Conduction, thermal resistance and material properties

Heat rate differs from heat amount

Conduction transfers energy through matter without requiring bulk motion of the material. Heat flows down a temperature gradient. For a flat slab with constant conductivity k, area A and thickness L in steady one-dimensional conduction, the magnitude is Q̇=kA(Th−Tc)/L. Q̇ is a rate in watts, while energy transferred during a constant-rate interval is Q=Q̇t in joules.

Conductivity has units W/(m·K). It is a material property under stated temperature, direction and structural conditions, not a property of a whole assembled wall. The minus sign in the general gradient form Q̇=−kA dT/dx expresses flow toward decreasing T. For the simple slab, specifying hot and cold temperatures makes the direction explicit without calculus.

Add the relevant resistances

Define slab resistance R=L/(kA), measured in K/W, so Q̇=ΔT/R. Series layers carry the same steady heat rate when side losses and internal generation are absent. Their resistances add. Parallel paths share boundary temperature differences and their rates add; a conducting bridge can bypass an otherwise resistive layer.

Real contacts may have a temperature difference between the two nominal touching surfaces. An interface resistance Rc=ΔTcontact/Q̇ represents that unresolved contact behavior. Its value depends on surface and assembly conditions. It is different from a material conductivity. State whether a quoted resistance is for the complete component in K/W or normalised by area in m²·K/W.

Match the temperature boundary to the model

The plotted example uses specified solid-face temperatures, not distant air temperatures. If surrounding fluid temperatures are supplied instead, include convection between fluid and each face. A finite material with ongoing heat generation may have a curved temperature profile; the simple constant-rate layered model then needs revision.

A high conductivity layer can have a very small temperature drop even when another part of the same route has a large one. Adding insulation reduces heat transfer under fixed boundary temperatures. If a component generates a fixed power that must escape, increasing resistance can instead raise its temperature. The boundary condition decides which consequence the resistance change predicts.

Find the rate and each temperature drop

Use fictional layers of area0.02 m². R1=0.010/(0.2×0.02)=2.5 K/W and R2=0.002/(20×0.02)=0.005 K/W. Add a supplied contact resistance0.495 K/W to obtain Rtotal=3 K/W. With faces at 80 and 20 °C, Q̇=60/3=20 W.

The first-layer drop is 20×2.5=50 K, contact drop9.9 K and second-layer drop0.1 K. Temperatures are80,30,20.1 and 20 °C along the path. The vertical graph segment represents a contact model, not an invented finite-thickness layer. Over60 s at this steady rate,1200 J crosses each complete section.

Try the reasoning

1. Does doubling slab thickness double its conductivity?

   Show the answer and reasoning

   No. It doubles slab resistance in this model; conductivity remains the stated material property.

2. Why cannot distant air temperatures replace the specified face temperatures without another term?

   Show the answer and reasoning

   There can be temperature differences across the fluid-side convection layers. Their resistances must be included.

3. Does adding insulation always lower the temperature of a heat-generating component?

   Show the answer and reasoning

   No. Under fixed generated power, greater escape resistance can raise the component temperature.

What to carry forward

Follow the whole heat path and its boundary conditions. Conductivity, component resistance and contact resistance are different descriptions.