Chapter 11:The Boundary Layer

Example no:11.1,Page no:680

In [6]:
#Variable declaration
u_s=1  #Velocity of water
w=0.6  #Width of plane surface
l=1  #Length of plane surface
A=0.6*1  #Area of the surface
Re_xc=10**5    #Reynods group

#Calculation
#Taking
Meu=1e-3 #Viscosity of water
rho=1000  #Density of water
Re=rho*u_s*l/Meu
#Mean value of S/pw2 from equation 11.41
#X=R/(rho*u**2)
X=0.037*(Re)**-0.2+(Re**-1)*(0.646*(10**5)**0.5-0.037*(10**5)**0.8)
F=X*rho*u_s**2*A 

#Result
print"\n Total drag force =",round(F,2),"N (approx)"
 Total drag force = 1.3 N (approx)

Example no:11.2,Page no:680

In [4]:
#Variable declaration
x=150e-3  #Distance from leading edge where thicness is to be found
Meu_o=0.05  #viscosity of oil
rho_o=1000  #Density of oil
u=0.3  #Velocity of flow

#Calculation
Re_x=x*u*rho_o/Meu_o 
#For streamline flow:
#from equation 11.17
dell=4.64*x/Re_x**0.5 #thickness of the boundary layer
#from equation 11.20
del_star=0.375*dell 

#Result
print"\n The thickness of the boundary layer =",dell*1e3,"mm"
print"\n The displacement thickness =",del_star*1e3,"mm"
 The thickness of the boundary layer = 23.2 mm

 The displacement thickness = 8.7 mm

Example no:11.3,Page no:684

In [8]:
#Variable declaration
D=50e-3  #Diameter of the pipe
Q=2e-3  #Flow rate of benzene through pipe
rho_b=870  #Density of benzene
Meu_b=0.7e-3  #Viscosity of benzene

#Calculation
import math
G=Q*rho_b  #Mass flow rate of benzene
Re=4*G/(Meu_b*math.pi*D)  #Reynolds number
#From equation 11.49:
del_b=62*D*Re**(-7.0/8.0) 
area=math.pi/4*D**2  #Cross sectional are of pipe
u=G/(rho_b*area)  #mean velocity
#From equation 11.47:
u_b=2.49*u*Re**(-1.0/8.0) 
print"\n The thickness of the laminar sub-layer =",round(del_b*1e3,3),"mm"
print"\n The velocity of the benzene at the edge of the laminar sub-layer =",round(u_b,3),"m/s"
 The thickness of the laminar sub-layer = 0.195 mm

 The velocity of the benzene at the edge of the laminar sub-layer = 0.637 m/s