Re: more physics (short)

[kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39)]

At 12:59 AM 6/4/96, Timothy van der Linden wrote:
>>>Thus the energy per kg is:
>>>
>>>1.394E-12 / 1.993E-26 = 6.995E13 J/Kg
>>
>>So we used the same numbers (thou in a different order) and got about the
>>same number.
>
>I guess so.
>
>>One suprize, I had that the mass of Protons and neutrons as about 1.673
>>E-27 kg, where you have 1.661E-27.  Is that a typo?  We came up with the
>>same velocity, which seems odd if we were using a different constant.
>
>I used the definition for 'u', which finds it origin in the mass of a Carbon
>12 atom. The weight of 12C is defined exactly 12*u. A single proton or
>neutron is heavier than the protons and neutrons in a atom-core of multiple
>nuclei (hence fusion frees energy).
>
>But like I said before, the accuracy of this number only makes sense if you
>know the (nearly) exact numbers of atom-masses.

Well eiather way I can't remember where I got  my atomic mass constant
from.  Eiather one seems to get similar numbers, so its probably not a
biggy.

>>>(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.
>>
>>No the numbers were not based on an engineering output.  That was supposed
>>to be the direct output from the reactions.  Could be I misread something
>>and the 14E6 number was for something else (though I can't see what).
>
>I don't know, but I've been discussing this with Rex lately and he thinks
>that the efficiency of a DIRECT fusion engine may be low. With direct I mean
>that one doesn't convert the kinetic energy of the reaction products to
>electricity.

Why?  The velocities seem to be similar to the 'optimum' velocities in your
table.  I thought you figured that by the time you converted the particals
momentum to elec, and then convert that elec back to the exaust velocity of
a mass it would all come out even?  (Give or take a lot of tonage of power
equipment.)

Please CC me if you have anything.

>>>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
>>
>>??
>>Why are you geting larger (often much larger) numbers for 75E6m/s, 125
>>E6m/s, and 150 E6m/s, but smaller for 100 E6m/s?  Also the deltas don't
>>seem even.  Hum.  I suppose I should rerun my numbers with relativistic
>>equations.  With our luck I'll come up with a third set of numbers.
>
>I think the deltas are exponential, this causes rapid deviations when one
>neglects higher terms (ie. use classical instead of relativistic formulas)
>I believe the relativistic rocket equation is:
>
>M=Mo Exp[Vend/(g Vexh)]  where g=1/Sqrt[1-Vexh^2/c^2]
>
>(Vexh is relative to the ship)

Thanks for the equation.  As to my question about why are you geting larger
numbers for 75E6m/s, 125 E6m/s, and 150 E6m/s, but smaller for 100 E6m/s?
My guess would be you calculated for .3c not 1/3rd c again.

>>Pity you gave no way to use you magic number.  (Like the value for given
>>fuels?)  Also the F number table didn't give the numbers for the fuels,
>>which naturally  didn't have even F numbers.
>
>Well simply devide c^2 (9E16) by the J/kg numbers (eg. 9E16/6.995E13=1287)
>Compared to 257 for the best fusion fuel we have come up with.

Best is of course a relative term, and so obviously is the term simple.  ;)

>Somewhere in my document there are a few fusion reactions with f, Mev and J/kg.
>
>Here is something that looks like it:
>
>                                           J/kg     f
>2H  +  3He  ->   4He +   1H  + 18.3 MeV  3.51E14   257
>2H  +  3H   ->   4He +    n  + 17.6 MeV  3.38E14   267
>3He +  3He  ->   4He + 2 1H  + 12.9 MeV  2.06E14   437
>6Li +  6Li  -> 3 4He         + 20.0 Mev  1.60E14   564  (Combined)
>3He +  6Li  -> 2 4He +   1H  + 16.0 MeV  1.28E14   704
>1H  +  6Li  ->   4He +   3He +  4.0 MeV  5.47E13  1645
>1H  + 11B   -> 3 4He         +  8.7 Mev  6.99E13  1287

Thank you.  Thats a table that would have made your other table useful.

Kelly


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Kelly Starks                       Internet: kgstar@most.fw.hac.com
Sr. Systems Engineer
Magnavox Electronic Systems Company
(Magnavox URL: http://www.fw.hac.com/external.html)

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