Energy needed to get to orbit using various fuels from various planets.

EDIT: I made a big mistake on how I calculate bulk density. I’ll fix it. EDIT AGAIN: I fixed it. I think.

I will pick stoichiometric mixes, oxygen as oxidizer and fuels of hydrogen, methane, and carbon monoxide. The three most obvious ISRU fuels (to me anyway). Picking stoic, I will also assumption that mass fraction (sans payload) is inversely proportional to bulk density. And at water density, I’ll say a mass fraction of 30 is doable, i.e. if wet mass is 30 ton (not counting payload), dry mass (not counting payload) will be 1 ton.

oxygen density at 90K and 1MPa: 1144kg/m^3
hydrogen density at 20K and 1MPa: 72.41kg/m^3
methane density at 111K and 1MPa: 424.2kg/m^3
carbon monoxide density at 81K and 1MPa: 798.2kg/m^3

Oxygen atomic mass is 16
carbon is 12
hydrogen is 1
stoichiometric ratios by mass:

We’ll use this equation for bulk density (thanks those who commented!).
bulk density = 1/(MR/fuel_density+ (1-MR)/oxidizer_density)

H2 + (1/2)*O2 = H2O So, 1:8 fuel:oxidizer, bulk density: 1/((1/9)/72.41+(8/9)/1144)kg/m^3 = 432.6kg/m^3
CH4 + 2*O2 = 2*H2O + CO2 So, 1:4 fuel:oxidizer, bulk density: 1/((1/5)/424.2+(4/5)/1144)kg/m^3 = 854.1kg/m^3
CO + (1/2)*O2 = CO2 So, 7:4 fuel:oxidizer, bulk density: 1/((7/11)/798.2+(4/11)/1144)kg/m^3 = 896.8kg/m^3

Turns out, stoichiometric is probably a very bad assumption for bulk density as hydrogen looks better than everything else. (But this is an interesting and useful result anyway.) EDIT:Just kidding, I was super wrong about bulk density the first time I did this. I should’ve known better! The TRUE result means CO/O2 has a better bulk density than the other options, which is more like what I expected.

Second assumption I’ll make is that rocket engines are exactly 50% efficient at converting chemical energy to jet energy. This is a conservative assumption, I think.

And the specific energy of the fuels (not counting oxygen mass) is:
hydrogen: 142MJ/kg, with oxygen: 15.8MJ/kg
methane: 55.5MJ/kg, with oxygen: 11.1MJ/kg
carbon monoxide: 10.1MJ/kg, with oxygen:6.43MJ/kg

specific kinetic energy is: .5*mass*velocity^2/mass = .5*velocity^2
So if the propellant mix specific energy is F, but we’re only 50% efficient so the energy effectively put in the velocity of the exhaust is .5*F. Setting that equal to specific kinetic energy:
solving for velocity:
velocity = sqrt(F), so the effective exhaust velocities of the above fuels are:
hydrogen: sqrt(15.8MJ/kg) = 3975m/s
methane: sqrt(11.1MJ/kg) = 3330m/s
carbon monoxide: sqrt(6.43MJ/kg) = 2535m/s

So the burnout velocity of stages would be:
hydrogen: 3975m/s*ln(30*.4326) = 10.19km/s
methane: 3330m/s*ln(30*.8541) = 10.81km/s
carbon monoxide: 2535m/s*ln(30*.8968) = 8.34km/s

If instead we fix mission delta-v at 9km/s, 8km/s (we’ll assume we get a bonus from getting high altitude balloon-launch automatically at Venus…), and 4km/s for Earth, Venus, and Mars, respectively, the mass of payload as a multiple of the rocket dry mass is:

hydrogen: (30*.4326-e^(9/3.975))/(e^(9/3.975)-1) = 0.3890
methane: (30*.8541-e^(9/3.33))/(e^(9/3.33)-1) = 0.7715
carbon monoxide: (30*.8968-e^(9/2.535))/(e^(9/2.535)-1) = -0.2333

hydrogen: (30*.4326-e^(8/3.975))/(e^(8/3.975)-1) = 0.8476
methane: (30*.8541-e^(8/3.33))/(e^(8/3.33)-1) = 1.4533
carbon monoxide: (30*.8968-e^(8/2.535))/(e^(8/2.535)-1) = 0.1538

hydrogen: (30*.4326-e^(4/3.975))/(e^(4/3.975)-1) = 5.902
methane: (30*.8541-e^(4/3.33))/(e^(4/3.33)-1) = 9.604
carbon monoxide: (30*.8968-e^(4/2.535))/(e^(4/2.535)-1) = 5.741

But we want this in terms of energy, so we’ll start with expressing as a proportion of propellant and then as energy:

hydrogen: 0.3890/(30*.4326-1)=0.03248 (kgpayload/kgpropellant)
methane: 0.7715/(30*.8541-1)=0.03133
carbon monoxide: (negative)

hydrogen: 0.8476/(30*.4326-1)=0.07077 (kgpayload/kgpropellant)
methane: 1.4533/(30*.8541-1) =0.05902
carbon monoxide: 0.1538/(30*.8968-1) =0.00594

hydrogen: 5.902/(30*.4326-1)= 0.4927 (kgpayload/kgpropellant)
methane: 9.604/(30*.8541-1) = 0.3900
carbon monoxide: 5.741/(30*.8968-1) = 0.2216

Payload per unit energy (kg/MJ):
hydrogen: .03248kg/(15.8MJ) = .002056kg/MJ
methane: .03133kg/(11.1MJ) = .002823kg/MJ
1carbon monoxide: (negative)

hydrogen: .07077kg/(15.8MJ) = .004479kg/MJ
methane: .05902kg/(11.1MJ) = .005318kg/MJ
carbon monoxide: .00594kg/(6.43MJ) = .000924kg/MJ

hydrogen: .4927kg/(15.8MJ) = .03118kg/MJ
methane: .3900kg/(11.1MJ) = .03514kg/MJ
carbon monoxide: .2216kg/(6.43MJ) = .03447kg/MJ

Also, I’ll add the more convenient form of the above:

Energy per kilogram payload Efficiency
H2/O2 486.492201 MJ/kg 0.08324902212
CH4/O2 354.2839962 MJ/kg 0.1143150705
CO/O2 -713.873967 MJ/kg -0.05673270335

H2/O2 223.2713187 MJ/kg 0.1433233797
CH4/O2 188.0569743 MJ/kg 0.1701611978
CO/O2 1082.792097 MJ/kg 0.02955322642

H2/O2 32.06682927 MJ/kg 0.2494789844
CH4/O2 28.45933589 MJ/kg 0.2811028349
CO/O2 29.01123497 MJ/kg 0.2757552379

Methane appears to be optimal for the SSTO case (though I think hydrogen pulls strongly ahead as the upper stage in the two stage case), as you would expect because of the higher stage burnout velocity. Or methane would be most optimal on other planets (where you have to get propellants through electrolysis), if its production were more efficient. Because, although hydrogen and carbon monoxide can be produced directly from electrolysis of water and carbon monoxide (respectively), methane requires a Sabatier reaction step, which loses some energy in the form of heat. Additionally, CO2 is ubiquitous on both Mars and Venus.

This is obviously simplistic, but an interesting result. I should’ve just done this in a spreadsheet. I did this in a spreadsheet now.

It kind of gives impetus to Jon’s oxygen-rich hydrolox engine idea. When your run ox-rich, hydrolox actually has very high bulk density. (This was sooo wrong… thanks, commentators, again.) Really, you’d want to optimize your Isp as you ascend, particularly on Mars. It might even make sense to blend in some CO2 at first as just reaction mass.

The exhaust velocities here are pretty pessimistic (although stoic is not a very good assumption, either).

We looked at efficiency by dividing the specific energy by (.5*(9,8,4 km/s)^2).

The efficiency of rockets isn’t too bad. Partly that’s because we used stoic. Also, this is mission delta-v which includes some losses (i.e. aero and gravity). However, we can do much better by using multiple stages and especially interestingly by playing with mixture ratios.

I think I’ll next do an analysis looking at dry mass required with more realistic propellant mixes.

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5 Responses to Energy needed to get to orbit using various fuels from various planets.

  1. johnhare john hare says:

    I’m missing something on your first stoichiometric mix. One cubic meter of LH2 is 72.41 kilograms that would mix stoichiometric with nearly 580 kilograms of oxygen with a volume of just over* half a cubic meter. That would be about 1.5 cubic meters for ~652 kilograms of propellant for a density of about 435kg/m^3.

    *Too lazy to go get my calculator out of the truck.

  2. Jim Davis says:

    As John mentioned you’re calculating your bulk densities incorrectly.

    The correct way is

    rhobulk = (1 + MR) /(1/rhofuel + MR/rhooxid)


    MR – mixture ratio
    rhofuel – fuel density
    rhooxid – oxidizer density
    rhobulk – bulk density

  3. Chris Stelter says:

    Thanks, Jim and John! You are absolutely correct! I should’ve known better, too. I’ve fixed the post.

  4. Matter Beam says:

    Sorry, I am confused by the negative values for carbon monoxide on Earth.

    Does it mean that a CO/O2 rocket can never lift off in 1g gravity?

  5. Peterh says:

    If I understand the calculations correctly, it means that based on the assumptions made an CO/O2 rocket runs out of fuel before making Earth orbit even with no payload.

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