# 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*to a final pressure of

^{2}*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)