Heat Transfer
Heat transfer questions in GATE CH split into conduction (fins, resistance networks, lumped capacitance), convection (dimensionless groups and correlations), radiation (view factors, shields) and exchanger design (LMTD and effectiveness–NTU). Expect at least one exchanger sizing item.
Core rules
- Conduction. Thermal resistance L/kA for a slab, ln(r₂/r₁)/2πkL for a cylinder. Fins: efficiency η = tanh(mL)/mL with m = √(hP/kA_c) for a rectangular fin; effectiveness compares fin heat flow to the bare base.
- Lumped capacitance. Valid when Bi = hL_c/k < 0.1; then T − T∞ = (T₀ − T∞)e^(−t/τ) with τ = ρVc_p/hA.
- Convection. Turbulent tube flow: Dittus–Boelter Nu = 0.023Re^0.8 Pr^n (n = 0.4 heating, 0.3 cooling), valid for Re > 10⁴ and 0.7 < Pr < 160. Fully developed laminar: Nu = 4.36 (constant heat flux) or 3.66 (constant wall temperature).
- Radiation. q = εσA(T⁴ − T_sur⁴); n identical shields cut net radiation to 1/(n+1) of the unshielded value.
- Exchangers. LMTD uses the two terminal differences; for multipass and crossflow multiply by F (< 1). Effectiveness–NTU: ε = q/q_max, NTU = UA/C_min, C_r = C_min/C_max.
Worked example
Counterflow exchanger: hot stream 150 → 90 °C, cold stream 30 → 70 °C. ΔT₁ = 150 − 70 = 80 K and ΔT₂ = 90 − 30 = 60 K. LMTD = (80 − 60)/ln(80/60) = 20/0.2877 = 69.5 K. Note the arithmetic mean would give 70 K - close here, but badly wrong when the terminal differences diverge.
Traps
- Applying LMTD with a correction factor F > 1, or omitting F for a 1–2 exchanger.
- Using the lumped-capacitance formula when Bi exceeds 0.1.
- Forgetting that a condensing stream is isothermal, so one ΔT is the saturation approach.
- Using Dittus–Boelter outside its Re/Pr range.