House Wall Heat Conduction Calculator
Results
Heat Flow Rate (Q): 0 Watts
Understanding Heat Conduction through House Walls
Heat conduction is the process by which thermal energy is transferred through a solid material from a region of high temperature to a region of lower temperature. In the context of a home, this represents the heat escaping through your walls during winter or entering your home during a hot summer day.
The Physics: Fourier's Law
The rate of heat transfer is governed by Fourier's Law of Heat Conduction. The formula used in this calculator is:
Q = (k × A × ΔT) / d
- Q: The rate of heat flow (measured in Watts).
- k: Thermal conductivity of the material (W/m·K). This measures how well a material conducts heat.
- A: The total surface area of the wall (m²).
- ΔT: The temperature difference between the inside and outside (°C or K).
- d: The thickness of the wall material (meters).
Common Material Conductivity Values (k)
To use the calculator accurately, you need the thermal conductivity (k-value) of your wall material. Typical values include:
| Material | Conductivity (W/m·K) |
|---|---|
| Common Brick | 0.60 – 1.00 |
| Concrete | 1.00 – 1.70 |
| Softwood | 0.12 – 0.15 |
| Fiberglass Insulation | 0.04 |
| Expanded Polystyrene (EPS) | 0.033 |
How to Reduce Heat Loss
Looking at the formula, there are three primary ways to reduce the energy lost through your walls:
- Lower the Conductivity (k): Replace air gaps or solid materials with high-performance insulation. The lower the k-value, the better the insulator.
- Increase Thickness (d): This is why adding thicker layers of insulation or building thicker walls helps maintain internal temperatures.
- Manage Temperature Difference (ΔT): While you can't control the weather, using smart thermostats to avoid over-heating or over-cooling can reduce the delta, thereby slowing heat transfer.
Example Calculation
If you have a brick wall (k = 0.7) with an area of 25m², a thickness of 0.3m (30cm), an indoor temperature of 20°C, and an outdoor temperature of 0°C:
Q = (0.7 × 25 × 20) / 0.3 = 1,166.67 Watts
This means your heating system needs to provide roughly 1.17 kW of power just to offset the heat lost through that single wall to maintain the temperature.