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re: cool tower alternatives
23 nov 2005
abby normal wrote:
>i am well aware of dalton's law of partial pressures.
good :-)
>how about adding an elevation of 1117 ft asl, then you can use 28.73
>inches of mercury.
no thanks. it seems to me that we keep getting sidetracked from the main
issue (whether these indoor evaporation schemes can work at all) by this
and other side issues, like whether to count the heating effect of sun on
walls, how to calculate how much cooling a house requires, whether these
schemes can work in a humid jungle, and so on.
>when you use swimming pool equations, you have to realize that the heat
>to evaporate the water is coming from the water in the pool itself as
>well as a severely flooded permimeter around the pool.
sure. page 4.7 of the 1991 ashrae applications handbook has empirical
formula (2) for evaporation of water "from public pools at high to normal
activity, allowing for splashing and a limited area of wetted deck":
wp = 0.1a(pw-pa) lb/h, with pool surface a in ft^2 and pw sat pressure
at the water surface temp and pa at the room air dew point, both in "hg.
again, let's not argue about activity levels now. the indoor evaporation
can come from any number of sources: a dampened slab with a soaker hose
and a solenoid valve and a thermostat, a pond, a fountain, a portable
swamp cooler, misters, green plants, indoor clotheslines, and so on.
>>from experience, i can tell you that during some sweltering conditions
>during a prolonged power failure, i would let water stand in a bathtub.
>the tub would be a couple degrees cooler than the room air because
>water was evaporating from it. it did not cool off the room.
sounds like the tub didn't have enough surface to cool the room much, but
if you had taken careful enough measurements, you would have noticed that
the a ft^2 bathtub cooled the room by about 100a(pw-pa) btu/h. the water
loses heat to the room air by evaporation, and the water surface and all
the other tub surfaces gain sensible heat from the room (ie cool the room.)
there is no magic. energy is conserved.
>water evaporating from the slab will get heat from the slab, heat will
>conduct to the slab from below.
not much, if the ground below is dry. the slab might be over a vapor
barrier or foamboard insulation. again, let's not get sidetracked. let's
simplify this and say the slab gains no heat from the ground, for now.
>then your ceiling fans, which use energy...
slow ceiling fans use very little energy. let's say 0, for now.
how much heat moves from the room to the slab by radiation? grainger's
4c853 48" fan moves 21k cfm at full speed at 315 rpm with 86 watts. how
much air do we need to move up from the slab to the room to keep it comfy?
how much power does that require, according to fan laws? swamp coolers
put all their electrical heat power into the house...
>will force the air down towards the slab and be cooled by the slab.
right :-) with a room temp thermostat and an occupancy sensor.
>air does not 'flow into corners' and the majority of this air will
>not contact the slab but will flow 'parallel to the slab' and will not
>have the benefit of contacting the slab to transfer sensible heat to
>the slab. however this air does get to recieve the addtion of moisture
>without the benefit of being cooled sensibly.
let's avoid this sidetrack and say the air in the room is fully mixed.
and not clever enough to collect water vapor without collecting coolth.
>you will add an excessive amount of moisture to the air...
we would add exactly p pounds per hour of moisture to the room air.
>and a small portion of the heat used to evaporate water actually is
>actually sensible heat removed from the room air.
the p lb/h of water provides 1000p of total cooling, which both cools
the room to say 80 f at wi = 0.0120 and cools c cfm of outdoor air
(with an exhaust fan and a humidistat, plus some air infiltration)
at ta (f) and wa that flows into the room.
>then, to try and lower humidity you must exhaust air, and will be
>drawing in triple digit outside air.
ok. say it's 3 pm on an average june day in phoenix, with ta = 103.5 f and
wa = 0.0056, and we want 10k btu/h of net sensible cooling for a small well-
insulated house (with g = 10k/(103.5-80) = 425 btu/h-f) to keep it 80 f with
wi = 0.012. air weighs 0.075 lb/ft^3. p = 60c0.075(wi-wa) = 0.0288c makes
c = 34.7p, and 1000p = 10k+(103.5-80)c = 10k+23.5x34.7p makes p = 54 lb/h
and c = 1886 cfm. wow.
at the average 93.5 f outdoor temp, we can provide (93.5-80)425 = 5.7k btu/h
with 1000p = 5.7k + (93.5-80)34.7p, so p = 11 lb/h, and c = 372 cfm. this
works more efficiently with cooler outdoor air, so we might turn off the
system and use stored slab coolth during the warmest part of the day.
>... more water will evaporate than you are counting on
let's try this at 3 am, when ta = 72.9 with wa = 0.0056. say we want to
store 22h(93.5-80)425 = 126k btu of coolth in a 2 hours for the rest of
the average 93.5 f day. if we turn on an 800 cfm 100 w $12 20" chinese
window box fan for 2 hours and evaporate p lb/h of water from a 2000 ft^2
80 f slab with a 1/(1/800+2/3/2000) = 632 btu/h-f conductance to outdoor
air, 126k = 2h(1000p+(80-72.9)(800+632)) makes p = 53 lb/h, about 13 gallons
for the whole day, since evaporative cooling is more efficient with cooler
night air. if we splurge and use lasko's $50 90 w 2470 cfm fan, 126k
= 2h(1000p+(80-72.9)(2470+1355)) makes p = 36 lb/h, about 9 gallons.
and we might use less water if the house were unoccupied for
8 hours during the day, with a cool slab under warm house air.
>2) you assume that all the heat that will evaporate water comes from
>the room air when it will becoming from the outside of the residence.
no. not at all. that's the last term in 1000p = 10k+(103.5-80)c above.
i hope that's clear now. you may have been missing this over and over.
>3) you are adding untreated triple digit outside air directly to the space.
and properly accounting for that...
>4) the ceiling fans (note the plural) use energy and so will your
>"constant" bath room fan, yet these are written off as insignificant.
yup. i'd like to stop now. feel free to work on the rest of the details.
nick
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