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HVAC

Air Heating, Cooling and Condensate

Enter values in the stated units. Example inputs are provided to help you explore the method.

Inputs

Results

Your results will appear here after calculation. Changing an input clears the previous result.

Method & assumptions

  1. Uniformly heat or cool an ideal moist-air stream at constant pressure to a specified outlet temperature. If cooling reaches saturation, remove liquid water and return saturated outlet air. No bypass or added moisture is modeled.
  2. Dry-air mass flow must be positive. Inlet: −50 to 80 °C and 0–100% RH. Outlet: 0–80 °C, excluding ice/frost processing. Pressure: 20–120 kPa absolute and above saturation pressure at both temperatures.
  3. Wout = min(Win, Wsat(Tout)). Without condensation, W is unchanged and RH changes with temperature. With condensation, outlet RH is 100%. Condensate flow = mda(Win − Wout).
  4. Heat removed = mda(hin − hout) − mcond hw,out. Here h is in J/kg dry air, and liquid-water enthalpy is approximated as hw,out = 4186 Tout J/kg relative to liquid water at 0 °C. Condensate leaves at outlet temperature.
  5. Positive heatRemoved means cooling; negative means heating. It is thermal transfer, not electrical input, compressor power or a selected equipment capacity. No arbitrary sensible/latent split is imposed.
  6. Convert measured inlet volume flow to dry-air mass flow with mda = Qin/vin using the inlet specific volume. Moist-air mass flow is a different basis. The output shows both inlet and outlet volume flows.
  7. No coil bypass factor, apparatus dew point, fan heat, heat leakage, pressure drop, frost, re-evaporation, humidity control or real-gas correction. A real coil can produce an unsaturated outlet; this ideal saturation model does not predict its detailed performance.

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