Thermodynamics and Heat Transfer
Thermodynamics and Heat Transfer tests energy balances, property relationships, power cycles, refrigeration, and the three heat-transfer modes. Expect steady-flow devices, entropy generation, log-mean temperature difference, and radiation with absolute temperature.
Core rules
- Closed system: Q − W = ΔU. Steady-flow: q − w = Δh + ΔKE + ΔPE.
- Entropy: ΔS ≥ ∫δQ/T; equality for internally reversible processes, isolated-system ΔS ≥ 0.
- Carnot efficiency η = 1 − Tc/Th, with absolute temperatures.
- Conduction: q = kAΔT/L. Convection: q = hA(Ts − T∞). Radiation: q = εσA(Ts⁴ − Tsur⁴).
- Heat exchangers: q = UAΔTlm; for counterflow ΔTlm = (ΔT1 − ΔT2)/ln(ΔT1/ΔT2).
- COP_ref = QL/Win; COP_HP = QH/Win.
Worked example
A counterflow heat exchanger has hot fluid 120°C → 70°C and cold fluid 30°C → 60°C. Find ΔTlm and q if UA = 2000 W/K. At one end ΔT1 = 120 − 60 = 60 K; at the other ΔT2 = 70 − 30 = 40 K. ΔTlm = (60 − 40)/ln(60/40) = 20/0.4055 = 49.3 K. Then q = UAΔTlm = 2000 × 49.3 = 98,600 W, about 98.6 kW. The end temperatures must be paired correctly: hot inlet with cold outlet in counterflow.
Traps
- Using °C in radiation or Carnot efficiency instead of kelvin.
- Reversing the LMTD end differences or using arithmetic mean for large temperature change.
- Ignoring pump or fan work in a cycle energy balance.
- Confusing COP for refrigerator and heat pump.
- Assuming a process is reversible when entropy generation is present.