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