Jed: There are really 2 issues regarding the emissivity. When the Thermal Scanner takes a reading it is imaging from the object. In order to convert that image to temperature one must know the emissivity. The scanner has a formula based on the emissivity. You are absolutely right that by inputting an emissivity of 1 the calculated temperature is at the lowest level calculated by the scanner and thus the most conservative. Thus the temperature calculated in the study is conservative.
If that was the end of it, the use of 1 for emissivity would be quite conservative. However, for the report that isn't the end. To calculate the energy from the reactor this temperature is used in the Stefan boltzmann constant and emissivity has to again be input to calculate the energy. Using an emissivity in this formula of 1. At any given temperature gives an inflated value of energy for a body with an emissivity less than 1. In this calculation using an emissivity of 1 is not conservative but inflating. The bottom line using a different emissivity in the 2 estimates (calculations) would be crazy and in actuality for all intents they most likely offset each other. Ransom Sent from my iPhone On May 27, 2013, at 4:12 PM, Jed Rothwell <jedrothw...@gmail.com> wrote: > For people not following the discussion, Ekström misunderstood the "e" > (emissivity) ratio. He wrote: > > "The emissivity for stainless steel could have any value from 0.8 to 0.075 > [2]. The lower value would > obviously yield a much lower net power, in fact it could easily make COP=1." > > He has this backwards. The lower value would yield a much higher temperature, > meaning higher power. The most conservative setting is 1. > > Not only did Ekström get this wrong, so did Cude (it goes without saying), > some blogger named Motl, and Andrew. Andrew realized his mistake. Ekström, > Cude and Motl will never admit they were wrong. > > - Jed >
