Compressible Gas Flow  Entropy
Calculating the entropy in a compressible gas flow.
The entropy change in a compressible gas flow can be expressed as
ds = c_{v} ln(T_{2}/ T_{1} ) + R ln(ρ_{1} / ρ_{2}) (1)
or
ds = c_{p} ln(T_{2}/ T_{1} )  R ln(p_{2}/ p_{1} ) (2)
where
ds = change in entropy (kJ)
c_{v} = specific heat capacity at a constant volume process (kJ/kgK)
c_{p} = specific heat capacity at a constant pressure process (kJ/kgK)
T = absolute temperature (K)
R = individual gas constant (kJ/kgK)
ρ = density of gas (kg/m^{3} )
p = absolute pressure (Pa, N/m^{2})
Example  Entropy Change in an Air Heating Process
Air  10 kg  is heated at constant volume from temperature 20 ^{o}C and 101325 N/m^{2} to a final pressure of 405300 N/m^{2}.
The final temperature in the heated air can be calculated with the ideal gas equation :
p v = R T (3)
where
v = volume (m^{3} )
The ideal gas equation (3) can be transformed to express the volume before heating:
v_{1} = R T_{1} / p_{1} (4)
Since v_{1} = v_{2} the ideal gas equation (3) after heating can be expressed as:
p_{2}v_{1} = R T_{2} (5)
or transformed to express the final temperature:
T_{2} = p_{2}v_{1} / R (6)
Combining (5) and (6):
T_{2} = p_{2}(R T_{1} / p_{1} ) / R
= p_{2}T_{1} / p_{1} (7)
= (405300 N/m^{2}) (273 K + 20 K) / (101325 N/m^{2})
= 1172 K  the final gas temperature
The change in entropy can be expressed by (2)
ds = c_{p} ln(T_{2}/ T_{1} )  R ln(p_{2}/ p_{1} )
ds = (1.05 kJ/kgK) ln((1172 K) / (293 K))  (0.33 kJ/kgK) ln((405300 N/m^{2}) / (101325 N/m^{2}))
= 1 (kJ/kgK)
Total change in entropy:
dS = (1 kJ/kgK) (10 kg)
= 10 (kJ/K)
Related Topics

Fluid Mechanics
The study of fluids  liquids and gases. Involving velocity, pressure, density and temperature as functions of space and time.
Related Documents

2nd Law of Thermodynamics
Entropy and disorder. 
Steam Entropy
Basic steam thermodynamics  entropy diagram. 
Superheated Steam  Entropy
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Wet Bulb Globe Temperature (WBGT)
The Wet Bulb Globe Temperature can be used to measure the general HeatStress index.