A short tutorial to the basic physics behind flash steam generation
If the pressure of the condensate - the saturated water at the boiling point at the actual pressure - is reduced, the heat energy in the water is reduced to a level appropriate to the final pressure. The connection between the pressure and the boiling temperature can be found in the steam tables.
The energy - or enthalpy - made available when the pressure is reduced, will evaporate a part of the water, producing the flash steam.
Only a part of the condensate water evaporates as flash steam. How much depends on the enthalpy in the condensate at the initial and the final pressures.
The amount of flash steam produced during the pressure reduction can be expressed as:
w = (hil - hfl) / hfe (1)
w = ratio of flash steam generated (kg flash steam / kg condensate)
hil = initial liquid enthalpy (kJ/kg)
hfl = final liquid enthalpy (kJ/kg)
hfe = enthalpy of evaporation (kJ/kg)
Example - Flash Steam Generation
Condensate is produced inside an heat exchanger at a pressure of 5 bar gauge (6 bar absolute). The condensate containing 670.9 kJ/kg of heat energy at saturation temperature 159 oC.
The pressure is reduced to atmospheric pressure - 0 bar gauge (1 bar absolute) through the steam trap and the maximum heat energy in water at atmospheric pressure and 100 oC is 419.0 kJ/kg.
The evaporation energy of water at atmospheric pressure is 2,257 kJ/kg.
The flash steam generated can be calculated as:
w = ((670.9 kJ/kg)- (419.0 kJ/kg)) / (2257 kJ/kg)
= 0.11 (kg flash steam / kg condensate)