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Physics HL · Chapter 7: Thermal Energy Transfers

7.4 Conduction and Convection

Model conduction through solids and convection in fluids, then use thermal-resistance thinking for insulation design.

Estimated time: 24 minutes

Microscopic Picture of Conduction

Conduction transfers thermal energy through a material without bulk transport of matter. In metals, mobile electrons carry a large part of this transfer. In all solids, lattice interactions also transmit energy through collisions and vibrations. The key driver is a temperature gradient: energy drifts from hotter regions to colder regions.

A high thermal conductivity material establishes large transfer rates for the same gradient and geometry, while insulators resist transfer. This is not about one side being absolutely hot or cold; it is about how effectively a material converts a given ΔT\Delta T into power flow.

Fourier Law in One Dimension

P=kAΔTLP = \frac{kA\Delta T}{L}

Steady one-dimensional conduction power scales with conductivity and area, and is inversely proportional to thickness.

This relation explains practical insulation decisions immediately. Doubling thickness halves conductive heat flow, all else equal. Increasing area increases total heat flow. Replacing a high-k layer with a low-k layer can reduce losses far more effectively than modest thickness changes in conductive materials.

Rth=LkA,P=ΔTRthR_{th} = \frac{L}{kA},\quad P = \frac{\Delta T}{R_{th}}

Thermal resistance form makes multilayer walls additive, like series resistors in circuits.

Convection as Bulk Fluid Motion

Convection requires a fluid and involves actual mass motion. Warmer, less-dense fluid rises while cooler, denser fluid sinks, setting up circulation cells that transport energy. Conduction still acts at boundaries and between neighboring fluid parcels, but convection adds organized bulk transport that can dominate in gases and liquids.

Natural convection is driven by buoyancy from density differences. Forced convection uses external drivers such as fans or pumps to strengthen circulation and increase transfer rate. Engineering systems often combine conduction through walls and convection in adjacent fluids, so separating pathways conceptually is crucial.

Simulation: Convection Cell (Rising Hot, Sinking Cool)

Watch buoyancy-driven circulation in a fluid cell and tune heater/cooling strength to see how convection loops carry thermal energy.

Convection Cell Lab

Heated fluid cell (rising hot fluid, sinking cool fluid)

Heater zoneCooling boundary

Mean thermal index

0.54

Particles rising

31

Particles sinking

41

Rising hot fluid and sinking cool fluid create circulation cells. Increase heater power to strengthen buoyancy-driven updrafts.

Insulation and Composite Walls

Most real boundaries are multilayer systems: structural material plus insulating layer plus interior and exterior convection effects. Treating each layer as thermal resistance clarifies design tradeoffs. The largest resistance segment controls most of the temperature drop, which is why adding low-k insulation can shift interface temperatures dramatically.

Simulation: Two-Layer Wall Conduction (Fourier Law)

Tune conductivity, thickness, area, and boundary temperatures to read heat flow and the piecewise temperature profile across each wall layer.

Heat Transfer + Radiation Lab

Composite wall and Fourier temperature profile

Total heat rate q = 20.27 WLayer 1 (k1 = 0.78)Layer 2 (k2 = 0.04)Tstart = 21.0 deg CTmiddle = 20.1 deg CTend = -4.0 deg C

Total thermal resistance

1.2332 K/W

Interface temperature

20.13 deg C

Total heat rate q

20.27 W

Heat flux magnitude

8.45 W/m^2

Move between conduction and radiation views to compare Fourier temperature-gradient transport with fourth-power thermal radiation behavior.

Test Yourself

A wall panel has k=0.80Wm1K1k = 0.80\,\text{W}\,\text{m}^{-1}\,\text{K}^{-1}, area A=3.0m2A = 3.0\,\text{m}^2, thickness L=0.10mL = 0.10\,\text{m}, and temperature difference ΔT=20K\Delta T = 20\,\text{K}. Enter conductive heat-transfer rate in watts.

Hint: Use P=kAΔTLP = \dfrac{kA\Delta T}{L}.