B Energy production that converts Hydrogen to Iron?

[Devin-M]

Suppose a craft derived its energy from fusion— converting Hydrogen into Iron…

If it started with a kilo of Hydrogen, how much mass of Iron would it have at the end and how much energy would be extracted?

 

[mfb]

The number of nucleons won't change in the process. You can look how many your target iron isotope has, that's how many hydrogen atoms you need per iron atom (if you start with normal hydrogen, it's less with deuterium or tritium). You can look up the masses of hydrogen and iron and compare.

 

[Hornbein]

Maybe I shouldn't mess with this but am going to do it anyway.

It has to do with the "relative isotopic mass." First simplify things by assuming that all your hydrogens and irons are purely of the most common isotopes. Take the relative isotopic mass of your hydrogen and multiply by the number of nucleons in your iron. This number will be higher than the relative isotopic mass of your iron. The difference is the amount of mass converted to energy during fusion. Use E=mc^2 to figure out how much energy.

 

[ohwilleke]

Suppose a craft derived its energy from fusion— converting Hydrogen into Iron…

Not to rain on your parade, but while this fusion engine would not facially violate any laws of physics, as a practical matter, it is basically an engineering impossibility.

In the real world, iron is created through nuclear fusion processes that start with hydrogen in a multiple step process interrupted by supernovas at intermediate steps. Early stars converted hydrogen to intermediate sized elements, went supernova, and then had some of those intermediate sized elements that the supernova spits out end up in a new star where they are fused in the second star in the fusion sequence, and so on, until eventually Nth generation star in the sequence fuses medium sized elements into iron.

If the craft likewise used a multiple step process, the energy generated in the intermediate steps in a way that attempted to hold onto the fusion products produced would blow the craft apart before it could get to the next steps.

If the craft tried to do it in one step, the problem is that it is virtually impossible to get dozens of hydrogen atoms to all fuse and organize themselves into an iron nucleus in the same place at the same time. It is hard enough to get two component atoms in once place at the same time in the right conditions to fuse.

This said, most of the energy gains from nuclear fusion happen at the beginning of the nuclear fusion chain ending with iron, and not at the end of it, as illustrated by this chart:

The initial leap from deuterium to helium-4 produces about 80% of the energy per nucleon return of the entire chain from deuterium to iron, and is eminently possible from an engineering perspective.

 

[phyzguy]

In the real world, iron is created through nuclear fusion processes that start with hydrogen in a multiple step process interrupted by supernovas at intermediate steps. Early stars converted hydrogen to intermediate sized elements, went supernova, and then had some of those intermediate sized elements that the supernova spits out end up in a new star where they are fused in the second star in the fusion sequence, and so on, until eventually Nth generation star in the sequence fuses medium sized elements into iron.

While I agree that what the OP proposed is an engineering impossibility, the above paragraph is not correct. A star more than about 20 solar masses can fuse hydrogen all the way to iron in a single star. It's believed that the fusion processes take place in shells, like this Wikipedia diagram:

 

[DaveC426913]

... basically an engineering impossibility.

If the craft likewise used a multiple step process, the energy generated in the intermediate steps in a way that attempted to hold onto the fusion products produced would blow the craft apart before it could get to the next steps.

I think this is overly-cynical and underly-supported.

1. If we grant the premise of being able to artificially create H>Fe Fusion in the first place, then it sort of goes without saying that part of that solution is to contain the products so it doesn't blow up.
2. That "energy generated" you are speaking of is not a liability; it is the desired goal. It is the purpose of the device in the first place, and we likewise assume that it is harvested and directed toward propulsion.

I mean, your statement, spoken in the 1920s could just as easily have been (and surely was) said about controllable nuclear fission - that is, until we found a way to not have it blow up, and a way to safely harvest the energy it produced.

And in 1959 they started putting them in ships.

 

[Nik_2213]

{ Joke ...}
Handwavium:
"We use 'Red Mercury' as a catalyst-- More stable than Dilithium..."
/

 

[Vanadium 50]

I mean, your statement, spoken in the 1920s could just as easily have been (and surely was) said about controllable nuclear fission

What?

Nuclear fission was not discovered until 1938. Heck, the neutron wasn't discovered until 1932. Where are you getting this?

 

[DaveC426913]

What?

Nuclear fission was not discovered until 1938. Heck, the neutron wasn't discovered until 1932. Where are you getting this?

(Okay, so change '20s to '30s.**)

Like the OP's H>Iron fusion, it might have been effectively science fiction at the time (if not for the war).

My point being simply that "X can't work because we'll never be able control or contain it" is a very weak argument.

** Wow. I knew the atom bomb timeline was short but I did not know it went from unknown to employed in six years. I'd assumed it had been theorized well before being discovered in the lab.

 

[mfb]

The initial leap from deuterium to helium-4 produces about 80% of the energy per nucleon return of the entire chain from deuterium to iron, and is eminently possible from an engineering perspective.

Proton-proton fusion has such a low cross section that we still haven't measured it (in the lab, we see it in the Sun). Fusion reactors on Earth use deuterium and tritium, reducing the difference to ~3.5 MeV per nucleon.

 

[Vanadium 50]

This thread went in a bad direction, and I got yelled at a little for responding to some of what is no longer here.

The word "spacecraft" in the title is unnecessary, and probably unhelpful. I will replace it with "extremely advanced technology".

I don't know how seriously to take this, now that we know it is about such an extremely advanced technology but: d-d fusion goes from a binding of 1.1 MeV perr nucleon to about 2 (3He + n or t + p) or 7 (the rarer 4He process). So there is some reason to go past 3He or tritium.

But since we are postulating a technology that can easily fuse all the way to iron, one needs to ask "Why?" By the time you get to, say, oxygen, you have already extracted 90% of the energy you are ever going to get. Further, you need 4 extra neutrons to get to Fe-56/ Where do they come from?

Such an advanced technology one needs to assume is starting to look like magic.

 

[ohwilleke]

Proton-proton fusion has such a low cross section that we still haven't measured it (in the lab, we see it in the Sun). Fusion reactors on Earth use deuterium and tritium, reducing the difference to ~3.5 MeV per nucleon.

I don't disagree which is why I referenced starting with deuterium rather than hydrogen-1.

Getting a bit further into answering the original question.

Nuclear fusion generates 580,000,000 MJ of energy from a kilogram of heavy hydrogen (i.e. Hydrogen-2 and Hydrogen-3) by fusing it into Helium. In contrast, hydrogen fuel cell produces 142 MJ/kg. Natural gas (i.e. basically methane) produces 54 MJ/kg and gasoline produces 46 MJ/kg.

MJ = megajoule and kg = kilogram and 1 Joule [J] = 1 Watt-second [Ws] = 1 V A s (volt-ampere-second)= 1 N m (Newton meter) = 1 kg m²s⁻² (kilogram-meter squared per second squared).

In theory, if it went from heavy hydrogen straight to iron, it would generate 725,000,000 MJ of energy from a kilogram of heavy hydrogen (i.e. 7.25 * 10⁸ MJ).

The energy that would be created by converting 1 kg of matter directly to pure energy (via the E=mc² conversion) would be 89,875,517,878.0128 MJ (i.e. roughly 9 * 10¹⁰).

So, the energy produced in the fusion reaction would be equivalent to a mass almost exactly 8 grams converted to pure energy.

The mass of the end product of converting 1 kg of heavy hydrogen to iron in nuclear fusion would be 992 grams of iron.

These calculations have a roughly 1% uncertainty.

 

[snorkack]

Not to rain on your parade, but while this fusion engine would not facially violate any laws of physics, as a practical matter, it is basically an engineering impossibility.

In the real world, iron is created through nuclear fusion processes that start with hydrogen in a multiple step process interrupted by supernovas at intermediate steps. Early stars converted hydrogen to intermediate sized elements, went supernova, and then had some of those intermediate sized elements that the supernova spits out end up in a new star where they are fused in the second star in the fusion sequence, and so on, until eventually Nth generation star in the sequence fuses medium sized elements into iron.

If the craft likewise used a multiple step process, the energy generated in the intermediate steps in a way that attempted to hold onto the fusion products produced would blow the craft apart before it could get to the next steps.

If the craft tried to do it in one step, the problem is that it is virtually impossible to get dozens of hydrogen atoms to all fuse and organize themselves into an iron nucleus in the same place at the same time. It is hard enough to get two component atoms in once place at the same time in the right conditions to fuse.

Yet the reason any metals exist (other than some Li) is that three alpha particles get in the same place within 10^-16 s.
The moment you are as far as d, getting to Ni-56 is "trivial". Simply compress the matter to sufficient temperature and density, and the next reactions will go arbitrarily fast, at sufficient temperature.
It is fusing p that is problematic, even at arbitrarily high temperature!

Note that simple rearrangement of nucleons to Ni-56 gives you Ni-56, not Fe-56! There is some energy that won´t be released quickly, and that is slowly released in positron emission of Ni-56 (6 days) and then Co-56 (77 days).

What DO you get if you heat protium quickly?
Provided you have some C, N or O, you first get some cold CNO cycle. But it is the production of neutrons that is the rate controlling step of cold CNO cycle. The speed of cCNO-I is limited by the 10 minute halflife of N-13, that is not sped up by heating.

Heating can lead to hot CNO cycle by adding another proton to N-13 and producing O-14... halflife still 70 s. And hCNO-I is actually controlled by the halflife of O-15 at 122 s. Further heating is no help: F-16 and F-15 are unbound. Converting a proton to neutron will take minutes regardless of temperature.

So... If you rapidly heat some deuterium, so that all of it is converted into Ni-56 before it has time to expand, how much energy will be converted to mechanic velocity of the propagating Hugoniot shockwave? What are the detonation velocities of more common substances like He-4, C-12 or O-16?

 

[usernamess]

I stumbled upon this thread and was curious, so I took a crack at the math. I'm not a physicist so please point out if I've made any errors.

The above makes the assumptions that all gamma rays and intermediary reactions have their energy entirely absorbed in the reaction, and that the iron exhaust is at absolute zero relative to itself. Others have already commented on the infeasibility of such a design. Even so, if engines of such a design were incorporated into the Apollo Lunar lander, the kilogram of hydrogen would barely be sufficient to accelerate it to the moon's relatively low escape velocity.

 

[berkeman]

Welcome to PF.

I stumbled upon this thread and was curious, so I took a crack at the math. I'm not a physicist so please point out if I've made any errors.

View attachment 349140
The above makes the assumptions that all gamma rays and intermediary reactions have their energy entirely absorbed in the reaction, and that the iron exhaust is at absolute zero relative to itself. Others have already commented on the infeasibility of such a design. Even so, if engines of such a design were incorporated into the Apollo Lunar lander, the kilogram of hydrogen would barely be sufficient to accelerate it to the moon's relatively low escape velocity.

Please repost this using code tags to make it easier to read your code. Enclose your code in these tags: [ code ] ..... [ /code ] (but without the spaces). Thank you.

 

[usernamess]

Welcome to PF.



Please repost this using code tags to make it easier to read your code. Enclose your code in these tags: [ code ] ..... [ /code ] (but without the spaces). Thank you.

...That's the output of my calculator. I didn't feel the need to write up code for this, but if you'd like some:

#include <stdio.h>
#include <cmath>
#include <iostream>

using namespace std;

int main() {

float c = 299792458;
float protium = 1000/ (1.67 * pow(10,-24));
float MeV = protium * 8.791;
float joules = MeV / 62415090744461;
float massLost = joules / (c * c);
float vel = sqrt(joules / (.5 * (1 - massLost)));
float lsf = (vel / c) * 100;

cout << "1kg H1 into Fe56 results in "<< 1 - massLost << " kg Fe traveling at "<< lsf<< "% the speed of light."<<endl;

}

 

[usernamess]

I stumbled upon this thread and was curious, so I took a crack at the math. I'm not a physicist so please point out if I've made any errors.

View attachment 349140
The above makes the assumptions that all gamma rays and intermediary reactions have their energy entirely absorbed in the reaction, and that the iron exhaust is at absolute zero relative to itself. Others have already commented on the infeasibility of such a design. Even so, if engines of such a design were incorporated into the Apollo Lunar lander, the kilogram of hydrogen would barely be sufficient to accelerate it to the moon's relatively low escape velocity.

Because I was curious, I ran the calculations a few more times with different fuel payloads. This is not my area of expertise; but I imagine as the velocity increases relativity becomes an important factor, so true values are lower than my calculations. I was surprised just how much energy it takes to reach appreciable fractions the speed of light, so I thought I'd share.


1 Kg H1 can impart 2640.74 m/s of Delta V on 4920 kg spacecraft:
0.000009% the speed of light
49201 Kg H1 can impart 31157924.82 m/s of Delta V on 4920 kg spacecraft:
0.103932% the speed of light
98401 Kg H1 can impart 39559931.13 m/s of Delta V on 4920 kg spacecraft:
0.131958% the speed of light
147601 Kg H1 can impart 44620503.74 m/s of Delta V on 4920 kg spacecraft:
0.148838% the speed of light
196801 Kg H1 can impart 48253342.61 m/s of Delta V on 4920 kg spacecraft:
0.160956% the speed of light
246001 Kg H1 can impart 51089265.26 m/s of Delta V on 4920 kg spacecraft:
0.170415% the speed of light
295201 Kg H1 can impart 53415767.20 m/s of Delta V on 4920 kg spacecraft:
0.178176% the speed of light
344401 Kg H1 can impart 55388292.57 m/s of Delta V on 4920 kg spacecraft:
0.184755% the speed of light
393601 Kg H1 can impart 57100466.85 m/s of Delta V on 4920 kg spacecraft:
0.190467% the speed of light
442801 Kg H1 can impart 58613071.88 m/s of Delta V on 4920 kg spacecraft:
0.195512% the speed of light
492000 Kg H1 can impart 59967785.79 m/s of Delta V on 4920 kg spacecraft:
0.200031% the speed of light

I used the dry mass of the Apollo LM, which barely held 2 crew for less than 3 days, and made no correction for the mass of the hypothetically perfectly efficient protium engines nor the infrastructure to store the fuel.

Fuel mass of 100x the dry mass *barely* reaches 0.2% C. At 1000x, 0.299%. Really drives home how ridiculously hard interstellar travel is.

 

[Astronuc]

Because I was curious, I ran the calculations a few more times with different fuel payloads. This is not my area of expertise; but I imagine as the velocity increases relativity becomes an important factor, so true values are lower than my calculations. I was surprised just how much energy it takes to reach appreciable fractions the speed of light, so I thought I'd share.

One needs to include the mass of the tanks that are used to confine the liquid hydrogen.

 

[usernamess]

One needs to include the mass of the tanks that are used to confine the liquid hydrogen.

The rocket equation is what governs the maximum change in velocity of a self-propelled spacecraft.

Δv is dependent upon the velocity of the engine's exhaust, and the ratio of the wet mass (ship + fuel) to the dry mass (ship with no fuel). The numbers themselves are irrelevant, what matters is how much of the ship's mass is fuel.

Suppose a craft derived its energy from fusion— converting Hydrogen into Iron…

If it started with a kilo of Hydrogen, how much mass of Iron would it have at the end and how much energy would be extracted?

I read the original question asked as: If a futuristic spacecraft could extract all the energy available from fusion and use it for thrust, how would that go? I thought it was an interesting question to ponder, and no one else seemed to have that interpretation - and the answer surprised me. To answer it, you need to know the engine's exhaust velocity, and what proportion of the spacecraft's total mass is purely fuel used for thrust. To calculate the impulse of this hypothetical fusion engine, I made assumptions that are impossible in real life - look no further than 100% efficiency. The mass of the fuel tanks is a part of the dry mass, so when I was calculating with larger and larger fuel loads, it could be assumed that less and less of the dry mass was available as payload. I thought using an example of a very small spacecraft most people are familiar with would be easiest to visualize; but any spacecraft with that same mass ratio and exhaust velocity will have the same Δv.

 

[Astronuc]

Δv is dependent upon the velocity of the engine's exhaust, and the ratio of the wet mass (ship + fuel) to the dry mass (ship with no fuel). The numbers themselves are irrelevant, what matters is how much of the ship's mass is fuel.

While true in an ideal case, the value assigned to the 'numbers' are relevant, and each must represent a physically reality. The dry mass of the spacecraft excluding the propellant must include not only the payload, but the propellant storage systems and mass of the power conversion system. The Isp must be realistic (obtainable). The devil is in the details.

One used an exhaust velocity of ~1.3E7 m/s or ~0.04336 c. That does not seem realistic. The mass of the propulsion system is a function of Isp. Accelerating ~ 1kg of Fe will require quite a distance, and it would be impractical to convert H to Fe, only to accelerate the Fe. One might wish to calculate the distance required to accelerate 1 kg to 1.3 E7 m/s at g (=9.81 m/s²), 10g, 100g, . . . . , and determine the forces on whatever system is used to accelerate the mass. One may wish to consider relativistic effects.

I'm quite familiar with the 'rocket' equation.
https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/ideal-rocket-equation/

In a fusion based system, one will likely discover that mass(dry) >> mass(propellant), and so one's $\frac{m_0}{m_f}$ ~ 1, or the ln(~1) ~ very small.

 

[usernamess]

While true in an ideal case, the value assigned to the 'numbers' are relevant, and each must represent a physically reality. The dry mass of the spacecraft excluding the propellant must include not only the payload, but the propellant storage systems and mass of the power conversion system. The Isp must be realistic (obtainable). The devil is in the details.

One used an exhaust velocity of ~1.3E7 m/s or ~0.04336 c. That does not seem realistic. The mass of the propulsion system is a function of Isp. Accelerating ~ 1kg of Fe will require quite a distance, and it would be impractical to convert H to Fe, only to accelerate the Fe. One might wish to calculate the distance required to accelerate 1 kg to 1.3 E7 m/s at g (=9.81 m/s²), 10g, 100g, . . . . , and determine the forces on whatever system is used to accelerate the mass. One may wish to consider relativistic effects.

I encourage you to run the numbers yourself. I agree, they feel wrong: that's why I posted the equations, steps, and assumption I took.

I'm quite familiar with the 'rocket' equation.
https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/ideal-rocket-equation/

In a fusion based system, one will likely discover that mass(dry) >> mass(propellant), and so one's $\frac{m_0}{m_f}$ ~ 1, or the ln(~1) ~ very small.

Newton observed that every action has an equal and opposite reaction¹: as a result, in order to accelerate under your own power, you must leave some mass behind. As I'm sure you recall, the natural logarithm of 1 is 0; from this fact and your familiarity with the rocket equation, I'm confident you can understand why the mass of the fuel must be substantial relative the to mass of the craft (mass_wet >> mass_dry), for any high-velocity design.

My calculations describe the upper bound of a purely fusion-based propulsion system, not an implementable design.

 

[usernamess]

1 Kg H1 can impart 2640.74 m/s of Delta V on 4920 kg spacecraft:
0.000009% the speed of light
49201 Kg H1 can impart 31157924.82 m/s of Delta V on 4920 kg spacecraft:
0.103932% the speed of light
98401 Kg H1 can impart 39559931.13 m/s of Delta V on 4920 kg spacecraft:
0.131958% the speed of light
147601 Kg H1 can impart 44620503.74 m/s of Delta V on 4920 kg spacecraft:
0.148838% the speed of light
196801 Kg H1 can impart 48253342.61 m/s of Delta V on 4920 kg spacecraft:
0.160956% the speed of light
246001 Kg H1 can impart 51089265.26 m/s of Delta V on 4920 kg spacecraft:
0.170415% the speed of light
295201 Kg H1 can impart 53415767.20 m/s of Delta V on 4920 kg spacecraft:
0.178176% the speed of light
344401 Kg H1 can impart 55388292.57 m/s of Delta V on 4920 kg spacecraft:
0.184755% the speed of light
393601 Kg H1 can impart 57100466.85 m/s of Delta V on 4920 kg spacecraft:
0.190467% the speed of light
442801 Kg H1 can impart 58613071.88 m/s of Delta V on 4920 kg spacecraft:
0.195512% the speed of light
492000 Kg H1 can impart 59967785.79 m/s of Delta V on 4920 kg spacecraft:
0.200031% the speed of light

Some egg on my face: The percentages listed are fractions, not percents, as I forgot to multiply by 100. The velocity expressed in meters per second is unchanged. Here is the data with the correct units.


1 Kg H1 can impart 2640.74 m/s of Delta V on 4920 kg spacecraft:
0.000881% the speed of light
4921 Kg H1 can impart 9007911.16 m/s of Delta V on 4920 kg spacecraft:
3.004716% the speed of light
9841 Kg H1 can impart 14275988.87 m/s of Delta V on 4920 kg spacecraft:
4.761957% the speed of light
14761 Kg H1 can impart 18013841.69 m/s of Delta V on 4920 kg spacecraft:
6.008771% the speed of light
19681 Kg H1 can impart 20913184.24 m/s of Delta V on 4920 kg spacecraft:
6.975887% the speed of light
24601 Kg H1 can impart 23282139.45 m/s of Delta V on 4920 kg spacecraft:
7.766086% the speed of light
29521 Kg H1 can impart 25285074.09 m/s of Delta V on 4920 kg spacecraft:
8.434193% the speed of light
34441 Kg H1 can impart 27020102.30 m/s of Delta V on 4920 kg spacecraft:
9.012936% the speed of light
39361 Kg H1 can impart 28550510.56 m/s of Delta V on 4920 kg spacecraft:
9.523425% the speed of light
44281 Kg H1 can impart 29919510.87 m/s of Delta V on 4920 kg spacecraft:
9.980075% the speed of light
49200 Kg H1 can impart 31157684.73 m/s of Delta V on 4920 kg spacecraft:
10.3931% the speed of light

Since relativity becomes a dominant factor at roughly 10% the speed of light, I have truncated the data to that $\Delta$v value. It is worth noting however, that $\Delta$v is used to accelerate and as well as slow down, so a craft at these speeds would need to reserve half its $\Delta$v for the insertion burn.

 

[mfb]

With the assumptions made (which are pretty unrealistic), these new numbers are plausible.

Relativistic effects are still pretty small at 10% c.

 

[Devin-M]

Not bad, a trip to Alpha Centauri in a human lifespan with 1/10th the takeoff mass of a Falcon 9…

 

[mfb]

Good luck living in a 4900 kg capsule for 40 years. We can't slow down either.

Then replace H1 to iron fusion by DD fusion, add some more realistic assumption for the exhaust generation, increase the spacecraft mass and suddenly we look at millions of tonnes to reach 5% the speed of light or so.

 

[usernamess]

Not bad, a trip to Alpha Centauri in a human lifespan with 1/10th the takeoff mass of a Falcon 9…

There's a LOT of simplifications and unrealistic assumptions made in this thread. A significant number of the more impossible assumptions were made by me. I took a little time to make some graphs to (hopefully) make the topic a little clearer.

These graphs shows the maximum possible change in velocity for a self-propelled spacecraft at revalistic velocities. Reserving fuel to slow down requires calculating in terms of rapidity, which I felt would make the answer too technical for this context. The take away is that these graphs are valid if and only if the acceleration is unidirectional, using the following formula:

$$ \Delta v = c ~tanh~\frac {Velocity_{exhaust}}{c} ~ln ~\frac {mass_{wet}}{mass_{dry}}$$

Recall that c = 299,792,458 m/s, and $mass_{dry}$ is everything that is not the fuel. Engines, life support, shielding, fuel storage infrastructure, etc. Therefore, $mass_{wet} = mass_{dry} + mass_{fuel}$.

This first graph shows what I think is the most "realistic" data. Rocket exhaust between 0 and 0.5 C, with a mass fraction between 0 and 100. Notice the asymptotic behavior of the graph in two dimensions:

$\Delta$v depends on the natural logarithm of the mass fraction ($ln \frac{mass_{wet}}{mass_{dry}}$) so high velocity space travel requires a mass fraction of no less than $e$, or your craft can never accelerate up to the exhaust velocity.

$\Delta$v also is a function of hyperbolic tangent, which conveniently handles the pesky can't-travel-faster-than-light issues. As your mass fraction approaches infinity and your exhaust approaches c, your maximum $\Delta$v approaches c as well. This helps to underline why the graph is only valid for unidirectional acceleration: it's assumptions make it impossible to carry over c $\Delta$v.

Alright, but what if you really wanna go fast. Increasing the mass fraction, as show in the figure above, allows for much higher velocity travel with lower exhaust velocity. The downside is very little usable payload:

$$ \frac {mass_{wet}}{mass_{dry}} = \frac {mass_{dry} + mass_{fuel}}{mass_{dry}}$$

At a mass fraction of 1000, every 1kg of ship needs 999kg of fuel, and again: that 1kg must include the mass of whatever stores the 999kg of fuel, as well as every other component of the ship.

Given the impracticality of an ever-increasing mass fraction, what about increasing the exhaust velocity? The figure above shows the relation if the engines' exhaust velocity can be scaled up to .999c. Increasing impulse is substantially more effective than increasing the mass fraction. This implies that efforts made towards interstellar travel should focus more on increases in engine impulse than increases in fuel capacity.

This post has strayed a little far from the original question asked, so to bring it back to the relevant context, the purely fusion powered engine discussed has an impulse of ~ 0.14c If it is 100% efficient at turning protium into iron. Hopefully the cacophony of other posters balking at that demonstrates the folly of assuming that such an engine could be constructed.
If you care less about how possible it is to build, the figure below shows the maximum unidirectional $\Delta$v of a spacecraft powered by 100% efficient H₁ to Fe₅₆ fusion engines.

Again, these are theoretical values, and make no claims on the practicality or possibility of building such a spacecraft.
If you would like to learn about the various elements that go into designing a spacecraft, I strongly recommend The New SMAD. The book is a little pricey, but it's an amazing resource used throughout the industry. The section on spacecraft and payload design is about 430 pages long and walks you through the various elements and considerations.
If you'd like to learn about more realistic orbital maneuvers, the way current spacecraft move in space, I've found both Orbital Mechanics for Engineering Students and Interplanetary Astrodynamics to be great books. These are textbooks used to teach the material to undergraduate students and do a fantastic job covering the basics in an entertaining and (relatively) easy to follow manner. Getting where you're trying to go in space is substantially more complicated than pointing where you want to go and hitting the gas. That said, they assume a strong background in physics and calculus, as well as at least a moderate ability in programming.

 

[DaveE]

I took a little time to make some graphs

What program did you use for these plots?
(Sorry, off topic...)

 

[Devin-M]

Do the plots specifically assume hydrogen to iron fusion, or do they more generally show the relationship between mass fraction and exhaust velocity, regardless of energy source?

 

[usernamess]

What program did you use for these plots?
(Sorry, off topic...)

MATLab. Have you ever heard people joke how Python is a calculator masquerading as a language? MATLab quite literally is. Very powerful tool for data analysis and engineering, and (relatively) easy to use.

However it's runtime compiled and not very computationally efficient, so its usefulness comes down to the question: will this take more time to code or to run?
Need 3 dozen graphs and charts generated from 20 odd thousand multidimensional dataset? MATLab is the best tool you could use. Need to calculate the next largest prime number? Learn C.

This is all it took to calculate and plot the 2 graphs showing the H₁ to Fe₅₆ velocity.

 

[usernamess]

Do the plots specifically assume hydrogen to iron fusion, or do they more generally show the relationship between mass fraction and exhaust velocity, regardless of energy source?

I've edited the post to add information clarifying this. Most of the graphs show the relationship regardless of energy source. The 2 graphs I just added show specifically the craft powered by H₁ to Fe₅₆ perfectly efficient fusion engines.

 

[Devin-M]

Does that mean each 1kg of Hydrogen to Iron fusion generates enough energy essentially to accelerate 992 grams of Iron "out the back" at 0.043c (lets say on a static test stand)?

 

[usernamess]

Does that mean each 1kg of Hydrogen to Iron fusion generates enough energy essentially to accelerate 992 grams of Iron "out the back" at 0.043c (lets say on a static test stand)?

This is not my area of expertise. If I have made no errors in the math, it means that each 1 kg of H₁ becomes 999 g of Fe₅₆ and releases 84.33 Terajoules of energy.

IF 100% of that released energy is kinetic energy, the iron would be traveling at 0.043c. Keep in mind that heat is kinetic energy, and temperature is just a convenient way to say "average kinetic energy of the atoms in a material". This means that in order to reach 0.043c as exhaust, the iron atoms must be in a reference frame where they are completely stationary to each other, i.e. at absolute zero. If that sounds ridiculous, it is.
AND this ignores that in fusion, some energy is released as gamma radiation. This hypothetical engine takes all of that radiation and somehow uses it to accelerate and cool the reaction. It's beyond ridiculous to think it could ever be built.

But it is fun to think about.

 

[Devin-M]

If I have made no errors in the math, it means that each 1 kg of H1 becomes 999 g of Fe56 and releases 84.33 Terajoules of energy.

Earlier in the thread someone mentioned 8 grams is turned to energy.

The mass of the end product of converting 1 kg of heavy hydrogen to iron in nuclear fusion would be 992 grams of iron.

 

[usernamess]

Earlier in the thread someone mentioned 8 grams is turned to energy.

Yes, I read through that:

I don't disagree which is why I referenced starting with deuterium rather than hydrogen-1.

Getting a bit further into answering the original question.

Nuclear fusion generates 580,000,000 MJ of energy from a kilogram of heavy hydrogen (i.e. Hydrogen-2 and Hydrogen-3) by fusing it into Helium. In contrast, hydrogen fuel cell produces 142 MJ/kg. Natural gas (i.e. basically methane) produces 54 MJ/kg and gasoline produces 46 MJ/kg.

MJ = megajoule and kg = kilogram and 1 Joule [J] = 1 Watt-second [Ws] = 1 V A s (volt-ampere-second)= 1 N m (Newton meter) = 1 kg m²s⁻² (kilogram-meter squared per second squared).

In theory, if it went from heavy hydrogen straight to iron, it would generate 725,000,000 MJ of energy from a kilogram of heavy hydrogen (i.e. 7.25 * 10⁸ MJ).

The energy that would be created by converting 1 kg of matter directly to pure energy (via the E=mc² conversion) would be 89,875,517,878.0128 MJ (i.e. roughly 9 * 10¹⁰.

So, the energy produced in the fusion reaction would be equivalent to a mass almost exactly 8 grams converted to pure energy.

The mass of the end product of converting 1 kg of heavy hydrogen to iron in nuclear fusion would be 992 grams of iron.

These calculations have a roughly 1% uncertainty.

To start, the above calculations are based on deuterium, not protium. Protium makes up 99.98% of all hydrogen in the universe. However, protium fusion has an incredible low cross section (it happens very, very rarely), so realistic fusion designs don't consider it.

Secondly, the numbers begin by referencing a Quora post answering a different question.

I chose to do the math from scratch and show all the steps and assumptions I took. It is possible I've made an error, and I encourage others to check the math for themselves to see if I have.

 

[Devin-M]

(Edited to fix typos) I'm definitely no expert but here's my crack at it.

Deuteron = 1875.61 MeV
per nucleon= 937.8

Iron 56 has roughly 7.5Mev more binding energy per nucleon than a deuteron or 930.3MeV per nucleon.

930.3/937.8 = 0.992002

In other words a kilo of deuteron to iron-56 fusion releases 8 grams worth of energy.

 

[Devin-M]

releases 84.33 Terajoules of energy.

According to my calculator, 8 grams works out to about 719TJ per kilo of Deuterium.

 

[Devin-M]

It makes me wonder whether it would be better to accelerate all the iron, or simply dump some overboard and accelerate the rest to a higher velocity.

 

[usernamess]

(Edited to fix typos) I'm definitely no expert but here's my crack at it.

Deuteron = 1875.61 MeV
per nucleon= 937.8

Iron 56 has roughly 7.5Mev more binding energy per nucleon than a deuteron or 930.3MeV per nucleon.

930.3/937.8 = 0.992002

In other words a kilo of deuteron to iron-56 fusion releases 8 grams worth of energy.

Nuclear physics is not my area of expertise: It's possible I'm calculating incorrectly. Could you walk me through your process? This is the process I took:

I used the isotopic mass of protium and deuterium from this table, and converted them to grams using Google's built in unit conversion tool.

 

[Devin-M]

Could you walk me through your process?

I found the ratio of the mass per nucleon of Iron / Deuterium = 0.992, so I reasoned 1 kg of Deuterium produces 992 grams of Iron-56.

Deuterium mass per nucleon= 937.8 MeV/c^2

Iron 56 has 7.5MeV more binding energy per nucleon

Deuterium mass per nucleon 937.8 MeV/c^2 - Binding Energy 7.5MeV/c^2 = 930.3Mev/c^2 Mass Per Nucleon Iron 56

Ratio: 930.3Mev/c^2 / 937.8 MeV/c^2 = 0.992

Therefore 1 kg Deuterium Converts to 0.992 kg Iron-56 with an 8 gram energy yield.

 

[usernamess]

I found the ratio of the mass per nucleon of Iron / Deuterium = 0.992, so I reasoned 1 kg of Deuterium produces 992 grams of Iron-56.

Deuterium mass per nucleon= 937.8 MeV/c^2

Iron 56 has 7.5MeV more binding energy per nucleon

Deuterium mass per nucleon 937.8 MeV/c^2 - Binding Energy 7.5MeV/c^2 = 930.3Mev/c^2 Mass Per Nucleon Iron 56

Ratio: 930.3Mev/c^2 / 937.8 MeV/c^2 = 0.992

Therefore 1 kg Deuterium Converts to 0.992 kg Iron-56 with an 8 gram energy yield.

Thank you. I had fat fingered an extra '4' in the conversion from MeV to Joules. Our numbers still disagree by a lot; I'm double checking all my other values now to see if I can find another error. In the meantime, your fusion rocket just jumped from .0433c to .1375c now that I'm converting the energy properly.

 

[usernamess]

I had fat fingered an extra '4' into the conversion from MeV to J in my original calculations, as highlighted in the screenshot of my calculations below:

Correcting that mistake yields the following:

Since I used Joules to calculate both mass lost and the velocity, the conversion error had a profound impact on the final numbers I generated. As I stated, nuclear physics isn't my forte: I've been unable to find another source of error in my math, however I have no formal education in the subject and am not confident none exist. I would appreciate anyone deeply knowledgeable in the subject reviewing the math and identifying any further errors made.

Deuterium mass per nucleon= 937.8 MeV/c^2

Iron 56 has 7.5MeV more binding energy per nucleon

Deuterium mass per nucleon 937.8 MeV/c^2 - Binding Energy 7.5MeV/c^2 = 930.3Mev/c^2 Mass Per Nucleon Iron 56

I greatly appreciate you walking me through your process. Converting your units to mine identified the constant I had mistyped. However, I am unsure if your formula is accurate. By my understanding, the binding energy is an energy, in units MeV/nucleon. In your equation you seem to be subtracting it from a mass (units MeV/c^2) without first converting it. Am I misunderstanding your work?

 

[Devin-M]

However, I am unsure if your formula is accurate. By my understanding, the binding energy is an energy, in units MeV/nucleon. In your equation you seem to be subtracting it from a mass (units MeV/c^2) without first converting it. Am I misunderstanding your work?

As I understand it, the (approximately) 7.5MeV more binding energy the iron 56 has per nucleon than the deuteron is the amount of mass/energy you would have to add to the Iron-56 to break it apart into deuterium. So you could either start with the Iron-56 mass per nucleon, then add 7.5MeV to get the Deuteron mass per nucleon, or you could start with the Deuteron mass per nucleon, and subtract the 7.5MeV to get the Iron 56 mass per nucleon.

 

[Devin-M]

I am curious what the rocket equation says about dumping say 9/10ths of the iron and using the all energy to accelerate the remaining 1/10th to a higher velocity. Wouldn’t that lighten the craft while giving higher specific impulse, resulting in more top speed?

 

[usernamess]

As I understand it, the (approximately) 7.5MeV more binding energy the iron 56 has per nucleon than the deuteron is the amount of mass/energy you would have to add to the Iron-56 to break it apart into deuterium. So you could either start with the Iron-56 mass per nucleon, then add 7.5MeV to get the Deuteron mass per nucleon, or you could start with the Deuteron mass per nucleon, and subtract the 7.5MeV to get the Iron 56 mass per nucleon.

I understand the underlying logic you presented: although thank you for clarifying further. My concern is with the units used in your equation:

Deuterium mass per nucleon= 937.8 MeV/c^2

Iron 56 has 7.5MeV more binding energy per nucleon

Deuterium mass per nucleon 937.8 MeV/c^2 - Binding Energy 7.5MeV/c^2 = 930.3Mev/c^2 Mass Per Nucleon Iron 56

I'm an engineer: broadly speaking, I prefer to use SI base units when thinking through problems.

$$ 1 J = \frac{kg * m^2}{s^2} $$
$$ 1 J = 6241509074461 MeV $$
$$ c = 299,792,458 \frac{m}{s} $$
$$ \frac{J}{c^2} = \frac{\frac{kg * m^2}{s^2}}{\frac{8.98755179 *10^{16} m^2 } {s^2}} = \frac{kg}{8.98755179 * 10^{16}}$$
$$ 937.8 \frac{MeV}{c^2} * \frac{1 J}{6241509074461 MeV} = 1.502521247365143* 10^{-10} \frac{J}{c^2} = 1.671780350102579 * 10^{-27} kg $$
$$ 7.5 MeV * \frac{1 J}{6241509074461 MeV} = 1.201632475499954* 10 ^{-12} J $$

With my conversions, it appears to me that you're subtracting an energy (1.201632475499954e-12 J or 7.5 MeV) from a mass (1.671780350102579e-27 kg or 937.8 $\frac{MeV}{c^2}$). Am I misunderstanding something?

 

[usernamess]

I am curious what the rocket equation says about dumping say 9/10ths of the iron and using the all energy to accelerate the remaining 1/10th to a higher velocity. Wouldn’t that lighten the craft while giving higher specific impulse, resulting in more top speed?

No. You're trying to think of engineering work-arounds to the laws of physics. At the end of the day, the energy has to come from somewhere and go somewhere.

Think about how you'd "dump" the iron. In order for it to leave your spacecraft, you'd have to impart some acceleration to it, and the spacecraft would have an equal and opposite acceleration as governed by Newton's Laws of Motion. This is all the "engine" we've been discussing does; take the energy from fusion and use it to accelerate the resultant matter out the back. If you don't dump all the iron and hold onto some of it, you could get higher velocity exhaust, but overall a lower speed because you've just turned some of your $mass_{wet}$ into $mass_{dry}$.

 

[Devin-M]

With my conversions, it appears to me that you're subtracting an energy (1.201632475499954e-12 J or 7.5 MeV) from a mass (1.671780350102579e-27 kg or 937.8 MeVc2). Am I misunderstanding something?

In those units I believe you can express it either as energy (MeV) or mass (MeV/c^2) via mass energy equivalence principle.

 

[Devin-M]

Think about how you'd "dump" the iron. In order for it to leave your spacecraft, you'd have to impart some acceleration to it

Not necessarily, to dump a spent rocket stage you don’t have to impart an acceleration on the spent stage.

 

[usernamess]

Not necessarily, to dump a spent rocket stage you don’t have to impart an acceleration on the spent stage.

You do, actually. Explosive bolts are the common mechanism of use. You could just fire the second stage engines to accelerate away from the spent first stage, but the second stage exhaust can and will damage the first stage. At best this adds high velocity shrapnel to the orbit where the stages separated: at worst it blows up the second stage.

A stage separation differs from dumping spent fuel. Think of a car dragging a trailer filled with gasoline that crossfeeds to the vehicle. At first the car is very slow because $(car + gas) + (trailer + gas)$ is very heavy. After some time $t$, the gas in the trailer is used up, so you unhook it. The car still has the same engine producing the same power, but now it only weighs $car + gas_{\text{inside car}}$. In this example the engines output is a constant, $v_e$.

Stage 1 = $\Delta v = v_e * ln \frac{car + gas_{\text{inside car}} + trailer + gas_{\text{inside trailer}}}{car + gas_{\text{inside car}} + trailer}$

Stage 2 = $\Delta v = v_e * ln \frac{car + gas_{\text{inside car}}}{car}$

Replace "car" with "Space Shuttle" and "trailer" with "external fuel tank" for a real-life example.

Note in the rocket equation, the masses are called $m_O$ and $m_f$, not $mass_{wet}$ and $mass_{dry}$. This is $mass_{original}$ and $mass_{final}$, which is only equal to wet and dry if all the propellant is used. For a multi-stage rocket, the final mass of the first stage includes the initial mass of all subsequent stages, as well as the mass of the engines and fuel tanks of the first stage.

 

[usernamess]

In those units I believe you can express it either as energy (MeV) or mass (MeV/c^2) via mass energy equivalence principle.

Are you saying that $7.5 \frac{MeV}{nucleon} == 7.5 \frac{MeV}{c^2}$?

As I've said previously, this is not my area of expertise, so I'll let someone more knowledgeable than us chime in to answer. That being said, I strongly believe you must convert the energy to mass (or vice versa) before subtracting.

 

[Devin-M]

Suppose we have 2 skaters initially at rest, one pushes the other, so they go in opposite directions.

One skater is the craft (100kg), the other skater is the propellant (either 1kg or 0.1kg)

After the push, in both cases the total combined kinetic energy of the craft and propellant is the same (100J).

Does the craft move faster if the propellant is 1kg or 0.1kg?

I found that the craft picks up more speed if the propellant is 0.1kg.

Craft velocity is 0.15m/s with the 1kg propellant, and craft velocity is 0.42m/s with the 0.1kg propllant

a=v_craft_m/s
b=propellant_mass_kg
c=propellant_velocity_m/s
d=craft_mass_kg

a=(bc)/d

e=total_kinetic energy_J

e=((1/2)*b*c^2)+((1/2)*d*a^2)