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re: inexpensive chiller in the 2-3 ton range?
10 mar 2002
al youngwerth wrote:
>i am looking for a air-source chiller to supplement my whole house fan
>for cooling my house in the summer. the chiller will be used to cool
>water that will be pumped through my raidant floor heating system.
>don't bother with telling me that it won't work or that i'll get
>condensation, i've gone through all of that (our summertime humidity
>is < 15% and using the whole house fan to cool the slab at night works
>pretty well to keep the house cool all day).
since you live in an arid climate, how about pumping some water over the
roof at night from a pvc pipe with holes near the ridge and gathering it
up in a gutter and storing it in a tank and pumping it through the slab
during the day?
>if we started the fan at night after it cooled off and then shut
>if off in the morning and shut all of our windows and doors, we would
>peak at right about 80f in the early evening on the hottest days
>(100f). problem is sometimes we forget to turn on the fan until
>morning (cause it was still real hot outside when we went to bed) and
>sometimes our kids would leave a sliding glass door open all day. once
>the house got warm, there was no way to cool it until it got cool
>again outside. my wife does not like this feature.
children can be beaten. grainger sells differential thermostats...
>i was hoping to get away with this for less than $2k.
where do you live? an average 15% rh at 90 f makes the dew point tdp
= 1/(1/(460+90)-ln(0.15)/9621)-460 = 36 f, with vapor pressure pa
= 0.15exp(17.863-9621/(460+90)) = 0.217 "hg. night air temp to = 80 f
makes the wet bulb temp twb (r) = 9621/(22.47-ln(460+to+100pa-twb),
ie 56 f, after some iteration.
a t (f) pond loses 1.63x10^-9((t+460)^4-a(to+460)^4) btu/h-ft^2 by radiation,
where a = 0.002056tdp+0.7378 = 0.853 in this case, so a 70 f wet roof in 80 f
air loses 10.4 btu/h-ft^2. in v = 4 mph air, it gains qc = (0.74+0.3v)(t-to)
= 19.4 btu/h-ft^2 by convection, and loses qe = b(t-twb)-qc by evaporation,
where b = 3.01(0.74+0.3v)((t+twb)/65-1) = 3.01(1.94)((70+56)/65-1) = 5.48,
so qe = 5.48(70-56)-19.4 = 57.3. the total loss is 48.3 btu/h-ft^2, so you
need 36,000/48.3 = 745 ft^2 of roof for 3 tons of cooling...
nick
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