The following graph, developed using the H2o Properties for Excel add-in, shows the difference in enthalpy between the moisture in the air and 0% quality saturated liquid, both calculated at the ambient temperature, for various relative humidities. This is useful for purposes such as sizing of a wet cooling tower. The dryer the air, the greater the potential for evaporative cooling.

image Relative humidity is defined as RH=Pw/Pws where Pw is the water vapor pressure in the air and Pws is the vapor pressure in the air at 100% relative humidity, when the air cannot hold any more moisture.

Example: Assume an ambient temperature of 25 oC and a relative humidity of 45%, using metric bara units

  • The saturation pressure is calculated using the function H2oProperties function H2o.TP(25,2) where 25 is input temperature (in units of oC) and the 2 selects the metric bara units. This returns a value of 0.031697469 bara.
  • The water vapor pressure in the air is therefore: Pw = RH x Pws = 0.014263861 bara
  • The enthalpy of the 100% quality saturated vapor in the air is calculated using the function H2oProperties function H2o.PQH(0.014263861,1,2) = 2523.348 kJ/kg. Note. The first input parameter is Pw, the second is the quality (100%), and the third selects the metric bara units.
  • The enthalpy at 0% quality saturated liquid is calculated using the function H2oProperties function H2o.PQH(00.031697469,0,2) = 104.838 kJ/kg. Note. The first input parameter is Pws, the second is the quality (0%), and the third selects the metric bara units.
  • The evaporative cooling potential is the difference between these two enthalpies.
    Δenthalpy = (2523.348 – 104.838) kJ/kg = 2418.51 kJ/kg

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