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Close but no cigar?
- To: KellySt@aol.com, email@example.com, firstname.lastname@example.org, email@example.com, firstname.lastname@example.org, email@example.com, firstname.lastname@example.org, David@InterWorld.com, email@example.com, firstname.lastname@example.org, DotarSojat@aol.com
- Subject: Close but no cigar?
- From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)
- Date: Tue, 12 Mar 1996 23:14:32 +0100
OK here the (revised) letter that I wrote Kevin earlier this day about the
>> Hello Kevin,
>> Could not help myself, so started calculating if the sail would melt or not:
>That's cool, because I wasn't really sure how to do it myself yet.
>> Power out: P=s T^4
>Is this the proper blackbody radiation formula? Looks to simple to me.
This is the Boltzmann equation, the Planck equation looks much more ugly and
only tells the energy density per "delta wavelength". So one would need to
integrate the latter (which doesn't seem easy).
Kevin suggested using Titanium because it is easy to mine on the moon.
Since I haven't numbers about that I will use Wolfraam.
>> s = Bolzmann constant = 5.67E-8 W m^-2 K^-4
>> T = Temperature [K]
>> P = Energy density [W/m^2]
>> Power in: 7.5E5 W/m^2
>> Energy buffer: Wolfraam
>> Specific heat = 135 J kg^-1 K^-1
>> Melt temp = 3680 K
>> 0.05 Kg/m^2
>> 0.05 135 = 6.75 J K^-1 m^-2
>> Max energy: 3680 * 67.5 = 2.48E4 J m^-2
>> 2.48E4 < 7.5E5 Close but no cigar. (But this may be solved by decreasing
>> the beam density) We need a metal that radiates more than 7.5E5 W/m^2 at a
>> temperature less than its melting temp.
>> Hey, having done the calculation, I see that the product of melting
>> temperature and specific heat is important, so lets look in my tables again:
>> For copper we get : 0.05*387*1356=2.6E4 J/m^2
>> For magnesium : 0.05*1026*922=4.7E4 J/m^2
Tim (and Kevin)