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 = 2.8006E+04
Whitaker Nu = 320.6651
Ranz Nu = 185.7641
Coeff h = 7862.7089 W/m2K
Transfer Q = 926.3036 W
--- VELOCITY SWEEP ---
V[m/s] Re Nu_Whitaker Nu_Ranz h[W/m2K] Q[W]
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0.050 1.400E+03 60.582 43.091 1485.4673 175.0025
0.215 6.010E+03 134.502 87.125 3297.9815 388.5343
0.379 1.062E+04 184.970 115.156 4535.4715 534.3226
0.544 1.523E+04 226.714 137.507 5559.0251 654.9072
0.708 1.984E+04 263.418 156.661 6459.0058 760.9337
0.873 2.445E+04 296.711 173.691 7275.3566 857.1078
1.038 2.906E+04 327.489 189.178 8030.0334 946.0160
1.202 3.367E+04 356.308 203.478 8736.6807 1029.2659
1.367 3.827E+04 383.543 216.829 9404.4839 1107.9397
1.531 4.288E+04 409.461 229.396 10039.9858 1182.8080
1.696 4.749E+04 434.260 241.305 10648.0479 1254.4436
1.860 5.210E+04 458.091 252.649 11232.4031 1323.2863
2.025 5.671E+04 481.076 263.501 11795.9932 1389.6827
2.190 6.132E+04 503.311 273.920 12341.1874 1453.9119
2.354 6.593E+04 524.875 283.955 12869.9278 1516.2026
2.519 7.054E+04 545.833 293.644 13383.8310 1576.7454
2.683 7.515E+04 566.242 303.022 13884.2599 1635.7008
2.848 7.976E+04 586.149 312.116 14372.3769 1693.2058
3.013 8.437E+04 605.595 320.951 14849.1829 1749.3782
3.177 8.898E+04 624.615 329.548 15315.5476 1804.3204
3.342 9.359E+04 643.239 337.925 15772.2322 1858.1223
3.506 9.819E+04 661.497 346.098 16219.9080 1910.8629
3.671 1.028E+05 679.412 354.081 16659.1703 1962.6123
3.835 1.074E+05 697.005 361.888 17090.5504 2013.4330
4.000 1.120E+05 714.295 369.528 17514.5248 2063.3813
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:
$$Q = h A_s (T_s - T_\infty) = 39.9 \times 0.00785 \times (100-25) = 23.5\text{ W}$$