Fourier's law and steady one-dimensional conduction

Fourier's law, thermal conductivity and the steady one-dimensional plane-wall solution, including walls whose conductivity varies with temperature.

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Why it matters

Every furnace lining, reactor jacket, pipe insulation and heat-exchanger tube wall is first analysed as a conduction problem. Fourier's law is the rate equation that turns a temperature difference into a heat flow, and the steady one-dimensional solution is the building block for composite walls, insulation design, fins and the overall heat transfer coefficient used in exchanger design.

Key ideas

Fourier's law. In a stationary medium heat flows by conduction from higher to lower temperature, at a rate proportional to the local temperature gradient and to the area normal to the flow. In one dimension the heat flux is q″ = −k·dT/dx. The minus sign is not decoration: heat flows down the gradient, so if T falls with x (dT/dx < 0) the flux is positive in the +x direction.

Fourier's law is an empirical constitutive law, like Ohm's law. It holds at every instant and every point, in steady and transient conduction; what changes in transient problems is that the temperature field itself is found from the heat diffusion equation (energy balance + Fourier's law).

Thermal conductivity k (W/m·K). A transport property of the material. Typical values at room temperature: copper ≈ 400, aluminium ≈ 200, carbon steel ≈ 45–55, stainless steel ≈ 15, firebrick ≈ 1, water ≈ 0.6, glass wool ≈ 0.04, air ≈ 0.026. Metals conduct mainly by free electrons, non-metals by lattice vibrations, gases by molecular collisions. For gases k rises with temperature; for most pure metals it falls; for most insulating solids it rises. Over a moderate range a linear model k = k₀(1 + βT) is often used.

Steady state. Temperatures do not change with time, so energy stored in the solid is constant. With no heat generation, the heat flow through every cross-section is the same.

One-dimensional. Temperature varies in one coordinate only (x, r). This is good when the wall is thin compared with its other dimensions and the faces are at uniform temperatures, or for long cylinders and complete spheres with uniform surface conditions.

The plane-wall result. For a plane wall with constant k, no generation and steady state, the heat equation reduces to d²T/dx² = 0, so the temperature profile is a straight line between the face temperatures. The heat rate is q = k·A·(T₁ − T₂)/L, which can be written q = ΔT/R with a conduction resistance R = L/(k·A). This thermal–electrical analogy (ΔT like voltage, q like current, R like resistance) is how composite walls, cylinders and convection at surfaces are combined in the next topic.

Variable conductivity. If k = k₀(1 + βT), the heat rate is still found from the plane-wall formula if k is replaced by the conductivity at the arithmetic mean temperature, k_m = k₀[1 + β(T₁ + T₂)/2]. The profile is no longer linear: with β > 0 the curve bows upward (the hot side, with higher k, needs a smaller gradient).

Connections. Fourier's law + energy balance gives the general heat conduction equation ρc·∂T/∂t = ∇·(k∇T) + q̇_g. Dropping terms gives the special cases: steady with generation (Poisson), steady without generation (Laplace), transient without generation (diffusion equation, using thermal diffusivity α = k/ρc).

Formulas

q = −k·A·dT/dx q heat rate (W), k thermal conductivity (W/m·K), A area normal to heat flow (m²), dT/dx temperature gradient (K/m). Applies at any point, steady or unsteady.

q″ = q/A = −k·dT/dx q″ heat flux (W/m²).

q = k·A·(T₁ − T₂)/L Plane wall, steady, no generation, constant k; L thickness (m), T₁ and T₂ face temperatures (°C or K — only the difference matters).

T(x) = T₁ − (T₁ − T₂)·x/L Linear temperature profile in the same plane wall; x measured from face 1 (m).

R = L/(k·A) Conduction resistance of a plane wall (K/W); q = ΔT/R.

k_m = k₀·[1 + β·(T₁ + T₂)/2], q = k_m·A·(T₁ − T₂)/L Plane wall with k = k₀(1 + βT); k₀ in W/m·K, β in 1/K (or 1/°C, consistent with the T used).

ρ·c·∂T/∂t = ∂/∂x(k·∂T/∂x) + q̇_g 1-D heat conduction equation; ρ density (kg/m³), c specific heat (J/kg·K), q̇_g volumetric generation (W/m³).

α = k/(ρ·c) Thermal diffusivity (m²/s).

Worked examples

Example 1 (standard). A furnace wall of area 15 m² is 0.20 m thick and made of brick with k = 1.0 W/m·K. The inner face is at 900 °C and the outer face at 150 °C. Find the heat loss and the temperature at the middle of the wall.

Given: A = 15 m², L = 0.20 m, k = 1.0 W/m·K, T₁ = 900 °C, T₂ = 150 °C.

  1. Heat rate: q = k·A·(T₁ − T₂)/L.
  2. q = 1.0 W/m·K × 15 m² × (900 − 150) K / 0.20 m = 1.0 × 15 × 750 / 0.20 W = 56 250 W.
  3. The profile is linear, so at x = L/2: T = T₁ − (T₁ − T₂)/2 = 900 − 375 = 525 °C.

Answer: q = 56.25 kW; mid-plane temperature = 525 °C.

Example 2 (GATE level). A plane wall 0.15 m thick has k = 0.8(1 + 0.002T) W/m·K, with T in °C. The faces are held at 500 °C and 100 °C. Find the heat flux and the mid-plane temperature.

Given: k₀ = 0.8 W/m·K, β = 0.002 1/°C, L = 0.15 m, T₁ = 500 °C, T₂ = 100 °C.

  1. Mean conductivity: k_m = k₀·[1 + β·(T₁ + T₂)/2] = 0.8 × (1 + 0.002 × 300) = 0.8 × 1.6 = 1.28 W/m·K.
  2. Heat flux: q″ = k_m·(T₁ − T₂)/L = 1.28 × 400 / 0.15 = 3413.3 W/m².
  3. Mid-plane: integrate q″·dx = −k₀(1 + βT)·dT from face 1 to x = L/2: k₀·[(T₁ − T) + (β/2)·(T₁² − T²)] = q″·L/2. Right side: 3413.3 × 0.075 = 256.0 W/m. Divide by k₀: 320.0 °C.
  4. (500 − T) + 0.001 × (250 000 − T²) = 320 → 0.001·T² + T − 430 = 0.
  5. T = [−1 + √(1 + 4 × 0.001 × 430)] / (2 × 0.001) = (−1 + 1.6492)/0.002 = 324.6 °C.

Answer: q″ ≈ 3.41 kW/m²; mid-plane temperature ≈ 324.6 °C (not 300 °C — the profile is curved because k varies).

Common mistakes

  • Dropping the minus sign and then getting the heat-flow direction wrong when setting up energy balances.
  • Using the temperature difference as if it were the gradient: divide by the thickness (in metres, not mm or cm).
  • Assuming a linear profile when k varies with temperature or when there is heat generation; the heat rate uses k_m, but the profile does not.
  • Believing Fourier's law is only for steady state. It always holds; only the simple plane-wall result q = kAΔT/L needs steady, 1-D, no generation.
  • Mixing °C and K inside k = k₀(1 + βT): β must match the temperature scale used.
  • Using the wrong area: A is normal to the heat flow (the face area of a wall, not its edge).

For GATE CH

Expect direct heat-rate and flux calculations for a plane wall, locating an intermediate temperature from the linear profile, and variable-conductivity walls where you use the mean k for the heat rate and integrate for a local temperature. Conceptual one-markers test the sign convention, the dependence of k on temperature for metals, gases and insulators, and which terms survive in the heat conduction equation under given assumptions. Practise writing the energy balance first and simplifying the general equation rather than memorising special cases.

Quick check

  1. A wall has T falling from 80 °C to 20 °C across 0.1 m with k = 2 W/m·K. What is the heat flux?
  2. Does Fourier's law apply during transient conduction?
  3. For k = k₀(1 + βT) with β > 0, is the mid-plane temperature above or below the arithmetic mean of the face temperatures?
  4. What are the SI units of thermal diffusivity?
  5. Which governing equation applies for steady 1-D conduction without generation and constant k?

Answers: 1. 1200 W/m². 2. Yes, it is a constitutive law valid at every instant. 3. Above. 4. m²/s. 5. d²T/dx² = 0 (linear profile).

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is Fourier's law of heat conduction?Concept

    Fourier's law states that the conductive heat flux is proportional to the negative of the local temperature gradient: q = −k·A·dT/dx in one dimension, with k the thermal conductivity (W/m·K) and A the area normal to the flow. The minus sign means heat flows from hot to cold, down the gradient. It is an empirical constitutive law, valid at every point and instant for isotropic materials.

  2. 2.Explain the concept of steady one-dimensional heat conduction.Concept

    Steady one-dimensional heat conduction refers to a situation where the temperature distribution in a material does not change with time, and heat flows in only one direction. This implies that the temperature gradient and heat transfer rate remain constant over time. It is a simplification used in many engineering problems to make analysis more manageable.

  3. 3.Why is thermal conductivity important in Fourier's law?Application

    Thermal conductivity is a material property that indicates how well a material can conduct heat. In Fourier's law, it is a proportionality constant that relates the heat transfer rate to the temperature gradient and area. A higher thermal conductivity means the material can transfer heat more efficiently, which is crucial in applications like heat exchangers and insulation.

  4. 4.What happens to the heat transfer rate if the temperature gradient increases?Application

    If the temperature gradient increases, the heat transfer rate also increases, assuming the thermal conductivity and area remain constant. This is because Fourier's law states that the heat transfer rate is directly proportional to the temperature gradient. A steeper gradient means a larger difference in temperature over a given distance, driving more heat flow.

  5. 5.How does the cross-sectional area affect heat conduction in a material?Application

    The cross-sectional area affects heat conduction by determining the amount of material available for heat to flow through. According to Fourier's law, the heat transfer rate is directly proportional to the area. A larger area allows more heat to be conducted, while a smaller area restricts the flow of heat.

  6. 6.In what scenarios would you assume steady one-dimensional conduction in engineering problems?Application

    When boundary temperatures and heat flow have stopped changing with time (steady state, not thermal equilibrium, since heat is still flowing) and temperature varies appreciably in only one coordinate. Typical cases are a furnace or building wall whose thickness is small compared with its height and width, the wall of a long pipe with uniform inside and outside conditions (radial only), and a complete spherical shell. Edge effects, corners and non-uniform surface conditions break the 1-D assumption.

  7. 7.What is the effect of increasing thermal conductivity on the temperature distribution in a material?Application

    For a given heat flux, a higher k needs a smaller temperature gradient (q″ = −k·dT/dx), so the temperature across the solid becomes more uniform; that is why heat sinks and spreaders are made of copper or aluminium. If instead both face temperatures are fixed in a plane wall with constant k, the profile stays linear regardless of k and only the heat rate rises in proportion to k. With temperature-dependent k the profile curves, bowing toward the side where k is higher.

  8. 8.Calculate the heat transfer rate through a wall with an area of 5 m², a thermal conductivity of 0.8 W/m·K, and a temperature gradient of 10 K/m.Numerical

    Using Fourier's law, q = -kA(dT/dx), we substitute the given values: q = -0.8 W/m·K * 5 m² * 10 K/m = -40 W. The negative sign indicates the direction of heat flow, but the magnitude of the heat transfer rate is 40 W.

  9. 9.A metal rod has a length of 2 m and a cross-sectional area of 0.01 m². If the thermal conductivity is 200 W/m·K and the temperature difference between the ends is 50 K, what is the heat transfer rate?Numerical

    First, calculate the temperature gradient: dT/dx = 50 K / 2 m = 25 K/m. Then, use Fourier's law: q = -kA(dT/dx) = -200 W/m·K * 0.01 m² * 25 K/m = -50 W. The heat transfer rate is 50 W.

  10. 10.Is Fourier's law valid for transient heat conduction?Concept

    Yes. Fourier's law q″ = −k·∇T is a constitutive (rate) law that relates the local heat flux to the local temperature gradient at every instant, in steady or unsteady problems alike. In transient conduction it is combined with an unsteady energy balance to give the heat diffusion equation ρc·∂T/∂t = ∇·(k∇T) + q̇_g. What is not valid in transient problems is the simple steady result q = kAΔT/L, because the heat rate then varies with position as the solid stores energy. (Fourier's law breaks down only in exotic cases such as extremely fast pulses or nanoscale conduction.)

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