🗄️ Material Properties Database
Thermal and physical properties of metals, alloys, polymers, and fluids. Conductivity, density, specific heat, viscosity.
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| Material | Category | ρ [kg/m³] | Cp [J/kg·K] | k [W/m·K] | α [m²/s] | μ [Pa·s] |
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📝 Notes
- Properties are at 25°C (298 K) and 1 atm unless noted otherwise
- ρ = density, Cp = specific heat at constant pressure
- k = thermal conductivity, α = thermal diffusivity = k/(ρ·Cp)
- μ = dynamic viscosity (for fluids and gases only)
- Values are representative; actual properties vary with composition, temperature, and pressure
📘 Calculation Methodology: Solid Material Thermophysical Properties
Mathematical Model & Theory
Retrieves temperature-dependent thermal conductivity $k$, mass density $\rho$, specific heat capacity $c_p$, and thermal diffusivity $\alpha = k/(\rho c_p)$ for engineering metals, polymers, refractories, and insulation:
$$\alpha = \frac{k}{\rho c_p} \quad [\text{m}^2/\text{s}], \quad R'' = \frac{L}{k} \quad [\text{m}^2\cdot\text{K/W}], \quad C'' = \rho c_p L \quad [\text{J/m}^2\cdot\text{K}]$$
Assumptions
- Isotropic solid materials without microstructural voids or cracks.
- Temperature range within valid non-melting solid phase limits.
Academic References
- Touloukian, Y. S.: Thermophysical Properties of Matter (TPRC Data Series), Plenum.
- Incropera, F. P. et al.: Heat Transfer, Appendix A.
Worked Engineering Example
Problem Statement:
Calculate the thermal diffusivity $\alpha$ of pure copper at $300\text{ K}$ ($k = 401\text{ W/m}\cdot\text{K}$, $\rho = 8933\text{ kg/m}^3$, $c_p = 385\text{ J/kg}\cdot\text{K}$).
Step-by-step Solution:
1. $\alpha = \frac{401}{8933 \times 385} = \frac{401}{3,439,205} \approx 1.166 \times 10^{-4}\text{ m}^2/\text{s} = 116.6\text{ mm}^2/\text{s}$.
Final Result:
Thermal diffusivity of copper is $\alpha = \mathbf{1.166 \times 10^{-4}\text{ m}^2/\text{s}}$ ($\mathbf{116.6\text{ mm}^2/\text{s}}$).
Calculate the thermal diffusivity $\alpha$ of pure copper at $300\text{ K}$ ($k = 401\text{ W/m}\cdot\text{K}$, $\rho = 8933\text{ kg/m}^3$, $c_p = 385\text{ J/kg}\cdot\text{K}$).
Step-by-step Solution:
1. $\alpha = \frac{401}{8933 \times 385} = \frac{401}{3,439,205} \approx 1.166 \times 10^{-4}\text{ m}^2/\text{s} = 116.6\text{ mm}^2/\text{s}$.
Final Result:
Thermal diffusivity of copper is $\alpha = \mathbf{1.166 \times 10^{-4}\text{ m}^2/\text{s}}$ ($\mathbf{116.6\text{ mm}^2/\text{s}}$).