Window Heater

You say 2350kW (kW*hr assumed) billed but you used 1692kW*hr for your calculation. And what purpose does the 2.35kW calculation serve?

Reply to
John S
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I do not think it matters.

I tried to find the source document , but failed. I believe it was a gov p ublication on insulation for buildings. It was about a forty page pamphlet and covered pretty much everything. It had soil temperature measurements of various places in the U.S. as well as air temp measurements. The materi al would still be good, but it was published before 1965 and is probably no longer . Dan

Reply to
dcaster

What you just described is foil agnostic. You described the heat flow from conduction and convection. Take the foil out of the picture and you get the same results except there will also be radiation in both cases while with foil there will be very little radiation in both cases.

You have not shown a difference in functioning of the foil in the two cases.

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Rick C
Reply to
rickman

And what happens when you ass-ume? I wrote 2350 kW average because that is what I meant. The average usage over that month was 2350 kW. The same number I wrote in the next sentence, "2.35 kW".

The surprise to me is that James Arthur is able to live in a way that he only heats a few cubic feet surrounding his body. I don't have a way to do that, but my window heater helps. To otherwise get the same warmth I would need to turn the heat up at least two degrees.

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Rick C
Reply to
rickman

Opps, I see my mistake. It should have been 2350 W average. I don't think I would like a bill of 2350 kW for a month.

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Rick C
Reply to
rickman

Keep thinking. You might get what I said.

Dan

Reply to
dcaster

I don't know why you think the foil has any impact other than reducing the radiation. If you replace the foil with any other material it works exactly as you describe. The only difference is that the foil prevents radiation which otherwise happens in both directions using some other film.

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Rick C
Reply to
rickman

Although it is true that in summer the effect of coating the inner surface of the roof with a thin layer of insulator and a reflective coating is extremely effective at stopping radiative heat getting in and since the warm air stays trapped against the upper parts a simple vent at the top will be very effective. The main reason though is that the temperature difference is huge. Roof in strong sunshine can be ~70C - well above the point where thermal radiation flow is important.

In winter the radiation barrier is still effective provided that the temperature difference across it is ~30K or so but the air convection of heat from below upwards is now the dominant process and summer vents at the top of the roof are invariably in exactly the wrong place.

You can gain a lot by insulating the floor of the loft with thick glass fibre in what amounts to space blanket aluminised polythene bag. This is the more common way to do it in the UK since we hardly ever have days so sunny that the roof space over heating is a problem. Even if the aluminising doesn't do much the bag makes it a lot nicer to handle. Typical winter inside to outside temperature seldom exceeds 30K here and with the mild winters recently 15K is about average. So most of the time the radiative contribution to winter losses is small.

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Regards, 
Martin Brown
Reply to
Martin Brown

If you have air present.

ITYM it does not *convect* down. It remains trapped as a thin barrier layer against the warm foil surface and heat transfer downwards is only by radiation and slow diffusion (and not by bulk convective motion).

If you have an idealised radiation balance problem with source temperatures T, t either side of a flat plate in vacuum then at equilibrium the plate temperature x is given by energy balance. Unless the plate is a perfect idealised mirror it will absorb a small amount of radiation from each source and emit the same power at equilibrium.

2x^4 = t^4 + T^4

Hence x = 0.841T[1 +(t/T)^4)]^(1/4)

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Regards, 
Martin Brown
Reply to
Martin Brown

So I got the power supply in and hooked it up. This is a laptop type supply with a switch for 12 to 24 volts output. Driving a 10 ohm load that produces 14.4 to 57.6 watts. With the old unit at 20 volts it was more than warm enough for a cold night. This one will surely do the job if it doesn't blow up. I ordered more resistors which I will connect to get a 6 ohm load for the full rated 96 watts. We'll see if the power pack dies with that high a load. There are some colds nights ahead in spite of the present 70 degree days. Very strange weather for February.

The supply is about the same size as my Toshiba brick, but MUCH lighter, so obviously not so much magnetics. I wonder what the quality is like inside? How would you make a 96 watt DC power supply that weighs less than 6 oz including the corded EMI filter?

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Rick C
Reply to
rickman

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