7.01 x 4.09 x 0.71 inches; 0.07 ounces
TF8 2g-d6
April 26, 2022
Thermalright
B09YT51LV9
#67 in Silicon Grease
4.8 out of 5 stars (58 ratings)
Of course. The thermal conductivity value is given, but the thermal resistance is incomplete because it lacks critical dimensional information.
Thermal resistance depends on both the material’s property (conductivity, k) and its physical dimensions (thickness, L, and area, A).
The formula that links them is:
Thermal Resistance (R) = Thickness (L) / (Thermal Conductivity (k) × Area (A))
Or, more simply:
R = L / (k × A)
Units:
- R: Kelvin per Watt (K/W) or °C/W
- L: Meters (m)
- k: Watts per meter-Kelvin (W/m·K)
- A: Square Meters (m²)
Improved and Complete Statements
Here are several improved versions, depending on what you want to communicate.
Option 1: Defining the Material Property (Most Common)
This is best for specifying the material itself.
“The material has a thermal conductivity (k) of 13.8 W/m·K.”
Option 2: Providing a Sample Calculation
This is best for giving context. You must assume or state standard dimensions.
“For a material with a thermal conductivity (k) of 13.8 W/m·K and a thickness (L) of 0.01 m (1 cm), the thermal resistance per unit area would be:
R = L / k = 0.01 m / 13.8 W/m·K ≈ 0.000725 m²·K/W
To find the total thermal resistance for a specific area (e.g., A = 0.5 m²), use R_total = (L / k) / A = 0.000725 / 0.5 ≈ 0.00145 K/W.”
Option 3: Creating a General Formula
This is the most precise and technically correct way to present it.
“The thermal resistance (R) for this material is calculated using the formula:
R = L / (13.8 W/m·K × A)
where L is the thickness in meters (m) and A is the cross-sectional area in square meters (m²).”
Option 4: Stating Resistance for a Standard Component
This is used in electronics or engineering for specific, common parts.
“The silicone pad has a thermal conductivity of 13.8 W/m·K. For a standard pad size of 30x30mm (0.0009 m²) and a 1mm thickness (0.001 m), its thermal resistance is approximately:
R = 0.001 / (13.8 × 0.0009) ≈ 0.0805 °C/W.”
In summary:
You cannot state a thermal resistance value from conductivity alone. To complete the statement, you must include the thickness (L) and area (A).
Original, Incomplete Statement:
Thermal conductivity 13.8W/m.k Thermal resistance:
Corrected and Improved Statement:
Thermal conductivity (k): 13.8 W/m·K. Thermal resistance (R) is calculated as R = L / (k × A), where L is the material thickness and A is the cross-sectional area.
7.01 x 4.09 x 0.71 inches; 0.07 ounces
TF8 2g-d6
April 26, 2022
Thermalright
B09YT51LV9
#67 in Silicon Grease
4.8 out of 5 stars (58 ratings)