Yes, it can, if you want to use more water over a day's time, but turning it off in the hottest part of the day can save water with no loss in comfort.
That requires a lot more slab and insulation and airflow.
No Abby, exactly the same, or less.
That is incorrect.
I disagree. An assertion demands no more than a counterassertion :-)
Collects or loses? Who cares, in the southwest?
I suggest you open your mind and examine the numbers.
Nick
Article 759892 of alt.home.repair: From: snipped-for-privacy@ece.villanova.edu Newsgroups: alt.home.repair,alt.energy.renewable,alt.architecture.alternative,sci.engr.heat-vent-ac Subject: Cool tower alternatives Date: 29 Oct 2005 16:23:08 -0400 Organization: Villanova University
Nader Chalfoun and Christopher Trumble had an interesting Tucson cool tower story in the Spring 2005 U Oregon "Connector" architecture newsletter.
It's nice to avoid the energy of a swamp cooler blower. Can we also avoid the large structure (sacrificing architectural drama) and use less water, based on weather conditions, with constant comfort and better controls?
How about testing an alternative? Evaporate water inside a house and also run a small exhaust fan as needed to remove water vapor from the house. The most efficient corner for evaporative cooling in the ASHRAE 55-2004 comfort zone is at 80 F and w = 0.012, approximately.
We (not me, with recent flooding in PA :-) might turn on a small indoor swamp cooler with a thermostat when the indoor temp rises to 80 F and turn on an exhaust fan with a humidistat when the indoor RH rises to
54% (w = 0.012 at 80 F.)
With enough green plants in a house, the cooler might seldom turn on. With enough air leaks, the exhaust fan might seldom turn on.
Or run a soaker hose with pressurized water from a solenoid valve (which might come from a dead washing machine) over a floorslab in an existing house or under a floorslab with a vapor barrier under the hose in a new house. The slab's thermal mass might store coolth for more efficient cooling with cooler night air below the comfort temp near the floor.
During the day, a slow ceiling fan with a room temp thermostat and an occupancy sensor could provide efficient cooling as needed. The fan could provide comfort cooling and raise the acceptable RH all the way to 100% at
81 F with v = 0.5 m/s, according to ASHRAE-55's BASIC program, altho that might cause mold, on a continuous basis. The slab could also lower the mean radiant temp. A low-e ceiling and walls could radiate less to the slab when nobody's home, conserving stored coolth.
NREL says Tucson has an average humidity ratio wo = 0.0054 in June, with a 67.9/99.6 F daily min/max. An 80 F house with a 400 Btu/h-F thermal conductance and 4K Btu/h of internal gain might need (99.6-80)400+4K = 8240 Btu/h of cooling at 3 PM.
Evaporating P lb/h of water makes 1000P Btu/h, and cooling C cfm from 99.6 to 80 F to make up for required exhaust air takes about (99.6-80)C Btu/h, and 1000P = 8240+(99.6-80)C with 0.075 lb/ft^3 air and P = 60C0.075(wi-0.0054) and wi = 0.0120 makes P = 0.0297C and 29.7C = 8240 + 19.6C, so C = 816 cfm and P = 24.2 lb/h of water, with a net cooling of 8240/24.2 = 340 Btu/lb.
How many pounds of water per hour would a cool tower need to achieve the same 80 F at 54% RH inside this house?
Ps = e^(17.863-9621/(460+80)) = 1.047 "Hg at 80 F and 100% RH, so A ft^2 of 80 F damp floorslab in 80 F air at 54% RH might evaporate 0.1APs(1-0.54) = 0.048A lb/h of water, (mis)using an ASHRAE swimming pool formula, ie
502 ft^2 of slab might evaporate 24.2 lb/h.
At 81 F and 100% RH indoors, 1000P = (99.6-81)400+(99.6-81)C and wi = 0.0233 and P = 0.0808C, so 80.8C = 7440 + 18.6C, so C = 120 cfm and P = 9.7 lb/h with 7440/9.7 = 770 Btu/lb of net cooling. This could work even in August, when conventional swamp cooling wouldn't, with wo = 0.0117 and Tdp = 61 F. It might precool a slab faster and more efficiently than simple AC.
An 80 F slab under 67.9 F air with wo = 0.0054 and Pa = 29.921/(0.62198/wo+1) = 0.257 "Hg might evaporate 0.1A(Ps-Pa) = 0.0789A lb/h and lose (80-67.9)1.5A = 18.2A Btu/h of sensible heat, for a total of 97.1A Btu/h. With enough air, a 1000 ft^2 slab might lose 24hx8240Btu/h = 198K Btu in 198K/97100 = 2 hours on a June night, with 198K/158 = 1255 Btu/lb of net cooling.
With that being said, it is not necessary to calculate how the slab will maintain air temperature in the space, as the whole goal of the discussion was how to improve evaporative cooling and not make it worse :)
The numbers has shown that as a design load and ambient temperature start to APPROACH something that is realistic, that air flow and water consumption increase with the flooded floor scheme to exceed those required for a swamp cooler.
Even with a 2,000 square foot unoccupied and superinsulated structure with no windows and no appliances creating internal heat, the swamp cooler uses less water and airflow.
Should a person just happen to go inside this theoretical stucture for a moment to take a temperature measurement, he would notice of the effect of having cool wet feet and hot dry 103.5F make up air infiltrating in as leaning towards the uncormfortable, however a cool breeze of of 74F would seem refreshing and the dry floor would be nice.
Put the 10,000 Btu/hr load in proper context, a small home maybe 500 to
800 sq feet in size in size, and that hot make up air will be very noticable.
Cool the outside air directly to the point that it is cooler than the temperature being maintained in the space is a rather efficient operation. Trying to cool a slab with water that evaporates into the space then, using the slab to keep the space cool, seems inefficient.
So there is no real point in investigating how to deal with the mold growth, and how air temperature would eventually stabalize over a cool slab when the effects of heat gain from the soil etc are when realisitcally factored in.
The flooded floor relies on heat loss from the room to the slab to keep the space cool. In a cold environment, below grade slabs with well insulated walls in contact with cool soil have a heat loss of perhaps 2 Btu/hr per square foot of floor space.
Has the flooded floor scheme improved anything in a realistic situation? A space occupied by humans, no. A small enclosure for mallard ducks, or amphibians not native to Arizona but on display in the Phoenix Zoo, yes.
Nope. The floor isn't flooded, and the numbers we came up with are very close. It's simple physics. The air is treated exactly the same inside the box or inside the house, so the numbers cannot be different, except for the swamp cooler's motor heat and lack of controls.
Lol the numbers were close, but we were ignoring all the other flaws in your scheme that you arbitrarily ignore. You would have ended up using even more water,exhausting more air, using ceiling fans to blast the air down to the slab, exhaust fans all over the place.
The air is not treated the same, that is what you do not seem to understand. You were trying to humidify then sensibly cool the air, - I used evaporative cooling.
My advice, admit you don't have such as good a grasp of physics as you like to think and take a refresher course in pyschrometrics.
Well if it was treated exactly the same, you should be able to come up with the exact same answer. Even when you use the correct specific heat of air and latent heat of evaporation for water, you will still be high on airflow and water consumption.
So, like I said Nick, go study pyschrometrics, you are obviuosly a little rusty on the concepts.
Would it help with your physics if you consider the specific heat of water vapour?
Pyschrometric charts have that covered automatically so I had an unfair advantage. Not sure if ole Clausius, Clapeyron and Bowen had that accounted for. Just a thought, I never factored it in when I looked at your scheme under a 15,550 Btu/hr load or the changing mass of the room air.
Maybe we could both try it again and this time, base it on mass not volumetric flows of AIR/MOISTURE MIXTURES?
Maybe see what you come up with if you considered the enthalpy of the outside air per pound of dry air (maybe respect to 0F for some easier math) and keep it constant. Reduce the temperature of that mixture we call outside air by evaporating some water, but keep in mind the mass of the mixture is changing, you have a pound of dry air, but you have more water mixed in with it. Humidifying would make it 'less dense' but the drastic change in temperature over compensates for that. The specific volume of that 'pound of dry air' would reduce.
Keep the enthalpy constant, pick an arbitrary temperature cooler than what you want to maintain the space at, using an equation for enthalpy and work backwards to get the final W, hence the water needed to do the job, take a careful look at how the change of mass impacts everything. Got to end up being pretty close those darn wet bulb lines must be more or less parallel to the enthalpy lines. Then you just need to go and work out the flow of air.
Damn, an uncontrolled urge for sarcasm.
You won't do this because it cools the outside air directly, you still feel your slab is the key, the slab somehow improves on this process.
I was thinking that once we got the first half of your physics straightened out, we could go back to the other flaws of the flooded floor. The mold, the stratified air vs destratification by blasting down a high volume air down towards the slab to actually cause some air to contact the slab and transfer some heat because without it, you would be lucky if you could get 2 Btu/hr per sq ft of heat transfer to that slab and the only mixing you would get would be the effect of turbulence from the Sirocco.
I saw some cool mist spray nozzles typically for outdoor use. Maybe you could put them by each make up air opening in the home. That way you could cool off the sirocco before it creates too much discomfort, and the excess water then floods the floor. Nope, can't do that, that is an attempt to directly cool the outside air as well.
At least you did not completely waste your time, you learned evaporative cooling, I hope. "Prove its 80% effective" LMAO.
We might see that it is with a mental experiment. Take 2 identical houses with identical outdoor and indoor air conditions. One evaporates P lb/h of water in a box on the roof. The other evaporates P lb/h inside the house. Put each in a large enclosure, including the box on the roof. C cfm of air flows into each enclosure and C cfm of cooler humidified house air at 80 F and w = 0.012 flows out of each. How can you tell which enclosure has the box on top of the house? :-)
Or put the swamp cooler on a stand outside a window... If we gradually slide it through the window into the living space, still drawing outdoor air in through the window, when does the air treatment "become different"?
The one with the box on the roof supplies 74 degree air to the space. It cools the outside air directly and uses an air process called evaporative cooling.
You want to put one inside the duct the outdoor air to it. Having the swamp cooler act on the indoor air is feasible but not as stupid as the flooded floor.
The other scheme involves flooding the floor and removes heat from a floor slab to himidify the air. Then it relies on a yet-to-be-thought-out sensible cooling scheme between the room air and the slab. It causes the home to be a mixing box where a high volumetric flow rate triple digit ambient air is mixed into a relatively small volume of humid indoor air. There are other problems such as mold and heat flow into the slab from other than the house itself.
The air process is evaporative cooling whether the swamp cooler is outside of the house or inside the house, however once you move the cooler inside and have it treat room air, it is more of a humidifier than a cooler. Lets just examine the problem of not directly treating the outside air with evaporative cooling. It will show the same spiral of airflow and water consumption when trying to deal with untreated make up air.
Try a swamp cooler at 75% effectiveness, volumetric calc based on standard air, not true mass flow rates, sea level elevation, it will evaporatively cool the outside air down below room temperature yet the dewpoint of the air being supplied is a little lower than what you are trying to maintain in the space.
Air flow required = 10,000/(1.08x(80-75.91))=2264 CFM using
2264/7000x4.5x(83.4-39.2)=64.3 lb of water
The problem is, when you make it act on inside air, the inside air is too humid to begin with. It is not the classic hot air with low dewpoint anymore.
When you put it inside the house, it gives less cooling per unit airflow and it just escalates as it tries to deal with the added sensible heat of the make up air. But at least it can be verified how much 'cooling' it will actually provide. Your hypothetical scheme 'just has to work' because your physics are never wrong.
At best, to try and deal with 10,000 Btu/hr of sensible heat, the indoor humidifier can turn 80 degree 84 grain air into 68.13F 103.3 grain air at 100% effective.This will minimize air flow through the humidifier.
quick numbers again based on volume not mass and sea level
10,000/1.08/(80-68.13)= 780 CFM using 9.68 lb of water per hour.
780x4.5x(103.3-84)/7000
but there is also the heat of the make up air
Exhaust to get rid of 9.68 pounds per hour
9.68x7000/(4.5 x(84-39.2))=336 CFM
the spiral is just starting now
additional heat of the make up air
336 x 1.08x (103.5-80)=8528 Btu/hr
oops better get a bigger adibiatic humidifier
18528/1.08/(80-68.13)=1445 CFM using 17.9 lb/hr
double check that exhaust
17.9 x 7000/ 201.6= 622 CFM
this darn spiral, its like an energizer bunny it just keeps going and going. We are stuck with a slightly higher wet bulb to be evaporating water into as well.
Round three of the spiral
(10,000 +622x1.08( 103.5-80))/(1.08x(80-68.13))=2011 CFM for the humidifier using 25 lbs of water an hour
Now need to exhaust 25x7000/4.5x(84-39.2)=868 CFM and we still have some latent internal gains. Humm two fans, one moves 2011 CFM, the second one moves 868 and it still is not working out. Is it an inherent flaw not to directly treat the outside air first?
Spiral up some more I guess
(10,000 + 868x1.08 x (103.5-80))/(1.08 x (80-68.13)) = 2499 CFM using
31 lb of water, still no allowance for any internal latent gain either
Is exhaust getting close yet?
31x7000/ (4.5x(84-39.2)=1076 CFM, nope it still just wants to spiral up, damn problem, this room air is just too humid to begin with.
(10,000 + 1076 x 25.38)/12.8196= 2910 CFM through humidifier now using
36.1 pounds of water
Exhaust rate needed 36.1 x 7000/ 201.6= 1253 CFM, still does not really allow for any internal latent gains
Convinced of the spiral yet?
Moving the cooler inside is going to use more fan power for sure you have two fans, so far you would need to move 2910 CFM through an indoor cooler, plus run another fan to exhaust 1253 CFM, still have a little indoor latent heat. And the system is no where close to stabalizing.
Must be some kind of inherent flaw here.
( 10,000 + 1253x25.38)/12.8196 = 3261 CFM through humidifier using 40.5 lbs of water need 40.5x7000/201.6=1406 CFM of exhaust. Still got some internal latent,the sirocco is starting to intensify.
What happens if we were to use 64.3 pounds of water?
64.3 x 7000/ 201.6= 2233 CFM of exhaust needed to avoid humidity build up from water used for cooling.
Lets see what the humidifer does to allow for this much make up air.
(10,000 + 2233x25.38)/12.8196 =5201 CFM wow. Lets check the water needed, 5201/(4.5x(103.3-84))/7000= 64.5
There you go, at least the water balances out, because I minimized the air flow of the indoor humidifier and we do not have to deal with all the water that a flooded floor will waste. Interesting tho is that a swamp cooler acting directly on the outside air supplies air a little below the dewpoint you need to maintain, where as putting the swamp cooler inside only adds air at a higher dewpoint than what you are trying to maintain so it really is just a humidifier.
So therefore what is better, have the evaporative cooler directly treating the outside air, or put it the swamp cooler inside use more fan power, and live through a sirocco.
Now consider the flooded floor, you have not even shown how the heat will transfer from the room air to the floor slab yet, it just has to work because you of course are never wrong. If the slab is cooler than the room air it will be cooler than the soil. You have heat flowing into the slab from above, from below, in from the edges.
A below grade slab in the winter is a 2 btu/hr per square foot of slab heat loss but you are going to cool a house in Arizona.
So now that maybe you have a handle on evaporative cooling are you ready to give up on the flooded floor?
So having an evaporative cooler treat the outside air directly is the best way. Trying to use a protable evaporative cooler to 'over treat' indoor air to compensate for the make up air is not a good idea as numbers have shown.
No real point in trying to see how much worse a flooded floor is to an evaporative cooler treating the outside air, as the flooded floor will be worse than using a portable swamp cooler indoors.
Go buy one and try it. They are sold all over the west and are not worth a damn. Those that use them are usually apartment dwellers who don't have other cooling. They end up setting them by the window so it draws in outside air and open a winow on the other side of the apartment to let the air out.
"We might see that it is with a mental experiment. Take 2 identical houses with identical outdoor and indoor air conditions. One evaporates P lb/h of water in a box on the roof. The other evaporates P lb/h inside the house. Put each in a large enclosure, including the box on the roof. C cfm of air flows into each enclosure and C cfm of cooler humidified house air at
80 F and w = 0.012 flows out of each. How can you tell which enclosure has the box on top of the house? :-)
Or put the swamp cooler on a stand outside a window... If we gradually slide it through the window into the living space, still drawing outdoor air in through the window, when does the air treatment "become different"? "
Let's read these words. First paragraph sort of says what is the difference if we use an evaporative cooler on the roof, which implies the conventional proven system that directly treats the outside air, vs an evaporative cooler inside the home which would treat indoor air.
The second paragraph is trying to see when the situation would change, but it does show that an open window could be a make up air inlet to the space. I guess the instant it stopped treating 100% outside air, it all goes down hill. Maybe model the momentum of the sicrocco coming in through the window vs the fact that the indoor evaporative cooler would draw air from the path of least resistance. Use some calculus on the equation, take the derivative, set it equal to zero, maybe you could come up with the optimum distance inside the home as being a negative number meaning it should be outside drawing in 100% outside air.
You could always have the 'box' inside and run a duct to connect it to the outside.
proven unhealthy, ignoring that , a dripping wet ceiling would work better than a flooded floor. At least natural convection would work with you.
Yes they do, in particular with what you were proposing
Yes and hopefully now, you now understand evaporative cooling. You clearly did not before yet you smuggly tell us 'hvac crimianls' how superior your intellect and understanding of physics is. Too funny.
heat gain by sun is significant and matters in a realistic situation.
Some species may find it comfortable. Its pushing the edge, scroll down and see further comments on this, keep an eye open for the keywords 'underwear' and 'rubber boots'
fan speed is relative to blade geometry and the amount of air being moved. You have taken this out of context as you have no clue as to the magnitude of air that you would have to drive down from the ceiling to the floor.
air moving parallel to the slab, is not making contact with a slab. Consider a DX cooling coil, air either makes contact with the coil and loses heat (sensible & latent), or it sails through the fins and tubes without making contact and leaves at the same condition it entered. Mixture of the air that makes contact with that which avoids making contact, is the leaving air condition.
The majority of the air being driven down by the ceiling fan does not contact the slab, therefore a high flow is needed to get the cooling.
The flooded floor humidifies the air, it adds latent heat without directly taking using sensible heat from the air. A portion of the heat that evaporates water will come from the soil and ultimately the ambient air and sun.
I suggested you try to use a flux plot several times to work out the heat flow into the slab but this was ignored.
Maybe you gain a Btu/hr per square foot of slab from the soil.
Convection and radiation can be of similar magnitude
You used a formula based on mean radiant temperature.
out of context rhetoric, however a lower mean temperature would indicate a more comfortable room
yes heat that will evaporate water from the flooded floor will come from outside of the house in addition to heat inside the house. The flooded floor increases heat flow into the house from the outside.
Concrete has thermal mass, heat will flow into it,. You like to call it 'coolth' and blindly use the concept to plug holes in your schemes. You like numbers but arbitrarily write off the important details.
Vapor barriers are not a significant insulator, as you were implying. At best it prevents your scheme from losing water to the earth. You were calling it an insulator but most likely you meant it was keeping water from the slab from dampening the soil beneath, and increasing the conductivity of the soil . By stopping water leaving the slab to the soil, it also reduced the amount of water the flooded floor wastes.
Heat will conduct through the soil to the slab. Maybe you meant lose 'coolth'
Attract was a poor choice of words, like saying a vacuum cleaner sucks in air. Water is driven towards the slab, atmospheric pressure pushes air into the vacuum cleaner.
I made no mention of deep artesian wells, again out of context rhetoric.
Not directly treating the outside air is an inherent flaw. You were also trying to use a non-realistic load and a low average ambinet temperature before. The effect of the make up air was not properly accounted for in that the flow rate and the temperature of the make up air were being kept artifically low.
An indoor swamp cooler does require a high air flow, a lot higher air flow than what a swamp cooler directly treating outside air will require. If you were to actually address the airflow needed to cause the room air to convect heat to the slab below you would have realized the 'high airflow' needed for the sensible cooling portion of your scheme.
The air flow of the ceiling fans, to get the sensible cooling process of your scheme to work will be orders of magnitude higher than what an evaporative cooler would require. The problem is you were evaporating water to cool a slab, you were not proposing an AIR PROCESS known as 'evaporative cooling'.
You were cooling a concrete slab and humidifying room air, then trying to get a mixture of room air and make up air to stabalize in the extreme corner of temperature/humidity that some people walking around in their underwear and wearing rubber boots may find comfortable
A missaplication of concepts you read about did not pan out this time.
You want respect and professional discussion? Then give some. Quit crossposting stuff to rile up us 'HVAC Criminals" in an attempt to boost your ego and show us 'how smart you are'. At least this round you did not have to demonstrate how to derive pi not intrinsic to GW Basic.
I am gloating now but what were your words? Was it, 'obtuse', 'arrogant' ? Just look at the man in the mirror.
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