Entropy Balances
Entropy balances look like mass/energy balances — except they include an entropy generation term that can never be negative.
⏱️ About 16 min
Mass balances don't have a 'generation' term — why does entropy?
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The big idea: For control volumes, the second law appears as an entropy balance with a nonnegative generation term Ṡ_gen that accounts for irreversibility.
Open-System Entropy Accounting
Entropy can cross a control-volume boundary by two mechanisms:
- with heat transfer at the boundary (entropy transfer rate ≈ Q̇/T at the boundary location), and
- with mass flow (entropy carried in/out with streams: ṅ s).
Unlike mass and total energy, entropy can also be generated internally by irreversibilities.
\[ \frac{dS_{\text{CV}}}{dt}=\sum_k \frac{\dot{Q}_k}{T_{b,k}}+\sum_{\text{in}}\dot{n}\,s_{\text{in}}-\sum_{\text{out}}\dot{n}\,s_{\text{out}}+\dot{S}_{\text{gen}} \]
General entropy rate balance for a control volume (open system). Each heat-transfer interaction k uses the boundary temperature T_b,k where the heat crosses.
\[ \dot{S}_{\text{gen}} \ge 0 \]
This inequality is the mathematical statement of the second law for the control volume (strictly > 0 for irreversible processes; = 0 for reversible limit).
⚠️ Pitfall: A negative Ṡ_gen is a red flag
Because the second law requires Ṡgen ≥ 0, a negative computed value almost always means:
- a sign mistake (especially in Σin − Σout),
- using the wrong temperature for Q̇/T,
- inconsistent units (e.g., J vs kJ), or
- an impossible process specification.
Steady-State Simplification (Most Common in Process Equipment)
At steady state, dSCV/dt = 0, so the balance becomes:
0 = Σ(Q̇/T) + Σin(ṅ s) − Σout(ṅ s) + Ṡ_gen
Rearranged to compute entropy generation rate:
Ṡ_gen = Σout(ṅ s) − Σin(ṅ s) − Σ(Q̇/T)
📝 Worked example: A steady-state heater has one inlet and one outlet stream. Data: ṅ = 2.00 mol/s (same in and out), s_in = 120 J/(mol·K), s_out = 150 J/(mol·K). Heat is transferred into the control volume at a boundary temperature T_b = 500 K at a rate Q̇ = +20.0 W. Compute the entropy generation rate Ṡ_gen.
- Steady state: dS_CV/dt = 0, so: Ṡ_gen = Σout(ṅ s) − Σin(ṅ s) − Σ(Q̇/T_b)
- Compute stream entropy flow terms:
- Σout(ṅ s) = (2.00 mol/s)(150 J/(mol·K)) = 300 J/(s·K)
- Σin(ṅ s) = (2.00 mol/s)(120 J/(mol·K)) = 240 J/(s·K)
- Compute heat-entropy transfer term (Q̇ is into CV): Σ(Q̇/T_b) = (20.0 J/s) / (500 K) = 0.0400 J/(s·K)
- Therefore: Ṡ_gen = 300 − 240 − 0.0400 = 59.96 J/(s·K)
- Check: Ṡ_gen > 0, consistent with an irreversible real heater.
✓ Ṡ_gen = 59.96 J/(s·K)
✏️ Practice: A steady-state device has one inlet and one outlet with ṅ = 1.50 mol/s. Entropy values: s_in = 80.0 J/(mol·K), s_out = 110 J/(mol·K). Heat transfer is into the device: Q̇ = +12.0 W at boundary temperature T_b = 400 K. Compute Ṡ_gen (J/(s·K)).
J/(s·K)
Solution
- Steady state: Ṡ_gen = Σout(ṅ s) − Σin(ṅ s) − (Q̇/T_b)
- Σout = (1.50)(110) = 165 J/(s·K)
- Σin = (1.50)(80.0) = 120 J/(s·K)
- Q̇/T_b = (12.0 J/s)/(400 K) = 0.0300 J/(s·K)
- Ṡ_gen = 165 − 120 − 0.0300 = 44.97 J/(s·K)
Check your understanding
1. Which term is unique to entropy balances (i.e., not present in mass and total-energy balances)?
Entropy has a generation term Ṡ_gen due to irreversibility; mass/energy do not have an analogous 'generation' from dissipation.
2. For any real process, the second law requires:
The mathematical form of the second law is Ṡ_gen ≥ 0 (zero only in the reversible limit).
✅ Key takeaways
- Entropy crosses boundaries via heat transfer (≈ Q̇/T_b) and mass flow (ṅ s)
- Entropy can be generated internally: Ṡ_gen accounts for irreversibility
- Second law (math): Ṡ_gen ≥ 0 always
- At steady state: Ṡ_gen = Σout(ṅ s) − Σin(ṅ s) − Σ(Q̇/T_b)
➡️ To use entropy balances, you need ways to compute entropy changes and stream entropies. Next we derive a workhorse formula: ideal-gas entropy change as a function of T and P.
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