Chapter 7 : Flow Through Open Channels

Example 7.1 Page No : 240

In [1]:
import math 
		
#initialisation of variables
i = 1./4500
w =3. 		#ft
d = 3. 		#ft
k = 0.003
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
A = 0.5*math.pi*d**2/4
P = math.pi*d/2
m = A/P
f = k*(1+(0.1/m))
C = math.sqrt(2*g/f)
V = C*math.sqrt(m*i)
Q = A*V
		
#RESULTS
print  'Discharge = %.2f cuses'%(Q)

Discharge = 6.28 cuses

Example 7.2 Page No : 241

In [2]:
import math 
		
#initialisation of variables
b = 40. 		#ft
d = 4.   		#ft
k = 0.004
g = 32.2 		#ft/sec**2
Q = 500. 		#cuses
		
#CALCULATIONS
A = b*d
P = b+2*d
m = A/P
f = k*math.sqrt(1+(0.2/m))
C = math.sqrt(2*g/f)
V = Q/A
i = V**2/(C**2*m)
D = 5280*i
		
#RESULTS
print  'fall in feet per mile = %.1f ft'%(D)

# rounding off error

fall in feet per mile = 1.0 ft

Example 7.3 Page No : 242

In [6]:
import math 
		
#initialisation of variables
b = 40. 		#ft
d = 4.   		#ft
n = 1.
k = 0.005
i = 1./3250
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
A = (b+d)*d
P = round(b+2*d*math.sqrt(n**2+1),2)
m = round(A/P,2)
f = k*(1+(0.8/m))
C = round(math.sqrt(2*g/f),2)
V = C*math.sqrt(m*i)
Q = V*A

#RESULTS
print  'Discharge = %.f cuses'%(Q)

# book answer is wrong. please check.

Discharge = 584 cuses

Example 7.4 Page No : 243

In [4]:
import math 
		
#initialisation of variables
Q = 400. 		#cuses
V = 2.   		#ft/sec
d = 3. 	    	#ft
n = 1.
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
A = Q/V
w = A/d
W = w-d
P = W+2*d*math.sqrt(n**2+1)
m = A/P
f = 0.006*(1+(4/m))
C = math.sqrt(2*g/f)
i = (V/C)**2/m
		
#RESULTS
print  ' slope  = %.5f '%(i)

		#ANSWER IN TEXTBOOK IS NOT GIVEN IN DECIMALS

 slope  = 0.00033

Example 7.5 Page No : 244

In [5]:
import math 
		
#initialisation of variables
Q = 600. 		#cuses
V = 3.   		#ft/sec
n = 1.
i = 1./3200
C = 80.
d = 6. 		#ft
		
#CALCULATIONS
A = Q/V
m = V**2/(C**2*i)
b = (A/d)-d
		
#RESULTS
print  'width = %.1f ft'%(b)

width = 27.3 ft

Example 7.6 Page No : 245

In [6]:
		
#initialisation of variables
Q = 20. 		#gallons / day
i = 50000. 		#inhabitants
p = 10. 		#percent
t = 24. 		#hrs
T = 0.25 		#in
a = 2000. 		#acres
		
#CALCULATIONS
q = Q*i*p/(100*60*60*6.24)
A = T*43560*a/12
Q1 = A/(t*60*60)
Q2 = q+Q1
		
#RESULTS
print  'total discharge = %.2f cuses'%(Q2)

total discharge = 25.46 cuses

Example 7.7 Page No : 249

In [7]:
import math 
		
#initialisation of variables
Q = 400. 		#cuses
V = 8. 	    	#ft/sec
C = 150.
		
#CALCULATIONS
A = Q/V
d = math.sqrt(A/2)
i = V**2/(C**2*(d/2))
		
#RESULTS
print  'slope %.4f '%(i)

slope 0.0011

Example 7.8 Page No : 250

In [8]:
import math 
		
#initialisation of variables
Q = 100. 		#cuses
V = 2.   		#ft/sec
n = 1.5
k = 0.006
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
A = Q/V
d = math.sqrt(A/((2*math.sqrt(n**2+1))-n))
m = A/d
mb = m-n*d
bt = m+n*d
m1 = d/2
f = k*(1+(4/m1))
C = math.sqrt(2*g/f)
i = V**2/(C**2*m1)
		
#RESULTS
print  'slope %.5f '%(i)

slope 0.00040

Example 7.9 Page No : 251

In [9]:
import math 
		
#initialisation of variables
i = 1./1000
d = 4. 		#ft
C = 125.
k = 0.95
o = 5.372
		
#CALCULATIONS
h = k*d
A = d**2*(o- math.sin(math.radians(o*180/math.pi)))/8
P = (d/2)*o
m = A/P
V = C*math.sqrt(m*i)
Q = V*A
		
#RESULTS
print  'Discharge = %.2f cuses'%(Q)

Discharge = 52.18 cuses

Example 7.10 Page No : 254

In [10]:
#initialisation of variables
Cd = 0.95
m = 300. 		#ft
V = 8. 		#ft/sec
d = 6. 		#ft
n = 6.
s = 40. 		#ft
g = 32.2 		#ft/sec**2
dh = 0.11
		
#CALCULATIONS
h = (V**2/(g+(d/3)))*(1.1*(m/(s*n))**2-1)
h1 = (V**2/(2*g))*(1.1*(m/(s*n))**2-(d/(s/n)))+dh
		
#RESULTS
print  'afflux upstream = %.2f ft'%(h1)

afflux upstream = 0.92 ft

Example 7.11 Page No : 255

In [11]:
import math 
		
#initialisation of variables
V = 8.   		#ft/sec
g = 32.2 		#ft/sec**2
d = 10. 		#ft
l = 2. 	    	#ft
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
a = math.sqrt(((l*g*l/V**2)+(d/12)**2)/1.1)
V1 = V*d/12
va = math.sqrt(2*g*0.69)
v1 = math.sqrt(2*g*(l+0.69))
		
#RESULTS
print  'total head producing velocity = %.1f ft/sec'%(v1)

total head producing velocity = 13.2 ft/sec

Example 7.13 Page No : 261

In [20]:
import math 
		
#initialisation of variables
d = 8. 		#ft
V = 6. 		#ft/sec
g = 32. 		#ft/sec**2
		
#CALCULATIONS
h = (V*d/4)**2/g
#d2 = (-4/2)+math.sqrt((2*(d/2)*(V*(d/2))/g)+((d/2)**2/4))
d2 = round((-4./2) + math.sqrt(2*4*12**2/g + 4**2/4. ),3)
x = (d/2)/d2
l = ((1/(x**1.5))-1)**0.81
Lw = l*(d/2)*10.3

#RESULTS
print  'height of standing wave = %.1f ft'%(Lw)

# note : answer are different because of rounding off error. this answer is accurate.

height of standing wave = 7.6 ft

Example 7.14 Page No : 264

In [14]:
import math 		

#initialisation of variables
w = 9. 		#in
wc = 6. 		#in
d = 8. 		#in
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
Q = 3.09*(wc/12)*(d/12)**1.5
V = Q*144/(w*d)
H = (d/12)+(V**2/(2*g))
Q = 3.09*(wc/12)*H**1.5
		
#RESULTS
print  'Discharge = %.2f cuses'%(Q)

Discharge = 0.93 cuses

Example 7.15 Page No : 265

In [38]:
import math 
from sympy import Symbol,solve
		
#initialisation of variables
i = 1./6400
b = 40. 		#ft
d = 5. 		#ft
C = 140.
h = 6. 		#ft
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
A = b*d
P = b+2*d
m = A/P
v = C*math.sqrt(m*i)
V = v*(d/h)
Q = v*b*d
x = h-(Q/(3.09*(b/2)))**(2./3)-(V**2/(2*g))

#RESULTS
print  'height of pump = %.2f ft'%(x)

# rounding off error. please check.

height of pump = 0.82 ft

Example 7.16 Page No : 268

In [16]:
import math 
		
#initialisation of variables
w = 40. 		#ft
h = 5. 		#ft
P =50. 		# lb/ft**2
i = 1./6400
h1 = 10. 		#ft
H = 100. 		#ft
g = 32.2 		#ft/sec**2

#CALCULATIONS
m = w*h/P
v = 140*math.sqrt(m*i)
v1 = v*h/h1
h2 = w*h1/(H-w)
a = v1**2/(140**2*h2)
s = (i-a)*1000/(1-(v1**2/(g*h1)))
dh = h1-s
		
#RESULTS
print  'depth of water = %.3f ft'%(dh)

depth of water = 9.866 ft

Example 7.17 Page No : 269

In [48]:
import math 
		
#initialisation of variables
h = 9.  		#ft
h1 = 9.5 		#ft
i = round(1./6400,6)
h2 = 40. 		#ft
h3 = 59. 		#ft
h4 = 5. 		#ft
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
m = round(h2*h1/h3,1)
v = 3.5 * h4/h1 #140*math.sqrt(m*i)*(h4/h1)
a = round(v**2/(140**2*m),6)
s = (i-a)/(1-0.11)
x = 1/s
 
#RESULTS
print  'distance upstream from the dam = %.f ft'%(x)

# answer is different because value of s is 0.000156 and in book it is taken as 0.00013 so rounding off error"

distance upstream from the dam = 6899 ft