hello all, if i have water at 135DegC (2.15bar.g) being supplied into a vessel that is open to the atmosphere (0bar.g) how much water will actually be boiled off (flash steam if you like)? I think it will be:
(569.4kj/kg - 419.04kj/kg)/ 2257kj/kg = 0.066 = 6.6% flash steam produced at atmospheric pressure. So if i send 1000L @135DegC (2.15Bar.g) into the vessel i will only have 1000-6.6% = 934L of water (@100degC) in the vessel.
Can someone confirm if this is correct (or at least my thought process). This is not a homework question, i have started a new position within my company (i only have 2 years work experience so any assistance would be appreciated)
thanks
mike