On Mon, May 27, 2013 at 4:12 PM, Jed Rothwell <[email protected]> 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. > Both temperature and emissivity enter the equation for power. So, higher temperature, yes, but lower emissivity. The net result can be both higher or lower power depending on the effective exponent use by the instrument's software. And we don't know what this would be for an emissivity of 0.2. We only know that for 0.8 and 0.95, the correction gives higher power. > The most conservative setting is 1. > That's not obvious from the company literature, even for grey bodies, and it can go either way for bodies that are not grey -- that have wavelength dependent emissivities. Metals are examples of this, and presumably there are paints that can emulate metals. > > 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. > > > I don't think you've actually grasped how emissivity comes into the final calculation of power. Fletcher has, or at least he's much closer than you. Start by reading the company's literature on temperature calculation. I have agreed from the beginning that if the emissivity were 0.8 or 0.95, and the object behaved as a grey body, then using e = 1 would underestimate the power. You can check my first posting on the subject. What happens for much lower emissivities and non grey bodies is far from obvious is all I've said. And from the description in the literature, it can go either way.
