Subject: Clever => Re: Cooling a house with ice

From: "Arthur E. Sowers" <[email-masked]>
Date: Fri, 4 Aug 1995 22:42:30 -0400

Offbeat but clever idea (below), but I was in a wooden icehouse in West 
Virginia many years ago and that icehouse was built before there was 
electricity. Yes, they carried in ice in the wintertime for use in the 
summertime. I don't know how far into the summer they got with it, or 
what they used the ice for, but thats what they told me. No envelope, no 
calculations, no math, no science, no engineering.

On a tangent, some of the finest refractor optical telescope lenses were 
ground by hand BEFORE the equations for lensmaking were understood. Its 
obvious, they ground, then looked at a star, then ground, ... and when 
they couldn't get the image sharper, they stopped.

AES.

======
On 15 Jul 1995, Nick Pine wrote:

> It's 85 F so far here today, and it's supposed to hit 100...
> How much ice does it take to keep a house cool in the summer?
> 
> Suppose it's a completely shaded two-story house, 32' on a side, 16' tall,
> with R20 walls and R40 ceiling, and 20% of the walls have R2 windows, and the
> house leaks 1 air change per hour, and two people live in the house, and use
> 500 kWh/month of electricity, ie 500 kWh/30 days/month/24 h/day = 700 Watts,
> all of which goes into heating the house, so the house itself is heated in
> summertime by about (2 people x 100 W + 700 W) x 3.4 Btu/W = 3K Btu/hour. 
> 
> The thermal conductance of the house is the sum of each area divided by its
> R-value. The walls contribute 4 x 0.8 x 512 ft^2/R20 = 82 Btu/hour/F, and
> the windows 4 x 0.2 x 512/R2 = 205 (more than twice as much heat gain, even
> with no sun...) The roof only adds 32' x 32'/R40 = 26 to that number, and
> the volume of the house is 32' x 32' x 16' = 16384 ft^3, so air infiltration 
> adds another 16384 ft^3/55 ft^3/Btu/F = 298 Btu/F, the biggest heat gainer
> of all. The total above is about 600, so when it's 100 F outside and 80 F
> inside, it takes (100-80) x 600 = 12K Btu/hr to keep the house cool, plus the
> 3K of internal heat gain in the house, ie 15K Btu/hr total, ignoring humidity.
> 
> It takes about 144 Btu/pound to melt ice, and warming the water from 32 F
> to say, 72, requires another 40 Btu. Say 200 Btu/lb in round numbers. So
> each hour of summer AC requires 75 pounds of ice to begin with, ignoring the
> heat leaks to the ice battery itself. A month of AC requires 30 x 24 x 75
> pounds of ice, 54K pounds or 27 tons, with a volume of 54K/62 = 870 ft^3,
> a cube 9.5 feet on a side, not counting insulation, or a 32' x 32' basement
> with 10" of ice in a perfectly insulated tank, under the floor, a tank in
> the corner, 8' high x 10.43' square, not counting insulation.
> 
> I live near Philadelphia, which has average daily minimum temperatures of 
> -5.1C, -4.0 and -2.2 in Jan, Feb and Dec of each year, according to NREL.
> Say it's this cold, ie 25 F, average, for 4 hours a day for 90 days, ie
> 2,440 cooling degree hours, with a 32 F base temperature. How large would
> our low-thermal mass, shallow anti-freeze pond have to be, ie how much shaded 
> surface area do we need, with an R1 still-air film resistance, to somehow
> collect 54,000 pounds of ice, or 7.8 million Btu in the winter? This looks
> like Ohm's law for heatflow to me, ie Q = delta T x delta t x Area/R-value,
> with Q = 7.8 million = 2,440 Area/R1, so Area = 7,800,000/2440 = 3200 ft^2,
> eg a square pond, 56 feet on a side... This would work better on a night
> with some wind, or on a clear night with no clouds and no wind. 
> 
> Or perhaps some snowmaking machines in a tent, over an insulated pit, with open
> tent flaps in the winter, or... Let's see, if an auto radiator with a fan can
> get rid of the heat from a gallon of gasoline in an hour, ie 100K Btu/hr, with
> 200 F water and 100 F air, that's 1000 Btu/hour/degree F, 1000 times better
> than a square foot of pond surface, so we'd need 320 of them. 
> 
> Once again, some numbers on the back of an envelope seem helpful.
> 
> Nick
> 
> 
>