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Heat Transfer Mechanisms

Level 1 - Physics topic page in Temperature and Heat.

Principle

Heat transfer can occur by conduction, convection, or radiation.

The mechanism matters because each one depends on different physical quantities: temperature gradient, fluid motion, or electromagnetic emission.

Notation

\(H\)
heat current, the rate of heat transfer
\(\mathrm{W}\)
\(k\)
thermal conductivity
\(\mathrm{W\,m^{-1}\,K^{-1}}\)
\(A\)
area through which heat is transferred
\(\mathrm{m^{2}}\)
\(L\)
thickness or conduction length
\(\mathrm{m}\)
\(T_H,T_C\)
hot-side and cold-side temperatures
\(\mathrm{K}\)
\(h_c\)
empirical convection coefficient
\(\mathrm{W\,m^{-2}\,K^{-1}}\)
\(e\)
emissivity
1
\(\sigma\)
Stefan-Boltzmann constant
\(\mathrm{W\,m^{-2}\,K^{-4}}\)

Method

Conduction is energy transfer through matter without bulk motion. In a uniform slab at steady state, the temperature gradient is approximately \((T_H-T_C)/L\), so Fourier's law gives a heat current proportional to area and conductivity.

Temperature gradient
\[\left|\frac{dT}{dx}\right|\approx\frac{T_H-T_C}{L}\]
Conduction current
\[H=kA\frac{T_H-T_C}{L}\]
Thermal resistance
\[R=\frac{L}{kA}\]
Resistance form
\[H=\frac{T_H-T_C}{R}\]

Convection transfers energy by bulk motion of a fluid. The detailed flow can be complicated, so simple convection laws use an empirical coefficient.

Convection model
\[H=h_cA(T_s-T_f)\]
\(h_c\) depends on the fluid, surface shape, flow speed, and whether convection is natural or forced.

Radiation transfers energy by electromagnetic waves and does not require matter between the emitter and absorber. Absolute temperature is essential because the power scales with the fourth power of temperature.

Emitted power
\[H_{\mathrm{emit}}=e\sigma AT^4\]
Absorbed surroundings
\[H_{\mathrm{abs}}=e\sigma AT_s^4\]
Net radiative loss
\[H_{\mathrm{net}}=e\sigma A(T^4-T_s^4)\]

Rules

These are the compact heat-transfer relations.

Conduction
\[H=kA\frac{T_H-T_C}{L}\]
Thermal resistance
\[R=\frac{L}{kA}\]
Convection model
\[H=h_cA(T_s-T_f)\]
Radiation emitted
\[H=e\sigma AT^4\]
Net radiation
\[H_{\mathrm{net}}=e\sigma A(T^4-T_s^4)\]

Examples

Question
A wall panel has area
\[2.0\,\mathrm{m^2}\]
thickness
\[0.040\,\mathrm{m}\]
and thermal conductivity
\[0.080\,\mathrm{W\,m^{-1}\,K^{-1}}\]
The two sides differ by
\[18\,\mathrm{K}\]
Find the conduction heat current.
Answer
\[H=kA\frac{\Delta T}{L}=0.080(2.0)\frac{18}{0.040}=72\,\mathrm{W}\]

Checks

  • Conduction heat current increases with area and decreases with thickness.
  • Convection requires fluid motion; conduction does not.
  • Radiation calculations must use kelvins because of the \(T^4\) dependence.
  • A shiny low-emissivity surface is both a poor emitter and a poor absorber.

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Equations of State