Chapter 4 : Conservation Principle of Momentum

Example 4.1 Page No : 111

In [1]:
import math 
			
#Initialization of variables
Q = 0.2 			#m**3/s
v = 30. 			#m/s
angle = 120. 			#degrees
rho = 1000. 			#kg/m**3
			
#calculations
Rx = rho*Q*(v-v*math.cos(math.radians(angle)))
Ry = rho*Q*v*math.sin(math.radians(angle))
R = math.sqrt(Rx**2 +Ry**2)
			
#results
print "Resultant force  =  %.2f kN"%(R/1000)

Resultant force  =  10.39 kN

Example 4.3 Page No : 112

In [2]:
import math 
			
#Initialization of variables
angle  = 45. 		#degrees
p1 = 150.*10**3 	#pressure at inlet - N/m**2
Q = 0.5 			#rate of flow - m**3/s
d1 = 60. 			#cm
d2 = 30. 			#cm
rho = 1000. 		#kg/m**3
g = 9.81 			#m/s**2
			
#calculations
V1 = Q/(math.pi/4 *(d1/100)**2)
V2 = V1*(d1/d2)**2
P2 = rho*g*(p1/(rho*g) + V1**2 /(2*g) -V2**2 /(2*g))
Rx = p1*math.pi/4*(d1/100)**2 - P2*math.pi/4 *(d2/100)**2 *math.cos(math.radians(angle)) -rho*Q*(V2*math.cos(math.radians(angle)) -V1)
Ry = P2*math.pi/4 *(d2/100)**2 *math.sin(math.radians(angle)) + rho*Q*(V2*math.sin(math.radians(angle)))
R = math.sqrt(Rx**2 + Ry**2)
			
#results
print "resultant force  =  %.2f kN"%(R/1000)

resultant force  =  35.58 kN

Example 4.4 Page No : 113

In [5]:
import math 
			
#Initialization of variables
Q = 20.*10**3 			#discharge - cc/s
depth = 4.   			#m
d = 5.   	    		#cm
g = 9.81 		    	#m/s**2
rho = 10.**3 			#kg/m**3
			
#calculations
V1 =  Q/(math.pi/4 *d**2) /100
V2 =  math.sqrt(2*g*(V1**2/(2*g) + depth))
W = rho*Q*(V2-V1)/10**6
			
#results
print "weight of water  =  %d N"%(W)

weight of water  =  66 N

Example 4.5 Page No : 116

In [6]:
import math 
			
#Initialization of variables
g = 9.81 	    		#m/s**2
rho = 10.**3 			#kg/m**3
V = 50. 		    	#velocity - m/s
u = 20. 			    #m/s
A = 6./10**4 			#area - m**2
angle = 180. 			#degrees
			
#calculations
Vr = V-u
rq = rho*A*Vr
Rx = -rq*(Vr*math.cos(math.radians(angle)) - Vr)
Rx2 = -rho*A*V*(Vr*math.cos(math.radians(angle)) -Vr)
power = Rx2*u
			
#results
print "Force exetred on fluid  =  %d N"%(Rx)
print " Force transferred in case 2  =  %d N"%(Rx2)
print " Power transferred in case 2  =  %d kW"%(power/1000)

Force exetred on fluid  =  1080 N
 Force transferred in case 2  =  1800 N
 Power transferred in case 2  =  36 kW

Example 4.6 Page No : 119

In [8]:
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 		#kg/m**3
Vr = 10. 			#m/s
u = 8.5 			#m/s
A = 250./10**4 		#m**2
			
#calculations
V = Vr-u
Q = A*Vr
R = rho*Q*V
P = R*u
eth = 1/(1+ V/(2*u))
			
#results
print "Power required  =  %.3f kW"%(P/1000)
print " Efficiency of jet propulsion  =  %.0f percent"%(eth*100)

Power required  =  3.188 kW
 Efficiency of jet propulsion  =  92 percent

Example 4.7 Page No : 123

In [5]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 		#kg/m**3
v1 = 20. 			#m/s
v2 = 5. 			#m/s
r1 = 50./100 		#cm
r2 = 30./100 		#cm
a1 = 20. 			#degrees
a2 = 80. 			#degrees
N = 300. 			#rpm
Q = 5. 			    #m**3/s
			
#calculations
u1 = round(math.pi*2*r1*N/60,1)
u2 = round(math.pi*2*r2*N/60,2)
T = rho*Q*(r1*v1*math.cos(math.radians(a1)) - r2*v2*math.cos(math.radians(a2)))
H = 1/g *(u1*v1*math.cos(math.radians(a1)) - u2*v2*math.cos(math.radians(a2)))
power = rho*g*Q*H
	
  
#results
print "torque  =  %d N m"%(T)
print " Heat  =  %.1f m"%(H)
print " Power  =  %d kW"%(power/10**3)

# Note :The answers given in textbook are a bit different due to rounding off error. please check using calculator.

torque  =  45682 N m
 Heat  =  29.2 m
 Power  =  1434 kW

Example 4.8 Page No : 124

In [10]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 		#kg/m**3
d1 = 0.05 			#m
d2 = 0.3 			#m
N = 1800. 			#rpm
Q = 0.425/60 		#m**3/s
			
#calculations
u1 = math.pi*d1*N/60
u2 = math.pi*d2*N/60
T = rho*Q*(d2*u2 - d1*u1)/2
			
#results
print "Torque supplied  =  %.1f Nm"%(T)

Torque supplied  =  29.2 Nm