Chapter 4 : Bernoulli's Equation and Measurement of Flow of Incompressible Fluids

Example 4.1.1 page no : 74

In [1]:
#initialisation of variables

H= 33. 			#ft lbf/lbf
Q= 100. 			#ft**3/min
w= 62.4 			#lbf/ft**3
s= 0.8
			
#CALCULATIONS
P= s*w*Q*H/33000.
			
#RESULTS
print  ' power required= %.2f h.p'%(P)
    

 power required= 4.99 h.p

Example 4.2.2 page no : 78

In [2]:
#initialisation of variables

import math 
g= 32.2 			#ft/sec**2
h= 1. 			    #in
ww= 62.4 			#lbf/ft**3
w= 0.0764 			#lbf/ft**3
			
#CALCULATIONS
u= math.sqrt(2*g*h*(1./12)*(ww/w))
			
#RESULTS
print  ' speed of air through the tunnel= %.1f ft/sec'%(u)

 speed of air through the tunnel= 66.2 ft/sec

Example 4.3.1. page no : 82

In [3]:
#initialisation of variables
import math 
za= 0. 			#ft
zb= 12. 			#ft
w= 62.3 			#lbf/ft**2
pa= 750. 			#lbf/in**2
p= 700. 			#lbf/in**2
ua= 3. 			#ft/sec
g= 32.2 			#ft/sec**2
d= 2. 			#in**2
			
#CALCULATIONS
ub= 4*ua
Hl= (za-zb)+((pa-p)*144/w)+(ua**2-ub**2)/(2*g)
P= (w*ua*(math.pi/4)*Hl*d**2)/(144*550.)
			
#RESULTS
print  ' horse-power expended in over coming losses= %.2f h.p'%(P)

 horse-power expended in over coming losses= 0.75 h.p

Example 4.4.1 pageno : 85

In [5]:
#initialisation of variables
import math 
d= 1. 			#in
d1= 3. 			#in
h= 9. 			#in
p= 3. 			#percent
g= 32.2 			#ft/sec**2
s= 13.6 			#gm/cm**3
a= 0.97
			
#CALCULATIONS
Ka= 1./(1-(d/d1)**2)
C= Ka*math.pi*(d/2)**2*math.sqrt(2*g*(s-1))/144
C1= a*C
Q= C1*h/12.
			
#RESULTs
print  ' flow rate = %.2f ft**3/sec '%(Q)

 flow rate = 0.13 ft**3/sec

Example 4.4.2 page no : 86

In [1]:
#initialisation of variables
import math 
Q= 1.4 			#ft**3/sec
d= 6. 			#in
d1= 3. 			#in
h= 9. 			#in
s= 13.6/0.78
Cd= 0.96
g= 32.2 			#ft/sec**2
w= 62.3 			#lb/ft**3
			
#CALCULATIONS
h1= (Q*4.*12**2/(Cd*math.pi*d1**2))**2*(1-(d1/(2*d))**2)/(2*g*(s-1))
dpbyw= (h/12.)+((s)-1)*h1
dp= dpbyw*h1*w/144.
			
#RESULTS
print  ' pressure difference = %.2f lbf/in**2 '%(dp)

# Note : answer may be vary because of rounding error.

 pressure difference = 4.60 lbf/in**2

Example 4.5.1 page no : 88

In [7]:
#initialisation of variables
import math 
C= 0.6
s= 0.0767 			#lbf/ft**3
g= 32.2 			#ft/sec**2
w= 62.4 			#lbf/ft**3
Hw= 0.7 			#in
			
#CALCULATIONS
Ha= Hw*w/(s*12)
Q= C*math.pi*math.sqrt(2*g*Ha)/144.
			
#RESULTS
print  ' volumetric flow rate = %.3f ft**3/sec '%(Q)

 volumetric flow rate = 0.724 ft**3/sec

Example 4.5.3 pageno : 89

In [9]:
#initialisation of variables
import math 
g= 32.2 		#ft/sec**2
h= 5. 			#ft
Q= 0.6 			#ft**3/sec
Cd= 0.6
d= 2. 			#in
Q1= 0.315 		#ft**3/sec
h1= 8. 			#ft
h2= 2. 			#ft
A= 9. 			#ft**2
			
#CALCULATIONS
H= Q1**2./((Cd*math.pi*(d/24.)**2.)**2.*2*g)
T= A*2*(math.sqrt(h1)-math.sqrt(h2))/(Cd*math.sqrt(2.*g)*60.*(d/24.)**2)
dhbyt= (Q-Cd*math.pi*(d/24.)**2*math.sqrt(2*g*h))*60*12/(math.pi*A)
			
#RESULTS
print  'depth of the water = %.f ft '%(H)
print  'time taken = %.1f min '%(T)
print  'rate of rise in water = %.1f in/min '%(dhbyt)

depth of the water = 9 ft 
time taken = 12.7 min 
rate of rise in water = 9.3 in/min

Example 4.6.1 page no : 91

In [6]:
#initialisation of variables
import math 
d= 8. 			#in
d1= 1.5 		#in
Cd= 0.65
w= 62.3 		#lbf.ft**3
W= 25. 			#tonf
u= 5. 			#miles/hour
u1= 20.5        #miles/hour
			
#CALCULATIONS
ds= W*2240*d1**4*Cd**2*math.log(u1/u)/(w*d**4*math.pi*(d/24)**2)
T= W*2240*d1**4*Cd**2*((5/(u*7.33))-(20/(u1*29.35)))/(w*d**4*math.pi*(d/24)**2)


#RESULTS
print  ' Distance that piston moves= %.2f ft '%(ds)
print  ' time taken = %.4f sec '%(T)
print " Answers may vary because of rounding error."

 Distance that piston moves= 1.90 ft 
 time taken = 0.1388 sec 
 Answers may vary because of rounding error.

Example 4.7.1 page no : 94

In [11]:
#initialisation of variables

c= 0.002378 			#slug/ft**3
u= 420. 			#mile/hour.
			
#CALCULATIONS
P= 0.5*c*u*616**2/420.
			
#RESULTS
print  ' Dynamic pressure= %.f lbf/ft**2 '%(P)

 Dynamic pressure= 451 lbf/ft**2

Example 4.8.2 page no : 97

In [13]:
#initialisation of variables
import math 
g= 32.2 			#ft/sec**2
A= 13. 			#in**2
l= 10. 			#in**1.5
			
#CALCULATIONS
Q= 2*math.pi*1.05*math.sqrt(2*g*12.)*A*l/1728.
			
#RESULTS
print  ' Rate of flow= %.1f ft**3/sec '%(Q)
print 'Answer may vary because of rounding error'

 Rate of flow= 13.8 ft**3/sec 
Answer may vary because of rounding error