A new approach to field modelling


RSS Twitter YouTube Blogger Facebook Linkedin

Main > Application > Sample problems

Natural convection from the inclined plate surface

Inclined steel plate release its heat to the environment.

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


Length of the plate l = 1000 mm, thickness of the plate d = 2 mm, angle of slope θ = 50°.
Plate surface area is 1 m2.

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 upper surface of the plate and the heat flux passing through the surface.

Lower surface of the plate is heated to the 20 °C and the temperature above the upper surface is -10 °C so that the heat flux inside the plate is directed to the upper surface. Upper surface of the plate is hotter than the surrounding air so that the heat flux of natural convection is directed to the cold environment. Convection coefficient depends on many factors such as environment temperature, shape and orientation of the surfaces.

For the inclined plate 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 Tplate = (T0+T)/2 = 5 °C for calculation of the convection coefficient. The model is the thermal conductivity problem in the plate with the boundary condition of the natural convection on the upper 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·cos(θ)·ΔT·β / μ2, where L - characteristic size (in our case it is the plate length l), ρ - air density, g - acceleration of gravity, ΔT - temperature difference, β - coefficient of the air thermal expansion;
         Ra = Gr·Pr;
         Nu = [0.825 + 0.387·Ra1/6 / (1 + (0.492/Pr)9/16)8/27]2.

    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;
    Gr = 1.74·109;
    Ra = 1.21·108;
    Nu = 129.7.

    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 = 129.7 · 0.027 / 1 = 3.5 W/(m2·K).
  3. QuickField problem is solved and the resulting temperatures of the surfaces and the heat flux through the surfaces are calculated.

Temperature of the upper surface of the plate: Tsurf = 293.14 K (19.99 °C).
Heat flux: F = 104.98 W (plate surface area is 1 m2).

 Download simulation files and parameter calculation spreadsheet.

*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.