrt(d8/d2)
sf =  (d2**2/d**(4./3))
	
#results
print "Diameter  =  %.2f m"%(d)
print " slope  =  %.5f "%(sf)
#The answer given in textbook is wrong. please check

Diameter  =  1.53 m
 slope  =  0.00040

Example 8.8 Page No : 315

In [2]:
import math 
			
#Initialization of variables
b = 0.5 			#m
y = 0.35 			#m
sf = 0.001          #slope
nc = 0.016
gam = 9.81*1000
Q = 0.15 			#m**3/s
			
#calculations
A = b*y
P =  b+ 2*y
R = A/P
ng = 1/Q *A*R**(2./3) *sf**(1./2)
n =  (b*nc**(3./2) + 2*y*ng**(3./2))**(2./3) /(P**(2./3))
Q2 = 1/n *A*R**(2./3) *sf**(1./2)
			
#results
print "flow in case 2  =  %.3f m**3/s"%(Q2)

# note : rounding off error

flow in case 2  =  0.120 m**3/s

Example 8.9 Page No : 316

In [3]:
import math 
			
#Initialization of variables
b1 = 8. 			#m
b2 = 5. 			#m
y = 5.  			#m
b5 = 15. 			#m
b3 = 3. 			#m
b4 = 3. 			#m
y2 = 2. 			#m
y3 = 3. 			#m
n1 = 0.025
n2 = 0.035
sf = 0.0008
			#calcuations
A =  (b1+b2)*y
P =  b1+ math.sqrt(b2**2 +y**2) + math.sqrt(b3**2 +b4**2)
R = A/P
Q1 = 1/n1 *A*R**(2./3) *sf**(1./2)
A2  =  b5*y2 - 0.5*y2*y2 + 0.5*y3*y2
P2 =  b5 + math.sqrt(b4**2 + y3**2)
R2 = A2/P2
Q2 =  1/n2 *A2*R2**(2./3) *sf**(1./2)
Q = Q1+Q2
			
#results
print "Total discharge  =  %.f m**3/s"%(Q)

# rounding off error

Total discharge  =  200 m**3/s

Example 8.10 Page No : 320

In [10]:
import math 
			
#Initialization of variables
Q = 12. 			#m**3/s
n = 0.023
A = 2.472
b = 0.472
sf = 1./8000
			
#calculations
y8 =  Q*n/A *2**(2./3) /sf**(1./2)
y = y8**(3./8)
b2 =  b*y
			
#results
print "depth  =  %.3f m"%(y)
print " width  =  %.2f m"%(b2)

depth  =  2.819 m
 width  =  1.33 m

Example 8.11 Page No : 320

In [11]:
import math 
			
#Initialization of variables
Q = 30. 
V = 1.
			
#calculations
A = Q/V
y  =  math.sqrt(A/(math.sqrt(2) + 0.5))
b =  (A- 0.5*y**2)/y
			
#results
print "width  =  %.2f m"%(b)
print " depth  =  %.2f m"%(y)

width  =  5.60 m
 depth  =  3.96 m