Flash Steam Percentage

Also known as flash steam loss

%F=hf1−hf2hfg2×100\%F = \frac{h_{f1} - h_{f2}}{h_{fg2}} \times 100

Worked example: 100 psig condensate to atmosphere → 13.27% flash — press Try an example to run it live, then adjust anything.

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Flash Steam Percentage explained

hf1%Fhfg2hf2

Condensate at 100 psig sits at 338 °F, and it can only do that under pressure. Open a trap to an atmospheric receiver and the water is suddenly 126 °F hotter than it is allowed to be, so it borrows its own excess sensible heat as latent heat and part of it boils instantly. The bookkeeping is pure energy balance: the surplus liquid enthalpy hf1−hf2h_{f1} - h_{f2} divided by the latent heat available at the lower pressure. With hf1=309h_{f1} = 309, hf2=180.2h_{f2} = 180.2 and hfg2=970.3h_{fg2} = 970.3 BTU/lb, 128.8/970.3 = 13.3% of the condensate flashes.

Thirteen percent by mass is more than a thousand percent by volume, which is why a receiver vent that "blows steam" is usually normal flash and not a failed trap — the classic misdiagnosis on a steam survey. Tell them apart by watching the plume: flash pulses with each trap discharge, live steam blows continuously. Better still, do not vent it. A flash tank recovering that steam into a low-pressure header, or a heat exchanger putting it into the makeup, pays back in months, and the same arithmetic applied to boiler blowdown is where blowdown heat recovery gets its numbers.

Flash Steam Percentage formula

%F=hf1−hf2hfg2×100\%F = \frac{h_{f1} - h_{f2}}{h_{fg2}} \times 100
Where
  • %F\%F= Flash steam percentage (%)
  • hf1h_{f1}= Liquid enthalpy at high pressure (J/kg)
  • hf2h_{f2}= Liquid enthalpy at low pressure (J/kg)
  • hfg2h_{fg2}= Latent heat at low pressure (J/kg)

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