Gary R. Lloyd wrote:

Let's assume the same house. The indoor conditions are to be
maintained at 70F and 40% RH (because the homeowner likes it that
way). This translates to an indoor dewpoint temperature of about 44F.

Td = 530/(1-530ln(0.4)/9621)-460 = 44.5. Pi =0.4e^(17.863-9621/530)
= 0.2994 "Hg... wi = 0.62198/(29.921/Pi-1) = 0.006286 pounds of water
per pound of dry air.

When the outdoor dewpoint drops below 44F, that 160CFM of infiltration
air will need to be humidified, right?

Somehow...

But wait, we have our family of four providing 2 gallons per day...

2x8.33/24/60 = 0.0116 lb/min.

Thus we can drop to a lower outdoor dewpoint before extra
humidification is needed for the infiltration air.

Sure.

At what outdoor dewpoint temperature do we need to start adding
moisture in order to maintain our 70F and 40% RH?

When 160x0.075(wi-wo) 0.0116, ie wo wi - 0.000966 = 0.005319 and Po
= 29.921/(0.62198/wo+1) = 0.2357 "Hg and Td = 9621/(17.863-ln(Pi))-460
= 40.2 F.

What if the house were half the square footage (1200 sq ft)?

When 80x0.075(wi-wo) 0.0116, ie wo wi - 0.001933 = 0.004353 and Po
= 29.921/(0.62198/wo+1) = 0.2079 "Hg and Td = 9621/(17.863-ln(Pi))-460
= 35.1 F, if it has half the air leakage.

Nick