💧 Fluid Properties vs. Temperature
Lookup and interpolate temperature-dependent density, viscosity, specific heat, thermal conductivity, Prandtl number, and expansion coefficient for 20+ fluids.
⚡ Fortran 90 Engine
Double Precision (IEEE 754)
✓ ISO / ASME Validated
📊 Solver Telemetry
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📦 Code Fortran
11 KB
📅 Mise en service
Jun 2026
⏱️ Latence
< 1 ms
📝 Input Selection
📊 Fluid Outputs
Density ($\rho$)
—
kg/m³
Specific Heat ($C_p$)
—
J/(kg·K)
Thermal Conductivity ($k$)
—
W/(m·K)
Dynamic Viscosity ($\mu$)
—
Pa·s
Kinematic Viscosity ($\nu$)
—
m²/s
Prandtl Number ($Pr$)
—
dimensionless
Thermal Diffusivity ($\alpha$)
—
m²/s
Expansion Coeff. ($\beta$)
—
1/K
📈 Property Curves vs. Temperature
Interpolation Model:
Calculated using high-precision linear spline interpolation over tabulated database data points sourced from Incropera & DeWitt (Appendix A) and Cengel.
Expansion Coefficient:
• Gas: $\beta = 1/T_K$ (ideal gas law)
• Liquid: numerical derivative $\beta = - \frac{1}{\rho}\frac{\partial\rho}{\partial T}$
Calculated using high-precision linear spline interpolation over tabulated database data points sourced from Incropera & DeWitt (Appendix A) and Cengel.
Expansion Coefficient:
• Gas: $\beta = 1/T_K$ (ideal gas law)
• Liquid: numerical derivative $\beta = - \frac{1}{\rho}\frac{\partial\rho}{\partial T}$
📘 Calculation Methodology: Thermophysical Fluid Property Formulations
Mathematical Model & Theory
Calculates temperature- and pressure-dependent density $\rho(T)$, dynamic viscosity $\mu(T)$, thermal conductivity $k(T)$, and specific heat $c_p(T)$ using high-accuracy polynomial and IAPWS correlations:
$$\rho(T) = \sum_{i=0}^n a_i T^i, \quad \mu(T) = A \exp\left(\frac{B}{T + C}\right) \quad (\text{Vogel-Fulcher-Tammann})$$
$$k(T) = \sum_{i=0}^m b_i T^i, \quad c_p(T) = \sum_{i=0}^k c_i T^i$$
Assumptions
- Single-phase liquid or gas within calibrated temperature bounds.
- Standard atmospheric pressure reference unless high-pressure EOS is selected.
Academic References
- IAPWS R7-97: Industrial Formulation for the Properties of Water and Steam.
- Yaws, C. L.: Thermophysical Properties of Chemicals and Hydrocarbons.
Worked Engineering Example
Problem Statement:
Evaluate the kinematic viscosity $\nu$ and Prandtl number $Pr$ of liquid water at $T = 60^\circ\text{C}$ ($\rho = 983.2\text{ kg/m}^3$, $\mu = 4.66 \times 10^{-4}\text{ Pa}\cdot\text{s}$, $c_p = 4185\text{ J/kg}\cdot\text{K}$, $k = 0.654\text{ W/m}\cdot\text{K}$).
Step-by-step Solution:
1. $\nu = \mu / \rho = 4.66 \times 10^{-4} / 983.2 \approx 4.74 \times 10^{-7}\text{ m}^2/\text{s} = 0.474\text{ cSt}$.
2. $Pr = (\mu c_p) / k = (4.66 \times 10^{-4} \times 4185) / 0.654 = 1.950 / 0.654 \approx 2.98$.
Final Result:
Kinematic viscosity is $\mathbf{0.474\text{ cSt}}$ and Prandtl number is $\mathbf{2.98}$.
Evaluate the kinematic viscosity $\nu$ and Prandtl number $Pr$ of liquid water at $T = 60^\circ\text{C}$ ($\rho = 983.2\text{ kg/m}^3$, $\mu = 4.66 \times 10^{-4}\text{ Pa}\cdot\text{s}$, $c_p = 4185\text{ J/kg}\cdot\text{K}$, $k = 0.654\text{ W/m}\cdot\text{K}$).
Step-by-step Solution:
1. $\nu = \mu / \rho = 4.66 \times 10^{-4} / 983.2 \approx 4.74 \times 10^{-7}\text{ m}^2/\text{s} = 0.474\text{ cSt}$.
2. $Pr = (\mu c_p) / k = (4.66 \times 10^{-4} \times 4185) / 0.654 = 1.950 / 0.654 \approx 2.98$.
Final Result:
Kinematic viscosity is $\mathbf{0.474\text{ cSt}}$ and Prandtl number is $\mathbf{2.98}$.