Thermal Conductivity (k) of Materials and Its Impact on Insulation

The real secret behind keeping our buildings warm in winter and cool in summer isn't just thick walls, but correctly chosen insulation materials. The most fundamental physical property that determines a material's insulation performance is its thermal conductivity (k) coefficient. Often denoted by the symbol "k" (or "lambda, λ") in engineering calculations, this value is a measure of how well (or how poorly) the microscopic structure of the material conducts thermal energy.

In this guide, we will explore what thermal conductivity is, compare the k-values of common construction and insulation materials, and examine how these values affect the total thermal resistance. Moreover, through simulations using the Thermal Resistance Calculator tool, we will see practically how changing the material type can make a massive difference in insulation performance.

What is Thermal Conductivity (k)?

Thermal conductivity (k) is a thermophysical property that expresses how easily heat passes through a material. According to the International System of Units (SI), its unit is W/(m·K) (Watts per meter-Kelvin).

In physical terms, the k-value is the amount of heat (Joules/second = Watts) that passes through a material 1 meter thick and 1 square meter in surface area in 1 second, when there is a temperature difference of 1 Kelvin (or 1°C) between its two surfaces.

• Materials with High k-Values: They conduct heat very quickly. Metals (copper, aluminum, steel) are the best examples. These materials are not used for insulation, but rather in places where we want to dissipate heat rapidly (such as processor heatsinks or radiators).
• Materials with Low k-Values: They offer high resistance to heat flow. Still air conducts heat very poorly. Most insulation materials (styrofoam, fiberglass, rockwool) have very low k-values thanks to the still air pockets trapped inside them.

Comparison of k-Values of Common Materials

Below are the approximate thermal conductivity (k) values in W/(m·K) for some common materials encountered in daily life and the construction industry:

• Copper: ~385.0 (Excellent conductor)
• Aluminum: ~205.0 (Good conductor)
• Steel: ~40.0 - 50.0
• Concrete: ~1.4 - 1.7 (Poor insulator)
• Glass: ~0.9 - 1.05
• Standard Brick: ~0.6 - 0.8
• Wood (Pine): ~0.12 - 0.15
• Rockwool / Fiberglass: ~0.035 - 0.040 (Good insulator)
• EPS / XPS (Foam): ~0.030 - 0.035 (Good insulator)
• Polyurethane Foam (PUR): ~0.022 - 0.028 (Excellent insulator)

As can be seen, there is a difference of about 50-70 times in thermal conductivity between a concrete wall and polyurethane foam! This means that a thin insulation board can block more heat than a concrete wall that is several meters thick.

The Contribution of a Low k-Value to a High R-Value

Thermal resistance (commonly known as R-value) is essentially the mathematical inverse of thermal conductivity. In a plane wall model, the thermal resistance per unit area is found using the following formula:

$R'' = \frac{L}{k}$

Here, $R''$ represents the resistance per unit area (m²·K/W), $L$ is the material thickness (m), and $k$ is the thermal conductivity (W/(m·K)). As is very clear from the formula, as the $k$ value decreases, the resulting $R$ value of the division will increase. Thus, the smaller the k, the stronger the insulation (the higher the R).

Practical Variations with the Calculation Tool

Let's test the theory above with a practical problem. We want to build a wall $L = 0.10$ m (10 cm) thick and $A = 20$ m² in size using two different materials. What is the difference in total thermal resistance ($R_{th}$) between standard brick and EPS foam?

Our total thermal resistance formula: $R_{th} = \frac{L}{k \times A}$

Let's open our Thermal Resistance Calculator tool and enter the scenarios:

Scenario 1: 10 cm Thick Brick Wall

• L = 0.10 m
• k = 0.70 W/(m·K) (Average value for brick)
• A = 20 m²
• When you enter these values into the tool, the Total Thermal Resistance ($R_{th}$) will come out as 0.0071 K/W.
• The Unit Area R-Value will be 0.1429 m²·K/W.

Scenario 2: 10 cm Thick EPS Foam Insulation

• L = 0.10 m
• k = 0.035 W/(m·K) (Average value for EPS)
• A = 20 m²
• When you update these values in the tool and calculate, the Total Thermal Resistance ($R_{th}$) will come out as 0.1429 K/W.
• The Unit Area R-Value will be 2.8571 m²·K/W.

Comparison Result: Even though their thicknesses and areas are exactly the same, because the k-value dropped from 0.70 to 0.035 (a 20-fold decrease), the thermal resistance of EPS foam is exactly 20 times higher than the resistance of brick. The mathematical truth of why trying to heat homes or keep the inside warm using only brick or concrete is a massive waste of energy is hidden in this fact.

Is the K-Value Always Constant?

In engineering calculations, assuming the k-value as a constant number (as in our tool) is sufficient for most standard applications and simplifies matters. However, in advanced building physics analyses, it should be remembered that the thermal conductivity of a material can vary depending on the following factors:

1. Moisture Content: Water is a much better heat conductor than air. If a porous material like fiberglass or brick gets wet, the air pockets inside fill with water. This rapidly increases the material's k-value and causes it to lose its insulating properties to a large extent. Therefore, waterproofing and thermal insulation are an inseparable pair.
2. Temperature: The k-value of most insulation materials tends to increase slightly as the temperature rises. (For example, the k-value measured at 10°C and the k-value measured at 50°C show minor differences). However, this difference is generally ignored in standard building applications.
3. Density: The density (kg/m³) of the material during production directly affects the k-value. In very low-density materials, heat transfer by radiation increases, while in very high-density ones, solid conduction increases. There is an "optimum density" level for every insulation material.

In summary, merely thickening the wall is not enough to make the right decisions in your insulation projects. You must know the k-value of the material used. By utilizing our Thermal Resistance Calculator tool, you can input the (k) values from the technical data sheets obtained from material suppliers, test the most ideal insulation thickness for your home or project yourself, and optimize your energy costs.