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Entropy generation for a control mass

Short answer

For a control mass, entropy generation is S_gen = S2 – S1 – ∫δQ/Tb ≥ 0. It is zero for reversible processes and positive for irreversible ones; for an adiabatic system, S2 – S1 = S_gen.

Derive expression of entropy generation for a control mass.

A control mass is a closed system (fixed mass, no mass crossing the boundary). Entropy generation S_gen measures the entropy created inside it by irreversibilities.

Derivation

Consider a closed system that goes from state 1 to state 2 by an irreversible process A, with heat exchanged at boundary temperature T_b. It returns from 2 to 1 by a reversible process B. Together they form a cycle that is irreversible overall.

Clausius inequality: ∮ δQ/T ≤ 0 (strictly < 0 for an irreversible cycle).

   P
   |  1 ______ A (irreversible)
   |   \          \
   |    \__ B ____2
   |   (reversible)
   +----------------> V

∫₁² (δQ/T_b)_A + ∫₂¹ (δQ/T)_B < 0

For the reversible path B, ∫₂¹ (δQ/T)_B = S₁ − S₂ (definition of entropy). So

∫₁² δQ/T_b + S₁ − S₂ < 0

S₂ − S₁ > ∫₁² δQ/T_b

Convert the inequality to an equation by adding a non-negative term S_gen:

S₂ − S₁ = ∫₁² δQ/T_b + S_gen, with S_gen ≥ 0

or in rate form

dS/dt = Σ Q̇_k/T_k + Ṡ_gen

Meaning of the terms

  • S₂ − S₁ is the entropy change of the system.
  • ∫ δQ/T_b is the entropy transferred by heat (positive, negative or zero).
  • S_gen = S₂ − S₁ − ∫ δQ/T_b is the entropy generated:
    • S_gen = 0 for a reversible process
    • S_gen > 0 for an irreversible process
    • S_gen < 0 is impossible

For an adiabatic system (δQ = 0), S₂ − S₁ = S_gen ≥ 0. If the system rejects heat Q_out to surroundings at T₀, taking the boundary at T₀ gives S_gen = ΔS_system + Q_out/T₀, i.e. the sum of the entropy changes of the system and the surroundings.

✓ Answer checked by Claude Opus 5.5 (high effort)

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