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

>again this 91.1 is like an average temperature, that you could use for
>easy math for energy/water consumption.

nrel says it was the 24-hour average from 1961-1990 for july in las vegas.

>you cannot design on averages though to maintain comfort.

that depends on the extremes and averaging period and house time constant.

>my hometown has an average year round temperature of 32f, but we need
>furnaces designed for at least -24f.

sure. a year is very large compared to a house time constant. i suppose
-24 is a 99 or 97.5 percentile, one sort of average. which to use depends
on the house time constant. maybe 97.5 for well-insulated masonry vs 99
for frame with poor insulation. you probably wouldn't design a heating
system capacity based on a 1-hour in 100-year extreme (99.99999%) temp :-)

>>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.
>unrealistic in that to me, it seems as if you, a solar heating
>advocate, was not allowing for this in the cooling load...

i'm not sure what you mean by "this." 

>...for cooling, the solar portion through windows, ceilings etc is very

it doesn't have to be. windows can be shaded. a roof with a little water
under some stone or a pile of old tires or old keds might be 65 f in july.
when i collect an old tire from a pa gas station, the owner gives me $1. 

>>...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.
>78 hour time constant?

sure. a 10k btu/f slab and a 128 btu/h-f conductance. rc = c/g = 78 hours. 

>a time period of 3 to 5 times this to reach a steady state temperature,

depends how steady you want your state... e^-5 = 0.007 < 1% :-)

>you could crunch a million numbers here and only convince me of
>the limited benefit of natural thermal storage.

pity. this  is  300-year-old physics.

>...the 453 will keep the house at 80, with a sensible gain of 3328.
>> >are we arguing some other air flow than that required to maintain
>the 1200 sq ft home at 80f?
>> yes...

>> >specific humidity at 106/65 is 26.9 grains,
>>so w = 26.9/7000 = 0.00384... again comparing apples and apples,
>>p = 60c0.075(wc-wi) = 0.03671c makes 36.71c = (106-80)(128+c) above,
>>so 1.412c = 128+c, and c = 311 cfm and p = 11.41 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.
>yes the supply air temp, to maintain 80 f. it tends to follow the wet
>bulb line on the psychrometric chart from 106 to 73.2 along the 65f wb
>line. the '80%' rate was what one manufacturer recommended. in this
>case its dewpoint was slightly in excess of what ashrae defines as
>neutral air. too keep the dewpoint down would require more air.

i usually work with formulae vs charts.

>> >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
>i was not quite following the 'a'. it is a wetted slab area to
>evaporate water?

yes. a minimum.

>this is the portabella slab ? :)

i think most slabs will evaporate water well, but there are other means. 
>let's just assume then that you were going to maintain the home at 80f
>and 54% rh.


>you are using an indoor evaporative cooler. its the design condition of
>106f db, 65 wb outside.

ok, with w = 0.00384, altho we might more economically cool the house
at night and button it up during the day. 
>your indoor wet bulb is about 67.88f.

it is? ok. i'll take your word for that. does that matter?

>we have 3328 btu/hr conducting into the home (neglecting solar).

ok. btw, that includes air leaks, which are double-counted here.

>your indoor wet bulb is higher than the outdoor, it compensates for
>internal latent gains. it is significantly higher than the outdoor
>specific humidity, but with evaporative cooling the key is constant wet

i'm not sure i understand or agree with all that.

>with evaporative cooling, the coldest you could make the air would be 67.88f.

well, who cares, in this case? we only need 80 f.

>realistically perhaps it will be 70.3f supply temperature with
>a dewpoint well over the ashrae recommended 60f max.

what do you mean by "supply temperature" and how is it relevant here?

>3328/1.08/(80-70.3)= 318 cfm to control space temperature.

where does the 70.3 come from and why is it relevant here?
>or if the machine were to blow fog into the space, air saturated at 67.88
>3328/1.08/(80-67.88) = 254 cfm

nobody's blowing fog. why do we need 67.88 f air?

>but you will still have to exhaust air to lower humidity and this adds
>to the load. you need to get rid of 0.002257x318x4.5=3.23 pounds per
>hour of moisture.

not me. i would exhaust 11.41 pounds per hour of water vapor in 311 cfm.

>the outdoor air at 106/65 has a specific humidity 26.9/7000. at 80f &
>54%rh the specific humidity is 82.7/7000
>3.23 = 4.5 x cfm x(82.7-26.9)/7000
>cfm = 90

curiouser and curiouser...

>so you would have the load conducting in 3328 btu/hr less your  0.2
>infiltration rate (32 cfm) and then add your net ventialtion load of
>90-32= 68 cfm. ventialtion  would over power infiltration.

i can almost understand and agree with that.

>ventilation sensible load equals = 68x1.08x(106-80)= 1909 btu/hr.

but not that.

my house conductance included 32 cfm of air leakage, but that's also
included in the exhaust cfm. in a sufficiently air-leaky house (311
cfm, above), the exhaust fan would never turn on. with enough green
plants in the house, the slab might never get dampened.

>you pretty much have to use the box on the roof.
>which you have already realized.

no way.

>> >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.
>again if this was the box on the roof...

but it isn't.

>the decrease in dry bulb would occur along a constant wet bulb line,
>in this case the outside wet bulb. if you are using a wetted slab which
>is impractical [who says?] or just a portable evaporative spot cooler,
>the evaporative cooling process would follow a constant wet bulb line.

i don't see much relevance here.

>you need to program in the physics of an adiabiatic (sp?) saturator.

i doan need no steeekeeng adiabattiacal sackturators.

>> >if so, to me it suggests that you are cooling ventilation 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...

so the sensible heat gain is included above. 

we could do this again at 106 f if you like. cooling 310 cfm of air from
106 to 80 f takes about (106-80)310 = 8060 btu/h. and remove (106-80)128
= 3328 of sensible heat from the house. the total is 11388 btu/h (mostly
for cooling air, which is why night cooling is better.) evaporating 11.41
lb/h of water takes about 11411 btu/h.

so again, the sensible heat gain is included above. 
>lets look at it with 106 db 65 wb, since it is more or less a published
>situation that coincides with one another. again for the box on the roof.

why do we need to talk about boxes on roofs?

>to cool from 106 to 80 would need roughly 108 cfm x 1.08 x 26 = 3033

yawn. why 108 vs 311?

>you would end up with air at 80 db and 65 wb (ignoring fan heat). the
>enthalpy would be more or less constant. we have added 0.0059 to the
>specific humidity, or 4.5 x .0059 x 108 = 2.8674 pounds of moisture.
>you will still argue close enough.

not me.

>but, all that is being done here, is to 'cool' outside air so as not to
>add to the space sensible cooling load.

let's try again: cooling 310 cfm of air from 106 to 80 f takes about
(106-80)310 = 8060 btu/h. then we remove (106-80)128 = 3328 btu/h of
sensible heat from the house. the total is 11388 btu/h. evaporating
11.41 lb/h of water takes about 11411 btu/h. 

so again, the sensible heat gain ***is*** included above. 

>you have already said this was a theoretical exhaust rate. it would
>overpower your infiltration rate of 0.2 ach (32 cfm).

counting that infiltration makes indoor evap more efficient.

>so all you are doing is creating ventilation air that does not add
>sensible heat to the space. you still need to do something about the
>sensible heat conducting into the space.

the sensible heat gain --***is***-- included above. 

>physics dictate that evaporative cooling requires very high airflows
>because of the low temperature differentials.

i disagree, altho physics requires conservation of energy. if we evaporate
11.41 pounds of water inside the house, ie 11,411 btu/h, where do you think
that energy comes from? i think 8060 btu/h comes from cooling 311 cfm of
106 f outdoor air to 80 f and 3328 btu/h comes in through the house walls
(and 32 cfm of double-counted air infiltration.)

>to me you have modelled a make up air system that adds no sensible heat
>to the room load. 

you seem like a reasonable person. perhaps you will change your mind.

>the main way to reduce the airflow is to reduce the load by insulating.

good idea. you've reminded me that a little air leakage is ok for cooling.

>some people from my home town are down in arizona designing systems for
>insulated concrete domes.

i spoke on solar heating monolithic domes at their first convention.


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