Short Answer:
To derive the heat conduction equation for a plane wall we apply an energy balance to a small one-dimensional slab (thickness dx) and use Fourier’s law. For a general case with internal heat generation and time dependence the result is
If thermal conductivity is constant this simplifies to , where is the thermal diffusivity.
For steady one-dimensional conduction (no time dependence) the equation reduces to
or if there is no internal heat generation , giving a linear temperature profile across the wall.
Detailed Explanation :
Heat conduction equation for a plane wall
Physical model and assumptions
Consider a homogeneous plane wall (flat slab) of infinite extent in y and z so heat flow is only in the x-direction (one-dimensional). Material properties may depend on position or temperature in the most general derivation, but we first present the usual case with constant density , specific heat and conductivity . Allow possible internal volumetric heat generation (W/m³) and allow temperature to vary with time .
Step 1 — Select a differential control volume
Take a thin slice of the wall between and with cross sectional area . The volume is .
Step 2 — Write the energy balance
Rate of increase of thermal energy stored in the slice = rate of heat in by conduction at minus rate of heat out by conduction at + heat generated within the slice.
Mathematically:
Left side (storage):
Conduction at position (Fourier’s law):
and at :
So net conduction in minus out:
Simpler: expand the outflow in Taylor series and retain first derivative:
Volumetric generation in slice:
Step 3 — Combine terms and divide by
Energy balance:
Divide by :
If cross-section is constant (common slab case), the cancels:
This is the general one-dimensional heat conduction equation for a plane wall with possible spatially varying and internal generation.
Step 4 — Special case: constant thermal conductivity
If is constant (independent of and ), the derivative simplifies:
Divide by and introduce thermal diffusivity :
This is the standard transient heat conduction (diffusion) equation in one dimension for a plane wall.
Step 5 — Steady-state form
For steady state , the transient term drops:
If there is no internal generation (), this reduces to:
a linear temperature profile; constants found from boundary conditions.
Boundary and initial conditions
To solve either transient or steady PDE you need:
- Two boundary conditions in for steady problem (e.g., specified temperatures , ; or one temperature and one flux/convective condition).
- For transient problems you need the initial temperature distribution plus boundary conditions in for all .
Interpretation and physical meaning
- The term governs diffusion of temperature (how curvature of temperature profile drives change).
- The source term adds temperature rise due to internal generation per unit mass and heat capacity.
- Larger → faster thermal response; small → slow thermal penetration.
Common solutions and uses
Analytical solutions exist for simple BCs (constant surface T, constant heat flux, convective surface) using separation of variables or eigenfunction expansions. These solutions are widely used to predict temperature evolution in walls, electronic packages, furnace linings, and transient heating/cooling problems.
Conclusion
By applying an energy balance to a differential slab and using Fourier’s law, we obtain the one-dimensional heat conduction equation for a plane wall. The general transient form with internal heat generation is . For constant this becomes . At steady state this reduces to , and if there is no internal heat generation a simple linear temperature profile results. Correct boundary and initial conditions are required to obtain specific temperature solutions for design and analysis.