# CHAPTER04:INCOMPRESSIBLE FLOW OVER AIRFOILS¶

## Example E01 : Pg 126¶

In :
# All the quantities are expressed in SI units
import math
c = 0.64;                                # chord length of the airfoil
V_inf = 70.;                              # freestream velocity
L_dash = 1254.;                           # lift per unit span L'
rho_inf = 1.23;                          # density of air
mu_inf = 1.789*10.**-5.;                       # freestream coefficient of viscosity
q_inf = 1./2.*rho_inf*V_inf*V_inf;         # freestream dynamic pressure

# thus the lift coefficient can be calculated as
c_l = L_dash/q_inf/c;

# for this value of C_l, from fig. 4.10
alpha = 4.;

# the Reynold's number is given as
Re = rho_inf*V_inf*c/mu_inf;

# for the above Re and alpha values, from fig. 4.11
c_d = 0.0068;

# thus the drag per unit span can be calculated as
D_dash = q_inf*c*c_d;

print"c_l =",c_l
print"\nfor this c_l value, from fig. 4.10we get alpha =",alpha
print"\nRe =",Re/1000000.
print"\nfor this value of Re, from fig. 4.11 we get c_d =",c_d
print"\nD=",D_dash,"N/m"

c_l = 0.650199104032

for this c_l value, from fig. 4.10we get alpha = 4.0

Re = 3.08015651202

for this value of Re, from fig. 4.11 we get c_d = 0.0068

D= 13.114752 N/m


## Example E02 : Pg 126¶

In :
# All the quantities are expressed in SI units

c = 0.64;                                # chord length of the airfoil
V_inf = 70.;                              # freestream velocity
rho_inf = 1.23;                          # density of air
q_inf = 1./2.*rho_inf*V_inf*V_inf;         # freestream dynamic pressure
c_m_ac = -0.05                           # moment coefficient about the aerodynamic center as seen from fig. 4.11

# thus moment per unit span about the aerodynamic center is given as
M_dash = q_inf*c*c*c_m_ac;

print"The Moment per unit span about the aerodynamic center is is M=",M_dash,"Nm"

The Moment per unit span about the aerodynamic center is is M= -61.71648 Nm


## Example E04 : Pg 127¶

In :
# All the quantities are expressed in SI units
from math import pi
alpha = 5.*pi/180.;            # angle of attack in radians

# from eq.(4.33)according to the thin plate theory, the lift coefficient is given by
c_l = 2.*pi*alpha;

# from eq.(4.39) the coefficient of moment about the leading edge is given by
c_m_le = -c_l/4.;

# from eq.(4.41)
c_m_qc = 0;

# thus the coefficient of moment about the trailing can be calculated as
c_m_te = 3./4.*c_l;

print"(a)Cl =", c_l
print"(b)Cm_le =",c_m_le
print"(c)m_c/4 =",c_m_qc
print"(d)Cm_te =",c_m_te

(a)Cl = 0.548311355616
(b)Cm_le = -0.137077838904
(c)m_c/4 = 0
(d)Cm_te = 0.411233516712


## Example E06 : Pg 131¶

In :
# All the quantities are expressed in SI units

alpha1 = 4.;
alpha2 = -1.1;
alpha3 = -4.;
cl_1 = 0.55;                # cl at alpha1
cl_2 = 0;                   # cl at alpha2
c_m_qc1 = -0.005;           # c_m_qc at alpha1
c_m_qc3 = -0.0125;          # c_m_qc at alpha3

# the lift slope is given by
a0 = (cl_1 - cl_2)/(alpha1-alpha2);

# the slope of moment coefficient curve is given by
m0 = (c_m_qc1 - c_m_qc3)/(alpha1-alpha3);

# from eq.4.71
x_ac = -m0/a0 + 0.25;

print"The location of the aerodynamic center is x_ac =",x_ac

The location of the aerodynamic center is x_ac = 0.241306818182


## Example E07 : Pg 139¶

In :
# All the quantities are expressed in SI units
import math
c = 1.5;            # airfoil chord
Re_c = 3.1e6;       # Reynolds number at trailing edge

# from eq.(4.84), the laminar boundary layer thickness at trailing edge is given by
delta = 5*c/math.sqrt(Re_c);

# from eq(4.86)
Cf = 1.328/math.sqrt(Re_c);

# the net Cf for both surfaces is given by
Net_Cf = 2*Cf;

print"(a)delta =",delta,"m"
print"(b)Cf =",Cf*10000
print"Net Cf =",Net_Cf

(a)delta = 0.00425971375685 m
(b)Cf = 7.5425331588
Net Cf = 0.00150850663176


## Example E08 : Pg 150¶

In :
# All the quantities are expressed in SI units
import math
c = 1.5;            # airfoil chord
Re_c = 3.1e6;       # Reynolds number at trailing edge

# from eq.(4.87), the turbulent boundary layer thickness at trailing edge is given by
delta = 0.37*c/(Re_c**0.2);

# from eq(4.86)
Cf = 0.074/(Re_c**0.2);

# the net Cf for both surfaces is given by
Net_Cf = 2*Cf;

print"(a)delta =",delta,"m"
print"(b)Cf =",Cf
print"Net Cf =",Net_Cf

(a)delta = 0.0279267658904 m
(b)Cf = 0.00372356878539
Net Cf = 0.00744713757078


## Example E09 : Pg 162¶

In :
# All the quantities are expressed in SI units
from math import sqrt
c = 1.5;                # airfoil chord length
Rex_cr = 5e5;           # critical Reynold's number
Re_c = 3.1e6;           # Reynold's number at the trailing edge

# the point of transition is given by
x1 = Rex_cr/Re_c*c;

# the various skin friction coefficients are given as
Cf1_laminar = 1.328/sqrt(Rex_cr);
Cfc_turbulent = 0.074/(Re_c**0.2);
Cf1_turbulent = 0.074/(Rex_cr**0.2);

# thus the total skin friction coefficient is given by
Cf = x1/c*Cf1_laminar + Cfc_turbulent - x1/c*Cf1_turbulent;

# taking both sides of plate into account
Net_Cf = 2*Cf;

print"The net skin friction coefficient is Net Cf=",round(Net_Cf,5)

The net skin friction coefficient is Net Cf= 0.00632