Chapter 03 · T
Heat transport
Heat moves through a porous medium three ways at once: it conducts through both phases, it rides along with the flowing fluid, and — if the fluid is free to overturn — it drives its own circulation. Which one dominates is decided by two dimensionless numbers, and the answers they give are often surprising.
One temperature for two phases
The standard simplification is local thermal equilibrium: within a representative volume, solid and fluid are assumed to share a single temperature. It is an excellent assumption for geological materials — the grain-scale conduction time $d^2/\kappa_s$ is seconds, while the macroscopic processes take days to millennia — and it collapses two energy balances into one:
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- $(\rho c)_m$ Storage in the mixture J m⁻³ K⁻¹ $(1-\phi)(\rho c)_s + \phi(\rho c)_f$ — an exact volume average, no geometry involved. For saturated rock it is around 2.2–2.6 MJ m⁻³ K⁻¹, and the solid supplies most of it.
- $(\rho c)_f\,\vec q$ Advection — the H→T coupling Note that it is the fluid heat capacity multiplying the Darcy flux — not the pore velocity, and not the mixture capacity. Getting this pairing wrong is a classic modelling bug and it makes fronts arrive at the wrong time.
- $\lambda_m\nabla T$ Conduction Through solid and fluid together, using the effective conductivity from chapter 01. In fast flows a mechanical dispersion term $\lambda_{\text{disp}} = (\rho c)_f\,\alpha_L |\vec q|$ is added, because pore-scale velocity variations spread heat exactly as they spread a solute.
- $Q$ Sources Radiogenic heat production in the crust ($\sim$1–3 µW m⁻³), decay heat from waste canisters, frictional heating on a slipping fault, or latent heat if freezing or boiling is in play.
Why the thermal front lags
Push water through a hot porous medium and two fronts set off. The water itself moves at the pore velocity $\vec v = \vec q/\phi$. The temperature front moves much more slowly, because every metre of solid it passes must be cooled — and the solid holds most of the heat. Equating the enthalpy carried in with the enthalpy stored gives the thermal front velocity directly:
A geothermal doublet exploits exactly this. The injected water reaches the producer in months; the cold front, retarded by a factor of five or so and blunted by conduction, may take decades. That gap is the resource lifetime.
Thermal front versus tracer front
Curves are the Ogata–Banks analytic solution $\tfrac12\big[\mathrm{erfc}\tfrac{x-v t}{2\sqrt{Dt}} + e^{vx/D}\,\mathrm{erfc}\tfrac{x+v t}{2\sqrt{Dt}}\big]$ applied to both fronts with their own velocities and dispersion coefficients. Heat has a much larger effective diffusivity than a dissolved tracer, so its front is both slower and smoother — at low Péclet number it barely resembles a front at all.
$\mathrm{Pe} = (\rho c)_f |\vec q| L / \lambda_m$ compares advective to conductive transport over a length $L$. Below about 0.5 the flow field barely matters and you can model the thermal problem as pure conduction. Above about 10 the front is sharp and you must resolve it — a coarse grid will smear it numerically, and the numerical smearing can easily exceed the physical dispersion.
When the fluid drives itself
So far the flow has been imposed. But hot water is buoyant, and if a permeable layer is heated from below, a fluid parcel that rises finds itself lighter than its new surroundings and keeps rising. Conduction and viscosity fight back. The competition is measured by the Rayleigh number for a porous medium:
The critical value $4\pi^2$ is a genuine analytic result — Horton, Rogers and Lapwood derived it by linear stability analysis in the 1940s, and the fastest-growing mode has square cells, one cell width per layer thickness. Below it, a perturbation decays and heat moves by pure conduction. Above it, the layer overturns. The simulation below solves the full nonlinear problem, so you can watch the transition happen.
Horton–Rogers–Lapwood convection, solved live
The Darcy–Boussinesq system in stream-function form: $\nabla^2\psi = -\mathrm{Ra}\,\partial T/\partial x$ solved by successive over-relaxation each step, with the temperature advected by $\vec u = (\partial_z\psi,\,-\partial_x\psi)$ and diffused explicitly. The Nusselt number is the measured heat flux through the bottom boundary divided by the pure-conduction value — it sits at 1 below the critical Rayleigh number and climbs steeply above it.
Free convection sets the natural state of geothermal fields, controls whether a saline aquifer holds injected CO₂ stably or fingers it downwards, and decides whether the near-field of a nuclear-waste repository can be modelled with conduction alone. The permeability enters $\mathrm{Ra}$ linearly, so a fracture network that raises $k$ by two orders of magnitude moves a layer from firmly conductive to vigorously convecting.
When one temperature is not enough
Local thermal equilibrium fails when heat cannot diffuse across a grain faster than the fluid carries it past. The comparison is between the grain-scale conduction time $d^2/\kappa_s$ and the residence time $d/v$; equilibrium holds while
For millimetre grains and metre-per-day Darcy fluxes this is comfortably satisfied. It fails in three places worth knowing: open fractures, where the fluid bypasses the matrix entirely and the correct model is a fracture–matrix exchange coefficient rather than an averaged conductivity; coarse gravels under strong pumping; and rapid transients such as the first hours of a thermal-response test. In those cases a two-equation (local thermal non-equilibrium) model is needed, with a heat-transfer coefficient $h_{sf}$ coupling the phases.
Thermal properties worth memorising
| Material | λ (W m⁻¹ K⁻¹) | ρc (MJ m⁻³ K⁻¹) | κ (mm² s⁻¹) | αT (10⁻⁶ K⁻¹) |
|---|---|---|---|---|
| Water | 0.60 | 4.18 | 0.14 | 210 (volumetric) |
| Air | 0.026 | 0.0012 | 21 | 3400 |
| Quartz | 7.7 | 2.0 | 3.9 | 11 |
| Granite | 2.5–3.4 | 2.2 | 1.3 | 8 |
| Sandstone (sat.) | 2.0–3.5 | 2.4 | 1.1 | 10 |
| Shale / clay (sat.) | 1.1–2.1 | 2.4 | 0.7 | 10 |
| Dry sand | 0.3–0.6 | 1.3 | 0.35 | 10 |
| Rock salt | 5.5 | 1.9 | 2.9 | 40 |
Two numbers are worth carrying in your head. Saturated soil and rock both have a thermal diffusivity of about $1\ \mathrm{mm^2\,s^{-1}}$, which means a thermal signal penetrates roughly $\sqrt{\kappa t}\approx 6$ m in a year and 60 m in a century. And the volumetric heat capacity of saturated ground is about 2.4 MJ m⁻³ K⁻¹, so cooling a cubic metre by one kelvin releases enough energy to boil about seven litres of water.