Subject: Re: Cooling a house with ice

From: [email-masked] (Sherwood Botsford)
Date: 23 Jul 1995 16:36:16 GMT

My grandfather homesteaded in the Peace River country in the early
thirties.  The second building they made was the icehouse. While they
didn't use it for airconditioning, they did use it for cold storage.  
They cut blocks off the pond in midwinter.

On the east side of Lake Winnipeg is the remnants of a fish house.  They
harvested enough ice each winter to pack ice for the entire summer fish
catch.

More on numbers:  If an acre of ice farm saves 20K in electrical costs,
then this is what you should use to measure the profit of the ice farm.
Looking at the street price is a bit misleading, because that includes
all the marketing overhead.

--
Sherwood Botsford          #  Contract Unix system admin & troubleshooting 
Physics Dept               #  Tech writing, Desktop Publishing.
University of Alberta      #  Image processing.        
Edmonton, AB, T6G 2J1      #  492-3713  [email-masked]

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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.

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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
> 
> 
>

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From: [email-masked] (Harry H Conover)
Date: 8 Aug 1995 15:56:53 GMT

Doug Beatty ([email-masked]) wrote:

: regards to all i always used 12'000 btu per ton or 6btu per lb of ice 
: for cooling calculationsIn article 

Then you are getting incorrect results.  As Nick has pointed out, it requires
the removal of approximately 288,000 Btu to freeze one ton of water into 
one ton of ice.  This works out to be approximately 144 Btu/pound.  You can
check this out in any engineering handbook.

The point of your confusion may be this:  One ton of refrigeration is defined
as the cooling capacity required to remove 288,000 Btu PER DAY.  This 
translates to 12,000 Btu per hour.  Since most of todays refrigeration 
equipment is sized in terms of its Btu capacity per hour (not per day), 
you may be equating 12,000 Btu to a ton, where in reality, its 288,000 
Btu per ton.  

                                         Harry C.\