Heat Transfer: Resistances in Series, Fins, Radiation and Heat Exchangers

Heat transfer is the sub-item where one analogy does most of the work: steady conduction obeys the same algebra as a resistor network. Write Q = ΔT/R with R = L/kA for a slab and R = 1/hA for a surface, put resistances in series along the path and in parallel across it, and composite walls, insulated pipes and finned surfaces all become the same calculation. The analogy also tells you where to look — in any series chain the largest resistance dominates, and the practical consequence is that a well-insulated wall’s performance is set almost entirely by the insulation and hardly at all by the metal on either side of it. Fins work by trading area for temperature: adding surface always adds area, but the added surface is cooler than the base, and the fin efficiency tanh(mL)/mL is exactly the size of that penalty. Unsteady problems split on the Biot number Bi = hL_c/k: below 0.1 the body is essentially isothermal and one exponential describes it; above 0.1 the internal gradient matters and you need charts or a series solution. Convection is where correlations live, and they are all statements of the form Nu = f(Re, Pr) — the useful discipline is to know what each group means physically rather than to memorise coefficients. Radiation goes as T⁴, which makes it negligible at room temperature and dominant in a furnace, and grey surfaces exchange less than black ones by a factor that is not a simple product of emissivities. Heat exchangers finally have two methods for two situations: LMTD when you know all four temperatures, effectiveness–NTU when you do not.

Resistances in Series, and Why the Insulation Is the Whole Answer

The resistance concept is the electrical analogy taken seriously: Q plays the part of current, ΔT of voltage and a thermal resistance R of ohms, so Fourier’s law q = −k dT/dx becomes Q = ΔT/R once you define R = L/kA for a slab, R = ln(r₂/r₁)/2πkL for a cylinder and R = 1/hA for a convective surface. The whole point of the form is that resistances add along the path, so a wall made of layers is one sum, and the sum tells you immediately which layer matters. Take a wall of 20 mm steel (k = 50), 50 mm insulation (k = 0.05) and 10 mm aluminium (k = 20), per square metre. The resistances are 0.0004, 1.0000 and 0.0005 K·m²/W, totalling 1.0009 — and the insulation is 99.91 per cent of it. At a 200 K temperature difference the wall passes 199.82 W/m², which is within 0.09 per cent of what the insulation would pass on its own. Doubling the steel thickness changes the answer in the fourth decimal place; shaving 5 mm off the insulation changes it by 10 per cent. That asymmetry is not a curiosity — it is the reason a heat-transfer calculation should always start by ranking the resistances, because the ranking usually decides the answer before any arithmetic.

Composite wall, per m²: the ranking is the answer
LayerL, kR = L/kA per m²Share of total
Steel20 mm, 50 W/m·K0.00040.04 %
Insulation50 mm, 0.05 W/m·K1.000099.91 %
Aluminium10 mm, 20 W/m·K0.00050.05 %
Total—1.0009q = 199.82 W/m² at ΔT = 200 K
🎯 A fin adds area and loses temperature, and the efficiency is the trade
Adding a fin always adds surface, but the added surface is not at the base temperature — it cools along its length, so each extra square millimetre is worth less than the last. Two numbers capture this. The fin parameter m = √(hP/kA) sets how fast the temperature decays, and the fin efficiency η = tanh(mL)/(mL) is the ratio of the heat actually removed to the heat an isothermal fin at base temperature would remove. Work one: an aluminium pin fin of diameter 5 mm and length 50 mm, k = 200 W/m·K, in air with h = 25 W/m²·K, base 80 K above ambient. Then P = 0.01571 m, A = 1.9635 × 10⁻⁵ m², and m = √(25 × 0.01571/(200 × 1.9635 × 10⁻⁵)) = 10 /m exactly, so mL = 0.5. Efficiency is tanh(0.5)/0.5 = 92.42 per cent and the heat removed is √(hPkA) θ_b tanh(mL) = 1.452 W. Now the design reading, which is the part worth carrying. An infinitely long fin of the same section would remove √(hPkA) θ_b = 3.142 W, so this 50 mm fin gets only 46.2 per cent of the infinite-fin limit — and lengthening it further buys progressively less, because tanh saturates. Past about mL = 3 a fin is essentially all wasted metal. The other half of the story is when a fin helps at all: the fin effectiveness here is 36.97, meaning the fin removes 37 times what the bare base area would — enormous, and it is enormous precisely because h is small. Fins pay in gases and hardly ever in liquids, and never on the high-h side of a heat exchanger.

Unsteady problems are decided by the Biot number Bi = hL_c/k, which is the ratio of the internal conduction resistance to the external convection resistance. If Bi < 0.1, internal gradients are negligible and the whole body follows one exponential — the lumped solution, (T − T_∞)/(T_i − T_∞) = e^(−t/τ) with τ = ρVc/hA. A steel sphere of diameter 10 mm cooling in air with h = 50 W/m²·K has L_c = V/A = d/6 = 1.667 mm and Bi = 0.00167, comfortably lumped, and a time constant of 119.6 s — so it needs about 3τ = 359 s to travel 95 per cent of the way to ambient. Above Bi = 0.1 the body is not isothermal, the surface leads the centre, and the one-term or chart solutions are needed. Getting the branch right first is the entire skill: applying the lumped formula to a large Bi problem is not a small error, because it predicts the centre reaching ambient long before it does.

Convection Groups and the Fourth Power

Every convection correlation in the syllabus has the same shape — Nu = f(Re, Pr) for forced convection, Nu = f(Gr, Pr) for free — and the marks are in knowing what the groups mean, because that is what lets you predict the direction of a change without a chart. The Nusselt number Nu = hL/k is the ratio of convective to pure-conductive heat transfer across the same layer, so Nu = 1 means the fluid is doing nothing but conducting. The Prandtl number Pr = ν/α compares how fast momentum diffuses with how fast heat does, which is exactly the ratio of boundary-layer thicknesses: Pr ≈ 0.7 for air means the thermal layer is slightly thicker than the velocity layer, Pr ≈ 7 for water means it is thinner, and Pr ≈ 0.01 for liquid metals means heat runs far ahead of momentum. The Reynolds number decides laminar or turbulent as always. Stanton and Peclet are combinations of these rather than new ideas. Two habits follow. First, quote correlations with their range attached — Dittus–Boelter is for turbulent pipe flow at moderate Pr, and using it at Re = 500 is not conservative, it is wrong. Second, when a question changes one variable, work through the groups rather than the correlation: doubling velocity raises Re, and in a Nu ∝ Re^0.8 regime that raises h by 2^0.8 = 1.74, not by 2.

⚠️ The fourth power, and why grey surfaces do not simply multiply
Radiation follows q = σ(T₁⁴ − T₂⁴) with σ = 5.67 × 10⁻⁸ W/m²·K⁴, and the fourth power is the whole character of the mode: it is negligible next to convection at room temperature and dominant in a furnace, and the crossover is somewhere around 400–500 K for typical surfaces. Temperatures must be absolute — using °C here is not an approximation but a different physical claim, and it is the single most common error in radiation problems. Now the part that catches people who have the law right. For two large parallel grey plates, the net exchange is not σ ε₁ ε₂ (T₁⁴ − T₂⁴); it is σ F_ε (T₁⁴ − T₂⁴) with F_ε = 1/(1/ε₁ + 1/ε₂ − 1). With ε₁ = ε₂ = 0.8 that factor is 1/(1.25 + 1.25 − 1) = 0.6667, whereas the naive product ε₁ε₂ would give 0.64 — close here, but the two diverge badly at low emissivity, and the reason is physical rather than algebraic: radiation leaving plate 1 that is reflected by plate 2 comes back, is partly reflected again, and the infinite series of re-reflections is exactly what the −1 accounts for. At 800 K and 300 K the two plates exchange 15.18 kW/m², against 22.77 kW/m² if both were black. And notice which temperature dominates: 800⁴ is 50 times 300⁴, so the cold plate’s temperature is almost irrelevant to the answer — a useful sanity check, and the reason a furnace loses much the same heat to a cold room as to a colder one.

Heat Exchangers: Two Methods, and Why Counterflow Always Wins

When all four terminal temperatures are known, the design equation is Q = U A × LMTD with LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂). The log mean, not the arithmetic mean, because the temperature difference varies exponentially along the exchanger. Take a hot stream cooling from 100 °C to 60 °C against a cold stream heating from 20 °C to 50 °C. In counterflow the ends see differences of 100 − 50 = 50 and 60 − 20 = 40, giving LMTD = 10/ln 1.25 = 44.81 °C. In parallel flow the same four temperatures give 100 − 20 = 80 and 60 − 50 = 10, so LMTD = 70/ln 8 = 33.66 °C. Counterflow’s driving force is 33.1 per cent larger, so for the same duty and the same U it needs 24.9 per cent less area — the same metal moves a quarter more heat purely because of which way the streams run. And parallel flow has a hard limit counterflow does not: the cold outlet can never exceed the hot outlet, so a parallel-flow exchanger simply cannot reach the 50 °C cold outlet in some duties that counterflow reaches comfortably.

Counterflow effectiveness at two NTU values, against the capacity ratio
C_r = C_min/C_maxε at NTU = 1.5ε at NTU = 3What this case is
00.77690.9502Condenser or evaporator — one stream at constant T
0.50.69080.8744Typical unbalanced exchanger
10.60000.7500Balanced — the hardest case, ε = NTU/(1+NTU)
🧠 Use effectiveness–NTU when you do not know the outlet temperatures
LMTD needs all four temperatures, so if two of them are what you are trying to find, using it means iterating. The effectiveness–NTU method removes the iteration by working in ratios instead. Define ε = Q/Q_max where Q_max = C_min(T_h,in − T_c,in) is the most any exchanger of that configuration could possibly transfer, NTU = UA/C_min as a dimensionless size, and C_r = C_min/C_max. Then ε is a known function of NTU and C_r, and one lookup gives you the duty directly. Read the table above for the two design lessons. First, diminishing returns: at C_r = 0.5, doubling NTU from 1.5 to 3 — twice the area, twice the cost — lifts effectiveness only from 0.691 to 0.874. Past about NTU = 3 extra area is nearly wasted, which is why real exchangers cluster in that range rather than chasing ε → 1. Second, the capacity ratio matters as much as the size: at the same NTU = 1.5, effectiveness runs from 0.777 down to 0.600 as C_r goes from 0 to 1. The C_r = 0 row is a condenser or evaporator, where one stream changes phase and so stays at constant temperature — its heat capacity rate is effectively infinite, and that is the easiest exchanger to build. C_r = 1 is the balanced case, where the two streams have equal capacity rates, the temperature difference is constant along the whole length, and ε = NTU/(1 + NTU) — the hardest case, and the one a regenerator has to live with.

Key takeaways

  • Rank the resistances before computing. In a steel / insulation / aluminium wall the insulation is 99.91 per cent of R, so q = 199.82 W/m² is within 0.09 per cent of the insulation acting alone.
  • Fin efficiency tanh(mL)/mL is the price of a cooler surface. The 5 mm × 50 mm pin fin has m = 10 /m, mL = 0.5, η = 92.4 per cent and removes 1.452 W — only 46.2 per cent of the infinite-fin limit. Past mL ≈ 3 extra length is wasted metal.
  • Fins pay where h is small. The same fin has an effectiveness of 37 in air; on the high-h side of a liquid exchanger the same geometry is barely worth its cost.
  • Bi = hL_c/k < 0.1 licenses the lumped solution. A 10 mm steel sphere in air has Bi = 0.0017 and a time constant of 119.6 s, so 359 s covers 95 per cent of the approach to ambient.
  • Radiation needs absolute temperatures, and grey plates use 1/(1/ε₁ + 1/ε₂ − 1), not ε₁ε₂. At ε = 0.8 both, that is 0.6667, and two plates at 800 K and 300 K exchange 15.18 kW/m² against 22.77 for black bodies.
  • Counterflow always beats parallel flow. On hot 100 → 60 and cold 20 → 50, LMTD is 44.81 against 33.66 °C — 33.1 per cent more driving force, so 24.9 per cent less area for the same duty.
  • Effectiveness saturates: at C_r = 0.5, doubling NTU from 1.5 to 3 lifts ε only from 0.691 to 0.874. C_r = 0 is a condenser (the easiest case), C_r = 1 the balanced one with ε = NTU/(1 + NTU).

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A wall consists of 20 mm steel (k = 50 W/m·K), 50 mm insulation (k = 0.05 W/m·K) and 10 mm aluminium (k = 20 W/m·K). Doubling the steel thickness would change the heat flux by approximately

    1. 0.04 per cent
    2. 25 per cent
    3. 50 per cent
    4. 2 per cent
    Show answer

    Answer: A — 0.04 per cent

    About 0.04 per cent. Rank the resistances first, per square metre: steel 0.02/50 = 0.0004, insulation 0.05/0.05 = 1.0000, aluminium 0.01/20 = 0.0005 K·m²/W. The total is 1.0009, of which the insulation is 99.91 per cent. Doubling the steel adds another 0.0004 to a total of 1.0009, a change of 0.04 per cent — and the heat flux changes by the same fraction, from 199.82 to 199.74 W/m² at ΔT = 200 K. Options B, C and D are the answers you get by reasoning about the steel in isolation, and the point of the question is that you cannot reason about a series element in isolation: what matters is its share of the total, not its own size. This is the single most useful habit in the topic. Before any arithmetic, list the resistances and see which one dominates; if one term is a thousand times the others, the calculation is already essentially done and the rest is decimal places. The same logic explains why adding fins to the water side of a water-to-air exchanger is a waste of money — the air-side resistance is the whole resistance, and improving the small term improves nothing.

  2. An aluminium pin fin of diameter 5 mm and length 50 mm (k = 200 W/m·K) is exposed to air with h = 25 W/m²·K. Its fin efficiency, in per cent, is ______ (round to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 92.42

    92.42 per cent. Compute the fin parameter: P = πd = 0.015708 m, A = πd²/4 = 1.9635 × 10⁻⁵ m², so m = √(hP/kA) = √(25 × 0.015708/(200 × 1.9635 × 10⁻⁵)) = √100 = 10 /m exactly — the numbers were chosen so that this comes out clean and you can see the structure. Then mL = 10 × 0.05 = 0.5, and for an insulated-tip fin η = tanh(mL)/(mL) = 0.462117/0.5 = 0.92423. A 92 per cent efficiency sounds like a well-designed fin, and here is the trap worth understanding: high efficiency is not the same as high heat removal. Efficiency measures how close the fin comes to being isothermal at base temperature, and a short fin scores well on that precisely because it barely cools — a fin of zero length would be 100 per cent efficient and remove nothing. The heat actually removed is √(hPkA) θ_b tanh(mL) = 1.452 W, against 3.142 W for an infinitely long fin of the same section, so this fin captures only 46.2 per cent of the available limit. Lengthening it improves the heat removed and worsens the efficiency, which is the trade the two numbers exist to express. Past about mL = 3, tanh has saturated and further length adds mass without adding duty.

  3. A hot stream cools from 100 °C to 60 °C while a cold stream heats from 20 °C to 50 °C. The counterflow LMTD, in °C, is ______ (round to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 44.81

    44.81 °C. In counterflow the streams meet end to end, so pair the hot inlet with the cold outlet: ΔT₁ = 100 − 50 = 50 and ΔT₂ = 60 − 20 = 40. Then LMTD = (50 − 40)/ln(50/40) = 10/0.22314 = 44.814 °C. Two checks worth running. The log mean always lies below the arithmetic mean — here 44.81 against 45.00 — and equals it only when the two end differences are equal; if your answer comes out above the arithmetic mean, a term is inverted. And the log mean lies between the two end differences, which rules out gross slips. Now the comparison that makes this a design result rather than a formula: the same four temperatures in parallel flow pair the hot inlet with the cold inlet, giving ΔT₁ = 80, ΔT₂ = 10 and LMTD = 70/ln 8 = 33.66 °C. Counterflow’s driving force is 33.1 per cent larger, so for the same duty and overall coefficient it needs 24.9 per cent less area. The same metal, the same fluids, a quarter less exchanger, purely from which way the streams run — and parallel flow has a hard ceiling counterflow does not, since its cold outlet can never rise above its hot outlet.

  4. Two large parallel grey plates each have emissivity 0.8. The net radiation exchange between them is proportional to which factor times σ(T₁⁴ − T₂⁴)?

    1. 0.64
    2. 0.6667
    3. 0.8
    4. 1.6
    Show answer

    Answer: B — 0.6667

    0.6667. For two large parallel grey plates the factor is F_ε = 1/(1/ε₁ + 1/ε₂ − 1) = 1/(1.25 + 1.25 − 1) = 1/1.5 = 0.6667. Option A, ε₁ε₂ = 0.64, is the intuitive guess and it is wrong — close here, but the two diverge badly at low emissivity, and the reason is physical: radiation that leaves plate 1 and is reflected by plate 2 comes back, is partly reflected again, and so on. The −1 in the denominator is exactly the sum of that infinite series of re-reflections, which the naive product ignores entirely. Option C uses one emissivity as though the other surface were black. Option D is dimensionally impossible — the factor can never exceed 1, since two grey surfaces cannot exchange more than two black ones. At 800 K and 300 K the plates exchange 0.6667 × 5.67 × 10⁻⁸ × (800⁴ − 300⁴) = 15.18 kW/m², against 22.77 kW/m² for black bodies. And note the asymmetry the fourth power produces: 800⁴ is over 50 times 300⁴, so the cold plate’s temperature barely affects the answer — a furnace loses nearly the same heat to a cold room as to a much colder one, which is a useful check on any radiation result.

  5. Which of the following are correct about the lumped-capacitance analysis of an unsteady cooling problem?

    1. It is valid when Bi = hL_c/k is below about 0.1
    2. The characteristic length L_c is the volume divided by the exposed surface area
    3. The time constant is ρVc/hA, so a larger body responds more slowly
    4. A low Biot number means the body conducts heat poorly compared with the convection at its surface
    Show answer

    Answer: A — It is valid when `Bi = hL_c/k` is below about 0.1; B — The characteristic length `L_c` is the volume divided by the exposed surface area; C — The time constant is `ρVc/hA`, so a larger body responds more slowly

    A, B and C. A is the criterion and 0.1 is the conventional threshold; B is the definition, L_c = V/A, which for a sphere gives d/6 and for a plate of thickness 2L cooled on both faces gives L; C is the time constant, and its physical reading is right — mass and specific heat store energy while hA removes it, so a bigger or heavier body lags. D has the ratio exactly backwards, and that is what the question is testing. Bi = hL_c/k is the internal conduction resistance divided by the external convection resistance, so a low Biot number means conduction inside the body is easy compared with getting heat across the surface — which is precisely why the interior stays nearly isothermal and one lumped temperature describes it. A body that conducted poorly would have a high Biot number, a hot surface and a cold core, and the lumped model would be exactly wrong for it. Worked: a 10 mm steel sphere in air with h = 50 has L_c = 1.667 mm, Bi = 0.0017 — deeply lumped — and τ = ρcL_c/h = 119.6 s, so about 3τ = 359 s gets it 95 per cent of the way to ambient. The practical value of checking Bi first is that the two branches disagree most where it matters: for a large Bi body the lumped formula predicts the centre reaching ambient long before it actually does.

  6. In a counterflow heat exchanger with C_min/C_max = 0.5, increasing NTU from 1.5 to 3 raises the effectiveness from 0.691 to 0.874. The best reading of this is that

    1. effectiveness is roughly linear in NTU, so area buys duty proportionally
    2. doubling the area gives strongly diminishing returns, so real exchangers rarely go far past NTU ≈ 3
    3. the capacity ratio has no effect once NTU exceeds 1
    4. effectiveness above 0.8 is impossible in counterflow
    Show answer

    Answer: B — doubling the area gives strongly diminishing returns, so real exchangers rarely go far past NTU ≈ 3

    Diminishing returns. Doubling NTU means roughly doubling the area — and therefore the cost — while effectiveness rises by only 0.183, from 0.691 to 0.874. Going further makes it worse: the remaining 0.126 to perfection would take several more doublings, because ε approaches 1 asymptotically. That is why practical exchangers cluster around NTU = 1 to 3 rather than chasing ε → 1, and why the economic optimum is a design decision rather than a thermodynamic one. Option A misreads the trend as linear, which would make ε = 1.38 at NTU = 3 — impossible, since effectiveness is a fraction of the maximum possible transfer and cannot exceed 1. That impossibility is itself the fastest way to reject A. Option C is wrong and the numbers say so: at the same NTU = 1.5, effectiveness runs from 0.777 at C_r = 0 down to 0.600 at C_r = 1, so the capacity ratio matters throughout. The C_r = 0 case is a condenser or evaporator — one stream changes phase and holds constant temperature, the easiest exchanger there is — while C_r = 1 is the balanced case with a constant temperature difference along the whole length and ε = NTU/(1 + NTU). Option D is contradicted by the figure in the question itself.

  7. The Prandtl number of a fluid is 0.01, typical of a liquid metal. This implies that

    1. the thermal boundary layer is much thicker than the velocity boundary layer
    2. the thermal boundary layer is much thinner than the velocity boundary layer
    3. the flow must be turbulent
    4. convection is negligible compared with conduction
    Show answer

    Answer: A — the thermal boundary layer is much thicker than the velocity boundary layer

    Much thicker. Pr = ν/α compares momentum diffusivity with thermal diffusivity, and the boundary-layer thicknesses follow the same ratio: roughly δ_thermal/δ_velocity ≈ Pr^(−1/3). A Prandtl number of 0.01 means heat diffuses about a hundred times faster than momentum, so the thermal layer runs far ahead of the velocity layer — which is exactly the physical situation in a liquid metal, whose free electrons conduct heat while its viscosity stays modest. That is why liquid sodium is used as a reactor coolant and why liquid-metal correlations are a separate family: the usual Nu = f(Re, Pr) fits, calibrated near Pr ≈ 1, do not extrapolate down to 0.01. Option B is the answer for a high Pr fluid such as oil, and water at Pr ≈ 7 already sits mildly on that side. Option C confuses Pr, a property of the fluid alone, with Re, which describes the flow — a fluid has a Prandtl number whether it is moving or not. Option D describes what a Nusselt number near 1 would mean, not a Prandtl number: Nu = hL/k is the group that compares convection with conduction, and conflating the two is the error the question is built around. Air’s Pr ≈ 0.7 is worth holding as the reference point, because it is close enough to 1 that the two layers are comparable and most textbook correlations were fitted there.

  8. A steel sphere of diameter 10 mm (ρ = 7800 kg/m³, c = 460 J/kg·K) cools in air with h = 50 W/m²·K. Its lumped time constant, in seconds, is ______ (round to one decimal)

    Numerical answer — type the value.

    Show answer

    Answer: 119.6

    119.6 s. The characteristic length is L_c = V/A, which for a sphere is (πd³/6)/(πd²) = d/6 = 1.6667 mm. Then τ = ρVc/(hA) = ρcL_c/h = 7800 × 460 × 1.6667 × 10⁻³/50 = 5980/50 = 119.6 s. Check the licence first, because it is the step that decides whether the formula applies at all: with k = 50 W/m·K for steel, Bi = hL_c/k = 50 × 1.6667 × 10⁻³/50 = 0.0017, far below 0.1, so the sphere really is essentially isothermal and one exponential describes it. The practical reading of the time constant is that (T − T_∞)/(T_i − T_∞) = e^(−t/τ), so the sphere covers 63 per cent of its journey to ambient in one τ and about 95 per cent in 3τ = 359 s. The common slip is using the radius or the diameter as L_c: d/6 is 1.67 mm, while d = 10 mm would inflate the time constant six-fold to 718 s. The definition L_c = V/A is worth trusting over a remembered shape factor, and it also generalises — for a plate of thickness 2L cooled on both faces it gives L, and for a long cylinder of diameter d it gives d/4.