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