One-dimensional steady conduction and fins

Why it matters

Conduction through walls, pipe insulation, motor windings, PCB layers and heat-sink bases is usually close to one-dimensional, and the thermal-resistance method turns it into simple circuit arithmetic. Fins — on motor housings, IGBT and CPU heat sinks, radiators and air-cooled engines — are how we beat a low convection coefficient. Knowing when a fin helps, how long to make it, and how to compute its heat flow is core thermal design for any mechatronics product.

Key ideas

Fourier's law. Heat conducts down a temperature gradient at a rate proportional to the gradient and to thermal conductivity k. The minus sign says heat flows from hot to cold. Typical k (W/m·K): copper about 400, aluminium about 200, steel 15–50, glass about 1, brick about 0.7, insulation 0.02–0.05, still air about 0.026.

Steady, one-dimensional, no generation, constant k. Temperature then depends on one coordinate only, and the heat rate Q is the same through every layer in series.

• Plane wall: the temperature profile is linear.
• Cylinder (pipe wall, insulation): the profile is logarithmic in r; the area grows with radius.
• Sphere: the profile varies as 1/r.

Thermal resistance. Writing Q = ΔT / R makes conduction and convection behave like resistors: layers in series add; parallel paths combine like parallel resistors. Contact resistance between pressed surfaces adds another series term (thermal interface materials reduce it). The overall heat-transfer coefficient U is defined by Q = U A ΔT_overall.

Critical radius of insulation. Adding insulation to a small pipe or wire increases conduction resistance but also increases outer surface area, which lowers convection resistance. Heat loss is a maximum at r_c = k_ins/h (cylinder). If the bare radius is below r_c, a thin layer of insulation increases heat loss — useful for electric cables, harmful for small steam lines.

Heat generation. With uniform volumetric generation q̇ (W/m³), as in a current-carrying conductor or a fuel rod, the profile becomes parabolic and the maximum temperature occurs at the centre (or at the insulated face).

Fins (extended surfaces). A fin adds area where h is low. Assumptions: steady, one-dimensional conduction along the fin (thin fin, low Biot number), constant k and h, no generation. The excess temperature θ = T − T∞ obeys d²θ/dx² − m²θ = 0, with m² = hP/(kA_c). Tip conditions:

• infinitely long fin (mL > about 3);
• insulated (adiabatic) tip;
• convective tip, usually handled by the insulated-tip result with a corrected length L_c = L + A_c/P (L + D/4 for a pin, L + t/2 for a thin plate).

Fin efficiency η_f = actual heat ÷ heat if the whole fin were at base temperature (always ≤ 1). Fin effectiveness ε_f = heat with fin ÷ heat from the same base area without the fin. A fin is justified only if ε_f is well above 1 — practically above 2 — which requires kP/(hA_c) ≫ 1: high-k material, thin fins, low h (gases, natural convection). Fins on the liquid side of a heat exchanger seldom pay. Beyond mL ≈ 2–3, extra length adds almost nothing, because tanh(mL) saturates.

Formulas

Q = −k A dT/dx — Fourier's law; Q (W), k (W/m·K), A: area normal to flow (m²), dT/dx (K/m). R_wall = L / (k A) — plane wall, K/W; L: thickness (m). R_cyl = ln(r₂/r₁) / (2π k L) — cylindrical shell, L: length (m). R_sph = (r₂ − r₁) / (4π k r₁ r₂) R_conv = 1 / (h A) — h: convection coefficient (W/m²·K). Q = (T_∞1 − T_∞2) / ΣR, U A = 1 / ΣR r_c = k_ins / h (cylinder), r_c = 2 k_ins / h (sphere) — critical radius (m). T_max = T_s + q̇ L² / (2k) — plane wall of half-thickness L with generation q̇ (W/m³), both faces at T_s. T_max = T_s + q̇ R² / (4k) — solid cylinder of radius R. m = √(h P / (k A_c)) — fin parameter (1/m); P: fin perimeter (m), A_c: cross-section area (m²). For a pin fin m = √(4h/(kD)). θ_b = T_b − T_∞, M = √(h P k A_c) · θ_b (W) Q_fin = M — infinitely long fin. Q_fin = M tanh(mL) — insulated tip. Q_fin = M [sinh(mL) + (h/mk) cosh(mL)] / [cosh(mL) + (h/mk) sinh(mL)] — convective tip. θ/θ_b = cosh(m(L − x)) / cosh(mL) — insulated-tip profile. L_c = L + A_c / P — corrected length. η_f = tanh(mL) / (mL) — insulated tip (use L_c for a convective tip). ε_f = Q_fin / (h A_c θ_b)

Worked examples

Example 1 (standard, composite wall). The wall of an air-conditioned room is 0.2 m brick (k = 0.72 W/m·K) with 0.05 m of insulation (k = 0.04 W/m·K). Inside air is at 25 °C (h_i = 10 W/m²·K); outside air is at 40 °C (h_o = 25 W/m²·K). Find the heat gain through 10 m² of wall and the inner surface temperature.

1. Resistances per m²: 1/h_i = 0.100; L/k (brick) = 0.2/0.72 = 0.278; L/k (insulation) = 0.05/0.04 = 1.250; 1/h_o = 0.040. Total = 1.668 m²·K/W.
2. Flux q = ΔT/ΣR″ = (40 − 25)/1.668 = 8.99 W/m²; Q = 8.99 × 10 = 89.9 W.
3. Inner surface: T = 25 + q × (1/h_i) = 25 + 8.99 × 0.1 = 25.9 °C.
4. U = 1/1.668 = 0.60 W/m²·K. The insulation carries 75 % of the total resistance.

Q ≈ 90 W, inner surface ≈ 25.9 °C, U ≈ 0.60 W/m²·K

Example 2 (GATE level, pin fin). An aluminium pin fin (k = 200 W/m·K), D = 5 mm and L = 50 mm, sits on a base at 100 °C in air at 25 °C with h = 25 W/m²·K. Find m, the heat flow (convective tip via corrected length), fin efficiency and effectiveness.

1. m = √(4h/(kD)) = √(4 × 25/(200 × 0.005)) = √100 = 10 m⁻¹.
2. A_c = π(0.005)²/4 = 1.963 × 10⁻⁵ m²; P = π × 0.005 = 0.01571 m.
3. M = √(h P k A_c) θ_b = √(25 × 0.01571 × 200 × 1.963 × 10⁻⁵) × 75 = 0.03927 × 75 = 2.945 W.
4. Corrected length L_c = L + D/4 = 0.05125 m; mL_c = 0.5125; tanh(0.5125) = 0.4719.
5. Q_fin = M tanh(mL_c) = 2.945 × 0.4719 = 1.390 W (the exact convective-tip formula also gives 1.390 W).
6. η_f = tanh(mL_c)/(mL_c) = 0.4719/0.5125 = 0.921.
7. ε_f = Q_fin/(h A_c θ_b) = 1.390/(25 × 1.963 × 10⁻⁵ × 75) = 1.390/0.0368 = 37.8.

m = 10 m⁻¹, Q ≈ 1.39 W per fin, η_f ≈ 92 %, ε_f ≈ 38

Example 3 (critical radius). A cable is insulated with PVC (k = 0.05 W/m·K) in air with h = 10 W/m²·K. r_c = k/h = 0.05/10 = 0.005 m = 5 mm. A conductor of 2 mm radius loses more heat as insulation is added up to 5 mm outer radius. r_c = 5 mm

Common mistakes

• Using an arithmetic-mean area for a thick cylinder instead of the logarithmic resistance.
• Adding conductivities of layers in series instead of adding resistances.
• Using fin efficiency to decide whether a fin is worth adding. Effectiveness answers that question.
• Taking the perimeter of a thin rectangular fin as 2(w + t) when the problem intends 2w, or the reverse; follow the problem's assumption.
• Forgetting the corrected length when the tip convects.
• Assuming insulation always reduces heat loss; check the critical radius for small diameters.
• Mixing mm and m inside √(hP/kA_c).

For GATE ME

Expect composite plane-wall and cylinder resistance problems, interface temperatures, critical radius, maximum temperature with heat generation, and fin questions: m, heat flow for long or insulated-tip fins, efficiency, effectiveness, and the ratio of heat flow between two fins of different k or D. Practise rearranging the fin formulas symbolically; many questions ask for a ratio, so constants cancel.

Quick check

1. A plane wall has L = 0.1 m, k = 1 W/m·K and A = 1 m². What is its resistance?
2. What is the critical radius for insulation with k = 0.1 W/m·K and h = 20 W/m²·K?
3. For an infinitely long fin, how does heat flow scale with k?
4. What is the efficiency of an insulated-tip fin with mL = 1? (tanh 1 = 0.762)
5. When is adding a fin not worthwhile?

Answers: 1. 0.1 K/W. 2. 5 mm. 3. As √k. 4. 0.762. 5. When effectiveness is near or below 1–2, e.g. with high h or low-k fin material.