Chapter 18 : Turbulence

Example 18.2 page no : 549

In [1]:
#Calculate the energy per unit mass and heat dissipation rate

#variables
v=0.82                             #m/s
energy_per_unit_mass=v**2/2.0      #J/Kg

# calculation
#Let dissipation rate be denoted by eta
#Let D denote d/dL
DP=0.0286                          #Pa/m
rho=1.2                           #Kg/m^3
eta=DP*v/rho                      #m^2/s^3 or J/Kg/s

# result
print "The energy per unit mass is %f J/Kg\n"%energy_per_unit_mass
print "The heat dissipation rate is %f J/Kg/s"%eta

The energy per unit mass is 0.336200 J/Kg

The heat dissipation rate is 0.019543 J/Kg/s

Example 18.3 page no : 550

In [7]:
#Calculate the value of k

# variables
Vx_rms=9.5                            #cm/s
Vy_rms=5.0                            #cm/s

# calculation
k=(1/2.0)*((Vx_rms/100.0)**2+(Vy_rms/100.0)**2)                #J/Kg

# result
print "k = %f J/Kg"%k

k = 0.005763 J/Kg

Example 18.4 page no : 551

In [8]:
#Calculate the Kolmogorov scale

# variables
v=1.613*10**(-4)                        #ft^2/s
eta=0.21                                #ft^2/s^3

# calculations
kolmogorov_scale=(v**3/eta)**0.25       #ft

# result
print "The Kolmogorov scale is %f ft"%kolmogorov_scale

The Kolmogorov scale is 0.002114 ft

Example 18.7 page no : 557

In [2]:
#Calculate the value of turbulent kinematic viscosity

# variables
K=0.00576                               #m^2/s^2
eta=0.0196                              #m^2/s^3
C_mew=0.09                              #dimentionless

# declarations
v_t=C_mew*(0.00576)**2/(0.0196)         #m^2/s

# results
print "The value of turbulent kinematic viscosity is %f m^2/s"%v_t

The value of turbulent kinematic viscosity is 0.000152 m^2/s