Chapter 10 : The Boundary Layer

Example 10.1 Page No : 379

import math 
			
#Initialization of variables
v = 30.      			#velocity - m/s
nu = 1.5e-5 			#m**2/s
			
#calculations
Re = 5*10**5
xc =  Re*nu/v
			
#results
print "Transistion region  =  %.2f m"%(xc)

Transistion region  =  0.25 m

Example 10.2 Page No : 388

import math 
			
#Initialization of variables
u = 2. 			    #velocity - m/s
x = 0.15 			#length - m
nu = 1.5e-5 		#m**2/s
B = 0.5 			#m
rho = 1.22 			#kg/m**3

#calcualtions
Rx = u*x/nu
delta =  4.91*x/math.sqrt(Rx)
deltas = 1.729*x/math.sqrt(Rx)
Cf = 1.328/math.sqrt(Rx)
Ff = Cf*0.5*rho*u**2 *2*B*x
			
#results
print "Boundary layer thickness  =  %.2f cm"%(delta*100)
print " Displacement thickness  =  %.2f cm"%(deltas*100)
print " Average drag coeffcient  =  %.4f"%(Cf)
print " Drag force  =  %.4f N"%(Ff)

Boundary layer thickness  =  0.52 cm
 Displacement thickness  =  0.18 cm
 Average drag coeffcient  =  0.0094
 Drag force  =  0.0034 N

Example 10.5 Page No : 400

import math 
			
#Initialization of variables
U = 172.*1000/3600 			#m/s
w = 3. 	        		#m
h = 3. 		        	#m
L = 100. 			    #m
nu = 1.5e-5 			#m**2/s
rho = 1.22 	    		#kg/m**3
			
#calculations
Rl = U*L/nu
Cf = 0.074 /(Rl**(1./5))
Ff = Cf*0.5*rho*U**2 *w*h*L
power =  Ff*U
			
#results
print "power required  =  %.1f kW"%(power/1000)
#The answer is a bit different due to rounding off error

power required  =  88.3 kW

Example 10.6 Page No : 400

import math 
			
#Initialization of variables
U = 4000. 		    	#m/s
L = 8. 		        	#m
nu = 3600.e-6 			#m**2/s
rho = 1000. 			#kg/m**3
b = 5.       			#m
			
#calculations
Rl = U*L/nu
Cf =  0.074/Rl**(1./5) -1700/Rl
Ff = Cf*0.5*rho*(U/3600)**2 *L*b
			
#results
print "Skin friction drag  =  %.2f N"%(Ff)

# note : rounding off error.

Skin friction drag  =  69.75 N