From: nick@vu-vlsi.ee.vill.edu (Nick Pine)
Date: 15 Jul 1995 11:34:54 -0400
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 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 3.2 of them.
Once again, some numbers on the back of an envelope seem helpful.
Nick