This is an ongoing project I suppose. I will check the Oct 20 data again, but I believe I got roughly the same amount of excess. That seems too good to be true.
What do you mean by considering that the reactor vessel is capturing 60% of the heat? Are you referring to the idea that the water and reactor have a combined capture of 100%? That would seem logical if the thermal capacity of each is considered. It is good to continue with the calibrations and I hope that the ambient is better controlled at this phase. We can make the calorimeter work well once it is properly compensated to take out the signal droop. The main issue is to keep the ambient variations to a minimum. One thought to consider. If something happens to disrupt the system, you can wait a few time constants until the ambient average changes work their way out of the calorimeter system. A good test of that condition is to measure the difference between the ambient and the coolant. Once that difference is equal to the nominal value set by the product of the thermal resistance and the leakage powers, you are good to go. Of course, you must ensure that the ambient is not varying too far since that immediately impacts the heat flowing into and out of the thermal capacity. It is assumed that the average ambient is constant which will prevent any major transients. This system should work well with the right precautions and compensations. Dave -----Original Message----- From: Jed Rothwell <[email protected]> To: vortex-l <[email protected]> Sent: Fri, Jan 30, 2015 8:48 pm Subject: Re: [Vo]:Alternate Calculation and Calibration Method for Mizuno Report David Roberson <[email protected]> wrote: I do not get anywhere near to the 3 to 1 excess power out over input power that is reported. I am seeing a 1.25 ratio instead of 3.0 or so. Oh, I think you did see the 3 to 1 excess, and the 6:1 as well. Keep looking! Look at Table 1. You got 1:1.25 for the water on Oct. 21. Right? I estimate 1:1.35, which is close enough. Now look at Oct. 20, which got a much better ratio with the water, 1:2.29. NOW add in the heat captured by the stainless steel reactor vessel. The thermal mass of the reactor is larger than that of the water, so I estimated that the vessel captures 60% of the heat, and the water 40%. So, the combined ratio for Oct. 20, for the water and stainless steel, is 1:5.69 Try applying your method to the Oct. 20 data and see if you get an answer in the same ballpark. Additional calibrations are underway. I believe I will soon be able to confirm that the reactor vessel is, in fact, capturing 60% of the heat. - Jed
