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
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INCLINED & HORIZONTAL PLATE CONVECTION
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Rayleigh Number Ra = 5.9360E+08
Nusselt Number Nu = 104.6523
Coeff h = 5.5047 W/m2K
Transfer Q = 45.4139 W
--- DELTA-T SWEEP ---
dT[C] Ra Nu h[W/m2K] Q[W]
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1.00 1.079E+07 31.897 1.6778 0.2517
7.83 8.454E+07 58.088 3.0554 3.5901
14.67 1.583E+08 70.065 3.6854 8.1079
21.50 2.320E+08 78.626 4.1357 13.3377
28.33 3.058E+08 85.482 4.4964 19.1096
35.17 3.795E+08 91.287 4.8017 25.3289
42.00 4.533E+08 96.366 5.0689 31.9338
48.83 5.270E+08 100.910 5.3079 38.8801
55.67 6.008E+08 105.039 5.5251 46.1344
62.50 6.746E+08 108.837 5.7248 53.6702
69.33 7.483E+08 112.362 5.9102 61.4664
76.17 8.221E+08 115.658 6.0836 69.5051
83.00 8.958E+08 118.759 6.2467 77.7714
89.83 9.696E+08 121.690 6.4009 86.2523
96.67 1.043E+09 124.474 6.5474 94.9366
103.50 1.117E+09 127.128 6.6869 103.8145
110.33 1.191E+09 129.665 6.8204 112.8771
117.17 1.265E+09 132.097 6.9483 122.1167
124.00 1.338E+09 134.435 7.0713 131.5261
130.83 1.412E+09 136.687 7.1898 141.0990
137.67 1.486E+09 138.861 7.3041 150.8296
144.50 1.560E+09 140.963 7.4146 160.7124
151.33 1.633E+09 142.998 7.5217 170.7427
158.17 1.707E+09 144.972 7.6255 180.9157
165.00 1.781E+09 146.889 7.7264 191.2274
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 $h$:
$$h = \frac{\text{Nu}_L k}{L_c} = \frac{95.8 \times 0.028}{0.5} = 5.36\text{ W/m}^2\text{K}$$