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re: inexpensive chiller in the 2-3 ton range?
22 mar 2002
al youngwerth wrote:
>last summer you helped me determine that i need my floor to be about
>7f cooler than the max air temp i want in the house to remove enough
>heat in the peak of the day (about 45k btu/h). so to move 45k btu/h
>from the air to the floor i need a 7f delta.
i dimly recall that. where did the 45k come from?
>what about moving that 45k btu all the way through that 3" of
>concrete, 2400' of 1/2" pex tubing and into the water. how much cooler
>does the water pumping through floor have to be to overcome the r
>value of the concrete and pex tubing? am i looking at this correctly?
well, the floor surface to room-air conductance might be about
2 btu/h-f-ft^2 for slowly moving air (or 2.7 for 130 fpm air
(the flying paper limit) with a ceiling fan), and it radiates
coolth at about 1 btu/h-f-ft^2, so you might have an equivalent
r0.33 conductance at the surface...
and concrete has an r-value of about 0.2 per inch, so 3" of concrete
is about r0.6, in series with r0.33 at the surface. if a 7 f dt moves
45k btu/h at the surface, moving 45k to the underside of the concrete
might require a (0.6+0.33)/0.33x7 = 20 f delta. can you put the tubing
closer to the surface, with some rebar on top to keep it from floating?
you might also look into a radiant ceiling. lay poly film over the
rafters and drywall, and greenhouse poly film duct over that, under
the attic insulation, with a shallow layer of water in the duct?
i don't know much about the pex tubing. you might assume that the wall
resistance is small compared to the concrete resistance and the water
flow rate is large enough that the water hardly warms as it passes
through the tubing. or do some more calculations using the pex specs.
>...if i could cool water overnight to what you suggest, 56f and i
>wanted to store 250k btu and the water stops absorbing a useful
>amount of heat at say 66f (this is the q i'm asking in the previous
>paragraph, house temp = 78, floor surface temp = 71, what does water
>need to be to transmit 45k btu/h?)...
maybe 78-20 = 58 f...
>i'd store about 80btu per gallon and need about 3125 gallons...
that's 80x3125/45k = 5.6 hours at 45k btu/h, ignoring the heat capacity
of the slab (about 25 btu/f-ft^3) and drywall (1 btu/f-board foot) and
furnishings (450 btu/f per 55 gallon accent drum.) since you live in idaho,
don't forget to count your 5-year supply of bottled water (8 btu/f per
gallon) and canned goods (1 btu/f per pound) and weapons and ammunition.
>...that assumes i'd be doing no cooling during the day with this system,
it does? you might be doing no cooling at night.
>what i'm thinking is this. if my wet bulb temp is about 67f (100f,
>12%rh, is that about right?)
nrel says the average july temp in boise is only 74 f, with an average
min/max of 57.7 and 90.2. you may be overestimating your cooling need.
the average humidity ratio w = 0.0067, so pa = 29.921/(1+0.62198/w)
= 0.319 "hg and tdp = 9621/(17.863-ln(pa))-460 = 46.2 f. with a night
to = (57.7+74)/2 = 66 f, twb (r) = 9621/(22.47-ln(460+to+100pa-twb)
= 514.5 r or 54.4 f, after some iteration. for more detail (hourly vs
monthly averages), you might run a tmy2 spreadsheet simulation.
>and 67f water will be cool enough to transmit 45k btu through the pex
>when the top of the floor is 71f...
oh?
>i should be able to have an on demand cooling system with minimal
>storage, right?
what happened to the 3125 gallon tank? (you could make it smaller and
maybe heat the house in wintertime by sprinking licl over the south roof
under a layer of polycarbonate, and rehydrate the solution by blowing
house air through the upper part of the poly film attic ducts, but that's
another story.)
>can i really expect to get down to 67f during the middle of a 100f day?
i'd water the roof at night, and leave it dry during the day.
>what about my medium brown asphalt roof, it's probably going to add
>some heat back won't it?
not at night.
>the other thing i'd need to consider are separation of the evaporative
>water from the floor water (i'm a bit concerned about running bird
>droppings through my floor, some sort of heat exchanger sunk into a
>small insulated tank that stored the water going over the roof).
that could be a very large heat exchanger, with 45k btu and a small dt.
i'd just filter that water and run it through the tubing. dig up the
floor if it gets clogged :-)
>also, how much water a day might i be evaporating to eliminate 250k btu,
...250 pounds or 31 gallons.
>water is very expensive here.
probably cheaper than electricity.
>> >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.
sounds like that would work fine on average days, with no need for water
cooling. beat the kids, and get a differential thermostat for the fan...
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 = 0.833 in boise, so a 66 f wet roof in 66 f
air loses 20.8 btu/h-ft^2. in v = 8 mph air, qc = (0.74+0.3v)(t-to) = 0
btu/h-ft^2 for convection, and it loses qe = b(t-twb)-qc by evaporation,
where b = 3.01(0.74+0.3v)((t+twb)/65-1) = 3.01(3.14)((66+54.4)/65-1) = 8.1,
so qe = 8.1(66-54.4)-0 = 93.4. the total loss is 114.2 btu/h-ft^2, so you
need 36,000/114.2 = 315 ft^2 of roof for 3 tons of cooling...
nick
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