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re: evaporative cooler question
17 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???
 
>...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.

>people don't live in an average,they want to be cool at the 'peak'.

otoh, averages are good for estimating water and energy consumption,
and a house with a long time constant can average day and night temps.

>if they are trying to maintain 80f in the space, you need a different
>approach to calculate air flow.

maybe not.

>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

why would you think that's important in this case?

>if you want to maintain 80 then the temperature differential is only
>going to be 80-73.2= 6.8f.

why would you think that's important in this case?
 
>so for  'a ton' of sensible cooling you would need over 1600 cfm of
>evaporative cooling for the ambient.

but the house above only needs (105.9-80)128 = 3315 btu/h of peak
cooling and (91.1-80)128 = 1421 btu/h of average cooling. with a 78
hour time constant, it can cool itself more efficiently at night...

and it wouldn't be hard to add a few earthtubes on the north side and
sprinkle the ground above them to make 108 cfm of dry cool air, then
add moisture to that.

nick

it's a snap to save energy in this country. as soon as more people
become involved in the basic math of heat transfer and get a gut-level,
as well as intellectual, grasp on how a house works, solution after
solution will appear.
                                        tom smith, 1980




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