Inclined & Horizontal Plate Natural Convection
Analyze natural convection boundary layers on flat plates oriented at arbitrary angles from vertical.
Effective Gravity and Angle
For inclined plates ($\theta < 90^\circ$ from vertical), natural convection correlations are modified by replacing the standard gravity constant $g$ with the effective parallel component: $$g_{eff} = g \cos \theta$$ For horizontal plates ($\theta = 90^\circ$), fluid buoyancy forces form rising plumes (on upper hot/lower cold surfaces) or stable stratification layers (on lower hot/upper cold surfaces).
Parameters Setup
Results & Curves
📈 Nusselt number vs Temperature difference (dT)
Engine Output
============================================
INCLINED & HORIZONTAL PLATE CONVECTION
============================================
Rayleigh Number Ra = 3.9129E+06
Nusselt Number Nu = 24.0170
Coeff h = 6.7376 W/m2K
Transfer Q = 55.5850 W
--- DELTA-T SWEEP ---
dT[C] Ra Nu h[W/m2K] Q[W]
------------------------------------------------------------
1.00 7.114E+04 8.819 2.4741 0.3711
7.83 5.573E+05 14.754 4.1390 4.8634
14.67 1.043E+06 17.259 4.8417 10.6517
21.50 1.530E+06 18.991 5.3275 17.1811
28.33 2.016E+06 20.347 5.7080 24.2592
35.17 2.502E+06 21.476 6.0248 31.7810
42.00 2.988E+06 22.451 6.2983 39.6795
48.83 3.474E+06 23.313 6.5402 47.9071
55.67 3.960E+06 24.089 6.7579 56.4284
62.50 4.447E+06 24.797 6.9564 65.2160
69.33 4.933E+06 25.449 7.1392 74.2475
76.17 5.419E+06 26.054 7.3089 83.5046
83.00 5.905E+06 26.619 7.4676 92.9719
89.83 6.391E+06 27.151 7.6168 102.6363
96.67 6.877E+06 27.653 7.7577 112.4864
103.50 7.363E+06 28.130 7.8913 122.5122
110.33 7.850E+06 28.583 8.0184 132.7051
117.17 8.336E+06 29.015 8.1398 143.0570
124.00 8.822E+06 29.430 8.2560 153.5611
130.83 9.308E+06 29.827 8.3674 164.2109
137.67 9.794E+06 30.209 8.4746 175.0008
144.50 1.028E+07 32.616 9.1498 198.3217
151.33 1.077E+07 33.122 9.2918 210.9240
158.17 1.125E+07 33.613 9.4296 223.7174
165.00 1.174E+07 34.090 9.5635 236.6964
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 $h$:
$$h = \frac{\text{Nu}_L k}{L_c} = \frac{95.8 \times 0.028}{0.5} = 5.36\text{ W/m}^2\text{K}$$