The houses in the SW are usually built for the climate. Large overhangs
compared to eastern houses. There are many Adobe houses in the southwest.
Either packed dirt or Adobe brick. One foot thick walls provide very good
Anyway, water is not the problem. Electric is. Trying to squeeze more
cooling by recycling the already humid air will result in more electric use.
It is often already at 60% humidity when it comes out of the cooler.
Easy to save water. Just cut out a few lush lawns, golf courses and
(heaven forbid) football fields.
And the discussion has been mostly about worst case temperatures. All
those coolers are multiple speed and have thermostats. When the outside
temperature is only about 90 degrees the cooler normally is running on it's
lowest speed. Once the temperature drops to 80 or below the water is turned
5 GH328' Again with 106/65 db/wb
15 QS= GH + SOLAR HEAT GAIN
25 TS= 106 - 0.8*(106-65)
65 CFM = QS/((60*.075*.24)*(TC-TS))
With Solar Gain = 0, QS=GH
CFM to maintain the space at 80 is 453
Are we arguing some other air flow than that required to maintain the
1200 sq ft home at 80F?
Specific humidity at 106/65 is 26.9 grains,
Specific humidity at 73.2/65 is 79.2 grains
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
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?
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.
Or am I totally missing a concept here?
To maintain a space at 80 you have to supply air cooler than 80.
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?
What about some allowance for solar gain through windows.
With an unrealistically low value of 3315 Btu/hr, you would still need
over 400 CFM.
You have to calculate your airflow based on the temperature difference
between the room air and the supply air, not the ambient air and the
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.
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.
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.
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.
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.
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...
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.)
Looks like the box on the roof again...
That's a different 453...
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.
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.
...73.2 is the delivered airflow temp, right? But you added water.
Perhaps the wet bulb temp changed.
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 GH8'house conductance (Btu/h-F)
30 TA.1'average air temp (F)
40 WA=.0066'average humidity ratio
50 TC'comfort temp (F)
60 WC=.012'comfort humidity ratio
80 CFM=GH/(1000*PC/(TA-TC)-1)'exhaust fan size (cfm)
90 P=PC*CFM'water usage (lb/h)
100 GPD$*P/8.33'water usage (gpd)
110 PW=EXP(17.863-9621/(TC+460))'vapor pressure of wet surface ("Hg)
120 PH).921/(1+.62198/WC)'vapor pressure of house air ("Hg)
130 RC0*PH/PW'humidistat setting (%)
140 A*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 think so.
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...
80 degrees at 60% humidity is outside of the acceptable comfort level. 75
degrees is most desirable with that amount of humidity. Those that have
refrigeration keep their thermostats at about 80 but their humidity level in
the house is very low.
Perhaps you would be surprised at how many people in the SW don't have any
type of cooling. They do exactly what you refer to in letting the house
cool down overnight and then they close it up tight in the morning. To me
it is not a very nice way to live.
I didn't realize you were an expert on how houses are built in the SW.
Houses have been built there for hundreds of years taking into account the
need to withstand the heat without air conditioning. For example I have a
tile roof and about 18 inches of insulation in the attic. Thermopane
windows shaded by large overhang.
And, I still do not think you have calculated the required area for the pads
to evaporate the amount of water you specify. The initial drop taking very
dry outside air and dropping it to 75 degrees is an efficient process. To
maintain that temperature with the moisture content giving it a 40 to 50%
humidity level is not practical.
The systems are designed for worst case such as to take 120 degree at 30 %
humidity and get the air into the comfort zone. Those coolers have
thermostats and speed controls. Normally I find they are running at a very
low speed with only a light breeze to keep the house cool taking about the
same power as a fan.
The chart I referenced does not agree.
And it depends on how uncomfortable you are willing to be. After all even
in areas around Washington DC no one had air conditioning 50 years ago.
Cars including Cadillac did not have them.
Unimportant? You may need 10^9 square feet!! Of course your "numbers"
will still be valid. You just won't be able to get the cooler into a house.
Anyway, this is a rather unimportant study. Swamp coolers are a poor mans
air conditioner. They are so cheap to buy and operate it really down't
merit a study - $2 or $3 worth of water and .$30 worth of electric a month.
Now if you can come up with a way to vastly improved refrigeration, which
can cost well over $100 a month for electric, you will accomplish
something. Or perhaps a better way to heat your PA house in the winter.
Yes the usual box on the roof, 'cooling' dry outside air by converting
sensible heat to latent heat. Outside air is 'cooled' to a temperature
cooler than the space.
On the roof the ambients would be even higher.
Again this 91.1 is like an average temperature, that you could use for
easy math for energy/water consumption. You cannot design on averages
though to maintain comfort. My hometown has an average year round
temperature of 32F, but we need furnaces designed for at least -24F.
Yes insulation could be improved to reduce the load, but the airflow
required is still based on small temperature differentials with
For the sake of arguement, I left it equal to zero. Allowing for solar
gain adds a fair bit to the equation.
Unrealistic in that to me, it seems as if you, a solar heating
advocate, was not allowing for this in the cooling load. It appeared to
me that the cooling load you were hypothesising, was due only to a
difference in air temperature, similar to how heat loads are
calculated. In heating you design for no solar gain especially for
overnight lows, and the passive solar reduces run time of equipment
during the day. For cooling, the solar portion through windows,
ceilings etc is very signficant.
78 hour time constant? A time period of 3 to 5 times this to reach a
steady state temperature, so the structure is never too hot or too
cool? A thermal flywheel? You could crunch a million numbers here and
only convince me of the limited benefit of natural thermal storage. I
am not referring to frozen ice storage or 20 tons of rock in a crib
under the house, just typical building construction.
Yes the box on roof again
Yes but the 453 will keep the house at 80, with a sensible gain of
I should have read everything in full before starting my reply. I just
typed a whole bunch of stuff to say why you had to go with the box on
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 was not quite following the 'A'. It is a wetted slab area to
evaporate water? This is the portabella slab ? :)
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.
Your indoor wet bulb is about 67.88F. We have 3328 Btu/hr conducting
into the home (neglecting solar).
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
With evaporative cooling, the coldest you could make the air would be
67.88F. Realistically perhaps it will be 70.3F supply temperature with
a dewpoint well over the ASHRAE recommended 60F max.
3328/1.08/(80-70.3)= 318 CFM to control space temperature.
or if the machine were to blow fog into the space, air saturated at
3328/1.08/(80-67.88) = 254 CFM
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.
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
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.
Ventilation sensible load equals = 68x1.08x(106-80)= 1909 Btu/hr.
You could go back and refigure the cooling load without infiltration
but most likely it would be a similar result.
So it is starting to look impossible using evaporative cooling on the
room air, you might as well use the box on the roof. Take advantage of
how dry it is outside.
You pretty much have to use the box on the roof.
Which you have already realized.
Again if this was the box on the roof, 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 or just a
portable evaporative spot cooler, the evaporative cooling process would
follow a constant wet bulb line.
You need to program in the physics of an adiabiatic (sp?) saturator.
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
To cool from 106 to 80 would need roughly 108 CFM x 1.08 x 26 = 3033
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.
BUT, all that is being done here, is to 'cool' outside air so as not to
add to the space sensible cooling load. You have already said this was
a theoretical exhaust rate. It would overpower your infiltration rate
of 0.2 ACH (32 CFM).
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. Too cool the home you need to
supply air cooler than what the space is to be maintained at.
Physics dictate that evaporative cooling requires very high airflows
because of the low temperature differentials. To me you have modelled a
make up air system that adds no sensible heat to the room load.
The main way to reduce the airflow is to reduce the load by insulating.
Some people from my home town are down in Arizona designing systems for
insulated concrete domes.
NREL says it was the 24-hour average from 1961-1990 for July in Las Vegas.
That depends on the extremes and averaging period and house time constant.
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 :-)
I'm not sure what you mean by "this."
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.
Sure. A 10K Btu/F slab and a 128 Btu/h-F conductance. RC = C/G = 78 hours.
Depends how steady you want your state... e^-5 = 0.007 < 1% :-)
Pity. This is 300-year-old physics.
I usually work with formulae vs charts.
Yes. A minimum.
I think most slabs will evaporate water well, but there are other means.
OK, with w = 0.00384, altho we might more economically cool the house
at night and button it up during the day.
It is? OK. I'll take your word for that. Does that matter?
OK. BTW, that includes air leaks, which are double-counted here.
I'm not sure I understand or agree with all that.
Well, who cares, in this case? We only need 80 F.
What do you mean by "supply temperature" and how is it relevant here?
Where does the 70.3 come from and why is it relevant here?
Nobody's blowing fog. Why do we need 67.88 F air?
Not me. I would exhaust 11.41 pounds per hour of water vapor in 311 cfm.
Curiouser and curiouser...
I can almost understand and agree with that.
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.
But it isn't.
I don't see much relevance here.
I doan need no steeekeeng adiabattiacal sackturators.
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.
Why do we need to talk about boxes on roofs?
Yawn. Why 108 vs 311?
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.
Counting that infiltration makes indoor evap more efficient.
The sensible heat gain --***IS***-- included above.
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.)
You seem like a reasonable person. Perhaps you will change your mind.
Good idea. You've reminded me that a little air leakage is OK for cooling.
I spoke on solar heating Monolithic Domes at their first convention.
I do not doubt it is an accurate monthly average. Very accurate for
estimating energy consumption. Inappropriate for peak designs and
Lol, a high of zero with an overnight low of -24. I guess we design for
97.5% , residential design scenario
'This' meaning your original hypothetical heat gain, excluded solar.
When we allow for solar heat we will need to evap-cool a higher volume
of air. We are arguing why airflows are so high with evap-cooling.
Old tires are great for West Nile virus in PA. How much water is your
roof pond going to go through in the SW? :) 1200 sq ft at an inch per
day evaporation into the arid desert air, that's about 748 gallons per
Old Biot's about 300 now?
need some formulae to approximate wet bulb then.
portabella as in mushroom, something that would grow in a wet
Yes, I would say it does.
3328-32x1.08x(106-80)= 2429 Btu/hr
No infiltration allowed for in the calculation then. Will carry the
2429 later when I look again at the exhaust rate.
The ambient wet bulb is 65F. Evaporative cooling follows a constant wet
You have computed a 54% RH at 80F inside. I am treating that as your
design condition you are trying to maintain. It has a higher wet bulb
than the ambient air, so it sort of makes an allowance for internal
latent gains rather than a humidity level caused by evaporative
You need the air colder than 80, to maintain the room at 80. You
obviously realize the concept except you do not think it applies in
this case. You seem to insist that as water evaporates off of the floor
it absorbs the sensible heat that conducts into the house and the
temperature in the house remains constant.
The room is not going to stay at 80 while water evaporates from the
floor. You need an indoor evaporative cooler. This cooler will need to
supply air colder than 80 to handle the sensible gain that conducts in.
67.88 is the maximum limit as to how cool the air leaving the
evaporative cooler can be, based on 80F @54% RH entering. No one is
going to pour water all over their floor each night. So if you do not
use the 'unit in a box on the roof', you are going to have to use an
indoor unit, as you were talking about before you started with wetting
70.3 comes from lowering the drybulb to 80% the difference between 80
and 67.88, where 67.88 is the wet bulb temperature for 80 @ 54% RH. To
cool the space you would need the least amount of air if you could blow
the fog into it.
With evaporative cooling of this air, we are adding moisture, reducing
dry bulb temperature but the wet bulb is going to stay constant. We can
add moisture to this air until it is saturated at 67.88F.
Now you are talking about a realistic airflow. What happened to 108?
Evap cooling is all based on wet bulb temperatures which you are
neglecting. I was merely looking at how evaporating moisture into room
air 80 F @ 54% RH, would need 90 CFM of exhaust to get rid of moisture
therefore meaning 90 CFM of make up air. You had 32 CFM in your
calculation already, I was merely allowing for the other 68 CFM not
Please spare me on the horticulture. Again where is 311 CFM coming
from, it's a good airflow number. I would not say "air-leaky
house", I would say air being brought in mechanically and evap-cooled
and then air from the space is being exhausted mechanically or
You are going to need a box. Put it inside or outside.
Yah mon you sure do.
You have to deal with the wet bulb here. Evap-cooling is all about wet
bulbs. Is there even a published wet bulb that coincides with 91.1
average? Again averages are good for easy math for energy consumption
calculations, not comfort design. The evap-cool process follows a
constant wet bulb line.
You came up with 108 CFM, now you have revised it to 311. Are you
taking my word on the higher airflow now or trying to put it in 'my
terms'. If you are settled on the higher airflow and water
consumption per unit cooling then our argument is finished.
In a sensible cooling process yes 8705 Btu/hr. A horizontal line to the
left on the p-chart.
But we are doing this via evaporative cooling. So if it is then 310
CFM, we are changing it from 106/65 to 80/65.
We have added roughly 310x4.5x(68.2-26.9)/7000= 8.23 pounds per hour to
There will be no real infiltration as the exhaust dominates. So rather
than have this hot make up air hit the space first, you are cooling it
off with its own evap cooler. Dumping a gallon per hour on the floor
slab may help cool the desert ground more than the air. Heat would
conduct up from the slab more easily than it would flow 'down' from
the air above.
So you blow air through a wet cell media in a box. Put the box inside
or up on the roof.
Excluding infiltration and ventilation there is 2429 Btu of sensible
heat to deal with. So you can further cool this make up air, or you can
put in a second indoor evap cooler just for re-circulated air. It would
be more practical to continue to cool this make up air as it will have
a lower wet bulb temperature.
So to keep with the same 310 CFM, we allow further evap cooling and
determine then that we must cool the air down by 2429/1.08/310= 7.26F
The required supply condition to the space would be 80-7.26r.74F
db/65 wb (79.9 Grains)
So, 310 CFM of outdoor air has been evap-cooled down to 72.74F
We have used 310x4.5x(79.9-26.9)/7000.56 pounds of water per hour.
To check, assuming maintaining 80F @ 54% RH (82.7 Grains), with us
exhausting 82.7 grains and replacing with 26.9 grains then we are
getting rid of 11.12 pounds of moisture per hour with 310 CFM exhaust.
So it is removing an internal latent gain of 0.56 pounds of moisture
per hour or less as there have been some errors with the specific heat
of dry air carried in these calcs as well as using standard air
Now the sensible heat is included, before it was not.
Previously with a supply temp differential of only 6.8F, I mentioned
evaporative cooling would require over 1600 CFM per ton of sensible
cooling. With only 2429 Btu/hr conducting in, and the temperature
differential slightly higher than 6.8F, an airflow of 310 CFM is
Now your airflow has almost tripled and your water consumption has
increased four fold. You are now falling in line with the 'box on the
roof'. You were saying air flows and water consumption were excessive
in evap-coolers. Now your value of 310 CFM still seems low compared to
5000 CFM, but the only reason it is low is that the load we are
discussing is extremely low.
When you went from dealing with 91.1 to 106, everything should have
scaled up by 26/11.1 a factor of 2.34. Airflow has increased by a
factor of 2.87 and for water by my numbers at 10.56 lb/hr it is over
4 times the 2.6 pounds per hour you were previously proposing.
Double counting 32 CFM has been dealt with.
Now we are talking about a reasonable amount of air flowing to cool an
extremely small load. You seem to conclude now that you need the higher
flow rates and water consumption for evap-cooling.
Controlled ventilation is OK for cooling. Uncontrolled infiltration and
air leakage can be disastrous.
House design is one reason cooler volumes seem so high. Another is control.
Running during the day uses more water per day than only running at night
(with the water turned on) and buttoning up during the day, but some people
turn off the water at night and run the cooler during the day.
Maybe it's better to add some dirt, or leave out the water. Just white,
or shaded, with some air circulation under the shade.
I found this formula 33 on page 6.13 of the 1993 ASHRAE HOF:
W = (1093-0.556t*)Ws*-0.24(t-t*))/(1093+0.444t-t*), where t* and t are
the (F) wet and dry bulb temps, and Ws* corresponds to saturation at t*.
t = 106 and t* = 65 and Ws* = 0.013270 makes
W = (1093-0.556x65)0.01327-0.24(106-65))/(1093+0.444x106-65) = 0.003893,
and 7000W = 27.25.
I'm not sure exactly how this would work. Mushrooms need food. Lots of
decaying hay and straw and horse manure, around here. My neighbor's
basement slab seems fine, after 20 years of evaporation. I dunno. Maybe
the floor would excrete minerals, making a white dust or an undesirable
stain. Doom, aesthetically-speaking.
With w = 0.00384 or 0.003893 outdoors? It doesn't matter much.
I like the slab idea better. Less hardware and noise and electrical energy,
direct cooling of thermal mass, a larger mass temp swing with a cooler air
pool near the floor to store more coolth, good room temp control during the
day with a ceiling fan and a thermostat, effective cooling setbacks during
unoccupied times. Warm air rises, so massy cold floors and hot ceilings can
store natural coolth and solar heat well.
I figured 3328-32x1x(106-80)= 2496 Btu/hr for conduction alone...
20 GH'house conductance (Btu/h-F)
30 TA6'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).921/(1+.62198/WA)'vapor pressure in outdoor air ("Hg)
70 RA0*PA/PS'outdoor RH (%)
80 TC'comfort temp (F)
90 WC=.012'comfort humidity ratio
110 CFM=GH/(1000*PC/(TA-TC)-1)'exhaust fan size (cfm)
120 P=PC*CFM'water usage (lb/h)
130 GPD$*P/8.33'water usage (gpd)
140 PW=EXP(17.863-9621/(TC+460))'vapor pressure of wet surface ("Hg)
150 PH).921/(1+.62198/WC)'vapor pressure of house air ("Hg)
160 RC0*PH/PW'humidistat setting (%)
170 A*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
Exactly. A 1200 ft^2 dry slab and a ceiling fan making 1.5 Btu/h-F-ft^2 of
slow moving airfilm conductance in parallel with 4x0.1714x10^-8(460+80)^3
= 1.08 Btu/h-F-ft^2 of linearized radiation conductance could keep the room
80 F with 3328 Btu/h of total heat gain if (80-Ts)1200x2.58 = 3328, ie with
a slab temp Ts = 78.9 F. A damp slab can do a lot better, with a much higher
What do you mean by "supply temperature" and how is it relevant here?
That's one solution.
How about under the floor? I used a humidistat and a solenoid valve and
a soaker hose to automatically water tomato plants in a sunspace...
Not me. I would exhaust 8.73 pounds per hour of water vapor in 238 cfm
if running the cooler with 106 outdoor air (not a good idea, IMO.)
That used the 24-hour 91.1 average temp, vs 106 in the afternoon.
We could calculate one, based on NREL's w = 0.0066 average, but I'm
not sure that matters. Water use at 91.1 for 24 hours might be close
to the total daily use, with less at night and more during the day.
With the fan off, this house needs 24h(91.1-80)128 = 34K Btu/day of
cooling, right? We can store that in a 3.4 F temp swing in a 10K Btu/F
slab. If the average slab temp around 3 AM with 76.2 F outdoor air is
80-3.4/2 = 78.3, Pw is approximately 0.978 "Hg. At w = 0.012 indoors,
Ph = 0.566, so a 1200 ft^2 slab can evaporate about 0.1x1200(0.978-0.566)
= 49.4 lb/h of water... 49.4 = 60C0.075(0.012-0.384) makes C = 1339 cfm
of house exhaust air (including 32 cfm of air leaks.) If 34K Btu
= (49400+(76.2-78.3)(98+1339)dt, we can do all this in dt = 0.65 hours,
using much less water (4.1 gallons) and energy (32 Wh/day with Air King's
9166 2470 cfm 90 W fan) than a cooler that runs during the day.
We might also do this without water, with a much larger fan...
34K Btu = 0.65(78.3-76.2)(98+C) makes C = 24810 cfm :-)
The exhaust includes air infiltration.
I don't think so, if it's dry. The main mechanism for upward heat flow
in soil is evaporation from lower layers, upward water vapor migration
and condendsation above. We might insulate under the slab, but desert
ground also adds desirable thermal mass.
That's a plus for coolth storage and setbacks during unoccupied times.
It is high airflow per unit cooling. You run it during the day becuase
that is when it is the hottest. If it is 76 degrees at night then just
open the windows. That would cool down the house to help you for the
first while. Why even run it at night?
So we are arguing the specific humidity now of 106/65? Grains are easy
to work with as they are easy numbers. 26.9 or 27.5. The method of
deriving the value does not matter as long as you are consistent. Does
"0.556" make an allowance for absolute temperature.
portabella was sarcasm.
Is the specific heat of air only 0.2222222222222222222222222222 Btu/(lb
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.
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.
1000 is the latent heat of water here?
Now its 238 from 310 when all you changed is 32 CFM of infiltration.
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
Your wetted slab will experience a similar effect. Heat from the ground
is going to evaporate water.
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 and other
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. You can rationalize this all
you want with Keds on the roof, watering tomato plants, adding a
ceiling fan to forced convection from a floor slab, but your numbers
keep going up and up.
When you have a realistic load, you are going to end up with 1000s of
CFM in a real home.
If you want to ignore thermal mass and use lots of water and electricity...
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.
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.
With w = 0.012 indoors, by weight, Pc = 29.921/(1+0.62198/0.012) = 0.566 "Hg.
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.
No thanks. Feel free to do so yourself. What do you think you will find?
Sure. Round numbers, like 1 cfm = 1 Btu/h-F.
Yup. That helped, when cooling during the day (not a good idea, IMO.)
It has to, by energy conservation. Where else would the heat come from?
Not much, if the ground is dry or the slab has insulation beneath.
Exactly. They allow a larger mass temp swing and more precise room temp
control and setbacks during unoccupied times. Radiation helps too.
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.
Like I said before, if you got a rock bin or storing ice we have some
valid thermal capacitance. Typical house, no.
How many hours at 77,78, and 79 :)
Okay it's a column of mercury. Usually with heat transfer equations
when you use differentials it is fine to use F or C, but when you are
multiplying and dividing you usually require absolutes as in the
empirical exponential one in program line 50. It looked like you were
multiplying by degrees F with the .566
I lost my entire library recently to a flood and I cannot check all the
formulae you are quoting,
at 80 F.
You are assuming 0.012. You are not proving. I think in arid regions
with evap cooling indoor RH will be lower. The indoor condition will
have to be very close to the same wet bulb as the ambient.
You will find that the room temp will rise above 80.
Adding sensible and latent heat to air, is sort of like horizontal and
vertical vector components on a right angled triangle. The problem is
the angle of the resultant is fixed with evaporative cooling.
I think you will have problems coming up with 0.0012 unless it is for
internal latent gains such as from occupants, or the fact that your
design emulates a sauna. Don't forget sensible gains from occupants
You use round numbers here, but get very particular elsewhere,
especially in the empirical formulae. Why alter physical properties
such as the specific heat of air or the heat of evaporation of water?
You keep talking about super insulated homes, massive thermal storage.
My argument is evaporative cooling is high airflow per unit cooling.
You fictitious numbers do not change this. You are merely shaving the
Most of the heat comes from the fluid inside the bottle. The
evaporating water draws heat from the remaining liquid water which
cools off and draws most of its heat from the bottle.
Now of course the slab is insulated. I have seen a lot of insulated
slabs in northern homes with radiant floor heat. You keep changing the
house parameters, but evaporative cooling is still high airflow per
I guess its going to be the wetted ceiling slab next. You can model the
maximum size water droplets that surface tension will allow on the
underside of the ceiling without it raining indoors. Then you can
eliminate the ceiling fans and take advantage of the heat
stratification. It would the 8th wonder of the world, one up on the
Hanging Gardens of Babylon. You can name it the Inverted Pool of Pine
Let's see, 106 degree outdoor air is going to enter a room. Is it
better not to control and remove the sensible heat from this room air
first before it creates discomfort or to walk around on a wet floor?
Treating the air first before it can add its heat to the space is like
getting a vaccine to prevent a disease. Trying to deal with it later,
is like going to the hospital to be healed.
Again, people don't live in averages and not every home has both
parents working and the kids in daycare. People are at home on
weekends. I would still like to see you show how your 'comfort'
conditions will exist following psychrometric rules. It all revolves
around wet bulb temperatures.
Concrete stores about 25 Btu/F-ft^3, so a 1200 ft^2 x 4" thick slab
has C = 25x1200x4/12 = 10000 Btu/F of valid thermal capacitance, or
more, if it's in contact with soil with no insulation beneath.
The house needs 24h(91-80)128 = 34K Btu/day of cooling, no?
The slab can provide that with a 34K/C = 3.4 F temperature swing,
plus a smaller amount that comes from the slab airfilm resistance:
when it's 106 F outdoors, the house needs (106-80)128 = 3328 Btu/h of
cooling. A dry slab with a U1.1 radiation conductance under slow-moving
air from a ceiling fan with a U1.5 convection conductance can provide
3328 Btu/h of cooling with a 3328/1200/(U1.1+U1.5) = 1.1 F slab-air
temperature difference. To keep the house 80 F all day, the slab has to
start the day at 80-1.1-3.4 = 75.5 F.
More. It's a sine wave. With some indoor evaporation at night, the fan
can be a lot smaller.
Sorry to hear that. It's mostly just "Ohm's law for heatflow." Line 50
below is an approximation for the ("Hg) vapor pressure of water at temp
Ta (F). It's close to the numbers you'd find in table 3 in chapter 6 of
your ASHRAE HOF, if you still had one.
50 PS=EXP(17.863-9621/(TA+460))'outdoor vapor pressure at 100% RH
I'm not sure what you are getting at, Abby. We evaporate water inside
a house, with an exhaust fan that turns on when the indoor RH reaches
54%. The outdoor air is much drier. What's to prove? Would we need to
prove that a house with an AC and an 80 F thermostat setting is 80 F
indoors? We might wonder if the AC has sufficient capacity...
Why do you think that? Perhaps you imagine that an indoor wet surface
will just stay wet, without evaporating any water. I used a simple ASHRAE
swimming pool formula. A ft^2 of wet surface at temp Tw in air with Pa
"Hg loses 100A(Pw-Pa) Btu/h by evaporation, where Pw is the saturation vapor
pressure at Tw. It's a little surprising that this formula doesn't include
the air temp, but it's well-known and often used. An airtight house with
an indoor humidity source will eventually have 100% RH indoor air...
Brrr. Why can't it be warmer than 65 F? Especially if the water only
turns on when the room temp reaches 80 F, using a thermostat?
Why would you say that? Conservation of energy (energy in = energy out,
over a day) says otherwise.
I don't understand that. Perhaps it has more to do with a box on the roof
than this indoor scheme.
Why would you say that, exactly? We can always add more details, but
that won't change the performance much, IMO.
That's true, traditionally.
In what ways are they fictitious? OK, it's a tiny house. We could scale it
up and add lots of air leaks and remove the insulation and thermal mass :-)
I'd say we are discussing more than that.
Of course. I didn't see that.
Or on dry soil.
Sometimes, as potential improvements arise.
Nonono. Cool floor, warm low-e ceiling.
Not a good idea, IMO.
The floor should never be "wet," and the house should never admit
106 outdoor air. I wonder if Fu-Tung Cheng's fancy concrete floors
http://www.chengdesign.com are porous to water or water vapor.
Is indoor evaporation at night like Mother Theresa?
Evaporative cooling works on a constant wet bulb temperature. You need
to research this. Whether it is done realistically by blowing hot dry
air through a wetted media or blowing hot dry air, down from the
ceiling to a wet floor.
I even lost my Dectron Manuals in the flood. They taught ASHRAE!
Evaporative cooling works on a constant wet bulb temperature.
Try this, evaporate so much water into the air that the room goes to
71F db/ 65 wb. Then turn off the water and seal off everything. Let
sensible heat gain warm air to 80F @ 54%.
Constant wet bulb.
Or let room go to 91.1F db / 65 wb. Then shut off water and let
sensible heat warm air to 100F. Turn the water back on until you get
80F @54% RH.
Evaporative cooling works under a constant wet bulb temperature. Check
out the slope of a wet bulb line.
It's true period.
So you are now cooling an insulated slab. You are trying to reason that
evaporating water water all night now cools this slab off so when you
blow hot air down with your ceiling fans all day, it cools this air. By
the time the occupants who are not home all day get home the air has
cooled to 80?
They come home and flood the floors so they have a cool home when they
get up for breakfast?
Let's recommend concrete partition walls, concrete cabinets and no
baseboard trim. Gives some more thermal capacitance and not a source of
food for mold. :)
You are unsuccesfully trying to be non-traditional. Evaporative cooling
works, by depressing the dry bulb temperature of flowing air, to be
cooler than the desired space temperature you wish to maintain.
You add humidity directly to the air through a wetted media. You
depress the air temperatue of the supply to be cool enough below the
setpoint temperature to offset the sensible heat coming into the space.
It will be a small temperature differential compared to DX cooling and
it will have a high airflow per unit cooling.
Wetting a floor slab, can probably give an average dry bulb temperature
that falls in the comfort range, but the process to get there will take
occupants through discomfort of overly high humidity or dry bulb
temperature to get there.
It is also impractical, unhealthy and dangerous to keep wetting a floor
Inverted Pool of Pine needs no fan :)
concrete will be pourous but hopefully the insulation that you are now
using under your slab is a vapour barrier. If not maybe the excess
water being used would naturally irrigate a ring of lush vegetation
around the house to reduce the solar gain you have been neglecting.
No. The fans don't run when nobody's home. Why waste stored coolth?
The floors are automatically dampened beneath around 3 AM.
Evaporative cooling is not all the same. When water evaporates from
a slab, it can cool the slab with the help of air flowing inside a house.
Some people do.
Nope... 75.5 F with w = 0.012 is comfortable, and that's the min slab temp.
House air can be warmer.
Why would you say that? My neighbor's basement floor slab
has been slightly damp for 20 years with no problems.
Let's try logic. Would you agree that:
1. 80 F with w = 0.012 is comfortable.
2. At 80 F and 100% RH, Pw = 1.033 "Hg.
3. At Wc = 0.012, Pa = 0.566 "Hg.
4. P = 0.1A(Pw-Pa) lb/h will evaporate from an A ft^2 wet surface.
5. Evaporating P lb/h of water indoors makes about 1000P Btu/h of cooling.
6. Removing it takes C cfm of exhaust air, where P = 60C0.075(Wc-Wa) = 0.0367C.
7. Cooling the house and C cfm from Ta (F) to 80 F makes 1000P = (Ta-80)(98+C).
8. We can store 34K Btu of coolth in a 10K Btu/F slab with a 3.4 F temp swing.
What's the conclusion?
You might try this (from Lewis Carroll) for practice first:
(1) Every idea of mine, that cannot be expressed as a syllogism,
is really ridiculous.
(2) None of my ideas about bath-buns are worth writing down.
(3) No idea of mine, that fails to come true, can be expressed as a syllogism.
(4) I never have any really ridiculous idea, that I do not at once
refer to my solicitor.
(5) My dreams are all about bath-buns.
(6) I never refer any idea of mine to my solicitor,
unless it is worth writing down.
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