Andrea Rossi’s E-cat: Potential to Provide 2 Liters of Clean Drinking Water to 83 Persons in One Day
Dear Andrea Rossi,
Consider the problem of potable (drinkable) water. Assume water source is nearby the village but is contaminated with disease-borne agents. Boiling the water would kill the agents.
Assume a 5kW eCat running continuously with an input water temperature of 20 degC. Bringing the water to boiling: Specific Heat of Water: 4.18 J/ (gm * degC) Delta temperature = 80 degC Heat of Vaporization for Water: 2257 kJ /kg Raise 1 kg of water from 20 degC to 100 degC: = 4.18 J / (gm * degC) * 1000 gm/kg * 80 degC = 334 kJ Boil 1 kg of water: = 2257 kJ Total Energy required: = 2591 kJ / kg Amount of Water turned into steam in one hour: = 5 kW * 3600 sec /(2581 kJ / kg) = 18 MJ / (2581 kJ / kg) = 6.95 kg / hour Amount of Water per day = 166 kg = 166 liters / day Required drinkable water per person per day = 2 liters per day. Number of people one 5kW eCat could support = 83 persons Steam could be used to provide electricity (5kW @ 30% = 1.5kW electricity) Waste heat used for structure heating, etc. ~Steven N. Karels
Dear Steven N. Karels: Yes, is important what you say. This is an issue in which we can give an important contribution.
Warm Regards, A.R.