Energy Fundamentals, Thermodynamics and Heat Transfer
Domain 1 examines whether you can set up and check energy calculations before trusting a design or an audit result. The assessment probes unit discipline, steady-flow energy balances, second-law limits and the three heat-transfer modes, usually through short numerical items and follow-up judgement questions.
Core rules
- Convert to consistent SI or to the unit the answer requires: 1 kWh = 3.6 MJ, 1 toe = 41.868 GJ, 1 refrigeration ton (TR) = 3.517 kW. Steam tables use absolute pressure, so add atmospheric pressure to a gauge reading.
- For a steady-flow control volume, Q − W = Σm_out h_out − Σm_in h_in when kinetic and potential energy are negligible. For a closed system, Q − W = ΔU.
- The Carnot efficiency 1 − T_C/T_H uses absolute temperature and is an upper bound, not a prediction. Real devices are compared using isentropic efficiency.
- Conduction resistance is L/kA; convection is 1/hA; clean overall coefficient U is based on 1/UA = ΣR. Fouling adds a resistance and must be stated.
- For heat exchangers, use LMTD = (ΔT1 − ΔT2)/ln(ΔT1/ΔT2) in counterflow or parallel flow; the arithmetic mean overstates duty when the temperature difference changes a lot.
Worked example
A boiler raises 5 kg/s of water from 30 °C to saturated steam at 10 bar. At 10 bar, h_f = 763 kJ/kg and h_g = 2,778 kJ/kg; feedwater enthalpy at 30 °C is about 126 kJ/kg. Heat absorbed = 5 × (2,778 − 126) = 13,260 kW. At 85% boiler efficiency, fuel input = 13,260/0.85 = 15,600 kW. If fuel costs HK$0.45/kWh (thermal), the hourly fuel cost is 15,600 × 0.45 = HK$7,020. Note that using h_fg alone would omit the sensible heating and understate the duty.
Traps
Using gauge pressure in steam tables; adding h_fg to h_g; treating kW and kWh as interchangeable; applying arithmetic mean ΔT to a large temperature glide; forgetting that U changes with fouling and flow; and mixing gross and net energy when auxiliaries are involved.