# Chapter 9 : Conservation law for Momentum¶

## Example 9.1 Page no 87¶

In [1]:
from __future__ import division
print "Example 9.1 page no 87\n\n"
#a horizontal water jet impinges on avertical plate
rho=62.4#density of water
v=100#horizontal velocity of water
q=0.5#flow rate
g=32.2#gravitational constant
print "\n density rho=%0.3f lb/ft**3\n horizontal velocity of water v=%0.3f ft/s\n flow rate q=%0.3f ft**3/s"%(rho,v,q)#
M_in=(rho*q*v)/g#momentum rate of inlet water in the horizontal direction
print "\n momentum rate M_in=%0.3f lbf"%(M_in)#
M_out=0#momentum rate of water out
F=M_out-M_in
print "\n net horizontal force F=%0.3f lbf"%(F)#
#negative sign indicate that to hold the plate in place, a force must be exerted in a direction opposite to that of the water flow

Example 9.1 page no 87

density rho=62.400 lb/ft**3
horizontal velocity of water v=100.000 ft/s
flow rate q=0.500 ft**3/s

momentum rate M_in=96.894 lbf

net horizontal force F=-96.894 lbf


## Example 9.2 Page no 87¶

In [2]:
from __future__ import division
from math import sqrt
print "Example 9.2 page no 87\n\n"
#a horizontal line carries saturated steam
#water is entrained by the steam,and line is bend
#select the control volume as the fluid in the bend and apply amass balance
#since m1_dot=m2_dot,v1=v2
m_dot=0.15#mass flow rate
V_in_x=420#velocity in horizontal x direction
V_out_x=0#velocity out ,horizontal direction
print "mass flow rate m_dot=%0.3f kg/s\n velocity in x direction V_in=%0.3f m/s\n velocity out in  the x direction=%0.3f=m/s"%(m_dot,V_in_x,V_out_x)#
#applying linear horizontal balance in x direction
F_x=m_dot*V_out_x-m_dot*V_in_x#force in x-dir
print "\n force F_x=%0.3f N"%(F_x)#
#the x-dir force acting on the 90 deg elbow therefore,F_x=+63 N
V_in_y=0#velocity in vertical in y direction
V_out_y=420#velocity out vertical in y direction
print "velocity in y dir V_in_y=%0.3f m/s\n velocity out y dir V_out_y=%0.3f m/s"%(V_in_y,V_out_y)#
F_y=m_dot*V_out_y-m_dot*V_in_y#force in y dir
print "\n force in y dir F_y=%0.3f N"%(F_y)#
#y dir force is acting on the elbow is therefore F_y=-63 N
F_res=sqrt(F_x*F_x+F_y*F_y)#resultant force F_res
print "\n resultant force F_res=%0.3f N"%(F_res)#
#this is the force required to hold the elbow

Example 9.2 page no 87

mass flow rate m_dot=0.150 kg/s
velocity in x direction V_in=420.000 m/s
velocity out in  the x direction=0.000=m/s

force F_x=-63.000 N
velocity in y dir V_in_y=0.000 m/s
velocity out y dir V_out_y=420.000 m/s

force in y dir F_y=63.000 N

resultant force F_res=89.095 N


## Example 9.3 Page no 88¶

In [3]:
from __future__ import division
from math import pi,sqrt
print "Example 9.3 page no 88\n\n"
#water flow in a pipe
rho=62.4#density of water
D=0.167#diameter of pipe
g=32.174#gravitational constant
M_dot_out=0#momentum out in x dir
F_x=5#foce in the x dir
print "density  rho=%0.3f lb/ft**3\n diameter D=%0.3f ft\n momentum M_dot_out=%0.3f lbf\n forc in x dir F_x=%0.3f lbf"%(rho,D,M_dot_out,F_x)#
M_dot_in=M_dot_out+F_x#momentum in
print "\n momentum M_dot_in=%0.3f lbf"%(M_dot_in)#
S=(pi*D**2)/4#surface area
print "\n surface area S=%0.3f ft**2"%(S)#
v=sqrt((M_dot_in*g)/(rho*S))
print "\n velocity =%0.3f ft/s"%(v)#
q=S*v#volumatric flow rate
m_dot=rho*q#mass flow rate
print "\n volumatric flow rate q=%0.3f ft**3/s\n mass flow rate m_dot=%0.3f lb/s"%(q,m_dot)#

Example 9.3 page no 88

density  rho=62.400 lb/ft**3
diameter D=0.167 ft
momentum M_dot_out=0.000 lbf
forc in x dir F_x=5.000 lbf

momentum M_dot_in=5.000 lbf

surface area S=0.022 ft**2

velocity =10.849 ft/s

volumatric flow rate q=0.238 ft**3/s
mass flow rate m_dot=14.828 lb/s


## Example 9.4 Page no 89¶

In [4]:
from __future__ import division
from math import sqrt,pi
print "Example 9.4 page no 89 fig. 9.2\n\n\n"
#water is discharged through a fire hose
rho=1000#density of water
meu=0.001#viscosity of water
q=0.025#flow rate  at section 1
D1=.1#diameter at section 1
D2=.03#diameter at section 2
print "\n density rho=%0.3f kg/m**3\n viscosity meu=%3f kg/m.s\n volumatric flow rate q=%0.3f m**3/s\n diametetr at section1 D1=%0.3f m\n diameter at section2 D2=%0.3f m"%(rho,meu,q,D1,D2)#
S1=(pi*D1**2)/4
S2=(pi*D2**2)/4
print "\n surface area at section 1 S1=%0.3f m**2\n surface area at section 2 S2=%0.3f m**2"%(S1,S2)#
v1=q/S1#velocity at section1
v2=q/S2#velocity at section2
print "\n velocity at sec1 v1=%0.3f m/s\n velocity at sec2 v2=%0.3f m/s"%(v1,v2)#
#appuing bernoulli's equation between point 1 and 2
P2=0#pressure at point 2
P1=(rho/2)*(v2**2-v1**2)#pressure at point 1
print "\n pressure at point2 P2=%0.3f Pag(pascal gauge)\n pressure atpoint1 P1=%0.3f Pag"%(P2,P1)#
m_dot1=25#mass flow rate at section 1
m_dot2=25#mass flow rate at section 2
print "\n mass flow rate m_dot1=%0.3f kg/s\n mass flow rate m_dot2=%0.3f kg/s"%(m_dot1,m_dot2)#
M_dot1_x=m_dot1*v1#momentum rate in x dir at section 1
M_dot2_x=m_dot2*v2#momentum rate in x dir at section 2
print "\n momentum rate M_dot1_x=%0.3f N\n momentum rate M_dot2_x=%0.3f N"%(M_dot1_x,M_dot2_x)#
#applying momentum balance in the x direction
F_x=M_dot2_x-M_dot1_x-P1*S1#force from momentum balance
print "\n force from momentum balance F_x=%0.3f N"%(F_x)#

Example 9.4 page no 89 fig. 9.2

density rho=1000.000 kg/m**3
viscosity meu=0.001000 kg/m.s
volumatric flow rate q=0.025 m**3/s
diametetr at section1 D1=0.100 m
diameter at section2 D2=0.030 m

surface area at section 1 S1=0.008 m**2
surface area at section 2 S2=0.001 m**2

velocity at sec1 v1=3.183 m/s
velocity at sec2 v2=35.368 m/s

pressure at point2 P2=0.000 Pag(pascal gauge)
pressure atpoint1 P1=620373.346 Pag

mass flow rate m_dot1=25.000 kg/s
mass flow rate m_dot2=25.000 kg/s

momentum rate M_dot1_x=79.577 N
momentum rate M_dot2_x=884.194 N

force from momentum balance F_x=-4067.784 N