Chapter 06:Momentum Analysis of Flow Systems

Example 6.6-1, Page No:248

In [21]:
import math
import scipy.integrate

#Variable Decleration
a=1 #Lower limit of the intergral to be carried out
b=0 #Upper limit of the intergral to be carried out

#Intergration

func = lambda y:-4*y**2 #Decleration of the variable and the function to be integrated
X=scipy.integrate.quadrature(func, a,b)

#Result

print "The Momentum Flux correction factor becomes",round(X[0],2)
The Momentum Flux correction factor becomes 1.33

Example 6.6-2, Page No:251

In [51]:
import math

#Variable Decleration
m_dot=14 #Mass flow rate in kg/s
rho=1000 #Density of water in kg/m^3
theta=pi/6 #Angle at which the pipe is deflected w.r.t the horizontal
A1=0.0113 #Cross-sectional Area at the inlet of the elbow in m^2
A2=7*10**-4 #Cross-sectional Area at the outlet of the elbow in m^2
C=10**-3 #Conversion factor
g=9.81 #Acceleration due to gravity in m/s^2
z2=0.3 #Elevational difference betweem inlet and outlet in m
z1=0 #Considering Datum in m
beta=1.03 #Momentum correction factor 

#Calculations
V1=m_dot/(rho*A1) #Velocity at the inlet of the elbow in m/s
V2=m_dot/(rho*A2) #Velocity at the outlet of the elbow in m/s

#Part(a)
#Applying the Bernoulli Principle
#Simplfying the calculations in two steps
a=(V2**2-V1**2)/(2*g)
P1_gauge=(a+z2-z1)*g*rho*C #Gauge pressure at inlet in kPa

#Part(b)
#Applying the momentum equation
#Anchoring forces required
F_rx=-(P1_gauge*1000*A1)+(beta*m_dot*((V2*cos(theta))-V1)) #N
F_rz=beta*m_dot*V2*sin(theta) #N

#Result
print "The gauge pressure at the inlet is",round(P1_gauge,1),"kPa"
print "The anchoring forces required to hold it in place are",round(F_rx,),"N and",round(F_rz),"N"
The gauge pressure at the inlet is 202.2 kPa
The anchoring forces required to hold it in place are -2053.0 N and 144.0 N

Example 6.6-3, Page No:253

In [52]:
import math

#Variable decleration
beta=1.03 #Momentum Correction factor
m_dot=14 #mass flow rate in kg/s
V2=20 #Velocity at outlet in m/s
V1=1.24 #Velocity at inlet in m/s
P1_gauge=202200 #gauge pressure at inlet in N/m^2
A1=0.0113 #Area at the inlet in m^2

#Calculations
#Applying the momentum equation
F_rx=-beta*m_dot*(V2+V1)-P1_gauge*A1 #Horiznotal force in N

#Result
print "The horizontal force is",round(F_rx),"N"
The horizontal force is -2591.0 N

Example 6.6-4, Page No:253

In [53]:
import math

#Variable Decleration
beta=1 #Momentum correction factor
m_dot=10 #Mass flow rate in kg/s
V1=20 #Velocity of flow of water in m/s

#Calculations
#Applying the momentum equation
F_r=beta*m_dot*V1 #The force exerted in N

#Result
print "The force exerted is",round(F_r),"N"
The force exerted is 200.0 N

Example 6.6-5, Page No:254

In [59]:
import math

#Variable decleration
W_s=11 #Wind speed in km/h
C=3.6**-1 #Conversion from km/h  to m/s
D=9 #Diameter of the blade in m
rho1=1.22 #Density of air in kg/m^3
W_actual=0.4 #Actual power generated in kW

#Calculations
#Part(a)
V1=W_s*C #Velocity in m/s
m_dot=(rho*V1*pi*D**2)/4 #Mass flow rate of air in kg/s
W_dot_max=0.5*m_dot*V1**2*10**-3 #Work done in kW
n_windturbine=W_actual/W_dot_max #Efficiency of the turbine-generator 

#Part(b)
V2=V1*((1-n_windturbine)**0.5) #Exit velocity in m/s

#Applying momentun equation
F_r=m_dot*(V2-V1) #Force exerted in N

#Result
print "The efficiency of the turbine is",round(n_windturbine,3)
print "The horizontal force exerted is",round(F_r,1),"N"
#Answer differs by 0.5 due to floating point accuracy in second part
The efficiency of the turbine is 0.361
The horizontal force exerted is -145.5 N

Example 6.6-6, Page No:256

In [1]:
import math

#Variable Decleration
V_gas=3000 #Velocity of the gas exiting in m/s
m_dot_gas=80 #mass flow rate of gas escaping in kg/s
m_spacecraft=12000 #mass of the spacecraft in kg
delta_t=5 #Time in s
#Calculations
#Part(a)
a_spacecraft=-(m_dot_gas*V_gas)/m_spacecraft #Acceleration of the spacecraft in m/s^2

#Part(b)
dV=a_spacecraft*delta_t #Change in velocity of the spacecraft in m/s

#PArt(c)
F_thrust=-(m_dot_gas*V_gas)/1000 #Thrust force exerted in kN

#Result
print "The acceleration of the spacecraft is",round(a_spacecraft),"m/s^2"
print "The change in velocity is",round(dV),"m/s"
print "The thrust force exerted is",round(F_thrust),"kN"
The acceleration of the spacecraft is -20.0 m/s^2
The change in velocity is -100.0 m/s
The thrust force exerted is -240.0 kN

Example 6.6-7, Page No:257

In [12]:
import math

#Variable Decleration
V_dot=70 #Volumertic Flow rate in L/min
D=0.02 #Inner diameter of the pipe in m
C=60*10**3 #Conversion factor
rho=997 #Density of water in kg/m^3
P1_gauge=90000 #Pressure at location in Pa
X=57 #Total weight of faucet in N

#Calculations
V=((V_dot*4)/(pi*D**2))/C #Velocity of flow in m/s
m_dot=(rho*V_dot)/C #mass flow rate in kg/s

#Applying the Momentum equation
F_rx=-(m_dot*V)-((P1_gauge*pi*D**2)/4) #X-component of force in N
F_rz=-m_dot*V+X #z-Component of force in N

#Result
print "The net force exerted on the flange in vector notation is Fr",round(F_rx,2),"i+",round(F_rz,2),"k  N"
#NOTE:The answer in the textbook differs due to decimal point accuracy difference in computation
The net force exerted on the flange in vector notation is Fr -31.99 i+ 53.29 k  N

Example 6.6-9, Page NO:266

In [14]:
import math

#Variable Decleration
rho=1000 #Denisty of water in kg/m^3
D=0.10 #Diameter of the pipe in m
V=3 #Average velocity of water in m/s
g=9.81 #Acceleration due to gravity in m/s^2
m=12 #Mass of horizontal pipe section when filled with water in kg
r1=0.5 #Moment arm 1 in m
r2=2 #Moment arm 2 in m

#Calculation
m_dot=rho*((pi*D**2)/4)*V #Mass flow rate in kg/s
W=m*g #Weight in N

#Applying the momentum equation
M_A=r1*W-(r2*m_dot*V) #Momentum about point A in N.m

#Setting M as zero and using the momentum equation
L=((2*r2*m_dot*V)/W)**0.5 #Length in m

#Result
print "The bending Moment at A is",round(M_A,1),"N.m and the length required is",round(L,1),"m"
The bending Moment at A is -82.5 N.m and the length required is 1.5 m

Example 6.6-9, Page No:267

In [33]:
import math

#Variable Decleration
V_dot_nozzle=5 #Volumertic flow rate in L/s
D_jet=0.01 #Diameter of the jet in m
C=10**-3 #Conversion Factor
n_dot=300 #R.P.M of the nozzle
r=0.6 #Radial arm in m
m_dot=20 #Mass flow rate in kg/s
s=60**-1 #Conversion factor
#Calculations
V_jet_r=(V_dot_nozzle*4)/(pi*D_jet**2)*C #Velocity relative to the rotating nozzle in m/s
w=(2*pi*n_dot)*s #Angular speed in rad/s
V_nozzle=r*w #Tangential Velocity in m/s

#Applying thr relative velocity principle
V_jet=V_jet_r-V_nozzle #Velocity of the jet in m/s

#Applying the momentum Equation and using the torque concept
T_shaft=r*m_dot*V_jet #Torque transmitted through the shaft in N.m
W_dot=w*T_shaft*C #Power generated in kW

#Result
print "The sprinkler-type turbine has the potential to produce",round(W_dot,1),"kW"
The sprinkler-type turbine has the potential to produce 16.9 kW