Air Latent Heat (0.68 Rule)

Also known as 0.68 formula · latent cooling load

Q˙l=ρaV˙hfgΔW\dot{Q}_l = \rho_a \dot{V} h_{fg} \, \Delta W

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Latent heat is the energy a coil spends condensing water vapour out of the air, and it never shows on a thermometer — the leaving air can be the same temperature and vastly drier. The physics is ṁair × hfg × ΔW, where ΔW is the humidity ratio change (mass of water per mass of dry air) and hfg ≈ 1060 BTU/lb (2.466 MJ/kg) at coil conditions. That gives the North American shortcut 0.68 = 60 min/hr × 0.075 lb/ft³ × 1060 BTU/lb ÷ 7000 grains/lb, which rounds 0.681 down to two digits.

Because there is no grains-per-pound unit in the picker, enter ΔW as a mass ratio: divide grains per pound by 7,000, so 1 gr/lb = 142.86 ppm = 0.014286 %, and metric users can enter g/kg directly as 1 g/kg = 0.1 %. A 2,000 cfm coil dropping the air 30 gr/lb (0.4286 %) removes 0.68 × 2,000 × 30 ≈ 40,800 BTU/hr of latent load. The classic trap is oversizing: a big compressor satisfies the thermostat in six minutes, never runs long enough to wet the coil, and leaves a clammy 74 °F house at 65 % RH — the reason ASHRAE cares more about run time than raw tonnage. Sanity check your answer against the condensate pail: 12,000 BTU/hr of latent removal is about 1.3 US gallons of water per hour.

Air Latent Heat (0.68 Rule)
Q˙l=ρaV˙hfgΔW\dot{Q}_l = \rho_a \dot{V} h_{fg} \, \Delta W
Where
  • Q˙l\dot{Q}_l= Latent heat rate
  • V˙\dot{V}= Airflow
  • ΔW\Delta W= Humidity ratio change
Missing one of these? Work it out first, then come back