re: cool tower alternatives
23 nov 2005
abby normal wrote:
>nick you are sounding evasive, every 'side track' you avoid is an
>attempt to avoid realistic conditions.
>i gave you a better than average insulated home, that included solar
>gain which you keep ignoring.
>vapour barrier below a slab is insignificant in particular with respect
>to conduction. i gave you better than average ceiling and shading as is
>found in the sw, everything but an insulated slab.
>the whole arguement really is the amount of air needed to be moved or
>exhausted. if you really look at your flawed dream you will see that
>you are putting out fire with gasoline.
>you will be running a high rate of exhaust which means a high rate of
>triple digit infiltration or make up air, to be added on top of the
>external and internal gains of the structure.
>so sad you will not try it the other way, you have proven nothing. i do
>not get impressed when some one has to write a program to develop a
>value of pi, not intrinsic to gw basic.
>you are the one trying to prove you revolutionized the concept of
>evaporative cooling, yet your thinking is flawed and you plug
>significant holes in your scheme by saying you can get a cheap fan at
>grainger or you can store 'coolth' , or just set the exhaust fan to
>come on at a certain humdity level and all is well.
>when you want to use the swimming pool equations that dectron
>developed, you need a pool. the water evaporates from the pool, the
>greatest heat loss from a pool is evaporation, as the evaporating
>vapour draws heat from the water itself. sort of like how they define
>the activity level basically covers 'splashing' where the water is
>splashed up in the air and actually gets its heat to evaporate from the
>air. if anything your scheme needs a high activity level.
>you want to rely on your ceiling fans then lets see you develop some
>some convection coefficients, then crunch some numbers with the
>multi-dimensional heat flux through the slab.
>at best your scheme will give conditions expected in a natatorium, not
>the best conditions for a residence.
>actually learn that evaporative cooling needs to rapidly change the
>indoor air with 'cooled' outside air, pay the penalty of a high air
>flow rate and you can maintain a liveable space for a couple months out
>of the year.
>> 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.
have a nice thanksgiving.