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re: evaporative cooler question
18 jan 2005
abby normal  wrote:

>> >>a 1200 ft^2 house with r30 walls and r40 ceiling and 96 ft^2 of r3
>> >>windows and 0.2 ach (i suppose nobody builds houses like that in
>> >>las vegas) would have a thermal conductance of 1200ft^2/r30 for
>> >>the ceiling + 96/2 for the windows + 1024/30 for the walls +
>> >>0.2x1200x8/60 for air leaks, totaling 128 btu/h-f. july looks like
>> >>the worst-case month for cooling, when it's 91.1 f over an average day,
>> >>with an average low and high of 76.2 and 105.9 and humidity ratio w
>> >>= 0.0066 pounds of water per pound of dry air.
>> >>
>> >>keeping that house 80 f while evaporating p lb/h of water
>> >>into c cfm of outdoor air means 1000p = (91.1-80)(128+c).
>> >>p = 60c(0.075)(0.012-0.0066) = 0.0243c makes c = 108 cfm
>> >>and p = 2.62 lb/h, ie 7.6 gallons per day.
>>
>> >> if the house has significant thermal mass (eg a floorslab), we can
>> >save more water and energy by only running the cooler at night.
>> >>
>> >> why do we need 5000 cfm???

this is the basic question i started with. so far, it seems to me that
swamp cooler cfms are inflated and their controls could be improved and
sw houses could be better designed for cooling, with lighter roofs and
walls and more window shading and fewer and better windows and more
insulation and thermal mass and airtightness. 

>> >...maybe i missed fenestration being factored in there. as expected
>> >the amount of glazing being assumed is 'frugally' less than 10 percent
>> >of the wall area.
>>
>> sure... 96 ft^2 of (shaded) windows is 8% of the floor area...

>> >assuming an ambient of 106f db and 65 wb. you could saturate this
>> >air to 65, but typically the temperature depression would possibly be
>> >80% of the difference between the wet bulb and dry bulb
>> >106-.8x(106-65)=73.2f

ok. sounds like the usual box on the roof. evaporating water indoors
seems thermally equivalent to me, except for a few details... better
controls, easier slab cooling, lower first cost, and an exhaust fan
that uses less power and moves the heat of the motor outdoors. 

>to maintain a space at 80 you have to supply air cooler than 80.

this 108 cfm is the average volume of 80 f air exhausted from the house,
figuring the outdoor temp is 91.1 and w = 0.0066, to use nrel's numbers
for july in las vegas. the slab in the house is close to 80 f, vs some
wet bulb temp, and when it's wet, it has good conductance to house air.

>it sounds like you are working on conductance that is driven by the
>difference between indoor and outdoor temperatures. what about the sun
>beating on the walls and the effect of a hot attic?

evaporative coolers need to be larger for stupid house designs :-) houses
with dark walls and roofs in the southwest, houses with lots of air leaks
and little insulation and poor window shading, and so on. house design is
a somewhat different subject.

>what about some allowance for solar gain through windows.

you didn't below. but it seems to me that can be reduced with shading.
we might also add some indoor electrical heat gain, and minimize that
with cfs, etc.
 
>with an unrealistically low value of 3315 btu/hr, you would still need
>over 400 cfm.

perhaps 3315 btu/h is "unrealistic" because most existing sw houses have
lots of air leaks and dark walls and roofs and not enough shading or
insulation... but it's realistic in the sense that there's no reason new
ones can't be built more airtight, with light walls and roofs and r48 sips
and so on. 

your 400 cfm isn't the same as my 108 cfm...

>5 gh=3328' again with 106/65 db/wb

when it's 106 outdoors, the house needs (106-80)128 = 3328 btu/h of cooling,
but a reasonably airtight house on a slab with good insulation like the one
above with a 78 hour time constant can cool by evaporation at night and stay
buttoned up during the day, with no outdoor air exchange during the day. (i
used gh = 128 btu/h-f as the house conductance, vs cooling load in btu/h.)

>15 qs= gh + solar heat gain
>25 ts= 106 - 0.8*(106-65)
>65 cfm = qs/((60*.075*.24)*(tc-ts))

looks like the box on the roof again...

>with solar gain = 0, qs=gh
>cfm to maintain the space at 80 is 453

that's a different 453...

>are we arguing some other air flow than that required to maintain the
>1200 sq ft home at 80f?

yes. to compare apples to apples at 106 f (even though it's more efficient
to do all the cooling at night), if c cfm of 106 f outdoor air enters the
house and it's cooled to 80 and then exhausted, and we evaporate p pounds
of water per hour inside the house, 1000p=(106-80)(128+c), ok?

second step: if the indoor humidity ratio wc = 0.012 pounds of water per
pound of indoor air and wa = 0.0066 outdoors and we evaporate p pounds of
water per hour into the house air, p = 60c0.075(wc-wi) = 0.0243c, right?

so we subsitute this p into the first equation to get 24.3c = (106-80)(128+c), 
ie 0.935c = 128+c, and c = -1957. ohoh. this won't work, and the box on the
roof will, in this case. that's a clear difference. but indoor evaporation
will work at night, with a slab, and achieve the same overall 24-hour comfort
while using less water and energy.

>specific humidity at 106/65 is 26.9 grains,

so w = 26.9/7000 = 0.00384 by the ashrae hof vs 0.0066 by nrel. we are not
talking about the same outdoor air. i'm surprised, since the hof has a 97.5%
coincident dry/wet bulb temp and nrel lists monthly averages. but again
comparing apples and apples, p = 60c0.075(wc-wi) = 0.0367c makes 36.7c
= (106-80)(128+c) above, so 1.41c = 128+c, and c = 310 cfm and p = 11.7
pounds per hour, not much different from what you got below.

>specific humidity at 73.2/65 is 79.2 grains

...73.2 is the delivered airflow temp, right? but you added water.
perhaps the wet bulb temp changed.

>keeping with standard air constants then
>
>water required = 60*.075*453*(79.2-26.9)/7000= 15.23 pounds per hour,
>1.83 us gallons per hour

i wonder how to reconcile these numbers. the 0.8 depends on the design, no?
the numbers below have more to do with the basic physics.

20 gh=128'house conductance (btu/h-f)
30 ta=91.1'average air temp (f)
40 wa=.0066'average humidity ratio
50 tc=80'comfort temp (f)
60 wc=.012'comfort humidity ratio
70 pc=60*.075*(wc-wa)
80 cfm=gh/(1000*pc/(ta-tc)-1)'exhaust fan size (cfm)
90 p=pc*cfm'water usage (lb/h)
100 gpd=24*p/8.33'water usage (gpd)
110 pw=exp(17.863-9621/(tc+460))'vapor pressure of wet surface ("hg)
120 ph=29.921/(1+.62198/wc)'vapor pressure of house air ("hg)
130 rc=100*ph/pw'humidistat setting (%)
140 a=10*p/(pw-ph)'wet surface area (ft^2)
150 print cfm,p,gpd,a,rc

exhaust       water use     water use     indoor wet    indoor rh
cfm           lb/h          gpd           surf ft^2     %

107.6363      2.615563      7.535836      54.36884      54.07039

>does this show that adding 2.615563/60 pounds of water per minute to
>108 cubic feet per minute of air at 91.1f? the water humidifies the air
>and drops the dry bulb temperature to 80f?

i think so.

>if so, to me it suggests that you are cooling ventialtion air so as not
>to add sensible heat to the space. the sensible heat is still
>conducting in. the indoor air temperature is going to rise.

i don't think so. we have to cool 108 cfm of air from 91.1 to 80 f, which
takes about (91.1-80)108 = 1199 btu/h and remove (91.1-80)128 = 1421 of
sensible heat from the house. the total is 2620 btu/h. evaporating 2.61
lb/h of water takes about 2610 btu/h. close enough...

nick




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