Abby Normal wrote:

We ran the numbers before and it shows that more airflow and water was
needed when you treat the indoor air.

Wrong.

Nick

Take another bong hit, everytime the load call smaller and smaller
until it worked. I think you even learned the term effectiveness on
that one lol.

How about a 105F db 65F wb ambient, 10,000 Btu/hr sensible gain. You
can even confer with your ficticious friends down under. Pretty small
load in a fairly dry place. Read up on wet bulb first.

Maintain 80F with an indooor evaporative cooler.

Guess Clauisus is not up to the challenge

Keep it 60% RH then at 80F dry bulb inside. You can do it Nick, your
fans are watching

Again, sensible space gain in a small home is 10,000 Btu/hr when trying
to maintain 80f db and 60% rh under an ambient of 105 DB and 65 WB.
This challenge neglects internal latent gains to simplify things and
lets consider sea level elevation.

A few problems with using an evaporative cooler to treat indoor air in
doing this. 80 db & 60% RH would have a wet bulb temperature of close
to 70F. The ambient air has a wet bulb temperature of 65F. The wet bulb
dictates what the supply air temperature can be, and treating the
outside air CAN provide, if needed, a supply air temperature almost 5
degrees F cooler than treating indoor air will allow.

If your down under data was in fact treating indoor air only, it still
seemed like the wet bulb was constant, and if remarkably they did not
run the exhaust, then there probably was farily high infiltration
occuring, keeping humidity down. They were up against some pretty
insignificant conditions at night as well, low 80s fairly low RH, so
it really proves squat except there is a wet bulb angle that your
missaplication of physics never accounts for.

When you treat the outside air, you are pressurizing the home with cool
air, and must provide pressure relief whether a dedicated opening or an
envelope with the integrity of swiss cheese. This will over power any
typical natural infiltration and cool air will be leaving the space.

Treating indoor air only will have to allow for natural infiltration
which would be in addition to the 10,000 gain, but you will end up with
a steady exhaust fan running and the space will be negative with
respect to outdoors so infiltration will be ruled out and you would be
dealing with the make up air volume into the space instead.

The make up air adds heat directly to the room air. The exhaust rates
will be higher than you initially think. In fact the exhaust will cause
the amount of air to be treated by the indoor unit and the water
consumption to increase. As per our last arguement on this subject, you
will find the airflow and water consumption to spiral upwards from what
your physics say.

You will be using two fans with the indoor scheme, move more air, use
more water, plus when you put the amount of make up air into
perspective of a small home, the amount of hot air rapidly infiltrating
in will feel like a sirocco wind. The solution to this problem is to
duct the make up air right to the evap cooler. Suddenly comfort will
increase, exhaust rate required drops, the amount of water needed drops
and on paper we have a system that works.

So what the exercise will prove is that you can put the swamp cooler
inside but it still works the best when it directly treats the outside
air. So outside of some effects of the sun blaring on a box on the
roof, you will find that there is no benefit of having the cooler
inside.

I will check back in a week or so, to see what you come up with

So sayeth Nick "Two potential improvements: 1) exhaust house air to an
attic ("upducts")
or some other cavity bordering an exterior surface, eg a garage or
sunspace
or storage space, and thereby reduce the usual conductive heatflow from

the warmer outdoor air into the living space, or 2) use a humidistat
and
a reversible fan like Lasko's $55 2155A 16" window fan (90 watts at
2470 cfm
on high speed) and Grainger's 2A179 $88.15 programmable cycle timer and

its $4.37 5X852 octal socket to periodically reverse the fan direction
when
it needs to run, making a "Shurcliff lung" that turns all the cracks
and
crevices in the house envelope into bidirectional air-air heat
exchangers. "

If you were still on the indoor evaporative cooling scheme and then
exuasted the humid room air into the attic. This would make the make
the space below the attic negative with respect to the outdoors and the
attic positive with respect to the outdoors . An attic could tend to be
a bit of a solar collector and this ventialtion shceme could push air
hotter and more humid than the ambient down around recessed light
fixtures, junction boxes for ceiling fans etc. and into the living
space below

Likewise exhausting to an attached garage could pressurize the garage
and make the home negative with respect to the garage. So with this
pressure differential, air could flow from the garage and into the
home. Not a good scheme should there happen to be a car idling in
there. Could be a CO risk.

Abby Normal wrote:
So sayeth Nick "Two potential improvements: 1) exhaust house air to an
attic ("upducts")
or some other cavity bordering an exterior surface, eg a garage or
sunspace
or storage space, and thereby reduce the usual conductive heatflow from

the warmer outdoor air into the living space, or 2) use a humidistat
and
a reversible fan like Lasko's $55 2155A 16" window fan (90 watts at
2470 cfm
on high speed) and Grainger's 2A179 $88.15 programmable cycle timer and

its $4.37 5X852 octal socket to periodically reverse the fan direction
when
it needs to run, making a "Shurcliff lung" that turns all the cracks
and
crevices in the house envelope into bidirectional air-air heat
exchangers. "

If you were still on the indoor evaporative cooling scheme and then
exuasted the humid room air into the attic. This would make the make
the space below the attic negative with respect to the outdoors and the
attic positive with respect to the outdoors . An attic could tend to be
a bit of a solar collector and this ventialtion shceme could push air
hotter and more humid than the ambient down around recessed light
fixtures, junction boxes for ceiling fans etc. and into the living
space below

Likewise exhausting to an attached garage could pressurize the garage
and make the home negative with respect to the garage. So with this
pressure differential, air could flow from the garage and into the
home. Not a good scheme should there happen to be a car idling in
there. Could be a CO risk.


He doesn't seem to be around anymore!! He needs to spend a summer in
Yuma to prove his theory instead of trying to sell it here. I can't see
any great gain in power savings or efficiency even if his assumptions
were valid.


A couple days ago I went up on the roof, taking off the cooler cover,
checking the pads and getting set for summer. The temperature was in
the 80s. Tonight it is snowing so will could have put it off a bit.

He is around its just there isn't really much he can say. The indoor
scheme is not an improvement on evap cooling, it makes it worse. It's
okay to dream, you just have to believe a person when he explains why
the dream will not work once in a while.

Abby Normal wrote:
He is around its just there isn't really much he can say. The indoor
scheme is not an improvement on evap cooling, it makes it worse. It's
okay to dream, you just have to believe a person when he explains why
the dream will not work once in a while.


And my question always is "Why bother?". Running a fan and a little
evaporated water is so cheap. I just can't see how the indoor fans and
water he talks about could do any better. In my opinion he is lost in
the math and not looking at reality.

Then too it is the comfort. My wife loves evaporative cooling where she
can keep windows open and operate ovens. With refrigerations I would
get somewhat upset to come home finding the air conditioner compressor
working it's tail off and she has two ovens operating.

As I recommend, if he spent a summer in Yuma to prove his theory I might
be willing to listen further.

Just checking weatherbase on Yuma:

Average High July: 107
Average Low : 81
Precip: Nil
Days above 90: 31

or he might try Blythe, CA

Average High: 108
low: 81
Record: 119


If too hot he could try a cooler place such as Phoenix where the high is
only 105.

Lol if that does get Nick out from hiding nothing will. Could bring on
the curse of NREL's average temperature lol.

Maybe if we each talk about him 3 times like Beetlejuice Beetlejuice
Beetlejuice he will bite

Rich256 wrote:

Abby Normal wrote:

The indoor scheme is not an improvement on evap cooling, it makes it worse.

Wrong.

I just can't see how the indoor fans and water he talks about could do
any better.

With no house mass, an indoor scheme would do no better than a swamp
cooler with perfect controls (more than an on/off switch :-) But swamp
coolers don't have those controls, and indoor schemes don't need
big blowers and boxes :-)

Just checking weatherbase on Yuma:

Average High July: 107
Average Low : 81
Precip: Nil
Days above 90: 31

We also need the humidity ratio. With wo pounds of water per pound of
dry air outdoors, we could keep a house with 400 Btu/h-F of thermal
conductance to 107 F outdoor air and no internal heat gains 82.9 F
with wi = 0.0121 indoors (an efficient corner of an ASHRAE 55-2004
extended comfort zone with clo = 0.5 and vel = 0.5 m/s) with a swamp
cooler with perfect controls or indoor evaporation by evaporating
P pounds per hour of water with C cfm of airflow if (107-82.9)(400+C)
= 1000P, since C cfm of airflow has an effective conductance of about
C Btu/h-F and it takes about 1000 Btu to evaporate a pound of water.
So P = 0.0241C + 9.64 pounds per hour.

And a cubic foot of air weighs about 0.075 pounds and there are
60 minutes in 1 hour (you guys might want to argue about that :-),
so P = 0.075x60C(0.0121-wo), which makes C = 2.14/(0.00674-wo), so
a perfect swamp cooler or indoor scheme would only work with 107 F
outdoor air if wo were less than 0.00674, which seems unlikely.
NREL's nearest weather stations are San Diego and Tucson, with
wo = 0.0116 and 0.0109 in July. Phoenix has wo = 0.0105 in July.

With 81 F dry ventilation at night, the house would be comfortable
if wo 0.0121. With LOTS of night ventilation and thermal mass and
insulation, it would stay comfortable all day. An underground house
might also be comfortable.

If the average outdoor temp in Yuma is (107+81)/2 = 94 F, the house
needs 24h(94-82.9)400 = 106.6K Btu/day of cooling. That might come
from a floorslab with C = 4"/12x40'x60'x25Btu/F-ft^3 = 20K Btu/F
warming about 5 F over the day. We might cool a slab over a layer of
hollow blocks with a perfect swamp cooler and an underfloor blower,
or cool a plain slab with a soaker hose. (The soaker hose or blower
could also make an AC more efficient with cooler night air, if wo
were too high for evaporative cooling.)

If the average outdoor temp is 85 for 6 hours at night and the average
slab temp is 78, we can remove 106.6K Btu from the slab in 6 hours
(at 17.8K Btu/h) if (85-78)(400+C)+17.8K = 1000P, ie P = 0.007C+20.6
= 4.5C(0.0121-wo), which makes C = 4.58/(0.0105-wo), so we can do perfect
swamp or indoor evaporative cooling if wo 0.0105, which might happen
50% of the time, if wo = 0.0105. With C = 2470 cfm (a $55 90 W 16"
Lasko 2155A window fan), wo = 0.00592 max, and P = 37.9, ie 4.5 gph.

But wo 0.00592 may be unlikely in July, and water costs money, esp
with "Precip: Nil," so we might decide to use AC during most of July.

Nick

Sensible heat gain of 10,000 Btu/hr -excludes make up air/infiltration

105F db 65F wb ambient

maintain it at 80F inside.

Abby Normal wrote:

Sensible heat gain of 10,000 Btu/hr -excludes make up air/infiltration

105F db 65F wb ambient

.... 100(Pw-Pa)/(105-65) = -1 (Bowen, 1926) makes the water vapor pressures
Pw-Pa = 0.4 "Hg, and Pw = e^(17.863-9621/(460+65)) = 0.6296, approximately,
using a Clausius-Clapeyron approximation, so Pa = 0.2296, and the absolute
outdoor humidity ratio wo = 0.62198/(29.921/Pa-1) = 0.00481 pounds of water
per pound of dry air.

maintain it at 80F inside.

.... (105-80)C+10K = 1000P makes P = 0.025C + 10 pounds per hour of water,
since C cfm of airflow moves about C Btu/h-F and evaporating each pound
of water takes about 1000 Btu. With wi = 0.0120 (an efficient corner of
the ASHRAE 55-2004 comfort zone), P = 0.075x60C(wi-wo) = 0.03236C, since
each cubic foot of air weighs about 0.075 pounds and there are 60 minutes
each an hour (want to argue about that? :-) so C = 1360 cfm and P = 44,
ie we evaporate 5.28 gallons per hour of water.

At 80 F, Pw = e^(17.863-9621/(460+80)) = 1.047 "Hg, approximately,
and Pa = 29.921/(1+0.62198/0.0120) = 0.5663, so 44 = 0.1A(Pw-Pa) makes
A = 915 ft^2, ie we might evaporate 44 lb/h of water from a 915 ft^2 damp
80 F floorslab (using an ASHRAE pool formula) with a soaker hose and
a solenoid valve from a dead washing machine and a thermostat that opens
the valve when the room temp rises to 80 F and a 1360 cfm window exhaust
fan with a humidistat that turns it on when the RH rises to 56%, or use
a swamp cooler with perfect controls.

It's more efficient to do this with cool night air, with 1) a damp slab or
2) a perfect swamp cooler and a hollow slab and a separate slab blower...

If we need 10K Btu/h with 105 and 80 F temps and the house conductance
G = 10K/(105-80) = 400 Btu/h-F and the 24-hour average outdoor temp is 95,
we might need 24h(95-80)G = 144K Btu/day of coolth. If it's 90 F for 6
hours per night and we need 144K/6h = 24K Btu/h (2 tons) of cooling with
lots of slab thermal mass, (90-80)C+24K = 1000P makes P = 0.01C+24, and
P = 0.03236C makes C = 1073 cfm and P = 34.7, ie 4.2 gallons per hour.

Nick

Give up already on the flooded floor, you are just going to germinate
spores.

Maybe try chapter 51 of ASHRAE's 2003 Applications Handbook for some
reference material. We could get away from usuing volumetric flow and
look at mass to see if it makes a difference

Abby Normal wrote:

... I guess 'perfect' means 100% effective.

Wrong again, Abby. Just better controls for the swamp cooler.

... you seem to avoid the little challenge I keep putting your way.
105F db, 65 wb ambient.

Nope. I met your challenge, but you failed to understand it again :-)

Nick

Your physics are missapplied again Nick.

The Pine splintered when he pondered


"105F db 65F wb ambient



.... 100(Pw-Pa)/(105-65) = -1 (Bowen, 1926) makes the water vapor
pressures
Pw-Pa = 0.4 "Hg, and Pw = e^(17.863-9621/(460+65)) = 0.6296,
approximately,
using a Clausius-Clapeyron approximation, so Pa = 0.2296, and the
absolute
outdoor humidity ratio wo = 0.62198/(29.921/Pa-1) = 0.00481 pounds of
water
per pound of dry air.


maintain it at 80F inside.


.... (105-80)C+10K = 1000P makes P = 0.025C + 10 pounds per hour of
water,
since C cfm of airflow moves about C Btu/h-F and evaporating each pound

of water takes about 1000 Btu. With wi = 0.0120 (an efficient corner of

the ASHRAE 55-2004 comfort zone), P = 0.075x60C(wi-wo) = 0.03236C,
since
each cubic foot of air weighs about 0.075 pounds and there are 60
minutes
each an hour (want to argue about that? :-) so C = 1360 cfm and P = 44,

ie we evaporate 5.28 gallons per hour of water. "

Well I can look at it from your newly chosen conditions then, you are
going to maintain 80F and W=0.012 or 84 grains, so you would NOW be
setting the 'improved controls' to 55% RH then rather 60%. To minimze
air flow and water use, then an indoor cooler could saturate this air
and produce air saturated at 68.13F, 103.3 Grains.

Would need to saturate 2901 CFM of room air through the indoor unit,
exhaust 1007 CFM and use 35.99 pounds per hour, there an improvement on
your 44 pounds per hour and a little less exhaust.

A 100% effective outdoor unit would still be 617 CFM and 25 pounds of
water.

A realistic outdoor unit at about 80% effectiveness would produce air
at 73 db/65 wb, 79.5 grains

Ignoring fan heat

CFM = 10,000/(1.08 x (80-73)) =1322.8

Water used = 4.5 x 1322.8 x (79.5-28.4)=43.5 lbs per hour, a similar
amount of air and water as you were guessing an indoor unit would move.

An indoor unit of comparable 80% effectiveness could produce air at
80-(.8x(80-68.13)=70.5 F (70.5Fdb, 68.13 WB, 99.5 Grains). Take a stab
at how much air and water the indoor scheme ends up using. Be a lot
more than what a comparable outdoor unit would use.

The outdoor unit is inherently superior as it directly treats the heat
of the outside air, and can supply cooler air to the space than what an
indoor unit can. Less air, less water, automatically deals with the
sensible heat of outside air, and only needs one fan. Indoor scheme you
need two fans, more water, need to move more air. FLAWED.

The problem is you refuse to factor in the importance of the wet bulb
temperature. I have given you a couple references now to this but you
refuse to consider this. So it is pretty hard to argue with someone who
does not understand an important concept. Typical of an electrical EIT,
perhaps the world's oldest one

Flooding the floor or even a 'dampened slab' is a stupid idea, of
similar magnitude of stupidity as exhausting the air to the attic or to
an attached garage.

Spore cases have an enzyme coating and the spores themselves do not
immediately extract moisture from the ambient air to germninate. What
spores need is a wet food source and dampening a slab and in particular
any dirt on that floor or say where the floor meets a baseboard trim or
water wicking up that trim or the paper in the sheet rock at the
bottom of the walls and you have mold. The moisture in the food
dissolves the enzymes on the spore case and creates a 'nutrient broth'
that enters the spore by osmosis. Osmosis being the same principle by
which I am trying to get the concept of wet bulb to enter your mind. So
mold does not really care what the RH of the air is, but mold does care
about what the moisture content of the food is.

If you want to improve evaporative cooler, look at indirect evap
cooling, been around for a while, just new to you. Keep trying to
re-invent the wheel Rube just really think about it when you give
people advice to blow air to an attached garage.

Abby Normal wrote:

"105F db 65F wb ambient

... 100(Pw-Pa)/(105-65) = -1 (Bowen, 1926) makes the water vapor
pressures Pw-Pa = 0.4 "Hg, and Pw = e^(17.863-9621/(460+65)) = 0.6296,
approximately, using a Clausius-Clapeyron approximation, so Pa = 0.2296,
and the absolute outdoor humidity ratio wo = 0.62198/(29.921/Pa-1)
= 0.00481 pounds of water per pound of dry air.

maintain it at 80F inside.

... (105-80)C+10K = 1000P makes P = 0.025C + 10 pounds per hour of water,
since C cfm of airflow moves about C Btu/h-F and evaporating each pound of
water takes about 1000 Btu. With wi = 0.0120 (an efficient corner of the
ASHRAE 55-2004 comfort zone), P = 0.075x60C(wi-wo) = 0.03236C, since each
cubic foot of air weighs about 0.075 pounds and there are 60 minutes in
each hour (want to argue about that? :-) so C = 1360 cfm and P = 44,
ie we evaporate 5.28 gallons per hour of water. "

Well I can look at it from your newly chosen conditions...

Nothing new about it. You merely said "maintain it at 80F inside."

Would need to saturate 2901 CFM of room air through the indoor unit,
exhaust 1007 CFM and use 35.99 pounds per hour, there an improvement on
your 44 pounds per hour and a little less exhaust.

This makes no sense to me. Care to explain more?

A 100% effective outdoor unit would still be 617 CFM and 25 pounds of water.

This makes no sense to me. Care to explain more?

A realistic outdoor unit at about 80% effectiveness would produce air
at 73 db/65 wb, 79.5 grains

So?

Ignoring fan heat

But the outdoor unit has a big blower, vs a small exhaust fan...

CFM = 10,000/(1.08 x (80-73)) =1322.8

Water used = 4.5 x 1322.8 x (79.5-28.4)=43.5 lbs per hour, a similar
amount of air and water as you were guessing an indoor unit would move.

I've been saying that a swamp cooler with appropriate controls would
achieve the same performance as an indoor scheme.

An indoor unit of comparable 80% effectiveness could produce air at
80-(.8x(80-68.13)=70.5 F (70.5Fdb, 68.13 WB, 99.5 Grains).

But "indoor units" are 100% effective.

Take a stab at how much air and water the indoor scheme ends up using.

I've done that in great detail, several times.

Be a lot more than what a comparable outdoor unit would use.

Wrong.

The outdoor unit is inherently superior as it directly treats the heat
of the outside air, and can supply cooler air to the space than what an
indoor unit can. Less air, less water, automatically deals with the
sensible heat of outside air, and only needs one fan. Indoor scheme you
need two fans, more water, need to move more air. FLAWED.

Wrong.

The problem is you refuse to factor in the importance of the wet bulb
temperature.

It seems to me that the problems are your arrogance and ignorance. If you
could set aside the arrogance, you might cure your ignorance by learning
more about the 300-year-old physics you talk about with no understanding.

Nick

I went through the methodolgy in a previous post, it sort of
illustrates how the wet bulb temperature of the air entering an
evaporative cooler will directly impact the leaving dry bulb
temperature. The difference between this leaving dry bulb and the room
dry bulb is used to determine how much air is required for the cooling
process. The outdoor unit CAN supply air with a cooler dry bulb
temperature than what an indoor evaporative cooler can. You do not
seem to think this matters.

The change in moisture content between the air entering and leaving the
cooler is how much water the process needs. Again look at Chapter 51 of
the 2003 handbook. I had your favourite Kreider and Rabl prior to the
hurricane, and if the book would have survived the storm, I would see
if I could give you a page or two there for your to read. How about
MacQusiton and Parker, I still have a copy of that in which I could
refer you to a few pages if you will not take the 2003 Handbook as
adequate reference?

You are actually describing yourself Nick "It seems to me that the
problems are your arrogance and ignorance. If you could set aside the
arrogance, you might cure your ignorance by learning more about the
300-year-old physics you talk about with no understanding." You keep
going on and on about evaporative cooling yet you ignore the most
important fundamental.

You are ignorant of the key fundamental of evaporative cooling and your
arrogance does not let you see how you could possibly be wrong. I
suggest you see what ASHRAE has to say and compare it to your 300 year
old physics so you can see where you went wrong.

Some people do HVAC for a living, others do not and want to dream.
Sometimes you need a dreamer to revolutionize things, but sometimes you
need to slap the dreamer in the head to wake him up.

Maybe learn the 100 year plus old physics of a guy who worked for
Buffalo Forge and came up with an apparaturs that controlled
temperature and humidity. He looked at dalton's law of partial
pressures, ideal gas law,saturated water vapour pressures and developed
a chart. It related the temperature of air to the amount of moisture in
the air. The most useful development ever in the history of controlling
the indoor environment.

Do you even acknowledge the problems of blowing the exhaust into an
attic or an attached garage?

The term effective relates to how the dry bulb temperature of air can
be lowered using evaporative cooling. If it was 100% effective, then
the dry bulb temperature would be depressed down to the wet bulb
temperature.

With air ambient air at 105 db and 65 wb, if you could apply
evaporative cooling that was 100% effective you would have saturated
air at 65F.

To further explain effectiveness as used when a process is 80%
effective the dry bulb temperature is depressed by 80% of the
theoretical maximum, in this case by 0.80 x (105-65)= 32 degrees. With
a typical 80% effectiveness, air entering at 105 db 65 wb, would leave
at 73 wb and 65 db.

You were using the term 'perfect' earlier and in a previous thread
using it to describe air that was saturated before being blown through
hollow blocks under a floor. This implied a 100% effective situation.

So in my ealier example I went through how much air a 100% effective
outdoor unit and indoor unit would go through, and previously I think
you were saying to set the dehumidistat at 60%. You then were talking
about 55% humidity so I re-did the indoor calc with 80F 55% air getting
humidified.

You should be able to follow through my example, just keep in mind the
conversion factor of 7000 when dealing with grains vs pounds and you
should do fine.

There will be some fan heat as an indoor unit pulls air through a media
as well. Plus then energy to run an exhaust fan. Indoor scheme will use
more energy to run fans. So combined you are moving a lot more air with
the indoor scheme and running two fans.

I think you will find that the indoor unit works best when the make up
air is ducted directly to it. Then when you explore the concept more,
you may realize that it will work the best when you just treat outside
air and do not recirculate.

The 2003 handbook goes into some mixed air applications, but you will
notice they are not using the conditioned space as a mixing box for
some strange reason.

They have an automatic on/off switch called a thermostat by the way. It
can trigger a fan to run, a pump to pump. It is an improvement over a
simple on/off switch due to the fact that it is a heat activated
switch.

The flooded floor scheme is a dog named Rube.

typo 73 db and 65 wb

Abby Normal wrote:

You were using the term 'perfect' earlier and in a previous thread
using it to describe air that was saturated before being blown through
hollow blocks under a floor.

Wrong again. This discussion is hopeless.

Nick

Abby Normal wrote:

You were using the term 'perfect' earlier and in a previous thread
using it to describe air that was saturated before being blown through
hollow blocks under a floor.

A "perfect swamp cooler," as I used the phrase, would have RH and temp
controls, which has nothing to do with how close it can cool air to
the wet bulb temp. I was thinking a swamp cooler like that could achieve
the same performance as any indoor scheme...

But thinking further, that isn't true, for swamp coolers that don't
recirculate indoor air. Swamp coolers with RH and temp controls may
still be less efficient than indoor schemes for houses with natural
air leakage, ie all houses :-)

For instance, in this case, the indoor scheme required
1360 cfm of exhaust air and 44 pounds per hour of water:

Sensible heat gain of 10,000 Btu/hr -excludes make up air/infiltration

105F db 65F wb ambient

.... 100(Pw-Pa)/(105-65) = -1 (Bowen, 1926) makes the water vapor pressures
Pw-Pa = 0.4 "Hg, and Pw = e^(17.863-9621/(460+65)) = 0.6296, approximately,
using a Clausius-Clapeyron approximation, so Pa = 0.2296, and the absolute
outdoor humidity ratio wo = 0.62198/(29.921/Pa-1) = 0.00481 pounds of water
per pound of dry air.

maintain it at 80F inside.

.... (105-80)C+10K = 1000P makes P = 0.025C + 10 pounds per hour of water,
since C cfm of airflow moves about C Btu/h-F and evaporating each pound
of water takes about 1000 Btu. With wi = 0.0120 (an efficient corner of
the ASHRAE 55-2004 comfort zone), P = 0.075x60C(wi-wo) = 0.03236C, since
each cubic foot of air weighs about 0.075 pounds and there are 60 minutes
each an hour (want to argue about that? :-) so C = 1360 cfm and P = 44,
ie we evaporate 5.28 gallons per hour of water.

Now suppose the house leaks 200 cfm of air (about average in the US.)
In the indoor scheme, the fan would only move 1160 cfm, and the cooler
might reduce its airflow to 1160 cfm (real vs inflated cooler cfm :-),
so the cooler and the indoor scheme would have equivalent performance.

But what can the cooler do if the house leaks more air or we need less
cooling? Suppose we only need 200 cfm of outdoor air? It can't reduce
the airflow to zero and still evaporate water, so it will have to move
excess outdoor air through the house and use excess water, ie the indoor
scheme will use less air and water in this case.

For equivalent performance, it seems we also have to add a motorized
bypass damper to the swamp cooler to allow indoor air recirculation.

Nick

Sensible heat gain of 10,000 Btu/hr -excludes make up air/infiltration

105F db 65F wb ambient

... 100(Pw-Pa)/(105-65) = -1 (Bowen, 1926) makes the water vapor pressures
Pw-Pa = 0.4 "Hg, and Pw = e^(17.863-9621/(460+65)) = 0.6296, approximately,
using a Clausius-Clapeyron approximation, so Pa = 0.2296, and the absolute
outdoor humidity ratio wo = 0.62198/(29.921/Pa-1) = 0.00481 pounds of water
per pound of dry air.

maintain it at 80F inside.

... (105-80)C+10K = 1000P makes P = 0.025C + 10 pounds per hour of water,
since C cfm of airflow moves about C Btu/h-F and evaporating each pound
of water takes about 1000 Btu. With wi = 0.0120 (an efficient corner of
the ASHRAE 55-2004 comfort zone), P = 0.075x60C(wi-wo) = 0.03236C, since
each cubic foot of air weighs about 0.075 pounds and there are 60 minutes
each an hour (want to argue about that? :-) so C = 1360 cfm and P = 44,
ie we evaporate 5.28 gallons per hour of water.

Now suppose the house leaks 200 cfm of air (about average in the US.)
In the indoor scheme, the fan would only move 1160 cfm, and the cooler
might reduce its airflow to 1160 cfm (real vs inflated cooler cfm :-),
so the cooler and the indoor scheme would have equivalent performance.

But what can the cooler do if the house leaks more air or we need less
cooling? Suppose we only need 200 cfm of outdoor air? It can't reduce
the airflow to zero and still evaporate water, so it will have to move
excess outdoor air through the house and use excess water, ie the indoor
scheme will use less air and water in this case.

Working backwards, if C = 200, P = 0.03236C = 6.472 pounds per hour,
and 1000P = 6472 Btu/h = (105-80)200 + Q makes cooling load Q = 1472
Btu/h. If the swamp cooler needs (say) 500 cfm min to evaporate water,
(105-80)500 + 1472 = 1000P makes P = 13.92 pounds per hour, over twice
the water required by the indoor scheme.

Nick

Abby Normal wrote:
I realize it is hopeless, you are too obtuse to see it any other way



This is moving away from the discussion but think you can answer my
question.

Many years ago, when I lived in the San Fernando Valley I had a
refrigeration unit. The humidity levels there were marginal for use of
a swamp cooler. However, with 30% RH they would provide some cooling.

What bothered me was the compressor/condenser sitting out in the sun at
115 degrees. I had thoughts about getting a window type swamp cooler
and sitting it on the ground by the unit and blowing cooled air on the
coils. I never have heard of anyone doing that.

The 30% RH came with temperatures in the 90s. At 115 the RH was usually
very low.

Do you think that set-up would be worthwhile?

Rheam at that time did make a unit with copper coils. It had pump that
sprayed water on the coils. It had a water tank like a swamp cooler.
About 1/3 the way down the coils was a small trough that caught a little
water which was drained off, keeping the mineral content down.

I remember an office building that had a fountain that was actually the
cooling pond for their air conditioner. They had to raise the fountain
nozzles to get more evaporation when the water got too warm.

The splintered pine pondered

A "perfect swamp cooler," as I used the phrase, would have RH and temp controls, which has nothing to do with how close it can cool air to the wet bulb temp. I was thinking a swamp cooler like that could achieve
the same performance as any indoor scheme...

But thinking further, that isn't true, for swamp coolers that don't recirculate indoor air. Swamp coolers with RH and temp controls may still be less efficient than indoor schemes for houses with natural air leakage, ie all houses :-)


For instance, in this case, the indoor scheme required 1360 cfm of exhaust air and 44 pounds per hour of water:"


As far as perfect controls, the outdoor system is simpler and
therefore superior. A thermostat turns it on, pressure relief gets rid
of the moisture. KISS baby.

Evaporative cooling has everything to do with a wet bulb temperature.
Until you understand this, you do not really know what the hell you are
talking about. Print off some posts and go see a PE friend or two.

When I first saw this thread, you cross posted to
"sci.engr.heat-vent-ac". You were going on about how some ozzies were
having wondrous results running an evaporative cooler that recirculated
air. Indoor cooler has a fan, powered exhaust. Two fans in the original
cross posted example. Go look at the indoor and outdoor wet bulbs in
that ozzie example, about the same

Recently you have morphed it back to your pathetic indoor scheme that
involves flooding the floor.

With the traditional outdoor evaporative cooler, the space is
pressurized. There is an air exchange between the home and the outside
and this exchange involves cool humid air (with repesct to the ambient)
migrating out of the home to the outside.

You are quite proud of your physics but do not pay much attention to
pressure differentials else you would not be suggesting to people to
pressurize an attic or an attached garage. See the problem with that
yet by the way? The garage idea is dangerous, a life safety issue.

An indoor evaporative cooler will be prone to natural air infiltration
until the exhaust fan runs. Under design conditions the exhaust will
run steady. The negative pressure caused by the exhaust will cause
outdoor air to transfer in at a much higher rate than it ever would
naturally.

The flooded floor will be prone to mold problems. There are several
steps involved to get this heat transfer and each step is an
inefficiency. You will use even more water and airflow than you
incorrectly calculate. A cool ceiling would be infinitely superior to a
flooded floor. But when you rethink the slab, think again that the
coldest you get ever get the slab would be the wet bulb temperature of
the room air.

I gave you an example of a northern home in a heating situation before.
Floor slab in contact with soil most likely 42F. Heat loss through the
slab 2 btu/hr per square foot and this slab will be significantly
cooler than anything flooding can do to it. But now with a warmer slab
you are going to transfer more heat per square foot out of the room.
Scheme is a dog pure and simple.


The pine splintered some more

... 100(Pw-Pa)/(105-65) = -1 (Bowen, 1926) makes the water vapor pressures Pw-Pa = 0.4 "Hg, and Pw = e^
(17.863-9621/(460+65)) = 0.6296, approximately, using a Clausius-Clapeyron approximation, so Pa = 0.2296,
and the absolute outdoor humidity ratio wo = 0.62198/(29.921/Pa-1) = 0.00481 pounds of water per pound of dry
air.

... (105-80)C+10K = 1000P makes P = 0.025C + 10 pounds per hour of water, since C cfm of airflow moves about
C Btu/h-F and evaporating each pound of water takes about 1000 Btu. With wi = 0.0120 (an efficient corner of
the ASHRAE 55-2004 comfort zone), P = 0.075x60C(wi-wo) = 0.03236C, since each cubic foot of air weighs about
0.075 pounds and there are 60 minutes each an hour (want to argue about that? :-) so C = 1360 cfm and P = 44,
ie we evaporate 5.28 gallons per hour of water."

So far you are dreaming that the flooded floor will approach the
performance of a typical swamp cooler on the roof with those numbers.
As I tried to explain to you the 1360 CFM and 44 pounds of water is
typical of an 80% effective swamp cooler. You are dreaming that it will
work as good, there are numerous other problems in the flooded floor
scheme that you will not acknowledge.

yet another splinter

Now suppose the house leaks 200 cfm of air (about average in the US.) In the indoor scheme, the fan would only move 1160 cfm, and the cooler might reduce its airflow to 1160 cfm (real vs inflated cooler cfm :-), so the cooler and the indoor scheme would have equivalent performance.

But what can the cooler do if the house leaks more air or we need less cooling? Suppose we only need 200 cfm of outdoor air? It can't reduce the airflow to zero and still evaporate water, so it will have to move excess outdoor air through the house and use excess water, ie the indoor scheme will use less air and water in this case.

For equivalent performance, it seems we also have to add a motorized bypass damper to the swamp cooler to
allow indoor air recirculation."

Lol Rube, it cycles off with the thermostat. Don't start pricing bypass
dampers or modualting exhaust fan controls at Grainger just yet, forget
buying a fanhandler

You run a 1360 CFM exhaust fan steady in a small home and that is going
to be your air exchange. Maybe some wind effect of from a tornando or a
hurricane will make a difference. :-)

Maybe don't cross post so much and spare your EIT ego a beating from
illiterate HVAC criminals . We do this for a living and we are
responsible for advise we give and the things we do. You spew out crap
with no consequences.

The traditional swamp cooler will over power natural infiltration, you
do not comprehend this. An indoor cooler with an exhaust fan will have
a make up air rate far greater than 200 CFM from natural causes such as
wind effect.

Maybe look at what your fellow solar geeks at the Florida Solar Center
say about the effects of Space Pressurization.
http://www.fsec.ucf.edu/bldg/science/mold/index.htm

Evaporative cooling could lower the temperature of the air entering a
condneser coil in almost any environment.

Roof is always hotter than documented in-the-shade dry bulb
temperatures. So a humif environment in the low 90s would be well over
100 on the roof top. Evaporative cooling could pull the air temp down
even in a humid environment.

A lawn sprinkler on a condenser coil makes a big difference on the
hottest of days

Working backwards it seems to show that you have taken another hit from
your bong and come up with a microscopic cooling load of 1.472 Btu/hr
in a futile attempt to prove something.

Outdoor swamp cooler, 80% effective on the microscopic load, supply air
temp 73, airflow 1472/(1.08x7)=195 CFM

Water required at 80% effective, 4.5 x 195 x (79.5-28.4)/7000= 6.4
pounds per hour. Your flawed scheme on paper tries to approach what a
typical outdoor system does. I have said this two times before. Maybe
3rd time is a charm.

When you put it under a real load, you make the house a mixing box with
either the flooded floor or using an indoor evaporative cooler. Just
another inherent flaw.

The occupants get to experience the hot drafts infiltrating in, unless
of course you duct the make up air right to the swamp cooler. As I have
also been trying to hammer into your head, once you realize the
advantage of ducting the make up air directly to the indoor evaporative
cooler, a light bulb should illuminate and you will soon find that it
is best just to run it on straight outside air.

At 100% effective, an outdoor unit under the microscopic load

1472/(1.08x 15)= 91 CFM

Water required 4.5 x 91 x 64.1/7000= 3.75 pounds per hour

To put 500 CFM in context then, it is inefficient as there is not a lot
of temperature differential between the supply and room air.

Temperature differential between room and supply temp = 1472/(1.08 x
500)= 2.73 degrees. therefore you need a supply dry bulb of 80-2.73=
77.27.

W @ 77.27 db 65 wb =72.6 Grains

Water required 4.5 x 500 x 44.2/7000=14.2 pounds per hour.

NIck, please, go read some of the references I have recommended.

Abby Normal wrote:
Evaporative cooling could lower the temperature of the air entering a
condneser coil in almost any environment.

Roof is always hotter than documented in-the-shade dry bulb
temperatures. So a humif environment in the low 90s would be well over
100 on the roof top. Evaporative cooling could pull the air temp down
even in a humid environment.

A lawn sprinkler on a condenser coil makes a big difference on the
hottest of days



My assumption was that if the temperature of the air cooling the
condenser was lowered to about 80 or below it should vastly improve the
efficiency of the unit, similar to having a cooling pond. These units
were on a slab on the ground but in full sun.

I would not want to have excessive water spraying on Aluminum coils.

yes spraying water on the coil could make you use more water than
necessary. I was agreeing with you that evaporatively cooling the
ambient air would improve the condenser performance, just using the
lawn sprinkler as an example.

I have a salt corrosion problem here, all aluminum spinefin is the
best, inherently superior to any type of corrosion protection coating
applied to a copper tube, aluminum fin coil.

If they wash uncoated aluminum fin/copper tube coils frequently they
last longer.

Abby Normal wrote:
yes spraying water on the coil could make you use more water than
necessary. I was agreeing with you that evaporatively cooling the
ambient air would improve the condenser performance, just using the
lawn sprinkler as an example.

I have a salt corrosion problem here, all aluminum spinefin is the
best, inherently superior to any type of corrosion protection coating
applied to a copper tube, aluminum fin coil.

If they wash uncoated aluminum fin/copper tube coils frequently they
last longer.



The Rheam unit with copper coils I talked about didn't use much water.
Most of the water was caught in a pan just like a swamp cooler. Just a
little was drained off through the little trough to keep the mineral
content down. Just thought that a swamp cooler sitting on the ground
next to the condenser would be a good compromise for an aluminum coil
unit. What puzzles me is that I never see it done.

Another item I thought of was to put a small evaporative unit to
somewhat cool the attic. That would more or less eliminate the ceilings
as a heat source.

Gets done directly on PTACS and window shakers all the time Rich they
sling the condensate at the condenser coil. It is being done
commercially treating the condenser air with a wet cell media.

Venting attics is a northern concept to stop ice dams. In the SE and
the SW I think a sealed attic with insulation on the under side of the
pitch and any gable walls above ceiling plane is the way to go. If you
air condition the attic space there would be no real overall gain on
the cooling load. Subtract the heat down though the ceiling plane and
in the situation here, heat gains on ductwork and equipment in the
attic space and the cooling in the attic subtracts from what would have
been needed in the space below.

I am now in a year round cooling environment, hot and humid. They vent
attics here because it 'flushes' out the heat. It is pretty difficult
to seal the ceiling plane here so while flushing out heat, you are also
pumping a lot of humidity into the attic due to the nature of the
outside air. So the humid air can short circuit insulation and enter
the home, whenever the home is depressurized. Nick mentioned exhausting
air to the attic put it could just come right back in.

So my gut reaction to using evap cooling in an attic is 'thumbs down'
don't want to add moisture. You could take take outside air, cool it
right down with evap cooling put it through an HRV type HX, run attic
air through the other could work, cost an arm and a leg tho.

Seen a hydronic fan coil advertised for attics, it capitalizes on the
fact that an attic is a solar collector, fan coil cools the attic and
the heat picked up in the attic gets dumped into a swimming pool. Nice
two birds with one shot, but swimming pools are a luxury.

If you don't have a pool and try a couple fan coils I wonder if the
energy of two fans gets you anywhere.

looks like some BS data in the report, they used 'physics' to correct
it lol

Abby Normal wrote:

Working backwards it seems to show that you have taken another hit from
your bong and come up with a microscopic cooling load of 1.472 Btu/hr

Wrong again, Abby :-) This time by a factor of 1000. With such careless
reading and such incredible arrogance, this discussion is hopeless.

Nick

wrote:
Abby Normal wrote:

Working backwards it seems to show that you have taken another hit from
your bong and come up with a microscopic cooling load of 1.472 Btu/hr

Wrong again, Abby :-) This time by a factor of 1000. With such careless
reading and such incredible arrogance, this discussion is hopeless.

Nick

Well you posted 1472 Btu/hr. Hey point out my spelling mistakes too,
it's about all you can do.

Trade in your Bowie knife for an AK47, you are out of your league on
this one Nick. Try the references I suggested and learn something.

For equivalent performance, it seems we also have to add a motorized
bypass damper to the swamp cooler to allow indoor air recirculation.

Maybe not, for an indoor swamp cooler, eg the WisperCool P300 ($154 at
Wal-Mart, but no longer being made) or the $298 Mastercool Mobile MMB10
(Grainger 5MU36), which has a garden hose connection and draws 3.5 amps
at 120 V. Adobe Air says it can cool 2000 cfm of 110 F 10% RH air 32 F,
about 69K Btu/h (6 tons), like 14 5K Btu/h window air conditioners :-)

For more cooling capacity with dry outdoor air, we might put one near
a window inside a house with a $55 Lasko 2155A 16" 90 W 2470 cfm intake
fan in the window and use the fan thermostat to turn on the cooler when
the room temp rises to 80 F and a humidistat to turn on the fan when
the indoor RH rises to 56%, with 1-way plastic film dampers in a box
between the cooler and the window to force outdoor fan air to flow
through the cooler pad when the window fan is running and make indoor
air flow through the cooler when the window fan is not running, like
this, viewed in a fixed font like Courier:

| |
| |
---------
| |llld| |
|c| d| | outdoors
|o| d|f|
==|o| d|a| == With the window fan off, indoor air
|l| d|n| would flow in through left and right
|e| d| | dampers lll and rrr. With the fan on,
|r| d| | ddd would open and the fan air would
| |rrrd| | force lll and rrr closed.
---------
| |

We might have 4 modes:

80 F 56%| window fan cooler fan cooler water
--------------|-----------------------------------------------
1. no no | off off off
2. no yes | on off off
3. yes no | off on on
4. yes yes | on off on

Case 3 would maintain indoor comfort with less water than an external
swamp cooler, for a house with significant natural air leakage, ie
for almost all houses.

Nick

Take a 600 square foot home with 8 foot ceilings, something perhaps
that could have a space sensible gain of 10,000 Btu/hr when you exclude
infiltration.

You have a volume of 4800 cubic feet.

Now considering this is a low building and the fact that it is not
winter, there will be no 'stack effect' caused infiltration to worry
about. It is not a multi-floor building so I would also doubt there
would be any sort of any inverted stack causing increased infiltration,
so for the most part be some wind driven infiltration.

So let's see what significant natural air leakage is, in the summer. If
you had something perhaps 2 air changes per hour in the winter due to
infiltration, then perhaps that could be considered very poor. So with
just some wind effect, from a slower, less dense summer wind compared
to a faster more dense winter wind, I will give you the benefit of the
doubt and say it could be one air change per hour.

So to look at this infiltration on a per minute basis, 4800 cubic
feet/60 minutes is approximately 80 CFM.

You have a scheme that could pressurize a home with 1360 CFM or perhaps
exhaust 1360 CFM. Or minimize things and have 617 CFM pressurization or
perhaps 660 CFM exhaust. Do you really think that 80 CFM of
infiltration is additive to this mechanical exchange? Ultimately a
natural pressure differential drives the natural exchange, in this case
there will be a mechanical differential. Maybe consider the
differential pressure required for 80 CFM of natural infiltration vs
1360 or 660 CFM of exhaust.

Hey argue that it is a super insultated 2000 square foot structure then
that has the envelope integrity of swiss cheese( never see this,
someone going to the extent to super insulate would pay some attention
to air tightness).

2000x8/60= 267 CFM infiltration from natural effects, The exhaust fan
or the outdoor evap cooler are still going to be able to over power
this even with the air flow rates established for a mere 10,000 Btu/hr
sensible gain.

Infiltration will only come into play when the system is cycled off or
perhaps some severe weather moved in like a hurricane.

Try the Florida Solar Energy Center link I gave you, go down and see
what they say about a 2 Pa pressure differential. They are fellow solar
geeks, maybe you will believe it if you hear it from them.

I pressurize to prevent humid air with a 79 to 81 dewpoint from
infiltrating in so what do I know.

I think if you thought about the pressure differentials you would not
be pressurizing the garage or attic while depressurizing the home at
the same time. So there is another piece of the puzzle for you.

Abby Normal wrote:

Take a 600 square foot home with 8 foot ceilings...

Kinda like Ted Kaczinski's? :-)

Nick