The dcx machine

[guest]

The DCX machine was mankinds first attempt at fusion. Basically it was a large water cooled discharge tube. This machine was built in the early fifties. I learned of it in those funny pamphlets that the then AEC put out (its the DOE now). All data we use today came out of the ten year run when that machine was running . It was a part of project Sherwood. The name was a smart ass pun. The pun originated due to this question "wouldn't it be great if fusion worked?". It sure would was the answer. Hense the Sherwood project.

I sense a great deal of blank looks when I post about
I^2 scaling. This machine was the first machine to drive home the concept of current scaling.

Plasma 101

All electrically driven machines wheather your toaster or Star Jone's vibrating joy pony all use watts to drive them. Electrically driven fusion machines are no different.

watts=(resistance)(amps)^2
w=(r)(I)^2 ( Look I know it not the position
you are used to)

fine and dandy....


In disharge tubes that are at a non conducting state
the resistance exists. The vacuum that makes up the
working enviroment of the tube causes it. But as soon as the tube starts to create ions the resistance vanishes.
So the equation becomes:

watts = (amps)^2

But what about this N=(amps)^2 bit .... I will
show it 's nothing to be afraid of... really.


I will show you how:

If I'm building a fusion machine neutrons would be a good thing to have.
I Know it takes energy to make neutrons.
That energy is in ev usually but I can convert it to joules.
If i have joules per second I've got watts.
So I can tie in the number of neutrons per second to watts.... cool.

More Physics...... Dont' Panic.... Very simple stuff.

We assume some really wacked out concepts in physics. Like the one I'm about to throw at you.
In order for the deuterium ion to fuse it must physically touch the other deterium ion. This causes electrostatic repulsion. It is the cost of overcoming this repulsion between the nuclii that is a cost of doing business.

so.... 2ke is about ke^2/(rd1+rd2)

( 9.0 x 10^9 Nm^2/C^2)(1.6 x10^-19 C^2)
--------------------------------------------------------
2(3.2 x 10^-15 m)

Works out to about 3.6 x 10^-14 joules/ reaction in which 1 neutron is formed about 1/2 the time.
(two competing reactions one with neutron one without)

so if we double the energy we are sure to get a neutron
so it is 7.2 x 10^-14 joules.

So in order to get watts we simply divide the joules of each neutron by a second to get watts.

So each neutron takes 7.2 x 10^-14 watts

so we put this into our equation to get (Tada!!!!)

7.2 x 10^-14 watts n = (amps)^2

but you might say we been robbed... There's more here.

Watts = (volts) (amps)

Amps = watts / volts
So the amps for one neutron is 7.6 x 10^-14 watts divided by the operating voltage
If the volts are 10 kev then the cost of doing business is 7.6 ^-14 watts/ 1x 10^4 = 7.6 x 10^ -18 amps

so now we got

n (Ix10-18 amps) = i^2 amps
n=i^2amps/1X10^18 amps
n=i^2.000000000000000001 amps
But in the days of slide rule three digits were king so...
n=i^2.00
But you guys were wondering where the extra powers come from there it is.

When you have large numbers of neutrons you have a corresponding cost of business That is tacked on to the original i^2 term.

Oh yeah before I forget ... no matter how high the powerof the i term the minimum power never falls below I^2... Huh?

Let me show you. 1neutron=(1 amp)^2,^3,^4,^5....

No matter how many neutrons you make it will always take 1 amp of current to make one neutron in fusion.
My cd player only draws .25 amps!

Class dismissed... Hope this was good for you too.
If you noticed the frequency of my posts... I'm partime right now.

Larry Leins
Physics Teacher