📉 Free Convection: Inclined & Horizontal Plates
Natural convection heat transfer on vertical, horizontal, and inclined flat plates with g*cos(theta) buoyancy scaling.
Heat Transfer
Parameters Setup
Results & Curves
Configure and run the calculator to see the computed results and temperature difference sweep curve.
Calculation Methodology
Mathematical Model & Theory
Natural convection on inclined flat plates is calculated using Churchill & Chu's vertical plate correlation with modified parallel gravity: $$g_{eff} = g \cos \theta$$ $$\text{Nu}_L = \left\{0.825 + \frac{0.387 \text{Ra}_L^{1/6}}{\left[1 + (0.492/\text{Pr})^{9/16}\right]^{8/27}}\right\}^2$$ For horizontal plates ($\theta = 90^\circ$), characteristic length is defined as: $$L_c = \frac{\text{Area}}{\text{Perimeter}} = \frac{L \cdot W}{2(L+W)}$$ Correlations for upper surface of hot plate / lower surface of cold plate: $$\text{Nu} = 0.54 \text{Ra}^{1/4} \quad (\text{Ra} < 10^7)$$ $$\text{Nu} = 0.15 \text{Ra}^{1/3} \quad (\text{Ra} \ge 10^7)$$ Correlations for lower surface of hot plate / upper surface of cold plate: $$\text{Nu} = 0.27 \text{Ra}^{1/4}$$
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. 7th Edition, John Wiley & Sons.
- Churchill, S. W., & Chu, H. S. (1975). Correlating Equations for Laminar and Turbulent Free Convection on a Vertical Plate. Int. J. Heat Mass Transfer.
Worked Engineering Example
A $0.5\text{ m} \times 0.3\text{ m}$ flat plate tilted at $30^\circ$ from vertical is maintained at $80^\circ\text{C}$ in ambient air at $25^\circ\text{C}$. Find the convection coefficient ($T_f = 52.5^\circ\text{C}$).
Step-by-step Solution:
1. Evaluate air properties at film temperature:
- $\nu = 1.83 \times 10^{-5}\text{ m}^2\text{/s}, \alpha = 2.6 \times 10^{-5}\text{ m}^2\text{/s}, k = 0.028\text{ W/m·K}, Pr = 0.71, \beta = 1/325.65 = 0.00307\text{ K}^{-1}$.
2. Calculate effective gravity:
$$g_{eff} = 9.81 \times \cos(30^\circ) = 8.495\text{ m/s}^2$$ 3. Compute Rayleigh number with $g_{eff}$ ($L_c = 0.5\text{ m}$):
$$\text{Ra}_L = \frac{g_{eff} \beta (T_s - T_a) L_c^3}{\nu \alpha} = \frac{8.495 \times 0.00307 \times (80-25) \times 0.5^3}{1.83 \times 10^{-5} \times 2.6 \times 10^{-5}} = 3.77 \times 10^8$$ 4. Compute Nusselt number using Churchill & Chu correlation:
$$\text{Nu}_L = \left\{0.825 + \frac{0.387 \times (3.77 \times 10^8)^{1/6}}{\left[1 + (0.492/0.71)^{9/16}\right]^{8/27}}\right\}^2 = \left\{0.825 + \frac{10.37}{1.157}\right\}^2 = 95.8$$ 5. Calculate convective coefficient $h$:
$$h = \frac{\text{Nu}_L k}{L_c} = \frac{95.8 \times 0.028}{0.5} = 5.36\text{ W/m}^2\text{K}$$