Alternative Architecture - Re: Swamp Cooler to Refrigeration A/C

Author

Re: Swamp Cooler to Refrigeration A/C

Abby Normal

2006-04-30, 7:21 pm

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.

nicksanspam@ece.villanova.edu

2006-05-01, 5:21 am

nicksanspam@ece.villanova.edu

2006-05-01, 9:21 am

Abby Normal <a_bee_normal@yahoo.com> 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

nicksanspam@ece.villanova.edu

2006-05-01, 10:21 am

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

Rich256

2006-05-01, 11:21 am

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.

Rich256

2006-05-01, 12:21 pm

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.

Rich256

2006-05-01, 3:21 pm

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.

nicksanspam@ece.villanova.edu

2006-05-02, 9:21 am

>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

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