Conduction, convection and radiation heat transfer fundamentals

Conduction by Fourier's law and thermal resistance, convection by Newton's law of cooling, black- and grey-body radiation by the Stefan–Boltzmann law, and combined-mode problems including spacecraft radiative equilibrium.

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Why it matters

Thermodynamics tells you how much heat must be transferred; heat transfer tells you how fast it happens and how big the surface must be. Turbine-blade cooling, cabin insulation, avionics cooling, engine oil coolers and the thermal control of satellites, which can reject heat only by radiation, all rest on the three basic modes covered here.

Key ideas

Conduction is energy transfer through a medium by molecular interaction (lattice vibrations and free electrons in solids, molecular collisions in fluids) with no bulk motion. It needs a material medium and a temperature gradient. Metals conduct well because of free electrons; gases and porous materials (insulation) conduct poorly. Thermal conductivity k ranges from about 0.025 W/(m·K) for air to about 400 W/(m·K) for copper.

Fourier's law. Heat flows down the temperature gradient: Q̇ = −k·A·dT/dx. The minus sign makes Q̇ positive in the direction of decreasing temperature. For steady one-dimensional conduction through a plane wall with constant k, the temperature profile is linear and Q̇ = k·A·(T₁ − T₂)/L.

Thermal resistance. Writing heat flow as ΔT/R gives an electrical analogy: resistances in series add, as for a composite wall with a convective film on each side. The same Q̇ flows through every layer in steady state, so the largest temperature drop occurs across the largest resistance — usually the insulation.

Convection is heat transfer between a surface and a moving fluid, combining conduction at the wall with transport by the bulk flow. Forced convection uses a fan, pump or vehicle motion; natural (free) convection is driven by buoyancy from density differences. It is described by Newton's law of cooling, Q̇ = h·A·(T_s − T_∞). The heat transfer coefficient h is not a property: it depends on flow velocity, geometry, fluid properties and regime, and is obtained from Nusselt-number correlations in a data book. Typical values: 2–25 W/(m²·K) for free convection in gases, 25–250 for forced convection in gases, up to 20 000 for liquids, higher still for boiling and condensation.

Radiation is energy carried by electromagnetic waves (mainly infrared at engineering temperatures). It needs no medium and works best in vacuum, which is why it is the only way a spacecraft can reject heat.

  • A black body emits the maximum possible: E_b = σ·T⁴.
  • A real (grey) surface emits E = ε·σ·T⁴, with emissivity 0 ≤ ε ≤ 1, and absorbs a fraction α of incident radiation. For a grey surface α = ε (Kirchhoff's law); for solar radiation the solar absorptivity α_s can differ from the infrared ε, which is exploited in spacecraft coatings.
  • Net exchange between a small surface and large surroundings at T_surr: Q̇ = ε·σ·A·(T_s⁴ − T_surr⁴).

Combined modes. A surface in air usually loses heat by convection and radiation in parallel; at moderate temperatures radiation is often comparable to free convection and must not be ignored.

Formulas

Q̇ = −k·A·dT/dx (Fourier) Q̇ = k·A·(T₁ − T₂)/L; R_cond = L/(k·A) (plane wall) Q̇ = h·A·(T_s − T_∞); R_conv = 1/(h·A) (Newton's law of cooling) Q̇ = ΔT_overall / ΣR (layers in series) E_b = σ·T⁴; E = ε·σ·T⁴ Q̇ = ε·σ·A·(T_s⁴ − T_surr⁴) (small body in large enclosure)

  • Q̇: heat transfer rate, W; A: area normal to heat flow, m²
  • k: thermal conductivity, W/(m·K); L: thickness, m
  • h: convection heat transfer coefficient, W/(m²·K)
  • R: thermal resistance, K/W
  • σ = 5.67 × 10⁻⁸ W/(m²·K⁴); ε: emissivity; T in K for radiation (always absolute)

Worked examples

Example 1 (standard). A 10 m² wall is 0.2 m brick (k = 0.7 W/(m·K)) with 0.05 m insulation (k = 0.04 W/(m·K)) on the outside. Inside air is at 25 °C with h = 10 W/(m²·K); outside air is at −5 °C with h = 25 W/(m²·K). Find the heat loss and the temperature of the brick–insulation interface.

  1. Resistances: inside film 1/(h·A) = 1/(10 × 10) = 0.0100 K/W; brick L/(k·A) = 0.2/(0.7 × 10) = 0.0286 K/W; insulation 0.05/(0.04 × 10) = 0.1250 K/W; outside film 1/(25 × 10) = 0.0040 K/W
  2. ΣR = 0.1676 K/W
  3. Q̇ = ΔT/ΣR = 30/0.1676 = 179.0 W
  4. Interface temperature: 25 − 179.0 × (0.0100 + 0.0286) = 25 − 6.9 = 18.1 °C

Answer: Q̇ ≈ 179 W; interface at about 18.1 °C (the insulation carries 22.4 K of the 30 K drop).

Example 2 (GATE level). A thin flat panel on a satellite faces the Sun, receiving 1360 W/m². Its sun-facing side has solar absorptivity 0.6; both sides have infrared emissivity 0.8 and see deep space (take 0 K). Neglecting conduction to the rest of the satellite, find the panel's equilibrium temperature. Also find the net radiative-plus-convective loss per m² from a 400 K pipe surface (ε = 0.8) in a room with air and walls at 300 K, h = 10 W/(m²·K).

  1. Panel energy balance per m²: absorbed = emitted from two sides: α_s·G = 2·ε·σ·T⁴
  2. T⁴ = 0.6 × 1360/(2 × 0.8 × 5.67 × 10⁻⁸) = 8.995 × 10⁹ K⁴
  3. T = 308 K (about 35 °C)
  4. Pipe: convection h·ΔT = 10 × 100 = 1000 W/m²; radiation εσ(T_s⁴ − T_surr⁴) = 0.8 × 5.67 × 10⁻⁸ × (400⁴ − 300⁴) = 793.8 W/m²
  5. Total = 1794 W/m² — radiation is almost half of the loss.

Common mistakes

  • Using °C in the T⁴ terms of radiation (only differences such as T₁ − T₂ may use °C).
  • Treating h as a material property; it depends on the flow.
  • Adding conductances instead of resistances for layers in series.
  • Writing net radiation as εσ(T_s − T_surr)⁴ instead of εσ(T_s⁴ − T_surr⁴).
  • Forgetting the minus sign convention in Fourier's law, then reporting a negative heat flow in the direction of flow.
  • Ignoring radiation in free-convection problems at moderate temperature.

For GATE AE

Expect one-mark questions identifying the mode or the governing law, and short numericals on plane-wall conduction, composite walls with convective films, Newton's law of cooling and Stefan–Boltzmann emission or net exchange. Radiation equilibrium of a surface (energy absorbed equals energy emitted) is a natural aerospace application. Practise the thermal-resistance network and keep all radiation temperatures in kelvin.

Quick check

  1. A 0.5 m² wall, k = 0.04 W/(m·K), 0.1 m thick, ΔT = 20 K: find Q̇.
  2. A surface with ε = 0.8 is at 300 K. How much does it emit per m²?
  3. By what factor does black-body emission rise if T (in K) doubles?
  4. Which mode of heat transfer works in a vacuum?
  5. Is the convection coefficient h a material property?

Answers: 1. 4 W. 2. About 367 W/m². 3. 16. 4. Radiation. 5. No — it depends on the flow and geometry.

Interview questions

All Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.Define conduction, convection, and radiation in the context of heat transfer.Concept

    Conduction is energy transfer through a medium — solid, liquid or gas — by molecular interaction (lattice vibrations, free electrons, molecular collisions) without bulk motion, governed by Fourier's law Q̇ = −kA dT/dx. Convection is heat transfer between a surface and a moving fluid, combining conduction at the wall with transport by the bulk flow; it is forced or natural and is described by Q̇ = hA(T_s − T_∞). Radiation is energy carried by electromagnetic waves emitted by every body above 0 K; it needs no medium, and a surface emits εσT⁴ per unit area.

  2. 2.Explain how Fourier's Law is used in conduction heat transfer.Concept

    Fourier's Law states that the rate of heat transfer through a material is proportional to the negative gradient of temperature and the area through which the heat flows. Mathematically, it is expressed as q = -kA(dT/dx), where q is the heat transfer rate, k is the thermal conductivity, A is the area, and dT/dx is the temperature gradient.

  3. 3.What is the difference between natural and forced convection?Concept

    Natural convection occurs when fluid motion is caused by buoyancy forces that result from density variations due to temperature differences in the fluid. Forced convection involves external forces, such as fans or pumps, to move the fluid and enhance heat transfer.

  4. 4.Why are metals generally good conductors of heat?Application

    Metals are good conductors of heat because they have free electrons that can move easily throughout the material. These electrons transfer energy quickly between atoms, facilitating efficient heat conduction.

  5. 5.What happens to the rate of heat transfer by radiation if the temperature of an object doubles?Application

    The rate of heat transfer by radiation is proportional to the fourth power of the absolute temperature, as described by the Stefan-Boltzmann Law. If the temperature of an object doubles, the rate of heat transfer increases by a factor of 2^4, or 16 times.

  6. 6.Explain why fins are used in heat exchangers.Application

    Fins are used in heat exchangers to increase the surface area available for heat transfer. By increasing the surface area, fins enhance the rate of heat transfer between the solid surface and the surrounding fluid, improving the efficiency of the heat exchanger.

  7. 7.How does the thermal conductivity of a material affect its insulating properties?Application

    The thermal conductivity of a material is a measure of its ability to conduct heat. Materials with low thermal conductivity are good insulators because they do not allow heat to pass through easily, thus reducing heat transfer.

  8. 8.Calculate the heat transfer rate through a 0.5 m² wall with a thermal conductivity of 0.04 W/m·K, a thickness of 0.1 m, and a temperature difference of 20 K across it.Numerical

    Using Fourier's Law: q = -kA(dT/dx). Here, k = 0.04 W/m·K, A = 0.5 m², dT = 20 K, and dx = 0.1 m. q = -0.04 * 0.5 * (20 / 0.1) = -4 W. The negative sign indicates the direction of heat flow.

  9. 9.A metal rod is heated at one end. Describe the process of heat transfer through the rod.Application

    Heat transfer through the metal rod occurs by conduction. The heat energy is transferred from the hot end to the cooler end by the vibration and movement of atoms and free electrons within the metal. This process continues until thermal equilibrium is reached.

  10. 10.If a surface has an emissivity of 0.8 and is at a temperature of 300 K, calculate the power radiated per unit area.Numerical

    By the Stefan–Boltzmann law for a grey surface, E = ε·σ·T⁴ with σ = 5.67 × 10⁻⁸ W/(m²·K⁴). T⁴ = 300⁴ = 8.1 × 10⁹ K⁴, so a black body would emit 459.3 W/m², and this surface emits 0.8 × 459.3 ≈ 367 W/m². This is the emitted power; the net loss also depends on the radiation it absorbs from its surroundings, Q̇/A = εσ(T_s⁴ − T_surr⁴).

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