Thermodynamics, Heat Transfer and Fluid Mechanics
What the exam tests
This domain tests the physical fundamentals behind mechanical design: steady-flow energy, entropy and Carnot limits, conduction and convection, radiation, pipe friction, pump and fan power, and heat-exchanger performance. Questions are usually short calculations with unit traps.
Core rules
- Steady-flow energy: q − w = Δh + Δke + Δpe. For a chiller or coil, q = m × cp × ΔT. For a pump, useful power is ρgQH.
- Carnot COP for cooling is Tc / (Th − Tc) with absolute temperatures. Real COP is lower; compare using the same temperature basis.
- Conduction: Q = ΔT / R, where R includes thickness/k and surface films. For a composite wall, add resistances in series.
- Convection: Q = hAΔT. Radiation: Q = εσA(Ts⁴ − Tsur⁴), with temperatures in kelvin.
- Reynolds number Re = ρVD/μ. Below about 2,300 is laminar, above 4,000 is turbulent, and the Moody chart gives friction factor.
- Darcy–Weisbach: hf = f (L/D)(V²/2g). Minor losses use K V²/2g.
- Heat exchangers: use LMTD or effectiveness–NTU; counterflow beats parallel flow.
Worked example
Chilled water flows at 12 L/s with a 6 K rise. Q = 4.18 × 12 × 6 = 301 kW. For a pump delivering 25 L/s at 30 m head with 72 % efficiency: P = 1000 × 9.81 × 0.025 × 30 / 0.72 = 10,220 W, or about 10.2 kW. The shaft power is 10.2 kW; motor input is higher by the motor efficiency. Always state which power you are calculating.
Traps
Using gauge instead of absolute pressure in gas laws; using °C in radiation or Carnot equations; using arithmetic mean temperature instead of LMTD; ignoring fouling resistance; mixing L/s and m³/s; forgetting that pump efficiency applies after hydraulic power.
Key points
- Use absolute temperatures in Carnot and radiation calculations.
- q = m cp ΔT is the core coil and water calculation.
- Pump hydraulic power is ρgQH; divide by efficiency for shaft power.
- Re determines flow regime and friction factor.
- LMTD, not arithmetic ΔT, applies to heat exchangers.