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Applied heat and thermodynamics. |
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| Oct4-10, 02:02 AM | #1 |
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Applied heat and thermodynamics.
Okay, to give a little contextual background, for one of my units we have been given an abandoned town in south-west queensland (AU). It was inhabited by the natives for a while, then Australia was colonised, now noone lives there and theres a few shitty buildings that are falling apart. The natives have recently been given the rights to the land, and we were tasked with the job of making the place habitable.
The largest building in the area, the shearing shed, is 40x9.5m and is just made from a layer of corrugated tin. Our proposal is to make a roof cavity, which is to insulate the roof preventing heat transfer and making the building livable. It should be noted that the temperature fluctuates from below freezing at night, to around 50C during the day. Whilst i think that the roof cavity should prevent the rest of the building from getting too hot during the day(since most of the heat a building is absorbed through the roof via the EM radiation from the sun). I was concerned that, at night when thermal conductivity between the thin walls with a large surface area and the air would cause the building to be cold at night. I am struggling to find information on my problem, and was wondering if someone with a heat/thermo background could help me out. Thanks! |
| Oct4-10, 11:16 AM | #2 |
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| Oct4-10, 01:36 PM | #3 |
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I WOULD worry about the walls especially if they constitute a large area....heat transfer occurs in three ways conduction, convection and radiation.
basics here: http://en.wikipedia.org/wiki/Heat_transfer You'd be most convcerned with conduction.....with inside and outside temperatures the same, there will be none...but if inside is say 70 degrees F and the outside 0 degrees, there will be a LOT. With a 70 degree temperature difference, there will be 70 times the heat loss of a 1 degree difference. Q = U [delta T], where U is heat exchanges, U is conductance and delta T the difference in temperature....where U is the inverse of R, the insulation value. whatever Q, heat loss you calculate, will be the heat required to maintain daytime temps at the temperature difference you select..... http://en.wikipedia.org/wiki/Heat_conduction R values for many materials are shown here: http://en.wikipedia.org/wiki/Insulation_value Your shed is almost certainly steel, not tin, but the R values are likely very close.. Here's another table, in different units..watts/ sq meter: http://en.wikipedia.org/wiki/Thermal_conductivity Looks like stainless steell might be 20 times as conductive as glass....1/20 the R value in other words, close enough for your calculation. |
| Oct4-10, 02:39 PM | #4 |
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Applied heat and thermodynamics.AM |
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