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Smarty Smarty is offline
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Default Honeywell Humidicalc Recommended Instead of Outdoor Sensor? (Automatic Humidity Control)

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