Training Conduction & Thermal Resistance Thermal Resistance Networks
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Thermal Resistance Networks

24 min Conduction & Thermal Resistance

Thermal Resistance Networks

Just as electrical circuits use Ohm’s law ($V = IR$), heat transfer uses a thermal analogy: $\dot{Q} = \Delta T / R_{\text{th}}$. This lets us solve composite walls and combined conduction/convection problems using simple series and parallel circuits.

Thermal Resistance (Conduction)

$$R_{\text{cond}} = \frac{L}{kA}$$

Units: K/W (or °C/W). This is the resistance to heat flow through a slab.

Thermal Resistance (Convection)

Newton’s law of cooling: $\dot{Q} = hA(T_s - T_\infty)$ gives:

$$R_{\text{conv}} = \frac{1}{hA}$$

$h$ = convection heat transfer coefficient (W/(m²·K)).

Series Resistance (Composite Wall)

$$R_{\text{total}} = R_1 + R_2 + R_3 + \cdots$$

$$\dot{Q} = \frac{T_{\text{hot}} - T_{\text{cold}}}{R_{\text{total}}}$$

Same $\dot{Q}$ flows through each layer. Use intermediate resistances to find interface temperatures.

Example 1 — Composite Wall

A wall has two layers: 10 cm brick ($k = 0.7$) and 5 cm insulation ($k = 0.04$). Area = 1 m². Inside = 25°C, outside = 0°C. Find $\dot{Q}$ and the interface temperature.

$R_{\text{brick}} = \frac{0.10}{0.7 \times 1} = 0.143$ K/W

$R_{\text{ins}} = \frac{0.05}{0.04 \times 1} = 1.25$ K/W

$R_{\text{total}} = 0.143 + 1.25 = 1.393$ K/W

$\dot{Q} = 25/1.393 = 17.9$ W

Interface: $T_{\text{int}} = 25 - \dot{Q} \times R_{\text{brick}} = 25 - 17.9 \times 0.143 = 22.4$°C

Notice: the insulation does 90% of the resisting!

Example 2 — With Convection Boundaries

Same wall as Example 1, but add convection: $h_{\text{in}} = 10$ W/(m²·K), $h_{\text{out}} = 25$ W/(m²·K). Find $\dot{Q}$.

$R_{\text{conv,in}} = 1/(10 \times 1) = 0.10$ K/W

$R_{\text{conv,out}} = 1/(25 \times 1) = 0.04$ K/W

$R_{\text{total}} = 0.10 + 0.143 + 1.25 + 0.04 = 1.533$ K/W

$\dot{Q} = 25 / 1.533 = 16.3$ W

Example 3 — Parallel Paths

A wall section has a steel beam ($k = 50$, $A = 0.01$ m², $L = 0.2$ m) in parallel with insulation ($k = 0.04$, $A = 0.99$ m², $L = 0.2$ m). $\Delta T = 30$°C. Find total $\dot{Q}$.

$R_{\text{steel}} = 0.2/(50 \times 0.01) = 0.40$ K/W

$R_{\text{ins}} = 0.2/(0.04 \times 0.99) = 5.05$ K/W

Parallel: $\frac{1}{R_{\text{eq}}} = \frac{1}{0.40} + \frac{1}{5.05} = 2.50 + 0.198 = 2.698$

$R_{\text{eq}} = 0.371$ K/W

$\dot{Q} = 30/0.371 = 80.9$ W

The steel beam (only 1% of area) carries $30/0.40 = 75$ W — a thermal bridge!

Practice Problems

1. Find $R_{\text{cond}}$ for 15 cm of concrete ($k = 1.4$), $A = 2$ m².
2. Three layers in series: $R_1 = 0.5$, $R_2 = 2.0$, $R_3 = 0.3$ K/W. $\Delta T = 40$°C. Find $\dot{Q}$.
3. Convection coefficient $h = 50$ W/(m²·K), $A = 3$ m². Find $R_{\text{conv}}$.
4. Two resistors in parallel: $R_1 = 2$ and $R_2 = 3$ K/W. Find $R_{\text{eq}}$.
5. A 3-layer wall: $R = 0.1 + 1.5 + 0.05$ K/W. Which layer controls the heat flow?
6. If $h$ increases from 10 to 50, does $R_{\text{conv}}$ increase or decrease? By what factor?
Show Answer Key

1. $R = 0.15/(1.4 \times 2) = 0.0536$ K/W

2. $R_{\text{total}} = 2.8$ K/W, $\dot{Q} = 40/2.8 = 14.3$ W

3. $R = 1/(50 \times 3) = 0.00667$ K/W

4. $R_{\text{eq}} = (2 \times 3)/(2 + 3) = 1.2$ K/W

5. The middle layer ($R = 1.5$) — it has the highest thermal resistance.

6. Decreases by a factor of 5 ($0.1 \to 0.02$).

Fourier's Law and Planar Conduction Radial Conduction — Pipes and Insulation