Heat Transfer: Conduction, Convection and Radiation

Thermodynamics tells you how much energy must move; heat transfer tells you how fast it moves. That rate decides how thick a cold-store wall should be, how big a car radiator must be and how hot a processor gets. In this article we cover the three modes of heat transfer, the equations behind each, and three solved engineering problems with units and sanity checks at every step.

The Three Modes at a Glance

Conduction: Fourier's Law

For steady one-dimensional conduction through a plane wall of thickness \( L \), area \( A \) and thermal conductivity \( k \):

$$ \dot{Q} = kA\,\frac{T_1 - T_2}{L} $$

Units: \( k \) in W/(m·K), \( A \) in m², \( L \) in m, temperatures in °C or K (a difference is the same in both), giving \( \dot{Q} \) in watts.

Typical conductivities show why materials are chosen as they are: copper about 400 W/(m·K), steel about 50, common brick about 0.7, and glass wool insulation about 0.04.

Convection: Newton's Law of Cooling

When a fluid flows past a surface at temperature \( T_s \), with the fluid far away at \( T_\infty \):

$$ \dot{Q} = hA\,(T_s - T_\infty) $$

The convection coefficient \( h \), in W/(m²·K), is not a material property; it depends on fluid speed, geometry and fluid type. Rough ranges are 5 to 25 for still air, 25 to 250 for forced air and 100 to 15 000 for water.

Radiation: The Stefan-Boltzmann Law

Every surface above absolute zero emits radiation. For a small surface of emissivity \( \varepsilon \) inside large surroundings at \( T_{\text{surr}} \):

$$ \dot{Q} = \varepsilon \sigma A\left(T_s^4 - T_{\text{surr}}^4\right) $$

with \( \sigma = 5.67 \times 10^{-8} \) W/(m²·K⁴). Temperatures must be in kelvin here, because the fourth power does not treat °C and K differences the same way.

Thermal Resistance: The Engineer's Shortcut

Each mode can be written like Ohm's law, with temperature difference playing the role of voltage and heat flow the role of current:

• Conduction: \( R_{\text{cond}} = \dfrac{L}{kA} \)
• Convection: \( R_{\text{conv}} = \dfrac{1}{hA} \)

Resistances in series simply add, so

$$ \dot{Q} = \frac{T_{\infty,1} - T_{\infty,2}}{R_{\text{total}}} $$

This lets you handle multi-layer walls, insulated pipes and heat sinks with the same simple arithmetic.

Solved Problem 1: Conduction Through a Brick Wall

A brick wall is 0.20 m thick with an area of 12 m² and \( k = 0.72 \) W/(m·K). The inner surface is at 22 °C and the outer surface at 4 °C. Find the heat loss.

Step 1: Temperature difference. \( \Delta T = 22 - 4 = 18 \) K.

Step 2: Apply Fourier's law.

$$ \dot{Q} = \frac{(0.72\ \text{W/m·K})(12\ \text{m}^2)(18\ \text{K})}{0.20\ \text{m}} = \frac{155.52}{0.20}\ \text{W} = 777.6\ \text{W} $$

Sanity check. About 780 W, comparable to a small electric heater, is plausible for an uninsulated brick wall on a cold winter night in Murree or Quetta. Doubling the thickness would halve the loss, as Fourier's law predicts.

Solved Problem 2: The Same Wall With Air Films

In reality we know the air temperatures, not the surface temperatures. Inside air is 22 °C with \( h_i = 10 \) W/(m²·K); outside air is 4 °C with \( h_o = 25 \) W/(m²·K) because of wind. Find the heat loss and the inner surface temperature.

Step 1: Individual resistances.

$$ R_i = \frac{1}{h_i A} = \frac{1}{(10)(12)} = 0.008333\ \text{K/W} $$

$$ R_{\text{wall}} = \frac{L}{kA} = \frac{0.20}{(0.72)(12)} = 0.023148\ \text{K/W} $$

$$ R_o = \frac{1}{h_o A} = \frac{1}{(25)(12)} = 0.003333\ \text{K/W} $$

Step 2: Total resistance.

$$ R_{\text{total}} = 0.008333 + 0.023148 + 0.003333 = 0.034815\ \text{K/W} $$

Step 3: Heat flow.

$$ \dot{Q} = \frac{18\ \text{K}}{0.034815\ \text{K/W}} = 517\ \text{W} $$

Step 4: Inner surface temperature. The same heat crosses the inside air film:

$$ T_{s,i} = 22 - \dot{Q} R_i = 22 - (517)(0.008333) = 22 - 4.31 = 17.7\ ^\circ\text{C} $$

Sanity check. Adding resistances must reduce heat flow, and it does (778 W to 517 W). The inner surface is cooler than the room air, which is why people feel chilly sitting next to a cold wall even when the thermostat reads 22 °C. Similarly, the outer surface works out to \( 4 + 517 \times 0.003333 = 5.7 \) °C, slightly above outdoor air, as it should be.

Solved Problem 3: Radiation From a Steam Pipe

An uninsulated steel steam pipe has an outer diameter of 100 mm and surface temperature 150 °C, with emissivity 0.8. The surrounding walls are at 25 °C. Find the radiation loss per metre of pipe.

Step 1: Convert to kelvin. \( T_s = 423.15 \) K, \( T_{\text{surr}} = 298.15 \) K.

Step 2: Surface area per metre.

$$ A = \pi D L = \pi (0.100\ \text{m})(1\ \text{m}) = 0.3142\ \text{m}^2 $$

Step 3: Fourth powers.

$$ T_s^4 = 3.206 \times 10^{10}\ \text{K}^4, \qquad T_{\text{surr}}^4 = 0.790 \times 10^{10}\ \text{K}^4 $$

Step 4: Radiation loss.

$$ \dot{Q} = (0.8)(5.67 \times 10^{-8})(0.3142)(3.206 - 0.790) \times 10^{10} \approx 344\ \text{W} $$

Sanity check. Each metre radiates about as much as three or four old incandescent bulbs. Natural convection would add a similar amount, so a 50 m run of bare pipe could waste over 30 kW, which is exactly why plants lag their steam lines. Had we wrongly used °C, we would have got \( 150^4 - 25^4 \), a result nearly 50 times too small.

Choosing Which Mode Matters

• In solids, conduction dominates.
• At a solid-fluid surface below about 100 °C, convection usually dominates radiation.
• At high temperatures (furnaces, the Sun, combustion), radiation dominates because of the \( T^4 \) dependence.
• In vacuum, such as a satellite or a thermos flask, radiation is the only mode left.

Did You Know?

• A thermos flask defeats all three modes at once: vacuum stops conduction and convection, and silvered walls reflect radiation.
• Your body loses roughly half its heat by radiation when you sit quietly indoors.
• Joseph Fourier published his theory of heat conduction in 1822, and the mathematics he invented, Fourier series, now powers everything from MP3 files to MRI scanners.

Key Takeaways

• Conduction follows \( \dot{Q} = kA\Delta T/L \), convection \( \dot{Q} = hA\Delta T \) and radiation \( \dot{Q} = \varepsilon\sigma A(T_s^4 - T_{\text{surr}}^4) \).
• Thermal resistances in series add, turning multi-layer problems into simple arithmetic.
• Always use kelvin in radiation calculations.
• Check that surface temperatures lie between the two fluid temperatures.

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