Thermal Conductivity Calculator:
Enter the values of thermal conductivity, k(W/m·K), cross-sectional area, A(m) and temperature gradient (ΔT(K) / Δx(m)) to determine the value of Heat transfer rate, q(W).
Thermal Conductivity Formula:
Thermal conductivity is a material property that indicates its ability to conduct heat. It quantifies how effectively heat energy is transmitted through a material due to a temperature gradient. High thermal conductivity materials, like metals, allow heat to flow quickly, while low thermal conductivity materials, like insulation, resist heat transfer.
This property is critical in applications involving heat management, such as in construction, electronics, and thermodynamics. The rate of heat transfer depends on the material’s thermal conductivity, the temperature difference across the material, and the material’s dimensions.
The formula for heat transfer rate considers these factors, expressed as q(W) = – k(W/m·K) * A(m) * (ΔT(K) / Δx(m)), where q(W) represents the heat transfer rate, k(W/m·K) is the thermal conductivity, A(m) is the cross-sectional area, ΔT(K)is the temperature difference, and Δx(m)is the material thickness. The negative sign indicates heat flows from high to low temperature.
Thermal conductivity is commonly measured in W/m·K watts per metre per kelvin. The property is vital in designing systems for heating, cooling, and thermal insulation. Engineers often optimize materials and structures to balance heat conduction and resistance as per specific requirements.
Heat transfer rate, q(W) in Watts equals the product of thermal conductivity, k(W/m·K) in Watts per metre Kelvin, the cross-sectional area, A(m) in square metres and the temperature gradient (ΔT(K) / Δx(m)) in kelvin per metre, with a negative sign indicating the direction of heat flow.
Heat transfer rate, q(W) = – k(W/m·K) * A(m) * (ΔT(K) / Δx(m))
q(W) = heat transfer rate in Watts, W.
k(W/m·K) = thermal conductivity in Watts per metre Kelvin, W/m.K.
A(m) = cross-sectional area in metres, m.
ΔT(K) = temperature difference in Kelvin, K.
Δx(m) = material thickness in metres, m
Thermal Conductivity Calculation:
1.A wall with a cross-sectional area of 2 m2, thermal conductivity 0.8W/m.K, and thickness 0.05m separates two environments with a temperature difference of 30K.
Given: k(W/m·K) = 0.8W/m.K, A(m) = 2m2, ΔT(K) = 30K, Δx(m) = 0.05m.
Heat transfer rate, q(W) = – k(W/m·K) * A(m) * (ΔT(K) / Δx(m))
q(W) = – 0.8 *2 * (30 / 0.05)
q(W) = – 0.8 * 2 * 600
q(W) = – 960W(negative sign indicates heat direction).
2.A material with thermal conductivity 0.5W/m.K, cross-sectional area of 1.5m², and thickness 0.02m transfers heat at a rate of 150W.
Given: k(W/m·K) = 0.5W/m.K, A(m) = 1.5m2, q(W) = 150W, Δx(m) = 0.02m.
Heat transfer rate, q(W) = – k(W/m·K) * A(m) * (ΔT(K) / Δx(m))
ΔT(K) = – (q(W) * Δx(m) / k(W/m·K) * A(m))
ΔT(K) = – (150 * 0.02 / 0.5 * 1.5)
ΔT(K) = – (3 / 0.75)
ΔT(K) = – 4K(ignoring the negative sign for absolute difference).