# Chapter 10 : Forces on Immersed Bodies¶

## Example 10.1 Page No : 301¶

In :
import math

#Initialization of variables
nu = 0.001 	#ft**2 /s
L = 1.5 	#ft
U = 2.   	#ft/s
s = 0.925
ro = 1.94
b = 6.

#calculations
R = L*U/nu
Cf = 1.328/math.sqrt(R)
Ff = Cf*s*ro*U*b/12 *L
delta = 4.91 *L/math.sqrt(R)
T0 = 0.332*nu*s*ro*U/L *math.sqrt(R)
print R
#Results
print "Friction drag  =  %.3f lb"%(Ff)
print " Thickness of boundary layer  =  %.4f ft"%(delta)
print " Shear stress  =  %.4f lb/ft**2"%(T0)

#Similar calculations are done for SI units case

3000.0
Friction drag  =  0.065 lb
Thickness of boundary layer  =  0.1345 ft
Shear stress  =  0.0435 lb/ft**2


## Example 10.2 Page No : 305¶

In :
import math

#Initialization of variables
nu = 0.00015 	#ft**2/s
L = 35.      	#ft
U = 88.     	#fps
g = 32.2    	#ft/s**2
b = 10. 	    #ft
w = 8.       	#ft
rho = 0.0725

#calculations
R = L*U/nu
Cf = 0.455 /(math.log10(R))**2.58
B = 2*b + w
Ff = Cf*rho/g *U**2 /2 *L*B
Rx = R/10**5
delta = L*0.377 /(b* Rx**(0.2))
T0 = 0.0587 *rho/g *U**2 /2 *(nu/(L*U))**(0.2)

#Results
print "Frictional drag  =  %.1f lb"%(Ff)
print " Thickness of  boundary layer  =  %.3f ft"%(delta)
print " Shear stress  =  %.4f lb/ft**2"%(T0)

Frictional drag  =  22.9 lb
Thickness of  boundary layer  =  0.455 ft
Shear stress  =  0.0176 lb/ft**2


## Example 10.3 Page No : 308¶

In :
import math

#Initialization of variables
nu = 0.0000166 	#ft**2 /s
U = 5.06 	#fps
L = 50. 	#ft
g = 32.2
dia = 10. 	#ft

#calculations
R = L*U/nu
Cf = 0.0028
Ff = Cf*64/g *U**2 /2 *math.pi*dia*L
Rx = R/L
ec = 26*nu/U *Rx**(0.25)
Rx2 = Rx*L/2
T02 = 0.0587*U**2 /2 /(Rx2)**(0.2)
delta2 = 60*nu/math.sqrt(T02)

#Results
print "Friction drag  =  %.f lb"%(Ff)
print " Critical roughness  =  %.4f ft"%(ec)
print " height of roughness  =  %.4f ft"%(delta2)

Friction drag  =  112 lb
Critical roughness  =  0.0020 ft
height of roughness  =  0.0056 ft


## Example 10.4 Page No : 314¶

In :
import math

#Initialization of variables
Cd = 0.45
rho = 0.0725/32.2
V = 88.  	#fps
A = 8*10.

#calculations
Fd = Cd*rho*V**2 /2 *A
Drag2 = 23
D = Fd-Drag2

#Results
print "Pressure drag  =  %.f lb"%(D)

Pressure drag  =  291 lb


## Example 10.5 Page No : 315¶

In :
from sympy import Symbol,solve
import math

#Initialization of variables
Cd = 0.2
D = 8.5/12 	#in
v = 1.57*10**-4
rho = 0.00238

#calculations
A = (math.pi*D**2)/4
#From trail and error method,
V = 412 	#fps
R = (D*V)/v

#Results
print "Free fall velocity  =  %.2e fps"%(R)
print "Since the values of R and Cd check with the figure, V = 412 fps"

Free fall velocity  =  1.86e+06 fps
Since the values of R and Cd check with the figure, V = 412 fps


## Example 10.6 Page No : 319¶

In :

#Initialization of variables
nu = 1.15*10**-5 	#m**2/s
D = 2.*10**-3 	    #m
V = 15.         	#m/s
T = -20. 	        #C

#calculations
R = D*V/nu
f = 0.2 *V/D *(1+T/R)

#Results
print "Frequency of oscillation  =  %d Hz"%(round(f,-2))

Frequency of oscillation  =  1500 Hz


## Example 10.7 Page No : 324¶

In :
import math

#Initialization of variables
n = 90. 	#rpm
R = 2.
rho = 0.0765/32.2
B = 25.
U = 120. 	#fps

#calculations
vt = 2*math.pi*R*n/60
T = 2*math.pi*R*vt
Fl = rho*B*U*T
theta = math.degrees(math.asin(-T/(4*math.pi*R*U)))

#Results
print "Peripheral velocity = %.2f fpx"%vt
print "Value of circulation  =  %.f ft62/s"%(T)
print " Transverse or lift force  =  %d lb"%(round(Fl,-1))
print " Position of stagnation points  =  %.1f degrees"%(180-theta)

Peripheral velocity = 18.85 fpx
Value of circulation  =  237 ft62/s
Transverse or lift force  =  1690 lb
Position of stagnation points  =  184.5 degrees


## Example 10.8 Page No : 331¶

In :
import math

#Initialization of variables
B = 36. 	#ft
c = 6.  	#ft
Cl = 0.8
tau = 0.175
rho = 0.001756
V = 300. 	#fps

#calculations
alphai = Cl/(math.pi*B/c) *(1+tau) *180/math.pi
alpha = 5.4
lift = -5.6 	#degrees
alphao = alpha-alphai
alphaod = alphao-lift
alphaor = alphaod*math.pi/180
eta = Cl/(2*math.pi*alphaor)
Fl = Cl*rho*V**2 /2 *B*c
Fd = 0.047/Cl *13680
HP = Fd*V/550

#Results
print "Friction coefficient  =  %.3f "%(eta)
print " weight of the wing  =  %d lb"%(round(Fl,-1))
print " Horsepower required  =  %d hp"%(HP)

# rounding off error

Friction coefficient  =  0.896
weight of the wing  =  13650 lb
Horsepower required  =  438 hp


## Example 10.9 Page No : 334¶

In :
import math

#Initialization of variables
k = 1.4
R = 287.
T = 249. 	#K
v = 600. 	#velocity - m/s
d = 0.2 	#diameter - m

#calculations
c = math.sqrt(k*R*T)
M = v/c
Cd = 0.62
rho = 47.22*10**3 /(R*T)
Fd = Cd*rho*v**2 /2 *math.pi*d**2 /4

#Results
print "Drag  =  %d N"%(Fd)

# note : answer is accurate

Drag  =  2316 N