## Chapter 5 : Fluid Momentum¶

### Example 5.1 Page no 192¶

In :
# Resultant force on the elbow

from math import *

# Given

Q = 0.3                 # Water flow rate in m**3/s

d1 = 0.3                # diameter at inlet in meters

A1 = pi*d1**2/4 # inlet area in m**2

d2 = 0.15               # diameter at outlet in m

A2 = pi*d2**2/4 # area at outlet in m**2

P1 = 175*10**3          # inlet pressure in kN/m**2

P2 = 160*10**3          # Pressure at outlet in kN/m**2

F1 = P1*A1              # Force at inlet

F2 = P2*A2              # Force at outlet

rho = 1000              # density of water in kg/m**3

V1 = Q/A1               # inlet velocity in m/s

V2 = Q/A2               # Velocity at outlet in m/s

theta = 45*pi/180  # angle in deg

# Soultion

# Applying the X momentum equation we get

Rx = F1 - F2*cos(theta)-rho*Q*(V2*cos(theta)-V1)

# Applying the Y momentum equation

Ry = F2*sin(theta)+rho*Q*(V2*sin(theta)-0)

R = sqrt(Rx**2+Ry**2)

print "Resultant force on the elbow = ",round(R,2),"N"

a = atan(Ry/Rx)*180/pi

print "Angle of resultant force = ",round(a,4),"deg"

Resultant force on the elbow =  9800.58 N
Angle of resultant force =  34.8516 deg


### Example 5.2 Page no 194¶

In :
# Force exerted by the jet on the vane

from math import *

# Given

V1 = 80            # Velocity in ft/s

A1 = 0.1            # area in ft**2

g = 32.2           # Acceleration due to gravity in ft/s**2

rho = 1.94         # density in lb/ft**3

a = pi/3 # angle of pipe bend

# Solution

Q = A1*V1          # Total discharge in m**3

# Applying bernoullis at point 1 and 2

V2 = sqrt((2*g*V1**2/(2*32.2))-3*2*g)

# Pressure at the end of the section are atmospheric and hence 0

# momentum equation in X direction

Rx = -(rho*Q*(V2*cos(a)-80))

# momentum equation in Y direction

Ry = (rho*Q*(V2*sin(a)-0))

R = sqrt(Rx**2+Ry**2)

print "Resultant force = ",round(R,0),"lbs"

ang = atan(Ry/Rx)*180/pi

print "Angle of resultant force = ",round(ang,4),"deg"

Resultant force =  1232.0 lbs
Angle of resultant force =  59.2396 deg


### Example no 5.3 Page no 195¶

In :
# Force  needed to hold the Y position

# Given

from __future__ import division

from math import *

Q1 = 0.5             # discharge from pipe 1 in m**3/s

Q2 = 0.3             # discharge from pipe 2 in m**3/s

Q3 = 0.2             # discharge from pipe 3 in m**3/s

d1 = 0.45            # diameter of pipe 1 in m

d2 = 0.3             # diameter of pipe 2 in m

d3 = 0.15            # diameter of pipe 3 in m

A1 = pi*d1**2/4  # area in m**2

A2 = pi*d2**2/4  # area in m**2

A3 = pi*d3**2/4  # area in m**2

P1 = 60*10**3        # Pressure at point 1 in kPa

gma = 9810

g = 9.81             # acceleration due to gravity in m/s**2

rho = 1000           # density in kg/m**3

# Solution

V1 = Q1/A1

V2 = Q2/A2

V3 = Q3/A3

P2 = gma*((P1/gma) + V1**2/(2*g) - V2**2/(2*g))

P3 = gma*((P1/gma) + V1**2/(2*g) - V3**2/(2*g))

F1 = P1*A1

F2 = P2*A2

F3 = P3*A3

Rx = rho*(Q2*V2*cos(pi/6)-Q3*V3*cos(pi/9)-0)+F3*cos(pi/9)-F2*cos(pi/6)

Ry = rho*((Q2*V2*sin(pi/6)+Q3*V3*sin(pi/9)-Q1*V1))+F3*sin(pi/9)-F2*sin(pi/6)-F1

R = sqrt(Rx**2+Ry**2)

a = atan(Ry/Rx)*180/pi

print "Resultant Force = ",round(R,0),"N"

print "Angle with horizontal = ",round(a,1),"deg with horizontal"

Resultant Force =  12489.0 N
Angle with horizontal =  69.2 deg with horizontal


### Example 5.4 Page no 199¶

In :
# Normal force on the plate

# Given
from math import *

d = 2            # diameter in inches

A = pi*d**2/(4*144)    # Area of jet

V = 100          # velocity of jet in ft/s

Q = A*V          # dischargge in ft**3/s

gma = 62.4       # mass

g = 32.2         # acceleration due to gravity in ft/s**2

# Solution

Rx = (gma*Q*V)/g   # horizontal force required to keep plate in position

print "Normal force on the plate = ",round(Rx,0),"lbs"

Normal force on the plate =  423.0 lbs


### Example 5.5 Page no 202¶

In :
# Force on the plate ; work doen per second; efficiency

# Given

from math import *

D = 0.075            # diameter in m

A =pi*D**2/4         # area of jet

V  =15               # velocity of jet in m/s

w = 9810             # specific weight

g = 9.81             # acceleration due to gravity in m/s^2

# Solution

Q =A*V              # Discharge in m**3/s

Vp = 10             # velocity of plate in m/s

Rx = w*Q*(V-Vp)/g   # force in X direction

print "Force on the plate = ",round(Rx,2),"N"

W = Rx*Vp

print "Work done per second = ",round(W,1),"N.m/s"

Eff = 2*(V-Vp)*Vp/V**2

E = 100*Eff

print "Efficiency = ",round(E,1),"%"

Force on the plate =  331.34 N
Work done per second =  3313.4 N.m/s
Efficiency =  44.4 %


### Example 5.6 Page no 204¶

In :
# Force exerted on the plate

# Given
from math import *

d = 3            # diameter in inches

A = pi*d**2/(4*144)    # Area of jet

Q = 2           # discharge in ft**3/s

rho = 1.94       # density in lbs/ft**3

# Solution

V = Q/A          # velocity in ft/s

alpha = pi/6     # inlet vane angle

bta = pi/6       # outlet vane angle

Rx = rho*Q*(V*cos(bta)+V*cos(alpha))   # force in X direction

Ry = rho*Q*(V*sin(bta)-V*sin(alpha))   # force in Y direction

print "Force exerted in X direction = ",round(Rx,1),"lbs"

print "Force exerted in Y direction = ",round(Ry,1),"lbs"

Force exerted in X direction =  273.8 lbs
Force exerted in Y direction =  0.0 lbs


### Example 5.7 Page no 207¶

In :
# Angle of blade tips at inlet and exit ; work done on the vane; efficiency of the vane

# Given

from math import *

V1 =40              # velocity in m/s

Vp = 20             # velocity of the plate in m/s

alpha = pi/6         # inlet vane angle

bta = pi/9 # outlet vane angle

g = 9.81

# Solution

V1x = V1*cos(alpha)

Vw1 = V1x;

V1y = V1*sin(alpha)

dV = V1x - Vp

theta = atan(V1y/dV)*180/pi

Vr1 = V1y/sin(theta*pi/180)

Vr2 = Vr1

# from trial and error we get the blade tip angle at inlet and outlet

print "a ) Angle of blade top at inlet and Outlet, Phi = 4 deg"

phi = 4*pi/180

V2 = Vr2*sin(phi)/sin(bta)

V2w = V2*cos(bta)

W = (V2w+V1x)*Vp/g

print "b ) Work done per N of fluid per second =  ",round(W,2),"N.m"

Eff = (1 - (V2/V1)**2)*100

print "c ) Efficiency = ",round(Eff,2),"%"

a ) Angle of blade top at inlet and Outlet, Phi = 4 deg
b ) Work done per N of fluid per second =   80.31 N.m
c ) Efficiency =  98.4 %


### Example 5.8 Page no 211¶

In :
# Thrust on the plane; propeller efficiency ; theoretical horsepower ; pressure difference acros the blades

# Given

from math import *

v = 220                     # velocity in ft/s

d = 6                       # diameter of the propeller

Q = 12000                   # discharge in ft**3/s

mf = 0.0022                 # mass flow rate in slugs/ft**3

# Solution

V1 = v*5280/3600            # velocity in ft/s

V = Q/(pi*d**2/4)           # velocity in ft/s

V4 = 2*V-V1

F = mf*Q*(V4-V1)              # thrust on the plane

print "a - Thrust on the plane = ",round(F,1),"lbs"

Eff = V1/V                 # efficiency

E = Eff*100

print "b - Theoretical efficiency = ",round(E,0),"%"

Thp = F*V1/(500*Eff)

print "c - Theoretical horsepower required = ",round(Thp,0),"hp"

dP = mf*(V4**2-V1**2)/2

print "d - Pressure difference across blades = ",round(dP,2),"lbs/ft**3"

a - Thrust on the plane =  5372.2 lbs
b - Theoretical efficiency =  76.0 %
c - Theoretical horsepower required =  4560.0 hp
d - Pressure difference across blades =  190.0 lbs/ft**3

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