Chapter 9 - Fluid flow about immersed bodies

Example 1a - Pg 389

In [1]:
#calculate the values of Fd
#Initialization of variables
import math
import numpy
rho=2.45 #slugs/ft^3
mu=9.2e-3 #lb-sec/ft^2
x=3.
v=3. #ft/s
B=6./12. #ft
L=36./12. #ft
#calculatons
Nr=v*x*rho/mu
y=([1.32, 1.46, 1.328])
Cd = numpy.zeros(len(y))
for i in range (0, len(y)):
	Cd[i]=y[i]*math.pow(Nr,(-0.5))
Fd = numpy.zeros(len(Cd))
for i in range (0, len(y)):
	Fd[i]=2*Cd[i]*B*L*(0.5*rho*v*v)

#results
print '%s' %("Drag on the plates using different formulae blasius, parabola and pohlhauser in order")
print '%.3f %.3f %.3f' %(Fd[0],Fd[1],Fd[2])

Drag on the plates using different formulae blasius, parabola and pohlhauser in order
0.892 0.986 0.897

Example 1b - Pg 390

In [2]:
#calcualte the boundary layer thickness and the shearing thickness
#Initialization of variables
import math
import numpy
x=36./12.
rho=2.45 #slugs/ft^3
mu=9.2e-3 #lb-sec/ft^2
v=3. #ft/s
#calculatons
Nr=v*x*rho/mu
z=([4.91, 5.48, 4.65])
x=36./12.
delta = numpy.zeros(len(z))
for i in range (0, len(z)):
	delta[i]=z[i] /math.sqrt(Nr) *x


f=([0.332, 0.365, 0.322])
T=numpy.zeros(len(f))
for i in range (0,len(f)):
	T[i]=f[i]*mu*v/x *math.sqrt(Nr)

#results
print '%s' %("Boundary layer thickness = ")
print '%s' %("In order of Blasius, parabola and pohlhauser")
print '%.3f %.3f %.3f' %(delta[0],delta[1],delta[2])
print '%s' %("Shearing stress = ")
print '%s' %("In order of Blasius, parabola and pohlhauser")
print '%.3f %.3f %.3f' %(T[0],T[1],T[2])

Boundary layer thickness = 
In order of Blasius, parabola and pohlhauser
0.301 0.336 0.285
Shearing stress = 
In order of Blasius, parabola and pohlhauser
0.150 0.164 0.145

Example 2a - Pg 398

In [3]:
#calculate the total frictional drag and horsepower required
#Initialization of variables
import math
e=0.01 #ft
rho=2 #slugs/ft^3
mu=2.6e-5 #lb sec/ft^2
speed=10. #knots
L=250. #ft
A=30000. #ft^2
#calculations
V=speed*1.69
Nrl=V*L*rho/mu
Cdf=1.32 /math.sqrt(Nrl)
Fd=Cdf*A*0.5*rho*V*V
hp=Fd*V/550.
#results
print '%s %d %s' %("Total frictional drag =",Fd,"lb")
print '%s %.1f %s' %("\n Horsepower required =",hp,"hp")

Total frictional drag = 627 lb

 Horsepower required = 19.3 hp

Example 2b - Pg 397

In [4]:
#calculate the total frictional drag and horsepower required
#Initialization of variables
import math
e=0.01 #ft
rho=2 #slugs/ft^3
mu=2.6e-5 #lb sec/ft^2
speed=10 #knots
L=250. #ft
A=30000. #ft^2
#calculations
V=speed*1.69
Nrl=V*L*rho/mu
Cdf=0.074/math.pow(Nrl,0.2) -1700./Nrl
Fd=Cdf*A*0.5*rho*V*V
hp=Fd*V/550.
#results
print '%s %d %s' %("Total frictional drag =",Fd,"lb")
print '%s %.1f %s' %("\n Horsepower required =",hp,"hp")
print '%s' %("The answer given in textbook is wrong. please use a calculator")

Total frictional drag = 12537 lb

 Horsepower required = 385.2 hp
The answer given in textbook is wrong. please use a calculator

Example 2c - Pg 398

In [1]:
#calculate the total frictional drag and horsepower required
#Initialization of variables
import math
e=0.01 #ft
rho=2. #slugs/ft^3
mu=2.6e-5 #lb sec/ft^2
speed=10. #knots
L=250. #ft
A=30000. #ft^2
#calculations
V=speed*1.69
Nrl=V*L*rho/mu
Cdf=1/math.pow((1.89 + 1.62*math.log10(L/e)),(2.5))
Fd=Cdf*A*0.5*rho*V*V
hp=Fd*V/550.
#results
print '%s %d %s' %("Total frictional drag =",Fd,"lb")
print '%s %.1f %s' %("\n Horsepower required =",hp,"hp")
print '%s' %("The answers are a bit different from textbook due to rounding off error")

Total frictional drag = 35117 lb

 Horsepower required = 1079.1 hp
The answers are a bit different from textbook due to rounding off error

Example 3 - Pg 398

In [7]:
#calculate the drag on the model
#Initialization of variables
import math
V=200. #ft/s
L=5. #ft
B=2. #ft
rho=0.00232 #slug/ft^3
mu=3.82e-7 #lb-sec/ft^2
p2=14.815 #psia
pa=14.7 #psia
#calculations
Nr=V*L*rho/mu
Cdf=0.0032
Fdf=Cdf*math.pi*L*B*0.5*rho*V*V
Fd=(p2-pa)*math.pi/4. *(B*12)*(B*12) -Fdf
#results
print '%s %.2f %s' %("Drag on the model =",Fd,"lb")

Drag on the model = 47.36 lb

Example 4 - Pg 405

In [8]:
#calculate the velocity of the flow
#Initialization of variables
import math
p1=14.7 #psia
z1=3 #ft
gam=62.4
rho=1.94 #slug/ft^3
pa=0.4 #psia
za=1 #ft
#calculations
v3=(pa-p1)*144 + (za-z1)*gam
V=math.sqrt(-v3*2/(3*rho))
#results
print '%s %.1f %s' %("Velocity of flow =",V," ft/s")
print '%s' %("The answer is a bit different due to rounding off error in textbook")

Velocity of flow = 27.4  ft/s
The answer is a bit different due to rounding off error in textbook

Example 5 - Pg 410

In [9]:
#calculate the horsepower required
#Initialization of variables
import math
rpm=60. 
rho=2. #slugs/ft^3
mu=3.5e-5 #lb-sec/ft^2
D=4./12. #ft
r=2. #ft
#calcualtions
V=rpm*2*math.pi/60. *2
Nr=V*D*rho/mu
Cd=1.1
Fd=Cd*math.pi/4. *(D)*D *0.5*rho*V*V
T=2*Fd*r
w=rpm*2*math.pi/60.
hp=T*w/550.
#results
print '%s %.2f %s' %("Horsepower required =",hp," hp")

Horsepower required = 0.69  hp

Example 6 - Pg 414

In [2]:
#calculate the terminal velocity
#Initialization of variables
import math
g=32.2 #ft/s^2
h=60000. #ft
F=2000. #;b
d=3. #ft
rho=0.00231
#calculations
V=math.sqrt(2*g*h)
print '%s' %("By trail and error")
Cd=0.25
Nm=0.87
A=math.pi/4. *d*d
Vt=math.sqrt(2*F/(Cd*A*rho))
#results 
print '%s %.1f %s' %("terminal velocity =",Vt," ft/s")
print '%s' %("The answers are a bit different from textbook due to rounding off error")

By trail and error
terminal velocity = 989.9  ft/s
The answers are a bit different from textbook due to rounding off error