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The entropy balance accounts for storage, heat transfer, mass streams and entropy generation by irreversibility.
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Formula
S in J/K; Ṡ and ṁs in W/K; T in K
| Symbol | Meaning | Unit |
|---|---|---|
| Entropy in the control volume | J/K | |
| Q̇j | Heat-transfer rate at boundary j, positive inward | W |
| Absolute boundary temperature | K | |
| Specific entropy of a stream | J/(kg·K) | |
| Ṡgen | Nonnegative entropy-generation rate | W/K |
\frac{dS_{CV}}{dt}=\sum_j\frac{\dot Q_j}{T_j}+\sum_{in}\dot m s-\sum_{out}\dot m s+\dot S_{gen}Heat entropy transfer uses the absolute temperature at each boundary location; separate terms are needed for multiple temperatures. The second law requires Ṡgen≥0. For steady adiabatic operation with one inlet and one outlet, Ṡgen=ṁ(s2−s1). A negative calculated value usually indicates inconsistent states, signs or data.
Heat entropy transfer uses the absolute temperature at each boundary location; separate terms are needed for multiple temperatures. The second law requires Ṡgen≥0. For steady adiabatic operation with one inlet and one outlet, Ṡgen=ṁ(s2−s1). A negative calculated value usually indicates inconsistent states, signs or data.
The entropy balance accounts for storage, heat transfer, mass streams and entropy generation by irreversibility.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Control-volume entropy balance” and check its unit and magnitude.
An adiabatic steady turbine has ṁ=1 kg/s, s1=6.5 and s2=6.65 kJ/(kg·K).
Ṡgen=1·0.15=0.15 kW/K≥0, consistent with the second law.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: S in J/K; Ṡ and ṁs in W/K; T in K.
Formula cluster
The entropy balance accounts for storage, heat transfer, mass streams and entropy generation by irreversibility.
An adiabatic steady turbine has ṁ=1 kg/s, s1=6.5 and s2=6.65 kJ/(kg·K). Ṡgen=1·0.15=0.15 kW/K≥0, consistent with the second law.
Assess irreversibility and process losses in open thermodynamic systems.
Here is how to work through a typical Control-volume entropy balance (Entropiespeicherung = Wärmeübertrag + Ströme + Erzeugung) task step by step:
Task
Work through the numerical example for “Control-volume entropy balance” and check its unit and magnitude.
Solution path
An adiabatic steady turbine has ṁ=1 kg/s, s1=6.5 and s2=6.65 kJ/(kg·K).
Ṡgen=1·0.15=0.15 kW/K≥0, consistent with the second law.
The second law says real irreversible processes generate entropy, while the reversible limit generates none. Friction, mixing and heat transfer across finite temperature differences contribute positive generation. A negative computed value therefore signals wrong signs, inconsistent properties or a misplaced boundary, not real “entropy consumption.” Entropy transfer uses absolute temperature; entropy generation cannot be negative. Application: Assess irreversibility and process losses in open thermodynamic systems. The numerical example An adiabatic steady turbine has ṁ=1 kg/s, s1=6.5 and s2=6.65 kJ/(kg·K). Ṡgen=1·0.15=0.15 kW/K≥0, consistent with the second law anchors the idea in a concrete measurement case.
Use the absolute temperature at the boundary where heat crosses, not automatically the mean fluid temperature. If heat crosses several regions at different boundary temperatures, each transfer gets its own term. Celsius values are not valid here; convert to kelvin first. Entropy transfer uses absolute temperature; entropy generation cannot be negative. The corresponding units are S in J/K; Ṡ and ṁs in W/K; T in K. An adiabatic steady turbine has ṁ=1 kg/s, s1=6.5 and s2=6.65 kJ/(kg·K). Ṡgen=1·0.15=0.15 kW/K≥0, consistent with the second law. This checks that the reference quantities are compatible.
An adiabatic throttle has no heat transfer and usually no shaft work; at steady state, enthalpy is approximately conserved. The entropy balance nevertheless gives s2>s1 in typical dissipative throttling. Isenthalpic therefore does not mean isentropic, a key distinction for valves and throttling devices. Entropy transfer uses absolute temperature; entropy generation cannot be negative. Scope: Heat entropy transfer uses the absolute temperature at each boundary location; separate terms are needed for multiple temperatures. The second law requires Ṡgen≥0. For steady adiabatic operation with one inlet and one outlet, Ṡgen=ṁ(s2−s1). A negative calculated value usually indicates inconsistent states, signs or data. A practical application is Assess irreversibility and process losses in open thermodynamic systems.
Write the quantity and unit from the variable list next to every value: S in J/K; Ṡ and ṁs in W/K; T in K. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example An adiabatic steady turbine has ṁ=1 kg/s, s1=6.5 and s2=6.65 kJ/(kg·K). Ṡgen=1·0.15=0.15 kW/K≥0, consistent with the second law. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. Entropy transfer uses absolute temperature; entropy generation cannot be negative.
The background states this starting condition: Heat entropy transfer uses the absolute temperature at each boundary location; separate terms are needed for multiple temperatures.. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. Entropy transfer uses absolute temperature; entropy generation cannot be negative.
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