STEADY-STATE HEAT CONDUCTION

1D Layered Geothermal Modeler

Edit layer boundaries, radioactive heat generation, and thermal conductivity. The analytical and finite-volume geotherms update immediately.

INPUT MODEL

Layer properties

A k Boundary
Selected layer Layer 1

Mouse: drag orange or blue vertical handles to change A or k; drag horizontal handles to change interface depth. Click a layer to select it.

MODEL RESPONSE

Temperature–depth profile

Analytical Finite volume
Surface heat flow—
Bottom temperature—
Maximum |ΔT|—
Thermal resistance—
Boundary conditions T(0) = Ts; k dT/dz at the model base = qb. Depth z is positive downward, and heat-flow values are reported as upward-positive magnitudes.

NUMERICAL VALUES

Layer model

Edits in the table and plots are synchronized.

LayerTop (km)Bottom (km)Thickness (km)A (µW m⁻³)k (W m⁻¹ K⁻¹)Top q (mW m⁻²)Top T (°C)

FORMULATION

Analytical and finite-volume solutions

The formulation is valid for an arbitrary number n of horizontal layers. Layer properties are piecewise constant, depth is positive downward, and heat flow is expressed as an upward-positive magnitude.

1. Layer notation and governing equation

Interfaces0 = z0 < z1 < ··· < zn = L
Layer thicknesshi = zi − zi−1
Layer propertiesAi, ki; i = 1, …, n
Numerical cellsM cells; Δz = L/M
ddz [k(z)dTdz] +A(z)=0 (1)
q(z)=k(z) dTdz (2)

The boundary conditions are T(0) = Ts and q(L) = qb. Temperature and heat flow remain continuous at every interface.

2. Exact piecewise solution for n layers

The heat flow at the top of layer i includes the basal contribution and the heat produced in layer i and all deeper layers:

qit=qb+ ∑j=in Ajhj (3)

Within layer i, define local depth ξ = z − zi−1, where 0 ≤ ξ ≤ hi. The exact solution is

qi(ξ)=qit−Aiξ Ti(ξ)=Tit+ qitξ−12Aiξ2 ki (4)

The next layer starts from the values obtained at ξ = hi:

qi+1t=qit−Aihi Ti+1t=Tit+ qithi−12Aihi2 ki (5)

3. Cell-centered finite-volume discretization

The interval 0 ≤ z ≤ L is divided into M cells. Each cell center receives the A and k values of its containing geological layer. Conductivity at an internal cell face is the harmonic mean:

kp+12 =2kpkp+1kp+kp+1 ,Gp+12=kp+12Δz (6)

For an interior cell p = 1, …, M−2, flux balance gives

−Gp−12Tp−1 +(Gp−12+Gp+12)Tp −Gp+12Tp+1 =ApΔz (7)

The surface half-cell conductance is Gs = 2k0/Δz. The first and last rows of the matrix are

(Gs+G12)T0 −G12T1 =A0Δz+GsTs −GM−32TM−2 +GM−32TM−1 =AM−1Δz+qb (8)

4. Calculation procedure

  1. Validate and order the n layer interfaces, then convert km, µW m⁻³, and mW m⁻² to SI units.
  2. Evaluate the exact piecewise solution by starting with qb and propagating heat flow and temperature through all layers.
  3. Assign A and k to M finite-volume cells, construct the tridiagonal system from Equations (6)–(8), and solve it with the Thomas algorithm.
  4. Interpolate the analytical solution at numerical cell depths and report max |TFV − Tanalytical|. Increase M to assess grid convergence.
Computational cost: the tridiagonal solve requires O(M) operations and O(M) storage. The analytical calculation requires O(n + P) operations, where P is the number of plotted sample points.