Calculator Inputs
Example Data Table
| Case | Method | Main Inputs | Specific Exergy | Exergy Rate |
|---|---|---|---|---|
| Steam line example | Direct property | h=850, h₀=300, s=2.4, s₀=1.2, T₀=298.15, V=50, z-z₀=12, ṁ=2.5 | 193.59 kJ/kg | 483.97 kW |
| Compressed air example | Ideal gas | cₚ=1.005, R=0.287, T=650, T₀=298.15, P=500, P₀=101.325, V=80, ṁ=1.8 | 259.95 kJ/kg | 467.91 kW |
Formula Used
For steady-flow engineering systems, specific physical exergy combines thermodynamic availability with kinetic and potential contributions.
ψ = (h − h₀) − T₀(s − s₀) + V²/2000 + g(z − z₀)/1000
ψ = cₚ[(T − T₀) − T₀ ln(T/T₀)] + R T₀ ln(P/P₀) + V²/2000 + g(z − z₀)/1000
Exergy Rate = ṁ × ψ
Total Exergy = Exergy Rate × Duration
Exergy Destroyed = Available Exergy for Work − Useful Work Output
Here, h is specific enthalpy, s is specific entropy, T₀ and P₀ are reference conditions, V is velocity, z is elevation, and ṁ is mass flow rate.
How to Use This Calculator
- Select the calculation mode. Use the direct property method when enthalpy and entropy are known. Use the ideal gas option when temperature and pressure are available.
- Enter the reference dead-state conditions. Ambient reference values commonly represent the environment used for exergy evaluation.
- Add mass flow rate, velocity, and elevation terms. These allow the tool to compute exergy rate and mechanical contributions.
- Optionally enter outlet specific exergy and useful work. This extends the calculation to component analysis and second-law efficiency.
- Click Calculate Exergy. The result section appears below the header and above the form, exactly as requested.
- Use the CSV or PDF buttons to export the result for reporting, audits, design studies, or classroom work.
FAQs
1) What does exergy measure?
Exergy measures the maximum useful work a system can deliver as it reaches equilibrium with its environment. It shows energy quality, not just quantity.
2) Why is the reference state important?
Exergy always depends on the chosen environment. Changing ambient temperature or pressure changes the dead state and therefore changes the calculated work potential.
3) When should I use the direct property method?
Use the direct property method when accurate thermodynamic property data are known. It is common for steam tables, refrigerants, and measured plant-state data.
4) When is the ideal gas approximation acceptable?
It is appropriate for gases that behave close to ideal conditions over the operating range. It is often useful for air-standard studies and preliminary engineering estimates.
5) Why can exergy destruction become negative here?
A negative value usually means the optional inputs are inconsistent. For example, useful work may exceed available exergy, or outlet exergy may be too low.
6) What units does this calculator expect?
Use kJ/kg for enthalpy and specific exergy, kJ/kg·K for entropy and gas constants, Kelvin for temperature, kPa for pressure, meters for elevation, and kg/s for mass flow.
7) Does the calculator include kinetic and potential exergy?
Yes. Velocity and elevation inputs are included automatically. Their contributions are shown separately, which is helpful for turbines, nozzles, and piping analyses.
8) Can I use this for component performance studies?
Yes. Enter outlet specific exergy and useful work output to estimate available work, destroyed exergy, and second-law efficiency for a component or control volume.