Rate of Heat Transfer Calculator (Conduction)
Calculate the rate of heat flow through a material using Fourier's Law of Thermal Conduction.
Calculation Results
Temperature Difference (ΔT): 0 °C
Rate of Heat Transfer (Q): 0 Watts
Heat Flux (Q/A): 0 W/m²
How to Calculate Rate of Heat Transfer
Understanding how heat moves through materials is fundamental in engineering, construction, and physics. The rate of heat transfer quantifies the energy moving from a region of higher temperature to a region of lower temperature per unit of time. While heat transfer occurs via conduction, convection, and radiation, the most common calculation for solid materials is based on Conduction.
Fourier's Law of Heat Conduction
The calculation used in the tool above is based on Fourier's Law. This law states that the rate of heat transfer through a material is proportional to the negative gradient in the temperature and to the area, at right angles to that gradient, through which the heat flows.
Where:
- Q = Rate of heat transfer in Watts (W) or Joules/second.
- k = Thermal conductivity of the material (W/m·K). This measures how well the material conducts heat. High values (metals) conduct well; low values (insulation) resist heat flow.
- A = Surface area perpendicular to heat flow (m²).
- ΔT = Temperature difference between the hot and cold sides (T_hot – T_cold). Since the interval of 1 Kelvin equals 1 degree Celsius, you can use either scale for the difference.
- d = Thickness of the material (m).
Common Thermal Conductivity (k) Values
To get accurate results, you need the correct 'k' value for your material. Here are some standard approximations:
- Copper: ~400 W/(m·K)
- Aluminum: ~205 W/(m·K)
- Steel: ~50 W/(m·K)
- Glass: ~0.8 W/(m·K)
- Concrete: ~0.6 – 1.0 W/(m·K)
- Fiberglass Insulation: ~0.04 W/(m·K)
- Air (still): ~0.024 W/(m·K)
Step-by-Step Calculation Example
Imagine you want to calculate the heat loss through a single pane glass window.
- Identify Parameters:
Area (A) = 1.5 m²
Thickness (d) = 3mm (which is 0.003 m)
Thermal Conductivity of Glass (k) = 0.8 W/(m·K)
Inside Temp = 20°C, Outside Temp = 5°C. - Calculate ΔT: 20 – 5 = 15°C.
- Apply Formula: Q = (0.8 × 1.5 × 15) / 0.003
- Compute: (18) / 0.003 = 6000 Watts.
Note: This theoretical result is very high because it ignores the thin layer of air on the surface of the glass (convection), which adds significant resistance. In real-world engineering, R-values (thermal resistance) are often used to account for these boundary layers.
Why Thickness Matters
As seen in the formula, the rate of heat transfer is inversely proportional to thickness. If you double the thickness of an insulation layer, you cut the conductive heat transfer rate in half (assuming the area and temperature difference remain constant). This is the fundamental principle behind building insulation.