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Cool tower alternatives
Abby Normal wrote:
You need to work out your scheme and plot it on a chart... 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 |
Cool tower alternatives
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. wrote: Abby Normal wrote: You need to work out your scheme and plot it on a chart... 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 |
Cool tower alternatives
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/Equi....aspx?prod=170 Or if you really want to go from first principles here is some MEASURED data on water http://www.imagewiz.net/usr/a_bee_no...er_Table_2.jpg wrote: Abby Normal wrote: Please demonstrate your claims on a psychrometric chart... No thanks. But go ahead, if that floats your boat. 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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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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Abby Normal wrote:
Nick You are someone trying to prove you have re-invented and 'improved' something... 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... The water vapour pressure for 80F at 84 grains is close enough. 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. I do not have the 2004 version of the ASHRAE standard... Too bad. but with your revolutionary discovery, 78F with RH under 60% should be attainable. 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 |
Cool tower alternatives
"Abby Normal" wrote in message oups.com... 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. Not to shoot holes in calculations, BUT. 106 with 20% RH does not last very long in Phoenix. We are by far dryer than a lot of the cities at 33 degrees. 106 and 40 is much more common, with night temps closer to triple digits. I had evaps for years, no more. They take to much to keep clean and the mineral build up from evaporation can be nasty to deal with. Especially when they really work a month or two out of a year. Unless of course you like 60 RH inside. 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. New home built in the valley of the sun have R-30 in the ceiling. Overhang Eves on Southwestern style homes are almost non existant. Direct sun on the windows/doors is common. R-19 is a typical rating for new construction today in the Phoenix area. 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. Slabs are 4 inches 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/Equi....aspx?prod=170 Or if you really want to go from first principles here is some MEASURED data on water http://www.imagewiz.net/usr/a_bee_no...er_Table_2.jpg wrote: Abby Normal wrote: Please demonstrate your claims on a psychrometric chart... No thanks. But go ahead, if that floats your boat. 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 |
Cool tower alternatives
"Abby Normal" wrote in message oups.com... 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/Equi....aspx?prod=170 Or if you really want to go from first principles here is some MEASURED data on water http://www.imagewiz.net/usr/a_bee_no...er_Table_2.jpg Or if you want some *really* definitive information about 'water' and 'steam', try www.iapws.org (these guys need to get a life ;-) (the math is not for the faint of heart!!) daestrom |
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I am giving him a chance with some better than average insulation
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at your elevation the 106F at 40% is a dewpoint just shy of 77F. I have
some worst case humidity data for there, but it shows nothing close to that. Maybe early in the morning when the temp is lower perhaps you would be seeing 40. Maybe higher RH at lower temps during the monsoons |
Cool tower alternatives
"Abby Normal" wrote in message oups.com... I am giving him a chance with some better than average insulation Ya I understand try this page for some information. http://ag.arizona.edu/azmet/data/1205em.txt Yearly temp and humidity data by month, with min and max. . Check July, the data is pretty clear the desert is not so dry any more. Not like it was 35 years ago when I moved here. |
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Thanks for the weather data, interesting the ground temperature I
guessed 80 before. You obviously get higher RH than 20, it just must coincide with the lower temperatures at night or first thing in the morning. Dewpoint seems to have peeked at two days at 70 and one day at 71 this year. SQLit wrote: "Abby Normal" wrote in message oups.com... I am giving him a chance with some better than average insulation Ya I understand try this page for some information. http://ag.arizona.edu/azmet/data/1205em.txt Yearly temp and humidity data by month, with min and max. . Check July, the data is pretty clear the desert is not so dry any more. Not like it was 35 years ago when I moved here. |
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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. From experience, I can tell you that during some sweltering conditions during a prolonged power failure, I would let water stand in a bathtub. 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? wrote: Abby Normal wrote: Nick You are someone trying to prove you have re-invented and 'improved' something... 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... The water vapour pressure for 80F at 84 grains is close enough. 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. I do not have the 2004 version of the ASHRAE standard... Too bad. but with your revolutionary discovery, 78F with RH under 60% should be attainable. 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 |
Cool tower alternatives
Abby Normal wrote:
I am well aware of Dalton's Law of Partial Pressures. Good :-) How about adding an elevation of 1117 ft ASL, then you can use 28.73 inches of mercury. 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. 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. 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. From experience, I can tell you that during some sweltering conditions during a prolonged power failure, I would let water stand in a bathtub. 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. 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. Water evaporating from the slab will get heat from the slab, heat will conduct to the slab from below. 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. Then your ceiling fans, which use energy... 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... will force the air down towards the slab and be cooled by the slab. Right :-) With a room temp thermostat and an occupancy sensor. 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. 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. you will add an excessive amount of moisture to the air... We would add exactly P pounds per hour of moisture to the room air. and a small portion of the heat used to evaporate water actually is actually sensible heat removed from 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. Then, to try and lower humidity you must exhaust air, and will be drawing in triple digit outside air. 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. ... More water will evaporate than you are counting on 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. 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. 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. 3) You are adding untreated triple digit outside air directly to the space. And properly accounting for that... 4) The ceiling fans (note the plural) use energy and so will your "Constant" bath room fan, yet these are written off as insignificant. Yup. I'd like to stop now. Feel free to work on the rest of the details. Nick |
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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. wrote: Abby Normal wrote: I am well aware of Dalton's Law of Partial Pressures. Good :-) How about adding an elevation of 1117 ft ASL, then you can use 28.73 inches of mercury. 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. 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. 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. From experience, I can tell you that during some sweltering conditions during a prolonged power failure, I would let water stand in a bathtub. 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. 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. Water evaporating from the slab will get heat from the slab, heat will conduct to the slab from below. 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. Then your ceiling fans, which use energy... 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... will force the air down towards the slab and be cooled by the slab. Right :-) With a room temp thermostat and an occupancy sensor. 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. 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. you will add an excessive amount of moisture to the air... We would add exactly P pounds per hour of moisture to the room air. and a small portion of the heat used to evaporate water actually is actually sensible heat removed from 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. Then, to try and lower humidity you must exhaust air, and will be drawing in triple digit outside air. 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. ... More water will evaporate than you are counting on 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. 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. 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. 3) You are adding untreated triple digit outside air directly to the space. And properly accounting for that... 4) The ceiling fans (note the plural) use energy and so will your "Constant" bath room fan, yet these are written off as insignificant. Yup. I'd like to stop now. Feel free to work on the rest of the details. Nick |
Cool tower alternatives
On 23 Nov 2005 09:21:21 -0800, "Abby Normal"
wrote: Nick you are sounding evasive, every 'side track' you avoid is an attempt to avoid realistic conditions. Nick's only actual computer program is a random number generator, which he uses to generate all his results :-) -- Click here every day to feed an animal that needs you today !!! http://www.theanimalrescuesite.com/ Paul ( pjm @ pobox . com ) - remove spaces to email me 'Some days, it's just not worth chewing through the restraints.' 'With sufficient thrust, pigs fly just fine.' HVAC/R program for Palm PDA's Free demo now available online http://pmilligan.net/palm/ |
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Abby Normal wrote:
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. wrote: Abby Normal wrote: I am well aware of Dalton's Law of Partial Pressures. Good :-) How about adding an elevation of 1117 ft ASL, then you can use 28.73 inches of mercury. 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. 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. 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. From experience, I can tell you that during some sweltering conditions during a prolonged power failure, I would let water stand in a bathtub. 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. 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. Water evaporating from the slab will get heat from the slab, heat will conduct to the slab from below. 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. Then your ceiling fans, which use energy... 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... will force the air down towards the slab and be cooled by the slab. Right :-) With a room temp thermostat and an occupancy sensor. 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. 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. you will add an excessive amount of moisture to the air... We would add exactly P pounds per hour of moisture to the room air. and a small portion of the heat used to evaporate water actually is actually sensible heat removed from 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. Then, to try and lower humidity you must exhaust air, and will be drawing in triple digit outside air. 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. ... More water will evaporate than you are counting on 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. 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. 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. 3) You are adding untreated triple digit outside air directly to the space. And properly accounting for that... 4) The ceiling fans (note the plural) use energy and so will your "Constant" bath room fan, yet these are written off as insignificant. Yup. I'd like to stop now. Feel free to work on the rest of the details. Have a nice Thanksgiving. Nick |
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Abby Normal wrote:
You want to rely on your ceiling fans then lets see you develop some some convection coefficients, then crunch some numbers... OK. 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? 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? 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? 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. How much power does that require, according to fan laws? ....(6K/21K)^3x86 = 2 watts. Nick |
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"Abby Normal" wrote in message oups.com... 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. 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. You will love convection coefficients, you get to play with so many dimensionless numbers. :) :-) 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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daestrom wrote: "Abby Normal" wrote in message oups.com... 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. 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. It is not a standrard practise in warm climates at all. Up north , for 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. You will love convection coefficients, you get to play with so many dimensionless numbers. :) :-) Ain't that the truth! Like Prandtl, Reynolds, throw in some Nusselt for the HT and things can be really 'fun'. Grashoff (sp?) too :) daestrom |
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Abby Normal wrote:
That does not appear like convection calculations to me. It's a conservative underestimate. 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... Oddly enough, I feel no such need :-) BTW, the vapor barrier under the slab was to keep the soil dry, so it's an excellent insulator for upward heatflow. Nick |
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"Abby Normal" wrote in message oups.com... daestrom wrote: "Abby Normal" wrote in message oups.com... 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. 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. It is not a standrard practise in warm climates at all. Too bad. Doesn't that lead to a lot of heat gain/loss where you don't want it? Up north , for 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. Getting to be pretty standard to insulate the inside of the walls/footers (around here, footers have to be 4 ft down to be below the 'frost line'), and then lay out foam board on the gravel bed for the slab. Rebar/grid on the foamboard, and pour the concrete on top of all that. Helps a lot with heating costs in the winter. daestrom |
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wrote in message ... Have a nice Thanksgiving. Ah, the Mayflower. I have sailed in her - literally. http://tinyurl.com/79svq |
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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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Abby Normal wrote:
It estimated nothing You have need otherwise your schemes remain unproven fantasy. A arrogant rule of thumb: "If Abby doesn't understand it. It won't work" :-) Mositure in the soil would be attracted to a cool slab, the insulation value of a vapour barrier is negligible. I disagree. Moisture is not "attracted" :-) Nick |
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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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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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Abby Normal wrote:
It is hard to understand your substitution for considering forced convection , when you are incorrectly applying small temperature differential radiant transfer shortcuts. 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! To be anal, I believe the constant is 0.1713E-8... 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. 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. 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. 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. Let's try to avoid this passive-agressive waffling. 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. This is promising, Abby. Confusion can be better than certainty :-) Pick a) or b) above, not both. ... the slab is cool, natural convection works against you That's good for daytime setbacks. in fact you are relying heavily on forced convection. In case b) with fans and low-e surfaces... ... the approach you are trying could possibly estimate the convective heat transfer from a HEATED slab to the air. 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. Or it just shows the net radiant heat transfer from a ceiling to a floor slab, stealing some of your 'coolth' available. Radiation melts more ice in a) like an ice rink with a plain roof, than b) like an ice rink with an aluminized roof. So trying to rationalize things out by assuming temperatures does not seem to work. 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. 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. Not needed, IMO. But knock your socks off, s'il te plait. So enough smoke and mirrors, if you want to see what is really happening, you need to establish what the slab temperature equalizes at. No I don't. More arrogant waffling... 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. More arrogant waffling words. The Inverted Pool of Pine, with a dripping wet ceiling slab... Please read more carefully. I've never suggested anything like that. ... What would work the best... 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. How would that work? Who said anything about 100% RH? 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. I don't understand. Ignoring your sarcasm, how would that work? You should investigate ground source heat pumps a little bit...' 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. particular the effect of MOISTURE MIGRATION in the soil. Typically soil moisture is attracted to cold pipes in heating mode. 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 ****, but nothing grows without it. --Carl Rogers |
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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. 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! 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? To be anal, I believe the constant is 0.1713E-8... 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. 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. 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. 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. 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. 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. Let's try to avoid this passive-agressive waffling. 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. This is promising, Abby. Confusion can be better than certainty :-) Pick a) or b) above, not both. ... the slab is cool, natural convection works against you That's good for daytime setbacks. in fact you are relying heavily on forced convection. In case b) with fans and low-e surfaces... ... the approach you are trying could possibly estimate the convective heat transfer from a HEATED slab to the air. 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. 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 :) Or it just shows the net radiant heat transfer from a ceiling to a floor slab, stealing some of your 'coolth' available. Radiation melts more ice in a) like an ice rink with a plain roof, than b) like an ice rink with an aluminized roof. Lol if it is melting ice, then it is stealing your 'coolth' So trying to rationalize things out by assuming temperatures does not seem to work. 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. 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. 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. Not needed, IMO. But knock your socks off, s'il te plait. 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. So enough smoke and mirrors, if you want to see what is really happening, you need to establish what the slab temperature equalizes at. No I don't. More arrogant waffling... I can't force you to do anything, let alone prove anything. 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. More arrogant waffling words. 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. :) The Inverted Pool of Pine, with a dripping wet ceiling slab... Please read more carefully. I've never suggested anything like that. ... What would work the best... 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. How would that work? Who said anything about 100% RH? 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. I don't understand. Ignoring your sarcasm, how would that work? Read through your HOF on evaporative cooling, try it on a pyschrometric chart. You seem weak in the area of pyschrometrics. You should investigate ground source heat pumps a little bit...' 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. particular the effect of MOISTURE MIGRATION in the soil. Typically soil moisture is attracted to cold pipes in heating mode. 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. 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 Missapplying convenient equations when not truly understanding them creates a natatorium environment. --Abby |
Cool tower alternatives
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. It is not a standrard practise in warm climates at all. Too bad. Doesn't that lead to a lot of heat gain/loss where you don't want it? No, because below a fairly shallow depth, ground temperature... Well, he from www.greenbuilder.com/sourcebook/groundsource/ "Throughout most of the U.S., the temperature of the ground below the frost line (about 3 to 5 feet below the surface) remains at a nearly constant temperature, generally in the 45 ° -50 ° F range in northern latitudes, and in the 50 ° -70 ° F range in the south." Ground temperatures are thus almost never warmer than you want the conditioned space to be, so you want to insulate when you're heating, and you don't when you're cooling. --Goedjn |
Cool tower alternatives
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Cool tower alternatives
Abby Normal wrote:
Nick You have a pattern of getting more and more obscure when you are losing an arguement. I disagree. Then again, you seem to get more obtuse when losing :-) ... 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. As far as you can tell :-) But that's incorrect... ... Pick one: a) an ordinary ceiling with radiation and no fans or setbacks, or b) a low-e ceiling with fans and possible setbacks. You need to pick one is the whole point. I do? Why? Owner's choice. OK, let's pick a)... You have a forced convection situation. Nope. No fans at all. No convection, just radiation. 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? It doesn't :-) You still seem confused. Maybe that's good. 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. 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. No not anal at all, very significant. 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 :-) 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. 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." But the point is, the approximation has nothing to do with the heat transfer from air to the floor slab. Agreed. It only concerns heat radiating from ceiling to slab, but that also bounds the room air temperature. 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. 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. in fact you are relying heavily on forced convection. In case b) with fans and low-e surfaces... ... the approach you are trying could possibly estimate the convective heat transfer from a HEATED slab to the air. 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. I don't have that figure, but why rely on some figure when you will not even use a pyschrometric chart? I'm more comfy with simple equations in plain ascii. 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. 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. Get a Big Ass Fan, like used in a barn and blast the air down... No thanks :-) 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. Would you have any evidence for this article of faith? 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. More arrogant waffling words. 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. You may be projecting here. This is science! Let's use numbers! 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. But the dry soil beneath is great insulation. And thermal mass. Wonder what the dewpoint of water in soil is when the when the soil at 42F is in contact with a pipe at 37F. Sounds irrelevant, in this case. Nick |
Cool tower alternatives
Goedjn wrote:
Well, he from www.greenbuilder.com/sourcebook/groundsource/ "Throughout most of the U.S., the temperature of the ground below the frost line (about 3 to 5 feet below the surface) remains at a nearly constant temperature, generally in the 45 ° -50 ° F range in northern latitudes, and in the 50 ° -70 ° F range in the south." Ground temperatures are thus almost never warmer than you want the conditioned space to be, so you want to insulate when you're heating, and you don't when you're cooling. Shallow perimeter ground may conduct lots of heat from outdoor air. Nick |
Cool tower alternatives
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Cool tower alternatives
Ground temperatures are thus almost never warmer than you want the conditioned space to be, so you want to insulate when you're heating, and you don't when you're cooling. Shallow perimeter ground may conduct lots of heat from outdoor air. In theory, yes. In practice, apparently not enough to worry about. Either way, the solution to that would be perimeter insulation, not under-slab. |
Cool tower alternatives
Me wrote:
wrote: Abby Normal wrote: Mositure in the soil would be attracted to a cool slab, the insulation value of a vapour barrier is negligible. I disagree. Moisture is not "attracted" :-) I've been watching your thread with Abby N. and I just have one question for him? Just which Law of Physics does this "Mositure in the soil would be attracted to a cool slab" Attraction conform to? .... Normals comment is completely unscientific but perhaps it's simply an observation from an unscientific observer. For example, moisture is not "attracted" to a cool glass on a warm day but beads of water form on the outside anyhow. Perhaps this is the phenomena that Abby was trying to express. Anthony |
Cool tower alternatives
When the moisture beads are condensed on a cold surface
the dry air surrounding it gets mixed with the normal damp air again. This continuous movement toward a cold surface is known as "attraction" George and Weiner have an "attraction" for each other. "Anthony Matonak" wrote in message ... Me wrote: wrote: Abby Normal wrote: Mositure in the soil would be attracted to a cool slab, the insulation value of a vapour barrier is negligible. I disagree. Moisture is not "attracted" :-) I've been watching your thread with Abby N. and I just have one question for him? Just which Law of Physics does this "Mositure in the soil would be attracted to a cool slab" Attraction conform to? ... Normals comment is completely unscientific but perhaps it's simply an observation from an unscientific observer. For example, moisture is not "attracted" to a cool glass on a warm day but beads of water form on the outside anyhow. Perhaps this is the phenomena that Abby was trying to express. Anthony |
Cool tower alternatives
"John P.. Bengi" JBengispam@spam@yahoo,com wrote in message ... When the moisture beads are condensed on a cold surface the dry air surrounding it gets mixed with the normal damp air again. This continuous movement toward a cold surface is known as "attraction" George and Weiner have an "attraction" for each other. And: http://www.concretenetwork.com/bob_cain/ snip The problems of moisture in and under a concrete slab-on-grade are a problem of vapor transmission through the slab. The attraction or flow of moisture to the surface is the normal flow from a point of higher vapor pressure to a point of lower vapor pressure to create equilibrium. By controlling or lessening the rate of moisture transmission in slabs-on-grade, we can successfully use impermeable systems on these surfaces. snip http://www.moisture-solutions.com/productpage.htm snip A similar phenomenon occurs in concrete and masonry walls/slab floors. Capillary action is a function of the natural attraction between water and the capillaries found in concrete and masonry. snip |
Cool tower alternatives
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Cool tower alternatives
"Anthony Matonak" wrote in message ... Me wrote: wrote: Abby Normal wrote: Mositure in the soil would be attracted to a cool slab, the insulation value of a vapour barrier is negligible. I disagree. Moisture is not "attracted" :-) I've been watching your thread with Abby N. and I just have one question for him? Just which Law of Physics does this "Mositure in the soil would be attracted to a cool slab" Attraction conform to? ... Normals comment is completely unscientific but perhaps it's simply an observation from an unscientific observer. For example, moisture is not "attracted" to a cool glass on a warm day but beads of water form on the outside anyhow. Perhaps this is the phenomena that Abby was trying to express. Or the fact that a clear sheet of plastic on the ground will become damp on the underside. In many areas, water is evaporating from soil on sunny days (and replenished on rainy ones), so anything like a tarp that traps that evaporation gives the appearance of 'attracting' moisture on the underside. daestrom |
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