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 = 7.1078E+08
Nusselt Number Nu = 44.0856
Coeff h = 288.2611 W/m2K
Transfer Q = 2378.1545 W
--- DELTA-T SWEEP ---
dT[C] Ra Nu h[W/m2K] Q[W]
------------------------------------------------------------
1.00 1.292E+07 16.188 105.8511 15.8777
7.83 1.012E+08 27.083 177.0851 208.0750
14.67 1.895E+08 31.680 207.1468 455.7230
21.50 2.778E+08 34.859 227.9319 735.0803
28.33 3.662E+08 37.349 244.2136 1037.9078
35.17 4.545E+08 39.422 257.7675 1359.7238
42.00 5.428E+08 41.212 269.4684 1697.6506
48.83 6.311E+08 42.794 279.8173 2049.6616
55.67 7.194E+08 44.219 289.1307 2414.2416
62.50 8.077E+08 45.517 297.6223 2790.2089
69.33 8.960E+08 46.713 305.4436 3176.6129
76.17 9.843E+08 47.824 312.7063 3572.6699
83.00 1.073E+09 48.863 319.4957 3977.7210
89.83 1.161E+09 49.839 325.8778 4391.2038
96.67 1.249E+09 50.760 331.9056 4812.6317
103.50 1.338E+09 51.635 337.6218 5241.5790
110.33 1.426E+09 52.467 343.0616 5677.6697
117.17 1.514E+09 53.261 348.2543 6120.5689
124.00 1.602E+09 54.021 353.2245 6569.9764
130.83 1.691E+09 54.750 357.9934 7025.6208
137.67 1.779E+09 55.452 362.5790 7487.2562
144.50 1.867E+09 56.127 366.9969 7954.6579
151.33 1.956E+09 56.779 371.2608 8427.6201
158.17 2.044E+09 57.410 375.3826 8905.9531
165.00 2.132E+09 58.020 379.3730 9389.4819
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 $h$:
$$h = \frac{\text{Nu}_L k}{L_c} = \frac{95.8 \times 0.028}{0.5} = 5.36\text{ W/m}^2\text{K}$$