I compared the volume and surface area of a nuclear fuel rod to that of a typical 1 liter water bottle. I assume that Rossi's device has at least as much surface area as a cylindrical water bottle.
I am hopeless at arithmetic so I cheat with an on-line calculator such as this one: http://www.calculatorsoup.com/calculators/geometry-solids/cylinder.php (Be sure you enter the radius, not the diameter.) Fuel rod dimensions: 1 cm diameter, 400 cm length V = 314.15927 cm3 L = 1256.63706 cm2 T = 0.7854 cm2 B = 0.7854 cm2 A = 1258.20786 cm2 Where: V = volume L = lateral surface area T = top surface area B = base surface area A = total surface area Liter bottle shaped cylinder: 4 cm diameter, 20 cm length V = 1005.30965 cm3 L = 502.65482 cm2 T = 50.26548 cm2 B = 50.26548 cm2 A = 603.18578 cm2 I assume that zirconium fuel rod cladding can transfer heat roughly as well as stainless steel. As I said, the reactor transfers 3 GW with 80,000 rods. That is approximately 37.5 kW per rod. (Previously I estimated per liter of rod.) That comes to 0.030 kW/cm^2. A liter-bottle shaped Rossi cell putting out 130 kW would be producing 0.216 kW/cm^2, an order of magnitude more. I will grant, it is unlikely nuclear reactors are designed to survive 30 GW of heat, but on other hand there is no reason to think that heat exchange from the rods is the limiting factor. Supposing the Rossi cell is tube with a 4 cm diameter, 80 cm long, surface area is 1030 cm^2, and power is 0.126 kW/cm^2, 4 times more than the fuel rod. At 16 kW, with the bottle-shaped version of the Rossi cell, it comes to 0.026 kW/cm^2. My teapot transfers 1.5 kW with a stainless steel heating disk at the bottom of the pot of ~10 cm diameter or ~79 cm^2 surface, 0.019 kW/cm^2. The expert Beene is talking to says this is impossible, but I am sure it is possible, since I drink tea every day. - Jed
