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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 = cv ln(T2/ T1 ) + R ln(ρ1 / ρ2) (1)


ds = cp ln(T2/ T1 ) - R ln(p2/ p1 ) (2)


ds = change in entropy (kJ)

cv = specific heat capacity at a constant volume process (kJ/kgK)

cp = 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/m3 )

p = absolute pressure (Pa, N/m2)

Example - Entropy Change in an Air Heating Process

Air - 10 kg - is heated at constant volume from temperature 20 oC and 101325 N/m2 to a final pressure of 405300 N/m2.

The final temperature in the heated air can be calculated with the ideal gas equation :

p v = R T (3)


v = volume   (m3 )

The ideal gas equation (3) can be transformed to express the volume before heating:

v1 = R T1 / p1 (4)

Since v1 = v2 the ideal gas equation (3) after heating can be expressed as:

p2v1 = R T2 (5)

or transformed to express the final temperature:

T2 = p2v1 / R (6)

Combining (5) and (6):

T2 = p2(R T1 / p1 ) / R

= p2T1 / p1 (7)

= (405300 N/m2) (273 K + 20 K) / (101325 N/m2)

= 1172 K - the final gas temperature

The change in entropy can be expressed by (2)

ds = cp ln(T2/ T1 ) - R ln(p2/ p1 )

ds = (1.05 kJ/kgK) ln((1172 K) / (293 K)) - (0.33 kJ/kgK) ln((405300 N/m2) / (101325 N/m2))

= 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.

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