What Changes When Cooling a Plate in Air Instead of Oil?

[capterdi]

I´ve found a webpage where the following problem is solved in detail, together with the substitution of numerical values into the pertinent formulas:

A plate is of cross-section thickness L = 0.1 m and has an initial temp of To = 250°C.
It is suddenly immersed into an oil bath of temperature Ta = 50°C.
The material properties are:
thermal conductivity k = 204W/m°C
heat transfer coefficient h = 80W/m^2°C
density rho = 2707 kg/m^3
specific heat Cp = 896 J/kg°C .
It is required to determine the time taken for the slab to cool to a temperature of 200°C.

The formula used to solve the problem is of the form
(T(t)-Ta)/(To-Ta) = e^-(mt)

where m is a function of h, rho, Cp and L.

The solution is then worked out for the conditions stated above. I had no problem in understanding the steps to solve the problem.

But now, suppose the same problem, but instead of submerging the plate into an oil bath at Ta = 50°C, leave the plate to cool into ambient air at Ta = 20°C.

My specific question is this:

Which coefficients and constants should I change to take care of the air instead of the oil
and which are their numeric values?

Your help will be greatly appreciated.

Thank you.

 

[Mapes]

You only need to replace the heat transfer (convection) coefficient with the one corresponding to air; the properties of the plate are obviously unchanged. However, the convection coefficient can vary depending on whether the air is being blown at the plate (called forced convection) or not (natural convection). In natural convection, even the orientation of the plate makes a difference, because air circulation is driven by buoyancy alone (i.e., hot air rising). Can you tell us a little more about the application?

As a rough estimate, though, you might use a convection coefficient of 10W/m²°C, which should be the right order of magnitude.

 

[capterdi]

Mapes,

Thank you. That was my guess...now I got the confirmation.

Well, about the application...In real life what I have are steel bars measuring each one 0.16 x 0.16 x 12.7 m
and I have to make piles of different size, placing them side by side, because what we have is a yard for
storing the bars. I´m guessing, without any background, that the wider and higher the pile, more time will it
take to cool the bars. But obviously, it´s impractical to place every bar widely separated one from the other,
because of physical space.