Chapter 5 : Elements of Similarity; Notches and Weirs

Example 5.1.1 page no : 104

In [1]:
#initialisation of variables

w= 1100. 			#rev/min
Q= 1. 			#ft**3/min
r= 2. 			#in
			
#CALCULATIONS
Q1= r**2*Q
N= w/r
			
#RESULTS
print  'allowable speed = %.f rev/min'%(N)

allowable speed = 550 rev/min

Example 5.2.2 page no : 107

In [2]:
#initialisation of variables
import math 

n= 15. 			#knots
			
#CALCULATIONS
Um= n/math.sqrt(36.)
			
#RESULTS
print  ' speed = %.1f knots'%(Um)

 speed = 2.5 knots

Example 5.3.1 page no : 109

In [3]:
#initialisation of variables
import math 

r= 1./64
t= 75. 			#sec
			
#CALCULATIONS
R= math.sqrt((1./r))
tfs= R*t/60.
			
#RESULTS
print  'time = %.f min'%(tfs)

time = 10 min

Example 5.5.2 page no : 115

In [2]:
#initialisation of variables
import math 
g= 32.2 			#ft/sec**2
			
#CALCULATIONS
C = 15.*0.305*(12)**(5./2)/(8.*8.05*60)


#RESULTS
print  'numerical value of proportional constant = %.2f '%(C)

numerical value of proportional constant = 0.59

Example 5.6.2 page no : 117

In [8]:
    
#initialisation of variables
import math 

A= 9. 			#ft**2
A1= 50000. 			#yd**2
l= 12. 			#ft
H1= 2. 			#ft
H2= 3. 			#in
g= 32.2 			#ft**2/sec
			
#CALCULATIONS
t= round((A*A1*2./(0.4*math.sqrt(2.*g)*l))*((H2/12.)**-0.5-(H1)**-0.5),-2)
			
#RESULTS
print  ' time required = %.f sec '%(t)

 time required = 30200 sec

Example 5.8.1 page no : 121

In [6]:
#initialisation of variables
import math 

A= 5. 			#ft**2
c= 0.6
g= 32.2 		#ft/sec**2
H= 6. 			#in
			
#CALCULATIONS
Q= 0.6*(8./15)*60*math.sqrt(2*g)*(H/12)**2.5
u= 0.455/A
h= u**2./(2*g)
			
#RESULTS
print  ' Discharge= %.1f ft**3/sec'%(Q)
print  '  mean approach velocity = %.3f ft/sec'%(u)
print  '  kinetic head = %.6f ft'%(h)

 Discharge= 27.2 ft**3/sec
  mean approach velocity = 0.091 ft/sec
  kinetic head = 0.000129 ft