🎓 Lesson 11
D5
Physical vs. Chemical Exergy: Definitions and Reference States
Physical exergy is the useful energy you can get from a substance just because it’s hotter, colder, or at higher pressure than its surroundings; chemical exergy is the useful energy you can get from its chemical composition—like how much energy is locked in coal or explosives before they react.
🎯 Learning Objectives
- ✓ Calculate physical exergy for compressed air and hot exhaust gases using reference state tables
- ✓ Explain the role of the reference environment (dead state) in determining chemical exergy values for fuels and blasting agents
- ✓ Apply standardized reference chemical exergy values (e.g., for ANFO, diesel, methane) to assess energy quality in mine ventilation and explosive selection
- ✓ Analyze and compare exergy destruction in detonation vs. controlled combustion processes using second-law efficiency concepts
📘 Core Principles
Exergy is the thermodynamic 'currency' of usefulness: it measures maximum theoretical work relative to a defined reference environment (the 'dead state'). Physical exergy arises from deviations in temperature (T), pressure (P), velocity, or elevation from this dead state—no chemical change needed. Chemical exergy arises from the substance’s molecular structure and its capacity to react exothermically with reference environment species (e.g., O₂, N₂, H₂O, CO₂, CaCO₃) to reach equilibrium. Crucially, chemical exergy depends on the *reference environment composition*: ISO 13602 and Szargut define standard atmospheric composition (78.9% N₂, 20.9% O₂, 0.039% CO₂, etc.) at 298.15 K and 101.325 kPa—but for underground mines, local air composition or post-ventilation gas mixtures may require site-specific adjustments. The separation is essential: physical exergy dominates in compressed air systems; chemical exergy dominates in explosives, fuel combustion, and spontaneous oxidation of sulfide ores.
📐 Key Calculation
The specific physical exergy of a fluid (neglecting kinetic/potential terms) is calculated using ideal gas relations or property tables. For chemical exergy, standardized values are typically used due to complexity—except for simple mixtures where formation-based calculation is feasible.
💡 Worked Example
Problem: Calculate the specific physical exergy of compressed air at 350 K and 700 kPa, relative to a dead state of 298.15 K and 101.325 kPa. Assume constant c_p = 1.005 kJ/kg·K and R = 0.287 kJ/kg·K.
1.
Step 1: Identify knowns: T = 350 K, T₀ = 298.15 K, P = 700 kPa, P₀ = 101.325 kPa, c_p = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K
2.
Step 2: Apply physical exergy formula: e_ph = c_p(T − T₀) − T₀ c_p ln(T/T₀) + R T₀ ln(P/P₀)
3.
Step 3: Compute: e_ph = 1.005(350−298.15) − 298.15×1.005×ln(350/298.15) + 0.287×298.15×ln(700/101.325) ≈ 52.1 − 48.7 + 165.3 = 168.7 kJ/kg
Answer:
The specific physical exergy is 168.7 kJ/kg, which falls within the typical range of 150–200 kJ/kg for compressed air at 7–10 bar used in underground drill rigs.
🏗️ Real-World Application
At the Bingham Canyon Mine (Rio Tinto), exergy analysis revealed that 68% of the chemical exergy in ANFO (ammonium nitrate/fuel oil) was converted to physical exergy (shock wave, rock motion, heat) during detonation—yet only ~12% contributed to effective fragmentation (mechanical work on rock). The remainder was exergy destruction due to irreversibilities (viscous dissipation, radiation, incomplete reaction). By contrast, the compressed air system powering raise-boring rigs had 42% exergy destruction in piping losses and throttling—highlighting that physical exergy optimization (e.g., variable-speed compressors, leak reduction) yielded faster ROI than chemical exergy upgrades. This insight shifted capital allocation toward air system retrofits and real-time blast monitoring to improve exergy utilization per ton of ore.
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