# pool pump

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• posted on November 18, 2005, 4:17 pm
It still sounds as though you are wetting a slab and or weting a ceiling above, and plan on using low rate exhaust which will not work.
Mr. G is putting on a photvoltaic seminar alond with the florida solar energy center down here. Listening to their philosphies I figured he would have to know you.
snipped-for-privacy@ece.villanova.edu wrote:

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• posted on November 18, 2005, 7:46 pm

Dampening...
I disagree. You might too, if you think outside the swamp cooler box.

I've never seen Drew use a calculator. As a Professional Engineer, he has a "mathematical license," like poetic license :-)
You can read some of our Solar Today stories at
http://www.ece.villanova.edu/~nick
Nick
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• posted on November 18, 2005, 10:02 pm
Your dampened slab cools the earth beneath it.
You need the high airflow, as the evaporative cooling process follows a constant wetbulb line. You are sensibly heating and humidifing the house with your train of thought. You can't program this one in BASIC, you have to plot it.
snipped-for-privacy@ece.villanova.edu wrote:

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• posted on November 18, 2005, 10:14 pm

But it's a good insulator, esp under a vapor barrier.

No. Think outta that swamp cooler box!

Nonsense. Of course you can, with a Clausius-Clapeyron approximation.
Nick
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• posted on November 18, 2005, 10:45 pm
The greatest heat loss from a swimming pool is evaporation.
The water evaporating draws the majority of its heat from the water that is left behind. Your evaporating water cools the slab and humidifies the indoor air.
Heat external to the residence, is evaporating a good portion of the water. The heat to evaporate is not all coming from the room air.
Evaporative Cooling is an adiabiatic process where the wet bulb is constant.
Then you use ceiling fans to try and blow down the warm air, to be cooled from contact with the slab.
You are trying to make the people live inside of a swamp cooler that does not work. You would be giving Rube Goldberg an allergy :)
You need to work out your scheme and plot it on a chart. I have pointed out to you before, with high latent loads and low SHR ratios, that cooling and dehumidying air can not always be down in a single process. You need to over cool air and then reheat, using much more energy than the difference in enthalpies of the starting and ending points calculate out as.
Clausius Claperon does not describe the adiabiatic conversion of sensible heat into latent heat.
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• posted on November 19, 2005, 7:25 am

I disagree. I just explain it with simple physics, step by step, eg
1. 80 F and wi = 0.012 is in the ASHRAE-55 2004 comfort zone.
2. Pa = 29.921/(1+0.62198/wi) = 0.566 "Hg...
which you don't seem to have understood so far, after several tries.
If you'd like to try again, I'm willing.
Would you agree with 1 and 2 above?
Nick
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• posted on November 19, 2005, 2:00 pm
Nick
You always like to push the comfort envelope and if you lived in a sweltering jungle, and inside your home you walked around naked, 80F @ 84 grains would feel pretty good especially with floor fans and ceiling fans operating.
Everywhere else. the occupants would be calling the AC service company saying the air conditioner is not working.
Here is the problem with physics, take that 80 degree air with 84 grains of moisture and change it to 75 degree air with 50% relative humidity.
The change in enthalpy per pound of dry air is approximately 4.25. The problem is the amount of mechanical cooling needed to perform this change is 9.12, almost double what physics would suggest.
Likewise when you deal with evaporative cooling, the air has a constant wet bulb. Wet bulb is pretty much indicative of total heat, so it remains constant as sensible heat gets converted to latent heat. You keep ignoring this.
You also deal with a conductance based on an air temperature differential alone, and make no allowance for the effect of hot sun beating down on a structure.
Please demonstrate your claims on a psychrometric chart and use standard pyschrometric equations. If your claims are valid, then you should get the same answer.
snipped-for-privacy@ece.villanova.edu wrote:

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• posted on November 19, 2005, 6:18 pm

1. 80 F and wi = 0.012 is in the ASHRAE-55 2004 comfort zone.
2. Pa = 29.921/(1+0.62198/wi) = 0.566 "Hg...
Would you agree with 1 and 2 above?
Nick
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• posted on November 21, 2005, 1:51 pm
Nick
You are someone trying to prove you have re-invented and 'improved' something so you should be able to plot it, in order to keep your dream boat from sinking.
The water vapour pressure for 80F at 84 grains is close enough.
I do not have the 2004 version of the ASHRAE standard, but with your revolutionary discovery, 78F with RH under 60% should be attainable.
Look at a realistic, well insulated home.
106F ambient at 20% RH, 33 degrees north latitude. At night it gets down to 81 on average.
35'x35'x9' home, R19 walls, R40 in ceiling, each wall has 47.24 sq ft of double glazed glass that is internally shaded. Eaves over hang glass by 2 ft, top of windows 1 ft below eaves. 4 occupants who cook and do their laundry inside their homes. The walls are internally insulated.
Since the home will be under negative pressure in your scheme we can neglect infiltration so a sensible heat gain could be 15,530 Btu/hr with a latent gain of 2,800 Btu/hr.
The 6 inch floor slab is not insulated.
You should be able to get the same results as phsyics using a pychrometric chart.
Here is a link to one that you can even read water vapour pressure in inches of mercury off of and it is free.
http://www.trane.com/Commercial/Equipment/ProductDetails.aspx?prod0
Or if you really want to go from first principles here is some MEASURED data on water
http://www.imagewiz.net/usr/a_bee_normal/24848_Water_Table_2.jpg
snipped-for-privacy@ece.villanova.edu wrote:

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• posted on November 21, 2005, 2:18 pm
With the slab on grade home 'above' , infiltration is neglected however you will have to add the sensible heat of the ventialtion air to the load as you are directly added this hot dry air to the indoor air without pre-conditioning it first without evaporative cooling.
When you pre-condition the air, it is like wearing a condom. I think you will find your scheme pans out as being analogous to getting an abortion.
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• posted on November 21, 2005, 5:02 pm

Nope. I've proved it, altho it's pretty obvious to people who understand basic physics. You seem to be trying to understand the "proof." Good luck. It seems to me you are having a hard time thinking beyond wet bulb temps and swamp coolers. I'd say it's mostly an attitude problem: arrogance...

In HVAC priesthood mumbo-jumbo, 84 grains makes the humidity ratio wi = 84/7000 = 0.012 pounds of water per pound of dry air, exactly. Close enough? :-) The vapor pressure inside the house Pi = 29.921/(1=0.62198/wi) = 0.566 "Hg. This "non-standard equation" (21) is on page 6.12 of the 1993 ASHRAE Handbook of Fundamentals. It is neither rocket science nor new... 0.62198 is the ratio of the molecular weight of water to the molecular weight of dry air... 29.921 "Hg is the standard atmospheric pressure. John Dalton (1766-1844) probably discovered this, as a part of his law of partial pressures.

Nothing revolutionary, I'd say, altho any sufficiently advanced technology is indistinguishable from magic, and people have different personal definitions for the word "advanced." Sure, 78 F at 60% is attainable in the right climate. So is 80 F at 54%, which is also within the ASHRAE 55-2004 comfort standard.
Shall we continue with this basic physics, as simple seekers of truth? Nick
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• posted on November 23, 2005, 1:48 pm
You are the one demonstrating arrogance, or should I say cowardice by refusing to verify your claims. If anyone can make me eat crow, it should be you. So come on, serve me up some dinner.
I am well aware of Dalton's Law of Partial Pressures. How about adding an elevation of 1117 ft ASL, then you can use 28.73 inches of mercury.
When you use swimming pool equations, you have to realize that the heat to evaporate the water is coming from the water in the pool itself as well as a severely flooded permimeter around the pool.

The tub would be a couple degrees cooler than the room air because water was evaporating from it. It did not cool off the room. Same principle as the boy scout trick of wrapping a wet newspaper around a bottle of water. The water in the newspaper evaporates and in doing so, it drew sensible heat from the bottle and hence the water inside of it.
Water evaporating from the slab will get heat from the slab, heat will conduct to the slab from below. Then your ceiling fans, which use energy, will force the air down towards the slab and be cooled by the slab.Air does not 'flow into corners' and the majority of this air will not contact the slab but will flow 'parallel to the slab' and will not have the benefit of contacting the slab to transfer sensible heat to the slab. However this air does get to recieve the addtion of moisture without the benefit of being cooled sensibly.
The evaporating water encourages heat to conduct up into the slab from below, you will add an excessive amount of moisture to the air, and a small portion of the heat used to evaporate water actually is actually sensible heat removed from the room air.
Then, to try and lower humidity you must exhaust air, and will be drawing in triple digit outside air.
The biggest problems with your scheme--
1) More water will evaporate than you are counting on 2) You assume that all the heat that will evaporate water comes from the room air when it will becoming from the outside of the residence. 3) You are adding untreated triple digit outside air directly to the space. 4) The ceiling fans (note the plural) use energy and so will your "Constant" bath room fan, yet these are written off as insignificant.
So come on Nick, work it out using pyschromtrics, not nickrometrics, -- if you can.
I will make it easier for you, see if your scheme can maintain 80 F @ 71% RH based on the new elevation I just threw in, forget comfort, let see if you can do it without using the high airflow of an evaporative cooler. How big is that exhaust fan going to be?
snipped-for-privacy@ece.villanova.edu wrote:

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• posted on November 23, 2005, 4:48 pm

Good :-)

No thanks. It seems to me that we keep getting sidetracked from the main issue (whether these indoor evaporation schemes can work at all) by this and other side issues, like whether to count the heating effect of sun on walls, how to calculate how much cooling a house requires, whether these schemes can work in a humid jungle, and so on.

Sure. Page 4.7 of the 1991 ASHRAE Applications handbook has empirical formula (2) for evaporation of water "from public pools at high to normal activity, allowing for splashing and a limited area of wetted deck": wp = 0.1A(Pw-Pa) lb/h, with pool surface A in ft^2 and Pw sat pressure at the water surface temp and Pa at the room air dew point, both in "Hg. Again, let's not argue about activity levels now. The indoor evaporation can come from any number of sources: a dampened slab with a soaker hose and a solenoid valve and a thermostat, a pond, a fountain, a portable swamp cooler, misters, green plants, indoor clotheslines, and so on.

Sounds like the tub didn't have enough surface to cool the room much, but if you had taken careful enough measurements, you would have noticed that the A ft^2 bathtub cooled the room by about 100A(Pw-Pa) Btu/h. The water loses heat to the room air by evaporation, and the water surface and all the other tub surfaces gain sensible heat from the room (ie cool the room.) There is no magic. Energy is conserved.

Not much, if the ground below is dry. The slab might be over a vapor barrier or foamboard insulation. Again, let's not get sidetracked. Let's simplify this and say the slab gains no heat from the ground, for now.

Slow ceiling fans use very little energy. Let's say 0, for now.
How much heat moves from the room to the slab by radiation? Grainger's 4C853 48" fan moves 21K cfm at full speed at 315 rpm with 86 watts. How much air do we need to move up from the slab to the room to keep it comfy? How much power does that require, according to fan laws? Swamp coolers put all their electrical heat power into the house...

Right :-) With a room temp thermostat and an occupancy sensor.

Let's avoid this sidetrack and say the air in the room is fully mixed. And not clever enough to collect water vapor without collecting coolth.

We would add exactly P pounds per hour of moisture to the room air.

The P lb/h of water provides 1000P of total cooling, which both cools the room to say 80 F at wi = 0.0120 AND cools C cfm of outdoor air (with an exhaust fan and a humidistat, plus some air infiltration) at Ta (F) and wa that flows into the room.

OK. Say it's 3 PM on an average June day in Phoenix, with Ta = 103.5 F and wa = 0.0056, and we want 10K Btu/h of net sensible cooling for a small well- insulated house (with G = 10K/(103.5-80) = 425 Btu/h-F) to keep it 80 F with wi = 0.012. Air weighs 0.075 lb/ft^3. P = 60C0.075(wi-wa) = 0.0288C makes C = 34.7P, and 1000P = 10K+(103.5-80)C = 10K+23.5x34.7P makes P = 54 lb/h and C = 1886 cfm. Wow.
At the average 93.5 F outdoor temp, we can provide (93.5-80)425 = 5.7K Btu/h with 1000P = 5.7K + (93.5-80)34.7P, so P = 11 lb/h, and C = 372 cfm. This works more efficiently with cooler outdoor air, so we might turn off the system and use stored slab coolth during the warmest part of the day.

Let's try this at 3 AM, when Ta = 72.9 with wa = 0.0056. Say we want to store 22h(93.5-80)425 = 126K Btu of coolth in a 2 hours for the rest of the average 93.5 F day. If we turn on an 800 cfm 100 W \$12 20" Chinese window box fan for 2 hours and evaporate P lb/h of water from a 2000 ft^2 80 F slab with a 1/(1/800+2/3/2000) = 632 Btu/h-F conductance to outdoor air, 126K = 2h(1000P+(80-72.9)(800+632)) makes P = 53 lb/h, about 13 gallons for the whole day, since evaporative cooling is more efficient with cooler night air. If we splurge and use Lasko's \$50 90 W 2470 cfm fan, 126K = 2h(1000P+(80-72.9)(2470+1355)) makes P = 36 lb/h, about 9 gallons.
And we might use less water if the house were unoccupied for 8 hours during the day, with a cool slab under warm house air.

No. Not at all. That's the last term in 1000P = 10K+(103.5-80)C above. I hope that's clear now. You may have been missing this over and over.

And properly accounting for that...

Yup. I'd like to stop now. Feel free to work on the rest of the details.
Nick
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• posted on November 23, 2005, 5:21 pm
Nick you are sounding evasive, every 'side track' you avoid is an attempt to avoid realistic conditions.
I gave you a better than average insulated home, that included solar gain which you keep ignoring.
Vapour barrier below a slab is insignificant in particular with respect to conduction. I gave you better than average ceiling and shading as is found in the SW, everything but an insulated slab.
The whole arguement really is the amount of air needed to be moved or exhausted. If you really look at your flawed dream you will see that you are putting out fire with gasoline.
You will be running a high rate of exhaust which means a high rate of triple digit infiltration or make up air, to be added on top of the external and internal gains of the structure.
So sad you will not try it the other way, you have proven nothing. I do not get impressed when some one has to write a program to develop a value of Pi, not intrinsic to GW BASIC.
You are the one trying to prove you revolutionized the concept of evaporative cooling, yet your thinking is flawed and you plug significant holes in your scheme by saying you can get a cheap fan at Grainger or you can store 'coolth' , or just set the exhaust fan to come on at a certain humdity level and all is well.
When you want to use the swimming pool equations that Dectron developed, you need a pool. The water evaporates from the pool, the greatest heat loss from a pool is evaporation, as the evaporating vapour draws heat from the water itself. Sort of like how they define Wet Bulb.
The activity level basically covers 'splashing' where the water is splashed up in the air and ACTUALLY GETS ITS HEAT TO EVAPORATE FROM THE AIR. If anything your scheme needs a high activity level.
You want to rely on your ceiling fans then lets see you develop some some convection coefficients, then crunch some numbers with the multi-dimensional heat flux through the slab.
At best your scheme will give conditions expected in a natatorium, not the best conditions for a residence.
Actually learn that evaporative cooling needs to rapidly change the indoor air with 'cooled' outside air, pay the penalty of a high air flow rate and you can maintain a liveable space for a couple months out of the year.
snipped-for-privacy@ece.villanova.edu wrote:

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• posted on November 23, 2005, 5:33 pm
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Nick's only actual computer program is a random number generator, which he uses to generate all his results :-)
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• posted on November 23, 2005, 7:48 pm

Have a nice Thanksgiving.
Nick
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• posted on November 24, 2005, 12:44 pm

OK.
If the linearized radiation conductance between an ordinary ceiling and slab is about 4x0.1714E-8x(460+80)^3 = 1.1 Btu/h-F-ft^2 at 80 F, a 2000 ft^2 slab could collect 10K Btu/h of heat by radiation with a 5 F temperature difference, so a 75 F slab might need no fans, but how would we turn the coolth off during unoccupied times?

With low-e ceiling and wall surfaces, we could keep the room 80 F with a 75 F slab with a 1.5x2000 = 3K Btu/h-F film conductance and C cfm of airflow if 10K = (80-75)(1/(1/C+1/3K)), which makes C = 6000 cfm.

...(6K/21K)^3x86 = 2 watts.
Nick
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• posted on November 24, 2005, 1:18 pm
That does not appear like convection calculations to me.
Radiation is a two way street, to the fourth power. The walls and the ceilings will be warmer than the room air and yes they will radiate heat at the slab, and the slab will radiate heat to them.
The slab is 35x35, 39% smaller than 2000 sq ft. The heat load inside of this space is 15,550 or 55% greater than you have allowed for.
You need to establish the slab temperature by balancing out the evaporation of water off of the slab and how this affects the conduction from the ground below and the edge of the slab, and then determine the forced convection that actually cools the air. You cannot just guess and say it is 75.
You will love convection coefficients, you get to play with so many dimensionless numbers. :)
snipped-for-privacy@ece.villanova.edu wrote:

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• posted on November 24, 2005, 3:24 pm

Up here in the north country, modern construction has insulation (typically 2 inches of foam board) put *under* the slab and between slab and outside walls/footers (only 1" there) to help prevent heat losses. Could be done in warm clients for the opposite reasons I think.

:-) Ain't that the truth! Like Prandtl, Reynolds, throw in some Nusselt for the HT and things can be really 'fun'.
daestrom
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• posted on November 24, 2005, 3:57 pm
daestrom wrote:

slab on grade homes without basements, you will see some perimeter insulation go down past the edge of the slab maybe 2 feet below grade, only noticed insulation below slab on radiant floor projects typically.

Grashoff (sp?) too :)

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