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Re: Physic help

At 11:29 AM 5/4/96, Timothy van der Linden wrote:
>I yesterday posted this by accident only to Kelly:
>- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
>To Kelly,
>>Hi all you physics types.  On my web page describing my explorer class'
>>fusion reactors.
>> I have a table where I list various fusion fuel cycles.  It lists the
>>resulting energy in Mev.  For those of us who arn't familure with
>>translating Mev into anything, could someone tell me what the speed of the
>>resulting particals is?  Given that all the power in those reactions is
>>contained in the resulting kinetic energy of the particals.  This should be
>>a simple calculation, and would certainly be a nice addition to the table.
>>Assuming of course your half of and undergrate physics degree wasn't over
>>15 years stale!
>I'm a bit amazed you ask this because 2 paragraphs later you explain what an
>electron Volt is.
>So I assume the problem is in the calculation of the velocities.

A definition is one thing, knowing how to plug real numbers into an
equation is another.

>I'll assume the velocities are relativistic, to make the formulas usable for
>all energies.
>Kinetic energy (in Joules) of a particle with mass m (in kg) moving with
>velocity v:
>               1
>        ---------------
>                    2
>gamma =            v        gamma >= 1
>         SQRT(1 - ----)
>                    2
>                   c
>       2
>K = M c (gamma - 1)   K=kinetic energy
>In fact this is all you need, but there is a bit of a problem, namely that
>there is more than one particle in each reaction, and that sometimes not all
>particles have the same mass. This makes that not all particles have the
>same final speed, ofcourse one could figure out a mean velocity, but I'm not
>sure how to do that best.

In the case of Li6 the resulting exaust is all 4He?  3 per 20 MeV.

>If I would do a very rough approximation, I would use the Watts/kg numbers
>in your table (by the way it should be Joule/kg).
>Since then the velocities wouldn't be relativistic at all I can simply use:
>E = 0.5 m v^2
>v = SQRT[2 E/m]

For example is E energy in Joules, watts, voltage...?  Is m in kilos,
proton mass, atomic weight?  Without this the equation isn't useable to me.
In physics as I remember only a few standard units are used.  In
Engineering anything is used.

>For 2.058E14 Joule/kg this would give 2E7 m/s or 0.067c

What are you using for mass?  Are you assuming a 4He and 2 p, or assuming
that the energy of the p's are transfered to the next reaction that are
involved in?  Or assumed the energy is lost?

>Oh to make the thing complete here the translation from eV to joule:
>1 eV = 1.6E-19 Joule
>Note that eV is a measure for energy and that in particle physics the mass
>of a particle is often given in eV also (this according to E=mc^2).

Damn, this is bad 2e7 would translate to a specific impulse of 2,000,000.
Which would mean the explorer class would need a fuel to weight ratio of
148 to 1 to get down from .3 c.  And given that you used 2.058E14, which is
the result for He3 not Li6, and Li6 reactions divide 1.596 E14 over about
twice as much mass, then a Li6 fueled ship would have a much lower specific
impulse.  I.E. the Explorer class couldn't work!!

Yeah I roughed out a calculation (I'm still not comfortable that I'm
following your math.) and get a specific impluse of about 1,600,000.  Which
would demand hundreds of times the ships weight in fuel!!!!




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