B Energy production that converts Hydrogen to Iron?

[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)

 

[PeterDonis]

Am I misunderstanding something?

Yes, you are. All of the terms in his subtraction have units MeV/c^2, so they're all nominally masses--but the key point is that the units are all the same, so the subtraction is fine.

 

[usernamess]

Yes, you are. All of the terms in his subtraction have units MeV/c^2, so they're all nominally masses--but the key point is that the units are all the same, so the subtraction is fine.

Interesting. Do you mind sharing a source for this? I don't normally work with electronvolts, but all of my references list it exclusively as an energy. I've been unable to find a source that indicates $1 MeV == 1\frac{MeV}{c^2}$, I'd love to learn more.

 

[PeterDonis]

I've been unable to find a source that indicates $1 MeV == 1\frac{MeV}{c^2}$,

That's not what I said. Go back and read what you quoted from my post again, carefully.

 

[PeterDonis]

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

Mass-energy equivalence does not mean you have to measure mass and energy in the same units. For example, kilograms are not the same units as Joules.

In particle physics, eV, keV, MeV, GeV, TeV, etc. units are convenient ones, and if you implicitly assume that $c = 1$, you can put a $/ c^2$ on any of the units without changing any numerical values. That's more or less what particle physicists do, but you have to be clear about what they are doing. They are not saying that mass units and energy units must be the same because of mass-energy equivalence. They are implicitly assuming that $c = 1$ in order to be able to not have to care about whether the units they are using are "mass" units or "energy" units. But that only works if you assume $c = 1$. If you don't assume $c = 1$, then you cannot say that, for example, MeV and MeV / c^2 are the same units, because if you don't assume $c = 1$, they're not.

In your actual equation, you used MeV / c^2 everywhere, which is fine and doesn't require you to say whether you are assuming $c = 1$ or not. But it also has nothing to do with mass-energy equivalence.

 

[usernamess]

That's not what I said. Go back and read what you quoted from my post again, carefully.

Yes, you are. All of the terms in his subtraction have units MeV/c^2, so they're all nominally masses--but the key point is that the units are all the same, so the subtraction is fine.

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.

In order for this equation to be accurate, $7.5 MeV == 7.5 \frac{MeV}{c^2}$. Every text I have thought to check has implied this relationship to be untrue. Could you please draw my attention to the error I am making?

 

[usernamess]

Mass-energy equivalence does not mean you have to measure mass and energy in the same units. For example, kilograms are not the same units as Joules.

In particle physics, eV, keV, MeV, GeV, TeV, etc. units are convenient ones, and if you implicitly assume that $c = 1$, you can put a $/ c^2$ on any of the units without changing any numerical values. That's more or less what particle physicists do, but you have to be clear about what they are doing. They are not saying that mass units and energy units must be the same because of mass-energy equivalence. They are implicitly assuming that $c = 1$ in order to be able to not have to care about whether the units they are using are "mass" units or "energy" units. But that only works if you assume $c = 1$. If you don't assume $c = 1$, then you cannot say that, for example, MeV and MeV / c^2 are the same units, because if you don't assume $c = 1$, they're not.

In your actual equation, you used MeV / c^2 everywhere, which is fine and doesn't require you to say whether you are assuming $c = 1$ or not. But it also has nothing to do with mass-energy equivalence.

Having never been taught particle physics, this is a new concept to me. Thank you for explaining. Could you elaborate on how this convenient notation avoids changing numerical values? I ran a dimensional analysis on the units and the conversion came out different.

 

[PeterDonis]

In order for this equation to be accurate, $7.5 MeV == 7.5 \frac{MeV}{c^2}$.

No, becase 7.5 MeV (as opposed to 7.5 MeV / c^2) does not appear in the actual equation. It only appears in the accompanying text. The actual equation where he does the subtraction has MeV / c^2 for every term, so it is correct.

Could you elaborate on how this convenient notation avoids changing numerical values?

It's not a "notation", it's a choice of units. If you choose units so that $c = 1$, then dividing, say, 7.5 MeV by $c^2$ gives 7.5 MeV / c^2, because $c^2 = 1$ if $c = 1$. In other words, if $c = 1$, then mass units are numerically the same as energy units. This is a common choice of units for convenience not just in particle physics but in relativity.

 

[Devin-M]

So would the craft get more delta v if it used all the energy to accelerate all the iron, or if it used the same total energy to accelerate 1/10th the iron with much higher exhaust velocity while continuously dumping 9/10ths of the produced iron overboard with low (or no) impulse?

In one scenario you’d have one exhaust stream with an exhaust velocity determined by the total iron mass and total available energy.

In the other scenario you’d have a high energy and low energy iron exhaust stream. The exhaust velocity of the high energy stream would be determined by the craft having the same total available energy as the 1st scenario. Since it’d be accelerating less propellant mass with the same amount of energy this option has higher exhaust velocity, with that high energy exhaust velocity being determined by the ratio of how much iron you dump continuously overboard as its produced at the lowest possible energy.

 

[Nugatory]

So would the craft get more delta v if it used all the energy to accelerate all the iron, or if it used the same total energy to accelerate 1/10th the iron with much higher exhaust velocity while continuously dumping 9/10ths of the produced iron overboard with low (or no) impulse?

Go back to your example of the two skaters in post #50. The math here is particularly easy because we're doing it with one shove: we have two masses stuck together, then we take some amount of add some amount of energy and the two masses are moving in opposite directions. We have two equations, one from conservation of momentum ($m_1v_1=-m_2v_2$) and one from energy. With these it's just a bit of algebra that you should do for yourself to get the basic principle of small masses at high speed or high masses at low speeds using classical ($v\ll c$) physics.
You should go through this exercise before you seriously take on the harder problem of a rocket in which the reaction mass is being released continuously instead of in one shove.

 

[Devin-M]

Go back to your example of the two skaters in post #50. The math here is particularly easy because we're doing it with one shove: we have two masses stuck together, then we take some amount of add some amount of energy and the two masses are moving in opposite directions. We have two equations, one from conservation of momentum (m1v1=−m2v2) and one from energy. With these it's just a bit of algebra that you should do for yourself to get the basic principle of small masses at high speed or high masses at low speeds using classical (v≪c) physics.
You should go through this exercise before you seriously take on the harder problem of a rocket in which the reaction mass is being released continuously instead of in one shove.

Assuming I did the math in post #50 right I think it shows the craft getting more delta v by accelerating 1/10th of the iron with the same total energy.