# Chapter 6 : Flow Through Open Channels¶

### Example 6.1 page no : 196¶

In [2]:
import math
#initialisation of variables
i = 0.000146
v = 2.8 			#ft/sec
m = 7. 			#ft
#CALCULAIONS
C = v/math.sqrt(m*i)
K = (157.6-C)*math.sqrt(m)/C
#RESULTS
print  'C  = %.3f  '%(C)
print  'K  = %.3f  '%(K)

# answers may vary because of rounding error

C  = 87.586
K  = 2.115


### Example 6.2 pageno : 197¶

In [9]:
import math
#initialisation of variables
b = 10. 			#ft
n = 1.
i = 1./1000
d = 1.5 			#ft
C = 110.
w = 62.4 			#lb/ft**3
#CALCULATIONS
L = math.sqrt(2*d**2)
P = b+2*L
A = d*(b+n*d)
m = A/P
v = round(C*math.sqrt(m*i),2)
A_v = round(A*v)
Q = A_v*w*60*60*24/10
#RESULTS
print  'Discharge  = %.2e gal/day '%(Q)

Discharge  = 3.56e+07 gal/day


### Example 6.3 page no : 197¶

In [3]:
import math
#initialisation of variables
b = 10. 			#ft
n = 2.
d = 3.5 			#ft
i = 1./625
#CALCULATIONS
A = d*(b+(d/n))
L = math.sqrt(d**2+(d/2)**2)
P = b+2*L
m = A/P
v = 1.486*m**(2./3)*i**0.5/0.03
Q = A*v
#RESULTS
print  'Discharge  = %.1f cuses '%(Q)

# Note : answer may vary because of rounding error.

Discharge  = 142.3 cuses


### Example 6.4 pageno : 198¶

In [4]:
import math
#initialisation of variables
d = 3. 			#ft
i = 1./4500
C = 80.
#CALCULATIONS
A = 0.5*(math.pi*d**2/4)
P = math.pi*d/2
m = A/P
v = C*math.sqrt(m*i)
Q = v*A
#RESULTS
print  'Discharge  = %.2f cuses '%(Q)

Discharge  = 3.65 cuses


### Example 6.5 page no : 198¶

In [6]:
import math
#initialisation of variables
A = 2500. 			#acres
n = 20.
C = 130.
i = 1./3000
p  = 7. 			#per cent
w = 62.4 			#lb/ft**3
#CALCULATIONS
Q1 = Q*50000*p/(60*100*60*w)
Q2 = Q1+(A*4840*9/(12*24*60*60))
d = (Q2*8*math.sqrt(4/i)/(math.pi*C))**0.4
#RESULTS
print  'Diameter  = %.3f ft '%(d)

# answer is different because of rounding error.

Diameter  = 8.754 ft


### Example 6.6 page no : 199¶

In [7]:
import math

#initialisation of variables
Qt = 150000. 			#cuses
i = 1/10000.
n1 = 1.
n2 = 2./3
d1 = 30. 			#ft
C1 = 100.
C2 = 75.
b1 = 600. 			#ft
b2 = 2000. 			#ft
r = 2.
A1 = (b1+d1)*d1
P1 = b1+(2*d1*math.sqrt(2))
m1 = A1/P1
v1 = C1*math.sqrt(m1*i)
Q1 = A1*v1
Q2 = Qt-Q1
v2 = v1/2
A2 = Q2/v2
d2 = (-b2+math.sqrt(b2**2+4*1.5*A2))/(2*1.5)
#RESULTS
print  'depth of water  = %.f ft '%(d2)

depth of water  = 10 ft


### Example 6.7 page no : 200¶

In [8]:
import math
#initialisation of variables
d = 3. 			#ft
i = 1./1000
C = 65.
Cd = 0.56
g = 32.2 			#ft/sec**2
h1 = 7.5 			#ft
h2 = 3. 			#ft
#CALCULATIONS
m = d
v = C*math.sqrt(m*i)
Q = v*d
H = (Q*d/(2*math.sqrt(2*g)*Cd))**(2./3)
h = h1+h2-H
#RESULTS
print  'Height of dam  = %.2f ft '%(h)

Height of dam  = 8.17 ft


### Example 6.8 page no : 207¶

In [9]:
import math
#initialisation of variables
Q =100.			#cuses
v = 2. 			#/ft/sec
n = 1.5
A = 50. 			#ft**2
C = 120.
#CALCULATIONS
d = math.sqrt((Q/v)/(2*math.sqrt(n**2+1)-n))
m = A/d
h1 = m-n*d
h2 = m+n*d
i = (v/C)**2*(2/d)
#RSULTS
print  'Depth  = %.2f ft '%(d)
print  ' Bottom width  = %.2f ft '%(h1)
print  ' Top width  = %.2f ft '%(h2)

Depth  = 4.87 ft
Bottom width  = 2.95 ft
Top width  = 17.57 ft


### Example 6.9 page no : 208¶

In [20]:
import math
#initialisation of variables
Q = 1100. 			#cuses
i = 1/1800.
C = 95.
n = 1.5
#CALCULATIONS
d = ((Q*math.sqrt(3600)/C)/(n+0.6))**0.4
b = 0.6*d
ht = b+2*(n*d)
#RESULTS
print  'Depth  = %.1f ft '%(d)
print  ' Bottom width  = %.2f ft '%(b)
print  ' Top width  = %.1f ft '%(ht)

Depth  = 10.2 ft
Bottom width  = 6.11 ft
Top width  = 36.7 ft


### Example 6.10 pageno : 209¶

In [21]:
import math
#initialisation of variables

n = 1.5
Q = 800. 			#cuses
i = 2.5/5280
n1 = 9.24
r = 0.6
k = 1.49
#CALCULATIONS
d = (k*10**7*4/n1)**(1/8.)
#RESULTS
print  'Depth of channel  = %.1f ft '%(d)

Depth of channel  = 7.1 ft


### Example 6.11 page no : 210¶

In [22]:
import math
#initialisation of variables
d = 8. 			#ft
i = 1./1200
C = 90.
a = 308. 			#degrees
#CALCULATIONS
h = 0.95*d
m = 0.29*d
Q = A*C*math.sqrt(m*i)
#RESULTS
print  'Discharge  = %.f cuses '%(Q)

Discharge  = 195 cuses


### Example 6.12 page no : 213¶

In [16]:
import math
#initialisation of variables
v = 5. 			#ft/sec
Q = 500. 			#cuses
w = 25. 			#ft
g = 32.2 			#ft/sec**2
#CALCULATIONS
h = (Q/v)/w
E = h+(v**2/(2*g))
he = round((400*2/64.4)**(1./3),2)
ve = round(20./2.32,2)
Emin = he + (ve**2 / (g*2))

#RESULTS
print  'Specific energy  = %.2f ft '%(E)
print "Critical Velocity of flow = %.2f ft/sec"%ve
print 'Minimum energy Emin = %.2f ft'%Emin

Specific energy  = 4.39 ft
Critical Velocity of flow = 8.62 ft/sec
Minimum energy Emin = 3.47 ft


### Example 6.13 page no : 216¶

In [24]:
import math
#initialisation of variables

i = 1./5000
C = 100.
b = 50. 			#ft
h = 10. 			#ft
Q = 1000. 			#cuses
g = 32.2 			#ft/sec**2
#CALCULATIONS
f = 2.*g/C**2
m = (b*h)/(b+2*h)
v = Q/(b*h)
r = (i-(f*4/(2*g*m)))/(1-(2**2/(g*h)))
s = i-r
#RESULTS
print  'Slope  = %.6f  '%(s)

Slope  = 0.000054


### Example 6.14 page no : 221¶

In [26]:
import math
#Initialization of variables
B =48. 			#ft
D =5. 			#ft
f =0.005
i =1./1000
g =32.2
#calculations
C =math.sqrt(2*g/f)
m =B*D/(B+2*D)
V =C*math.sqrt(m*i)
Q =B*D*V
Dc =(Q**2 /(g*B**2))**(1./3)
d1 =2.25 			#ft
Q1 =1*D*V
d2 =-d1/2 + math.sqrt(2*Q1**2 /(g*d1) + d1**2 /4)
hd =d2-d1
#results
print "height required  = %.1f ft"%(hd)
#The answer is a bit different due to rounding off error in textbook

height required  = 2.8 ft


### Example 6.15 page no : 222¶

In [29]:
import math
#initialisation of variables
Q = 360. 			#cfs
d1 = 1. 			#ft
B = 18. 			#ft
g = 32.2 			#ft/sec**2
w1 = 624. 			#lb/ft**3
d2 =4.5  			#ft
#CALCULATIONS
w = Q/B
v1 = w/d1
v2 = v1/d2
d2 = -0.5+math.sqrt((2*v1**2*d1/(g))+(d1**2./4))
El = (d1+(w**2/(2*g)))-(d2+(v2**2/(2*g)))
EL = round(w1*Q*El,-4)
#RESULTS
print  'loss in energy  = %.f lb '%(EL)

loss in energy  = 540000 lb


### Example 6.16 page no : 223¶

In [30]:
import math
#initialisation of variables
d1 = 4. 			#ft
v1 = 60. 			#ft/sec
g = 32.2 			#ft/sec**2
#CALULATIONS
d2 = d1*(math.sqrt(1+8*v1**2/(g*d1))-1)/2.
#RESULTS
print  'd2  = %.f ft '%(d2)

d2  = 28 ft


### Example 6.17 page no : 224¶

In [32]:
import math
#initialisation of variables
b = 150. 			#ft
d = 12. 			#ft
N = 0.03
i = 1./10000
h = 10. 			#ft
g = 32.2 			#ft/sec**2
#CALCULATIONS
A = b*d
P = b+2*d
m = A/P
v = m**(2/3.)*1.49*i**0.5/N
A1 = b*(h+d)
P1 = b+2*(h+d)
m1 = A1/P1
C1 = 1.49*m1**(1./6)/N
v1 = A*v/A1
s = (i-(v1**2/(C1**2*m1)))/(1-(v1**2/(g*(h+d))))
L = round(2*h/s,-3)
#RESULTS
print  'Length of back water  = %.f ft '%(L)

Length of back water  = 236000 ft


### Example 6.18 page no : 227¶

In [33]:
import math
#initialisation of variables
b1 = 3.2 			#ft
b2 = 1.3 			#ft
h1 = 1.86 			#ft
h2 = 1.63 			#ft
g = 32.2 			#ft/sec**2
#CALCULATIONS
a1 = b1*h1
a2 = b2*h2
Q = a1*a2*math.sqrt(2*g)*math.sqrt(h1-h2)/(math.sqrt(a1**2-a2**2))
#RESULTS
print  'Discharge  = %.1f cuses '%(Q)

Discharge  = 8.7 cuses


### Example 6.19 page no : 227¶

In [17]:
import math
#initialisation of variables
b1 = 4. 			#ft
b2 = 2. 			#ft
h1 = 2. 			#ft
g = 32.2 			#ft/sec**2
#CALCULATIONS
Qmax = 3.09*b2*h1**1.5
v1 = Qmax/(b1*h1)
H = h1+(v1**2/(2*g))
Qmax2 = 3.09*b2*H**1.5
h2 = 2*H/3
#RESULTS
print  'Qmax  = %.2f cfs '%(Qmax2)
print  'h2  = %.3f ft '%(h2)

Qmax  = 18.46 cfs
h2  = 1.383 ft


### Example 6.20 pageno : 229¶

In [20]:
import math
#initialisation of variables
h1 = 8. 			#ft
b1 = 32. 			#ft
h = 1. 			#ft
b2 = 24. 			#ft
g = 32.2 			#ft/sec**2

#CALCULATIONS
H = h1-h
h = 9.65
Q = 3.09*H**1.5*b2
v1 = Q/(b1*h1)
Q1 = 3.09*(H+(v1**2/(2*g)))**1.5*b2
hc = (Q1**2/(g*b2**2))**(1./3)
d2 = -(hc/2)+math.sqrt(9*hc**2/2)

#RESULTS
print  'Q  = %.f cfs '%(Q1)
print  'hc  = %.2f ft '%(hc)
print  'max depth  = %.2f ft '%(d2)
print  'Maximum depth of water downstream %.2f ft'%(d2+1)
print  'h = %.2f ft'%h

# answers may vary because of rounding error.

Q  = 1507 cfs
hc  = 4.97 ft
max depth  = 8.05 ft
Maximum depth of water downstream 9.05 ft
h = 9.65 ft

In [ ]: