Heat Transfer
Modes and resistance networks
Conduction uses q = kAΔT/L; convection q = hAΔT; radiation q = εσA(T1⁴ − T2⁴). Build thermal resistance networks in series and parallel: R_cond = L/kA, R_conv = 1/hA, R_rad = 1/h_r A. Then q = ΔT/R_total and U = 1/(A·R_total). Add a fouling factor on the waterside of coils and condensers.
Exchangers and extended surfaces
For a heat exchanger, q = UA·LMTD, with LMTD computed counterflow and a correction factor F applied for other arrangements. When outlet temperatures are unknown, use the effectiveness–NTU method: ε = q/q_max, where q_max = C_min(Th,in − Tc,in). Fins improve the air side of coils, where h is low; the fin efficiency accounts for the temperature drop along the fin.
Traps
Using arithmetic mean temperature difference instead of LMTD; forgetting the F correction for crossflow and multipass shells; applying radiation laws with °F instead of °R; treating a composite wall as parallel resistances when it is series; and neglecting that radiation can dominate in radiant slab or roof-solar problems (sol-air temperature).
Worked example
A wall assembly: inside film 1/h_i = 0.50, insulation L/k = 10.0, outside film 1/h_o = 0.25 (all in h·ft²·°F/Btu). R_total = 10.75. U = 1/10.75 = 0.093 Btu/h·ft²·°F. With 3,000 ft² and ΔT = 40°F, q = U·A·ΔT = 0.093 × 3,000 × 40 ≈ 11,160 Btu/h.