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# Natural convection from the horizontal plate surface

Horizontal steel plate divides two air environments with different temperatures.

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

Geometry:

Length of the plate l = 1000 mm, thickness of the plate d = 2 mm.
Plate surface area is 1 m2.

Given:
temperature of the air below the plate T1 = 20°C,
temperature of the air above the plate T2 = -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 temperatures of the lower and the upper surfaces of the plate and the heat flux passing through the surfaces.

Solution:
The heat flux flows from hot to cold areas (in our case from below of the plate, then through the plate toward air above). 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. Convection coefficients depend on many factors such as environment temperature, shape and orientation of the surfaces.

For the horizontal 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 = (T1+T2)/2 = 5 °C for calculation of the convection coefficients. The model is the thermal conductivity problem in the plate with the boundary conditions of the natural convection on the lower and the upper surfaces. 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 plate length l), ρ - air density, g - acceleration of gravity, ΔT - temperature difference, β - coefficient of the air thermal expansion;
Ra = Gr·Pr;
Nu = 0.15·Ra1/3, for 107≤Ra≤1011.

Parameters of the air (ρ, C, μ, β) are calculated for the average temperatures:
below plate: Tair1 = (T1 + Tplate)/2 = (20+5)/2 = 12.5 °C;
C1 = 1000 J/(kg·K), μ1 = 1.87·10-5 N·s/m2, ρ1 = 1.25 kg/m3, β1 = 0.003501 1/K;
above plate: Tair2 = (T2 + Tplate)/2 = (-10+5)/2 = -2.5 °C**.
C2 = 1000 J/(kg·K), μ2 = 1.87·10-5 N·s/m2, ρ2 = 1.32 kg/m3, β2 = 0.003695 1/K.

Numerical values of the parameters:
ΔT1 = 20 - 5 = 15 °C, ΔT2 = 5 - (-10) = 15 °C;
Pr1 = 0.69; Pr2=0.69;
Gr1 = 2.30·109, Gr2 = 2.71·109;
Ra1 = 1.59·108, Ra2 = 1.88·108;
Nu1 = 175, Nu2 = 185.

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
After the substitution of the values, numerical value could be obtained
α1 = 4.73 W/(m2·K) for the lower surface,
α2 = 5.00 W/(m2·K) for the upper surface.
3. QuickField problem is solved and the resulting temperatures of the surfaces and the heat flux through the surfaces are calculated.

Result:
Temperature of the lower surface of the plate: T1 = 277.73 K (4.59 °C).
Temperature of the upper surface of the plate: T2 = 277.74 K (4.58 °C).
Heat flux: F = 72.91 W (plate surface area is 1 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 each of the surfaces. 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 surfaces temperatures.

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