Closed system (physical exergy)
e_ph = (u - u0) + p0(v - v0) - T0(s - s0)
Where u, v, s are specific internal energy, volume, and entropy at system state; subscript 0 denotes environment state.
Definition and concept
- Formal definition: Maximum theoretical work obtainable as a system interacts reversibly with the environment until equilibrium is reached.
- State function: Depends on system state and reference environment state.
- Zero at equilibrium: No driving forces (ΔT, Δp, Δμ) → no work potential.
Energy vs exergy
- Energy: Conserved (First Law), quantity only.
- Exergy: Measures quality of energy; destroyed by irreversibility (Second Law).
- Example: 1 MJ of heat at 1000 K has more exergy than 1 MJ at 350 K, because it can produce more work relative to environment.
Reference environment
- Defined by temperature T0, pressure p0, and composition of surroundings.
- Often taken as ambient conditions (e.g., T0 = 298 K, p0 = 1 atm).
- All exergy calculations are relative to this state.
Types of exergy
- Physical exergy: Due to differences in temperature and/or pressure from environment.
- Chemical exergy: Due to differences in chemical composition from environment.
- Kinetic exergy: Due to velocity relative to environment.
- Potential exergy: Due to elevation in a gravitational field relative to environment.
- Mixing exergy: Potential work from separating mixed substances.
Key formulas
Flow exergy (steady-flow devices)
ψ = (h - h0) - T0(s - s0) + (V² - V0²)/2 + g(z - z0)
h = specific enthalpy, V = velocity, z = elevation.
Exergy of heat transfer at boundary temperature Tb
Ex_Q = Q * (1 - T0/Tb)
Exergy of work
Ex_W = W (for mechanical work fully convertible to work)
Gouy–Stodola theorem
Exergy destruction is proportional to entropy generation:
Ex_dest = T0 * S_gen
This links Second Law irreversibility directly to lost work potential.
Worked examples
Example 1: Physical exergy of steam
Given: Steam at 2 MPa, 400°C; environment T0 = 25°C, p0 = 100 kPa.
From steam tables: h = 3230.1 kJ/kg, s = 6.769 kJ/(kg·K); h0 = 104.83 kJ/kg, s0 = 0.3672 kJ/(kg·K).
Flow exergy: ψ = (3230.1 − 104.83) − 298.15 × (6.769 − 0.3672) ≈ 3125.27 − 1916.6 ≈ 1208.7 kJ/kg.
Example 2: Exergy of heat transfer
Given: Q = 500 kJ at Tb = 500 K, T0 = 300 K.
ExQ = 500 × (1 − 300/500) = 500 × 0.4 = 200 kJ.
Applications
- Identify and quantify inefficiencies in power plants, refrigeration, and industrial processes.
- Guide design improvements by locating major exergy destructions.
- Compare alternative processes on a second-law basis.
- Environmental impact assessment: quantify wasted work potential.
Exergy destruction and losses
- Destruction: Due to irreversibilities (finite ΔT, friction, mixing, chemical reaction).
- Loss: Exergy rejected to environment without use (e.g., exhaust gases, warm cooling water).
- Both reduce overall second-law efficiency.
Exam checklist
- Know definition and difference from energy.
- Be able to compute physical and flow exergy.
- Apply Gouy–Stodola to find lost work.
- Use correct reference environment values.
- Identify major exergy destructions in a system.