tag:blogger.com,1999:blog-7729093380675162051.post7517821722182475165..comments2023-01-27T16:07:01.495+11:00Comments on moyhu: Miracles of 2LoTNick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comBlogger22125tag:blogger.com,1999:blog-7729093380675162051.post-63993143427321150272012-03-19T12:25:21.373+11:002012-03-19T12:25:21.373+11:00"There was another post by Stephen Wilde ther..."There was another post by Stephen Wilde there on no backradiation etc. It got me looking for the earliest measurements I could find, and this 1837 paper by Pouillet turned up."<br /><br />Kevin McKinney has written an essay on <a href="http://www.skepticalscience.com/news.php?n=1350" rel="nofollow">The History of Climate Science - William Charles Wells</a> over at SkepticalScience concerning Wells' <a href="http://books.google.com/books?id=q8kOAAAAYAAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0" rel="nofollow">Essay On Dew</a> that dates to 1814 and may be of interest.Kevin O'Neillhttps://www.blogger.com/profile/06692943768484857724noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-75733219148755583002012-03-11T10:00:37.686+11:002012-03-11T10:00:37.686+11:00Doc,
"However, your solution can work if the ...Doc,<br /><i>"However, your solution can work if the rate the shell material conducts heat is far greater that the rate at which it radiate heat. "</i><br /><br />Your original spec at Tallblokes said the shell was 1000 atoms thick (could be hard to make). That's about 0.2 micron. I said there that 300W flowing through would produce a 0.00003C temp drop. That was working on 1μ, at 0.2 it's less than .00001C. That compares with the 228K or whatever needed to radiate 300W.<br /><br />In this problem, as stated, spherical symmetry means you don't have to worry about lateral conduction. Nor do you have to worry about the heat distribution within the sphere. That depends on the unstated conductivity anyway.Nick Stokeshttps://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-31783871797771309122012-03-11T09:27:08.299+11:002012-03-11T09:27:08.299+11:00Indeed,
"Inner sphere S1 emits 450 W"
...Indeed, <br /><br />"Inner sphere S1 emits 450 W"<br /><br />The sphere, 1 m2, has a steady state temperature of 298.4 K. It radiates 450 W so the inner shell receives 225 W/m2 from the sphere<br /><br />"the remaining 150W lands on S2"<br /><br />it also gets 75W/m2 from the shell; a total of 300 W/ms, thus the inner shell, at steady state has an influx 0f 300 W/m2 and the inner surface must radiate at 300 w/m2. So the twice what the outgoing flux should be. <br /><br />The influx and the efflux of the inner surface and sphere surface MUST balance.<br /><br />"S2 emits 300+300'<br /><br />The material in the shell does not know it has two surfaces, the inner surface is getting 300W/m2 and must heat up until it radiates at 300 W/m2; thus it should be at 269.5 K. <br /><br />However, your solution can work if the rate the shell material conducts heat is far greater that the rate at which it radiate heat. In that case, heat absorbed at the inner surface can either reradiate (slow) or migrate into the body (fast) and will have an identical likelihood of eventually being emitted on either face. <br /><br />If the rate of heat transfer in the shell is slow, then there will be a steady state temperature gradient in the shell. The Earth itself is an example of this. <br />If you have ever visited a hot steel rolling mill you would see this in action. The white hot steel cools quickly, by radiation and not by heat transfer into the air. you can observe the cooling and note that the inner core is much hotter than the outer surface. <br />Here is a video of a hot steel forge. Not how the yellow hot surface of the pigs is transformed into the white hot center on compression. <br /><br />http://www.youtube.com/watch?v=tLRkOupbARM<br /><br /><br />Using a guillotine forge cutter on ingots, you ofter observe a white/red center in the two halves even though the exterior surface is dull.DocMartynhttps://www.blogger.com/profile/03948603313788102830noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-64973894635449980772012-03-11T06:15:45.103+11:002012-03-11T06:15:45.103+11:00DocM,
I think it is:
Inner sphere S1 emits 450 W...DocM,<br /> I think it is:<br /> Inner sphere S1 emits 450 W, which all lands on the shell S2.<br /> It receives 150 W from the shell, and generates 300.<br /> <br /> S2 emits 300W out, and 300 W in. Of the 300 in, 150 W lands on S1 and the remaining 150W lands on S2. <br /> So the balance in W for S2 is<br /> emits 300+300<br /> receives 450 (from S1) plus 150 (from S2)<br /> And for S1<br /> emits 300W+150W<br /> receives 150W<br /><br />I think it is easier to think of balance in W, then divide by respective areas for W/m2.<br /><br /> <br /> Maybe the simplified numbers create confusion, since 150W has 2 different roles. Suppose S2 is bigger - surface 2.5 m2. Then it still has to emit 300 W out and in. But a smaller fraction of the 300 in goes to S1; only 120W. So 180W lands on S2. Then<br /> S1 receives 120W, emits 300+120W=420W<br /> S2 emits 600W, receives 420+180.Nick Stokeshttps://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-53409346291180403592012-03-11T02:38:00.932+11:002012-03-11T02:38:00.932+11:00Nick, I still cannot understand how the heat fluxe...Nick, I still cannot understand how the heat fluxes for your solution can possibly balance.<br />Initially the sphere has an output of 300 W and a flux of 300 W/m2.<br />When enveloped by the shell, the shell must radiate 150W/m2.<br />In your final steady state you state that the inner sphere emits 450 W/m2.<br />We know that the sphere cannot radiate to itself, so the total 450 W must be directed to the shell, so the inner surface of the shell gets a minimum of 225 W/m2. The shell will also radiate some fraction to itself.<br />So what are the heat fluxs?<br />Outer shell efflux is known to be 150W/m2. <br />Inner shell efflux is known to be 150W/m2.<br />What is the influx from the sphere to the inner shell?<br />What is the influx from the inner shell to the inner shell?DocMartynhttps://www.blogger.com/profile/03948603313788102830noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-84206143717013550642012-03-08T01:29:31.390+11:002012-03-08T01:29:31.390+11:00Nick, Once you have n spheres, then you can adjus...Nick, Once you have n spheres, then you can adjust for imperfection by making the spheres progressively more oblique. One thing I have noticed, is that the non-uniform distribution of water produces forcing response more like oblique spheres than concentric spheres. You math is much less rusty than mine, http://paoc.mit.edu/labweb/notes/chap5.pdf so you might have more fun with the potential temperature profile of moist air :)Recovering in the Florida Keyshttps://www.blogger.com/profile/07913299764512464597noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-12527988486300653082012-03-06T13:04:18.639+11:002012-03-06T13:04:18.639+11:00Anon writes "If you have a black body in fron...Anon writes "If you have a black body in front of another black body I cannot see (pun intended) how, at thermal eqilibrium, you would be able to dectect its presence."<br /><br />Admiral Ackbar would know.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-58664410835658221282012-03-06T09:41:49.603+11:002012-03-06T09:41:49.603+11:00oh hell, the whole point of thermal radiation is t...oh hell, the whole point of thermal radiation is that it is incoherent.EliRabetthttps://www.blogger.com/profile/07957002964638398767noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-87368422003200071802012-03-05T03:28:33.658+11:002012-03-05T03:28:33.658+11:00Nick,
Thanks for the good work. The Jelbring stuff...Nick,<br />Thanks for the good work. The Jelbring stuff is the new skydragon. Argue!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-29397777515765269552012-03-04T05:35:56.099+11:002012-03-04T05:35:56.099+11:00Dallas,
Yes, interesting thought. The "Update...Dallas,<br />Yes, interesting thought. The "Update" formula extends. The outer shell S3 emits P at emittance P/A3; The second receives irradiance P/A3 from S3 (again it gets the same as it would get at isothermal), total power P+A2*P/A3. And the innermost, S1, receives that emittance P*(1/A2+1/A3) that S2 emits inward (same as outward), which is power P*A1*(1/A2+1/A3). So balancing, S1 emits that plus the source P, with an emittance of P*(1/A1+1/A2+!/A3).<br /><br />For N spheres, P*(1/A1+....+1/AN)Nick Stokeshttps://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-80608237801490461012012-03-04T03:44:23.920+11:002012-03-04T03:44:23.920+11:00Very nice. If you add a third sphere?Very nice. If you add a third sphere?Recovering in the Florida Keyshttps://www.blogger.com/profile/07913299764512464597noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-32997509045697924852012-03-03T12:33:14.585+11:002012-03-03T12:33:14.585+11:00Doug,
You've been saying this everywhere, but ...Doug,<br />You've been saying this everywhere, but it's totally wrong. One thing is that you never in the real world get a standing wave in 3D over millions of wavelengths. Secondly, while an exact standing wave won't transmit power, it takes only a slight realignment of phase to do so.Nick Stokeshttps://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-15612045941879032932012-03-03T11:26:02.373+11:002012-03-03T11:26:02.373+11:00If one body is cooler than the other you can ignor...If one body is cooler than the other you can ignore its radiation to the warmer one because all that happens is that a <b>standing wave</b> is set up corresponding to the amount of radiation the cooler one could emit. You then deduct this amount which the cooler one could emit from that which the warmer one could emit. The difference represents that portion of the warmer one's radiation which is absorbed and converted to thermal energy in the cooler one. <br /><br />That's what actually happens physically. I know you can get the same result by assuming radiation from the cooler one is absorbed in the warmer one, but it isn't and the warmer one's absorptivity for radiation from the cooler one is zero. <br /><br />This is why radiation from a cooler atmosphere cannot transfer thermal energy to a warmer surface. <b>It can only slow the rate of radiative cooling by forming standing waves which do not transfer any energy between the atmosphere and the surface.</b> Other cooling processes like evaporation and diffusion followed by convection will increase to compensate. Carbon dioxide molecules can play no greater role than water vapour molecules. Hence their effect in regard to any greenhouse conjecture would be less than 4% of the effect of water vapour. Work out the consequences for yourself.Doug Cottonhttp://climate-change-theory.comnoreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-29446948037147845182012-03-03T08:32:05.609+11:002012-03-03T08:32:05.609+11:00Jeff,
I've installed the MathJax version of La...Jeff,<br />I've <a href="http://moyhu.blogspot.com.au/2011/05/latex-now-works-here.html" rel="nofollow">installed</a> the MathJax version of Latex. It works well in posts and comments. There's some overhead - it taken a little time to render the Latex, and adds a bit to the size of every page. <br /><br />There are some WYSIWYG <a href="http://en.wikipedia.org/wiki/Comparison_of_TeX_editors" rel="nofollow">Latex editors</a>; I haven't tried them.<br /><br />But I find that you can go a long way with the <a href="http://everything2.com/title/HTML%2520symbol%2520reference" rel="nofollow">symbols</a> in html, with <sub> and sup as well. It's trappy, though, as I found at TB's. Blog hosts often restrict their use in comments.Nick Stokeshttps://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-27619733092987212852012-03-03T07:36:48.445+11:002012-03-03T07:36:48.445+11:00Nick,
Is there an easy way to write equations on ...Nick,<br /><br />Is there an easy way to write equations on blogs? I've worked with LaTeX a little but it is such a pain in the butt that i usually resort to looking around the web for a screen grab.Jeffhttps://www.blogger.com/profile/00102232063298547788noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-35486197608568720172012-03-03T06:56:57.588+11:002012-03-03T06:56:57.588+11:00Jeff,
Yes, it can be frustrating. There was anothe...Jeff,<br />Yes, it can be frustrating. There was another post by Stephen Wilde there on no backradiation etc. It got me looking for the earliest measurements I could find, and this <a href="http://nsdl.org/archives/onramp/classic_articles/issue1_global_warming/n2-Poulliet_1837corrected.pdf" rel="nofollow">1837 paper</a> by Pouillet turned up.<br /><br />I hadn't posted a link as the discussion there seemed to have gone quiet, but your post showed up there.Nick Stokeshttps://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-10425878563928807662012-03-03T06:48:42.252+11:002012-03-03T06:48:42.252+11:00Anon,
Indeed so - if it's isothermal then it&...Anon,<br /><br />Indeed so - if it's isothermal then it's featureless. That's another way of seeing why you don't really have to analyze the geometry to know the answer.Nick Stokeshttps://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-29052425489086218622012-03-03T01:05:08.106+11:002012-03-03T01:05:08.106+11:00I haven't read tallblokes guest post but I use...I haven't read tallblokes guest post but I used to chase these guys away all the time from tAV. They just don't get that you ain't gonna prove AGW wrong with some basic theory about radiation or whatever. I've posted one once just to see how people reacted - it didn't go well for them. SOD left a comment asking why I would put a thing like that up if I didn't endorse it.<br /><br />More often than not, they are not technical people in general. I had one guy insist that in converting units, there was a division by 1 and that was where all of radiation theory went wrong. I told him over and over that he was in error. <br /><br />At least you set the record straight. Have you dropped a link to this over there?Jeffhttps://www.blogger.com/profile/00102232063298547788noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-59916433869028733742012-03-03T00:59:35.460+11:002012-03-03T00:59:35.460+11:00If you have a black body in front of another black...If you have a black body in front of another black body I cannot see (pun intended) how, at thermal eqilibrium, you would be able to dectect its presence.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-90651959084009014402012-03-01T07:29:02.767+11:002012-03-01T07:29:02.767+11:00Eli,
yes, the assumption (by HJ and I) was that bo...Eli,<br />yes, the assumption (by HJ and I) was that both spheres are black, ie black bodies.Nick Stokeshttps://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-67985432784336757532012-03-01T02:31:38.316+11:002012-03-01T02:31:38.316+11:00If you are going to engage in Lambert's law, y...If you are going to engage in Lambert's law, you also have to make some assumptions about the surface of the inner sphere and how the absorption and reflection vary with angle.EliRabetthttps://www.blogger.com/profile/07957002964638398767noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-83632783300363663462012-03-01T01:35:37.942+11:002012-03-01T01:35:37.942+11:00Oh dear. Hans doesn't do a very good job of un...Oh dear. Hans doesn't do a very good job of understanding you. What a waste of everyone's time.<br /><br />Though it *could* be useful. The people who comment there could read your words, realise that you're right, and realise that therefore Hans and Tallbloke are clueless.William M. Connolleyhttps://www.blogger.com/profile/05836299130680534926noreply@blogger.com