Aragon PAU Technology and Engineering II - independent engineering foundations study materials
Aragon PAU Technology and Engineering II - independent engineering foundations study resources

Aragon PAU Technology and Engineering II - independent engineering foundations: Energy, thermal machines and environmental assessment

Aragon PAU Technology and Engineering II - independent engineering foundations · Work through your exam, one topic at a time.

All guide sections

Energy, thermal machines and environmental assessment

This domain asks you to compare energy sources, apply conversion efficiencies with correct units and assess sustainability. Typical items are a hydraulic or wind power calculation, a thermal efficiency question and a comment on the environmental impact of the energy mix.

Sources and conversion

  • Non-renewable: coal, oil, natural gas, nuclear fission. Renewable: hydro, wind, solar PV and thermal, biomass, geothermal, tidal and wave.
  • Units: 1 kWh = 3.6 MJ; 1 tep ≈ 41.86 GJ.
  • Efficiency η = useful energy/primary energy (equivalently for power).
  • Thermal plant: η = W/Q_in = 1 − Q_out/Q_in; For a heat engine between hot reservoir T_hot and cold reservoir T_cold, the Carnot limit is η =1−T_cold/T_hot with absolute temperatures in kelvin.
  • Hydro: P = η·ρ·g·Q·H, with ρ = 1000 kg/m³ and g = 9.81 m/s².
  • In the ideal unconfined actuator-disk model, extracted wind power is P =½ρAv³C_p and the Betz upper bound is C_p =16/27≈0.593; actual systems and operating conditions require additional losses and constraints.
  • A supplied uniform-irradiance PV estimate is P =GAη. Standard-test irradiance does not establish field temperature, shading or annual yield.

Sustainability

  • CO₂ emitted per kWh, life-cycle analysis, carbon footprint, energy mix, savings and efficiency, waste management and recycling.

Worked example

A hydro plant with Q = 2 m³/s, H = 50 m and η = 0.85. P = 0.85 · 1000 · 9.81 · 2 · 50 = 833 850 W ≈ 834 kW. With 4000 equivalent hours per year, E = 834 kW · 4000 h = 3.34·10⁶ kWh = 3.34 GWh, which in joules is 3.34·10⁶ · 3.6 MJ = 1.20·10⁷ MJ ≈ 12 000 GJ. As a check on limits: a Carnot machine between 800 K and 300 K has η = 1 − 300/800 = 0.625, and no real cycle between those temperatures can beat it.

Marks lost

  • Using degrees Celsius in the Carnot formula.
  • Confusing kW (power) with kWh (energy).
  • Forgetting the efficiency factor in a power calculation.
  • Using v instead of v³ in the wind formula.
  • Adding energies without converting units first.