🌡️ Natural Convection Calculator
Analyze buoyancy-driven heat transfer for vertical and horizontal plates, horizontal cylinders, and spheres using Rayleigh and Grashof correlations.
🔥 Thermal Plume & Buoyancy Schematic
📝 Configuration
Key Equations:
Rayleigh Number:
Ra = g·β·(T_s - T∞)·Lc³·Pr / ν²
Grashof Number:
Gr = Ra / Pr
Vertical Plate (Churchill-Chu):
Nu = [0.825 + 0.387 Ra^(1/6) / [1 + (0.492/Pr)^(9/16)]^(8/27)]²
Rayleigh Number:
Ra = g·β·(T_s - T∞)·Lc³·Pr / ν²
Grashof Number:
Gr = Ra / Pr
Vertical Plate (Churchill-Chu):
Nu = [0.825 + 0.387 Ra^(1/6) / [1 + (0.492/Pr)^(9/16)]^(8/27)]²
📊 Results & Visualization
Configure inputs and click Calculate to view results.
ℹ️ About Natural Convection
Natural convection occurs when fluid motion is generated solely by buoyancy forces, resulting from temperature-driven density gradients within the fluid.
**Rayleigh Number (Ra)** determines the flow regime: - $Ra < 10^9$: Laminar buoyancy boundary layer. - $Ra \ge 10^9$: Turbulent buoyancy boundary layer.
Natural convection occurs when fluid motion is generated solely by buoyancy forces, resulting from temperature-driven density gradients within the fluid.
**Rayleigh Number (Ra)** determines the flow regime: - $Ra < 10^9$: Laminar buoyancy boundary layer. - $Ra \ge 10^9$: Turbulent buoyancy boundary layer.
📘 Calculation Methodology
Mathematical Model & Theory
Natural convection heat transfer is driven by buoyancy forces. The strength of the natural convection is measured by the Rayleigh number ($Ra_L$), which dictates the Nusselt number ($Nu_L$) via empirical relations (like the Churchill and Chu correlation for vertical plates):
$$Ra_L = Gr_L \cdot Pr = \frac{g \beta (T_s - T_{\infty}) L^3}{
u^2} \cdot Pr$$
$$Nu_L = \left\{0.825 + \frac{0.387 Ra_L^{1/6}}{[1 + (0.492/Pr)^{9/16}]^{8/27}}\right\}^2$$
Worked Engineering Example
Problem Statement:
A vertical plate 0.3 m high is maintained at 60°C in ambient air at 20°C. Air properties: $\rho = 1.13$ kg/m³, $\mu = 1.92 \times 10^{-5}$ Pa·s, $\beta = 0.0032$ K⁻¹, $Pr = 0.72$, $k = 0.026$ W/m·K. Calculate the convective heat transfer coefficient.
Step-by-step Solution:
1. Calculate kinematic viscosity: $ u = \mu / \rho = 1.70 \times 10^{-5}$ m²/s.
2. Calculate Rayleigh number $Ra_L$:
$$Ra_L = \frac{9.81 \times 0.0032 \times (60 - 20) \times 0.3^3}{(1.70 \times 10^{-5})^2} \times 0.72 = 8.45 \times 10^7$$ 3. Calculate Nusselt number $Nu_L$:
$$Nu_L = \left\{0.825 + \frac{0.387 \times (8.45 \times 10^7)^{1/6}}{[1 + (0.492/0.72)^{9/16}]^{8/27}}\right\}^2 = 56.45$$ 4. Calculate convection coefficient $h_c$:
$$h_c = \frac{Nu_L \cdot k}{L} = \frac{56.45 \times 0.026}{0.3} = 4.89 \text{ W/m}^2\text{K}$$
Final Result:
The average heat transfer coefficient is 4.89 W/m²·K.
A vertical plate 0.3 m high is maintained at 60°C in ambient air at 20°C. Air properties: $\rho = 1.13$ kg/m³, $\mu = 1.92 \times 10^{-5}$ Pa·s, $\beta = 0.0032$ K⁻¹, $Pr = 0.72$, $k = 0.026$ W/m·K. Calculate the convective heat transfer coefficient.
Step-by-step Solution:
1. Calculate kinematic viscosity: $ u = \mu / \rho = 1.70 \times 10^{-5}$ m²/s.
2. Calculate Rayleigh number $Ra_L$:
$$Ra_L = \frac{9.81 \times 0.0032 \times (60 - 20) \times 0.3^3}{(1.70 \times 10^{-5})^2} \times 0.72 = 8.45 \times 10^7$$ 3. Calculate Nusselt number $Nu_L$:
$$Nu_L = \left\{0.825 + \frac{0.387 \times (8.45 \times 10^7)^{1/6}}{[1 + (0.492/0.72)^{9/16}]^{8/27}}\right\}^2 = 56.45$$ 4. Calculate convection coefficient $h_c$:
$$h_c = \frac{Nu_L \cdot k}{L} = \frac{56.45 \times 0.026}{0.3} = 4.89 \text{ W/m}^2\text{K}$$
Final Result:
The average heat transfer coefficient is 4.89 W/m²·K.