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Liquid CO2 or LCO2, has a bulk density of 1,000kg/m^3 to 1,200kg/m^3. That means it's about as dense as seawater, per unit volume. Oddly enough, just like H2O, every 2CO2 (2CO + 1O2) molecules also contain the constituent atomic components for complete combustion in a rocket engine.
I posted this before in a topic about Carbon Monoxide rocket engines, but here it is again since we're talking about shipping large quantities of materials as rocket fuel:
NIST - Room Temperature Conversion of CO2 to CO: A New Way to Synthesize Hydrocarbons
Illustration of a novel room-temperature process to remove carbon dioxide (CO2) by converting the molecule into carbon monoxide (CO). Instead of using heat, the nanoscale method relies on the energy from surface plasmons (violet hue) that are excited when a beam of electrons (vertical beam) strikes aluminum nanoparticles resting on graphite, a crystalline form of carbon. In the presence of the graphite, aided by the energy derived from the plasmons, carbon dioxide molecules (black dot bonded to two red dots) are converted to carbon monoxide (black dot bonded to one red dot. The hole under the violet sphere represents the graphite etched away during the chemical reaction CO2 + C = 2CO.
...
Although the LSP reaction generates a poisonous gas — carbon monoxide — the gas readily combines with hydrogen to produce essential hydrocarbon compounds, such as methane and ethanol, that are often used in industry, said NIST researcher Renu Sharma.
This is room temperature electro-chemistry using sunlight and CO2. We only need to ship the device to Mars, which contains Aluminum and Carbon, rather than a fuel or oxidizer of any kind. We can use Moxie to produce pure Carbon black and O2 as well. Sunlight can provide our solar-pumped laser beam into the reaction chamber. That limits total complexity to stainless steel for the reactor pressure vessel, which we have already shipped in large quantities in the form of stainless steel Starship rocket carcasses, Aluminum nanoparticles, and the Carbon black coating for the Aluminum nanoparticles can come from Mars, which produces the CO. The water on Mars can then be reacted with the CO, in the water-gas shift reaction (CO + H2O ⇌ CO2 + H2), to form CO2 and H2. Carbon Monoxide is the feedstock to make any kind of hydrocarbon fuel, as well as the precursor to make any kind of plastic.
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The table produced on Sunday, and published here the same day, delivered an incorrect result...
I went back and asked ChatGPT(4) to take another look at it's work. This time it found a balance using atom counts with two glucose molecules.
Update 2023/11/06 (th) I'd like to give credit to kbd512 for suggesting the "Atom Count" method of explaining how glucose converts to methane and oxygen without loss of atoms along the way. The "secret" is to start with two glucose molecules:
This equation shows that two molecules of glucose (each with 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms) can produce four molecules of methane (each with 1 carbon and 4 hydrogens) and three molecules of oxygen (each consisting of two oxygen atoms), ensuring all the atoms from the reactants are conserved in the products.
The correct representation of the reaction in a tabular form with the final count of atoms explicitly shown would be:
+------------------------------+-----------+---------+----------+----------+
| Reactant/Product | Carbon | Hydrogen | Oxygen | Methane | Oxygen |
+------------------------------+-----------+---------+----------+----------+
| Initial Molecules | | | | | |
| Glucose | | | | | |
| (C6H12O6) x 2 | 12 | 24 | 12 | 0 | 0 |
+--------------------+--------+-----------+----------+----------+---------+
| After Reaction | | | | | |
| Methane (CH4)x4| 4 | 16 | 0 | 4 | 0 | <<<== this row does not match so I'll go back again for rework
| Oxygen (O2) x 3 | 0 | 0 | 6 | 0 | 3 |
+-------------------+---------+-----------+---------+-----------+---------+
| Final Atom Count | 12 | 24 | 12 | | |
+-------------------+---------+-----------+---------+-----------+---------+
| Molecule Count | | | | 4 | 3 |
+-------------------+---------+-----------+---------+-----------+---------+This table accurately reflects the atom economy of the process, showing that starting with two molecules of glucose can yield four molecules of methane and three molecules of oxygen, with all atoms accounted for.
Credit for this table: ChatGPT4
I appreciate the patience of all participants as we try to make our way through this learning process.
I believe the result will justify the investment of time and thought. As kbd512 pointed out in the Sunday Zoom, there are significant advantages to shipping the ingredients for the propellant to return home in the form of a solid that cannot leak or burst or corrode it's tanks or in any other way reduce chances of a successful voyage.
Energy is required to reconstitute the propellant at the turn-around point. The amount of energy to be invested is precisely equal to the amount of energy to be harvested. The amount of energy required is neither high nor low. It is what Nature requires, no more, and no less.
(th)
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In the latest round, we tried four glucose molecules to start with, and ChatGPT(4) ** thinks ** it has a balance.
By this point my head is spinning, but I'll put the result here in case someone has time to approach this with a fresh eye:
Certainly, we can try again with four molecules of glucose to make sure the carbon, hydrogen, and oxygen atoms are all fully used without any waste. Here's the atom tally starting with four glucose molecules:
Each glucose molecule (C6H12O6) contains:
6 carbon atoms
12 hydrogen atoms
6 oxygen atoms
So four glucose molecules contain:24 carbon atoms (6 C x 4)
48 hydrogen atoms (12 H x 4)
24 oxygen atoms (6 O x 4)
To find the minimum number of glucose molecules needed to create methane and oxygen with no waste, let's work through the chemistry:Methane (CH4) requires 1 carbon and 4 hydrogen atoms.
Oxygen (O2) requires 2 oxygen atoms.
With 24 carbon atoms, we can create 24 methane molecules. Since methane also requires 4 hydrogen atoms per molecule, we'll need 96 hydrogen atoms for 24 methane molecules, which we do not have; we only have 48 hydrogen atoms. This indicates that the number of methane molecules we can form is limited by hydrogen, not carbon. Since we can only use 48 hydrogen atoms to make methane, this allows us to form only 12 methane molecules (because 48 divided by 4 hydrogen atoms per methane molecule equals 12).With these 12 methane molecules accounted for, we've used 12 carbon atoms and 48 hydrogen atoms. We originally had 24 carbon atoms from four glucose molecules, so we can actually form more methane molecules. With the remaining 12 carbon atoms, we can form 12 more methane molecules (which would require 48 more hydrogen atoms).
Now, we must check if we have enough oxygen atoms to match the oxygen molecule formation:
We've used 24 carbon atoms in total to form methane, with no carbon atoms left.
We've used 48 hydrogen atoms for the first 12 methane molecules, and we can use the remaining 48 hydrogen atoms for the next 12 methane molecules.
This leaves us with 24 oxygen atoms (from the 4 molecules of glucose), which would form 12 molecules of oxygen (O2), using all oxygen atoms.
Here's the updated atom tally:+-----------------------------+---------+----------+--------+
| Reactant/Product | Carbon | Hydrogen | Oxygen |
+-----------------------------+---------+----------+--------+
| Glucose (C6H12O6) x 4 | 24 | 48 | 24 | <- Reactants
+-----------------------------+---------+----------+--------+
| Methane (CH4) x 24 | 24 | 96 | 0 | <- Products << where did 96 come from?
| Oxygen (O2) x 12 | 0 | 0 | 24 | <- Products
+-----------------------------+---------+----------+--------+
| Total Atom Count | 24 | 96 | 24 | <- Total
+-----------------------------+---------+----------+--------+This calculation shows that four glucose molecules can produce 24 methane molecules and 12 oxygen molecules, perfectly matching the atom counts with no waste.
We are not successful, once again.
(th)
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I concluded ChatGPT(4) cannot solve this problem ... I asked it to compose a Python program to solve the problem
def find_minimum_glucose_for_methane_oxygen():
# Starting with one molecule of glucose
glucose_count = 1
iteration_step = 1000
while True:
# Calculate the number of atoms for the given number of glucose molecules
carbon_atoms = glucose_count * 6
hydrogen_atoms = glucose_count * 12
oxygen_atoms = glucose_count * 6
# Calculate the maximum number of methane molecules that can be produced
methane_molecules = min(carbon_atoms, hydrogen_atoms // 4)
# Calculate the remaining hydrogen atoms after methane production
remaining_hydrogen = hydrogen_atoms - (methane_molecules * 4)
# Calculate the number of oxygen molecules that can be produced
oxygen_molecules = oxygen_atoms // 2
# Check if all atoms are used up without any waste
if remaining_hydrogen == 0 and carbon_atoms == methane_molecules and oxygen_atoms == (oxygen_molecules * 2):
# If all conditions are met, return the result
return glucose_count, methane_molecules, oxygen_molecules
# If not, increase the number of glucose molecules and try again
glucose_count += 1
# Print the progress every 1000 iterations
if glucose_count % iteration_step == 0:
print(f"Iteration: {glucose_count}")
print(f"Current glucose count: {glucose_count}, Methane: {methane_molecules}, Oxygen: {oxygen_molecules}")
response = input("Do you want to continue? (yes/no): ")
if response.lower() != 'yes':
return None# Run the function and print the result
result = find_minimum_glucose_for_methane_oxygen()
if result:
minimum_glucose, methane_produced, oxygen_produced = result
print(f"Minimum glucose molecules needed: {minimum_glucose}")
print(f"Methane molecules produced: {methane_produced}")
print(f"Oxygen molecules produced: {oxygen_produced}")
else:
print("Operation was stopped by the user.")
Update ... I asked ChatGPT to add code to stop every 1000 iterations.
Anyone in our membership can run this program, because Python runs on every OS including the latest Chromebook, which now supports Linux.
(th)
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After running the program shown in Post #29, I showed the results to ChatGPT(4) and it concluded the program would never converge.
The difference is very small, but the difference can be accounted for by producing a small amount of water as a byproduct.
This version of the Python program is intended to find a solution that produces the minimum amount of water while producing a matching quantity of methane and oxygen.
def compute_min_water_from_glucose(glucose_mass_kg):
# Molar masses in grams per mole (g/mol)
molar_mass_glucose = 180.156 # C6H12O6
molar_mass_methane = 16.04 # CH4
molar_mass_oxygen = 32.00 # O2
molar_mass_water = 18.01528 # H2O# Convert mass of glucose to moles
moles_glucose = (glucose_mass_kg * 1e3) / molar_mass_glucose# For each mole of glucose, we can produce up to 3 moles of methane.
# For each 1 mole of methane, we produce 2 moles of water and 2 moles of oxygen.
# Therefore, the reaction looks like: C6H12O6 → 3CH4 + 3O2 + 3H2O# Calculate the moles of methane, oxygen, and water produced
moles_methane = moles_glucose * 3
moles_oxygen = moles_glucose * 3
moles_water = moles_glucose * 3# Convert moles back to mass for each product
methane_mass_kg = moles_methane * molar_mass_methane / 1e3
oxygen_mass_kg = moles_oxygen * molar_mass_oxygen / 1e3
water_mass_kg = moles_water * molar_mass_water / 1e3return methane_mass_kg, oxygen_mass_kg, water_mass_kg
# Example: Compute the amount of methane, oxygen, and water produced from 1 ton of glucose
glucose_mass_kg = 1 * 1e3 # 1 ton of glucose
methane_mass_kg, oxygen_mass_kg, water_mass_kg = compute_min_water_from_glucose(glucose_mass_kg)print(f"Methane produced: {methane_mass_kg:.2f} kg")
print(f"Oxygen produced: {oxygen_mass_kg:.2f} kg")
print(f"Minimum water produced: {water_mass_kg:.2f} kg")
Here is the result of the code above:
C:\Temp>python MinimumWater.py
Methane produced: 267.10 kg
Oxygen produced: 532.87 kg
Minimum water produced: 299.99 kg
(th)
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In discussing the output of the Python program, ChatGPT(4) and I decided to try adding carbon powder to the glucose. This would allow the shipment to remain solid, and it should help to reduce the amount of water produced.
This program will attempt to find a minimum of water output when carbon is added:
def find_minimal_water_production():
glucose_count = 1 # Start with one molecule of glucose
carbon_added = 0 # Start with no additional carbon
iteration = 0while True:
# Calculate atoms from reactants
carbon_atoms = glucose_count * 6 + carbon_added
hydrogen_atoms = glucose_count * 12
oxygen_atoms = glucose_count * 6# Calculate molecules produced (using the balanced chemical reaction)
methane_molecules = min(carbon_atoms, hydrogen_atoms // 4)
water_molecules = (hydrogen_atoms - methane_molecules * 4) // 2
oxygen_molecules = min(oxygen_atoms // 2, water_molecules)# Check if we've reached the condition where water is minimal
if water_molecules <= methane_molecules // 2 and oxygen_molecules >= methane_molecules // 2:
print(f"Converged at iteration: {iteration}")
print(f"Glucose count: {glucose_count}, Additional Carbon: {carbon_added}, Methane: {methane_molecules}, Oxygen: {oxygen_molecules}, Water: {water_molecules}")
break# Increment counts for the next iteration
if water_molecules > methane_molecules // 2:
# Add more carbon to reduce water
carbon_added += 1
else:
# Add more glucose if we don't have enough reactants
glucose_count += 1
iteration += 1
if iteration % 1000 == 0:
print(f"Iteration: {iteration}")
print(f"Current glucose count: {glucose_count}, Additional Carbon: {carbon_added}, Methane: {methane_molecules}, Oxygen: {oxygen_molecules}, Water: {water_molecules}")
user_input = input("Do you want to continue? (yes/no): ").strip().lower()
if user_input != 'yes':
break# Run the function
find_minimal_water_production()
(th)
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tahanson43206,
I can answer the question without using an AI program, or even knowing much about chemistry, except how to count inputs and outputs.
C6H12O6
Produces 3CH4, 3C, 6O
Each CH4 requires 4O to completely combust
Each Glucose molecule has the Carbon for 6CH4
The Hydrogen deficit is 12. The Oxygen deficit is 18. Divide both by 6. We need a molecule with 2 Hydrogens for every 3 Oxygens to completely react the reactants with zero leftovers (assuming 100% conversion efficiency).
You need Sodium Chloride (NaCl) and water (H2O).
NaCl + 3H2O → NaClO3 + 3H2 <- For your missing Hydrogen to make Methane with your 3 spare Carbon atoms.
You need 2NaCl and 6H2O to supply the missing 12.
When you heat up Sodium Chlorate past a certain point, you get a chemical Oxygen generator (used aboard aircraft):
2NaClO3 → 2NaCl + 3O2 <- Your missing Oxygen atoms.
You need 3NaClO3 (implying 9H2O) to supply the missing Oxygen.
Does it look like we only need to ship Glucose and water to make subsequent batches of fuel (meaning the salt is reusable)?
It does to me.
If you do end up with some extra salt or water or sugar when all is said and done, what are the odds that a large group of humans can and will consume it?
Edit:
You need 1/3rd more salt to make the Oxygen than Hydrogen per batch of fuel. You end up with 6 extra Hydrogens per reaction to make enough Oxygen, which means you can make an extra 1.5X additional Methane molecules using CO2 from the Martian atmosphere and a Sabatier reactor.
Maybe there's some chemical inputs which will yield perfect outputs, but this reaction allows you to make extra Methane. Now that I think about this some more, there's a quantity of additional O2 and CH4 from the Sabatier side-reaction which will yield a "more perfect" output, relative to the required inputs.
Last edited by kbd512 (2023-11-06 14:28:14)
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For kbd512 re Glucose as a shipping method ...
Thank you for the inspiration to count atoms in working this problem...
I have discovered (with some dismay) that ChatGPT(4) has been making the same kind of elementary mistake all along.
One glucose molecule is good for three methane molecules and three oxygen molecules, with 3 carbon atoms left over.
The result is that for a ton of glucose, ? 300-400 ? kg can be sold as propellant, and there is 200 kg of carbon left over.
The lesson for me is on the painful side, but I think I've learned it.
From memory...
C6 H12 O6 yields 3 CH4 and 3 O2 with 3 C left over
Carbon counts are: 3 + 3 C >> 6 matches input
Oxygen counts are: 3*2 >> 6 matches 6 input
Hydrogen counts are: 3*4 >> 12 matches input
I don't know if a business can still make a profit with that amount of carbon left over, but it might be possible.
The competing systems would try to make propellant using on-planet resources, and that might work for Mars, but it is not an option for many sites that folks might want to visit.
This topic is now open for consideration of how to use that free carbon.
(th)
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This post offers a conversation with BARD.
It builds upon the earlier sessions with ChatGPT(4).
https://docs.google.com/document/d/1Q39 … sp=sharing
The bottom line is that only about 1/3 of a shipment of glucose to Mars (or anywhere) is available for immediate sale. However, at Mars, the local environment can supply oxygen to make the unsold methane available for sale, and the carbon can be consumed to make methane given water available.
This concept reduces the need to carry hydrogen as a gas, and it reduces the work required to make propellant from local resources.
In a worst case scenario, such as a destination (like Phobos) where no insitu resources are (likely to be available) the expedition has the capability of making propellant to return to Earth.
(th)
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A result of the analysis carried out in this topic is the discovery that while the proportion of a shipment that can be sold as rocket fuel is between 1/3 and a half, the remaining constituents may be of interest for other applications.
In particular, the free Carbon would be a feed stock for materials that need pure carbon as input, such as Graphene.
The methane is a potential feed stock for manufacture of plastic objects.
As noted previously, oxygen can be harvested from the atmosphere of Mars, if manufacture of fuel is the highest priority.
It seems to me the competition between entities using different approaches should be lively, to the benefit of customers.
(th)
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tahanson43206,
Did you read what I wrote in Post #32?
Send salt and water, in addition to sugar, so you can make LOX/LCH4 in the correct proportions to serve as rocket propellants. The entire point of your thought exercise was to make LOX/LCH4 propellant from Glucose. To do that, you need additional materials (salt and water). When last I checked, salt and water are as readily storable as sugar. All three materials are also non-toxic foodstuffs. Yes, eventually you need to source the water from Mars, along with all the other materials. However, to get started, the materials can be shipped to Mars.
Why did you immediately shift to using the excess Carbon to make Graphene when the point is to make and use rocket propellants?
The entire point of all my other suggestions was the input energy, complexity, materials sourcing, and cost.
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For kbd512 re #36
Thank you for staying with this topic!
I did not realize it at the time, but I was dealing with a mental block. ChatGPT did not help. All it did was to reinforce the mental block.
You were right in the beginning, when you pointed out there is not enough oxygen in glucose to make more than three methane molecules per glucose molecules.
However, in your posts, you provided a great deal of information, not realizing you were dealing with an unrecognized mental block.
As reported in recent posts, I have cleared the block.
I am ready to go back over the ground you have already covered, and will begin by asking why you include sodium chloride (NaCl) in your recommended shipment. What I would like for you to attempt is to see if you can restrain your outpouring of text, and just focus on one tiny issue at a time. Try to write no more than a paragraph. This will be a hardship, and I will not be surprised if you refuse, but please try.
(th)
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While waiting for kbd512 to return (he is (after all) working full time) I've decided to return to a session with ChatGPT, properly chastened.
We started over with one glucose molecule, and ChatGPT computed that four water molecules would allow for delivery of matched methane and oxygen pairs:
+-----------------------+------------------+------------------+------------------+----------------+
| Reactant/Product | Glucose (C6H12O6)| Water (H2O) | Methane (CH4) | Oxygen (O2) |
+-----------------------+------------------+------------------+------------------+----------------+
| Initial Molecules | 1 | To be determined | 0 | 0 |
| Pyrolysis Products | -1 | +2 | +2 | 0 |
| Additional Water Added| 0 | +4 | 0 | 0 |
| Electrolysis Products | 0 | -6 | 0 | +3 |
| Sabatier Reaction | 0 | 0 | +1 | 0 |
| Final Molecules | 0 | 0 | 3 | 3 |
+-----------------------+------------------+------------------+------------------+----------------+
The table above was written with molecules ... I asked for the same table using atoms.
(th)
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tahanson43206,
List of Major Processing Steps and Equipment Required:
1. Glucose reactor makes 3CH4 and 6O2, with 3C leftovers, 12H deficit 18O deficit (goal is to make and combust 6CH4, meaning full consumption of each Glucose molecule)
2. Sodium Chlorate reactor, combined with water, provides 12H and 18O, with 6H leftovers (goal is to obtain CO2 from atmosphere to make more CH4 and O2)
3. Sodium Chloride is both the starting point and the leftover from the Sodium Chlorate reactor (infinitely reusable)
4. Because the 6H leftover from Step #2 is a multiple of the required reaction, there is a multiple of the Methane fuel batch size which will completely consume all Hydrogen, while coming up a little short on Oxidizer.
5. You need 3 materials (sugar, salt, water) and 3 pieces of processing equipment (Glucose reverse fuel cell, Chlorate fuel cell, Sabatier reactor).
Detailed written explanation, which is not intended to confuse, but somehow does so anyway (my fault, not yours):
You have 3 Carbon atoms left over. You need 12 additional Hydrogen to avoid plating out your Glucose-splitting reaction chamber with pure Carbon (3 leftover Carbon atoms per Glucose molecule). Your Oxygen deficit to completely combust 6CH4 (the 3 we already have directly from the Glucose and the 3 we're about to produce using the reactions below) is 18, or 9O2.
2 Sodium Chloride (2NaCl, aka "2 salt molecules") reacted with 3 water molecules creates Sodium Chlorate and 6 free Hydrogen atoms (3H2). It splits the water. If you do that with 2NaCl molecules, then you get your 12 additional Hydrogen to combine with your 3 unreacted Carbon atoms from your Glucose-splitter, making 3 additional CH4. At this point, 100% of the Carbon from the Glucose is now Methane.
If you further heat up the 2 Sodium Chlorate (2NaClO3) created during the prior reaction, the one which produced the 3H2, they will release their 6 Oxygen atoms (3O2) merely by heating it past 300C to 400C, and then you're then left with Sodium Chloride (the "salt" that you started with). Do that 3 times and you get your missing 18 Oxygen atoms to completely combust the 6 Methane molecules (3CH4 made directly from the Glucose, 3CH4 made from the Hydrogen liberated using Sodium Chloride.
You end up with some extra H2. If you have a Sabatier reactor and some CO2, then you can make and store some extra Methane and extra Oxygen.
The salt is completely reusable (Sodium Chloride to Sodium Chlorate and back to Sodium Chloride), which means you don't have to ship more of it if you make fuel in batches.
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For kbdt12 re #39
Thanks for this reply! I'll return to it later today.
Update later: Thanks for explaining the need for salt, and how it is used!
In my earlier work on this topic, I was simply trying to come to grips with the chemistry for the shipment. Having achieved ** that ** goal, I am ready to (at least attempt to) take in the processing steps you've described.
***
Just FYI ... you may know this but some of our readers may not .... ChatGPT informed me that SpaceX burns methane a bit rich, in order to improve ISP. The ratio quoted by ChatGPT is 3.55:1 compared to the default of 4:1.
If they follow that practice on Mars, then there would be some oxygen left over after they've purchased an order, but I think any oxygen that might be available would be snapped up by other Mars customers.
All in all, I think this approach would support a viable business in competition with others attempting other approaches.
In particular, for support a customer arriving from Earth with empty tanks but a decent bank account, the option of buying propellant from a local vendor would be attractive.
The entire facility can be installed and operated remotely, so that propellant is available for delivery to customers well ahead of their launch from Earth.
(th)
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Here is the same table, showing atom counts...
+-----------------------+-----------+------------+------------+--------------+
| Reactant/Product | Carbon (C)| Hydrogen(H)| Oxygen (O) | Result |
+-----------------------+-----------+------------+------------+--------------+
| Initial Atoms | | | | |
| Glucose (C6H12O6) | 6 | 12 | 6 | |
| Water (H2O) x 6 | 0 | 12 | 6 | |
+-----------------------+-----------+------------+------------+--------------+
| Total Initial Atoms | 6 | 24 | 12 | |
+-----------------------+-----------+------------+------------+--------------+
| Pyrolysis & Processing| | | | |
| Methane (CH4) x 6 | 6 | 24 | 0 | 6 CH4 |
| Oxygen (O2) x 6 | 0 | 0 | 12 | 6 O2 |
+-----------------------+-----------+------------+------------+--------------+
| Total Final Atoms | 6 | 24 | 12 | |
+-----------------------+-----------+------------+------------+--------------+
(th)
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tahanson43206,
When I say, "you will come up a little short on Oxidizer", what I really mean is that you end up producing a little more Methane than is strictly necessary for a complete batch of fuel. You still get your 1t of Methane per 1t of Glucose, plus some extra Methane. That means you can run your rocket engine fuel-rich, or you can keep the excess Methane and use some other process to produce a minor amount of O2 for ground consumption of the Methane (to power a vehicle).
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For kbd512 re #42
Thanks for noting the possible customer preference to run "rich" !!!
If the process you've described yields what the customer wants, then things can't get much better than that!
(th)
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In this exchange with ChatGPT, I am attempting to arrive at a description of the complete process from initial input to final combustion.
ChatGPT seems to have a hard time getting to an accurate result, so I'm pasting this table with full expectation it is NOT correct.
+-----------------------+-----------+------------+------------+--------------+
| Reactant/Product | Carbon (C)| Hydrogen(H)| Oxygen (O) | Molecules |
+-----------------------+-----------+------------+------------+--------------+
| Initial Atoms | | | | |
| Glucose (C6H12O6) x 1 | 6 | 12 | 6 | 1 glucose |
| Water (H2O) x 24 | 0 | 48 | 24 | 24 water |
+-----------------------+-----------+------------+------------+--------------+
| Total Initial Atoms | 6 | 60 | 30 | |
+-----------------------+-----------+------------+------------+--------------+
| Conversion to CH4 & O2| | | | |
| Methane (CH4) x 6 | 6 | 24 | 0 | 6 methane |
| Oxygen (O2) x 6 | 0 | 0 | 12 | 6 oxygen |
+-----------------------+-----------+------------+------------+--------------+
| Total Midway Atoms | 6 | 24 | 12 | |
+-----------------------+-----------+------------+------------+--------------+
| Combustion | | | | |
| Carbon Dioxide (CO2)x6| 6 | 0 | 12 | 6 CO2 |
| Water (H2O) x 12 | 0 | 24 | 12 | 12 water |
+-----------------------+-----------+------------+------------+--------------+
| Total Final Atoms | 6 | 24 | 24 | |
+-----------------------+-----------+------------+------------+--------------+
| Waste Atoms | 0 | 0 | 0 | No waste |
+-----------------------+-----------+------------+------------+--------------+
I have learned to expect that when ChatGPT "thinks" there is no waste, it is right about 50% of the time.
And, by inspection, I see the same thing here ... initial atoms are 6-60-30, and final are 6-24-24
Is there a software that can perform this kind of analysis?
(th)
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Here is the three units of equipment.
Revisiting Chlor-Alkali Electrolyzers: from Materials to Devices
Sodium chlorate oxygen generation for fuel cell power systems
A clean and membrane-free chlor-alkali process with decoupled Cl2 and H2/NaOH production
By-Product Hydrogen to Fuel Cell Vehicles
Electricity generation by direct oxidation of glucose in mediator less microbial fuel cells
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tahanson43206,
GRFC = Glucose Reverse Fuel Cell (Glucose + heat + electricity = CH4 + CO2)
SCRFC = Sodium Chlorate Reverse Fuel Cell (2NaCl + 6H2O = 2NaClO3 + 6H2)
RHE = Resistive Heating Element (2NaClO3 + heat = 2NaCl + 3O2)
SR = Sabatier Reactor CO2 + 4H2 (400C heat + pressure + catalyst) → 1CH4 + 2H2O
MAC = Martian Atmospheric Collection of CO2 (Dyson / proper amount of suction → 3CO2)
Round 1 Reactions:
{GRFC} C6H12O6 → 3CH4 + 3CO2
{SCRFC} 4NaCl + 12H2O → 4NaClO3 + 12H2
{RHE} 4NaClO3 + heat → 4NaCl + 6O2
{SR} 3CO2 + 12H2 → 3CH4 + 6H2O
Round 1 Reaction Results:
GRFC: 3CH4 (into fuel tank) + 3CO2 (into SR)
SCRFC: 4NaClO3 + 12H2 (into SR)
RHE: 4NaCl (recycled for use on 62HO from SR in Round 2) + 6O2 (into oxidizer tank)
SR: 3CH4 (into fuel tank) + 6H2O (into SCRFC in Round 2)
Round 1 Final Molecular Products:
6CH4 + 6O2 + 4NaCl + 6H2O
Round 2:
{SCRFC} 4NaCl + 12H2O → 4NaClO3 + 12H2
{RHE} 4NaClO3 + heat → 4NaCl + 6O2
{MAC} Dyson → 3CO2
{SR} 3CO2 + 12H2 → 3CH4 + 6H2O
Round 2 Final Molecular Products:
3CH4 + 6O2 + 4NaCl + 6H2O
2 Rounds of chemical processing nets 9CH4, 12O2, 4NaCl (endlessly reusable Oxygen generator, with water and heat)
4 Oxygen required to combust 1 Methane, producing 1CO2 and 2H2O
36 Oxygen required to combust 9 Methane, but we have 24 Oxygen, so we're always 12 Oxygen short
Our "Oxygen shortness ratio" using Glucose is 2:3, however. We wind up with 2/3rds of the required O2 with only 1 round of MAC, so 1/3rd as much power to produce the proper amount of suction.
What if we eliminated the Glucose and just brought water and salt instead, sourcing all of our CO2 from the MAC?
Round 1 Reactions:
{SCRFC} 4NaCl + 12H2O → 4NaClO3 + 12H2
{RHE} 4NaClO3 + heat → 4NaCl + 6O2
{MAC} Dyson → 3CO2
{SR} 3CO2 + 12H2 → 3CH4 + 6H2O
Round 1 Final Molecular Products:
3CH4 + 6O2 + 4NaCl + 6H2O
3CH4 means 12 Oxygen atoms are required for complete combustion. 6O2 means 12 Oxygen atoms have been delivered. We still end up with 6 spare H2O molecules. Big deal. We need water for lots of other uses. O/F ratio is now in equilibrium to achieve a stoichiometric burn. If we need to run an engine "oxidizer rich", then we have to make more Methane than we can use at any given time, which means we should run the engines "fuel rich", to take advantage of the lower temperatures and longer engine life. Running engines oxidizer rich is a great way to destroy them if your super alloy steel is not up to snuff. Watch the old Soviet NK-33 videos of oxidizer rich stainless steel engines melting like butter on the test stand.
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For SpaceNut re #45
Thank you for these interesting looking links added to the Glucose topic!
It's late here so I'll come back to them tomorrow.
I just spend a knockdown dragout session with ChatGPT trying to get Python to solve the problem. I learned about advanced features of Python, and installed software I had no idea existed. We didn't find a solution that balances, but I'm not discouraged. Even the solutions that have some excess atoms are good enough for a business, because whichever atoms are excess are valuable in their own right.
Plus! kbd512 put a post into the topic that seems to show how to process the ingredients on site to make fuel for customers, which is the whole point of the exercise.
All in all, I'm happy about the way the topic is shaping up!
One wrinkle that ChatGPT came up with is similar to kbd512's suggestions, but I don't recall if kbd512 suggested the use of paraffin.
That is a hydrocarbon that is solid at room temperature, so is a candidate for shipment in our scenario. Amusingly (to me at least) ChatGPT was unable to write software to use the paraffin. Something for another day!
In the mean time, thanks ** again ** for your interesting looking post!
(th)
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For anyone ... this little Python program is what ChatGPT calls a "brute force" approach ...
It did not find a solution, but it reported a number of near misses...
# We will look for a simple integer solution within a reasonable range
max_range = 20 # This is an arbitrary range limit for demonstration purposesfor glucose in range(max_range):
for water in range(max_range):
for paraffin in range(max_range):
# Calculate the balance of each element
c_balance = glucose * 6 + paraffin * 25
h_balance = glucose * 12 + water * 2 + paraffin * 52
o_balance = glucose * 6 + water# We want to find when we have twice as many hydrogen atoms as oxygen
# And at least one carbon atom for every methane molecule we want to produce
if c_balance >= 1 and h_balance == o_balance * 2:
print(f"Possible solution found: {glucose} glucose, {water} water, {paraffin} paraffin")
print(f"Balance: C: {c_balance}, H: {h_balance}, O: {o_balance}")
break # Remove this break if you want to find all possible solutions
(th)
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My adventures with ChatGPT and Python over the past few days have resulted in a realization that I've been on a wild goose chase all along, as I'm pretty sure kbd512 recognized almost immediately.... there is ** never ** going to be a time when all the atoms balance.
Instead, the process is always out by three hydrogen or three oxygen at every cycle. That is (probably) why the Python programs never found a solution, despite running for thousands of iterations.
Tomorrow I'll approach the problem with an different approach, taking into account that balance will always be out by a few atoms, but that won't matter, because an atom from a stash can be added to remove whatever out-of-balance condition exists in any iteration.
It will be interesting to see what ChatGPT makes of ** that ** idea!
This evening's foray into scientific computing and numerical methods to solve simultaneous equations was eye opening ....
It was a reminder that a bit more effort in study of calculus would have had long term benefits.
(th)
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Tom,
During our Zoom call Sunday, Brian asked about a storable that doesn't require cryogenic cold and has everything required to make both fuel and oxidizer for rocket fuel. The goal was propellant for a return vehicle to return from Mars back to Earth. As I said, you'll never find that in one molecule. It is possible, but anything that does will be explosive. Nitroglycerin has both fuel and oxidizer in one molecule, and with the right balance. But if you shock it too much, such as landing on Mars, it will explode. RDX has the right balance, but it's an even more powerful explosive. Nitroglycerin can be stabilized by mixing it with clay to form dynamite. Or mix it with cellulose for the explosive in hand grenades. RDX can be mixed with plastic to make it less likely to explode, the result is called C4. You don't want to carry large quantities of this on a human mission.
Bi-propellant separates fuel and oxidizer in order to make them safe. Apollo used hydrazine as fuel and di-nitrogen tetra-oxide as oxidizer. That was sometimes called nitrogen tetra-oxide, in the 1960s the acronym was NTO. In the 21at century it's commonly called by its formula N2O4. That is liquid at room temperature so easy to store. It was used as oxidizer for Titan rockets used to launch Gemini. The main engine for the Apollo service module used a 50:50 mixture of hydrazine and UDMH as fuel, a mixture called aerozine-50. Oxidizer was N2O4. Thruster quads used MMH as fuel and N2O4. Shuttle used MMH/N2O4. Russian Soyuz uses UDMH/N2O4. Russian modules of ISS have thrusters to maintain position, they also use UDMH/N2O4. SpaceX Dragon uses MMH/N2O4. MMH is hydrazine with a methyl group added to make it more stable. UDMH is hydrazine with two methyl groups. Mono-Methyl Hydrazine vs Unsymmetrical Di-Methyl Hydrazine. Raw hydrazine can explode. The explosion is more powerful if mixed with N2O4, but it's dangerous. MMH and UDMH are safer. All forms of hydrazine are slightly toxic. However, N2O4 is very toxic.
Solid rocket fuel uses a type of rubber called polybutadine. It's the rubber used for car tires. Gaskets or wet suits use neoprene rubber, partially because neoprene doesn't burn well. Polybutadine will burn. For rocket fuel it's mixed with aluminum powder, which burns very hot. Aluminum oxide is also a powder so it doesn't generate the gas required to produce pressure. That gas pressure is how a rocket produces thrust. Polybutadine rubber will become CO2 and steam and partially burned hydrocarbons. A pinch of iron oxide (rust) is added to control the rate that aluminum burns. Oxidizer is ammonium nitrate. Ammonium nitrate is white granules, a solid. It's nitrogen fertilizer, used in agriculture. When I was a kid some neighbors and relatives applied ammonium nitrate to their lawns for a very green lawn. It's safe to handle.
Brian asked me to think about something that is easy to transport, not cryogenic, and can be used as return propellant. I said the Mars Direct plan made return propellant from stuff found on Mars. That's most effective. But if you want an easy to transport oxidizer, consider ammonium nitrate. Chemical formula is NH4NO3. It's a salt, ammonium is NH4+, nitrate is NO3-. Years ago, before the great crash, I posted how to make it on Mars. Not as oxidizer for propellant, but just fertilizer for greenhouses. Since the Oklahoma bombing I won't post that again. A simple method that requires air, water, electricity, and equipment you can buy in a hardware store. Yes, Earth has a thick atmosphere with 78% nitrogen while Mars has a thin atmosphere with 2.7% nitrogen.
You need a multistage pump, freeze CO2 out, and use a warmed rhodium catalyst to combine carbon monoxide with oxygen to form more CO2. There's little O2 in Mars atmosphere, but more than enough to decompose all CO. The result is gas that can be mixed with oxygen as breathing gas for a habitat. It's mostly N2 and argon with the same trace gasses as Earth's atmosphere: neon, krypton, xenon. Mars has a trace of ozone, but the same rhodium catalyst will break it down into oxygen: 2 O3 -> 3 O2. This diluent gas can be further processed to make ammonium nitrate.
In the last quarter of 1989 and first half of 1990, Robert Zubrin and his partner David Baker came up with Mars Direct. It included ISPP. This didn't come out of nowhere. NASA was already talking about making carbon monoxide on Mars as fuel for a return vehicle. CO with LOX will burn, but is a very poor rocket fuel. NASA was also talking about Sabatier Reactor for life support. Robert Zubrin put these together and came up with LCH4/LOX as return propellant. NASA had looked at this rocket fuel, although hadn't done a lot of work. It was already known that rocket engines could use this; it would just require development of a new engine. ISPP would convert hydrogen brought from Earth with CO2 from Mars atmosphere into propellant. 1kg of H2 becomes 18kg of LCH4/LOX. The Sabatier Reactor produces heat once it's started, so primary energy is the electrolysis tank to split water into H2 and O2. The Sabatier combines H2 with CO2 to form CH4 and H2O. The process doesn't quite produce enough O2, so Moxie is required to produce O2 to top up the tank. Moxie requires 3 times as much energy per kg of O2, but it's needed.
You could just harvest ice from Mars. Melt, filter the water, put in an electrolysis tank for LH2/LOX. But in 1989/'90 we didn't have any idea how much ice was on Mars or where. You could use a Sabatier to produce LCH4/LOX because LCH4 doesn't require the deep cold of LH2. But again Mars Direct was devised before we knew where to find ice.
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