CHAPTER16:SOME SPECIAL CASES COUETTE AND POISEUILLE FLOWS

Example E01 : Pg 524

In [2]:
# All the quantities are expressed in SI units
import math 
mue = 1.7894*10**5;                    # coefficient of viscosity
ue = 60.96;                         # velocity of upper plate
D = 2.54*10**4;                        # distance between the 2 plates
T_w = 288.3;                        # temperature of the plates
Pr = 0.71;                          # Prandlt number
cp = 1004.5;                        # specific heat at constant pressure

# (a)
# from eq.(16.6)
u = ue/2;

# (b)
# from eq.(16.9)
tow_w = mue*ue/D;

# (c)
# from eq.(16.34)
T = T_w + Pr*ue**2/8/cp;

# (d)
# from eq.(16.35)
q_w_dot = mue/2*ue**2/D;

# (e)
# from eq.(16.40)
T_aw = T_w + Pr/cp*ue**2/2;

print"(a)u =",u,"m/s"
print"(b)tow_w =",tow_w,"N/m2"
print"(c)T =",T,"K"
print"(d)q_w_dot =",q_w_dot,"Nm-1s-1"
print"(e)Taw =","K",T_aw
(a)u = 30.48 m/s
(b)tow_w = 429.456 N/m2
(c)T = 288.628328315 K
(d)q_w_dot = 13089.81888 Nm-1s-1
(e)Taw = K 289.613313258

Example E02 : Pg 524

In [1]:
# All the quantities are expressed in SI units
from math import sqrt
mue = 1.7894*10**5;                    # coefficient of viscosity
Me = 3.;                             # mach number of upper plate
D = 2.54*10**4;                        # distance between the 2 plates
pe = 101000.;                        # ambient pressure
Te = 288.;                           # temperature of the plates
Tw = Te;
gam = 1.4;                          # ratio of specific heats
R = 287;                            # specific gas constant
Pr = 0.71;                          # Prandlt number
cp = 1004.5;                        # specific heat at constant pressure
tow_w = 72.;                         # shear stress on the lower wall
# the velocity of the upper plate is given by
ue = Me*sqrt(gam*R*Te);
# the density at both plates is
rho_e = pe/R/Te;
# the coefficient of skin friction is given by
cf = 2.*tow_w/rho_e/ue**2.;
# from eq.(16.92)
C_H = cf/2/Pr;
# from eq.(16.82)
h_aw = cp*Te + Pr*ue**2/2;
h_w = cp*Tw;
q_w_dot = rho_e*ue*(h_aw-h_w)*C_H;
print"The heat transfer is given by:q_w_dot =",q_w_dot/1e4,"W/m2"
The heat transfer is given by:q_w_dot = 3.67387998933 W/m2