One-dimensional steady conduction and fins

Why it matters

Furnace and kiln linings, insulation on steam lines, die and mould walls, cylinder fins on air-cooled engines and heat sinks on drives and electronics are all designed with one-dimensional conduction and fin theory. These models are simple enough to use by hand, yet accurate enough to size insulation thickness, estimate heat losses and choose fin geometry.

Key ideas

Fourier's law. Heat conducted through a solid is proportional to the area and the temperature gradient: q = −k·A·dT/dx. The minus sign shows that heat flows down the temperature gradient. Thermal conductivity k (W/m·K) is a material property: about 400 for copper, 200 for aluminium, 15–50 for steels, about 1 for refractory brick, and 0.03–0.1 for insulation.

Steady, one-dimensional conduction. Steady means temperatures at every point do not change with time (this is not thermal equilibrium; heat is still flowing). One-dimensional means temperature varies in only one coordinate. For a plane wall with constant k and no internal heat generation, the temperature profile is linear, so the gradient is constant through the thickness. In a cylinder or sphere the area changes with radius, so the profile is logarithmic (cylinder) or hyperbolic (sphere) even though the heat rate is constant at every radius.

Thermal resistance. Heat flow behaves like current: Q = ΔT/R. Plane wall R = L/(kA); cylindrical shell R = ln(r₂/r₁)/(2πkL); spherical shell R = (r₂ − r₁)/(4πk·r₁·r₂); convection at a surface R = 1/(hA). Resistances in series add (composite walls, inside and outside films); parallel paths combine like parallel resistors. The overall heat-transfer coefficient U is defined by Q = U·A·ΔT_overall, so 1/(UA) = ΣR. Contact resistance between layers may also need to be added (from data).

Critical radius of insulation. Adding insulation to a small pipe or wire increases the outer surface area while adding conduction resistance. Heat loss is maximum at the critical radius r_c = k/h (cylinder) or 2k/h (sphere). If the bare radius is smaller than r_c, a thin layer of insulation increases heat loss — useful for cooling electrical cables, harmful for small steam tubes.

Heat generation. For a plane wall with uniform volumetric generation q̇ (W/m³), the temperature profile becomes parabolic; for a solid cylinder the centre-line temperature rise is q̇R²/(4k). This applies to electrical conductors and nuclear fuel rods.

Fins (extended surfaces). When the convection coefficient h is low (air), adding fins increases the area for convection. A fin loses heat along its length, so its temperature falls from the base to the tip. The governing parameter is m = √(hP/(kA_c)), where P is the perimeter and A_c the cross-sectional area. Common cases: infinitely long fin; fin with insulated (adiabatic) tip; fin with convection at the tip (approximated by using a corrected length L_c = L + A_c/P).

Fin efficiency and effectiveness.

• Efficiency η_f = actual heat transfer / heat transfer if the whole fin were at the base temperature. It is always ≤ 1 and falls as mL increases.
• Effectiveness ε_f = heat transfer with the fin / heat transfer from the base area A_c without the fin. Fins are worthwhile only if ε_f is well above 1 (a guideline is ε_f ≥ 2).
• For a long fin, ε_f = √(kP/(hA_c)). So effectiveness is higher for high-k materials, thin closely spaced fins (large P/A_c) and low h. This is why fins are used on the air side of heat exchangers, not on the water side.
• Beyond about mL ≈ 2–3, extra length adds almost nothing (tanh(mL) → 1).

Formulas

Q = −k·A·dT/dx

• Q (W); k (W/m·K); A area normal to flow (m²); dT/dx (K/m).

Q = (T₁ − T₂)/R; R_wall = L/(k·A); R_conv = 1/(h·A)

• R (K/W); L thickness (m); h convection coefficient (W/m²·K).

R_cyl = ln(r₂/r₁)/(2·π·k·L); R_sph = (r₂ − r₁)/(4·π·k·r₁·r₂)

• Cylinder of length L; sphere.

r_c = k/h (cylinder), r_c = 2k/h (sphere)

• Critical radius of insulation (m); k of insulation.

m = √(h·P/(k·A_c)); for a circular pin m = √(4h/(k·D))

• m (1/m); P perimeter (m); A_c cross-section (m²).

Q_fin = √(h·P·k·A_c) · θ_b · tanh(m·L)

• Insulated tip; θ_b = T_b − T∞ (K). Long fin: replace tanh(mL) by 1.

η_f = tanh(m·L)/(m·L) (insulated tip)

ε_f = Q_fin/(h·A_c·θ_b); long fin ε_f = √(k·P/(h·A_c))

Worked examples

Example 1 (standard): composite furnace wall. Given: a furnace wall has 200 mm of firebrick (k = 1.2 W/m·K) and 100 mm of insulating brick (k = 0.15 W/m·K). Hot gas inside at 900 °C with h = 50 W/m²·K; outside air at 30 °C with h = 10 W/m²·K. Find the heat loss per m² and the temperatures at the surfaces and interface.

1. Resistances per m²: inside film 1/50 = 0.020; firebrick 0.2/1.2 = 0.1667; insulation 0.1/0.15 = 0.6667; outside film 1/10 = 0.100 (all m²·K/W).
2. ΣR = 0.9533 m²·K/W
3. q = (900 − 30)/0.9533 = 912.6 W/m²
4. Inner surface = 900 − 912.6 × 0.020 = 881.7 °C; interface = 881.7 − 912.6 × 0.1667 = 729.6 °C; outer surface = 729.6 − 912.6 × 0.6667 = 121.3 °C; check 121.3 − 912.6 × 0.1 = 30 °C ✓. Answer: q ≈ 913 W/m²; inner face ≈ 882 °C, interface ≈ 730 °C, outer face ≈ 121 °C

Example 2 (GATE level): aluminium pin fin. Given: aluminium pin fin, k = 200 W/m·K, D = 5 mm, L = 50 mm, base at 100 °C, air at 25 °C, h = 20 W/m²·K, tip insulated. Find the heat loss, fin efficiency and effectiveness.

1. m = √(4h/(kD)) = √(4 × 20/(200 × 0.005)) = √80 = 8.944 m⁻¹; mL = 0.4472; tanh(mL) = 0.4196.
2. P = π × 0.005 = 0.015 71 m; A_c = π × 0.005²/4 = 1.963 × 10⁻⁵ m²
3. √(hPkA_c) = √(20 × 0.015 71 × 200 × 1.963 × 10⁻⁵) = 0.035 12 W/K
4. Q = 0.035 12 × 75 × 0.4196 = 1.105 W
5. η_f = 0.4196/0.4472 = 0.938
6. ε_f = 1.105/(20 × 1.963 × 10⁻⁵ × 75) = 37.5 Answer: Q ≈ 1.11 W per fin; η_f ≈ 94%; ε_f ≈ 37.5

Example 3 (quick): critical radius. Insulation with k = 0.1 W/m·K on a wire in air with h = 10 W/m²·K: r_c = k/h = 0.01 m = 10 mm. A 3 mm radius wire covered up to 10 mm radius loses more heat than when bare.

Common mistakes

• Adding conductivities instead of resistances for layers in series.
• Using the plane-wall formula for a thick pipe; use ln(r₂/r₁).
• Confusing fin efficiency (≤ 1) with effectiveness (usually much greater than 1).
• Using the fin's cross-sectional area as its convecting area, or vice versa.
• Believing more insulation always reduces heat loss; check the critical radius for small pipes and wires.
• Saying fins help most when h is high; they help most when h is low.
• Mixing °C and K is fine for differences, but not for radiation or ratios.

For GATE PI

Expect composite-wall and composite-cylinder numericals with interface temperatures, overall U, critical radius of insulation, and fin questions (heat loss, efficiency, effectiveness, the effect of k, h, P/A_c and length). Conceptual questions test the linear versus logarithmic profile and when fins are justified. Practise drawing the resistance network first — it organises every problem in this topic.

Quick check

1. What is the thermal resistance of a 0.1 m thick wall with k = 0.5 W/m·K, per m²?
2. In which coordinate system is the steady temperature profile logarithmic?
3. What is the efficiency limit of a fin as mL → 0?
4. Should fins be placed on the air side or the water side of an air–water exchanger?

Answers: 1. 0.2 m²·K/W 2. Cylindrical (radial conduction in a tube) 3. 1 (100%) 4. The air side, where h is low