I realize that I am not actually talking about a dehumidifier.

My house suffers from too much humidity. The wood floors my DH
installed (his trade) are beginning to buckle. So the situation has
now become somewhat critical.

As I understand it I should build able to build basically an exhaust
fan that is controlled by humidity.

But I have no clue exactly how to go about it. I intend initially to
fit it into a window (similar to the way a window air conditioner is
installed in a window). I don't know where to obtain the fan (get a
bathroom exhaust fan and rip the guts out?) and I definitely don't know
where to get the humidity control.

Any help?

Take care,
Melody

Melody,

The controller you want is called a humidistat. You can buy it at most
Heat and AC stores but these stores may not sell retail. I'm not sure that
your plan will work however. First, why do you think that humidity is
causing problems? Have you measured the humidity? A cheap hygometer can be
bought at Radio Shack for about $30. I would think that an average humidity
above 70% would be needed to cause problems and you should be seeing mold
and mildew as well. Second, Your plan is the replace the air in your house
with air from outside so measure the humidity outside. Third, figure out
where the source of the humidity is and try and fix the problem. Finally, if
all else fails go with your fan, though AC or a dehumidifier would be much
better. You'll need a whole house fan such as a large attic fan wired to the
humidistat rather than a bathroom vent fan.
All in all I think that one fan would be fairly expensive and also noisy.

Good luck,
Dave M.

songofruth wrote:

My house suffers from too much humidity. The wood floors my DH
installed (his trade) are beginning to buckle. So the situation has
now become somewhat critical.

As I understand it I should build able to build basically an exhaust
fan that is controlled by humidity.

Sure, if it's less humid outdoors, in the absolute sense, ie if wo wi,
ie ln(Ro/Ri) 9621(1/(460+To)-1/(460+Ti)). For example, if it's 70 F with
80% RH indoors and 80 F outdoors, ventilate when Ro 0.8e^-0.336 = 0.57,
ie the outdoor RH is less than 57%. This could use 100X less energy than
a dehumidifier with a compressor. For efficient house heating and cooling,
you'd want to ventilate at night in summertime and during the day in
wintertime, every few days, when outdoor air happens to be dry.

But I have no clue exactly how to go about it. I intend initially to
fit it into a window (similar to the way a window air conditioner is
installed in a window). I don't know where to obtain the fan (get a
bathroom exhaust fan and rip the guts out?) and I definitely don't know
where to get the humidity control.

You might put Lasko's 2155A reversible window fan (2470 cfm, 90 W, $53
from Ace hardware stores) in a partition wall that divides the house into
two "airtight" spaces and run it whenever the house RH exceeds 50%, along
with the calc above, and reverse from time to time to turn all the exterior
house wall cracks and crevices into efficient bidirectional heat exchangers.
You can do this automatically with a repeat cycle timer like Grainger's
$83.90 2A179 (with its $4.26 5X582 socket), with adjustable off and cycle
times from 1.2 seconds to 300 hours.

Herbach and Rademan (800) 848-8001 http://www.herbach.com sell a nice
$4.95 Navy surplus humidistat, their item number TM89HVC5203, with a
20-80% range, a 3-6% differential, and a 7.5A 125V switch that can be
wired to open or close on humidity rise.

We may see a wireless home automation controller that can do all this
and more automatically at the U Mass DOE Decathlon contest house on
the Washington DC Mall in October :-)

Nick

Melody,

It depends on where you are. If the air outside has a higher absolute
humidity than the air inside, the fan will make it worse. To determine
that you need a sling psychrometer and a psychrometric chart. In the
humid south, what you are suggesting would be a very bad idea, because
the air you blow out creates a vacuum inside the house. Then humid
outside air is sucked in to the house to replace the missing air and
the humidity goes up!

Stretch

songofruth wrote:
I realize that I am not actually talking about a dehumidifier.

My house suffers from too much humidity. The wood floors my DH
installed (his trade) are beginning to buckle. So the situation has
now become somewhat critical.

As I understand it I should build able to build basically an exhaust
fan that is controlled by humidity.

But I have no clue exactly how to go about it. I intend initially to
fit it into a window (similar to the way a window air conditioner is
installed in a window). I don't know where to obtain the fan (get a
bathroom exhaust fan and rip the guts out?) and I definitely don't
know where to get the humidity control.

Any help?

Take care,
Melody

I think you are doing this backwards. Don't think getting the humidity
out, think keep it out. Where is the moisture coming from? If it is the
humid air outside, you won't fix it with a fan.


--
Joseph Meehan

Dia duit

I think I've seen line voltage dehumidistats from Graingers, or Johnstone.

If you find a dehum on the curb, you might be able to scavenge the
dehumidistat off of it. I used to have a couple I'd pulled off old
equipment.

Bathroom fan is good, or perhaps a window fan from Walmart.

--

Christopher A. Young
Learn more about Jesus
www.lds.org
www.mormons.com


"songofruth" wrote in message
ups.com...
I realize that I am not actually talking about a dehumidifier.

My house suffers from too much humidity. The wood floors my DH
installed (his trade) are beginning to buckle. So the situation has
now become somewhat critical.

As I understand it I should build able to build basically an exhaust
fan that is controlled by humidity.

But I have no clue exactly how to go about it. I intend initially to
fit it into a window (similar to the way a window air conditioner is
installed in a window). I don't know where to obtain the fan (get a
bathroom exhaust fan and rip the guts out?) and I definitely don't know
where to get the humidity control.

Any help?

Take care,
Melody

Melody,

Are you certain you want to build and install this? Stating that
you don't know what a humidistat is, how to obtain one or how to
obtain a basic exhaust fan gives me the impression that you
might be better off purchasing what you need and/or considering
a different approach.

The exhaust fan approach works well only when the outdoor
humidity is acceptable. Even then, you are exhausting indoor
air which you may have paid to heat or cool. Whenever your
exhaust system is running, you need to keep a window cracked
open somewhere to relieve negative pressure in your house.
Otherwise, you can greatly increase the radon levels in your home
since the negative pressure will draw radon from the basement
floor (this is especially bad if you have a sump pump or you
have cracks in the basement floor or the basement walls.)

If your windows are open, then your device will give you the same
inside humidity as the ambient (outdoor) air, which may not be
desirable. If you are running AC, then you are getting the air
dehumidified automatically by the AC and you probably don't
need your device. If you are running the furnace, then you can
also run a $100-$150 dehumidifier, which has automatic controls.
If needed, you can also use a dehumidifier in a problem room while
running the AC, if needed. (Usually, low humidity levels are the
problem in furnace season.)

You seem intent upon reinventing the wheel but you don't sound
like an experienced inventor. Buy what you need and avoid a
lot of hassle. I know how to acquire the materials and build what
you want, but I'd never consider wasting my time when the $100
dehumidifier would work better and be a much easier solution. If
the initial cost was a concern, then I'd look for a dehumidifer in a
garage sale, or better yet I'd grab one of the many that sit on the
curb on trash day. One half hour cleaning the coils will usually
restore one to excellent operating condition.

A dehumidifer uses more electricity per hour than an exhaust
fan, but the dehumidifer is going to run a lot less than a fan
set to run automatically by a humidistat which isn't aware of
the ambient humidity. When you consider the heated or air
conditioned air which is wasted by the exhaust fan, then the
dehumidifer is going to be much less expensive to operate.

FYI - my next door neighbors have a humistatically controlled
basement exhaust system that was installed by a basement
waterproofing company. It is extremely rare to walk by the
house and not hear it running. They paid a small fortune to
have it installed (about $1000 !) and they are continuing to
pay dearly as it runs even when the furnace or the AC is running.
That's like leaving a front and back window open when running
the furnace or the AC.

Good luck,
Gideon

=======

songofruth wrote in message
. com...
I realize that I am not actually talking about a dehumidifier.

My house suffers from too much humidity. The wood floors my DH
installed (his trade) are beginning to buckle. So the situation has
now become somewhat critical.

As I understand it I should build able to build basically an exhaust
fan that is controlled by humidity.

But I have no clue exactly how to go about it. I intend initially to
fit it into a window (similar to the way a window air conditioner is
installed in a window). I don't know where to obtain the fan (get a
bathroom exhaust fan and rip the guts out?) and I definitely don't know
where to get the humidity control.

Any help?

Take care,
Melody

songofruth wrote:
I realize that I am not actually talking about a dehumidifier.

My house suffers from too much humidity.

Not clear why you think a dehumidifier is not the best solution.

songofruth wrote:

As I understand it I should build able to build basically an exhaust
fan that is controlled by humidity.

You seem to believe the outside air to be less humid than the inside air.

Gideon wrote:

I'd grab one of the many that sit on the
curb on trash day. One half hour cleaning the coils will usually
restore one to excellent operating condition.

hear, hear

Start going to a few garage sales and find a dehumidifier there.
They are cheap enough and will solve the problem.

If it were winter and you had a humidity problem, i'd recommend
looking for the source before searching for the temporary fix.

DH is a hardwood flooring guy... been in the business for over 26
years. He says the problem is the humidity. That's why we have a
hygrometer on every floor. This morning it was 51% on the top floor
and 64% in the basement. It's never been that high. DH says that for
hardwood floors it's best to keep the humidity around 35-40%, even
lower if possible.

At one point DH was wanting a Humidex - which is just a big exhaust fan
hooked up to a humidistat but for the wonderful price of like over $1K.

We have someone coming out tomorrow to see about a whole house
dehumidifier hooked up to the central HVAC. I suspect it will be mega
bucks and I'd be happier with something less expensive that would do
the job.

The AC in the house is geothermal and doesn't seem to pull as much
moisture out of the air as a regular system.

Take care,
Melody

David Martel wrote:
First, why do you think that humidity is
causing problems? Have you measured the humidity? A cheap hygometer
can be
bought at Radio Shack for about $30. I would think that an average
humidity
above 70% would be needed to cause problems and you should be seeing
mold
and mildew as well. Second,

Thank you very much, Gideon. My gut feeling was that such a system
would be practical only a certain amount of time during the year. The
humidity is due to the general tightness of the house and (imo) the
geothermal HVAC system not pulling as much moisture out of the air as a
standard system would have.

We would have to manually disable the system I'm talking about during
times of high humidity outside. The radon is of course a concern as
well. Although I suspect that air would be drawn in thru the bathroom
exhaust fans... but then could add the whole new problem of drawing in
septic air since I suspect that our bathroom exhaust fans are tied into
the waste system vents.

My gut feeling is that dehumidifiers are the only practical way to go
and we just live with the expense of running them (they can add around
$10-$20 a month around here to the electricity bill). We have a small
40pint portable unit in the basement that keeps freezing up. I wish I
knew what size we would really need to do the job properly.

Take care,
Melody

Gideon wrote:
A dehumidifer uses more electricity per hour than an exhaust
fan, but the dehumidifer is going to run a lot less than a fan
set to run automatically by a humidistat which isn't aware of
the ambient humidity. When you consider the heated or air
conditioned air which is wasted by the exhaust fan, then the
dehumidifer is going to be much less expensive to operate.

FYI - my next door neighbors have a humistatically controlled
basement exhaust system that was installed by a basement
waterproofing company. It is extremely rare to walk by the
house and not hear it running. They paid a small fortune to
have it installed (about $1000 !) and they are continuing to
pay dearly as it runs even when the furnace or the AC is running.
That's like leaving a front and back window open when running
the furnace or the AC.

Matt wrote:
songofruth wrote:
I realize that I am not actually talking about a dehumidifier.

My house suffers from too much humidity.

Not clear why you think a dehumidifier is not the best solution.

Mostly I've been keying in on the solution I suggested because my DH
had a very strong preference for the Humidex unit. This would give him
the same thing but at a (hopefully) greatly reduced out-of-pocket cost.

I do prefer a humidifier. Sure it adds bucks to the electricity bill
but hey if that's the best solution, then that's the best solution. We
have someone from our HVAC company coming out tomorrow with literature
on ones that would connect to our HVAC system. (Since they service our
machine one would think this could all be handled over the phone but
noooo... they feel they must come out.) It looks like this solution
too could run to near $1000.

I'm starting to think that I need to change my thread title (or start a
new one) on how to figure out just how much dehumidifier I would need.

Take care,
Melody

stretch wrote:

...If the air outside has a higher absolute humidity than the air inside,
the fan will make it worse. To determine that you need a sling
psychrometer and a psychrometric chart.

Or a calculator, or a home automation computer.

Nick

Gideon wrote:

A dehumidifer uses more electricity per hour than an exhaust
fan, but the dehumidifer is going to run a lot less than a fan
set to run automatically by a humidistat which isn't aware of
the ambient humidity. When you consider the heated or air
conditioned air which is wasted by the exhaust fan, then the
dehumidifer is going to be much less expensive to operate.

A properly controlled exhaust fan can dehumidify with 100X
less energy than a dehumidifier.

Nick

wrote in message
...
Gideon wrote:

A dehumidifer uses more electricity per hour than an exhaust
fan, but the dehumidifer is going to run a lot less than a fan
set to run automatically by a humidistat which isn't aware of
the ambient humidity. When you consider the heated or air
conditioned air which is wasted by the exhaust fan, then the
dehumidifer is going to be much less expensive to operate.

A properly controlled exhaust fan can dehumidify with 100X
less energy than a dehumidifier.

Nick


Do you have any links to information about these units being 100X more
efficient? That would be interesting reading for me!
Thanks, Rich

Nick. a calculator or computer will not measure wet bulb or relative
humidity, but once the data is collected could convert it with the
proper software. Paul Milligan has software to do that. But most
people need someone with a strong HVAC background to determine what
they need. Even most a/c contractors don't use psychrometrics enough
to figure it out.

Stretch

A properly controlled exhaust fan can dehumidify with 100X
less energy than a dehumidifier.

Nick


Nick, that depends on the outside ambient. If it is very humid
outside, running a fan will make it worse. What you say may be true in
Massachusetts, but here in South Carolina, a residential exhaust fan
does more harm than good about 8-9 months per year.

Stretch

Melody,

If you can tell us something about your location, the size and
construction of your house, We might be able to give better advice.

Stretch

Melody,

I have written a few articles on humidity and mold issues that were
published. Here are links to some of them.


Crawlspace problems:
http://www.contractingbusiness.com/C...S=&NI L=false

We're in the mold business:
http://www.contractingbusiness.com/C...S=&NI L=false

Humidity Control:
http://www.contractingbusiness.com/C...S=&NI L=false





Stretch
Kevin O'Neill

"songofruth" wrote in message
oups.com...
DH is a hardwood flooring guy... been in the business for over 26
years. He says the problem is the humidity. That's why we have a
hygrometer on every floor. This morning it was 51% on the top floor
and 64% in the basement. It's never been that high. DH says that for
hardwood floors it's best to keep the humidity around 35-40%, even
lower if possible.

At one point DH was wanting a Humidex - which is just a big exhaust fan
hooked up to a humidistat but for the wonderful price of like over $1K.

We have someone coming out tomorrow to see about a whole house
dehumidifier hooked up to the central HVAC. I suspect it will be mega
bucks and I'd be happier with something less expensive that would do
the job.

The AC in the house is geothermal and doesn't seem to pull as much
moisture out of the air as a regular system.

Take care,
Melody

David Martel wrote:
First, why do you think that humidity is
causing problems? Have you measured the humidity? A cheap hygometer
can be
bought at Radio Shack for about $30. I would think that an average
humidity
above 70% would be needed to cause problems and you should be seeing
mold
and mildew as well. Second,

Thank you for explaining who DH is. I was wandering ... Dear Husband or
Deceised Husband but did not want to ask.
:)) MG

Rich wrote:

A properly controlled exhaust fan can dehumidify with 100X less energy
than a dehumidifier.

Do you have any links to information about these units being 100X more
efficient? That would be interesting reading for me!

I measured a new efficient dehumidifier, which produced 1.6 kWh of heat
for every kWh consumed, ie 1 kWh from motors plus 0.6 kWh to condense
about 2 pints of water, ie it consumed 0.5 kWh per pound of water, ie
5 cents/pint at 10 cents/kWh.

A 70 F basement with 60% RH has humidity ratio w = 0.00947 pounds of water
per pound of dry air. NREL says w = 0.008 for outdoor air on an average May
day in Phila with a 62.9 F 24-hour average temp and 52.7 and 73.1 daily min
and max. On an _average_ (vs dry) May day, we might run Lasko's 16" 2470 cfm
90 watt 2155A window fan ($53 from Ace Hardware) for an hour and remove
60x2470x0.075(0.00947-0.008) = 16.34 pounds of water at 90/16.34 = 5.5 Wh/pint,
ie $0.1x90/1000/16.33 = 0.055 cents/pint, 91X less than the dehumidifier.

But some days are drier than average. Running the fan longer on dry days
vs running it every day would be more efficient. We might invent a nice
adaptive algorithm with the help of a simulation using TMY2 weather data.

And the outdoor humidity ratio (vs the RH) doesn't change much over a day,
so we might as well run the fan when it's warmer outdoors if the house needs
heat or when it's cooler outdoors if the house needs cooling.

Nick

Stretch wrote:

...a calculator or computer will not measure wet bulb or relative humidity,

A computer with sensors can. Check out the U Mass team house at the 2005
DOE DC Decathlon contest in October.

but once the data is collected could convert it with the proper software.

Radio Shack sells "data collectors." Here's some "proper software" for an $8
Casio FX-260 calculator: ventilate when Ro Rie^(9621(1/(460+To)-1/(460+Ti)),
ie Ro Rie^(9621(1/(460+To)-1/530) if it's 70 F indoors. For example, if it's
70 F with 80% RH indoors and 80 F outdoors, ventilate when Ro 0.8e^-0.336
= 0.57, ie the outdoor RH on the $20 Radio Shack display is less than 57%.

...most people need someone with a strong HVAC background to determine what
they need. Even most a/c contractors don't use psychrometrics enough
to figure it out.

I've noticed that few a/c contractors have strong HVAC backgrounds.

A properly controlled exhaust fan can dehumidify with 100X less energy
than a dehumidifier.

...that depends on the outside ambient. If it is very humid outside,
running a fan will make it worse. What you say may be true in Massachusetts,
but here in South Carolina, a residential exhaust fan does more harm than
good about 8-9 months per year.

Raleigh looks fairly humid, but a properly controlled exhaust fan might help
from October (w = 0.0081) through May (w = 0.0099). July is dampest, with
w = 0.0149 on a 78.1 F average day with 68.1 and 88.0 daily min and max. An
"airtight" house with 15 cfm of natural air leakage (vs. 2.5 cfm in Canada)
and w = 0.0120 indoors would need about 24hx60x15x0.075(0.0149-0.0120) = 4.7
lb/day of dehumidification from air leakage plus about 2 gallons per day from
human activities (Andersen's estimate for a family of 4), about 21 lb/day.

On an average July day, 1920 Btu/ft^2 falls on the ground and 750 falls on
a south wall in Raleigh. We might have an EPDM rubber liner with a passive
greenhouse-type solar still with shallow LiCl lakes separated by dry EPDM
beds to act as water collectors and parasitic air heaters, like this,
viewed in a fixed font like Courier:

| 2' |
carbo
poly s nate
flat p clear
clear a flat
c poly nate...
e carbo
epdm LiCl r lake epdm heater dry bed epdm
epdm 2x4 epdmepdmepdm 2x4 epdmepdmepdmepdmepdmepdmepdm 2x4 epdm
--------------------------------------------- top of SIP ------

(What's a good lake to heater area ratio?)

How many square feet of 80 F LiCl solution (precooked to 160 F) are needed
to remove 20 pounds of water from 80 F house air with w = 0.012 in 12 hours?
Here's a 9-pound calc for Miami, based on some crude assumptions:

1) The LiCl still operates at a constant temp for 12 hours per day.
2) The solar energy that enters the R1 glazing with 90% transmission
equals the sensible and latent heat energy needed for concentration.
3) The solution cools to 25 C at night.
4) The solution gains heat like an ASHRAE pool loses heat.

The next step might be a simple TMY2 simulation.

10 A1=12.7409'LiCl vapor pressure constants from the 1993 Hawlader paper
20 A2=-.065536
30 A3=-8.2416E-04
40 B1=-4675.4
50 B2=+29.31
60 B3=+.66911
70 C1=372690!
80 C2=-1689.8
90 C3=-187.1
100 TA=82.8'average ambient August temperature in Miami (F)
110 SG=1770'average August sun on ground in Miami (Btu/ft^2-day)
120 H=12'distillation day length (hours)
130 W=.0176'average ambient August humidity ratio in Miami
140 PV=25.4*29.921/(1+.62198/W)'ambient vapor pressure (mmHg)
150 P=9'dehumidification load (lb H2O/day)
160 FOR TC=60 TO 90 STEP 10'solution temp (C)
170 TK=273.1+TC'solution temp (K)
180 C=A1+B1/TK+C1/TK^2-LOG(PV)/LOG(10)
190 B=A2+B2/TK+C2/TK^2
200 A=A3+B3/TK+C3/TK^2
210 CONC=(-B-SQR(B^2-4*A*C)/(2*A))'equilibrium soln conc (wt%)
220 TF=1.8*TC+32'solution temp (F)
230 CONCSURF=1000*P/(.9*SG-H*(TF-TA))'LiCl surf needed for conc (ft^2)
240 TK=298.1'solution temp (25 C)
250 AP=A1+A2*CONC+A3*CONC^2
260 BP=B1+B2*CONC+B3*CONC^2
270 CP=C1+C2*CONC+C3*CONC^2
280 PVC=10^(AP+BP/TK+CP/(TK^2))'vapor pressure at 25 C (mmHg)
290 PVI=29.921/(1+.62198/.012)'indoor vapor pressure ("Hg)
300 PVL=PVC/25.4'LiCl vapor pressure ("Hg)
310 DRYRATE=.1*(PVI-PVL)'lb/h/ft^2 H2O (like an ASHRAE pool)
320 DRYSURF=P/(12*DRYRATE)'LiCl surface needed to dry P lb H2O in 12 h (ft^2)
330 PRINT TC,CONC,PVC,DRYSURF,CONCSURF
340 NEXT

still solution LiCl Pv drying concentrating
temp (C) conc (wt%) (mmHg) surf (ft^2) surf (ft^2)

60 39.15389 5.493444 21.42437 9.927201
70 45.89019 2.75522 16.3801 13.03215
80 52.33653 1.244954 14.49746 18.96333
90 58.57794 .5091767 13.72873 34.80277

If the still temp is too low, it looks like we need lots of drying surface.
If it's too high, we need lots of concentrating surface.
A 70 or 80 C still temp seems good...

Nick

Just curious, are you a high school student that happen to be doing
a science project on this subject?

wrote:
Stretch wrote:


...a calculator or computer will not measure wet bulb or relative humidity,


A computer with sensors can. Check out the U Mass team house at the 2005
DOE DC Decathlon contest in October.


but once the data is collected could convert it with the proper software.


Radio Shack sells "data collectors." Here's some "proper software" for an $8
Casio FX-260 calculator: ventilate when Ro Rie^(9621(1/(460+To)-1/(460+Ti)),
ie Ro Rie^(9621(1/(460+To)-1/530) if it's 70 F indoors. For example, if it's
70 F with 80% RH indoors and 80 F outdoors, ventilate when Ro 0.8e^-0.336
= 0.57, ie the outdoor RH on the $20 Radio Shack display is less than 57%.


...most people need someone with a strong HVAC background to determine what
they need. Even most a/c contractors don't use psychrometrics enough
to figure it out.


I've noticed that few a/c contractors have strong HVAC backgrounds.


A properly controlled exhaust fan can dehumidify with 100X less energy
than a dehumidifier.


...that depends on the outside ambient. If it is very humid outside,
running a fan will make it worse. What you say may be true in Massachusetts,
but here in South Carolina, a residential exhaust fan does more harm than
good about 8-9 months per year.


Raleigh looks fairly humid, but a properly controlled exhaust fan might help
from October (w = 0.0081) through May (w = 0.0099). July is dampest, with
w = 0.0149 on a 78.1 F average day with 68.1 and 88.0 daily min and max. An
"airtight" house with 15 cfm of natural air leakage (vs. 2.5 cfm in Canada)
and w = 0.0120 indoors would need about 24hx60x15x0.075(0.0149-0.0120) = 4.7
lb/day of dehumidification from air leakage plus about 2 gallons per day from
human activities (Andersen's estimate for a family of 4), about 21 lb/day.

On an average July day, 1920 Btu/ft^2 falls on the ground and 750 falls on
a south wall in Raleigh. We might have an EPDM rubber liner with a passive
greenhouse-type solar still with shallow LiCl lakes separated by dry EPDM
beds to act as water collectors and parasitic air heaters, like this,
viewed in a fixed font like Courier:

| 2' |
carbo
poly s nate
flat p clear
clear a flat
c poly nate...
e carbo
epdm LiCl r lake epdm heater dry bed epdm
epdm 2x4 epdmepdmepdm 2x4 epdmepdmepdmepdmepdmepdmepdm 2x4 epdm
--------------------------------------------- top of SIP ------

(What's a good lake to heater area ratio?)

How many square feet of 80 F LiCl solution (precooked to 160 F) are needed
to remove 20 pounds of water from 80 F house air with w = 0.012 in 12 hours?
Here's a 9-pound calc for Miami, based on some crude assumptions:

1) The LiCl still operates at a constant temp for 12 hours per day.
2) The solar energy that enters the R1 glazing with 90% transmission
equals the sensible and latent heat energy needed for concentration.
3) The solution cools to 25 C at night.
4) The solution gains heat like an ASHRAE pool loses heat.

The next step might be a simple TMY2 simulation.

10 A1=12.7409'LiCl vapor pressure constants from the 1993 Hawlader paper
20 A2=-.065536
30 A3=-8.2416E-04
40 B1=-4675.4
50 B2=+29.31
60 B3=+.66911
70 C1=372690!
80 C2=-1689.8
90 C3=-187.1
100 TA=82.8'average ambient August temperature in Miami (F)
110 SG=1770'average August sun on ground in Miami (Btu/ft^2-day)
120 H=12'distillation day length (hours)
130 W=.0176'average ambient August humidity ratio in Miami
140 PV=25.4*29.921/(1+.62198/W)'ambient vapor pressure (mmHg)
150 P=9'dehumidification load (lb H2O/day)
160 FOR TC=60 TO 90 STEP 10'solution temp (C)
170 TK=273.1+TC'solution temp (K)
180 C=A1+B1/TK+C1/TK^2-LOG(PV)/LOG(10)
190 B=A2+B2/TK+C2/TK^2
200 A=A3+B3/TK+C3/TK^2
210 CONC=(-B-SQR(B^2-4*A*C)/(2*A))'equilibrium soln conc (wt%)
220 TF=1.8*TC+32'solution temp (F)
230 CONCSURF=1000*P/(.9*SG-H*(TF-TA))'LiCl surf needed for conc (ft^2)
240 TK=298.1'solution temp (25 C)
250 AP=A1+A2*CONC+A3*CONC^2
260 BP=B1+B2*CONC+B3*CONC^2
270 CP=C1+C2*CONC+C3*CONC^2
280 PVC=10^(AP+BP/TK+CP/(TK^2))'vapor pressure at 25 C (mmHg)
290 PVI=29.921/(1+.62198/.012)'indoor vapor pressure ("Hg)
300 PVL=PVC/25.4'LiCl vapor pressure ("Hg)
310 DRYRATE=.1*(PVI-PVL)'lb/h/ft^2 H2O (like an ASHRAE pool)
320 DRYSURF=P/(12*DRYRATE)'LiCl surface needed to dry P lb H2O in 12 h (ft^2)
330 PRINT TC,CONC,PVC,DRYSURF,CONCSURF
340 NEXT

still solution LiCl Pv drying concentrating
temp (C) conc (wt%) (mmHg) surf (ft^2) surf (ft^2)

60 39.15389 5.493444 21.42437 9.927201
70 45.89019 2.75522 16.3801 13.03215
80 52.33653 1.244954 14.49746 18.96333
90 58.57794 .5091767 13.72873 34.80277

If the still temp is too low, it looks like we need lots of drying surface.
If it's too high, we need lots of concentrating surface.
A 70 or 80 C still temp seems good...

Nick

unknown user wrote:

Just curious, are you a high school student that happen to be doing
a science project on this subject?

Possibly. Are you a nubile senior with similar interests? :-)

...some "proper software" for an $8 Casio FX-260 calculator: ventilate
when RoRie^(9621(1/(460+To)-1/(460+Ti)), ie RoRie^(9621(1/(460+To)-1/530)
if it's 70 F indoors. For example, if it's 70 F with 80% RH indoors and
80 F outdoors, ventilate when Ro0.8e^-0.336 = 0.57, ie the outdoor RH
on the $20 Radio Shack display is less than 57%...

That's a Clausius-Clapeyron approximation.

Nick

Nick wrote:
I measured a new efficient dehumidifier, which produced 1.6 kWh of heat

for every kWh consumed, ie 1 kWh from motors plus 0.6 kWh to condense
about 2 pints of water, ie it consumed 0.5 kWh per pound of water, ie
5 cents/pint at 10 cents/kWh.







Nick,

How can a dehumidifier produce mor heat than the equivalent of the
electricity it uses. A heat pump does that because it extracts heat
from outside. A dehumidifier is usually self contained...the whole
thing is all in the same space. All electricity consumed should be
converted to heat. Even though it converts the water vapor to liquid
water, the total heat is still there because it never left the room,
only changed form.


Stretch

stretch wrote:

Nick wrote:

I measured a new efficient dehumidifier, which produced 1.6 kWh of heat
for every kWh consumed, ie 1 kWh from motors plus 0.6 kWh to condense
about 2 pints of water, ie it consumed 0.5 kWh per pound of water, ie
5 cents/pint at 10 cents/kWh.

How can a dehumidifier produce mor heat than the equivalent of the
electricity it uses.

It's an indoor heat pump that converts latent to sensible heat,
with a COP of 1.6, in the case above.

A heat pump does that because it extracts heat from outside.

Not always. The dehum above is 60% more efficient than electric resistance
house heating in wintertime. Bill Shurcliff suggests air conditioning damp
basements in wintertime. The AC might be mounted in a stairwell, with the
hot coils in the living space.

...All electricity consumed should be converted to heat.

It is, but beyond that, each pint of water condensed adds about 1000 Btu
of sensible heat to the room air. Evaporating water requires heat energy.
Condensing water releases heat energy.

Nick

Have you been voted most boring nerd on campus yet?

=======================

wrote in message ...
Stretch wrote:

...a calculator or computer will not measure wet bulb or relative humidity,

A computer with sensors can. Check out the U Mass team house at the 2005
DOE DC Decathlon contest in October.

but once the data is collected could convert it with the proper software.

Radio Shack sells "data collectors." Here's some "proper software" for an $8
Casio FX-260 calculator: ventilate when Ro Rie^(9621(1/(460+To)-1/(460+Ti)),
ie Ro Rie^(9621(1/(460+To)-1/530) if it's 70 F indoors. For example, if it's
70 F with 80% RH indoors and 80 F outdoors, ventilate when Ro 0.8e^-0.336
= 0.57, ie the outdoor RH on the $20 Radio Shack display is less than 57%.

...most people need someone with a strong HVAC background to determine what
they need. Even most a/c contractors don't use psychrometrics enough
to figure it out.

I've noticed that few a/c contractors have strong HVAC backgrounds.

A properly controlled exhaust fan can dehumidify with 100X less energy
than a dehumidifier.

...that depends on the outside ambient. If it is very humid outside,
running a fan will make it worse. What you say may be true in Massachusetts,
but here in South Carolina, a residential exhaust fan does more harm than
good about 8-9 months per year.

Raleigh looks fairly humid, but a properly controlled exhaust fan might help
from October (w = 0.0081) through May (w = 0.0099). July is dampest, with
w = 0.0149 on a 78.1 F average day with 68.1 and 88.0 daily min and max. An
"airtight" house with 15 cfm of natural air leakage (vs. 2.5 cfm in Canada)
and w = 0.0120 indoors would need about 24hx60x15x0.075(0.0149-0.0120) = 4.7
lb/day of dehumidification from air leakage plus about 2 gallons per day from
human activities (Andersen's estimate for a family of 4), about 21 lb/day.

On an average July day, 1920 Btu/ft^2 falls on the ground and 750 falls on
a south wall in Raleigh. We might have an EPDM rubber liner with a passive
greenhouse-type solar still with shallow LiCl lakes separated by dry EPDM
beds to act as water collectors and parasitic air heaters, like this,
viewed in a fixed font like Courier:

| 2' |
carbo
poly s nate
flat p clear
clear a flat
c poly nate...
e carbo
epdm LiCl r lake epdm heater dry bed epdm
epdm 2x4 epdmepdmepdm 2x4 epdmepdmepdmepdmepdmepdmepdm 2x4 epdm
--------------------------------------------- top of SIP ------

(What's a good lake to heater area ratio?)

How many square feet of 80 F LiCl solution (precooked to 160 F) are needed
to remove 20 pounds of water from 80 F house air with w = 0.012 in 12 hours?
Here's a 9-pound calc for Miami, based on some crude assumptions:

1) The LiCl still operates at a constant temp for 12 hours per day.
2) The solar energy that enters the R1 glazing with 90% transmission
equals the sensible and latent heat energy needed for concentration.
3) The solution cools to 25 C at night.
4) The solution gains heat like an ASHRAE pool loses heat.

The next step might be a simple TMY2 simulation.

10 A1=12.7409'LiCl vapor pressure constants from the 1993 Hawlader paper
20 A2=-.065536
30 A3=-8.2416E-04
40 B1=-4675.4
50 B2=+29.31
60 B3=+.66911
70 C1=372690!
80 C2=-1689.8
90 C3=-187.1
100 TA=82.8'average ambient August temperature in Miami (F)
110 SG=1770'average August sun on ground in Miami (Btu/ft^2-day)
120 H=12'distillation day length (hours)
130 W=.0176'average ambient August humidity ratio in Miami
140 PV=25.4*29.921/(1+.62198/W)'ambient vapor pressure (mmHg)
150 P=9'dehumidification load (lb H2O/day)
160 FOR TC=60 TO 90 STEP 10'solution temp (C)
170 TK=273.1+TC'solution temp (K)
180 C=A1+B1/TK+C1/TK^2-LOG(PV)/LOG(10)
190 B=A2+B2/TK+C2/TK^2
200 A=A3+B3/TK+C3/TK^2
210 CONC=(-B-SQR(B^2-4*A*C)/(2*A))'equilibrium soln conc (wt%)
220 TF=1.8*TC+32'solution temp (F)
230 CONCSURF=1000*P/(.9*SG-H*(TF-TA))'LiCl surf needed for conc (ft^2)
240 TK=298.1'solution temp (25 C)
250 AP=A1+A2*CONC+A3*CONC^2
260 BP=B1+B2*CONC+B3*CONC^2
270 CP=C1+C2*CONC+C3*CONC^2
280 PVC=10^(AP+BP/TK+CP/(TK^2))'vapor pressure at 25 C (mmHg)
290 PVI=29.921/(1+.62198/.012)'indoor vapor pressure ("Hg)
300 PVL=PVC/25.4'LiCl vapor pressure ("Hg)
310 DRYRATE=.1*(PVI-PVL)'lb/h/ft^2 H2O (like an ASHRAE pool)
320 DRYSURF=P/(12*DRYRATE)'LiCl surface needed to dry P lb H2O in 12 h (ft^2)
330 PRINT TC,CONC,PVC,DRYSURF,CONCSURF
340 NEXT

still solution LiCl Pv drying concentrating
temp (C) conc (wt%) (mmHg) surf (ft^2) surf (ft^2)

60 39.15389 5.493444 21.42437 9.927201
70 45.89019 2.75522 16.3801 13.03215
80 52.33653 1.244954 14.49746 18.96333
90 58.57794 .5091767 13.72873 34.80277

If the still temp is too low, it looks like we need lots of drying surface.
If it's too high, we need lots of concentrating surface.
A 70 or 80 C still temp seems good...

Nick