Heat Transfer · interactive worked example

Solar Heat Exchanger: Distributed vs. Lumped

Fluid flows through a flat-plate heat exchanger absorbing solar heat. How hot does it come out? One model treats the temperature profile exactly (an infinite series); a much simpler model assumes it's uniform across the channel. When can you get away with the simple one?

▶ Play 📖 Design background

Based on Edward Lumsdaine, "Comparison of Solar Heat Exchangers," Solar Energy, Vol. 17, pp. 269–275 — solved originally by hand, then one of the first engineering problems demonstrated in Mathematica.

βn tan βn = Nu    ψ = 1 − 2Σ Nu²e−βn²/φ / [βn²(Nu(Nu+1)+βn²)]
The eigenvalues βₙ come from a transcendental equation — no closed form, so they're found numerically. φ is the Graetz number (dimensionless position); Nu is a Biot-like number weighing convection against conduction.

▸The toolbox this problem actually needs

Mathematical tools, in the order you'd reach for them

  1. Nondimensionalize the energy balance. Strip out units by dividing through by characteristic scales — this is what turns "mass flow rate, channel dimensions, conductivity, convection coefficient" into just two numbers: the Graetz number φ (how far along the channel, scaled) and the Nusselt-like number Nu (how strong the convection is relative to conduction).
  2. Separate variables on the governing PDE. Same move as the heat equation lab — assume the solution is a product of a function of position and a sum over modes, and the PDE splits into an ODE for each mode.
  3. Apply a convective (Robin) boundary condition. Unlike a fixed-temperature wall, here heat leaves the fluid by convection, so the boundary condition involves both the temperature and its derivative. Plugging that into the ODE's general solution is exactly what produces the transcendental equation β tan β = Nu — there's no boundary condition simple enough to give a clean closed-form βₙ.
  4. Find the eigenvalues numerically. The same job as the root-finding lab: bracket each root (one lives in every interval (nπ, nπ+π/2), since tan shoots to +∞ at the right end) and bisect or Newton your way in.
  5. Sum the eigenfunction series. Like a Fourier series, the full solution is a sum over all the modes, each weighted by how strongly the initial/inlet condition excites it. The e-β²/φ factor means higher modes die out fast as φ grows — usually only the first few terms matter far from the inlet.
  6. Check whether the lumped shortcut is good enough. The lumped model assumes the temperature is uniform across the channel at every cross-section — the same "ignore the spatial profile" assumption as lumped capacitance in transient conduction, governed by the same kind of small-parameter criterion (there, Biot number; here, Nu).

None of these steps are unique to solar collectors — they're the standard toolkit for any 1-D diffusion problem with a convective boundary: a fin tip losing heat to air, a cooling fluid in a pipe, a slab quenched in a bath. Once you've solved this one, you've effectively solved that whole family.

Dimensionless exit temperature ψ vs. Graetz number φ

ψ distributed (full eigenfunction series) ψ lumped (single exponential)
Nu
0.1
β₀ (first eigenvalue)
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Max |distributed − lumped|
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What you're seeing

—

Nusselt-like number

-1.00

First eigenvalues βₙ (live)

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▸Reading the comparison

For small Nu (weak convection relative to conduction), the lumped and distributed curves sit right on top of each other — the temperature really is nearly uniform across the channel, so averaging it away costs almost nothing. As Nu grows past about 1, convection starts creating real temperature gradients across the channel that the lumped model can't see, and the two curves visibly separate. This matches the original paper's finding: lumped is "a suitable approximation" specifically for Nu < 1.

EngineeringCandy · Worked example from a real published solar-collector paper · eigenvalues found live by root-bracketing