One-dimensional steady conduction and thermal resistance

Why it matters

Heat leaking through a car's roof and doors loads the air-conditioner; heat flowing through cylinder liners, exhaust pipes, brake pads and battery-pack walls decides their temperatures. The thermal-resistance network turns all of these into simple series and parallel circuits, and it is the backbone of every steady heat-transfer problem in GATE, including fins, heat exchangers and insulation.

Key ideas

Fourier's law. The conduction heat flux is proportional to the temperature gradient and flows from hot to cold: q″ = −k dT/dx. The minus sign makes heat flow positive in the direction of decreasing temperature. Thermal conductivity k (W/m·K) ranges from about 400 for copper, around 45–50 for carbon steel, about 1 for glass, down to 0.03–0.05 for insulation foams.

One-dimensional, steady. "Steady" means temperatures do not change with time; "one-dimensional" means temperature varies in one coordinate only (x for a wall, r for a pipe or sphere). With no heat generation and constant k:

• In a plane wall the temperature profile is linear and the heat rate is the same at every x.
• In a cylinder or sphere the heat rate is constant with r but the flux falls as area grows, so the profile is logarithmic (cylinder) or hyperbolic (sphere), not linear.

Thermal resistance. Since Q = ΔT/R, every mode of heat transfer can be written as a resistance (K/W):

• Conduction through a plane wall: L/(kA).
• Convection at a surface: 1/(hA).
• Radiation (linearised): 1/(h_r A). Layers in series add their resistances; parallel paths (for example a stud and the insulation beside it) combine like parallel resistors. The same heat rate passes through each series layer, so the largest temperature drop occurs across the largest resistance.

Overall heat transfer coefficient. For a wall between two fluids, U = 1/(R_total A), so Q = U A ΔT_overall. For a pipe, U must be referred to a specific area (inner or outer).

Contact resistance. Real surfaces touch only at asperities, so a joint adds a resistance R″_c (m²·K/W) that depends on pressure, roughness and interface material. Thermal paste in electronics and battery modules reduces it. Values come from data books.

Critical radius of insulation. Adding insulation to a small pipe or wire increases the conduction resistance but also increases the outer surface area, lowering the convection resistance. Heat loss is maximum when the outer radius equals r_c = k/h (cylinder) or 2k/h (sphere). For pipes with r_o > r_c insulation always reduces heat loss; for thin electric wires, a coating up to r_c actually helps cooling.

Variable conductivity. If k = k₀(1 + βT), the plane-wall result still holds with k evaluated at the mean temperature.

Formulas

Q = −k A dT/dx — Q: heat rate, W; k: W/m·K; A: area normal to flow, m²; dT/dx: K/m.

R_wall = L / (k A) — plane wall of thickness L (m).

R_conv = 1 / (h A) — h: convection coefficient, W/m²·K.

R_cyl = ln(r₂/r₁) / (2π k L) — cylindrical shell, length L (m).

R_sph = (r₂ − r₁) / (4π k r₁ r₂) — spherical shell.

Q = (T_∞1 − T_∞2) / ΣR — series network.

1/R_parallel = 1/R_a + 1/R_b — parallel paths.

U = 1 / (A ΣR) — W/m²·K; for a plane wall per unit area: 1/U = 1/h₁ + Σ L/k + 1/h₂.

r_c = k / h (cylinder), r_c = 2k / h (sphere) — critical radius of insulation, m.

k_m = k₀ [1 + β (T₁ + T₂)/2] — mean conductivity for linearly varying k.

Worked examples

Example 1 (standard, composite car door). A door panel consists of 1 mm steel (k = 45 W/m·K), 20 mm foam insulation (k = 0.04 W/m·K) and 5 mm plastic trim (k = 0.2 W/m·K). Outside air is at 45 °C with h = 25 W/m²·K (moving vehicle); cabin air is at 25 °C with h = 8 W/m²·K. Find the heat gain per m² and the U value. Neglect radiation and contact resistance.

1. Resistances per m² (A = 1 m²): outside convection 1/25 = 0.0400; steel 0.001/45 = 0.00002; foam 0.020/0.04 = 0.5000; trim 0.005/0.2 = 0.0250; inside convection 1/8 = 0.1250 K/W.
2. ΣR = 0.0400 + 0.00002 + 0.5000 + 0.0250 + 0.1250 = 0.6900 K/W.
3. Q = ΔT/ΣR = (45 − 25)/0.6900 = 29.0 W per m².
4. U = 1/ΣR = 1.45 W/m²·K.
5. Temperature drop across the foam = 29.0 × 0.5 = 14.5 K, about 72% of the total; the steel's drop is negligible (0.0006 K). Insulation dominates, so adding steel thickness is pointless.

Example 2 (GATE level, insulated pipe). A steel pipe (k = 45 W/m·K) of inner radius 25 mm and outer radius 30 mm carries a fluid at 150 °C (h_i = 500 W/m²·K). It is covered with 20 mm of insulation (k = 0.06 W/m·K) and exposed to air at 30 °C (h_o = 10 W/m²·K). Find the heat loss per metre, the outer surface temperature, and the loss if the pipe were bare.

1. Inside convection: 1/(h_i 2π r_i L) = 1/(500 × 2π × 0.025 × 1) = 0.01273 K/W.
2. Pipe wall: ln(r₂/r₁)/(2πkL) = ln(30/25)/(2π × 45) = 0.00064 K/W.
3. Insulation: ln(50/30)/(2π × 0.06) = 1.35501 K/W.
4. Outside convection: 1/(10 × 2π × 0.05) = 0.31831 K/W.
5. ΣR = 1.68669 K/W; Q = (150 − 30)/1.68669 = 71.1 W per metre.
6. Outer surface temperature = 30 + 71.1 × 0.31831 = 52.6 °C.
7. Bare pipe: outside resistance 1/(10 × 2π × 0.03) = 0.53052 K/W; Q = 120/(0.01273 + 0.00064 + 0.53052) = 220.6 W/m. Insulation cuts the loss by about 68%.
8. Critical radius r_c = k/h = 0.06/10 = 6 mm, far below the 30 mm pipe, so any thickness of this insulation reduces heat loss.

Common mistakes

• Using the plane-wall formula L/(kA) for a thick pipe; use ln(r₂/r₁)/(2πkL).
• Dropping the convection resistances, which are often the largest terms.
• Using diameters inside ln when the formula has radii (the ratio is the same, but mixing a diameter with a radius is not).
• Applying the critical-radius idea to large pipes, where insulation always helps.
• Forgetting that the heat rate, not the heat flux, is constant through cylindrical layers.
• Mixing mm and m in L/k.

For GATE ME

Expect composite-wall and composite-cylinder numericals (heat rate, interface temperatures, overall U), critical radius of insulation and the effect of adding insulation to a wire or pipe, walls with variable conductivity, and MCQs on temperature profiles in walls, cylinders and spheres. Practise drawing the resistance network first; most errors come from leaving out a resistance.

Quick check

1. What is the resistance of a 0.1 m thick wall with k = 0.5 W/m·K and area 2 m²?
2. In a composite wall, which layer has the largest temperature drop?
3. A wire has k_insulation = 0.2 W/m·K and h = 10 W/m²·K. What is the critical radius?
4. Is the steady temperature profile in a cylindrical shell linear?

Answers: 1. 0.1/(0.5 × 2) = 0.1 K/W. 2. The one with the largest thermal resistance. 3. 0.2/10 = 0.02 m = 20 mm. 4. No, it is logarithmic in r.