## Time to Equilibrium

## Time to Equilibrium

(OP)

Its been a long time since my last thermo class, and I am trying to brush up on my basics.

If I had a steel plate(7°C) that is A x B x C in a room of air (20°C), how would I go about calculating the time for the steel plate to reach equilibrium with the air (within say 1°C)? Assuming the plate is mounted in such a way to simulate floating and the air temp is constant.

If I had a steel plate(7°C) that is A x B x C in a room of air (20°C), how would I go about calculating the time for the steel plate to reach equilibrium with the air (within say 1°C)? Assuming the plate is mounted in such a way to simulate floating and the air temp is constant.

## RE: Time to Equilibrium

= = = = = = = = = = = = = = = = = = = =

P.E. Metallurgy, Plymouth Tube

## RE: Time to Equilibrium

## RE: Time to Equilibrium

## RE: Time to Equilibrium

Assume all of the following are placed in simple room of uniform still air = T_air_initial.

Now, assume you have flat thin plate: 100x100x1.0 = 10,000 units volume & 10,000 mass but surface area = 20,400 sq units. That horizontal plate will cool (literally) with different film coefficients and radiation losses than a equal volume cube of 21.54 units per side. A horizontal plate will require different equations and approximations than a vertical thin plate. The relative plate, air and room wall temperatures will directly control the thermal equations as well: Hot plate, cool air; colder plate and hotter air.

A vertical box 10x10x100 will have 4x sides cooling as vertical walls, and two as small (near negible) horizontal plates. Turn that box horizontal, and you have two vertical walls, and two horizontal sides, but the four act more like a horizontal cylinder. Rotate the horizontal long box by 45 degrees, and it begins acting like a fin with low flow resistance.

A horizontal sheet 400x400x0.625 will act exactly opposite the vertical box.