Horace Heffner <hheff...@mtaonline.net> wrote:

> We can thus see from this table that if 1 percent by volume of the steam is > entrained water micro-droplets, easily not seen in tubing or exhaust ports, > that only 5.6 percent of the heat of vaporization is required to produce > that mixture. > I do not think so. See: http://www.engineeringtoolbox.com/wet-steam-quality-d_426.html QUOTE: Example - Enthalpy and Specific Volume of Wet Steam Steam at pressure <http://www.engineeringtoolbox.com/pressure-d_587.html> *5 bar gauge* has a dryness fraction of *0.95*. Total enthalpy can be expressed as: *ht** = (2085 kJ/kg) 0.95 + (1 - 0.95) (670.4 kJ/kg)* * = 2,014 kJ/kg* Specific volume can be expressed as: *v** = (0.315 m3/kg) 0.95* * = 0.299 m3/kg* That says 2,014 kJ/kg (not 2.0). Almost as much enthalpy as dry steam. Anyway, we can test this at home. Get something like an electric frying pan, rated at 1.5 kW. Boil some water in. That produces wet steam; you can see the vapor condensing about the pan. The power is half of the 2.97 kW Heffner estimates the Rossi device may be producing (assuming 1% wet steam). So, it should be able to boil away the water in about twice the time it takes Rossi. Take a gallon of water (3.8 L) starting at 15 deg C, and see if you can boil away the entire gallon in 26 minutes. - Jed