Heat transfer fundamentals for the PE Mechanical exam

Heat transfer underpins more of the PE Mechanical exam than its own line in the spec suggests — every heat exchanger, coil, condenser, and insulation problem rests on it. The exam doesn't test derivations; it tests whether you can pick the right relation, build a resistance network, and keep your units straight under time. Three modes, one bookkeeping system — with radiation the nonlinear exception (T⁴).

Conduction — Fourier's law

Through a plane wall, q = kA·ΔT/L, which is cleaner written as a resistance: q = ΔT/R with R_cond = L/(kA). Through a pipe or insulation (cylindrical), R_cond = ln(r₂/r₁)/(2πkL) — a ratio of radii inside a natural log, not (r₂ − r₁).

Layers in series add resistances (R_total = ΣR) and pass the same heat flow; parallel paths add conductances (1/R_total = Σ1/Rᵢ). The move that saves time: in series, find the flux once, then take the temperature drop across whichever layer you care about.

Two-layer wall: inside 100 °C, outside 20 °C, per-area R₁ = 0.10 and R₂ = 0.30 m²·K/W. Flux = ΔT/ΣR = 80/0.40 = 200 W/m²; the drop across the first layer is 200 × 0.10 = 20 °C, so the interface sits at 80 °C.

Convection — Newton's law of cooling

q = h·A·ΔT (surface-to-fluid), and a film is just another resistance: R_conv = 1/(hA).

h = 25 W/m²·K, A = 4 m², ΔT = 30 °C → q = 25 × 4 × 30 = 3,000 W.

Combining conduction and convection — the overall U

A wall with fluid on both sides is resistances in series: 1/(UA) = 1/(h₁A) + L/(kA) + 1/(h₂A), and q = U·A·ΔT_overall.

h₁ = 10, h₂ = 40 W/m²·K, wall L/k = 0.04 → 1/U = 0.10 + 0.04 + 0.025 = 0.165 → U ≈ 6.06 W/m²·K; at ΔT = 50 °C, q ≈ 303 W/m².

The teaching point the exam rewards: the largest resistance dominates the total — here the inside film is ~60% of it (0.10 of 0.165), so it's the biggest lever on U. Improving a small resistance (the wall) moves U far less than changing the dominant one. Which way you want U to go sets the move: to cut heat loss, add resistance (insulate); to enhance heat transfer, attack the dominant resistance (the film).

Radiation — Stefan–Boltzmann

q = ε·σ·A·(T₁⁴ − T₂⁴), σ = 5.67×10⁻⁸ W/m²·K⁴, and temperatures absolute (K or °R). The fourth power and the absolute temperature are the problem — get those right and radiation is straightforward.

ε = 0.80, A = 2 m², 500 K to 300 K surroundings → q = 0.8 × 5.67×10⁻⁸ × 2 × (500⁴ − 300⁴) ≈ 4,935 W.

(Transient/lumped-capacitance and fins show up occasionally — know they exist, but the three modes above are where the points are.)

Where people lose points

• Series vs. parallel resistances — adding resistances when paths are parallel (or vice versa). In series the heat flow is common; in parallel the ΔT is common.
• Radius vs. diameter in cylindrical conduction — the log term is ln(r₂/r₁); using (r₂ − r₁), or mixing a radius with a diameter, is wrong.
• Radiation in °C/°F, or dropping the 4th power — T must be absolute, and ΔT⁴ ≠ (ΔT)⁴.
• Dropping a film resistance — using only the wall (k, L) and ignoring the convective films, so U comes out far too high.
• Unit mismatch — mixing SI and USCS across k, h, A, L. Pick one system and stay in it.

Drill it

Practice these on real problems — the tutor walks any you miss, grounded in the worked solution. Where this lives: TFS Module 1 — Thermal/Fluid Principles (this area) and, where the modes combine into LMTD/ε-NTU, Heat exchangers — LMTD & ε-NTU (the shared deep-dive). Pair it with Thermodynamics fundamentals.