Thermal Resistance Calculator
Enter thickness, conductivity, area and temp difference to compute thermal resistance R=Δx/(kA), heat-flow rate Q=ΔT/R and conductance G=1/R. Copper k=401, A=1 cm², Δx=1 m → R≈24.94 K/W.
Input Data
Results
At a glance:Thermal resistance: in 1-D steady-state Fourier conduction, a material's ability to impede heat flow is R=Δx/(kA), units K/W (temperature drop per watt). Δx is thickness (m), k is thermal conductivity (W/(m·K)), A is cross-sectional area (m²). It is the thermal analog of electrical resistance: temperature drop ΔT ↔ voltage, heat flow Q ↔ current, R ↔ resistance, giving Ohm's-law form Q=ΔT/R. Thermal conductance G=1/R=kA/Δx ↔ electrical conductance. Series layers add R_total=R₁+R₂+…; parallel conductances add G_total=G₁+G₂+… . Classic example: copper (k=401), area 1 cm²=1e-4 m², thickness 1 m → R=1/(401×1e-4)≈24.94 K/W; polystyrene (k=0.033) 5 cm thick, 1 m² → R=0.05/(0.033×1)≈1.52 K/W — for the same resistance the insulation is only 1/12000 the thickness of copper. History: Fourier's law (1822), Ohm's analogy (1827), Carnot's heat-engine theory (1824). Applications: (1) building insulation — higher wall R-value insulates better (Hong Kong energy code sets wall R requirements); (2) electronics cooling — CPU heatsink R<0.3 K/W keeps the die below 85°C; sum thermal grease, heatsink and heat-pipe resistances; (3) insulation materials — vacuum flasks, fridges, cold-chain (polystyrene, polyurethane, vacuum panels); (4) process industries — heat exchangers, boilers; (5) spacecraft thermal control; (6) animal fur and fat layers. Note: R applies to steady 1-D conduction; transient needs the diffusion equation ρc(∂T/∂t)=k∇²T; multidimensional geometry needs FEA.
Formula
Resistance: R = Δx/(k·A) (K/W)
Heat flow: Q = ΔT/R = k·A·ΔT/Δx (W)
Conductance: G = 1/R = k·A/Δx (W/K)
Fourier's law: Q = −k·A·(dT/dx)
Series: R_total = R₁+R₂+…+R_n
Parallel: G_total = G₁+G₂+…+G_n
$$R = \frac{\Delta x}{kA}, \quad Q = \frac{\Delta T}{R} = \frac{kA \Delta T}{\Delta x}$$How to Use
- Enter thickness Δx (m), conductivity k (W/(m·K)), area A (m²) and ΔT (K).
- The tool computes R=Δx/(kA), Q=ΔT/R and G=1/R.
- Common: copper 1 cm² ×1 m → R=24.94 K/W; polystyrene 5 cm ×1 m² → R=1.52 K/W.
Case Studies
Hong Kong residential wall insulation
Summer indoor-outdoor ΔT=10 K, concrete wall k=1.7, Δx=0.15 m, A=10 m².
R=0.15/(1.7×10)=0.0088 K/W; Q=10/0.0088=1136 W (heavy A/C load).
Add 5 cm polystyrene R'=0.05/(0.033×10)=0.152 K/W; total R=0.161 K/W, Q'=62 W — 95% energy saved.
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.