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# Natural convection from the spherical surface

Steel sphere divides two air environments with different temperatures.

Problem type:
Axisymmetric problem of steady-state heat transfer.

Geometry:

Sphere radius R = 75 mm, thickness of the sphere's wall d = 3 mm.

Given:
temperature of the lower surface of the plate T0 = 20 °C,
temperature of the air above the plate T = -10 °C
thermal conductivity of the plate λplate = 40 W/(m·K),
thermal conductivity of the ambient air λair = 0.027 W/(m·K).

Calculate the temperature of the outer surface of the sphere and the heat flux passing through the surface.

Solution:
The heat flux flows from hot to cold areas (in our case from inner surface of the sphere, then through the sphere toward air outside). Heat flux in the sphere is regulated by the sphere thermal conductivity. The heat exchange between air and the sphere is regulated by natural convection. Convection coefficient depends on many factors such as environment temperature, shape and orientation of the surface.

For the sphere an average convection coefficient could be calculated using the similarity theory of heat transfer. We estimate that the temperatures of the upper and the lower surfaces of the plate equal to the average temperature of the plate which is Tsphere = (T0+T)/2 = 5 °C for calculation of the convection coefficient. The model is the thermal conductivity problem in the sphere with the boundary condition of the natural convection on the outer surface. The procedure of the solution is:

1. Prandtl number (Pr), Grashof number (Gr), Relay number (Ra) and Nusselt number (Nu) are calculated by the known formulae*:
Pr = μ·C / λair, where μ - dynamical viscosity of the air, C - thermal capacity of the air;
Gr = L3ρ2g·ΔT·β / μ2, where L - characteristic size (in our case it is the diameter 2R), ρ - air density, g - acceleration of gravity, ΔT - temperature difference, β - coefficient of the air thermal expansion;
Ra = Gr·Pr;
Nu = 2 + 0.589·Ra1/4 / (1 + (0.469/Pr)9/16)4/9.

Parameters of the air (ρ, C, μ, β) are calculated for the average temperature:
Tavg = (T0 + T)/2 = (-10+5)/2 = -2.5 °C**.
ρ = 1.32 kg/m3, C = 1000 J/(kg·K), μ = 1.87·10-5 N·s/m2, β = 0.003695 1/K.

Numerical values of the parameters:
ΔT = 5 - (-10) = 15 °C;
Pr=0.69;
Gr = 9.14·106;
Ra = 6.33·106;
Nu = 24.7.

Calculations are automated using the spreadsheet prepared by Harlan Bengtson and Lamar Stonecypher. Spreadsheet for the inclined plane convection is attached.

2. Using the numerical values, we have
α = Nu·λair /L = 24.7 · 0.027 / 0.15 = 4.45 W/(m2·K).
3. QuickField problem is solved and the resulting temperature of the surface and the heat flux through the surface is calculated.

Result:
Temperature of the external surface of the sphere: Tsurf = 293.14 K (19.99 °C).
Heat flux: F = 9.43 W (sphere surface area is 0.707 m2).

*CIBSE Guide C: Reference Data (2010). Butterworth-Heinemann, ISBN: 0750653604. Table 3.5.

**Remark: Since the formulas of the similarity theory relate to the big plates, in our case formula above gives the average value of the Nusselt number for the surface. This approach is very approximate by its nature, so there is no reason to make iterations to calculate the accurate Nusselt number related to actual surface temperature.