tag:blogger.com,1999:blog-1500584444083499721.post2114417056977897238..comments2024-02-29T11:22:30.853+01:00Comments on CJ on Mathematics and Science: Answer to Question by Roy SpencerClaes Johnsonhttp://www.blogger.com/profile/07411413338950388898noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-1500584444083499721.post-88037118937626467682011-07-20T17:49:10.674+02:002011-07-20T17:49:10.674+02:00I agree with Cementafriend. Also if you look at t...I agree with Cementafriend. Also if you look at the extreme example of Ta = Te then W(E) is zero. Not one iota of energy is passing from anyone to anyone. Now if that is true if absorber is same temperature as emitter then how can absorber added energy to emitter when at a lower temperature. It adds nothing at same temperture.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1500584444083499721.post-82419771270357744722011-07-20T07:18:49.447+02:002011-07-20T07:18:49.447+02:00Claes, this time I agree with you and this is used...Claes, this time I agree with you and this is used in most engineering texts books on heat transfer (for example you will see a graph of h(r) as Fig. 5-7 in Perry's Chemical Engineering Handbook). The thing missing is the emissivity factors. The formula is then used in a heat transfer coefficient h(r) which can be added to the convective heat transfer coefficient h(c). The emissivities of combustion products at various temperatures have been extensively studied. <br />Those people who just use the black body Stefan-Boltzman relation have no understanding or experience with heat transfer.<br />keep strong<br />CementafriendAnonymousnoreply@blogger.com