Comparing insulation and energy-saving claims

Heat leaves a building by three paths: conduction through the wall material, convection in air gaps and pores, and radiation from surfaces. Insulation lowers conductivity (small $\lambda$), breaks up large convective cells (fibres, closed pores) and, with a foil facing, reduces radiation. To compare claims, convert everything to thermal resistance: for a layer of thickness $d$ and conductivity $\lambda$, $R = d/\lambda$; for a multilayer wall $R_{\mathrm{tot}} = R_{si} + \sum_i d_i/\lambda_i + R_{se}$, and the thermal transmittance is $U = 1/R_{\mathrm{tot}}$. Then the heat flow through area $A$ at temperature difference $\Delta T$ is $\Phi = U A \Delta T$, and the energy over time $\Delta t$ is $Q = \Phi\,\Delta t$. A claim such as "saves 30 % of heat" can only be checked against a baseline, an area, a climate and a $\Delta T$: without those, the percentage is not a measurable quantity.

A mineral-wool layer has thickness $d = 100$ mm and conductivity $\lambda = 0.040$ W/(m·K). Find the thermal resistance $R$ of the layer and the thermal transmittance $U = 1/R$ without surface resistances. What is this value in imperial units (R-value)?

A wall consists of brick $d_1 = 250$ mm ($\lambda_1 = 0.70$), insulation $d_2 = 150$ mm ($\lambda_2 = 0.035$) and plaster $d_3 = 15$ mm ($\lambda_3 = 0.80$); $\lambda$ in W/(m·K). The surface resistances are $R_{si} = 0.13$ and $R_{se} = 0.04$ m²·K/W. Find $R_{\mathrm{tot}}$ and $U$, then the heat flow and the daily energy loss through a wall of area $A = 100$ m² at $\Delta T = 30$ K.

A vendor claims: "Add 100 mm of insulation with $\lambda = 0.035$ and the heat loss through the wall will be halved." Check this for the wall from the previous example ($R_{\mathrm{tot}} = 4.83$ m²·K/W). What is the real reduction, and what thickness would truly halve the loss if thermal bridges and infiltration are neglected?

Test a claim in a spreadsheet (safe computer work). Open a spreadsheet and build a wall model step by step.

1. Create columns: layer, $d_i$ (m), $\lambda_i$ (W/(m·K)), $R_i = d_i/\lambda_i$.
2. Add rows for $R_{si} = 0.13$ and $R_{se} = 0.04$; compute $R_{\mathrm{tot}}$ and $U = 1/R_{\mathrm{tot}}$.
3. Set $A$, $\Delta T$ and $\Delta t$; compute $\Phi = U A \Delta T$ and $Q = \Phi\,\Delta t$.
4. Take any insulation advertisement, convert it into $\lambda$ and $d$, enter it into the model and compare the advertised percentage with your calculation.
5. Write down which data are missing (thermal bridges, infiltration, climate) — these define the limits of the claim.

Work only with numbers in the spreadsheet: no heating or cooling of samples is required.

Show that wall layers behave as series thermal resistances: the heat flow is the same in every layer, while the total temperature difference is divided among them.

Diagram: on the left, a cross-section of the wall with three layers labelled $d_i$, $\lambda_i$; on the right, the equivalent electrical circuit of series resistances $R_{si}$, $R_1$, $R_2$, $R_3$, $R_{se}$; an arrow for the heat flow $\Phi$ runs through the whole circuit, with the temperature profile from $T_i$ to $T_e$ plotted along it.

You compare two rolls of insulation for a (50\ \text{m}^2) attic. Roll A has thickness (100\ \text{mm}) and thermal conductivity (\lambda = 0.035\ \text{W/(m·K)}). Roll B has thickness (80\ \text{mm}) and (\lambda = 0.040\ \text{W/(m·K)}). Indoor temperature is (20\ ^\circ\text{C}), outdoor (0\ ^\circ\text{C}).

Compute thermal resistance (R = d/\lambda): for A, (R = 0.100/0.035 \approx 2.86\ \text{m}^2\text{K/W}); for B, (R = 0.080/0.040 = 2.00\ \text{m}^2\text{K/W}). Heat flow (P = A\Delta T/R) gives (P_A = 50\cdot20/2.86 \approx 350\ \text{W}), (P_B = 50\cdot20/2.00 = 500\ \text{W}). The claim 'A saves more' is true by (150\ \text{W}), but you must check both (\lambda) and thickness, not just one.

A jacket claim says it keeps you warm at (-10\ ^\circ\text{C}). The fabric has thickness (10\ \text{mm}), (\lambda = 0.040\ \text{W/(m·K)}), and covers (1.2\ \text{m}^2). Your skin is at (30\ ^\circ\text{C}).

Using conduction, (U = \lambda/d = 0.040/0.010 = 4.0\ \text{W/(m}^2\text{K)}). Heat loss (P = U A \Delta T = 4.0\cdot1.2\cdot40 = 192\ \text{W}). At rest your body produces about (100\ \text{W}), so the jacket alone cannot keep you warm; you need extra layers or higher activity. The claim ignores convection and radiation from the outer surface.

A thermos claims to keep (0.5\ \text{L}) of coffee at (80\ ^\circ\text{C}) for (6\ \text{hours}). The stopper is a plastic disk of thickness (2\ \text{cm}), diameter (5\ \text{cm}), (\lambda = 0.15\ \text{W/(m·K)}). Room temperature is (20\ ^\circ\text{C}).

Conduction through the stopper: area (A = \pi (0.025)^2 \approx 0.00196\ \text{m}^2), (U = 0.15/0.02 = 7.5\ \text{W/(m}^2\text{K)}), (P = 7.5\cdot0.00196\cdot60 \approx 0.88\ \text{W}). Energy lost in (6\ \text{h}): (Q = 0.88\cdot6\cdot3600 \approx 19\ \text{kJ}). For (0.5\ \text{kg}) water ((c = 4186\ \text{J/(kg·K)})), (\Delta T = 19000/(0.5\cdot4186) \approx 9\ ^\circ\text{C}), so coffee drops to about (71\ ^\circ\text{C}) — still hot. The claim is plausible only if radiation and convection through the vacuum are negligible.