# Chapter 13 : Varying Flow in Open Channels¶

### Example 13.1.1 pageno : 413¶

In [2]:
#initialisation of variables

b= 15. 			#in
h= 1.25 			#in
h1= 2.75 			#in
g= 32.2 			#ft/sec**2

#CACULAIONS
Q= 3.09*(b/12.)*(h/12.)**1.5
u1= Q*144./(b*h1)
H= (u1**2./(2.*g))*12.
h2= H+h
Q1= 3.09*(b/12)*(h2/12.)**1.5

#RESULTS
print ' Q = %.2f ft**3/sec'%Q
print  ' Rate of flow= %.4f ft**3/sec'%(Q1)

 Q = 0.13 ft**3/sec
Rate of flow= 0.1359 ft**3/sec


### Example 13.3.1 pageno : 418¶

In [3]:
import math
#initialisation of variables
C= 100. 			#ft**0.5/sec
m= 2. 			#ft
i= 0.0003
y1= 2.6 			#ft
y2= 2.5 			#ft

#CALCULATIONS
u= C*math.sqrt(m*i)
f= u**2./(2.*32.2)
x= i/(1-f)
x1= round((y1-y2)/x,-1)

#RESULTS
print  ' Distance= %.f ft'%(x1)

 Distance= 300 ft


### Example 13.4.1 pageno : 420¶

In [4]:
import math
#initialisation of variables

u1= 0.5 			#ft**3/sec
b= 5. 			#ft
w= 4. 			#ft
g= 32.2			#ft/sec**2

#CALCULATIONS
u= u1*12.*12/(b*w)
s= math.sqrt(g*w/12.)
F= u/s
r= 0.5*(math.sqrt(1.+8*F**2)-1)
y= r*w
yc= (((w*y*(y+w)))/2.)**(1/3.)

#CALCULATIONS
print  ' critical depth= %.2f in'%(yc)

# answer may vary because of rounding error

 critical depth= 4.26 in


### Example 13.4.2 page no : 421¶

In [5]:
import math
#initialisation of variables

w= 2. 			#ft
F= 3.
d= 2. 			#ft
g= 32.2 			#ft/sec**2
w1= 62.3 			#lbf/ft**3

#CALCULATIONS
r= 0.5*(math.sqrt(1+8.*F**2.)-1)
y1= w/r
dy= w-y1
h1= dy**3/(4*w*y1)
u1= F*math.sqrt(g*y1)
W= w1*y1*u1*d*h1/550.

#RESULTS
print  ' Horse-power dissipated = %.2f h.p'%(W)

 Horse-power dissipated = 1.11 h.p


### Example 13.5.1 page no : 425¶

In [7]:
import math
#initialisation of variables

Q= 20. 			#ft/sec
h= 12. 			#in
g= 32.2 			#ft/sec**2

#CALCULATIONS
F= Q/math.sqrt(g*h/12.)
r= 0.5*(math.sqrt(1+8.*F**2)-1)
y= h*r/12.
s=(y-(h/12.))**3*12./(4.*h*y)
Q1= s*62.3*Q/550.

#RESULTS
print  ' Rate of flow= %.2f in'%(Q1)

# Answers may vary because of rounding error

 Rate of flow= 5.43 in


### Example 13.6.1 pageno : 427¶

In [9]:
import math
#initialisation of variables

d= 0.94
b= 20. 			#ft
h= 5. 			#ft
w= 40. 			#ft
g= 32.2 			#ft/sec**2

#CALCULATIONS
Q= 0.309*d*b*h**1.5
u=Q/(h*w)
h1= h+(u**2/(2*g))
Q1= 0.309*d*b*h1**1.5

#RESULTS
print  ' Rate of flow= %.1f ft**3/sec'%(Q1)

# Answer may vary because of rounding error.

 Rate of flow= 65.0 ft**3/sec