# Chapter 17 : The boundary layer¶

### Example 17.1 page no : 519¶

In [1]:
#Calculate the boundary layer thickness
#for the aeroplane

# variables
v=1.61*10**(-4)                 #ft**2/s
x=2.                            #ft
V=200.                          #miles/hr

# calculation
#1 mile = 5280 ft
#1 hr = 3600 sec
delta_aeroplane=5*(v*x/(V*5280/3600.0))**0.5

# result
print "The boundary layer thickness for the aeroplane is %f ft\n"%delta_aeroplane
#for the boat
v1=1.08*10**(-5)                #ft**2/s
x1=2.                           #ft
V1=10.                          #miles/hr
#1 mile = 5280 ft
#1 hr = 3600 sec
delta_boat=5*(v1*x1/(V1*5280/3600.0))**0.5
print "The boundary layer thickness for the boat is %f ft"%delta_boat

The boundary layer thickness for the aeroplane is 0.005239 ft

The boundary layer thickness for the boat is 0.006068 ft


### Example 17.2 page no : 521¶

In [2]:
#Calculate the force required to tow a square metal plate by a boat

# variables
rho_water=998.2                     #Kg/m**3
V=15.                               #km/hr
v=1.004*10**(-6)                    #m**2/s
l=1.                                #m length of plate

#Calculations
#1 km = 1000 m
#1 hr = 3600 s
Rx=(V*1000/3600.)*l/v               #dimentionless (reynold's number)
Cf=1.328/Rx**0.5                    #dimentionless
F=Cf*rho_water*(V*1000/3600.0)**2   #N

#Result
print "The force required to tow the square plate is %f N"%F

The force required to tow the square plate is 11.297065 N


### Example 17.3 page no : 528¶

In [2]:
#Calculate the distance between the wall and edge of the laminar sublayer and buffer layer

#Variables
V=10.                          #ft/s
l=0.25                         #ft
v=1.08*10**(-5)                #ft**2/s
f=0.0037                       #dimentionless (fanning friction factor)
u01=5.                         #dimentionless
y01=5.                         #dimentionless

#Calculations
R=V*l/v                        #dimentionless (reynold's number)
u1=V*(f/2.0)**0.5              #ft/s
r1=y01*v/u1                    #ft

#for buffer layer
u02=12.                        #dimentionless
y02=26.                        #dimentionless
r2=y02*v/u1                    #ft

#Results
print "the distance between the wall and edge of the laminar sublayer is %f ft\n"%r1
print "the distance between the wall and edge of the buffer layer is %f ft"%r2

the distance between the wall and edge of the laminar sublayer is 0.000126 ft

the distance between the wall and edge of the buffer layer is 0.000653 ft


### Example 17.4 page no : 530¶

In [4]:
#Calculate the boundary layer thickness and the drag on the plate

# variables
V=50.                          #ft/s
l=20.                          #ft
b=1.                            #ft
v=1.08*10**(-5)                #ft**2/s

# calculation and result
R=V*l/v                        #dimentionless (reynold's number)
delta=0.37*l/R**0.2            #ft
print "The boundary layer thichness at the end of the plate is %f ft\n"%delta
Cf=0.072/R**0.2                #dimentionless
rho_water=62.3                 #lbm/ft**3
V=50.                          #ft/s
#let A be the area of contact
A=2*l*b                        #ft**2
#1 lbf.s**2 = 32.2 lbm.ft
F=(1/2.0)*Cf*rho_water*V**2*A/32.2             #lbf
print "The drag on the plate is %f lbf"%F

The boundary layer thichness at the end of the plate is 0.188763 ft

The drag on the plate is 177.672178 lbf