[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

*To*: T.L.G.vanderLinden@student.utwente.nl, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmitl.ippt.gov.pl, hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com, David@interworld.com, lparker@destin.gulfnet.com, bmansur@oc.edu*Subject*: Re: Close but no cigar? again*From*: DotarSojat@aol.com*Date*: Thu, 14 Mar 1996 03:41:54 -0500

I know. I promised no more remarks until I did my homework. The Stefan-Boltzmann law of radiation is P = s T^4, where s (the Stefan-Boltzmann constant, usually written as the greek lower-case letter sigma) does indeed have the value 5.67 E-8 W/(m^2 K^4). A black-body surface absorbing a power of 7.5E5 W/m^2 (your number) will be heated to an ultimate equilibrium temperature Teq at which it will reradiate all the absorbed power. Therefore, 7.5E5 = 5.67E-8 Teq^4 . So Teq equals 1907 K, below the melting temperature of titanium (2073 K) and well below the melting temperature of tungsten. Unless I missed something, I believe you (I don't know who, because I haven't yet learned to read these quotes of quotes of quotes) equated power to energy in your calculation of no cigar. Rex Finke

- Prev by Date:
**Thanks** - Next by Date:
**Re: Thanks** - Prev by thread:
**Re: Thanks** - Next by thread:
**Re: Close but no cigar? again** - Index(es):