How this calculation works
Relative humidity can rise while air loses water because cooling lowers its saturation pressure. A moisture-removal calculation must compare humidity ratios, not simply subtract the two RH percentages.
Condensate kg/h = dry-air kg/s × (inlet humidity ratio − outlet ratio) × 3600; enthalpy reduction kW = dry-air kg/s × (hin − hout)A worked example
The example cools 30 °C air at sixty-percent RH to 15 °C at ninety-percent RH, at 101.325 kPa and one kilogram per second of dry air. Despite the higher outlet RH, the outlet humidity ratio is lower. The state difference produces roughly 23.25 kg/h of ideal condensate and about 31.87 kW of moist-air enthalpy reduction. Keeping inlet and outlet states identical should return zero removal and zero enthalpy reduction.
Good to know
Entered-state ideal moist-air balance only; no coil selection, bypass factor, condensate piping capacity or equipment guarantee. Saturation pressure is approximated over liquid water within the supported temperature range. Real property data and the full equipment balance may be needed for design.
Reference: ASHRAE — psychrometric definitions and ideal-gas relations (opens in a new tab) ↗
What the model answers
Relative humidity can rise while air loses water because cooling lowers its saturation pressure. A moisture-removal calculation must compare humidity ratios, not simply subtract the two RH percentages.
This page accepts both air states rather than predicting what a coil will produce. Dry-air mass flow then connects the per-kilogram state differences to a water-removal rate and an energy-rate estimate.
The Humidity ratio and moist air properties calculator explains each state’s vapor pressure, dry-air basis and enthalpy convention before the two states are compared here.
Measure and enter consistently
Use total absolute pressure for both states and a dry-air mass flow, not an unconverted moist-air mass or cubic-metre flow. The companion property calculator can help interpret the dry-air basis.
The supported cooling scenario cannot increase dry-bulb temperature, humidity ratio or moist-air enthalpy. Reheat, humidification and other processes need a different combination of state changes and energy terms.
Interpret the result
Enthalpy reduction uses the conventional moist-air reference. It excludes a separate sensible-enthalpy correction for leaving liquid condensate and any fan heat. It should not be presented as a complete coil plant duty under all conditions.
A sensible approximation is shown using inlet humidity ratio. The difference between total enthalpy reduction and that approximation is a residual, not an exact latent-heat decomposition. Actual apparatus dew point and bypass are not solved.
Work through the numbers
The example cools 30 °C air at sixty-percent RH to 15 °C at ninety-percent RH, at 101.325 kPa and one kilogram per second of dry air. Despite the higher outlet RH, the outlet humidity ratio is lower.
The state difference produces roughly 23.25 kg/h of ideal condensate and about 31.87 kW of moist-air enthalpy reduction. Keeping inlet and outlet states identical should return zero removal and zero enthalpy reduction.
Where this model stops
Entered-state ideal moist-air balance only; no coil selection, bypass factor, condensate piping capacity or equipment guarantee.
Saturation pressure is approximated over liquid water within the supported temperature range. Real property data and the full equipment balance may be needed for design.
A few questions, answered
Can outlet RH exceed inlet RH while water is removed?
Yes. RH is temperature-dependent; humidity ratio is the relevant mass comparison.
Does this predict the coil leaving temperature?
No. Both states are supplied, so the page calculates the consequences of that scenario.