# pool pump

Page 5 of 8
• posted on November 24, 2005, 7:43 pm
It estimated nothing
You have need otherwise your schemes remain unproven fantasy.
Mositure in the soil would be attracted to a cool slab, the insulation value of a vapour barrier is negligible.
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<%-name%>
• posted on November 24, 2005, 8:13 pm

A arrogant rule of thumb: "If Abby doesn't understand it. It won't work" :-)

I disagree. Moisture is not "attracted" :-)
Nick
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<%-name%>
• posted on November 24, 2005, 10:49 pm
It is hard to understand your substitution for considering forced convection , when you are incorrectly applying small temperature differential radiant transfer shortcuts.
To be anal, I believe the constant is 0.1713E-8, however the temperature raised to the power of 3 is the absolute MEAN temperature of the ceiling and the slab. You state the slab is 75 so that would make the ceiling 85. The air would be quite warm then wouldn't you think. Its already established that it is humid so this side track of yours seems to point to a natatorium again.
I will admit you have confused me as you appear to be using a net radiant heat transfer from the ceiling to the slab, to estimate FORCED convection from the air to the slab.
I have seen the radiative heat transfer coefficient used in examples of hot water radiators. Heat transfered by NATURAL convection to room is roughly equal to the heat transfered by radiation. But the slab is cool, natural convection works against you, in fact you are relying heavily on forced convection.
Anyway you look at it, the approach you are trying could possibly estimate the convective heat transfer from a HEATED slab to the air. Or it just shows the net radiant heat transfer from a ceiling to a floor slab, stealing some of your 'coolth' available.
So trying to rationalize things out by assuming temperatures does not seem to work.
This would be a good one to program as you would have to do quite a few iterations to get it all balanced out. Increased heat flux from warm earth into slab, now radiation from ceiling and walls putting heat into the slab balanced out against the heat removed by water evaporating to humidify the home, and the forced convection of warm room air ( mixture of room air and triple digit air infiltrating in) contacting the slab. You would have to keep iterating until you could equate everything and finally come out with the room air temperature and the slab temperature. At least the rising indoor temperature would reduce the sensible heat gain of the space. :)
Sure looks like the conditions found adjacent to an indoor swimming pool in the making to me.
So enough smoke and mirrors, if you want to see what is really happening, you need to establish what the slab temperature equalizes at. Your whole scheme revolves around it. It is the foundation of everything you are hoping to prove so without establishing this, it is yet one more pipe dream.
The Inverted Pool of Pine, with a dripping wet ceiling slab would work better, be unhealthy but more practical at least you would have natural convection working with you. What would work the best, (without DX air conditioning) is a system that cools outside air to a temperature lower than the desired room temperature and then displaces room air outside of the structure to avoid a build up of 100% relative humidity. What a great idea, reduce the sensible heat in triple digit ambient air to the point where it is cooler than the room air, and then calculate the air flow required based on the sensible heat gain of the space.
You should investigate ground source heat pumps a little bit, in particular the effect of MOISTURE MIGRATION in the soil. Typically soil moisture is attracted to cold pipes in heating mode. In the cooling mode, warm pipes dissipating heat below ground tend to drive moisture away from the pipes.
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<%-name%>
• posted on November 25, 2005, 1:05 am
OK You guys. On this same vein...
I have to do a heat loss calculation for my home design with slab heat.
How do I calculate the total equiv R factor for the earth under the slab? I am planning R10 (2 inches of styroboard) and the law says I have to insulate the perimeter somewhat.
I realize convection doesn't really happen down below the 2 feet of styroboard but the heat sinking is infinite after the R10 underlay.
Is there a rule of thumb to use for this?
Let it shine boys...LOL
Thanx
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<%-name%>
• posted on November 25, 2005, 6:04 am

Arrogance again. It might be, were I so doing :-) Pick one: a) an ordinary ceiling with radiation and no fans or setbacks, or b) a low-e ceiling with fans and possible setbacks. Don't pick both!

That's extraordinarily anal, and wrong, according to the ASHRAE HOF and lots of other sources which say the Stefan-Boltzman constant is 0.1714x10^-8 Btu/h-ft^2-R^4.

Also anal. The radiation conductance is a way to discover the temperature diff required to move heat in this case, vs starting with a known diff... 77.5 or 80 F changes the answer infinitesimally. Try it, if you like.

Let's try to avoid this passive-agressive waffling.

This is promising, Abby. Confusion can be better than certainty :-) Pick a) or b) above, not both.

That's good for daytime setbacks.

In case b) with fans and low-e surfaces...

It "could possibly do so" because it does :-) With cool slabs as well. Figure 1 on page 22.1 of the 1993 ASHRAE HOF, "Surface conductance for different 12-inch square surfaces as affected by air movement," shows 1.5+V/5 Btu/h-F-ft^2 smooth surface airfilm conductance, with V in mph. I conservatively assumed V = 0.

Radiation melts more ice in a) like an ice rink with a plain roof, than b) like an ice rink with an aluminized roof.

Maybe I wasn't clear enough in explaining that a) and b) were mutually exclusive cases, altho there are in-betweens, with moderate emissivity. Foil walls and ceilings (b) can save more water with daytime setbacks, but they aren't everyone's cup of tea.

Not needed, IMO. But knock your socks off, s'il te plait.

No I don't. More arrogant waffling...

More arrogant waffling words.

How would that work? Who said anything about 100% RH?

I don't understand. Ignoring your sarcasm, how would that work?

How arrogant to assume that I haven't. PE Norman Saunders writes that the main mechanism for upward heat flow in soil is evaporation from lower layers and condensation above...
All soil contains some moisture. When heat is moving downward, the conductivity of the water speeds the temperature change slightly, even as water's thermal capacitance slows it. A warm wet strata below is another matter. Vapor is always forming, and because of its low density, it rises through the inevitable pores until the lower temperature causes it to condense, releasing the considerable heat from change of state. See Penrod, "Measurement of the Thermal Diffusivity of a Soil by the Use of a Heat Pump," J. App. Phys. 21 May 1950.

Water vapor is only "attracted" to COLD pipes, below the dew point. It's hard to get to the dew point with a dampened slab and "attract" water vapor up through 800' of sand below in the southwest, but it's useful to avoid dampening soil with a vapor barrier below the slab, if only to avoid wasting water.
Nick
Confusion is like fertilizer. It feels like shit, but nothing grows without it. --Carl Rogers
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<%-name%>
• posted on November 25, 2005, 3:05 pm
Nick
You have a pattern of getting more and more obscure when you are losing an arguement.
In a case of something such as a hot water radiator, radiant heat transfer and natural convection are of roughly the same magnitude, therefore some one incapable of calculating the convective heat transfer can do a quick calculation of the radiant heat transfer, to be in the ballpark of the NATURAL convection.
So what you are trying to do as far as I can tell is to quickly estimate the net heat transfer from the ceiling to the floor and say this is equal to the convective heat transfer of the room air to the slab. So the value of 1.08 that you round off to 1.1 seems analogous to 1/.92 , heat flow down through a convective air film on the underside of the ceiling. Quite a coincidence how this works out, 80F could be a MEAN temperature of the underside of a ceiling and an air conditioned room.
You go on and on but have not been able to prove a single thing, all you can do is abitrarily pick a temperature that SEEMS logical out of your head to plug holes in your scheme or use 'stored coolth', turn on a grainger fan. Everytime.

You need to pick one is the whole point. You have a forced convection situation. Explain how your attempt to work out the radiative heat transfer ceofficient has anything to do with FORCED AIR convecting heat to a cool slab?

you made me blow the dust off of a couple books,my sources are 50/50 on that one. I said I was being anal in the first place.

No not anal at all, very significant. The insignificant changes in MEAN temperature seem to come up with the same air flim resistance for heat flow down from a warm surface to the air beneath it.
But the point is, the approximation has nothing to do with the heat transfer from air to the floor slab. It does point out however the loss of 'coolth' (I love your buzzword) as the ceiling radiates more heat to the floor slab than the the floor slab radiates back.

I don't have that figure, but why rely on some figure when you will not even use a pyschrometric chart?
Since you want to avoid calcualting convection coeffcients why not take the easy way out. Treat it as an air flim resistance with moving air, R0.17. Hey just need the air blasting down from the ceiling at 15 mph and the required differential is 2.16F to get rid of the space sensible gain of 15,550 Btu/hr in a 1225 sq ft buidling.
Get a Big Ass Fan, like used in a barn and blast the air down. Only need 1320 FPM at a 1225 square foot slab.What do the fan laws say about power requirements now at 1.6 million CFM? Plus you need to deal with the sensible heat from the make up air, so you can do the math on how it increases the required temperature differential maybe 75 is a valid slab temperature after all.
Sure looking more plausible once again to directly cool the outside air before it is introduced to the space using a swamp cooler. Air flow is only a thousandth of what your scheme potentially requires. OSHA regualtions would not make the occupants wear eye protection either. , just non-slip foot wear :)

Lol if it is melting ice, then it is stealing your 'coolth'

The case has always been forced convection heat transfer between air and a cool slab, your estimate of NATURAL convection by equating it to radiant transfer has back fired so please move on.

I think if you invested the time to do so you would become disillusioned with your scheme and come to the conclusion that you are replicating the environment found in a natatorium.

I can't force you to do anything, let alone prove anything.

Lol, sticks and stones. The cooled slab is the cornerstone of your plan and you have no real clue how cool it will be or the rate of convection between the air and the slab. So you are there with your beach towell enjoying the natatorium.
You are like the marketting guys in Dilbert. :)

Read through your HOF on evaporative cooling, try it on a pyschrometric chart. You seem weak in the area of pyschrometrics.

I pointed out that your cool slab will increase heat transfer from the ground into the slab and therefore the space. You countered with a vapour barrier which is negligible with respect to insulation value.
So if anything the vapour barrier stops loss a 'coolth' with water flowing from the slab to the earth, but has virtually no effect on the rate that heat conducts into the slab from the ground. Actually from the surface of the surrounding earth through the ground and into the slab.
Wonder what the dewpoint of water in soil is when the when the soil at 42F is in contact with a pipe at 37F.
Mositure is drawn to the cold is the point.

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<%-name%>
• posted on November 25, 2005, 5:35 pm

I disagree. Then again, you seem to get more obtuse when losing :-)

As far as you can tell :-) But that's incorrect...

I do? Why? Owner's choice. OK, let's pick a)...

Nope. No fans at all. No convection, just radiation.

It doesn't :-) You still seem confused. Maybe that's good.

OK. I'll try it. Gr = 4x0.1714E-8(460+80)^3 = 1.079573184 Btu/h-F-ft^2 at 80 F, so we need dT = 10K/2K/Gr = 4.631459983 F to radiate 10K Btu/h from ceiling to slab. Using a closer mean Tm = 80-dT/2 makes Gr = 4s(460+Tm)^3 = 1.065743771 Btu/h-F-ft^2, which makes dT = 4.691559206 F... also < 5 F. If you consider that 0.06 F temperature difference "very significant," you might enjoy doing another iteration :-)

I have no idea what you are talking about. Do you? Air is transparent to radiation, so there is zero "heat flow down from a warm surface to the air beneath it."

Agreed. It only concerns heat radiating from ceiling to slab, but that also bounds the room air temperature.

Exactly. BTW, one special benefit of this system is the reduction of the mean radiant temperature for a person inside the room, which makes for more comfort than cool air alone.

I'm more comfy with simple equations in plain ascii.

That's a bad implementation of plan b). Plan a) has no fans. BTW, my example concerns 10K Btu/h and 2K ft^2, vs 15,550 Btu/h and 1225 ft^2.

No thanks :-)

You may be projecting here. This is science! Let's use numbers!

But the dry soil beneath is great insulation. And thermal mass.

Sounds irrelevant, in this case.
Nick
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<%-name%>
• posted on November 27, 2005, 3:55 am
snipped-for-privacy@ece.villanova.edu wrote:

Lol, losing, I am just sitting here watching you tap dance.
I keep challenging you to prove your scheme in a realistic situation. I gave you a small home with a modest heat gain, it included gains which you constantly ignore in your cooling schemes, heat from the sun. Cooling is not based on the difference in outdoor/indoor air temperature alone as you have a habit of assuming.
Over and over you have unrealistic low loads to deal with and base your schemes on average temperatures. If you used an appropriate load, then your use of averages would allow you to properly estimate energy consumption. There will be no comfort under peak conditions because your system cannot handle it.
It is a good thing you do not make your living by designing for comfort.
You keep trying to make it a larger home with even less of a cooling load. 10K in a 2000 sq ft home may be theoretically possible, but extremely rare, not many windows. A sensible gain of 15,500 Btu/hr in a 1225 sq ft home with a 106 ambinet exceeds 'good' when compared to insualtion levels and shading typically in use in the SW.
So a 15,550 sensible gain on a 1225 sq ft house in the sunbelt with a triple digit ambient is a low load for you to prove your scheme by.

Your previous schemes had you cooling a slab, and then arbitrarily using a 21000 CFM ceiling fan to cool the air by blowing it at a slab.
My arguements are
1) The heat that evaporates the water does not directly come from the air, it comes from the slab 2) The cool slab, draws in heat from the earth 3) Only a small portion of the air being blown down to the floor makes contact with the slab and is cooled by the slab. 4) You allow the heat of the make up air directly to the room air 5) You plug holes in your scheme with unproven claims like 'stored coolth' or randomly picking fans out of a grainger catalog without validating they are even close to being sufficient.
You then suddenly change this to radiant cooling, I realize the thread is called "Cool Towers" but you were the one mentioning your wet slab scheme again, and you claim you have a scheme that provides evaporative cooling with minimal air flow.
The main arguement is evaporative cooling can work but you need a swamp cooler to cool the outside air directly. You are making the house a swamp cooler, and ignoring the unhealthy environment you will create, you are going to make the inside of this home as warm and as humid as a natatorium unless you move some significant air.

So you are showing that the ceiling radiates heat to the floor. What does the floor do with this heat? It evaporates water to humidify the space. I already conceded a well insulated ceiling at R40 so the gain here is pretty small, the ceiling will not be cooler than the indoor air.
The only time I have ever seen a ceiling cooler than the space was a dropped ceiling, below an insulated attic, with Airtex Modular Radiant Ceiling Panels. Just had to keep them above the room dewpoint by a few degrees and they did remove some sensible heat from the space, air naturally convected up to them. I guess you would radiate some heat directly to them as well
As I was trying to explain to you earlier, as can be the case with a hot water radiator, the magnitude of heat radiated into the room is a similar magnitude of the heat that convects into the room. So using the temperatures you can quickly calculate the radiant heat transfer coefficient based on the MEAN temperature and equate it as being roughly equal to the convective coefficient.
So for an e=0.9 ceiling, the air film resistance for heat flow down is R0.92, which is roughly equal to 1/1.079573184=0.926. Is this a pure coincidence or have you verified the air film resitance for heat flow down into a space from a ceiling?

see above

The heat through the ceiling would be approximately 2100 btu/hr based on R40 for 1225 sq ft.
Walls - 1600 (forgot to add walls for the 15500 total sensible gain) Glass - 8600 People/Appliances -4800
106 F ambient ,20% outdoor RH, 78 inside, 33N lattitude, 80 F soil
35x35x9 house, slab on grade, R19 walls, R40 ceiling, scroll back for the square footage of double glazing I gave you before on each wall, eaves overhang by 2 ft, eaves one foot above windows, internal shades on windows drawn.
People cook and do laundry in their home. I gave you everything but an insulated slab as I do not think there are many of those in the SW.
You want 80F inside then neglect the walls and call it 15,500 that I was saying before.
A lower than average load. 10,000 sensible gain in 2000 square feet, no way. Look at the fenestration and load from occupants and appliances. Bigger home more glass. Sq Lit also states that there is no real overhang above windows in his area and I am allowing for some external shading.

People sitting by a window when it is -40F outside lose heat by radiation and feel cold and those who leave the blinds open in the summer will feel hot, in this situation it may offer them some relief from what you are trying to tell them is comfortable.

Oh you like equations yet you quote a figure. Like I said I do not have that one, perhaps you could scan it and give a link so I can see if you are not taking something newly discovered out of context again. Is it for air flow over a cool plate or a hot plate?
So sticking with equations, then use some equations that show how heat will convect from the air DOWN to the slab.
I used a simple published value, but it shows an extreme amount of air moving. A very reliable air film value too. R=0.17 for 15 mph wind, maybe you want to try 7.5 mph but the R-value increases. So you would need a bigger differential temperature. Maybe for still air it will work out to R=0.92.
15550/1225/0.92= a temp differential of 13.8F. Here let's use this temperature, I will give you this one.
Room is 80 therefore slab is 66.2 degrees and you have to evaporate water to get the temperature down to that, water likes to evaporate at its wet bulb temperature so you need to consider that more than you think, plus you get more heat conducting up from the ground. Surrounding ground probably 80 degrees even a couple feet down, a simple flux plot could work out the increased conduction there. But that is for you to figure out, its your proof that is lacking.
AT 66F water will probably require about 1056 Btu/lb. Ignoring the extra load of heat into the slab from the soil and the radiation from the ceiling, for 15,550 Btu/hr sensible gain (excluding the make up air) you would need to evaporate 14.68 pounds of water in an hour.
Don't get excited , water evaporating off of a 66.2 degree slab evaporating into 80 db/66.2wb air sounds like your extreme worst corner of the comfort zone but then you have to exhaust air to avoid the humidity building up. Here is where your ship sinks.
At 80db/66.2wb indoor and 106F 20% RH oustide, for every CFM you exhaust, the make up air adds 1.08x26= 28.08 Btu of sensible heat each hour. So for each CFM, you would have to evaporate an additional 28.08/1056x70006 grains of moisture each hour. However each CFM of exhaust only gets rid of 5.8 grains of moisture. NOT GOING TO HAPPEN- natatorium for sure. Swamp cooler going to have a problem here as well.
Take SQ lits data then for June 27. http://ag.arizona.edu/azmet/data/1205em.txt
Shows a max temp of 106F and a min RH of 7%. Assume they coincide. Your scheme you have 15,500 Btu/hr + 1.08xCFMx(106-80) of heat to get rid of. Each CFM of exhaust gets rid of 50.4 grains each hour. But you need to evaporate 1056/7000 x grains + 4.5x50.4xCFM x grains each hour. Two equations, 2 unknowns maybe 2527 CFM of exhaust to maintain 80F, 81.9 pounds of moisture per hour. Let me guess, you are poke a hole in the ceiling in each room and put a 100W grainger mushroom fan on the roof and turn the window into a motorized damper with a 25va actuator and show me how you will save energy.
Take a swamp cooler drawing in 106 air at 7% RH. Add 53.1 grains of moisture per pound of dry air and the air is cooled to 72.7. Conservatively assume the fan/motor adds a degree of heat, now you have 73.7 degree air. To maintain 80F, 15500/1.08/(80-73.7)=2,278 CFM. To maintain 78 F 17100/1.08/(78-73.7)= 3682 CFM with 126 pounds of water each hour.
Less air, less water, outside air is filtered and don't have 2527 CFM of triple digit air infiltrating in, moving by like a sirrocco wind. Hope that attic is air tight! The floor is dry, no mold spores 'germinating' on the floor, swamp cooler has a chance of working.
This assumes you can maintain a 66F slab, and you neglect the extra heat coming in from the soil below. Heat into the slab from the soil will require more water evaporated into the air, then the upward spiral of the sensible heat of even more triple digit make up air and the amount of ADDITIONAL water needed to compensate for it.
A swamp cooler will move less air and use less water to maintain a lower temperature. A swamp cooler will have a dry floor and the outside air has been filtered. Ever go camping and get your feet cold and wet? Try and get a good nights sleep? Lol, just make them wear golashes inside with wool socks when it is 106 outside.
Just seems so much easier and so much more plausible to simply cool the outside air first and then blow it into the house, displace air to avoid the high humidity buildup rather than make people live inside of an evaporative cooler that does not work so good.
This is all on sea level, Pheonix is about 1100 ft but, making the home itself into a swamp cooler wastes water, creates discomfort, and will grow mold. Hot wind convecting heat directly to the occupants or a 74 degree breeze. You will be using extra energy for ceiling fans. If they could have done it with a fart fan and a dehumidistat they would of already.

Well lets just stick to your plan with the 21K grainger fan and see what it does.

Too bad you are going to be needing a wind tunnel in there.

Well what can I say, swamp coolers can work although posters from the SW already point out the maintenance problems and their limitations. This is the proven system, you keep proposing a revolutionary system that cuts down on the air flow,but you have not proven anything yet.

Well yes lets stick to numbers, calculate the temperature that the slab will equalize at. So far you have been assigning arbitrary values to everything. To spare you the double iterations, take the room air as an average temperature of 80F, averaged from 6 inches above floor, middle of 9' high room and 6 inches below ceiling. Or work out the convection coefficient and keep in mind that water will evaporate at its wet bulb temperature.

Sq lit's link shows that soil is easily is 80F 20 inches down. The house is slab on grade, you may have to use a flux plot .

You can ignore the moitsture migration then, assume conductivity of dry compacted sand.

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<%-name%>
• posted on November 27, 2005, 6:03 am

True. (BTW, argument only has one "e.")

Not much.

I disagree.

And properly account for that.

You are surprised that concrete can store coolth? :-)

That's low-e case b) again, for which 6K cfm out of 21K cfm sufficed. But for some reason, you said we had to choose a) or b). We chose a).

Nope. That's how the slab cools the room in case a).

Yup. As proven, with numbers. It's pretty obvious. Simple physics.

Nope. (BTW, argument only has one "e.")

Yup.
There is no unhealthy environment.

Nope.
It loses that heat as water evaporates from the slab.

That happens, but that's not the goal.

That's irrelevant in this case, with an 80 F ceiling and 75 F floor.

The ceiling's air film resistance to downward heat flow is irrelevant.

Let's keep talking about case a), since you insisted we choose one.

Where is this town you call Pheonix? :-)

It's just simple physics, Abby.
Nick
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<%-name%>
• posted on November 27, 2005, 6:35 am
Funny how you completely ignore the calculations that prove you wrong.
All you have left is spelling mistakes to point out.
Pathetic.
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<%-name%>
• posted on November 27, 2005, 1:30 pm

Just 2. Then again, you've imagined that:
this is "a revolutionary system," the indoor environment is "unhealthy," with "dripping wet ceiling slabs," 100% RH, actual swimming pools, and mold and mildew, bathtub experiments say a lot about house cooling, psych charts and wet bulb temps are needed for understanding, heat gain by sun on walls matters for understanding, ASHRAE's 80 F with w = 0.012 and a cool slab isn't comfortable, case a) with no fans and b) with fans are the same, slow ceiling fans use lots of power, air moving parallel to a cool slab loses no heat, air can collect moisture without losing heat, ceiling radiation and convection are linked, using radiation conductance requires knowing a temp diff, an 80 vs 78 F mean temp makes a "very significant difference." heat required to evaporate water doesn't come from a house, concrete cannot store coolth, vapor barriers don't keep soil dry, cool slabs lose lots of heat to dry soil, they "attract water" up through 800 feet of dry soil, cooling incoming air is unaccounted for here, indoor swamp cooling requires "a high airflow," outdoor swamp coolers use "only a thousandth" of that, and 300-year-old physics is "unproven fantasy" :-)
Nick
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<%-name%>
• posted on November 27, 2005, 2:33 pm
Pointing out spelling mistakes was lame so now you are resorting to paraphrasing me out of context.
You said you like numbers, yet you have two posts now and you keep avoiding my calculations that prove the problem with your scheme.
I produced numbers, based on a low but realistic cooling load on a 1225 square foot house in a 106 ambient. This house was better insulated and more shaded than what is deemed 'good'.
It shows the flaw of your scheme -- the sensible heat of the make up air. Your scheme adds this sensible heat directly to the room air and this creates an 'upward spiral' in the amount of air that needs to be exhausted and the amount of water consumed.
Without even considering the additional heat from the soil into your cooled your cooled slab, the numbers show a swamp cooler will maintain the temperature using less air flow and less water.
When you factor in the extra heat gain and the extra water needed because you are cooling the earth, it will only show that the swamp cooler will maintain a lower temperature using less air and less water than your flawed scheme.
Think about how they measure wet bulb temperature with a sling. You cause water to evaporate. What temperature does the water evaporate at? What temperature will the surface of the slab be?
snipped-for-privacy@ece.villanova.edu wrote:

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<%-name%>
• posted on November 27, 2005, 6:41 pm

For the umpteenth time, that sensible heat is accounted for.

I disagree.

About 9621/(22.47+ln(460+80+56.66-Twb)) = 526.6, 528.0, 527.4, 527.7, 527.6 R, ie 67.6 F in 80 F air with Pa = 0.566 "Hg.

About 80-4.6 = 75.4 F, with suitable controls.
Nick
Abby's song:
I am the very model of a modern HVAC criminal. I've information vegetable, animal, and mineral. I know the charts psychrometral and quote wet bulbs historical From Freon to swamp coolerdom in order categorical. ...
In short, when I've a smattering of elementary physicstry, You'll say a better criminal has never pumped a gas refridgery :-)
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<%-name%>
• posted on November 27, 2005, 7:23 pm
snipped-for-privacy@ece.villanova.edu wrote:

Well, please clearly account for it one more time. Again, I did a complete set of calculations based on some realistic conditions, and they show you scheme needs to move more air and evaporate more water.
So take the same load, same ambinet conditions if you dare.

Yes because you would have to admit you were wrong otherwise :)

Well you are going to have to clarify this one for me, seems like you are close to coming up with the wet bulb temperature of what you are trying to maintain the room at. Perhaps the difference is an error in the accuracy of the empirical formula?

Again take the loads I have established and prove how your suitable controls will do this.

Like I said, you are avoiding my numbers like the plague, come on take them on you can do it. You have avoided them three times in a row now. Must be "No joy in Pineville, tricky Nicky has struck out."

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<%-name%>
• posted on November 27, 2005, 8:48 pm

OK. It's 93.5 F on an average June day in Phoenix, with wa = 0.0056.
We want to make 5.7K Btu/h of NET house cooling to keep it 80 F with wi = 0.0120. If we add P pounds per hour of water to the house air and bring C cfm of outdoor air into the house, that outdoor air needs to be cooled to 80 F with approximately (93.5-80)C Btu/h, no?
Air weighs 0.075 lb/ft^3, so P = 60C0.075(wi-wa) = 0.0288C lb/h of water move out of the house, so we need C = 34.7P cfm of outdoor air to make that happen, right?
If P lb/h of evaporation provides 1000P = 5.7K+(93.5-80)34.7P of total cooling, including the required makeup air cooling, P = 11 lb/h and C = 34.7P = 372 cfm, no?
This is more efficient with cooler outdoor air, so we might well turn off the system and use stored slab coolth during the warmest part of the day. This is easier with a slab than a swamp cooler, because the slab is cooled directly and daytime setbacks are possible, since the slab can stay cool while the room air is hot.

That's an iterative wet bulb approximation based on a Clausius (1822-1888)- Clapeyron (1799-1864) approximation and Bowen's 1926 equation, mere 80-year-old simple physics :-)
Nick
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<%-name%>
• posted on November 28, 2005, 12:31 pm

Oops. An average June day is 88.2.

So the house might need (88.2-80)425 = 3485 Btu/h.
If we add P pounds per hour of water to the house air and bring C cfm of outdoor air into the house, that outdoor air needs to be cooled to 80 F with approximately (88.2-80)C Btu/h.

If P lb/h of evaporation provides 1000P = 3485+(88.2-80)34.7P of total cooling, including the required makeup air cooling, P = 4.9 lb/h and C = 34.7P = 169 cfm.
This would be a good time for you to smack yourself on the forehead and say "We might do exactly the same thing with an external swamp cooler if we turned on the pump with a room temp thermostat and turned on the blower with a room humidistat, using exactly the same water and air flows" :-) But swamp coolers aren't usually controlled that way, and the motor heat ends up in the house, and it's nice to avoid the big box.
It might make sense to add these controls to existing swamp coolers.

This might also be more efficient and comfortable with a hollow slab that allows indirect cooling with a higher RH between 2 vapor barriers. If we evaporate water at close to 100% RH at 80 F with wi = 0.0223 and move C cfm of outdoor air under the slab, P = 60C0.075(wi-wa) = 0.0752C lb/h of water, so we only need C = 13.3P cfm of outdoor air... 1000P = 3485+(88.2-80)13.3P of total cooling makes P = 3.9 lb/h and C = 52 cfm. The RH of the 80 F house air would be about 100x0.0056/0.0233 = 24%.
Nick
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• posted on November 28, 2005, 4:26 pm
snipped-for-privacy@ece.villanova.edu wrote:

Lol, again with even a lower average number and once again not comprehending that cooling loads depend on more than just the difference between indoor and out door temperature.
Conveniently neglects internal and solar gains yet one more time. But then again, your scheme only maintains comfort after the sun has set for several hours and then again after the sun has only risen for a few hours.

But swamp coolers aren't usually controlled that way, and the motor heat ends up in the house, and it's nice to avoid the big box.
Under your microscopic cooling loads, fan/motor heat from a swamp cooler would be would be negligible. Under REAL conditions, as I already have shown, even when allowing for fan/motor heat, the swamp cooler will maintain a lower temperature and use less water, than your scheme.

You could have a floor made from hollow blocks and use a swamp cooler to blow air through the core. Have you ever poured a hollow slab? :)
Looks like you are finally seeing the benefit of not making the inside of the home a swamp cooler :) Wow a light bulb finally illuminates.
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• posted on November 28, 2005, 5:15 pm
Here you go Nicky
The chance for the Inverted Pool of Pine. A concrete roof, beams infilled with hollow blocks. Be three inches of concrete with mesh above this along with inappropriate red barrel tiles in an environment that does not really need a big overhead thermal mass.
http://www.imagewiz.net/usr/a_bee_normal/25022_P1010016.JPG
In the UK, you get a flooring system similar to this roof and they use it for radiant heat. Nice to walk on a warm floor in winter, no one will like a 68 degree floor in the summer.
Instead of the dripping wet cathedral ceiling, maybe I could put PEX tubing through the roof cores and heat some hot water. I will put the storage tanks which I will also paint black on top of the peak so that I can turn the roof into a solar collector and have a thermosyphon system set up.
Then I can use this hot water to drive an absorption chiller (mounted on the ground) utilizing the solar heat you neglect in your cooling calculations. I can syphon hot water out of the roof storage tanks to the chiller below but I am afraid I will need some mechanical energy to run a pump to deliver chilled water through some tubing in the slab.
I can put a donkey on a treadmill and transfer this mechanical energy to run a pump. I just have to work out the 'suitable controls' that dangles a carrot on a string in front of the donkey when the space temperature rises.
I was thinking of ordering some thermal links from grainger to trigger the relaes from the carrot storage hopper, but I can seem to find any that are set for 78F. Can you help me out with the controls?
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• posted on November 28, 2005, 5:30 pm

This "have to" comment is just worn-out sneaky net.rhetoric :-)
But if we can store coolth in the slab more easily than with a swamp cooler, why operate inefficiently during the hot part of the day?

1000P = 5.7K+(93.5-80)34.7P makes 531.55P = 5.7K, so P = 10.72 lb/h.

That is incorrect, for the scheme I suggested.

Not. This is case a), as you insisted we choose.

This "impractical" comment is just more arrogant sneaky net.rhetoric :-)
A swamp cooler with appropriate controls could do exactly the same thing as the indoor scheme, with the same water and airflows.

It might, if the only control were a power switch.

No. It might, but that's not a requirement.

Not much.

Maybe dampish...

Nope. They need more than 60% RH continuously for about 2 weeks.

False. I never said anything like that.

Sure. Same equation, but the Twb on the left makes it more understandable.

It's 100 times the vapor pressure, from Bowen's equation.

I know :-) Then again, you seem to know very little about physics.

No thanks. But I'll take a look, if you'd like to do that.

Using real weather data.

More arrogant snotty sneaky slimy rhetoric. I have no such miscomprehension.

I'm afraid you have erred again, my good man.

Then again, you might be too pompous and obtuse to do that :-)

Doh :-) How close can we get to 100% RH under the slab?

Sure, or evaporate water there, with a lot less hardware.
Nick
"I am the Monarch of AC, the Ruler of HVAC, whose praise PJM loudly chants, And we are his sisters and his cousins and his aunts..." :-)
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• posted on November 28, 2005, 8:41 pm
snipped-for-privacy@ece.villanova.edu wrote:

This would "have to" appear as "the truth hurts". Average temps are great for estimating fuel consumption, but that is it. A system designed for an average temperature will not maintain comfort when it gets warmer than average.
You don't like published values, then use the average high temperature.

Your stored coolth is the convenient rhetoric you use all the time. Your average system has to run 24/7 and cannot keep up.
A system designed for a peak temperature (ie published design dry bulb) could perhaps store some 'coolth' and limit the indoor temperature when the outdoors exceeds published design ambinets if wanted.
Use the electrical power and water when you need it. Trying to store an excess wastes energy everytime, because there will be storage losses ( remember the extra heat into the slab). In addition, what you need to do is to ultimately cool the air. Your system makes the slab a middle man, everytime there is an exchange there is a loss.

Well historically you never correctly allow for the specific heat of air and I was looking at the actual amount of heat required for evaporation of water. Perhaps that is why your numbers are a little on the high side. I thought you would like a low exhaust rate and less water.

Thanks I guess were are back to the natural convection of heat to the floor, or is it the house and contents radiate all their heat to the floor, and then the objects inside and the structure of the house cool the room air.

Well for about 5 times you have avoided putting your system up against some REAL situations. You keep talking about loads of 5,700 Btu/hr and LESS and use low average temperatures. The only practicality in this stance is it provides you with rhetoric and a means to stall in an attempt to save your battered ego.
Arrogance, pride, obtuse -- just look in a mirror Nicky.

Thermostat, line voltage or 24 volt, take your pick

It is based on an average.

dampish coolth lol, you will be constantly adding water to it.

Sorry Nicky, maybe try Chapter 7 of ASHRAE's Humidity Design Control Manual - it is an excellent book http://www.masongrant.com/pdf/design_guide/ASHRAE_HCDG_C7_Mold.pdf
All it takes is one spore and a wet food source and you goose could be cooked. Mold does not eat concrete but there will be plenty to eat off of that floor. At least the swamp cooler keeps the floor dry and does provide some filtration of the outside air.
High RH in the space could cause some localized condensation in a cool corner or place where air gets trapped, then the mold starts. It needs to be wet to get going and your slab is the perfect place. No baseboard trim I hope, no capillary action up exterior and partion walls.
Here is a place that was flooded with 46 inches of water. Driving rain got through to the ceiling as well. The photo is about two weeks after the flood. no electrical power. Rampant heat and humidity.
http://www.imagewiz.net/usr/a_bee_normal/Mold/18811_RuinedDrawings.JPG
Note the mold only made it four feet up the wall, a little capillary action above the flood level. The dry sheetrock above shows no mold. The ceiling was infested as well although the photo does not show it.

You said you can shut it off during hot weather and the 'stored coolth' will take care of it.

I keep substituting 527.6R into the left side of the equation and the Twb term that you are taking the natural log of keeps coming up as (80+56.6)F. That is why I asked for the clarification. Is there a typo error?
Maybe try newton-raphson for the iteration. I am wondering about the 56.66 term, you confirmed its based on the water vapour pressure but the constant you multiplied it by changes its units to degrees R?

well your ego must be hurting then

No not snotty, you just cannot base the load on temperature alone like you always do.

So far you have shown that you can theoretically maintain a temperature in a house with a load so small that it is pathetic, during an average outdoor temperature. It is the only way your numbers can work.
You yourself keep harping on how you are going to 'store coolth', so the only way you will store excess cooling with this scheme is to get the house cooler than it has to be when the outdoor temperature is cooler than the average, and then hopefully use this 'stored coolth' to help to limit how much warmer the space will be when the outside temperature is hotter than the average. Your system design for an average temperature cannot keep up in the afternoon and so conditions are going to rise up out of the comfort zone even with whatever 'stored coolth' you have.
At best the indoor condition you will maintain 'averages out' to the extreme corner of the comfort zone. You will have to get cooler inside overnight with high RH and by afternoon you will be warmer and more humid than the comfort zone.
Under a real scenario your scheme has more holes than Swiss cheese, you will be growing mold, you will be moving more air through the home than an evaporative cooler,you will be using more water than an evaporative cooler and, you will be maintaining a higher temperature than an evaporative cooler.

Your ego must be really suffering.

Well you could try cooling your house by spraying water on the floor.

Should we referred to you as "Your Majesty", I can't say I ever witnessed PJM praising you. You are PJM's sister, or do you just feel like trapped in a male body?