Tutorial to the basic physics behind flash steam generation
If the pressure of 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 relation pressure and the boiling temperature can be found in steam tables.
The energy - or enthalpy - made available when pressure is reduced, will evaporate a part of the water, producing flash steam.
Only a part of the condensate 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 calculated:
w = (hil - hfl) / hfe (1)
w = ratio of flash steam generated (kg flash steam / kg condensate)
hil = initial condensate enthalpy (kJ/kg, Btu/lb)
hfl = final condensate enthalpy (kJ/kg, Btu/lb)
hfe = enthalpy of evaporation of condensate at final condition (kJ/kg, Btu/lb)
It is common that the final condition is atmospheric pressure 0 bar gauge (1 bar absolute) and temperature 100 oC. But closed condensate receivers with some pressure elevated condensate temperature are also used.
Example - Flash Steam Generation
Condensate is produced inside an heat exchanger with steam pressure 5 bar gauge (6 bar absolute) on the heating surfaces. The condensate at 5 bar gauge (6 bar absolute) contains 670.9 kJ/kg of heat energy - the enthalpy - at saturation temperature 159 oC.
The pressure in the condensate is reduced to atmospheric pressure - 0 bar gauge (1 bar absolute) when passing through the steam trap. The maximum heat energy - enthalpy - in the water (condensate) at atmospheric pressure and 100 oC is 419.0 kJ/kg.
Evaporation energy of water at the final condition at atmospheric pressure is 2257.9 kJ/kg.
The flash steam generated can then be calculated:
w = ((670.9 kJ/kg)- (419.0 kJ/kg)) / (2257.9 kJ/kg)
= 0.11 (kg flash steam / kg condensate)
= 11 %
Volume of Flash Steam
The volume of condensate as flash steam is much larger than the volume of condensate as condensate.
Example - Volume of Flash Steam
By using the values from the example above we can calculate
- the volume of the condensate Vc
Vc = (1 - 0.11) (0.00104 m3/kg)
= 0.00093 m3/kg
- the volume of the flash steam Vf
Vf = 0.11 (1.67 m3/kg)
= 0.1837 m3/kg
This is a flash steam to condensate liquid rate of
(0.1837 m3/kg) / (0.00093 m3/kg)
Recovery of Flash Steam
Flash steam can be used to heat consumers with temperature demand lower than 100 oC (212 oF) - like HVAC systems, hot water service systems and similar.