CHAPTER18:LAMINAR BOUNDARY LAYERS¶

Example E01 : Pg 595¶

In [1]:
# All the quantities are expressed in SI units
from math import sqrt
p_inf = 101000.;                    # freestream pressure
T_inf = 288.;                       # freestream temperature
c = 2.;                             # chord length of the plate
S = 40.;                            # planform area of the plate
mue_inf=1.7894*10.**5.;               # coefficient of viscosity at sea level
gam=1.4;                         # ratio of specific heats
R=287.;                           # specific gas constant
# the freestream density is
rho_inf = p_inf/R/T_inf;
# the speed of sound is
a_inf = sqrt(gam*R*T_inf);
# (a)
V_inf = 100.;
# thus the mach number can be calculated as
M_inf = V_inf/a_inf;
# the Reynolds number at the trailing is given as
Re_c = rho_inf*V_inf*c/mue_inf;
# from eq.(18.22)
Cf = 1.328/sqrt(Re_c);
# the friction drag on one surface of the plate is given by
D_f = 1./2.*rho_inf*V_inf**2.*S*Cf;
# the total drag generated due to both surfaces is
D = 2.*D_f;
print"The total frictional drag is:(a)D =",D,"N"
# (b)
V_inf = 1000.;
# thus the mach number can be calculated as
M_inf = V_inf/a_inf;
# the Reynolds number at the trailing is given as
Re_c = rho_inf*V_inf*c/mue_inf;
# from eq.(18.22)
Cf = 1.2/sqrt(Re_c);
# the friction drag on one surface of the plate is given by
D_f = 1./2.*rho_inf*V_inf**2.*S*Cf;
# the total drag generated due to both surfaces is
D = 2.*D_f;
print"(b) D =",D,"N"

The total frictional drag is:(a)D = 17563872.6566 N
(b) D = 501884115.614 N


Example E02 : Pg 596¶

In [2]:
# All the quantities are expressed in SI units
from math import sqrt
Pr = 0.71;                    # Prandlt number of air at standard conditions
Pr_star = Pr;
Te = 288.;                     # temperature of the upper plate
ue = 1000.;                    # velocity of the upper plate
Me = 2.94;                    # Mach number of flow on the upper plate
p_star = 101000.;
R = 287.;                      # specific gas constant
T0 = 288.;                     # reference temperature at sea level
mue0 = 1.7894*10**-5;             # reference viscosity at sea level
c = 2.;                        # chord length of the plate
S = 40.;                       # plate planform area

# recovery factor for a boundary layer is given by eq.(18.47) as
r = sqrt(Pr);

# rearranging eq.(16.49), we get for M = 2.94
T_aw = Te*(1+r*(2.74-1));

# from eq.(18.53)
T_star = Te*(1 + 0.032*Me**2. + 0.58*(T_aw/Te-1.));

# from the equation of state
rho_star = p_star/R/T_star;

# from eq.(15.3)
mue_star = mue0*(T_star/T0)**1.5*(T0+110.)/(T_star+110.);

# thus
Re_c_star = rho_star*ue*c/mue_star;

# from eq.(18.22)
Cf_star = 1.328/sqrt(Re_c_star);

# hence, the frictional drag on one surface of the plate is
D_f = 1./2.*rho_star*ue**2.*S*Cf_star;

# thus, the total frictional drag is given by
D = 2.*D_f;

print"The total frictional drag is: D =",D,"N"

The total frictional drag is: D = 4978.09594496 N


Example E03 : Pg 600¶

In [3]:
# All the quantities are expressed in SI units
from math import sqrt
Pr = 0.71;                    # Prandlt number of air at standard conditions
Pr_star = Pr;
Te = 288.;                     # temperature of the upper plate
ue = 1000.;                    # velocity of the upper plate
Me = 2.94;                    # Mach number of flow on the upper plate
p_star = 101000.;
R = 287.;                      # specific gas constant
gam = 1.4;                    # ratio of specific heats
T0 = 288.;                     # reference temperature at sea level
mue0 = 1.7894*10**-5;             # reference viscosity at sea level
c = 2.;                        # chord length of the plate
S = 40.;                       # plate planform area

# recovery factor for a boundary layer is given by eq.(18.47) as
r = sqrt(Pr);

# from ex.(8.2)
T_aw = Te*2.467;
T_w = T_aw;

T_star = Te*(0.45 + 0.55*T_w/Te + 0.16*r*(gam-1)/2*Me**2.);

# from the equation of state
rho_star = p_star/R/T_star;

# from eq.(15.3)
mue_star = mue0*(T_star/T0)**1.5*(T0+110)/(T_star+110.);

# thus
Re_c_star = rho_star*ue*c/mue_star;

# from eq.(18.22)
Cf_star = 1.328/sqrt(Re_c_star);

# hence, the frictional drag on one surface of the plate is
D_f = 1./2.*rho_star*ue**2.*S*Cf_star;

# thus, the total frictional drag is given by
D = 2.*D_f;

print"The total frictional drag is: D =",D,"N"

The total frictional drag is: D = 5014.11379241 N