Math Fun

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Re: Math Fun

Postby James S Saint » Fri Jan 05, 2018 6:59 am

Carleas wrote:(Based on a true story)

Suppose you have an uninsulated hardwood floor that acts as a uniform heat sink for the room. You also have a blanket with an area \(A\) equal to the area of the floor. Will you keep the room warmer by laying the blanket flat across the whole floor, or folding it to double thickness and putting it across half the floor?

I have an intuition, but I don't know the right answer; I'm more interested how people reason about this. Will post my thoughts when I have a minute tomorrow.

Did you mean for this to be a puzzle or did you just want the answer, or the explanation?
Treat the materials as resistors
Floor = R0
Blanket = R1

Total floor area = A

Blanket unfolded;
Floor resistance alone = R0/A
Blanket resistance alone = R1/A
In series,
Ra = (R0+R1)/A

With blanket folded;
Floor resistance bare = R0/(0.5A) == R3
Blanket doubled on floor = (R0+2*R1)/(0.5A) == R4
Total bare floor and doubled blanket resistance;
Rb = R3*R4 / (R3+R4)

For any positive value of R0 and R1, Rb will be lower resistance to heat flow, thus less insulated.

But I suspect the more critical variable is what you are doing on that blanket. 8)
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Re: Math Fun

Postby Carleas » Thu Jan 11, 2018 6:21 pm

I did mean it to be a puzzler, because for me, with a poor knowledge of thermodynamics, I can only use my intuition and reason (assuming I am capable of using those :) . So pardon the rambling I'm about to do, and please point out where I'm wrong (not that you've ever held back).

Your answer matches my intuition, that the blanket spread out would do a better job of insulating, but I don't fully understand your math. Am I correct in reading you as calculating the total resistance of the floor? What is the equation you're using in the folded blanket example? I'm just not sure how you're calculating parallel resistance there.

And is that really the best way to calculate it? It doesn't seem obvious that changing insulation on one part of the floor will affect the cooling effect of the other parts, but isn't that the case for resistance, at least where the two halves are treated as parallel paths?

My intuition was to think of the problem in terms of rates, and ask how the blanket would affect the rate of heat flow. If an insulator decreases heat flow proportionally, i.e. by x%, then doubling the blanket would be less effective: The first layer of blanket would decrease the rate by x%, and then the second layer would decrease that lowered rate by x%. I suspect this is how insulation works.

If, however, an insulator reduced the rate by a fixed amount rather than a percent, then doubling up would be no different.

Finally, if insulation increased non-linearly, so that 2x the insulation produced >2x the reduction in rate, doubling could reduced total heat flow. The situation I've come up with where this might be the case is one where the blankets have a checkerboard pattern of open and closed spaces, so that when doubled they are fully closed. This could create a super-linear increase in insulation if a large obstruction to airflow is more effective than several small obstructions that sum to the same area.
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Re: Math Fun

Postby Serendipper » Mon Feb 12, 2018 6:58 am

Carleas wrote:I did mean it to be a puzzler, because for me, with a poor knowledge of thermodynamics, I can only use my intuition and reason (assuming I am capable of using those :) . So pardon the rambling I'm about to do, and please point out where I'm wrong (not that you've ever held back).

Your intuition is the best solution because a precise answer would have to account for radiative, convective, and conductive heat transfers requiring more knowledge of your blanket and the floor in terms of air leakage and thermal conductivity.

When insulating an area, your biggest bang will come with eliminating air leaks between insulating pockets of air, so even a sheet of saran wrap over the entire floor would do a better job than a blanket folded in half and it's why people put plastic over their windows. Anytime you can divide one area from another area to prevent the volumes of air from intermingling, that's half the insulation battle. The other half is slowing the conduction of heat from one air pocket to the other via the material used for the division. One way to accomplish that is to slow the turbulence of the air in the pockets which is the purpose of the fiberglass stuffing. If air cannot move around, it cannot convect heat and it slows conduction through the barrier considerably which allows a mere paper backing to be a good-enough barrier dividing the 2 volumes. (Plus glass happens to not pass IR light easily, which is insulative as well.)

Heat, as you think of it (and not the proper definition) is infrared light which is only different from visible light in that it oscillates a bit slower. All light comes from the vibration (acceleration) of charged particles. So keeping your vibrating charged particles inside your room is of utmost importance. As molecules vibrate, they are emitting IR radiation (ie cooling). When the IR meets another charge, it induces a vibration and continues on its way. The charge then emits its own IR in all directions resulting from the induced vibration and etc, etc, etc from particle to particle. Now if you let the particle float out of the room, you've lost all that energy. Also, if you let that particle touch another particle, it will transfer all its energy immediately to the other particle. The best strategy is to keep all your particles still and let them vibrate (cool) slowly over time. So reduce air turbulence and seal all leaks.

You could put a tarp down and then a blanket over it, but it would be much more effective to put the blanket down first and the tarp over it because you will have created a pocket of air between the floor and the tarp wherein the stuffing inside the blanket would reduce air turbulence and essentially mimic typical pink insulation. You see how it works?
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