Heat generation in solids

Steady conduction with uniform internal heat generation in slabs, cylinders and spheres: parabolic profiles, maximum temperature and surface energy balance.

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Why it matters

Electrical heating elements, current-carrying conductors, nuclear fuel rods, catalyst pellets with exothermic reactions, curing concrete and composting beds all produce heat inside the solid. The heat must be conducted to the surface before it can be removed, so the centre always runs hotter than the surface. Predicting that maximum temperature is how you size a heater, rate a cable, or avoid a runaway hot spot in a packed-bed reactor.

Key ideas

Volumetric heat generation q̇ (W/m³). Energy converted to heat per unit volume inside the solid: Joule heating I²R spread over the conductor volume, fission energy, or reaction heat (rate × heat of reaction). Absorption of radiation at a surface is a boundary condition, not generation.

Governing equation. An energy balance on a thin element with Fourier's law gives, for steady 1-D conduction and constant k:

  • plane wall: d²T/dx² + q̇/k = 0
  • cylinder: (1/r)·d/dr(r·dT/dr) + q̇/k = 0
  • sphere: (1/r²)·d/dr(r²·dT/dr) + q̇/k = 0 Generation turns the linear (or logarithmic) profiles of the no-generation case into parabolas.

Heat rate is no longer constant. With generation, the heat flowing through a section grows with distance from the centre because each layer adds its own heat: q(x) = q̇·A·x in a slab measured from the mid-plane. Thermal resistances in series therefore cannot be used inside the generating solid (they can still be used outside it).

Symmetry. In a symmetric slab with both faces at the same temperature, and in a solid cylinder or sphere, the maximum temperature is at the centre, where dT/dx = 0 (no heat crosses the centre line — an adiabatic plane). An insulated face behaves the same way, so a slab of thickness L insulated on one side is equivalent to half of a symmetric slab of thickness 2L.

Surface temperature from an overall balance. At steady state all heat generated must leave through the surface: q̇·V = h·A_s·(T_s − T∞). This gives T_s directly, without solving the differential equation. Then the centre temperature is T_s plus the internal rise.

Shape factors. The centre-to-surface rise is q̇·L²/(2k) for a slab of half-thickness L, q̇·R²/(4k) for a cylinder and q̇·R²/(6k) for a sphere. The divisor grows from 2 to 4 to 6 because the surface-to-volume ratio rises (1/L, 2/R, 3/R), giving the heat more exit area.

Limits. These results assume uniform q̇ and constant k. If k varies strongly with temperature, or if q̇ depends on temperature (exothermic reactions), the problem is non-linear and can have no steady solution — the basis of thermal runaway in catalyst pellets.

Formulas

T(x) = T_s + q̇·(L² − x²)/(2k) Symmetric plane wall of half-thickness L (m), faces at T_s; x from mid-plane (m); q̇ (W/m³); k (W/m·K).

T_max − T_s = q̇·L²/(2k) Plane wall (also a wall of thickness L insulated on one face, with L the full thickness).

T(r) = T_s + q̇·(R² − r²)/(4k), T_max − T_s = q̇·R²/(4k) Solid cylinder of radius R (m).

T(r) = T_s + q̇·(R² − r²)/(6k), T_max − T_s = q̇·R²/(6k) Solid sphere.

T_s = T∞ + q̇·L/h (slab, per face), T_s = T∞ + q̇·R/(2h) (cylinder), T_s = T∞ + q̇·R/(3h) (sphere) Surface temperature from the energy balance q̇·V = h·A_s·(T_s − T∞); h (W/m²·K), T∞ fluid temperature.

q̇ = I²·ρ_e/A_c² Joule heating in a wire carrying current I (A), electrical resistivity ρ_e (Ω·m), cross-section A_c (m²).

Worked examples

Example 1 (standard). A plane heating plate 40 mm thick (k = 25 W/m·K) generates q̇ = 2 × 10⁶ W/m³ uniformly. Both faces are cooled by a fluid at 30 °C with h = 500 W/m²·K. Find the surface and maximum temperatures.

Given: L = 0.020 m (half-thickness), q̇ = 2 × 10⁶ W/m³, k = 25 W/m·K, h = 500 W/m²·K, T∞ = 30 °C.

  1. Heat leaving each face per m²: q″ = q̇·L = 2 × 10⁶ × 0.02 = 40 000 W/m².
  2. T_s = T∞ + q̇·L/h = 30 + 40 000/500 = 30 + 80 = 110 °C.
  3. T_max − T_s = q̇·L²/(2k) = 2 × 10⁶ × 0.0004/(2 × 25) = 800/50 = 16 K.
  4. T_max = 110 + 16 = 126 °C at the mid-plane.

Answer: T_s = 110 °C, T_max = 126 °C.

Example 2 (GATE level). A stainless-steel wire 3 mm in diameter, 1 m long (k = 15 W/m·K, ρ_e = 7 × 10⁻⁷ Ω·m), carries 200 A. It is cooled by a fluid at 30 °C with h = 5000 W/m²·K. Find the generation rate, the surface temperature and the centre temperature.

Given: R = 0.0015 m, I = 200 A.

  1. Cross-section: A_c = π·R² = π × 0.0015² = 7.069 × 10⁻⁶ m².
  2. q̇ = I²·ρ_e/A_c² = 40 000 × 7 × 10⁻⁷/(7.069 × 10⁻⁶)² = 5.604 × 10⁸ W/m³. (Check: q̇·A_c = 3961 W per metre = I²R per metre.)
  3. T_s = T∞ + q̇·R/(2h) = 30 + 5.604 × 10⁸ × 0.0015/(2 × 5000) = 30 + 84.1 = 114.1 °C.
  4. T_max − T_s = q̇·R²/(4k) = 5.604 × 10⁸ × 2.25 × 10⁻⁶/(4 × 15) = 21.0 K.
  5. T_centre = 114.1 + 21.0 = 135.1 °C.

Answer: q̇ ≈ 5.60 × 10⁸ W/m³, T_s ≈ 114 °C, T_centre ≈ 135 °C.

Common mistakes

  • Using the full thickness instead of the half-thickness in q̇·L²/(2k) for a slab cooled on both faces.
  • Using the slab factor 2 for a cylinder (4) or a sphere (6).
  • Adding a conduction resistance for the generating solid in a series network; the heat rate is not constant inside it.
  • Forgetting the convective rise: the centre temperature is T∞ + (surface film rise) + (internal rise).
  • Writing the surface balance with the wrong area-to-volume ratio: V/A_s is L for a slab face, R/2 for a cylinder and R/3 for a sphere.

For GATE CH

Expect maximum-temperature questions for slabs, wires and spheres, problems where you must find q̇ from a current or reaction rate first, and slabs with one face insulated. Conceptual questions test the shape of the profile (parabolic), the location of the maximum and the zero-gradient condition at a symmetry plane. Practise setting up the differential equation, applying the two boundary conditions, and checking your answer with the overall energy balance.

Quick check

  1. What is the temperature profile shape in a slab with uniform generation?
  2. A sphere of radius 0.1 m, k = 10 W/m·K, q̇ = 6 × 10⁴ W/m³. Centre-to-surface rise?
  3. Where is dT/dx zero in a symmetric heated slab?
  4. A slab 50 mm thick is insulated on one face. Which length goes into q̇·L²/(2k)?
  5. In a long cylinder, what is V/A_s?

Answers: 1. Parabolic. 2. 10 K. 3. At the mid-plane. 4. The full 50 mm. 5. R/2.

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is heat generation in solids?Concept

    Heat generation in solids refers to the production of heat within a solid material due to internal sources such as chemical reactions, electrical resistance, or nuclear reactions. This heat generation can affect the temperature distribution within the solid and is an important factor in thermal analysis.

  2. 2.Explain the significance of heat generation in the context of thermal analysis of solids.Concept

    Heat generation is significant in thermal analysis because it influences the temperature distribution within a solid. Understanding this distribution is crucial for designing materials and systems that can withstand thermal stresses, prevent overheating, and ensure efficient thermal management in applications like electronics, reactors, and manufacturing processes.

  3. 3.Does internal heat generation change the thermal conductivity of a solid?Application

    No, not directly: k is a material property that depends on temperature (and composition), not on whether heat is generated. Generation changes the temperature field — it makes the profile parabolic and raises the centre temperature — so in materials whose k varies with temperature the local conductivity will differ from point to point. For accurate work you then evaluate k at the mean temperature or solve the non-linear problem.

  4. 4.Why is it important to consider heat generation in the design of electronic devices?Application

    In electronic devices, heat generation due to electrical resistance can lead to overheating, which may damage components or reduce their lifespan. Proper thermal management, including considering heat generation, is essential to ensure device reliability, performance, and safety by preventing excessive temperatures.

  5. 5.What happens if heat generation in a solid is not properly managed?Application

    If heat generation is not properly managed, it can lead to excessive temperatures, thermal stresses, and potential failure of the material or system. This can result in reduced efficiency, safety hazards, and increased maintenance costs. Effective thermal management strategies are necessary to mitigate these risks.

  6. 6.Explain how heat generation is modeled in the heat conduction equation.Concept

    It enters as a volumetric source term q̇ (W/m³) in the energy balance: ρc·∂T/∂t = ∇·(k∇T) + q̇. For steady 1-D conduction with constant k this becomes d²T/dx² + q̇/k = 0 for a slab, (1/r)·d/dr(r·dT/dr) + q̇/k = 0 for a cylinder, and the analogous spherical form. Integrating twice with the boundary conditions (zero gradient at the centre, known surface temperature or convection) gives parabolic profiles and maximum rises of q̇L²/2k, q̇R²/4k and q̇R²/6k respectively.

  7. 7.Why is heat generation considered in the design of nuclear reactors?Application

    In nuclear reactors, heat generation occurs due to nuclear fission reactions. It is crucial to consider this heat generation to ensure efficient heat removal and prevent overheating of the reactor core. Proper thermal management is essential for reactor safety, efficiency, and longevity.

  8. 8.What is the effect of heat generation on the thermal expansion of a solid?Application

    Heat generation can cause a solid to expand due to increased temperatures. This thermal expansion can lead to mechanical stresses and potential deformation. Understanding the relationship between heat generation and thermal expansion is important for designing materials and structures that can accommodate these changes without failure.

  9. 9.A plane slab of half-thickness 0.01 m (k = 20 W/m·K) generates heat uniformly at 5 × 10⁶ W/m³, with both faces at the same temperature. Find the rise in temperature from the surface to the centre.Numerical

    For a symmetric slab with uniform generation, T_max − T_s = q̇·L²/(2k), with L the half-thickness. ΔT = 5 × 10⁶ × (0.01)²/(2 × 20) = 500/40 = 12.5 K. The maximum is at the mid-plane, where the temperature gradient is zero.

  10. 10.A solid cylinder generates heat at a rate of 1000 W/m³. If the cylinder has a radius of 0.1 m and height of 0.5 m, calculate the total heat generated.Numerical

    The total heat generated is calculated by multiplying the heat generation rate by the volume of the cylinder. Volume = π·r²·h = π·(0.1 m)²·(0.5 m) = 0.0157 m³. Total heat = 1000 W/m³ · 0.0157 m³ = 15.7 W.

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