Conduction - Thermal Resistance

The opposition of a body to the flow of heat through it is called thermal resistance.

The conduction rate P_cond is the amount of energy transferred per unit time through a slab of area A and thickness x, where the two sides are at temperatures T₁ and T₂ (T₁ > T₂):

P_cond = \[\frac {Q}{t}\] = kA\[\frac {(T_1-T_2)}{x}\]

The value of k depends on the material of the slab. The graph below lists k for common materials.

In western countries, where temperature falls below 0 °C in winter, insulating the house from surroundings is very important. In our country, if we wish to carry cold drinks for a picnic or bring ice cream from a shop, we need to keep them in containers (made of thermocol, for example) that are poor thermal conductors.

Hence, the concept of thermal resistance, R_T, similar to electrical resistance, is introduced.

• The greater the thermal conductivity of a material, the smaller is its thermal resistance, and vice versa.
• Bad thermal conductors are those which have high thermal resistance.

Starting from Eq.,

\[\frac{T_1-T_2}{P_\mathrm{cond}}=\frac{x}{k\cdot A}\]

Analogy with Ohm's Law:

When current flows through a conductor, the ratio V/I is called the electrical resistance, where:

• V = electrical potential difference between the ends
• I = current (rate of flow of charge)

In Eq. 7.38, by direct analogy:

• (T₁ − T₂) plays the role of potential difference (temperature difference between the ends)

• P_cond plays the role of current (rate of flow of heat)

Therefore, T₁ − T₂ P_cond T₁ − T₂ is called the thermal resistance R_T:

\[{R_T=\frac{x}{k\cdot A}}\]

• SI Unit = °C s/kcal or °C s/J
• Dimensional Formula = [M⁻¹ L⁻² T³ K¹]

Problem: What is the rate of energy loss in watts per square metre through a glass window 5 mm thick if the outside temperature is −20 °C and the inside temperature is 25 °C?
($=1$ W/m·K)

Given

Solution

Step 1 — Temperature difference:

T₁ − T₂ = 25 − (−20) = 45 K

Step 2 — Formula for rate of energy loss per unit area:

\[\frac {P_cond}{A}\] = k ⋅ \[\frac {T_1−T_2}{x}\]

Step 3 — Substitute:

\[\frac {P_cond}{A}\] = 1 W m⁻¹K⁻¹ × \[\frac{45\mathrm{~K}}{5\times10^{-3}\mathrm{~m}}\]

Step 4 — Calculate

\[\frac{P_{\mathrm{cond}}}{A}=\frac{45}{5\times10^{-3}}\] = 9 × 10³ W/m²

• The lower the thermal conductivity kk, the higher the thermal resistance R_T.
• A material with high R_T is a poor thermal conductor and a good thermal insulator.
• Thermal resistivity ρ_T is the reciprocal of thermal conductivity k:
  ρ_T = \[\frac {1}{k}\]