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Natural convection from the vertical plate surface
Vertical steel plate divides two air environments with different temperatures.
Problem type:
Plane problem of steadystate heat transfer.
Geometry:
Height of the plate l = 1000 mm, thickness of the plate d = 2 mm.
Plate surface area is 1 m^{2}.
Given:
temperature of the air below the plate T_{1} = 20°C,
temperature of the air above the plate T_{2} = 10°C
thermal conductivity of the plate λ_{plate} = 40 W/(m·K),
thermal conductivity of the ambient air λ_{air} = 0.027 W/(m·K).
Task:
Calculate the temperatures of the right and the left 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 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. Convection coefficients depend on many factors such as environment temperature, shape and orientation of the surfaces.
For the vertical plate an average convection coefficients 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 T_{plate} = (T_{1}+T_{2})/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:
 Prandtl number (Pr), Grashof number (Gr), Relay number (Ra) and Nusselt number (Nu) are calculated by the known formulae^{*}:
Pr = μ·C / λ_{air}, where μ  dynamical viscosity of the air, C  thermal capacity of the air;
Gr = L^{3}ρ^{2}g·ΔT·β / μ^{2}, where L  characteristic size (in our case it is the plate height 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·Ra^{1/6}/(1 + (0.492/Pr)^{9/16})^{8/27}]^{2}.
Parameters of the air (ρ, C, μ, β) are calculated for the average temperatures:
below plate: T_{air1} = (T_{1} + T_{plate})/2 = (20+5)/2 = 12.5 °C;
C_{1} = 1000 J/(kg·K), μ_{1} = 1.87·10^{5} N·s/m^{2}, ρ_{1} = 1.25 kg/m^{3}, β_{1} = 0.003501 1/K;
above plate: T_{air2} = (T_{2} + T_{plate})/2 = (10+5)/2 = 2.5 °C^{**}.
C_{2} = 1000 J/(kg·K), μ_{2} = 1.87·10^{5} N·s/m^{2}, ρ_{2} = 1.32 kg/m^{3}, β_{2} = 0.003695 1/K.
Numerical values of the parameters:
ΔT_{1} = 20  5 = 15 °C, ΔT_{2} = 5  (10) = 15 °C;
Pr_{1} = 0.69; Pr_{2}=0.69;
Gr_{1} = 2.30·10^{9}, Gr_{2} = 2.71·10^{9};
Ra_{1} = 1.59·10^{8}, Ra_{2} = 1.88·10^{8};
Nu_{1} = 141, Nu_{2} = 149.
Calculations are automated using the spreadsheet prepared by Harlan Bengtson and Lamar Stonecypher. Spreadsheet for the vertical plane convection is attached.
 Using the numerical values, we have
α = Nu·λ_{air} /L
After the substitution of the values, numerical values could be obtained
α_{1} = 3.82 W/(m^{2}·K) for the left surface,
α_{2} = 4.02 W/(m^{2}·K) for the right surface.
 QuickField problem is solved and the resulting temperatures of the surfaces and the heat flux through the surface is calculated.
Result:
Temperature of the left surface of the plate: T_{1} = 277.77 K (4.62 °C).
Temperature of the right surface of the plate: T_{2} = 277.76 K (4.61 °C).
Heat flux: F = 58.756 W (plate surface area is 1 m^{2}).
Download simulation files and parameter calculation spreadsheet.
^{*}CIBSE Guide C: Reference Data (2010). ButterworthHeinemann, 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.
