re: evaporative coolers question
19 jul 2001
david hatunen wrote:
>>>>nick, would it be possible to build something to cool a single car
>>>>garage sized room to use evaproative cooling?...
>>>here in arizona we have evaporative coolers to cool whole houses.
>>i suggested an outdoor evaporator that cooled water for an indoor
>given the way they work, that's rather pointless...
it works better in june than july. in june, we might cool some outdoor
air to a temperature close to the wet bulb temp, without adding humidity,
then raise the humidity by evaporating water indoors to further lower
the temperature of that air...
10 ta=103.5'nrel's average daily max temp, phoenix in june (f)
20 w=.0056'average humidity ratio
30 pa=29.921/(1+.62198/w)'vapor pressure ("hg)
50 for i=1 to 10
70 next i
80 print tw'wet bulb temp (f)
90 r=.6'humidity (%/100)
100 print 9621/(17.863-log(pa/r))-460'raise humidity, lower temp
but july air cooled to the wet bulb temp already has 60% rh :-(
we might dehumidify it with a desiccant, as in the genius system
by walter albers...
>a better scheme would use an evaporative cooler to provide the
>cooling air for a regular air conditioner condensor unit.
how about a cop over 20?
>>nrel says phoenix is 93.5 f on average in july, with a 0.0105
>>humidity ratio (vs 0.0056 in june.)
>>how well would that indirect system work now? what would the average
>>temp be in a 16' cube with r16 insulation and an optimum ventilation
>>rate with 4'x8' wet surfaces inside and out and 60% indoor humidity?
in july, we might hang light-colored shadecloth over the north wall
of and attach two layers of poly film to the ceiling to make a cool
radiator... 6 mil poly will support about 15 psf on 4' centers, but
we could use less water...
say the cube's thermal conductance to outdoors is 6x16x16ft^2/r16 = 96
btu/h-f, and the radiator has 1.5 btu/h-f-ft^2 of conductance to still
air and a radiation conductance of 4x0.1714x10^-8(460+70)^3 = 1, for
a total of 16x16x2.5 = 640 btu/h-f. the conductance from the radiator
to the outdoors is then 1/(1/96+1/640) = 83.5 btu/h-f.
we might use bowen's equation to predict the water temp tw if we ran
this all day. if the ratio of evaporative to convective heat loss is
100(pw-pa)/(tw-ta) = -1, regardless of windspeed, where pw is the
saturated vapor pressure near the water surface and pa = 0.497 "hg
for the outdoor air in july, and the shadecloth gains 1.5 btu/h-f-ft^2
by convection, it loses l = 150(pw-0.497)-1.5(93.5-tw) = 150pw+1.5tw-215
btu/h-ft^2 net, right? or 256l for one side of the whole cloth.
so we have 256l = (93.5-tw)83.5, ie 460pw+5.6tw = 752. using a clausius-
clapeyron approximation, pw = exp(17.863-9621/(tw+460)). taking the log
of both sides, tw = 9621/(17.863-ln((752-5.6tw)/460))-460, which leads
to another iteration...
20 for i=1 to 5
40 print tw
50 next i
if the water is 70.9 f, (93.5-70.9)83.5 = 1887 btu/h of heat flows into
the room from the outdoors, making the room air 70.9+1887/640 = 73.8,
with 0.497/(exp(17.863-9621/(460+73.8)) = 0.58, ie 58% rh.
with a $25 20 watt pump, the cop is 1887/3.41/20 = 28.
this would probably work better running at night, with a lower wet
bulb and more sky radiation loss and a shaded pool to store coolth.