Moist Air Specific Volume (per kg DRY air)

Specific volume is the bridge between the volume a fan moves and the mass a coil has to condition. Fans are rated in cubic feet per minute or litres per second; every energy balance in thermodynamics is written in kilograms. This is the conversion, and like enthalpy it is expressed per kilogram of DRY air, for exactly the same reason: the dry air is the part that does not change as moisture is added or removed.

The equation is the ideal gas law wearing psychrometric clothes. $R_{da}T/p$ is the volume a kilogram of dry air alone would occupy, and $(1 + 1.6078W)$ is the swelling caused by the water vapour riding along with it. The 1.6078 is $1/0.62198$, the reciprocal of the molar-mass ratio, and it is the reason for a fact that surprises nearly everyone.

Moist air is LIGHTER than dry air at the same temperature and pressure. Damp air feels heavy, and the intuition is completely wrong. Avogadro settles it: at a fixed temperature and pressure, a cubic metre contains a fixed NUMBER of molecules regardless of what they are. Every water molecule that joins the mixture has displaced a nitrogen or oxygen molecule, and water at 18 g/mol is lighter than nitrogen at 28 or oxygen at 32. Add moisture, remove mass. This is why a low-pressure system is also a humid one, why baseballs carry further in humid air, and why aircraft performance charts include a humidity correction. The effect is real but modest — a few tenths of a percent at ordinary conditions — and it is dwarfed by the temperature and altitude terms.

Watch the basis when converting a fan's flow to a mass flow. $v$ is cubic metres of MIXTURE per kilogram of DRY air. So $\dot{V}/v$ gives you kilograms of dry air per second, which is the mass flow every enthalpy calculation in this shard wants. If you want the total mixture mass flow instead, that is $\dot{V}(1+W)/v$, and the mixture density is $(1+W)/v$ rather than $1/v$. Mixing these up is a small error at 9 g/kg and a real one in a humid climate.

The trade constants assume $v = 0.833$ m³/kg, which is 0.075 lb/ft³. Room air at 24 °C and 50 % RH is actually at 0.854, so a load figured with 1.08 is already about 2.5 % out before altitude is considered at all. This page also prints 0.287042 rather than the textbook's rounded 0.287, because that is genuinely what it computes with: $R_{da} = 8314.46/28.9645 = 287.042$ J/(kg·K). The rounding is worth 0.015 % and would never be noticed, but a displayed formula that disagrees with the arithmetic behind it is a small lie.

A note on where the ideal gas law starts to creak. Moist air is not quite ideal — the vapour and the dry air interact, and real-gas treatments introduce compressibility factors that depart from unity by a few tenths of a percent at atmospheric conditions. ASHRAE's own tables carry those corrections. At building pressures and temperatures the departure is smaller than the uncertainty in the barometric pressure you typed in, so this page uses the ideal form without apology. If you are working at high pressure, in a compressed-air dryer or a gas turbine inlet, that assumption is the first one to go looking at.