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Layers and U-values

Heat and matter · level 7, lesson 2 of 6

A question to start

A real habitat wall is a sandwich: a steel skin to hold the air in, a thick layer of foam, and a wooden panel inside. How do you find the heat loss through all three at once?

The idea

Each layer resists heat flow. For a square metre of a layer of thickness \(d\), its thermal resistance is

\[R = \frac{d}{k}, \quad \text{in m}^2\text{ K/W}.\]

Layers that heat must pass through one after another simply add, like resistors in series: \(R_\text{total} = R_1 + R_2 + R_3\). The U-value is the heat flow per square metre for each kelvin of difference:

\[U = \frac{1}{R_\text{total}}, \qquad P = UA\,\Delta T.\]
outsideinsidesteel 5 mmfoam 15 cmwood 2 cm
Layer \(d\) \(k\) \(R = d/k\)
Steel skin 0.005 m 50 0.0001
Foam 0.15 m 0.022 6.82
Wood panel 0.02 m 0.13 0.15
Total 6.97

So \(U = 1/6.97 = 0.14\) W/(m² K). The foam does almost all the work; the steel hardly counts.

100 m² of this wall Mars Earth
\(\Delta T\), inside 20 °C 79 K 5 K
Heat lost, \(UA\Delta T\) about 1,130 W about 70 W

Worked examples

1. The wall above. \(P = 0.14 \times 100 \times 79 = 1{,}130\) W.

2. Doubling the foam. \(R = 0.0001 + 13.64 + 0.15 = 13.8\), \(U = 0.072\): the loss roughly halves, to about 570 W.

3. A thermal bridge. Suppose a steel bolt runs straight through the foam. It lets heat bypass the foam. Engineers avoid such thermal bridges, because a small area of metal can leak as much as a large area of wall.

Try it

  1. Find \(R\) for 0.1 m of mineral wool (\(k = 0.04\)).
  2. A wall has layers with \(R\) = 0.2, 3.0 and 0.3. Find its U-value.
  3. Find the heat lost through 50 m² of that wall with 60 K across it.
  4. Why does the steel skin add so little resistance?
Answers
  1. \(0.1/0.04 = 2.5\) m² K/W.
  2. \(R = 3.5\), so \(U = 1/3.5 = 0.29\) W/(m² K).
  3. \(0.29 \times 50 \times 60 = 860\) W.
  4. Steel conducts heat very well and the skin is thin, so \(d/k\) is tiny.

Common mistake

Adding the U-values. For layers one after another, add the resistances, then take \(1/R\). Adding U-values would make a thicker wall seem to lose more heat.

Where it leads