Heat transfer from fins

The fin equation and its long-fin, insulated-tip and convective-tip solutions, fin efficiency and effectiveness, finned-surface efficiency and thermometer-well error, with worked numericals.

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Why it matters

Air-cooled motorcycle cylinders, radiator and intercooler cores, AC condensers, power-electronics heat sinks and EV battery cold plates all rely on fins. Because convection to air is weak (h of 10–100 W/m²·K), adding surface area is often the cheapest way to dump heat. Fin theory tells you how long, thick and conductive a fin should be, and when a fin is simply wasted metal.

Key ideas

Why fins. Q = h A (T_s − T_∞). If h cannot be raised much (air side), increase A. A fin conducts heat from the base along its length while convecting it from its surface, so its temperature falls from the base toward the tip; the far part of a long fin does little work.

Assumptions of the simple fin model. Steady state, one-dimensional conduction along the fin (the fin is thin, so the temperature across its thickness is uniform; Biot number h t/k ≪ 1), constant k and h, uniform cross-section, no heat generation and negligible radiation.

Fin equation. An energy balance on a slice gives d²θ/dx² − m² θ = 0, where θ = T − T_∞ is the excess temperature and m² = hP/(kA_c). Its solutions depend on the tip condition:

  • Infinitely long fin: θ = θ_b e^(−mx). A fin behaves as "infinite" once mL > about 2.5–3 (tanh mL > 0.99).
  • Adiabatic (insulated) tip: θ/θ_b = cosh m(L − x)/cosh mL. Most commonly used.
  • Convective tip: use the adiabatic-tip result with a corrected length L_c = L + A_c/P (L + t/2 for a thin plate fin, L + D/4 for a pin fin).

Fin efficiency η_f = actual heat transfer / heat transfer if the whole fin surface were at the base temperature. It falls as mL grows: long, thin, low-conductivity fins in high-h flows are inefficient.

Fin effectiveness ε_f = heat transfer with the fin / heat transfer from the base area A_c without the fin. A fin should give ε_f of at least about 2 to be worth it. For a long fin ε_f = √(kP/(hA_c)), which shows fins work best with high k (aluminium, copper), a high perimeter-to-area ratio (thin fins), and low h (gas side, not liquid side). If h t/(2k) > 1 roughly, a fin can even reduce heat transfer.

Finned surfaces. For an array, the overall surface efficiency η_o = 1 − (N A_f/A_t)(1 − η_f), and Q = η_o h A_t (T_b − T_∞), where A_t includes fins and the bare base between them.

Thermometer wells. A thermowell is a fin attached to a pipe wall. Heat conducts along it toward the cooler wall, so the thermometer at its tip reads between the wall and fluid temperatures. Error is reduced by making the well long and thin-walled, of low-k material, and by increasing h.

Formulas

m = √(h P / (k A_c)) — m in 1/m; P: perimeter, m; A_c: cross-sectional area, m². For a pin fin of diameter D: m = √(4h / (k D)). For a thin plate fin of thickness t: m ≈ √(2h / (k t)).

M = √(h P k A_c) θ_b — W; θ_b = T_b − T_∞.

Q = M — infinitely long fin.

Q = M tanh(mL) — adiabatic tip.

θ/θ_b = cosh m(L − x) / cosh(mL); tip: θ_L/θ_b = 1/cosh(mL) — adiabatic tip.

L_c = L + A_c/P — corrected length for a convective tip.

η_f = tanh(m L_c) / (m L_c) — straight or pin fin of uniform section.

ε_f = Q_fin / (h A_c θ_b); for a long fin ε_f = √(k P / (h A_c)).

η_o = 1 − (N A_f / A_t)(1 − η_f) — finned-surface efficiency.

Worked examples

Example 1 (standard, pin fin). An aluminium pin fin (k = 200 W/m·K) of diameter 5 mm and length 50 mm sticks out of a 100 °C surface into air at 25 °C with h = 40 W/m²·K. Find the heat transfer, the fin efficiency, effectiveness and the tip temperature.

  1. P = πD = 0.01571 m; A_c = πD²/4 = 1.963 × 10⁻⁵ m².
  2. m = √(4h/(kD)) = √(4 × 40/(200 × 0.005)) = √160 = 12.65 m⁻¹.
  3. Convective tip, so L_c = L + D/4 = 0.05 + 0.00125 = 0.05125 m; m L_c = 0.648.
  4. M = √(hPkA_c) θ_b = √(40 × 0.01571 × 200 × 1.963 × 10⁻⁵) × 75 = 0.04967 × 75 = 3.725 W.
  5. Q = M tanh(m L_c) = 3.725 × tanh(0.648) = 3.725 × 0.5706 = 2.13 W.
  6. η_f = tanh(m L_c)/(m L_c) = 0.5706/0.648 = 0.88.
  7. Without the fin, the base area would lose h A_c θ_b = 40 × 1.963 × 10⁻⁵ × 75 = 0.0589 W, so ε_f = 2.13/0.0589 = 36.
  8. Tip temperature (adiabatic-tip estimate with L): θ_L = 75/cosh(12.65 × 0.05) = 75/1.207 = 62.2 K, so T_tip ≈ 87 °C.

Example 2 (GATE level, thermometer well). A thermometer sits at the closed end of a steel well (k = 50 W/m·K, wall thickness 1 mm, length 120 mm) projecting into a gas stream with h = 50 W/m²·K. The thermometer reads 150 °C and the pipe wall at the well's base is at 100 °C. Find the true gas temperature. Treat the well as a fin with an insulated tip.

  1. For a thin tube wall, P/A_c ≈ 1/t, so m = √(h/(k t)) = √(50/(50 × 0.001)) = √1000 = 31.62 m⁻¹.
  2. mL = 31.62 × 0.12 = 3.795; cosh(mL) = 22.24.
  3. Tip relation: (T_L − T_∞) = (T_b − T_∞)/cosh(mL), i.e. (150 − T_∞) = (100 − T_∞)/22.24.
  4. Rearranging: T_∞ (1 − 1/22.24) = 150 − 100/22.24 → T_∞ × 0.95504 = 150 − 4.496 = 145.504.
  5. T_∞ = 152.4 °C. The reading is low by about 2.4 K. Halving the wall thickness or lengthening the well would reduce the error.

Common mistakes

  • Using Q = h A_fin θ_b, which assumes 100% fin efficiency and overestimates heat transfer.
  • Using P/A_c of a solid rod (4/D) for a hollow thermowell; for a thin tube it is about 1/t.
  • Forgetting the corrected length when the tip is not insulated.
  • Thinking a longer fin always helps; beyond mL ≈ 3 extra length adds almost nothing.
  • Putting fins on the liquid side of a heat exchanger, where h is already high; they belong on the gas side.
  • Computing m with diameter where radius is needed (m = √(4h/(kD)) uses diameter).

For GATE ME

Expect numericals on heat loss from a pin or plate fin with insulated or long-fin tips, tip temperature, fin efficiency and effectiveness, and the thermometer-well error. Conceptual MCQs ask how efficiency and effectiveness change with k, h, length and thickness, and which tip condition applies. Practise computing m first and deciding the tip model from mL.

Quick check

  1. For h = 25 W/m²·K, P = 0.02 m, k = 200 W/m·K and A_c = 10⁻⁴ m², what is m?
  2. What is the fin efficiency when mL is very small?
  3. Should fins be placed on the air side or the water side of a radiator?
  4. Beyond roughly what value of mL does a fin behave as infinitely long?

Answers: 1. √(25 × 0.02/(200 × 10⁻⁴)) = √25 = 5 m⁻¹. 2. Close to 1 (100%). 3. The air side, where h is low. 4. About 2.5–3.

Interview questions

All Thermodynamics and Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is a fin in the context of heat transfer?Concept

    A fin is an extended surface used to increase the rate of heat transfer from a solid surface to a fluid or from a fluid to a solid surface. Fins are typically made of materials with high thermal conductivity to efficiently transfer heat.

  2. 2.Explain the purpose of using fins in heat exchangers.Concept

    Fins add surface area on the side of a heat exchanger where the convection coefficient is low, usually the gas or air side, so that its thermal resistance 1/(hA) approaches that of the liquid side. In a car radiator, the coolant-side h is perhaps tens of times the air-side h, so thin louvred aluminium fins on the air side multiply its area many times. This makes the exchanger compact for a given duty; adding fins on the liquid side would add cost and pressure drop for little gain.

  3. 3.How does the material of a fin affect its performance in heat transfer?Application

    The material of a fin affects its performance by influencing its thermal conductivity. Materials with higher thermal conductivity, such as aluminum or copper, are more effective at transferring heat. The choice of material also affects the weight, cost, and corrosion resistance of the fins.

  4. 4.Why are fins often used in automobile radiators?Application

    Fins are used in automobile radiators to increase the surface area for heat dissipation from the coolant to the air. This helps in maintaining the engine temperature within the optimal range, preventing overheating and ensuring efficient engine performance.

  5. 5.What happens if the fin efficiency is low in a heat transfer application?Application

    Low fin efficiency means the fin temperature drops steeply from base to tip, so much of the fin surface is close to the ambient temperature and contributes little; the material, weight and pressure drop are being spent for little extra heat transfer. It happens when mL is large: fins that are too long, too thin, made of low-conductivity material or placed in a high-h flow. The remedy is shorter or thicker fins, a higher-conductivity material such as aluminium or copper, or more fins of smaller height.

  6. 6.Explain the concept of fin efficiency.Concept

    Fin efficiency is the ratio of the actual heat transfer from the fin to the maximum possible heat transfer if the entire fin were at the base temperature. It indicates how effectively a fin is performing its function of heat transfer.

  7. 7.What is the effect of fin length on heat transfer?Application

    Increasing the fin length generally increases the surface area available for heat transfer, which can enhance the heat dissipation rate. However, beyond a certain length, the additional surface area may not significantly contribute to heat transfer due to the temperature gradient along the fin.

  8. 8.A straight fin of uniform section (k = 200 W/m·K, length 0.1 m, perimeter 0.02 m, cross-section 1 × 10⁻⁴ m²) has its base at 100 °C in air at 25 °C with h = 50 W/m²·K. Estimate the heat transfer assuming an insulated tip.Numerical

    First m = √(hP/(kA_c)) = √(50 × 0.02/(200 × 10⁻⁴)) = √50 = 7.07 m⁻¹, so mL = 0.707. Then √(hPkA_c) = √(50 × 0.02 × 200 × 10⁻⁴) = 0.1414 W/K. Heat transfer Q = √(hPkA_c)·θ_b·tanh(mL) = 0.1414 × 75 × tanh(0.707) = 10.61 × 0.609 ≈ 6.5 W. The fin efficiency is tanh(mL)/mL ≈ 0.86.

  9. 9.What is the role of the heat transfer coefficient in the performance of fins?Application

    A higher h increases the heat transferred from each square metre of fin surface, so total heat transfer rises. But it also raises m = √(hP/(kA_c)), so the temperature falls faster along the fin and fin efficiency drops. Fin effectiveness, roughly √(kP/(hA_c)) for a long fin, also falls as h rises. That is why fins are most worthwhile where h is low, as on the air side, and add little where h is already high, as in boiling or liquid flow.

  10. 10.A 10 mm diameter pin fin (k = 150 W/m·K, length 0.2 m) has its base at 120 °C in air at 30 °C with h = 20 W/m²·K. Estimate the tip temperature, assuming an insulated tip.Numerical

    For a pin fin m = √(4h/(kD)) = √(4 × 20/(150 × 0.01)) = √53.3 = 7.30 m⁻¹, so mL = 1.46 and cosh(mL) = 2.27. For an insulated tip, (T_tip − T_∞)/(T_b − T_∞) = 1/cosh(mL), so T_tip − 30 = 90/2.27 = 39.6 K. The tip temperature is about 69.6 °C, well below the base, which shows the outer part of this fin is only moderately effective.

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