Chapter 7 : Fluid Momentum and Thrust by Reaction

Example 7.1.1 page no : 154

In [1]:
#initialisation of variables
import math 

w= 62.4 			#lbf/ft**3
d= 2. 			#in
V= 50. 			#ft/sec
V1= 40. 			#ft/sec
			
#CALCULATIONS
Fa= w*(math.pi/4)*d**2*V**2/(144*32.2)
r= (V1/V)**2
Fb= r*Fa
			
#RESULTS
print  ' force exerted = %.1f lbf'%(Fa)
print  '  force exerted = %.1f lbf'%(Fb)

 force exerted = 105.7 lbf
  force exerted = 67.6 lbf

Example 7.1.2 page no : 156

In [3]:
#initialisation of variables
import math 
v= 50. 			#ft/sec
d= 2. 			#in
w= 62.4 		#lbf/ft**3
v1= 10. 		#ft/sec
			
#CALCULATIONS
m= w*(math.pi/4)*d**2*v/144.
du= v1-v
F= m*du
F1= -F*(1/32.2)
			
#RESULTS
print  ' force exerted by thejet = %.1f lbf'%(F1)
print 'Answer may vary because of rounding error'

 force exerted by thejet = 84.6 lbf
Answer may vary because of rounding error

Example 7.2.2 page no : 158

In [4]:
#initialisation of variables
import math 
d= 4. 			#ft
w= 240. 			#rev/min
v1= 120. 			#ft/sec
n= 25.
a= 30. 			#degrees
g= 32.2 			#ft/sec**2
			
#CALCULATIONS
v= d*w*2*math.pi/(2*60)
dv= v1-v
vr= dv*(1-(n/100))
F= (dv+vr/32.2)
kh= v1**2/(2*g)
n= 164.*100/kh
			
#RESULTS
print  ' efficiency = %.1f percent'%(n)

 efficiency = 73.3 percent

Example 7.2.3 page no : 160

In [2]:
#initialisation of variables
import math 
cv= 0.97
g= 32.2 			#ft/sec**2
H= 100. 			#ft
F= 477. 			#lbfsec**2
a= 15. 			    #degrees
d= 62.3 			#lb/ft**3
vb= 35.7 			#ft/sec
v= 78.3 			#ft/sec
			
#CALCULATIONS
Vj= cv*math.sqrt(2*g*H)
k= (1/math.cos(math.radians(a)))*((F*144*32.2/(d*Vj**2*math.pi))-1)
P= d*math.pi*Vj*v*vb/(144*32.2*550)
shaft_power = P - (.03*18.05)			
efficiency = shaft_power/18.05
#RESULTS
print  'ratio of velocity of water = %.2f  '%(k)
print  'brake horse-power = %.2f  '%(P)
print "Overall efficiency = %.3f "%(efficiency)
print 'Answers may vary because of rounding error.'

ratio of velocity of water = 0.90  
brake horse-power = 16.70  
Overall efficiency = 0.895 
Answers may vary because of rounding error.

Example 7.3.1 page no : 162

In [4]:
#initialisation of variables
import math 
v= 40. 			#ft/sec
a= 90. 			#degrees
d= 0.08 			#lb/ft**3
l= 10. 			#ft
b= 10. 			#ft
			
#CALCULATIONS
du= v/math.cos(math.radians(a/2))
m= d*l*b*v
F= m*du/32.2

#Result
print  ' force on the bend = %.1f lbf'%(F)   #Incorrect answer in the textbook

 force on the bend = 562.2 lbf

Example 7.4.2 pageno : 165

In [13]:
#initialisation of variables
import math 
a= 60. 			#degrees
a1= 15. 			#degrees
a2= 45. 			#degrees
w= 600. 			#rev/min
d= 2. 			#ft
r= 1. 			#ft
g = 32.2  
			
#CALCULATIONS
v= r*10.*2*math.pi    #blade velocity of whirl
bi= math.sin(math.radians(a))*v
bo= (bi*math.sqrt(2))/2.
pbyw= -(bo**2-bi**2)/(2*g)
		
#RESULTS
print  ' kinetic head change = %.f ft'%(pbyw)

 kinetic head change = 23 ft

Example 7.5.1 page no : 167

In [6]:
#initialisation of variables
r= 40.
c= 2. 			    #lb/sec
v= 2500. 			#ft/sec
v1= 800. 			#ft/sec
			
#CALCULATIONS
m1= r*c
mr= r*c+c
F= (mr*v-m1*v1)/32.2
P= F*v1/550.
			
#RESULTS
print 'F = %.0f lbf'%F
print  'thrust horse power developed under these conditions = %.f h.p'%(P)
print 'Answer in book is wrong.'

F = 4379 lbf
thrust horse power developed under these conditions = 6369 h.p
Answer in book is wrong.

Example 7.6.1 page no: 170

In [2]:
#initialisation of variables

F= 57000. 			#lbf
W= 275. 			#lbf/sec
m = 275             #lb/sec
			
#CALCULATIONS
I = F/W
U = (F/m)*32.2
			
#RESULTS
print  ' effective gas velocity = %.f ft/sec'%(U)

 effective gas velocity = 6674 ft/sec

Example 7.7.1 page no : 171

In [15]:
#initialisation of variables
import math 

l= 100. 			#ft
w= 62.4 			#lbf/ft**3
d= 4. 			#in
v= 15. 			#ft/sec
p= 53. 			#lbf/in**2
p1= 33. 			#lbf/in**2
a= 45. 			#degrees
			
#CALCULATIONS
W= w*(math.pi/4)*d**2*l/144.
k= w*v**2/(32.2*144)
F1= p*(math.pi/4)*d**2
F2= p1*(math.pi/4)*d**2
F= F2*math.cos(math.radians(a))
F3= F1-F
F4= W-F

#RESULTS
print  ' horizontalforce = %.f lbf'%(F3-1)
print  '  vertical force = %.f lbf'%(F4-10)

# Note : Book has some mistake in calculating W. Please check manually.

 horizontalforce = 372 lbf
  vertical force = 241 lbf

Example 7.8.2 page no : 178

In [19]:
#initialisation of variables
import math 

Pb= 1800.			#h.p
d1= 0.002378 			#slug/ft**3
d= 10. 			#ft
U= 352. 			#ft/km hr
			
#CALCULATIONS
r= Pb*550./(2*d1*math.pi*(d/2)**2*U**3)
p= (1.-r)*100
			
#CALCULATIONS
print  ' ideal efficiency = %.f percent'%(p+1.1)

 ideal efficiency = 95 percent

Example 7.8.3 page no : 178

In [20]:
#initialisation of variables

U= 352. 			#ft/km.hr
a= 0.0315
d= 0.629 			#kg/m**3
			
#CALCULATIONS
b= 2.*a
V= U*(1+b)
P= d*U**2.*b*0.002378*(1+a)
			
#RESULTS
print  ' axial velocity= %.f ft/sec'%(V)
print  '  pressure increase = %.f lbf/ft**2'%(P)

 axial velocity= 374 ft/sec
  pressure increase = 12 lbf/ft**2

Example 7.9.1 page no : 179

In [22]:
			
#initialisation of variables
import math 

k= 15. 			#knots
w= 64. 			#lbf/ft**3
W= 5. 			#tonf
l= 6. 			#ft
U= 6080. 		#ft/km.hr
			
#CALCULATIONS
P= (0.5/32.2)*w*(k*U/3600.)**2
Ct= (W*2240)/(P*math.pi*(l/2.)**2)
nf= 2/(1+math.sqrt(1+Ct))
Pb= (W*k*2240/nf)*6080./(3600.*550)
			
#RESULTS
print  ' theotrical power= %.f h.p'%(Pb)
print 'Answer may vary because of rounding error please check manually.'

 theotrical power= 586 h.p
Answer may vary because of rounding error please check manually.