Is there a consensus whether Honeywell's Humidicalc is better or worse
than using an actual outdoor sensor, as with Aprilaire humidifiers that
automatically adjust the humidity?

The following is from a Honeywell PDF manual:
"The HumidCalc+™ software inside your automatic humidity control is
designed to measure or infer outdoor temperature and automatically
adjust the humidity based on the frost factor setting that the homeowner
sets to allow for variances in furnace size, window type and insulation."

In trying to find info on the web, I found this:

"Honeywell humidistat wins 'Seven Wonders' Engineering Award.
Contracting Business, April, 1999

"No outdoor sensor needed for most applications Honeywell's new
humidistat has earned a Seven Wonders of Engineering Award from the
Minnesota Society of Professional Engineers. The H1008 Automatic
Humidity Control with HumidiCalc+[TM] Software is the first standalone
humidistat to control to dewpoint instead of relative humidity."

http://findarticles.com/p/articles/m...4/ai_n13102640

Thank you very much for any helpful information.
--
Please respond to the newsgroup only. Email sent to this account goes
unread.

Since the dew point and condensation on your windows and inner walls is
directly related to outdoor temperature, I personally can see no compelling
reason to prefer a software algorithm which attempts to "infer" when the
humidifier is called for or not.

The Aprilaire, in addition to using the outdoor sensor, has a consumer
adjustable control to correct for the individual effects of furnace size /
blower speed / insulation / etc. This in concert with the outdoor sensor
measurement very nicely compensates for all of the external and internal
effects of importance.

It is my personal belief that the Honeywell solution has one and only one
virtue.....it allows the installer to not have to deal with running a wire
to the outside, drilling a tiny hole for the wire to pass through, mounting
the sensor, and thus spending another hour putting it in.

Having lived with older, non automatic humidifiers, and now using the
Aprilaire and outdoor sensor in a widely varying northeast climate, I would
never want to do it any other way. This Aprilaire just works superbly and
with no issues whatsoever. The extra hour spent running the outdoor sensor
was very well worth it.

Smarty

"poster3814" wrote in message
ink.net...
Is there a consensus whether Honeywell's Humidicalc is better or worse
than using an actual outdoor sensor, as with Aprilaire humidifiers that
automatically adjust the humidity?

The following is from a Honeywell PDF manual:
"The HumidCalc+™ software inside your automatic humidity control is
designed to measure or infer outdoor temperature and automatically
adjust the humidity based on the frost factor setting that the homeowner
sets to allow for variances in furnace size, window type and insulation."

In trying to find info on the web, I found this:

"Honeywell humidistat wins 'Seven Wonders' Engineering Award.
Contracting Business, April, 1999

"No outdoor sensor needed for most applications Honeywell's new
humidistat has earned a Seven Wonders of Engineering Award from the
Minnesota Society of Professional Engineers. The H1008 Automatic
Humidity Control with HumidiCalc+[TM] Software is the first standalone
humidistat to control to dewpoint instead of relative humidity."

http://findarticles.com/p/articles/m...4/ai_n13102640

Thank you very much for any helpful information.
--
Please respond to the newsgroup only. Email sent to this account goes
unread.

Smarty wrote:

Having lived with older, non automatic humidifiers, and now using the
Aprilaire and outdoor sensor in a widely varying northeast climate, I would
never want to do it any other way. This Aprilaire just works superbly and
with no issues whatsoever...

It raises the fuel bill. Airsealing can raise
the indoor RH, while lowering the fuel bill.

Nick

It really doesn't raise the fuel bill, since the furnace cycle is exactly
the same whether or not the humidifier is installed. The humidifier has a
very small fan inside which does draw a negligible amount of electricity,
but this is the only "fuel" required. Since the Honeywell humidifier also
draws electricity, I don't see how the Aprilaire using an outdoor sensor can
in any way whatsoever use any additional energy / fuel when compared to the
Honeywell.

If your point is that adding ***ANY*** humidifier requires energy, (in this
case a very small amount) versus better sealing of the house, then I agree
that airsealing the house does raise the relative humidity. In my case, I
still need a humidifier even when the house has been carefully insulated,
weather-stripped, and sealed.

The original poster is merely trying to compare the merits of the Honeywell
which does not use an outdoor sensor versus the Aprilaire which does use an
outdoor sensor. I know it would be very attractive not to use any humidifier
whatsoever, but in the area where I live where subzero temperatures occur
commonly in the long winters, the only way I have been able to get a
comfortable house which has no dew on the windows but is nicely humidified
to avoid dry throats, dry noses, itchy skin, huge electric shocks when
walking on carpet, sticky doors, sticky drawers, etc. is to use a powered
humidifier and to control it with an outdoor temperature sensor.

Smarty


wrote in message
...
Smarty wrote:

Having lived with older, non automatic humidifiers, and now using the
Aprilaire and outdoor sensor in a widely varying northeast climate, I
would
never want to do it any other way. This Aprilaire just works superbly and
with no issues whatsoever...

It raises the fuel bill. Airsealing can raise
the indoor RH, while lowering the fuel bill.

Nick

Smarty wrote:

It really doesn't raise the fuel bill, since the furnace cycle is exactly
the same whether or not the humidifier is installed...

Wrong. It takes 1000 Btu to evaporate a pound of water.

The humidifier has a very small fan inside which does draw a negligible
amount of electricity, but this is the only "fuel" required.

Wrong. That's a tiny part of the fuel requirement.

If your point is that adding ***ANY*** humidifier requires energy, (in this
case a very small amount)

Wrong. How much energy do you need to keep your house RH 50% at 70 F
with an indoor humidity ratio wi = 0.00787 pounds of water per pound
of dry air when the outdoor humidity ratio wo = 0.0025, with 200 cfm
of natural air leakage?

... In my case, I still need a humidifier even when the house has been
carefully insulated, weather-stripped, and sealed.

Sounds like you need more airsealing. How much fresh air would you need
for wintertime DEhumidification, given Andersen's estimate that an average
family of 4 evaporates 2 gallons per day of water indoors by breathing,
showering, cooking, washing floors, and some green plants?

... I know it would be very attractive not to use any humidifier whatsoever,
but in the area where I live where subzero temperatures occur commonly in
the long winters, the only way I have been able to get a comfortable house
which has no dew on the windows but is nicely humidified... is to use
a powered humidifier and to control it with an outdoor temperature sensor.

Wrong. Given your natural indoor humidity sources, you might keep a more
airtight house at a comfortable indoor RH with a lower fuel bill with
simple ventilation controls.

Nick

Nick,

Let's go through this point by point.

It really doesn't raise the fuel bill, since the furnace cycle is exactly
the same whether or not the humidifier is installed...

Wrong. It takes 1000 Btu to evaporate a pound of water.


I disagree. Since the furnace cycles exclusively on temperature and the
burner cycle is determined by the thermostat, the only way your claim would
have the potential to be correct would be if the moist humidified air
requires more energy to heat than the same air if it were to be dry. If this
is indeed your argument, then your claim is wrong and specious, since your
alternative "airsealed" humidification approach would, in fact, require this
additional energy as well. Whether the moisture was added by my humdifier or
retained naturally by better weathersealing, the incremental cost of energy
to heat the same moisture-laden air which is aritificially humidified with
the humidifier should therefore be ***the same** as the energy to heat the
moist air resulting from better airsealing.


The humidifier has a very small fan inside which does draw a negligible
amount of electricity, but this is the only "fuel" required.

Wrong. That's a tiny part of the fuel requirement.

Again, I disagree. This is the ***ONLY*** incremental energy use, since
moist air requires the same energy to heat regardless of where the moisture
comes from.


If your point is that adding ***ANY*** humidifier requires energy, (in
this
case a very small amount)

Wrong. How much energy do you need to keep your house RH 50% at 70 F
with an indoor humidity ratio wi = 0.00787 pounds of water per pound
of dry air when the outdoor humidity ratio wo = 0.0025, with 200 cfm
of natural air leakage?

... In my case, I still need a humidifier even when the house has been
carefully insulated, weather-stripped, and sealed.

Sounds like you need more airsealing. How much fresh air would you need
for wintertime DEhumidification, given Andersen's estimate that an average
family of 4 evaporates 2 gallons per day of water indoors by breathing,
showering, cooking, washing floors, and some green plants?

... I know it would be very attractive not to use any humidifier
whatsoever,
but in the area where I live where subzero temperatures occur commonly in
the long winters, the only way I have been able to get a comfortable house
which has no dew on the windows but is nicely humidified... is to use
a powered humidifier and to control it with an outdoor temperature sensor.

Wrong. Given your natural indoor humidity sources, you might keep a more
airtight house at a comfortable indoor RH with a lower fuel bill with
simple ventilation controls.

Nick

Nick,

I have used and applied all conventional methods of both blown-in and
stapled-in insulations, weather-stripped everyplace, have excellent windows
throughout, and still come up extremely dry. Your "typical" example is
worthless in my case.........I have no family of 4 since my kids moved out
decades ago, do very little cooking since I eat out daily, vent my dryer to
an external outside vent, have no plants, no pets,.....and most important,
know from very practical experience that without supplemental humidification
that this place is dry as a bone in the winter months.

I therefore reject your entire set of counter-arguments entirely. Even if I
were to accept your premise that the evaporation of a gallon or so of water
per day was an incremental cost of a five or ten thousand BTU daily, this is
still on the order of way less than 1% of my daily heating consumption.

Smarty

Smarty wrote:

... It takes 1000 Btu to evaporate a pound of water.

I disagree.

Really? :-)

How much energy do you need to keep your house RH 50% at 70 F with
an indoor humidity ratio wi = 0.00787 pounds of water per pound
of dry air when the outdoor humidity ratio wo = 0.0025, with 200 cfm
of natural air leakage?

Nick

Smarty wrote:

... It takes 1000 Btu to evaporate a pound of water.

I disagree.

Really? :-)

How much energy do you need to keep your house RH 50% at 70 F with
an indoor humidity ratio wi = 0.00787 pounds of water per pound
of dry air when the outdoor humidity ratio wo = 0.0025, with 200 cfm
of natural air leakage?

Hint: 70 F air weighs about 0.075 lb/ft^3.

Nick

Nick,

You answered your own question several times now. It takes 1000 BTU of
energy to evaporate a pound of water. It matters not whether this energy is
expended in my hot water tank, dryer, or any other contributor to the
moisture in my house.

If my humidifier supplies moisture rather than my cooking, drying, etc. the
net energy penalty is precisely and exactly the same.

The house is a closed system, as closed as I can make it with all sorts of
"airsealing"." Any incremental energy needed to raise the humidity has to
come from someplace, and it matters not whether this is the humidifier,
bathing hot water, or any other friggin source.

I have 2 degrees in engineering, and have plenty of thermodynamics in my
education, so if you want to discuss this in terms of entropy, enthalpy,
sensible heat, or any terms, let's go at it.

Smarty


wrote in message
...
Smarty wrote:

... It takes 1000 Btu to evaporate a pound of water.

I disagree.

Really? :-)

How much energy do you need to keep your house RH 50% at 70 F with
an indoor humidity ratio wi = 0.00787 pounds of water per pound
of dry air when the outdoor humidity ratio wo = 0.0025, with 200 cfm
of natural air leakage?

Hint: 70 F air weighs about 0.075 lb/ft^3.

Nick

Smarty wrote:

... It takes 1000 Btu to evaporate a pound of water.

I disagree.

It takes 1000 BTU of energy to evaporate a pound of water.

So now you disagree with yourself? :-)

The house is a closed system...

If that were true, you would need DEhumidification in wintertime.

I have 2 degrees in engineering...

So this should be a piece of cake:

How much energy do you need to keep your house RH 50% at 70 F with
an indoor humidity ratio wi = 0.00787 pounds of water per pound
of dry air when the outdoor humidity ratio wo = 0.0025, with 200 cfm
of natural air leakage?

Hint: 70 F air weighs about 0.075 lb/ft^3.

A typical US house leaks about 200 cfm. An "airtight" US house might
leak 60 cfm. A 2400 ft^2 house that meets the Canadian IDEAS standard
might leak 1 cfm...

Nick

Nick,

I did not state nor have I assumed that my house is an idealized, perfectly
closed system with no flows in or out. There are, however, no longer
economical ways to further reach a 60X improvement to go from my present
"airtight" house to the Canadian 1 cfm "ideal" you refer to.

My original reply talked directly to the question raised by this specific
thread, namely, whether the Honeywell Humidicalc algorithm is recommended
versus the use of an outdoor temperature sensor. Your series of attacks and
comments have not offered any insights or contributions whatsoever on this
thread's topic. Since you have chosen to hijack this thread on another topic
with remarks which are critical and disparaging to my original reply, I have
felt obliged to respond. I do so reluctantly since my basic opinion is that
this exchange does not serve the original poster in answering his question
in any way, but instead, leads off in the path of your digression.

My prior comments center on the premise that energy to evaporate moisture
into the home is expended regardless of whether the humidifier, the furnace,
the hot water tank, the clothes dryer, or other appliance provides it.

A house with a less than perfect seal does indeed require more energy to
heat and humidify, and neither of us needs to further elaborate on such an
obvious distinction.....no doubt homes built to the newer Canadian standards
will use less energy and require less humidification.

I have never disagreed with the physical fact that energy is required to
evaporate water. My only disagreement was, and is, your assertion that a
humidifier inherently demands more heating fuel, and my reasons for so
believing are as simple as the observation that moisture evaporated into the
air requires energy from someplace. A family of 4 (to use your example)
burns additional energy in the activities you cite (cooking, bathing,
washing and drying clothes, etc.) to evaporate equivalent humidity that a
humidifier would provide. And in a perfect world where true adiabatic homes
with no flows or losses exist, the need for either would be moot.

Perhaps in some home in the future where there is only 1 cfm of 'leak' it
will be, as you assert, possible to avoid a humidification method entirely,
but I, for one, will reserve judgment until much more is known about the
consequential issues of mold, oxygen deprivation, smells, radon effects, and
other poor air quality issues.

Since the original topic remains unanswered by opinions other than my own, I
welcome your thoughts from Villanova's Electrical and Computer Engineering
Department as to how a Honeywell humidistat using only a software algorithm
might, as the original poster asks, compare to the method used by several
other manufacturers who add an outdoor temperature sensor to allow their
algorithms to adjust to outside changes.


Smarty



wrote in message
...
Smarty wrote:

... It takes 1000 Btu to evaporate a pound of water.

I disagree.

It takes 1000 BTU of energy to evaporate a pound of water.

So now you disagree with yourself? :-)

The house is a closed system...

If that were true, you would need DEhumidification in wintertime.

I have 2 degrees in engineering...

So this should be a piece of cake:

How much energy do you need to keep your house RH 50% at 70 F with
an indoor humidity ratio wi = 0.00787 pounds of water per pound
of dry air when the outdoor humidity ratio wo = 0.0025, with 200 cfm
of natural air leakage?

Hint: 70 F air weighs about 0.075 lb/ft^3.

A typical US house leaks about 200 cfm. An "airtight" US house might
leak 60 cfm. A 2400 ft^2 house that meets the Canadian IDEAS standard
might leak 1 cfm...

Nick

Smarty wrote:

I did not state nor have I assumed that my house is an idealized, perfectly
closed system with no flows in or out. There are, however, no longer
economical ways to further reach a 60X improvement to go from my present
"airtight" house to the Canadian 1 cfm "ideal" you refer to.

Have you sought professional help with a blower door test?

Why 1 CFM? How many min CFM do we need to keep a house RH 50% at 70 F with
an indoor humidity ratio wi = 0.00787 pounds of water per pound of dry air
when the outdoor humidity ratio wo = 0.0025 (Phila in January) and you and
your pint-a-day green plants naturally evaporate 1 gallon of water per day?

Hint1: 8lb/24h = 60CFMx0.075(wi-wo).

Hint2: ASHRAE suggests 15 cfm of fresh air per full-time occupant.

I have never disagreed with the physical fact that energy is required to
evaporate water.

You may well find that you did, if you re-examine your words carefully,
but it's nice to see you reagreeing with yourself :-)

... moisture evaporated into the air requires energy from someplace.

Sure. Aprilaire's advertising ignored that for years and claimed people
could save energy by turning the thermostat down, but they dropped that
claim after I pointed out that the thermostat savings were 10X less than
the heat required to evaporate the water, in a typical US house.

Perhaps in some home in the future where there is only 1 cfm of 'leak' it
will be, as you assert, possible to avoid a humidification method entirely,

It's perfectly doable today with more than 1 CFM.

but I, for one, will reserve judgment until much more is known about the
consequential issues of mold, oxygen deprivation, smells, radon effects, and
other poor air quality issues.

"Build it tight and ventilate it right" with a mechanical system, eg
a bathroom or kitchen exhaust fan with a humidistat that turns it on
when the indoor RH rises to 50% in wintertime. For extra credit, you
might figure out how to automatically decrease that 50% setpoint when
it's colder outdoors to avoid condensation on indoor window surfaces.

... a Honeywell humidistat using only a software algorithm might...
compare to the method used by several other manufacturers who add
an outdoor temperature sensor to allow their algorithms to adjust to
outside changes.

Clever Honeywell thermostats measure indoor air and wall temps and
temporarily raise the air temp to compensate for initially colder walls,
which can save energy by prolonging night setbacks, compared to air-temp-
only thermostats. Maybe clever Honeywell humidistats are mounted indoors
on exterior walls, so they can get an idea of the outdoor temp, which
would be colder with a larger air-wall temp difference, but how would
it know the wall or window insulation values? It might get calibration
help from an owner who pushes a button when there is condensation.

Nick

Nice try but a failing grade, Nick. Go to the back of the class.

The Honeywell Humidicalc is a duct mounted device, and does not use the
exterior walls in any way whatsoever.

Smarty

http://electronicaircleaners.com/dat...ode=H1008A1008




Clever Honeywell thermostats measure indoor air and wall temps and
temporarily raise the air temp to compensate for initially colder walls,
which can save energy by prolonging night setbacks, compared to air-temp-
only thermostats. Maybe clever Honeywell humidistats are mounted indoors
on exterior walls, so they can get an idea of the outdoor temp, which
would be colder with a larger air-wall temp difference, but how would
it know the wall or window insulation values? It might get calibration
help from an owner who pushes a button when there is condensation.

Nick

Smarty wrote:

Nice try but a failing grade, Nick. Go to the back of the class.

No thanks, asshole :-) Then again, my two hints were insufficient for you?

The Honeywell Humidicalc is a duct mounted device, and does not use the
exterior walls in any way whatsoever.

http://electronicaircleaners.com/dat...ode=H1008A1008

Maybe clever Honeywell humidistats are mounted indoors on exterior walls,
so they can get an idea of the outdoor temp, which would be colder with
a larger air-wall temp difference, but how would it know the wall or
window insulation values? It might get calibration help from an owner
who pushes a button when there is condensation.

So they have something like that button, in the "frost control" knob.
Maybe their patented Humidicalc software (what's the patent number?)
estimates the outdoor temp by the furnace duty cycle, ie the percentage
of time the duct air is more than (say) 120 F.

... How many min CFM do we need to keep a house RH 50% at 70 F with
an indoor humidity ratio wi = 0.00787 pounds of water per pound of dry air
when the outdoor humidity ratio wo = 0.0025 (Phila in January) and you and
your pint-a-day green plants naturally evaporate 1 gallon of water per day?

Hint1: 8lb/24h = 60CFMx0.075(wi-wo).

Hint2: ASHRAE suggests 15 cfm of fresh air per full-time occupant.

Hint3: CFM = 8/24/60/0.075/(wi-wo) = 13.8.

Nick

Nick,

Your original reply to my one (and only) attempt by others on this newsgroup
to answer the stated question was dismissive, not at all to the point of how
the two humidistats compare, and digressive since it answered only the
question which you (yourself) posed...regarding water evaporation.

When I answered the original reply, you found it necessary to treat my
answer (which was directly on the poster's topic) in a sophomoric, pedantic,
and rude manner with your "I'll give you a hint" form of arrogance.

When I went to engineering school and for the 40 years thereafter, I was
taught to do research to solve a problem. A mere 90 seconds of research on
the poster's question revealed Honeywell's description and ads for the
Humidicalc duct-mounted humidistat.....clearly something incapable of making
any inference about external wall temperatures.

So now your reply, rather than addressing the topic is a personal attack. Is
name-calling part of what they teach at Villanova Computer and Electrical
Engineering School? It is very juvenile, and displays how angry and
embarrassed you truly are.

Smarty


wrote in message
...
Smarty wrote:

Nice try but a failing grade, Nick. Go to the back of the class.

No thanks, asshole :-) Then again, my two hints were insufficient for you?

The Honeywell Humidicalc is a duct mounted device, and does not use the
exterior walls in any way whatsoever.

http://electronicaircleaners.com/dat...ode=H1008A1008

Maybe clever Honeywell humidistats are mounted indoors on exterior
walls,
so they can get an idea of the outdoor temp, which would be colder with
a larger air-wall temp difference, but how would it know the wall or
window insulation values? It might get calibration help from an owner
who pushes a button when there is condensation.

So they have something like that button, in the "frost control" knob.
Maybe their patented Humidicalc software (what's the patent number?)
estimates the outdoor temp by the furnace duty cycle, ie the percentage
of time the duct air is more than (say) 120 F.

... How many min CFM do we need to keep a house RH 50% at 70 F with
an indoor humidity ratio wi = 0.00787 pounds of water per pound of dry air
when the outdoor humidity ratio wo = 0.0025 (Phila in January) and you and
your pint-a-day green plants naturally evaporate 1 gallon of water per
day?

Hint1: 8lb/24h = 60CFMx0.075(wi-wo).

Hint2: ASHRAE suggests 15 cfm of fresh air per full-time occupant.

Hint3: CFM = 8/24/60/0.075/(wi-wo) = 13.8.

Nick

Smarty wrote:

When I went to engineering school and for the 40 years thereafter, I was
taught to do research to solve a problem. A mere 90 seconds of research on
the poster's question revealed Honeywell's description and ads for the
Humidicalc duct-mounted humidistat...

"When I was a lad I served a term as office boy to an attorneys firm..."

...clearly something incapable of making any inference about external
wall temperatures.

You never learned to think? :-)

So now your reply, rather than addressing the topic...

And you never learned to read? :-)

So they have something like that button, in the "frost control" knob.
Maybe their patented Humidicalc software (what's the patent number?)
estimates the outdoor temp by the furnace duty cycle, ie the percentage
of time the duct air is more than (say) 120 F.

That would be consistent with their statement that we need the outdoor
temp sensor with multizone systems...

Nick

wrote:
Smarty wrote:

When I went to engineering school and for the 40 years thereafter, I was
taught to do research to solve a problem. A mere 90 seconds of research on
the poster's question revealed Honeywell's description and ads for the
Humidicalc duct-mounted humidistat...

"When I was a lad I served a term as office boy to an attorneys firm..."

...clearly something incapable of making any inference about external
wall temperatures.

You never learned to think? :-)

So now your reply, rather than addressing the topic...

And you never learned to read? :-)

So they have something like that button, in the "frost control" knob.
Maybe their patented Humidicalc software (what's the patent number?)
estimates the outdoor temp by the furnace duty cycle, ie the percentage
of time the duct air is more than (say) 120 F.

That would be consistent with their statement that we need the outdoor
temp sensor with multizone systems...

Nick


Thank you very much for your replies, and the time, etc., spent on them.
I appreciate it. Other helpful opinions from knowledgeable people are
also welcome.

Thank you.
--
Please respond to the newsgroup only. Email sent to this account goes
unread.

poster3814,

Very glad to try to help you answer your question. In researching this
matter, I also found your post in the home do-it-yourself web site forum,
and hope you have been able to get some additional insights.

Having installed humdistats of both the "outdoor sensor" type (the later
model Aprilaire) as well as those of the earlier type (Sears, Aprilaire,
Autoflow), no doubt it is much simpler to install a humidistat directly in
the furnace plenum with no wiring going to either a separate humidistat
upstairs in the heated space or wiring going to an outdoor sensor. The time
saved in not wiring, and also time saved in not mounting an outdoor sensor,
adds up to at least an hour or two, maybe more. It is my belief that the
Honeywell solution appeals to installers for this reason.

The extra labor invested in installing an outdoor sensor and wiring it seems
to pay off well, based on the limited sample of systems I have had in this
house. Not until I installed the outdoor sensor Aprilaire did I truly reach
the total automation I was seeking, where the dew point is dynamically
adjusted and there is never, ever, ever,.....any moisture, frost, or dew
forming on my windows or anyplace else, yet the humidification is always
extremely comfortable with none of the problems arising when there is too
little humidity. My kids were prone to allergies when they were young, my
wife complained on dry skin, we frequently drew big sparks as we walked
across the carpets, and drawers and doors would begin to stick. The outside
sensor eliminated any misadjustment issues entirely.

As to whether a Honeywell Humidicalc with no outdoor sensor sitting in a
basement cold-air return plenum can infer enough from the surrounding air
and plenum temperature / humidity to make really appropriate guesses about
what the dew point is remains to be seen in my opinion. Given the option of
a direct measurement of outdoor temperature versus a software algorithm
which, at best, knows only current and past temperatures, and current and
past humidity local to the duct, it is hard for me to imagine a superior
outcome, particularly since the Aprilaire humidistat also sits in the same
location when installed and also has humidity and temperature data to work
with, or so it appears. Moreover, Honeywell also adds an optional outdoor
temperature sensor when controlling either multizone or heat pump systems,
since in either / both cases the local duct measurements are insufficient.

Good luck with your decision and glad to be of help. I am sorry for the
digression which occurred in this thread earlier regarding energy needed to
evaporate water.

Smarty






"poster3814" wrote in message
link.net...
wrote:
Smarty wrote:
Thank you very much for your replies, and the time, etc., spent on them. I
appreciate it. Other helpful opinions from knowledgeable people are also
welcome.

Thank you.
--
Please respond to the newsgroup only. Email sent to this account goes
unread.

Smarty wrote:

As to whether a Honeywell Humidicalc with no outdoor sensor sitting in a
basement cold-air return plenum can infer enough from the surrounding air
and plenum temperature / humidity to make really appropriate guesses about
what the dew point is remains to be seen in my opinion...

It's easily seen, given the frost control knob feedback, if it measures
the furnace duty cycle, ie how often the duct air is moving. A system with
an outdoor temp sensor and no user feedback might do a lot worse, with
no knowledge of window R-values.

... I am sorry for the digression which occurred in this thread earlier
regarding energy needed to evaporate water.

That's the important part, you pompous ass :-) Caulking and humidification
can both increase indoor humidity, but caulking reduces fuel consumption,
and humidification can dramatically increase it.

Nick

Nick,

Here we go again.

For starters, had you replied to the original poster's question, and had you
done even the slightest amount of research before expounding on how external
wall temperatures were being used by the Humidicalc, I would have respected
you.

Maybe 40 years of professional engineering seems like an old man's tale to
you since you find it necessary to mock my experience, but I am here to tell
you 2 crucial and basic things which you apparently have not been taught yet
in your Villanova engineering program which were basic to my engineering
education and even more important to my eventual success in managing
thousands of engineers in a large aerospace company:

1. Listen to the stated question, in this case how do 2 humidistats compare,
and reply to it.

2. Do the research..........and only then offer an opinion. A 90 second
Google search would have revealed to you (as it did to me) that the
Honeywell Humidicalc humidstat is a duct-mounted device, is *** N O T P L
A C E D O N A N E X T E R N A L WA L L *** to indirectly measure
outdoor temperature, and furthermore, actually needs an outdoor Honeywell
temperature sensor to be installed in some circumstances. You would also
find that Aprilaire and other duct mounted humdistats with outdoor sensors
also have the very same "Frost Control" to apply manual feedback, and thus
the Humidicalc has absolutely nothing novel or different in this regard,
contrary to your unresearched opinion. Offering this group an uninformed
and unresearched position such as you did damages your credibility.

There is a 3rd rule which is never taught in school, but is simple basic
manners. Don't rely on name-calling. It looks and is childish, and
undermines the integrity of your thinking process and your up-bringing.

It gives engineering a bad reputation when its' trained practitioners ignore
the question, provide the wrong answers based on opinion rather than
published facts, and use name-calling as a discussion tactic.

Smarty




wrote in message
...
Smarty wrote:

As to whether a Honeywell Humidicalc with no outdoor sensor sitting in a
basement cold-air return plenum can infer enough from the surrounding air
and plenum temperature / humidity to make really appropriate guesses about
what the dew point is remains to be seen in my opinion...

It's easily seen, given the frost control knob feedback, if it measures
the furnace duty cycle, ie how often the duct air is moving. A system with
an outdoor temp sensor and no user feedback might do a lot worse, with
no knowledge of window R-values.

... I am sorry for the digression which occurred in this thread earlier
regarding energy needed to evaporate water.

That's the important part, you pompous ass :-) Caulking and humidification
can both increase indoor humidity, but caulking reduces fuel consumption,
and humidification can dramatically increase it.

Nick

Smarty wrote:

Don't rely on name-calling. It looks and is childish, and undermines
the integrity of your thinking process...

My thinking process is fine, thanks :-)

As to whether a Honeywell Humidicalc with no outdoor sensor sitting in a
basement cold-air return plenum can infer enough from the surrounding air
and plenum temperature / humidity to make really appropriate guesses about
what the dew point is remains to be seen in my opinion...

It's easily seen, given the frost control knob feedback, if it measures
the furnace duty cycle, ie how often the duct air is moving. A system with
an outdoor temp sensor and no user feedback might do a lot worse, with
no knowledge of window R-values.

... I am sorry for the digression which occurred in this thread earlier
regarding energy needed to evaporate water.

That's the important part, you pompous ass :-) Caulking and humidification
can both increase indoor humidity, but caulking reduces fuel consumption,
and humidification can dramatically increase it.

If your average US house naturally leaked 224 cfm on an average 30.4 F
Philadelphia January day with an outdoor humidity ratio wo = 0.0025,
how would the fuel consumption change if you

a) airsealed it to reduce the natural air leakage to 24 cfm, or

b) humidified it to 50%, with no airsealing?

Your thinking process doesn't seem up to these simple calcs,
even with 2 engineering degrees. Consider your bluff called.
Perhaps you should give up now :-)

Nick

If your average US house naturally leaked 224 cfm on an average 30.4 F
Philadelphia January day with an outdoor humidity ratio wo = 0.0025,
how would the fuel consumption change if you

a) airsealed it to reduce the natural air leakage to 24 cfm,

This way, you can raise the RH to 50% and reduce the fuel consumption
by about 24h(70-32)200 = 182.4K Btu/day, saving something like
2 therms of gas or 2 gallons of oil per day.

b) humidified it to 50%, with no airsealing?

This way, with no reduced air leakage savings, the fuel consumption
INcreases by 24hx224x60x0.075(0.00787-0.0025)1000 = 129.9K Btu/day.

The difference is 312K Btu/day, over $6/day at $2/therm.

Nick

Nick,

As earlier, you want to answer your own question and not that of the poster.

Since you think I am, to use your word, "bluffing", I will answer your
digression. It is patently obvious without doing the "calc" as you call it
that a sealed house will require less evaporation energy than an unsealed
house. There has never been either a question or a dispute in this regard.
An order of magnitude drop in water / evaporation energy consumption occurs
when a corresponding order of magnitude drop in air leak takes place (from
224 to 24), and the exact water used / evaporation energy delta depends on
what assumptions you want to make about desired indoor temperature and
desired indoor humidity, neither of which you specified. For 70% indoor temp
and 30% indoor humidity, you avoid evaporating about .09 gallons of water
per day with the correspondingly tiny drop in energy consumption.

Conversely, if you wanted to raise it to 50% humidity inside, you need more
water / energy, the amount of which is again determined by what initial
indoor temperature and humidity you specify. Since you do not specify the
initial amounts, I used 70% and 30% once again, and, on this basis, see an
increase of .15 gals of water to be evaporated per day.

I have no desire to discuss or debate with you the relative merits of
airsealing versus active humidification, and indeed this is the digression
for which I previously apologized, even though you were (and continue to be)
the one who pushes for it in this thread.

Enough of your nonsense!

*****Plonk*****

Smarty








wrote in message
...
Smarty wrote:

Don't rely on name-calling. It looks and is childish, and undermines
the integrity of your thinking process...

My thinking process is fine, thanks :-)

As to whether a Honeywell Humidicalc with no outdoor sensor sitting in a
basement cold-air return plenum can infer enough from the surrounding
air
and plenum temperature / humidity to make really appropriate guesses
about
what the dew point is remains to be seen in my opinion...

It's easily seen, given the frost control knob feedback, if it measures
the furnace duty cycle, ie how often the duct air is moving. A system
with
an outdoor temp sensor and no user feedback might do a lot worse, with
no knowledge of window R-values.

... I am sorry for the digression which occurred in this thread earlier
regarding energy needed to evaporate water.

That's the important part, you pompous ass :-) Caulking and
humidification
can both increase indoor humidity, but caulking reduces fuel
consumption,
and humidification can dramatically increase it.

If your average US house naturally leaked 224 cfm on an average 30.4 F
Philadelphia January day with an outdoor humidity ratio wo = 0.0025,
how would the fuel consumption change if you

a) airsealed it to reduce the natural air leakage to 24 cfm, or

b) humidified it to 50%, with no airsealing?

Your thinking process doesn't seem up to these simple calcs,
even with 2 engineering degrees. Consider your bluff called.
Perhaps you should give up now :-)

Nick

The difference is 312K Btu/day, over $6/day at $2/therm.

Nick

Very wrong analysis. 312K Btu/day would evaporate almost 40 gallons of water
per day (~8000 BTU/gallon). This is many times more than really evaporated.
Savings are much smaller.

Lin Kwang wrote:

The difference is 312K Btu/day, over $6/day at $2/therm.

Very wrong analysis. 312K Btu/day would evaporate almost 40 gallons of water
per day (~8000 BTU/gallon). This is many times more than really evaporated.
Savings are much smaller.

No. The choices were a) airseal, reducing the existing fuel consumption
by about 2 therms per day, or b) humidify, increasing the existing fuel
bill by about 1 extra therm per day, with no airsealing.

You might reread that posting.

Nick

Smarty wrote:

Since you think I am, to use your word, "bluffing", I will answer...
An order of magnitude drop in water / evaporation energy consumption occurs
when a corresponding order of magnitude drop in air leak takes place (from
224 to 24)...

Wrong problem :-) I asked:

If your average US house naturally leaked 224 cfm on an average 30.4 F
Philadelphia January day with an outdoor humidity ratio wo = 0.0025,
how would the fuel consumption change if you

a) airsealed it to reduce the natural air leakage to 24 cfm, or

b) humidified it to 50%, with no airsealing?

but Smarty talks about airsealing WITH humidification, which isn't needed,
given enough airsealing and natural indoor humidity sources like people
and green plants, but let's solve the problems he poses...

For 70% indoor temp and 30% indoor humidity, you avoid evaporating about
.09 gallons of water per day with the correspondingly tiny drop in energy
consumption.

Smarty measures temperatures as percentages? :-) On my planet, 70 F air
at 30% RH has a humidity ratio wi = 0.0047 pounds of water per pound of
dry air, so keeping a 224 cfm house 70 F at 30% with an outdoor humidity
ratio wo = 0.0025 requires evaporating 224x60x0.075(wi-wo) = 2.22 pounds
of water per hour or 53.2 pounds of water per day, ie 6.39 gallons.

A 24 cfm house requires 0.68 gallons per day, and the difference is
5.71 gallons, at an energy cost of 47.5K Btu/day, about 0.5 therms,
or $1/day at $2/therm.

Conversely, if you wanted to raise it to 50% humidity inside, you need more
water / energy, the amount of which is again determined by what initial
indoor temperature and humidity you specify... I used 70% and 30% once
again, and, on this basis, see an ncrease of .15 gals of water to be
evaporated per day...

So Smarty is perfectly capable of providing different wrong answers
for the same 70F/30% problem :-)

... Enough of your nonsense!

*****Plonk*****

When a man is wrong and he won't admit it, he always becomes angry :-)

Nick

wrote:
Smarty wrote:

Since you think I am, to use your word, "bluffing", I will answer...
An order of magnitude drop in water / evaporation energy consumption occurs
when a corresponding order of magnitude drop in air leak takes place (from
224 to 24)...

Wrong problem :-) I asked:

If your average US house naturally leaked 224 cfm on an average 30.4 F
Philadelphia January day with an outdoor humidity ratio wo = 0.0025,
how would the fuel consumption change if you

a) airsealed it to reduce the natural air leakage to 24 cfm, or

b) humidified it to 50%, with no airsealing?

but Smarty talks about airsealing WITH humidification, which isn't needed,
given enough airsealing and natural indoor humidity sources like people
and green plants, but let's solve the problems he poses...

For 70% indoor temp and 30% indoor humidity, you avoid evaporating about
.09 gallons of water per day with the correspondingly tiny drop in energy
consumption.

Smarty measures temperatures as percentages? :-) On my planet, 70 F air
at 30% RH has a humidity ratio wi = 0.0047 pounds of water per pound of
dry air, so keeping a 224 cfm house 70 F at 30% with an outdoor humidity
ratio wo = 0.0025 requires evaporating 224x60x0.075(wi-wo) = 2.22 pounds
of water per hour or 53.2 pounds of water per day, ie 6.39 gallons.

A 24 cfm house requires 0.68 gallons per day, and the difference is
5.71 gallons, at an energy cost of 47.5K Btu/day, about 0.5 therms,
or $1/day at $2/therm.

Conversely, if you wanted to raise it to 50% humidity inside, you need more
water / energy, the amount of which is again determined by what initial
indoor temperature and humidity you specify... I used 70% and 30% once
again, and, on this basis, see an ncrease of .15 gals of water to be
evaporated per day...

So Smarty is perfectly capable of providing different wrong answers
for the same 70F/30% problem :-)

... Enough of your nonsense!

*****Plonk*****

When a man is wrong and he won't admit it, he always becomes angry :-)

Nick


Thanks again for the replies. For what it's worth, I'm going to start a
new message thread that is somewhat related but not totally related.
--
Please respond to the newsgroup only. Email sent to this account goes
unread.

poster3814,

Your reply to the numerous comments from me and others never indicated any
opinions or conclusions on your part as to which of the humidistat
approaches you prefer. As I imagine you are aware, both the Aprilaire and
the Honeywell humidifiers can be used with other humidistats, and in fact
the Aprilaire humidifier could be used with the Honeywell humidistat or vice
versa. I bring this up because you may not be aware of this based on your
more recent thread, once again comparing Aprilaire to Honeywell.

In an earlier reply I did indeed make a mistake, and erroneously typed a
percent sign % when I intended to type a notation for degrees °. I think it
was still apparent that my references to temperature and humidity of 70 and
30 respectively were pretty obvious despite the typo. Sorry for the wrong
keystroke.

Smarty



"poster3814" wrote in message
ink.net...
wrote:
Smarty wrote:

Since you think I am, to use your word, "bluffing", I will answer...
An order of magnitude drop in water / evaporation energy consumption
occurs when a corresponding order of magnitude drop in air leak takes
place (from 224 to 24)...

Wrong problem :-) I asked:

If your average US house naturally leaked 224 cfm on an average 30.4 F
Philadelphia January day with an outdoor humidity ratio wo = 0.0025,
how would the fuel consumption change if you

a) airsealed it to reduce the natural air leakage to 24 cfm, or

b) humidified it to 50%, with no airsealing?

but Smarty talks about airsealing WITH humidification, which isn't
needed,
given enough airsealing and natural indoor humidity sources like people
and green plants, but let's solve the problems he poses...

For 70% indoor temp and 30% indoor humidity, you avoid evaporating about
.09 gallons of water per day with the correspondingly tiny drop in
energy
consumption.

Smarty measures temperatures as percentages? :-) On my planet, 70 F air
at 30% RH has a humidity ratio wi = 0.0047 pounds of water per pound of
dry air, so keeping a 224 cfm house 70 F at 30% with an outdoor humidity
ratio wo = 0.0025 requires evaporating 224x60x0.075(wi-wo) = 2.22 pounds
of water per hour or 53.2 pounds of water per day, ie 6.39 gallons.

A 24 cfm house requires 0.68 gallons per day, and the difference is
5.71 gallons, at an energy cost of 47.5K Btu/day, about 0.5 therms,
or $1/day at $2/therm.

Conversely, if you wanted to raise it to 50% humidity inside, you need
more water / energy, the amount of which is again determined by what
initial indoor temperature and humidity you specify... I used 70% and
30% once
again, and, on this basis, see an ncrease of .15 gals of water to be
evaporated per day...

So Smarty is perfectly capable of providing different wrong answers
for the same 70F/30% problem :-)

... Enough of your nonsense!

*****Plonk*****

When a man is wrong and he won't admit it, he always becomes angry :-)
Nick


Thanks again for the replies. For what it's worth, I'm going to start a
new message thread that is somewhat related but not totally related.
--
Please respond to the newsgroup only. Email sent to this account goes
unread.

On Jun 13, 4:43 pm, "Smarty" wrote:
poster3814,

Your reply to the numerous comments from me and others never indicated any
opinions or conclusions on your part as to which of the humidistat
approaches you prefer. As I imagine you are aware, both the Aprilaire and
the Honeywell humidifiers can be used with other humidistats, and in fact
the Aprilaire humidifier could be used with the Honeywell humidistat or vice
versa. I bring this up because you may not be aware of this based on your
more recent thread, once again comparing Aprilaire to Honeywell.

In an earlier reply I did indeed make a mistake, and erroneously typed a
percent sign % when I intended to type a notation for degrees °. I think it
was still apparent that my references to temperature and humidity of 70 and
30 respectively were pretty obvious despite the typo. Sorry for the wrong
keystroke.

Smarty

"poster3814" wrote in message

ink.net...



wrote:
Smarty wrote:

Since you think I am, to use your word, "bluffing", I will answer...
An order of magnitude drop in water / evaporation energy consumption
occurs when a corresponding order of magnitude drop in air leak takes
place (from 224 to 24)...

Wrong problem :-) I asked:

If your average US house naturally leaked 224 cfm on an average 30.4 F
Philadelphia January day with an outdoor humidity ratio wo = 0.0025,
how would the fuel consumption change if you

a) airsealed it to reduce the natural air leakage to 24 cfm, or

b) humidified it to 50%, with no airsealing?

but Smarty talks about airsealing WITH humidification, which isn't
needed,
given enough airsealing and natural indoor humidity sources like people
and green plants, but let's solve the problems he poses...

For 70% indoor temp and 30% indoor humidity, you avoid evaporating about
.09 gallons of water per day with the correspondingly tiny drop in
energy
consumption.

Smarty measures temperatures as percentages? :-) On my planet, 70 F air
at 30% RH has a humidity ratio wi = 0.0047 pounds of water per pound of
dry air, so keeping a 224 cfm house 70 F at 30% with an outdoor humidity
ratio wo = 0.0025 requires evaporating 224x60x0.075(wi-wo) = 2.22 pounds
of water per hour or 53.2 pounds of water per day, ie 6.39 gallons.

A 24 cfm house requires 0.68 gallons per day, and the difference is
5.71 gallons, at an energy cost of 47.5K Btu/day, about 0.5 therms,
or $1/day at $2/therm.

Conversely, if you wanted to raise it to 50% humidity inside, you need
more water / energy, the amount of which is again determined by what
initial indoor temperature and humidity you specify... I used 70% and
30% once
again, and, on this basis, see an ncrease of .15 gals of water to be
evaporated per day...

So Smarty is perfectly capable of providing different wrong answers
for the same 70F/30% problem :-)

... Enough of your nonsense!

*****Plonk*****

When a man is wrong and he won't admit it, he always becomes angry :-)
Nick

Thanks again for the replies. For what it's worth, I'm going to start a
new message thread that is somewhat related but not totally related.
--
Please respond to the newsgroup only. Email sent to this account goes
unread.- Hide quoted text -

- Show quoted text -


Heh, Smarty, just a side note. I didn't see your interchange with
the genius from Villanova until today. Just wanted to say, like
others here, I agree with you. He's well known for hijacking
threads. Someone asks a simple, practical, real world question and he
answers with equations and calcs trying to show how smart he is. But
in reality, all he shows is that he has a complete lack of practical
experience. He thinks because he has an equation, means it's
applicable to the question or that someone can re-engineer a 50 year
old house. The reality is, he couldn't even install a humidifier and
just gives V a bad reputation.

Smarty wrote:

In an earlier reply I did indeed make a mistake, and erroneously typed a
percent sign % when I intended to type a notation for degrees °.

Let's not forget your 100x6.39/0.09 = 7100% computational error :-)

For 70% indoor temp and 30% indoor humidity, you avoid evaporating about
.09 gallons of water per day...

Smarty measures temperatures as percentages? :-) On my planet, 70 F air
at 30% RH has a humidity ratio wi = 0.0047 pounds of water per pound of
dry air, so keeping a 224 cfm house 70 F at 30% with an outdoor humidity
ratio wo = 0.0025 requires evaporating 224x60x0.075(wi-wo) = 2.22 pounds
of water per hour or 53.2 pounds of water per day, ie 6.39 gallons.

Nick

trader4,

Your kind comments are very much appreciated. Because I am a retired
engineer and have worked with literally thousands of other engineers in my
career since the 1960's, I do especially dislike a simple technical matter
becoming complicated when it doesn't need to be, and a specific technical
question being answered with an unresearched and inappropriate answer. The
fellow at Villanova U. was asked about comparing humidistats, and replied
with clearly wrong information on a distantly related subject, and then
assumed the Honeywell could somehow magically infer outdoor temperature by
being mounted on an external wall even though it is clearly a cold air
return plenum mounted device with no physical proximity whatsoever to the
periphery of the house.

It doesn't take an engineering degree to know or use the equations he
mis-applies, and either psychometric charts or simple
spreadsheet/calculators are used by ASHRAE heating/cooling consultants and
contractors all the time to size these equipments.

I actually think the gentleman at MIT said it best when he corrected Nick at
Villanova by saying:

"Very wrong analysis. 312K Btu/day would evaporate almost 40 gallons of
water
per day (~8000 BTU/gallon). This is many times more than really evaporated.
Savings are much smaller."

I apologized once before to this newsgroup for the digression and confusion
which arose, and do so once again now. I sincerely hope we can get the
original poster to be both well informed and fully comfortable with the
choice of a new humidifier and humidistat per his/her stated inquiries.

Smarty










wrote in message
ups.com...
On Jun 13, 4:43 pm, "Smarty" wrote:
poster3814,

Your reply to the numerous comments from me and others never indicated any
opinions or conclusions on your part as to which of the humidistat
approaches you prefer. As I imagine you are aware, both the Aprilaire and
the Honeywell humidifiers can be used with other humidistats, and in fact
the Aprilaire humidifier could be used with the Honeywell humidistat or
vice
versa. I bring this up because you may not be aware of this based on your
more recent thread, once again comparing Aprilaire to Honeywell.

In an earlier reply I did indeed make a mistake, and erroneously typed a
percent sign % when I intended to type a notation for degrees °. I think
it
was still apparent that my references to temperature and humidity of 70
and
30 respectively were pretty obvious despite the typo. Sorry for the wrong
keystroke.

Smarty

"poster3814" wrote in message

ink.net...



wrote:
Smarty wrote:

Since you think I am, to use your word, "bluffing", I will answer...
An order of magnitude drop in water / evaporation energy consumption
occurs when a corresponding order of magnitude drop in air leak takes
place (from 224 to 24)...

Wrong problem :-) I asked:

If your average US house naturally leaked 224 cfm on an average 30.4
F
Philadelphia January day with an outdoor humidity ratio wo = 0.0025,
how would the fuel consumption change if you

a) airsealed it to reduce the natural air leakage to 24 cfm, or

b) humidified it to 50%, with no airsealing?

but Smarty talks about airsealing WITH humidification, which isn't
needed,
given enough airsealing and natural indoor humidity sources like people
and green plants, but let's solve the problems he poses...

For 70% indoor temp and 30% indoor humidity, you avoid evaporating
about
.09 gallons of water per day with the correspondingly tiny drop in
energy
consumption.

Smarty measures temperatures as percentages? :-) On my planet, 70 F air
at 30% RH has a humidity ratio wi = 0.0047 pounds of water per pound of
dry air, so keeping a 224 cfm house 70 F at 30% with an outdoor
humidity
ratio wo = 0.0025 requires evaporating 224x60x0.075(wi-wo) = 2.22
pounds
of water per hour or 53.2 pounds of water per day, ie 6.39 gallons.

A 24 cfm house requires 0.68 gallons per day, and the difference is
5.71 gallons, at an energy cost of 47.5K Btu/day, about 0.5 therms,
or $1/day at $2/therm.

Conversely, if you wanted to raise it to 50% humidity inside, you need
more water / energy, the amount of which is again determined by what
initial indoor temperature and humidity you specify... I used 70% and
30% once
again, and, on this basis, see an ncrease of .15 gals of water to be
evaporated per day...

So Smarty is perfectly capable of providing different wrong answers
for the same 70F/30% problem :-)

... Enough of your nonsense!

*****Plonk*****

When a man is wrong and he won't admit it, he always becomes angry :-)
Nick

Thanks again for the replies. For what it's worth, I'm going to start a
new message thread that is somewhat related but not totally related.
--
Please respond to the newsgroup only. Email sent to this account goes
unread.- Hide quoted text -

- Show quoted text -


Heh, Smarty, just a side note. I didn't see your interchange with
the genius from Villanova until today. Just wanted to say, like
others here, I agree with you. He's well known for hijacking
threads. Someone asks a simple, practical, real world question and he
answers with equations and calcs trying to show how smart he is. But
in reality, all he shows is that he has a complete lack of practical
experience. He thinks because he has an equation, means it's
applicable to the question or that someone can re-engineer a 50 year
old house. The reality is, he couldn't even install a humidifier and
just gives V a bad reputation.

psychrometric replaced by my spell checker with psychometric...sorry


"Smarty" wrote in message
...
trader4,

Your kind comments are very much appreciated. Because I am a retired
engineer and have worked with literally thousands of other engineers in my
career since the 1960's, I do especially dislike a simple technical matter
becoming complicated when it doesn't need to be, and a specific technical
question being answered with an unresearched and inappropriate answer. The
fellow at Villanova U. was asked about comparing humidistats, and replied
with clearly wrong information on a distantly related subject, and then
assumed the Honeywell could somehow magically infer outdoor temperature by
being mounted on an external wall even though it is clearly a cold air
return plenum mounted device with no physical proximity whatsoever to the
periphery of the house.

It doesn't take an engineering degree to know or use the equations he
mis-applies, and either psychometric charts or simple
spreadsheet/calculators are used by ASHRAE heating/cooling consultants and
contractors all the time to size these equipments.

I actually think the gentleman at MIT said it best when he corrected Nick
at Villanova by saying:

"Very wrong analysis. 312K Btu/day would evaporate almost 40 gallons of
water
per day (~8000 BTU/gallon). This is many times more than really
evaporated.
Savings are much smaller."

I apologized once before to this newsgroup for the digression and
confusion which arose, and do so once again now. I sincerely hope we can
get the original poster to be both well informed and fully comfortable
with the choice of a new humidifier and humidistat per his/her stated
inquiries.

Smarty










wrote in message
ups.com...
On Jun 13, 4:43 pm, "Smarty" wrote:
poster3814,

Your reply to the numerous comments from me and others never indicated
any
opinions or conclusions on your part as to which of the humidistat
approaches you prefer. As I imagine you are aware, both the Aprilaire and
the Honeywell humidifiers can be used with other humidistats, and in fact
the Aprilaire humidifier could be used with the Honeywell humidistat or
vice
versa. I bring this up because you may not be aware of this based on your
more recent thread, once again comparing Aprilaire to Honeywell.

In an earlier reply I did indeed make a mistake, and erroneously typed a
percent sign % when I intended to type a notation for degrees °. I think
it
was still apparent that my references to temperature and humidity of 70
and
30 respectively were pretty obvious despite the typo. Sorry for the wrong
keystroke.

Smarty

"poster3814" wrote in message

ink.net...



wrote:
Smarty wrote:

Since you think I am, to use your word, "bluffing", I will answer...
An order of magnitude drop in water / evaporation energy consumption
occurs when a corresponding order of magnitude drop in air leak takes
place (from 224 to 24)...

Wrong problem :-) I asked:

If your average US house naturally leaked 224 cfm on an average 30.4
F
Philadelphia January day with an outdoor humidity ratio wo = 0.0025,
how would the fuel consumption change if you

a) airsealed it to reduce the natural air leakage to 24 cfm, or

b) humidified it to 50%, with no airsealing?

but Smarty talks about airsealing WITH humidification, which isn't
needed,
given enough airsealing and natural indoor humidity sources like
people
and green plants, but let's solve the problems he poses...

For 70% indoor temp and 30% indoor humidity, you avoid evaporating
about
.09 gallons of water per day with the correspondingly tiny drop in
energy
consumption.

Smarty measures temperatures as percentages? :-) On my planet, 70 F
air
at 30% RH has a humidity ratio wi = 0.0047 pounds of water per pound
of
dry air, so keeping a 224 cfm house 70 F at 30% with an outdoor
humidity
ratio wo = 0.0025 requires evaporating 224x60x0.075(wi-wo) = 2.22
pounds
of water per hour or 53.2 pounds of water per day, ie 6.39 gallons.

A 24 cfm house requires 0.68 gallons per day, and the difference is
5.71 gallons, at an energy cost of 47.5K Btu/day, about 0.5 therms,
or $1/day at $2/therm.

Conversely, if you wanted to raise it to 50% humidity inside, you
need
more water / energy, the amount of which is again determined by what
initial indoor temperature and humidity you specify... I used 70% and
30% once
again, and, on this basis, see an ncrease of .15 gals of water to be
evaporated per day...

So Smarty is perfectly capable of providing different wrong answers
for the same 70F/30% problem :-)

... Enough of your nonsense!

*****Plonk*****

When a man is wrong and he won't admit it, he always becomes angry :-)
Nick

Thanks again for the replies. For what it's worth, I'm going to start a
new message thread that is somewhat related but not totally related.
--
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Heh, Smarty, just a side note. I didn't see your interchange with
the genius from Villanova until today. Just wanted to say, like
others here, I agree with you. He's well known for hijacking
threads. Someone asks a simple, practical, real world question and he
answers with equations and calcs trying to show how smart he is. But
in reality, all he shows is that he has a complete lack of practical
experience. He thinks because he has an equation, means it's
applicable to the question or that someone can re-engineer a 50 year
old house. The reality is, he couldn't even install a humidifier and
just gives V a bad reputation.

Smarty wrote:

It doesn't take an engineering degree to know or use the equations he
mis-applies...

I applied. Smarty misapplied, with a 7100% error :-)

Airsealing and humidification can both raise the indoor RH, but
airsealing lowers the fuel bill and humidification raises it.

Nick, enjoying repeating obvious truths to people who won't listen.