Chapter 4 : Orifices and Notches

Example 4.1 Page No : 51

In [1]:
import math 

#initialisation of variables
Cd= 0.98    # velocity
g= 32.2 	#ft/sec**2
H= 2.    	#ft

#CALCULATIONS
v= math.sqrt(2*g*H)
t= H/v
h= 0.5*g*t**2

#RESULTS
print  'Vertical distance fallen in this ttime = %.3f ft'%(h) 

#Note : The answer given in textbook is wrong.

Vertical distance fallen in this ttime = 0.500 ft

Example 4.2 Page No : 51

In [2]:
import math 

#initialisation of variables
r= 53.4
T= 60. 	    #F pressure of air
h= 29.7 	#in of mercury
sm= 13.6
w= 62.4 	#lb/ft**3
d= 1.5  	#in diameter
Qin= 2. 	#cuses air
g=32.2 	    #ft/s**2

#CALCULATIONS
W= h*sm*w/(r*(460+T)*12)
dP= 0.75*w/(12*W)
Q= math.sqrt(2*g*dP)*math.pi*d**2/(4*144)
W= Q*W*60
Cd= Qin/W

#RESULTS
print  'coefficient of discharge = %.2f '%(Cd) 

coefficient of discharge = 0.62

Example 4.3 Page No : 52

In [3]:
import math 

#initialisation of variables
H1= 34. 	#ft height
H2= 8. 	    #ft head
H3= 7.   	#ft pressure head
g= 32.2 	#ft/sec**2
d= 1.5 	    #in

#CALCULATIONS
v2= math.sqrt(2*g*(H1+H2-H3))
Q= v2*math.pi*d**2/(4*144)
v3= (2*v2+math.sqrt(4*v2**2-4*6*(v2**2-H2*2*5*g)))/12
dr= math.sqrt(v2/v3)

#RESULTS
print  'ratio of diameteres = %.1f '%(dr) 
print " Maximum discharge = %.3f cusec"%(Q)

ratio of diameteres = 1.6 
 Maximum discharge = 0.583 cusec

Example 4.4 Page No : 54

In [4]:
import math 

#initialisation of variables
Q1= 8./15 	#cuses
Q2= 2./15 	#cuses

#CALCULATIONS
A= math.degrees(math.atan(Q2/Q1))

#RESULTS
print  'Angle of inclination = %.2f degrees'%(A) 

# rounding off error

Angle of inclination = 14.04 degrees

Example 4.5 Page No : 56

In [6]:
import math 

#initialisation of variables
g= 32.2 	#ft/sec**2

#CALCULATIONS
r= g**2/((math.sqrt(2))**2*g**2)

#RESULTS
print  'coefficient of contraction = %.1f '%(r) 

coefficient of contraction = 0.5

Example 4.6 Page No : 56

In [5]:
import math 

#initialisation of variables
B= 3. 	    #ft long
H= 2.   	#ft depth of water
H1= 3.75 	#ft 
w= 4.    	#ft wide
g= 32.2 	#ft/sec**2

#CALCULATIONS
Q= 3.33*(B-(H1/5))*H**1.5
v= Q/(H*w)
kh= v**2/(2*g)
Q1= 3.33*(B-(H1/5)-kh)*(((H1/5)+kh)**1.5-kh**1.5)

#RESULTS
print  'Discharge = %.2f cuses'%(Q1) 


# NOte : ANSWER IN THE TEXTBOOK IS WRONG

Discharge = 5.42 cuses
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