What is the potential energy output of an Americium 242 Fission Engine?

[Intuitive]

http://en.wikipedia.org/wiki/Americium" 242 Fission Engine

http://www.spaceflightnow.com/news/n0101/19marsnuclear/"


this Americium Engine would be interesting to see get started.

I still think http://en.wikipedia.org/wiki/Terajoule" shield would out run it in the end.

Note: 9.0 × 1013 J = 90 TJ – Theoretical total mass-energy of one gram of matter, That's 90 Terajoules.

4.184 × 1015 J — energy released by explosion of 1 megaton of TNT
1.74 × 1017 J — total energy from the Sun that hits the Earth in one second

 

[Astronuc]

The Americium engine concept is still only on paper.

A major problem with this concept is the proximity of the superconducting magnetic cools to the thermal source (Am-reactor). In addition, the high temperature means low propellant density and therefore low thrust.

"The gas will be magnetically confined so temperatures of about 250,000 degrees can be reached," explained Ronen. "With such temperatures a velocity of 80 km per second can be obtained."

Ronen

The high exhaust velocity is only part of it. The other part is the mass flow rate with which one develops the momentum/thrust.

I am highly skeptical of the confining magnetic field in conjunction with the Am-reactor chamber.

 

[Morbius]

Note: 9.0 × 1013 J = 90 TJ – Theoretical total mass-energy of one gram of matter, That's 90 Terajoules.

Intuitive,

The above part of your post is misleading - because fission doesn't get you anywhere
NEAR the theoretical mass-energy conversion.

For example, the energy equivalent of 1 nucleon is about 930 MeV. [Neutrons are
a bit more massive than protons - but an approximate figure will suffice here. ]

Therefore, the Uranium-235 nucleus represents a theoretical mass-energy of about
218,550 MeV. However, the fission of a U-235 nucleus gives about 200 MeV.

So nuclear fission only nets you less than one-tenth of one percent [ 0.1% ] of the
theoretical maximum - so the 90 TJ figure is pretty worthless.

Dr. Gregory Greenman
Physicist

 

[Intuitive]

Intuitive,

The above part of your post is misleading - because fission doesn't get you anywhere
NEAR the theoretical mass-energy conversion.

For example, the energy equivalent of 1 nucleon is about 930 MeV. [Neutrons are
a bit more massive than protons - but an approximate figure will suffice here. ]

Therefore, the Uranium-235 nucleus represents a theoretical mass-energy of about
218,550 MeV. However, the fission of a U-235 nucleus gives about 200 MeV.

So nuclear fission only nets you less than one-tenth of one percent [ 0.1% ] of the
theoretical maximum - so the 90 TJ figure is pretty worthless.

Dr. Gregory Greenman
Physicist

My Apologies, I should of put the link to the information related to the energies, The energy mentioned on the next line wasn't for a fission reaction but was just stating the Theoretical total mass-energy of one gram of matter. Thanks for all your help.


I think what was ment was
9.0 × 1013 J = 90 TJ – http://en.wikipedia.org/wiki/1_E13_J"

4.184 × 1012 J = 4.184 TJ – http://en.wikipedia.org/wiki/Terajoules"

4.184 × 1015 J = 4.184 PJ - http://en.wikipedia.org/wiki/1_E15_J"

9.0×1016 J – http://en.wikipedia.org/wiki/1_E16_J"

1.74 × 1017 J – http://en.wikipedia.org/wiki/1_E17_J"

2.5 × 1017 J – http://en.wikipedia.org/wiki/1_E17_J"

_________________________
Humbly sits down to listen.

 

[Astronuc]

Intuitive,

The above part of your post is misleading - because fission doesn't get you anywhere NEAR the theoretical mass-energy conversion.

For example, the energy equivalent of 1 nucleon is about 930 MeV. [Neutrons are a bit more massive than protons - but an approximate figure will suffice here. ]

Therefore, the Uranium-235 nucleus represents a theoretical mass-energy of about 218,550 MeV. However, the fission of a U-235 nucleus gives about 200 MeV.

So nuclear fission only nets you less than one-tenth of one percent [ 0.1% ] of the theoretical maximum - so the 90 TJ figure is pretty worthless.

And the other part of that is the energy released in the fission products, as opposed to gamma and beta rays which mostly interact in the solid material, not the propellant.

The 141 yr half-life means a respectable radiological issue for several kg's, as well as heat removal issues prior to deployment.

 

[Morbius]

And the other part of that is the energy released in the fission products, as opposed to gamma and beta rays which mostly interact in the solid material, not the propellant.

Astronuc,

Yes - most of the energy goes into the fission product kinetic energy.

Some goes into radiation; gamma and beta; and about 10 MeV or about
5% of the total fission energy goes into anti-neutrinos. That energy is
just plain LOST - you can't recover the neutrinos because their interaction
cross-section is extremely low. [ The average distance a neutrino will
travel in solid lead before it interacts - is measured in light-years.]

Dr. Gregory Greenman
Physicist

 

[Astronuc]

I still think TeraJoule EMP Bursts reflected on a Diamagnetic shield would out run it in the end.

I would like to see the basis of this statement.

One has to look at the 'energy density' required and the pressures and stresses involved. One key factor, as the temperature of a solid increases, the tolerable stresses decrease. Solids are normally used in the elastic range (maximum principal stress less than yield). Then there is the issue of creep, and generally systems are designed on the basis of 1% plastic strain, for example.

Anything that man makes will be technologically limited by the fact that materials are limited. Materials can only handle so much. That is the challenge in propulsion systems whether they are chemical, nuclear, anti-matter or plasma.

Static magnetic fields are generally limited to about 15T. Transient magnetic fields can go much higher, but the issue is that they are transient (pulsed)!

 

[Astronuc]

Yes - most of the energy goes into the fission product kinetic energy.

Some goes into radiation; gamma and beta; and about 10 MeV or about
5% of the total fission energy goes into anti-neutrinos. That energy is
just plain LOST - you can't recover the neutrinos because their interaction
cross-section is extremely low. [ The average distance a neutrino will
travel in solid lead before it interacts - is measured in light-years.]

Thanks Greg. The point I was trying to make is that 'not all' the fission energy (~205 MeV/fission) is recoverable/useful. Rather the fission products of U-235 account for about 168 MeV of energy of the 205-207 MeV emitted from all processes including and subsequent to the fission. This matter seems lost on many people who are not intimately familiar with the fission process.