You are allowed a few typos under those conditions. ;) We will miss him.
Dave -----Original Message----- From: Jones Beene <jone...@pacbell.net> To: vortex-l <vortex-l@eskimo.com> Sent: Sat, Jun 22, 2013 10:28 pm Subject: RE: [Vo]:Rossi and DGT Similarity? Already see a couple oftypos – kW is power, not energy. Calculations, even with aidof Windows - are not recommend during a (Pinot Grigio fueled) wake for Sergio… From:Jones Beene Dave, In terms of kg of hydrogen per kW of energy, the rule of thumb isa gain of 200:1 (ratio) if the hydrogen goes to an average redundancy levelthat Mills apparently believes is correct. This would be on the low side - ifsome of that f/H then converts via a nuclear pathway. Thus, without redundancy compare to the 33 kWhr/kg or 33watt-hr/gm as energy density for regular hydrogen combustion, that becomesabout 6,600 kWhr/kg for f/H. Thus 1 kWhr requires ~ .15 g with f/H and nosecondary nuclear contribution. For 4000 hrs (half a year) for a continuous megawatt of thermaloutput (supposedly the big blue box) or 4 gigawatt-hrs - that would consumeover 600 kg. of hydrogen in half a year. This would be derived from 2,400 kg ofmethane. The energy density by weight of methane is published as 14 kWh/kg soif you merely burned the methane instead of removing the hydrogen for the ECatyou would face as much as a 500:1 deficit over the ECat. Apparently a HotCat which may be more efficient - at a kW outputwith only 5 grams of hydrogen available in a sealed capsule could not run forover 30 hours unless it was better than the 200:1. Caveat – this is a Saturday evening back of the envelopecalculation J and given that Robin is already Sunday morning, I was hoping hewould oblige. From:David Roberson I was hoping that you or someone else would have calculated theamount of hydrogen required to put out a reasonable amount of power for themandatory 1/2 year.
