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Re: more physics (short)
- To: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com, David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com, neill@foda.math.usu.edu, 101765.2200@compuserve.com, MLEN3097@Mercury.GC.PeachNet.EDU
- Subject: Re: more physics (short)
- From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)
- Date: Sun, 02 Jun 1996 18:37:46 +0100
>I went over the Bussard paper and found 14E6 as the upper exaust vel for p
>+ 11B. Which obviously isn't what I'm getting. Eiather I miscalculated
>the watts per Kg (I sent my equations for that in another letter) or I
>misread Bussards paper. (I suppose I could write him.) I'ld hate to write
>off about 15% of spec impulse if I didn't need to. (See table below.)
>If you know how to get the exaust speed from Mev numbers let me know. I'm
>sure those numbers from the papers are correct.
Watch carefully:
The reaction as found in my tables book:
H + 11B --> 3 4He + 8.7 MeV
Let's assume energy conversion is 100% and ALL reaction products will be
accelerated in the 100% effective lineac.
First turn MeV in to Joules:
8.7 MeV = 8.7E6 * 1.602E-19 = 1.394E-12 Joule
Then determine the weight of the particles in kg:
H + 11B is approx 12 u (u=atomic mass unit)
12 u = 12 * 1.661E-27 = 1.993E-26 Kg
Thus the energy per kg is:
1.394E-12 / 1.993E-26 = 6.995E13 J/Kg
(Don't use much more significant numbers, if you do use them you need to
define some things more accurate and cannot simply think that 11 H weighs as
much as 11B.)
Using E=0.5 m v^2 I still get v=11.8E6 m/s, so I really wonder how Bussard
can get a higher velocity, unless he dumps some of the reaction mass instead
of accelerating it.
>I was working up the following table of the fuel mass ratios needed to get
>to or from certain speeds given the exaust velocities.
>
>NEW numbers
>
>Fuel --> Exhaust
>Vexh 75E6m/s 100 E6m/s 125 E6m/s 150 E6m/s
>
>p + 11B --> 3 4He
> 11,800,000 m/s 576.0 4,790.0 39,900 332,000
>
>6Li + 6Li --> 3 4He
> (Combined)
> 17,800,000 m/s 67.6 275.0 1,120 4,570
>
>3He + 3He --> 4He + 2 p
> 20,300,000 M/s 42.5 138.0 472 1,620
>
>De + 3He --> 4He + p
> 26,500,000 m/s 16.9 43.5 112 287
Of course ;) I get completly different numbers (at least for higher Vend)
674 2,680 82,000 1,200,000
73 181 1,728 10,000
44 97 708 3,377
18 34 154 509
I assume this time relativistics is the origin (for 0.5c gamma=1.15 which
means its not almost equal to 1 anymore) for our differences
>Amazing how touchy the fuel ratios are to changes in exaust vel/specific
>impulse. Look at the difference between 6Li + 6Li and 3He + 3He! ;)
Well yes, that is because of the energy per kilogram of fuel. Which is 3
times higher for 3He + 3He. (This is what my "magic" number f says)
Timothy