tag:blogger.com,1999:blog-8282587.post116532010569621062..comments2024-03-19T01:48:39.709+00:00Comments on Power and Control: Reactor ScalingM. Simonhttp://www.blogger.com/profile/09508934110558197375noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-8282587.post-17353372592102457922020-09-15T02:25:14.895+00:002020-09-15T02:25:14.895+00:00Indeed.Indeed.adearthichttps://www.blogger.com/profile/13063618488208442643noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-20489196890027795012020-09-15T02:24:12.859+00:002020-09-15T02:24:12.859+00:00Indeed.Indeed.adearthichttps://www.blogger.com/profile/13063618488208442643noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-42801934853565624892007-09-07T11:49:00.000+00:002007-09-07T11:49:00.000+00:00jeremy,Cooling is going to be a very tough job.Not...jeremy,<BR/><BR/>Cooling is going to be a very tough job.<BR/><BR/>Not so much the reactor vessel which will be hard enough. <BR/><BR/>Keeping the magnet coils at 20 deg K with a heat flux estimated at 5 to 10 MW is going to be very difficult. <BR/><BR/>I'm thinking a 3 coolant system.<BR/>Water - vacuum - LN2 - vacuum - LHe.<BR/><BR/>In a lot of ways the heat flux resembles what is required for the Space Shuttle Main engines.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-24118261859406311542007-09-07T04:53:00.000+00:002007-09-07T04:53:00.000+00:00Hmmm. A more important question, IMHO, is how to c...Hmmm. A more important question, IMHO, is how to cool the reactor. There will be un upper limit to the power of the device, and this, from my understanding, won't be much bigger than break even sizing. Assuming a 100MW figure postulated at 95%alpha production, and ignoring anode losses, this leaves between 1 and approaching 5MW of heat encapsulated in an evacuated sphere no more than 5 metres across. Nasty. Even keeping the inevitable x-rays out of it. The happy reactor size between break even and meltdown I hereby designate the "M.Simon Region".adearthichttps://www.blogger.com/profile/13063618488208442643noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-61894834980615955882007-04-27T06:45:00.000+00:002007-04-27T06:45:00.000+00:003/1.55 = 1.933/1.55 = 1.93M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-77039209178839327672007-04-27T06:44:00.000+00:002007-04-27T06:44:00.000+00:00BTW Bussard says 100 MW for 3 m.Power MW Radius m1...BTW Bussard says 100 MW for 3 m.<BR/><BR/>Power MW Radius m<BR/>1000 4.17<BR/>100 3.00<BR/>10 2.16<BR/>1 1.55<BR/>0.1 1.12<BR/>0.01 0.80<BR/>0.001 0.58<BR/><BR/><BR/>Radius m Vol. m^3<BR/>4.17 303.4057465<BR/>3.00 113.0973355<BR/>2.16 42.15809177<BR/>1.55 15.71482381<BR/>1.12 5.857847857<BR/>0.80 2.183567688<BR/>0.58 0.81394532<BR/><BR/>I'll send you my spread sheet if you will send me an e-mail.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-81160944426413504792007-04-27T00:48:00.000+00:002007-04-27T00:48:00.000+00:00You assumed a 1MW RX for 1 m..Nope. Normalization ...<I>You assumed a 1MW RX for 1 m.</I>.<BR/><BR/>Nope. Normalization removes the units.<BR/><BR/>100 MW radius/1 MW radius = 1.93M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-32912788372858510312007-04-26T20:18:00.000+00:002007-04-26T20:18:00.000+00:00so a nominal 1 MW RX (for thermal testing, etc) wo...so a nominal 1 MW RX (for thermal testing, etc) would be 1.77 mTom Cuddihyhttps://www.blogger.com/profile/00998611274249390948noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-53790408163272099212007-04-26T20:15:00.000+00:002007-04-26T20:15:00.000+00:00Yep, understand you're normalized.You assumed a 1M...Yep, understand you're normalized.<BR/><BR/>You assumed a 1MW RX for 1 m. Just pointing out a 1 m reactor actually only gets you 20 kW using Bussard's reference: 3 m for 40 MW. A 1GW Rx is about 4.5 m.Tom Cuddihyhttps://www.blogger.com/profile/00998611274249390948noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-37271778498260017892007-04-26T19:32:00.000+00:002007-04-26T19:32:00.000+00:00Tom,You are using actual values.Mine are normalize...Tom,<BR/><BR/>You are using actual values.<BR/><BR/>Mine are normalized to 1 MW.<BR/><BR/>Notice how everything on the 1 MW line is 1.00?M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-45599091095932195402007-04-26T19:25:00.000+00:002007-04-26T19:25:00.000+00:00ah, ok.Well, using Bussard's numbers, I did a 5 mi...ah, ok.<BR/><BR/>Well, using Bussard's numbers, I did a 5 min BOE in excel.<BR/>V= (4/3)*pi*r^3 P ∞ V^7 <BR/> psf = v^7*40/(113.097)^7 <BR/> 1.69003E-07 <BR/>vol(m^3)rad(m) P (W) (MW)<BR/>0.52 0.50 142.89 0.00<BR/>1.77 0.75 2441.41 0.00<BR/>4.19 1.00 18289.90 0.02<BR/>8.18 1.25 87213.04 0.09<BR/>14.14 1.50 3.13E+05 0.31<BR/>23.27 1.77 1.00E+06 1.00<BR/>33.51 2.00 2.34E+06 2.34<BR/>65.45 2.50 1.12E+07 11.16<BR/>113.10 3.00 4.00E+07 40.00<BR/>268.08 4.00 3.00E+08 299.66<BR/>523.60 5.00 1.43E+09 1428.90<BR/>904.78 6.00 5.12E+09 5120.00<BR/>1436.76 7.00 1.51E+10 15062.52<BR/>2144.66 8.00 3.84E+10 38356.70<BR/>3053.63 9.00 8.75E+10 87480.02Tom Cuddihyhttps://www.blogger.com/profile/00998611274249390948noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-64473781908228311262007-04-26T18:55:00.000+00:002007-04-26T18:55:00.000+00:00Radius to the 7th power = powerSetting the radius ...Radius to the 7th power = power<BR/><BR/>Setting the radius for 1 MW = 1.00<BR/><BR/>1.39 to the 7th power = 10<BR/><BR/>1.39 to the 3rd power = 2.68<BR/><BR/>All the above are ratiometic numbers and not actual diameters or radii.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-90289696294072759032007-04-26T18:40:00.000+00:002007-04-26T18:40:00.000+00:00something's not quite right with that. In his goog...something's not quite right with that. In his google talk, Dr. Bussard said the first full-power model should be a 3-m 40MW reactor.<BR/><BR/>You have a 3-m reactor being over a GW.<BR/><BR/>I thought he said power scales as the 7th power of the size (p 26 of the IAC paper). (B^4R^3), with the B being proportional to the radius, thus P:r^7, not v^7?Tom Cuddihyhttps://www.blogger.com/profile/00998611274249390948noreply@blogger.com