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.
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
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.
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
Your scheme does not follow a constant wet bulb. Your constant exhaust
directly adds sensible heat to the room air.
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.
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?
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
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
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.
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
Here is a link to one that you can even read water vapour pressure in
inches of mercury off of and it is free.
Or if you really want to go from first principles here is some MEASURED
data on water
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
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?
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.
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
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?
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 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
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.
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.
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. :)
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'.
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.
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