Chapter 24:Air Movement¶

Example 24.1,PAGE NUMBER:281¶

In [1]:
# Variable declaration
Z=4500;# Altitude in m
p=575;# mbar barometric pressure
t=-10;# Temperature in °C

# Calculation
rho=1.2*(p/1013.25)*((273.15+20)/(273.15+t));# The density of dry air in kg/m**3
print"The density of dry air,rho=%0.2f kg/m**3"%rho

The density of dry air,rho=0.76 kg/m**3


Example 24.2,PAGE NUMBER:282¶

In [2]:
# Variable declaration
V=1;# The volume of air in m**3
t=20;# The dry bulb temperature in °C
H=60;# % saturation
p=101.325;# The pressure in kPa
v=7;# The velocity in m/s
v_s=0.8419;# The specific volume in m**3/kg

# Calculation
m=V/v_s;# Mass in kg
Ke=(m*v**2)/2;# Kinetic energy in kg/(m s**2)
print"Kinetic energy=%2.1f kg/(m s**2)"%Ke

Kinetic energy=29.1 kg/(m s**2)


Example 24.3,PAGE NUMBER:296¶

In [3]:
# Variable declaration
v_e=8;# The entering velocity of air in m/s
v_l=5.5;# The leaving velocity of air in m/s
fl=20;# Friction losses in %
m=1.2;# Masss in kg

# Calculation
P_e=(m*v_e**2)/2;# Velocity pressure entering expansion in Pa
P_l=(m*v_l**2)/2;# Velocity pressure leaving expansion in Pa
FL=fl*10**-2*(P_e-P_l);# Friction losses in Pa
Sr=(1-(fl*10**-2))*(P_e-P_l);# Static regain in Pa
print"The amount of Static regain=%2.1f Pa"%Sr

The amount of Static regain=16.2 Pa