Steady Conduction: Composite Walls & Fins

Heat Transfer contributes ~6–9 marks to GATE ME, and steady conduction is its most-used foundation. The thermal-resistance (electrical-analogy) method and fin analysis are reliable 1–2 mark sources.

Fourier's law & thermal resistance

Fourier's law: Q = −kA·(dT/dx); heat flows down the temperature gradient. Using the electrical analogy (Q ↔ current, ΔT ↔ voltage):

• Plane wall resistance R = L/(kA).
• Cylindrical shell R = ln(r₂/r₁)/(2πkL).
• Spherical shell R = (r₂ − r₁)/(4πk·r₁r₂).
• Convection film resistance R = 1/(hA).
  Composite walls: resistances in series add (R_total = ΣR); parallel paths combine as 1/R = Σ(1/R). Overall Q = ΔT_total/R_total. Contact resistance adds at imperfect interfaces.

Fins (extended surfaces)

Fins add surface area to boost convective heat loss. Fin parameter m = √(hP/kA_c) (P = perimeter, A_c = cross-section). For a long (infinite) fin: temperature θ/θ_b = e^(−mx); heat dissipated Q = √(hPkA_c)·θ_b.

• Fin efficiency η = actual heat/ideal (whole fin at base temp).
• Fin effectiveness = fin heat / no-fin heat; a fin is worthwhile only if effectiveness > 1 (practically > 2). Fins help most when h is low (gas side) and k is high.

Exam Tricks & Tips

• 🎯 Use the resistance network: series ΣR for layered walls, then Q = ΔT/R_total — far faster than solving the ODE.
• 🎯 Cylindrical R = ln(r₂/r₁)/2πkL — note the logarithm; don't use the plane-wall form for pipes.
• 🎯 Fin parameter m = √(hP/kA); infinite-fin Q = √(hPkA)·θ_b — the two most-used fin results.
• 🎯 Fins pay off when h is low and k is high (e.g. air-cooled aluminium) — putting fins on the high-h (liquid) side is pointless.
• 🎯 Critical radius of insulation r_c = k/h (cylinder): adding insulation below r_c actually INCREASES heat loss — a favourite trap.
• ❌ Common mistake: adding parallel thermal resistances like series ones — parallel paths combine reciprocally (1/R = Σ1/R), as in electrical circuits.

Expected exam pattern

A 1-mark single-wall or fin-parameter NAT, and a 2-mark composite-wall (series/parallel resistance) or fin heat-dissipation problem. The critical-radius-of-insulation concept and "when do fins help" are common MCQs.

Quick recap

Fourier Q = −kA dT/dx; resistances: wall L/kA, cylinder ln(r₂/r₁)/2πkL, film 1/hA. Series add, parallel reciprocal. Fins: m = √(hP/kA), infinite-fin Q = √(hPkA)θ_b; help when h low, k high. Critical insulation radius r_c = k/h.

Cover the answer, recall, then check. 11 cards on steady conduction for GATE ME.

Q1. State Fourier's law of conduction.
A1. Q = −kA·(dT/dx) — heat flows down the temperature gradient; k = thermal conductivity.

Q2. Thermal resistance of a plane wall?
A2. R = L/(kA), where L = thickness, A = area. Heat Q = ΔT/R.

Q3. Thermal resistance of a cylindrical shell?
A3. R = ln(r₂/r₁)/(2πkL) — note the logarithm (not L/kA).

Q4. Convective film resistance?
A4. R = 1/(hA), where h = convective coefficient.

Q5. How do series and parallel thermal resistances combine?
A5. Series: R_total = ΣR (layered walls). Parallel: 1/R_total = Σ(1/R) (parallel paths) — like electrical resistors.

Q6. Define the fin parameter m.
A6. m = √(hP/kA_c), with P = perimeter and A_c = cross-sectional area of the fin.

Q7. Temperature distribution and heat loss for a long (infinite) fin?
A7. θ/θ_b = e^(−mx); heat dissipated Q = √(hPkA_c)·θ_b.

Q8. Difference between fin efficiency and effectiveness?
A8. Efficiency = actual heat / heat if the whole fin were at base temperature. Effectiveness = fin heat / heat without the fin.

Q9. When are fins most beneficial?
A9. When the convective coefficient h is low (gas side) and the fin conductivity k is high — e.g. aluminium fins in air.

Q10. What is the critical radius of insulation for a cylinder?
A10. r_c = k/h. Below it, adding insulation increases heat loss (more surface area outweighs added resistance).

Q11. Contact resistance arises where?
A11. At an imperfect interface between two solids, where micro-gaps add extra resistance to heat flow.

Conduction through walls and fins is the bread-and-butter of heat transfer, and the thermal-resistance analogy makes it as easy as solving a DC circuit. GATE reliably asks for heat flow through composite walls/cylinders and fin performance.

Core concept: in steady conduction, heat flows down a temperature gradient (Fourier's law), and a series of layers behaves exactly like resistors in a circuit — add the thermal resistances and divide the temperature difference.

Deep explanation

Beginner — Fourier's law and thermal resistance

Fourier's law: Q = −kA(dT/dx). For a plane wall of thickness L:
Q = kA(T₁ − T₂)/L = ΔT/R, with thermal resistance R = L/(kA).
This is Ohm's law for heat: Q ↔ current, ΔT ↔ voltage, R ↔ resistance. Convection at a surface adds a resistance R_conv = 1/(hA).

Intermediate — composite walls and cylinders

• Composite wall (series): R_total = ΣLᵢ/(kᵢA) + convection resistances; Q = ΔT_overall/R_total. Parallel paths (e.g. studs and insulation) add as parallel resistances.
• Cylinder (pipe): radial resistance R = ln(r₂/r₁)/(2πkL). For a sphere R = (1/r₁ − 1/r₂)/(4πk).
• Critical radius of insulation: for a small pipe/wire, adding insulation can increase heat loss until r = r_cr = k/h (cylinder) because added surface area outpaces added resistance. Beyond r_cr, more insulation reduces loss.

Advanced — fins (extended surfaces)

Fins add area to boost convective heat transfer. For a long fin, the temperature decays exponentially; define fin parameter m = √(hP/kA_c) (P = perimeter, A_c = cross-section). Heat from a long fin:
Q_fin = √(hPkA_c) · θ_b (θ_b = base excess temperature).

• Fin efficiency η_f = actual heat/(heat if entire fin were at base temperature) — high for short, thick, high-k fins.
• Fin effectiveness ε = fin heat/no-fin heat; a fin is worthwhile only if ε > 1 (roughly kP/hA_c > 1), which is why fins use high-k metal and go on the low-h (gas) side.

Worked example

A composite wall has two layers: 100 mm brick (k = 0.7 W/mK) and 50 mm insulation (k = 0.05 W/mK), area 10 m². Inside 30°C, outside 0°C (neglect convection). Find the heat flow.
R₁ = L/kA = 0.100/(0.7×10) = 0.01429 K/W.
R₂ = 0.050/(0.05×10) = 0.100 K/W.
R_total = 0.01429 + 0.100 = 0.1143 K/W.
Q = ΔT/R_total = (30 − 0)/0.1143 = 262.5 W. (The insulation, though thinner, carries ~88% of the resistance.)

GATE relevance

Composite-wall/cylinder conduction with the resistance analogy, critical radius of insulation, and fin efficiency/effectiveness are recurring Heat Transfer questions. The R = L/kA and R = ln(r₂/r₁)/2πkL forms and the m = √(hP/kA_c) fin parameter are essential.

Exam tricks & shortcuts

• Treat layers as series resistors; add R and divide ΔT — no need to find interface temperatures unless asked.
• Critical radius r_cr = k/h (cylinder), 2k/h (sphere): small wires/pipes may lose more heat when first insulated.
• Fins pay off only when kP/hA_c ≫ 1 — high-k metal, low-h (gas) side.
• Mnemonic: "Heat is current, ΔT is voltage, L/kA is resistance."

Key formulas

• Fourier's law: Q = −kA(dT/dx); plane wall Q = kAΔT/L.
• Thermal resistance: R_cond = L/kA; R_conv = 1/hA; series resistances add.
• Composite wall: Q = ΔT_overall/ΣR.
• Cylinder: R = ln(r₂/r₁)/(2πkL).
• Fin (long): Q = √(hPkA)·θ_b; efficiency η_fin = tanh(mL)/mL, m = √(hP/kA).

Heat flow uses a thermal-resistance network; fins enhance heat transfer by increasing surface area.

Worked Example

Problem: A composite plane wall (area 1 m²) has two layers: layer 1 has k₁ = 0.5 W/m·K and thickness 0.1 m; layer 2 has k₂ = 2 W/m·K and thickness 0.2 m. The inner surface is at 200°C and the outer surface at 50°C. Find the heat transfer rate and the temperature at the interface between the layers.

Solution:
Treat the wall as thermal resistances in series (R = L/(kA)):
R₁ = L₁/(k₁·A) = 0.1/(0.5 × 1) = 0.20 K/W.
R₂ = L₂/(k₂·A) = 0.2/(2 × 1) = 0.10 K/W.
Total resistance: R_total = R₁ + R₂ = 0.30 K/W.

Heat transfer rate (driven by the overall temperature difference):
Q = ΔT/R_total = (200 − 50)/0.30 = 150/0.30 = 500 W.
(Heat flux = Q/A = 500 W/m².)

Interface temperature — apply the temperature drop across layer 1 only:
T_interface = T_inner − Q·R₁ = 200 − 500 × 0.20 = 200 − 100 = 100°C.

Answer: Q = 500 W and the interface temperature is 100°C.