In [1]:

```
# 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"
```

In [2]:

```
# 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"
```

In [3]:

```
# 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
```

In [6]:

```
# 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
```

In [7]:

```
# 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
```

In [8]:

```
# 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
```

In [9]:

```
# 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)
```