Forced Convection Over Spheres
Determine boundary layer heat transfer coefficients and total heat rate for flows over spherical surfaces.
External Flow Over Sphere
Similar to crossflow over cylinders, flow separates from the rear side of spheres, creating a low-pressure wake region. Convection is analyzed using empirical correlations that account for the fluid viscosity change near the surface ($\mu/\mu_s$).
Parameters Setup
Results & Curves
📈 Nusselt number vs Velocity
Engine Output
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CONVECTION OVER SPHERE ENGINE
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Reynolds Re = 5.2200E+02
Whitaker Nu = 521.5707
Ranz Nu = 110.8036
Coeff h = 730.1990 W/m2K
Transfer Q = 2752.7855 W
--- VELOCITY SWEEP ---
V[m/s] Re Nu_Whitaker Nu_Ranz h[W/m2K] Q[W]
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0.050 8.700E+01 197.765 46.419 276.8707 1043.7781
0.098 1.704E+02 283.747 64.160 397.2454 1497.5799
0.146 2.538E+02 352.002 77.860 492.8033 1857.8247
0.194 3.371E+02 410.786 89.439 575.0999 2168.0756
0.242 4.205E+02 463.401 99.654 648.7614 2445.7729
0.290 5.039E+02 511.576 108.898 716.2060 2700.0329
0.337 5.872E+02 556.350 117.404 778.8896 2936.3447
0.385 6.706E+02 598.408 125.324 837.7710 3158.3221
0.433 7.540E+02 638.230 132.766 893.5216 3368.4971
0.481 8.374E+02 676.167 139.806 946.6341 3568.7266
0.529 9.208E+02 712.488 146.504 997.4827 3760.4213
0.577 1.004E+03 747.400 152.904 1046.3597 3944.6833
0.625 1.088E+03 781.071 159.044 1093.4989 4122.3936
0.673 1.171E+03 813.636 164.953 1139.0904 4294.2695
0.721 1.254E+03 845.208 170.655 1183.2918 4460.9051
0.769 1.338E+03 875.882 176.171 1226.2354 4622.7984
0.817 1.421E+03 905.738 181.517 1268.0332 4780.3725
0.865 1.504E+03 934.844 186.708 1308.7815 4933.9900
0.912 1.588E+03 963.260 191.758 1348.5636 5083.9651
0.960 1.671E+03 991.037 196.676 1387.4522 5230.5716
1.008 1.754E+03 1018.222 201.473 1425.5111 5374.0504
1.056 1.838E+03 1044.855 206.158 1462.7969 5514.6145
1.104 1.921E+03 1070.971 210.737 1499.3598 5652.4534
1.152 2.005E+03 1096.603 215.218 1535.2449 5787.7368
1.200 2.088E+03 1121.780 219.607 1570.4925 5920.6171
Calculation Methodology
Mathematical Model & Theory
Convective heat transfer over a spherical object is evaluated using two primary models:
1. Whitaker Correlation:
$$\text{Nu}_D = 2 + \left(0.4 \text{Re}_D^{1/2} + 0.06 \text{Re}_D^{2/3}\right) \text{Pr}^{0.4} \left(\frac{\mu}{\mu_s}\right)^{1/4}$$
where fluid properties (except $\mu_s$) are evaluated at the free-stream temperature $T_\infty$.
2. Ranz-Marshall Correlation:
$$\text{Nu}_D = 2 + 0.6 \text{Re}_D^{1/2} \text{Pr}^{1/3}$$
This correlation is a classical simplification widely used for droplets and evaporating sprays.
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. 7th Edition, John Wiley & Sons.
- Whitaker, S. (1972). Forced Convection Heat Transfer Correlations for Flow in Pipes, Past Flat Plates, Single Cylinders, Single Spheres, and for Flow in Packed Beds and Tube Bundles. AIChE J.
Worked Engineering Example
Air at $25^\circ\text{C}$ flows at $5.0\text{ m/s}$ past a $50\text{ mm}$ sphere maintained at $100^\circ\text{C}$. Find the convective heat loss ($T_\infty = 25^\circ\text{C}$).
Step-by-step Solution:
1. Evaluate air properties at free-stream temp $25^\circ\text{C}$:
- $\rho = 1.177\text{ kg/m}^3, \mu = 1.85 \times 10^{-5}\text{ Pa·s}, k = 0.0263\text{ W/m·K}, Pr = 0.71$.
- At $T_s = 100^\circ\text{C}$, $\mu_s \approx 2.18 \times 10^{-5}\text{ Pa·s}$.
2. Calculate Reynolds number:
$$\text{Re}_D = \frac{\rho V D}{\mu} = \frac{1.177 \times 5.0 \times 0.05}{1.85 \times 10^{-5}} = 15,905$$ 3. Evaluate Nusselt number via Whitaker correlation:
$$\text{Nu}_D = 2 + \left[0.4 \times (15,905)^{1/2} + 0.06 \times (15,905)^{2/3}\right] \times (0.71)^{0.4} \times \left(\frac{1.85}{2.18}\right)^{1/4}$$ $$\text{Nu}_D = 2 + \left[50.4 + 37.9\right] \times 0.87 \times 0.96 = 75.8$$ 4. Convection coefficient $h$:
$$h = \frac{\text{Nu}_D k}{D} = \frac{75.8 \times 0.0263}{0.05} = 39.9\text{ W/m}^2\text{K}$$ 5. Surface area $A_s = \pi D^2 = \pi \times 0.05^2 = 0.00785\text{ m}^2$. Total heat loss:
$$Q = h A_s (T_s - T_\infty) = 39.9 \times 0.00785 \times (100-25) = 23.5\text{ W}$$