Recent content by TheH

Note that here it is possible for X to be much larger than Y (when I want to heat the water up rapidly) and Y to be much larger than X (when I want to heat the water up slowly).

I'm trying to size (in watts) a heater to heat a glass of water from ambient (Ti) to a certain temperature (Tf) in a given time period (say, 1 hour). The heat-loss from the glass at Ti is 0 watts. The heat-loss from the glass at Tf is Y watts. I'm assuming that the heat loss increases...

Alright then :)

hikaru, I don't completely understand the issue. The equation I am using for heat-loss by an object via radiation to it's surroundings is: q = ε σ (Th^4 - Tc^4) A from: http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html They do not indicate on that page anything about...

So since the heat loss by radiation is non-trivial, should we revise the system of equations to: P = (k_glass / d) * A * (Th-T) P = ( h_air * A * (T - Tl) ) + (em * sbc * A * (T^4 - Tl^4)) and use this to solve for P?

This is quite embarrassing, I can see my mistake now. Actually the first time I used the equation, I thought it was (Th - Tl)^4 and so I thought to myself celsius/kelvin wouldn't matter. I forgot to re-consider that when I corrected it to (Th^4 - Tl^4). Thanks for the correction.

Since a vacuum contains no particles, can it contain any heat? If not, then the heat lost by any object to a vacuum by radiation would be extreme right? For example, even a 1 meter cube of ice would lose significant heat: 6 * 0.000000056703 * 1 * (273.15^4) = ~1900 watts

hikaru, thanks again. I can't remember where I got 0.95 from but most the stuff I was seeing on the web was in that range. For example on this page they say 0.99: http://www.engineeringtoolbox.com/radiation-heat-emissivity-d_432.html From this page: http://physics.info/convection/ he...

I see, so the heat transferred by conduction through the pane must be equal to the heat transferred by convection from the pane into the air? I did not realize you were stating a system of equations, I thought they were two independent equations. I see that 0.1m of glass doesn't offer the...

hikaru1221, thank you for your reply. I now see that the temperature will not be a full four degree difference as you pointed out, but I am not quite clear how you settled on T = 28.4C ? Will this value be a function of the thickness and material of the pane and the overall difference in...

It may not be clear from the above, but I am considering heat loss from each side of a rectangular aquarium turn-by-turn. In the example I gave, I am only considering heat loss from one side. Also, is this perhaps the wrong sub-forum for this kind of question?

I am trying to understand how heat is lost from a glass aquarium tank maintained at some temperature above ambient. If we imagine a direct interface between the body of water and the surrounding air, I think heat transfer would be entirely due to radiation and convection. When a glass pane is...