re: evaporative cooler question
22 jan 2005
abby normal wrote:
>it is high airflow per unit cooling. you run it during the day becuase
>that is when it is the hottest.
if you want to ignore thermal mass and use lots of water and electricity...
>if it is 76 degrees at night then just open the windows.
and turn on your 25k cfm fan :-) on an average day with a sinewave temp
curve, it's only 76 for an hour or so.
>...why even run it at night?
to cool the thermal mass of the house so it keeps the house cool all day.
evaporative cooling is a lot more efficient with cooler night air. during
the day, most of the water just cools hot air:
with c cfm of exhaust flow and p lb/h of water and outdoor temp ta and wa
= 0.00384 and 80 f and wc = 0.012 indoors, p = 60c0.075(0.012-0.00384)
= 0.0367c lb/h. the net coolth stored is 1000p-(ta-80)(98+27.2p), ie 0 btu/h
at ta = 106 f and p = 2.62 lb/h, and 2548 btu/h at ta = 76 and p = 2.62.
>does "0.556" make an allowance for absolute temperature.
with w = 0.012 indoors, by weight, pc = 29.921/(1+0.62198/0.012) = 0.566 "hg.
>> 20 gh=98'house conductance (btu/h-f)
>> 30 ta=106'average air temp (f)
>> 40 wa=.00384'average humidity ratio
>> 50 ps=exp(17.863-9621/(ta+460))'outdoor vapor pressure at 100% rh ("hg)
>> 60 pa=29.921/(1+.62198/wa)'vapor pressure in outdoor air ("hg)
>> 70 ra=100*pa/ps'outdoor rh (%)
>> 80 tc=80'comfort temp (f)
>> 90 wc=.012'comfort humidity ratio
>try modelling this to predict what wc will actually end up being in
>this house. assume a set value for 2 occupants. latent heat from 2
>people, we can exclude food prep, laundry.
andersen says an average family of 4 evaporate 2 gallons per day,
including all that. that's minimal (and helpful) compared to the cooling
water flow. no need to predict. the total makes wc = 0.012 by design and
control. this corresponds to the 54% exhaust fan humidistat setting at 80 f.
>show the pychrometric process of how you go from 106db/65 wb, to what
>ever the evaporation rate of water works out to be, add the internal
>latent, and balance it with your exhaust. basically i guess a complete
>moisture balance of the home.
>for a starting point, assume the house is at the ambient condition with
>respect to how much moisture is in the indoor air.
>do this following evap cooling guidelines.
no thanks. feel free to do so yourself. what do you think you will find?
>> 100 pc=60*.075*(wc-wa)
>> 110 cfm=gh/(1000*pc/(ta-tc)-1)'exhaust fan size (cfm)
>1000 is the latent heat of water here?
sure. round numbers, like 1 cfm = 1 btu/h-f.
>> 120 p=pc*cfm'water usage (lb/h)
>> 130 gpd=24*p/8.33'water usage (gpd)
>> 140 pw=exp(17.863-9621/(tc+460))'vapor pressure of wet surface ("hg)
>> 150 ph=29.921/(1+.62198/wc)'vapor pressure of house air ("hg)
>> 160 rc=100*ph/pw'humidistat setting (%)
>> 170 a=10*p/(pw-ph)'wet surface area (ft^2)
>> 180 print ta,ra,tc,rc
>> 190 print cfm,p,gpd,a
>> 106 7.732043 80 54.07039
>> 237.6865 8.727848 25.14626 181.4229
>now its 238 from 310 when all you changed is 32 cfm of infiltration.
yup. that helped, when cooling during the day (not a good idea, imo.)
>a boy scout trick to cool a bottled beverage is to wrap the bottle in a
>wet newspaper. evaporating water from a newspaper cools the fluid inside
>the bottle but i suppose it grabs some heat from ambient air as well.
it has to, by energy conservation. where else would the heat come from?
>your wetted slab will experience a similar effect. heat from the ground
>is going to evaporate water.
not much, if the ground is dry or the slab has insulation beneath.
>all your cfms are exhausts, causing hot dry air to infiltrate in where
>it can or where allowed to put all of its sensible heat into the room
>air. it will not magically transfer to this air. the heat will stratify
>and elevate the room temperature. you can add ceiling fans...
exactly. they allow a larger mass temp swing and more precise room temp
control and setbacks during unoccupied times. radiation helps too.
>the most practical way to do this is to run this hot air through an
>evaporative cooler first.
>this is a high airflow per unit cooling system, your number keep
>getting higher and higher verifying this.
the first airflow (108 cfm) was for cooling 24 hours per day with average
91 f outdoor air. this increased to 170 cfm (iirc) if we only cool for 12
hours at night, but the total daily water and energy use decreased. you
asked how much air (238 cfm) we would need if we cooled when outdoor air
was 106 f (not a good idea, imo.) and the last 1339 cfm corresponds to
only cooling for 0.65 hours at night, with the lowest total daily water
and energy use (4.1 gallons and 32 wh.) this is an overestimate, since air
leaks will lower the indoor rh and wc = 0.012 while the house is closed up
during the day, which allows a max temp greater than 80 f and wc < 0.012
for equivalent comfort.