> HEAT FLOW THROUGH THE NICKEL CONTAINING STAINLESS STEEL COMPARTMENT > > If the stainless steel compartment has a surface area of > approximately S = 180 cm^2, as approximated above, and 4.39 kW heat > flow through it occurred, as specified in the report, then the heat > flow was (4390 W)/(180 cm^2) = 24.3 W/cm^2 = 2.4x10^5 W/m^2. > > The thermal conductivity of stainless steel is 16 W/(m K). The > compartment area is 180 cm^2 or 1.8x10^-2 m^2. If the wall thickness > is 2 mm = 0.002 m, then the thermal resistance R of the compartment is: > > R = (0.002 m)/(16 W/(m K)*(1.8x10^-2 m^2) = 1.78 °C/W > > Producing a heat flow of 4.39 kW, or 4390 W then requires a delta T > given as: > > delta T = (1.78 °C/W) * (4390 W) = 7800 °C > > The melting point of Ni is 1453°C. Even if the internal temperature > of the chamber were 1000°C above water temperature then power out > would be at best (1000°C)/(1.78 °C/W) = 561 W. > > Most of the input water mass flow necessarily must have continued on > out the exit port without being converted to steam.
That presumes that the heat exchange takes place on the surface of the core. But the heat is (supposedly) produced by thermalization of gamma rays, which could be anywhere nearby. Rossi has said that it is partly in the copper tubing and partly in the lead shielding. The total available area is easily 10 times that of the core, so the delta T could be 780C, not 7800C.
