In reply to Jones Beene's message of Mon, 18 Apr 2011 10:52:44 -0700: Hi, [snip] >You are using the wrong criteria, as I understand the situation. The >'volume' of the heater is relatively unimportant compared to the surface >area exposed to water flow, the time of exposure and the metal transferring >the heat. With a tubular reactor as described in the Rossi patent, there is >not enough surface area to provide heat transfer through stainless steel to >a straight-thru flow of cold water to transfer the level of heat claimed. If >it were copper, it would barely work. > We have been told that the volume of the cylinder in the E-kitten is 50 cc. Since there is a photo with a ruler next to it, we can see that the length is about 5 cm. This implies a radius of 1.78 cm, and consequently a surface area of about 56 cm^2 (not including the ends). If we add the not unreasonable assumption that the thickness of the stainless steel is about 1 mm, then combined with the known thermal conductivity of stainless steel [16.3 W/(m*K)], we can calculate a thermal conductivity of the whole cylinder at about 91 W/K. Thus in order to transfer 15 kW, the temperature differential across the steel would have to be about 165 K. Given that the reaction occurs at many hundreds of degrees, and steam production would limit the cold side temperature to about 100 degrees, this is well within the capability of even the E-kitten, and it isn't designed to output anywhere near 15 kW.
In short, the limit for the E-kitten would be about (500-100)*91*W/K=36.4 kW. However this is based upon an operating temperature of 500 ºC. The melting point of stainless is much higher than this, implying that even higher power outputs are possible if the operating temperature of the reaction can be higher, or the stainless is thinner than 1 mm, or if you include the surface area of the ends of the cylinder. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
