tag:blogger.com,1999:blog-326167780677397310.post3817604113774400130..comments2024-04-08T11:10:33.047-07:00Comments on EGO OUT: JUNE 14, 2016 QUARK X THE LITTLE LIGHTHOUSE OF LENR'S FUTUREGeorgina Popescuhttp://www.blogger.com/profile/04628821029016016988noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-326167780677397310.post-83503727131683985442016-06-15T09:21:00.517-07:002016-06-15T09:21:00.517-07:00Brian,
I'm the "anonymous" wh...Brian,<br /> I'm the "anonymous" who posted an estimate of the power radiated using the Stefan-Boltzmann equation. I'm not sure I understand your comment, but let me break-down my reasoning regarding my calculation of the area. Ignoring the end-pieces (which would only increase the total area) the area of a cylinder is equal to the circumference (pi*d where d is the diameter) times the height of the cylinder h. Here d = 1 mm or 0.001 m and h = 30 mm or 0.03 m so A = pi*d*h = 3.14*.001 m*.03 m = 9.43 e-5 m^2. If I multiply this by the Stefan-Boltzmann constant 5.67e-8 W/m^2-K^4 and again by (1773 K)^4 (since 1500 C = 1773 K) I obtain 53 W. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-46800349083584946292016-06-15T08:22:29.902-07:002016-06-15T08:22:29.902-07:00Celebrate Calibrate, Dance To the music.Celebrate Calibrate, Dance To the music.sam northhttps://www.blogger.com/profile/13268558018307793474noreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-46198783511810484082016-06-15T06:52:13.540-07:002016-06-15T06:52:13.540-07:00the anonymus Taliban calculated with and area off ...the anonymus Taliban calculated with and area off by a lot Area + piR*2Anonymoushttps://www.blogger.com/profile/10564497666847741316noreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-41704789071257601172016-06-14T21:04:07.012-07:002016-06-14T21:04:07.012-07:00As has also been pointed out on "www.e-catwor...As has also been pointed out on "www.e-catworld.com", it is likely that, given the small size of the filament, a non-negligible amount of additional heat is being lost through heat conduction through the leads or to the surroundings. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-13808037803438033562016-06-14T20:57:57.094-07:002016-06-14T20:57:57.094-07:00The Stefan-Boltzmann equation states that the tota...The Stefan-Boltzmann equation states that the total power P radiated is equal to the area A multiplied by the emissivity epsilon multiplied by the Stefan-Boltzmann constant 5.67e-8 W/m^2 K^4 multiplied by T^4 where T is the temperature in Kelvin. Assuming for simplicity epsilon = 1, T = 1773 K, A = pi*0.001 m * 0.03 m, I obtain P = 53 W which is 100 times the input power of 0.5 W for a COP of 100. Does pointing this out make me a member of the "Rossi Taliban"?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-77249825592646978612016-06-14T20:39:50.617-07:002016-06-14T20:39:50.617-07:00The blue like might be similar to the light that i...The blue like might be similar to the light that is produced by Sonoluminescence. In this process, recent experiments conducted by the University of Illinois at Urbana–Champaign indicate temperatures around 20,000 K (19,700 °C; 35,500 °F).<br /><br />Maybe the Quark reactor is not a black body but use the same processes that are at work in Sonoluminescence.Axilhttps://www.blogger.com/profile/07190120527431077518noreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-53378865075773965442016-06-14T19:56:41.826-07:002016-06-14T19:56:41.826-07:00The blue like might be similar to the light that i...The blue like might be similar to the light that is produced by Sonoluminescence. In this process, recent experiments conducted by the University of Illinois at Urbana–Champaign indicate temperatures around 20,000 K (19,700 °C; 35,500 °F).<br /><br />Maybe the Quark reactor is not a black body but use the same processes that are at work in Sonoluminescence.Axilhttps://www.blogger.com/profile/07190120527431077518noreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-77425192927506826102016-06-14T18:42:06.379-07:002016-06-14T18:42:06.379-07:00Would that be celebrate or calibrate? If it is a c...Would that be celebrate or calibrate? If it is a calibration, it is a first for Rossi, and reason to celebrate.Jed Rothwellhttps://www.blogger.com/profile/00179077151947615762noreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-37370871062835534232016-06-14T18:12:07.137-07:002016-06-14T18:12:07.137-07:00A message to all Rossi Taliban. The filament that ...A message to all Rossi Taliban. The filament that is 1mm in diameter and has 0.5 wats of input will glow at a temperature of 1500C. Welcome to the Stefan-Boltzmann equation.<br /><br />There is no magic along these lines since Edison.Brian Ahern, Physicist and Rossi denyer.noreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-45131208485999890172016-06-14T14:44:39.509-07:002016-06-14T14:44:39.509-07:00WE CELABRATED.
That's what I like to hearWE CELABRATED.<br />That's what I like to hearsam northhttps://www.blogger.com/profile/13268558018307793474noreply@blogger.com