Chapter8-Boundary Layers, Wakes and Other Shear Layers¶

Ex1-pg302¶

In [1]:
import math
#calculate Displacement thickness and Momentum thickness
delta=0.6; ## mm

delta1=delta/3.;

theta=2./15*delta;

print'%s %.1f %s'%("Displacement thickness =",delta1,"mm")

print'%s %.2f %s'%("Momentum thickness =",theta,"mm")

Displacement thickness = 0.2 mm
Momentum thickness = 0.08 mm


Ex2-pg309¶

In [2]:
import math
#Evaluate the constants A and B
import numpy
from numpy import linalg
import scipy
from scipy import integrate
## To determine the values of a1 & a2 following conditions must be satisfied

## Condition I - When n=0, u/um=0
## Condition II - When n=1, u/um=a1+a2=1
## Condition III - When n=1, d(u/um)/dn = a1+2a2=0

## By  satisfying these conditions, we have
## a1+a2=1;
## a1+2a2=0;

A=([[1,1],[1,2]]);
B=([[1],[0]]);
X=numpy.linalg.inv(A)*B;

a1=X[0];
a2=X[1];

print("a1=",a1[0])

print("a2=",a1[1])

print("Evaluate the constants A and B")

## A = integrate('(1-f(n))*f(n)','n',0,1)

def function(n):
fun=(1-(2*n-n**2))*(2*n-n**2)
return fun

print("A = ",A[0],"")

## B = differentiation of (2*n-n^2) at n=0, we get
B=2;

print("B =",B,"")

('a1=', 2.0)
('a2=', -1.0)
Evaluate the constants A and B
('A = ', 0.13333333333333333, '')
('B =', 2, '')


Ex3-pg310¶

In [3]:
import math
#Evaluate the constants A and B
import numpy
from numpy import linalg
import scipy
from scipy import integrate
## To determine the values of a1 & a2 following conditions must be satisfied

## Condition I - When n=0, u/um=0
## Condition II - When n=1, u/um=a1+a2=1
## Condition III - When n=1, d(u/um)/dn = a1+2a2=0

## By  satisfying these conditions, we have
## a1+a2=1;
## a1+2a2=0;

A=([[1,1],[1,2]]);
B=([[1],[0]]);
X=numpy.linalg.inv(A)*B;

a1=X[0];
a2=X[1];

print("a1=",a1[0])

print("a2=",a1[1])

print("Evaluate the constants A and B")

## A = integrate('(1-f(n))*f(n)','n',0,1)

def function(n):
fun=(1-(2*n-n**2))*(2*n-n**2)
return fun

print("A = ",A[0],"")

## B = differentiation of (2*n-n^2) at n=0, we get
B=2;

print("B =",B,"")

('a1=', 2.0)
('a2=', -1.0)
Evaluate the constants A and B
('A = ', 0.13333333333333333, '')
('B =', 2, '')


Ex4-pg312¶

In [4]:
import math
#calculate the velocity of the airstream and frictional drag of the plat
v=1.5*10**(-5); ## m^2/s
Re_t=5.*10**5;
x_t=1.2; ## m
rho=1.21; ## kg/m^3

u_m=v*Re_t/x_t;

print'%s %.2f %s'%("the velocity of the airstream =",u_m,"m/s")

theta=0.646*x_t/math.sqrt(Re_t);

F=rho*u_m**2*theta;

D_F=2.*F*x_t;
print'%s %.3f %s'%("the frictional drag of the plate, D_F =",D_F,"N")

the velocity of the airstream = 6.25 m/s
the frictional drag of the plate, D_F = 0.124 N


Ex5-pg316¶

In [5]:
import math
#calculate the boundary layer thickness at the rear of the train and the frictional drag acting on the train and the power required to overcome the frictional drag
u_m = 50.; ## m/s or 180 km/h
v=1.5*10**(-5); ## m^2/s
l=100.; ## m
rho=1.2; ## kg/m^3
b=8.3; ## m

delta = 0.37*(v/u_m)**(1/5.)*l**(4/5.);

print'%s %.3f %s'%("the boundary layer thickness at the rear of the train =",delta,"m")
Re_l = u_m*l/v;
C_F=0.074*(Re_l)**(-1/5.);
F=0.037*rho*u_m**2*l*Re_l**(-1/5.);

D_F = F*b;

print'%s %.4f %s'%("the frictional drag acting on the train, D_F =",D_F,"N")

P=D_F*u_m;
print'%s %.1f %s'%("the power required to overcome the frictional drag =",P/1000,"kW")

the boundary layer thickness at the rear of the train = 0.731 m
the frictional drag acting on the train, D_F = 1818.9691 N
the power required to overcome the frictional drag = 90.9 kW


Ex6-pg318¶

In [14]:
import math
#calculate the proportion of the plate occupied by the laminar boundary layer and   the skin friction coefficient CF evaluated at the trailing edge
Re_t=5.*10**5;
Re_l=5.*10**6;

r1=Re_t/Re_l; ## r1=x_t/l
r2=1-36.9*(1./Re_t)**(3/8.); ## r2=x_0/x_t

r=r1*r2; ## r=x_0/l;

print'%s %.2f %s'%("the proportion of the plate occupied by the laminar boundary layer =",r*100,"%")

C_F = 0.074/Re_l**(1/5.)*(1-r)**(4/5.);
print'%s %.4f %s'%("the skin friction coefficient CF evaluated at the trailing edge =",C_F,"")

the proportion of the plate occupied by the laminar boundary layer = 7.31 %
the skin friction coefficient CF evaluated at the trailing edge = 0.0032