Date: Sun, 5 Aug 2007 10:11:05 -0700
Reply-To: Michael Elliott <camping.elliott@GMAIL.COM>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: Michael Elliott <camping.elliott@GMAIL.COM>
Subject: Re: Comments wanted on this portable fridge
In-Reply-To: <5FE19128-2595-409D-8388-D75CA771E5F6@uvic.ca>
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So from this we take that while letting the cold air fall out of the
refrigerator will have no perceivable effect on the coldness of the
remaining beers, it is the removal of a cold beer and replacing it with
a warm one which will have a massively greater effect on the coldness of
the other beers.
So just leave the cold beers in the refrigerator and open the door
occasionally to look at them.
It's that simple.
[dusts hands and walks away, whistling tunelessly]
--
Mike "Rocket J Squirrel" Elliott
71 Type 2: the Wonderbus
84 Westfalia: Mellow Yellow ("The Electrical Banana")
74 Utility Trailer. Ladybug Trailer, Inc., San Juan Capistrano
KG6RCR
Alistair Bell typed:
> back of envelope calculations to original question...
>
> On 8/4/07, Mark Drillock <mdrillock@cox.net> wrote:
>
>>
>> OK, students on the list, snap quiz time. How much mass in one cubic
>> foot of air at sea level and 50% humidity? How much energy does it
>> take
>> to raise or lower that mass by 40 degrees F?
>>
>> Mark
>> too old to remember how to figure this out
>>
>>
>
> air weighs approx .0755 pounds per cubic foot at sea level at 60 or
> so degrees F.
>
> at same temp, 50% saturation would be approx 7 grams or water per kg
> of air.
>
>
>
> converting to same unit, 0.755 pounds is about 0.034 kg, so at 50%
> sat, one cubic foot of water holds (very approx) (0.034 * 7.0 g), or,
> 0.238 g of water.
>
>
>
> 1 calorie is amount of energy needed to raise one gram of water by on
> degree C. to raise 1 gram 40 degrees F (very approx 22 C) would take
> 22 Cal. To raise 0.238 g water same amount would take (0.238*22) = 5
> cal.
>
> all this is VERY approx estimations, forgive me if i have dropped any
> decimal points. Bottom line is that there is not much water in t he 1
> cubic foot of air at 50% sat, really inconsequential compared to the
> amount of beer in the fridge.
>
>
>
> Alistair
>
>
>
>
>
>
>
> On 4-Aug-07, at 8:36 PM, Pensioner wrote:
>
> Normalizing to 1cu ft is left as an exercise for the interested.
> Curves for
> the specific heat of one cc of air at various temperatures and RH
> values
> exist. So to figure out how many btus it takes to change the mass of
> air
> one degree either way (same value) becomes a plug and play exercise.
> If it
> takes n BTU to raise one cc of air at 50% rh and 50f one degree f,
> determined from the curve values then some multiplication to scale up
> to one
> cubic foot will yield the number N btus required to raise that mass one
> degree from 50F. A similar scaling not likely linear will allow a
> calculation of the heat required to be removed or added from the thermal
> mass to change it's temperature T degrees F. QED? No, you can't
> easily add
> with a Post Versalog but you can do the rest to within 1 RCH or so
> for three
> significant digits worth. The adding will require the rest of your
> significant digits.
>
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