Natural convection from the vertical plate surface - QuickField simulation example
Vertical steel plate divides two air environments with different temperatures.
How to find natural convection heat transfer from vertical plate?
Answer Typical applications Geometry
Height of the plate L = 1000 mm, thickness of the plate d = 2 mm.
Given Task Solution Result *Remark: the formulas of the similarity theory relate to the big plates and gives the average value of the convection coefficient. This approach is very approximate by its nature, so there is no reason to make iterations to calculate the accurate convection coefficient value related to actual surfaces temperatures.
Engineering question
Set up a plane-parallel QuickField Steady-state Heat Transfer problem for a vertical plate and evaluate natural convection heat transfer from computed field results.
vertical heated plates, external convection plates, natural convection surfaces
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Simulation problem
Problem Type
Plane-parallel problem of steady-state heat transfer.
temperature of the air on the left side of the plate T1 = 20°C,
temperature of the air on the right side of the plate T2 = -10°C
thermal conductivity of the plate λplate = 0.4 W/(m·K)
Calculate the temperatures of the right and the left surfaces of the plate and the heat flux passing through the surfaces.
The heat flux flows from hot to cold areas (in our case from the left surface of the plate then through the plate toward air on the right). Heat flux in the plate is regulated by the plate thermal conductivity. The heat exchange between air and the plate is regulated by natural convection.
We estimate that the temperatures of the left and the right surfaces of the plate equal to the average temperature of the plate which is
Tplate = (T1+T2)/2 = 5 °C.
This value is used for the convection coefficients calculation*:
α1 = 3.6 W/(m²·K) for the left surface,
α2 = 3.7 W/(m²·K) for the right surface.
Temperature of the left surface of the plate: Tleft = 4.9 °C.
Temperature of the right surface of the plate: Tright = 4.6 °C.
Heat flux: f = 54 W (plate surface area is 1 m²).
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