Perfectly energy-efficient rocket vehicle

So I was watching a video of a talk by Geoffrey Landis (my old mentor when I was an intern at Glenn), and he made a very interesting point.

If you’re trying to maximize energy efficiency for a rocket and you have the ability to change Isp, you should set the exhaust velocity equal to the rocket’s current velocity. Think about it for a little bit, and you’ll see that it’s true.

We make some simplifying assumptions here, such as assuming arbitrarily good mass fraction, and thrust occurring outside any gravity wells (or, equivalently, thrust that is arbitrarily higher than acceleration due to gravity at that point). But with these assumptions, this finding is true.

One thing that’s a little surprising is that you’d have PERFECT efficiency under this scenario. That means you’re wasting exactly no energy by accelerating all that propellant, which goes a little counter to the intuition. A perfectly efficient rail gun (with an arbitrarily low-mass acceleration cart) would not be any more energy efficient.

…the downside, here is that your initial exhaust velocity is zero, which implies an infinite amount of fuel. This isn’t so bad because we can just truncate the initial portion of the flight by picking a certain minimum Isp to keep mass fraction to a reasonable number. We lose some efficiency this way, but it’d small. Alternatively, you could use rail-launch for the low-speed portion of the flight and maintain arbitrarily-good efficiency. The other side of this is that it implies 700-900s Isp for the last portion of the flight if going all the way to Earth orbit. That implies use of non-chemical thermal rockets (such as NTR), but you’d be no better off energy-wise since the only practical propellant for those (while still reaching those Isps) is hydrogen, and producing hydrogen requires a lot of energy (energy which is not at all utilized in the rocket! Hydrogen is considered basically inert… you could also use helium). Besides, terrible mass fraction.

Anyway, there is also an energy-ideal exhaust velocity if you can’t adjust the Isp. I believe it’s about 5/8th the final delta-v, if I remember geoffrey’s talk. I may try to recreate this calculation.

Another point: It makes things like Aerojet’s thrust-augmented rocket look pretty interesting from an energy standpoint.

I have a bunch of thoughts related to this that I may blog later on. Here’s Landis’s talk. (He argues for NTR here. I disagree with him, but very interesting talk, and he definitely makes a convincing case that NTR is not some unobtainable technology.)

Overall, I appreciate the change in focus to energy. I think we have an (understandable) obsession with mass in the rocketry world, but that doesn’t tell the whole picture and can somewhat bias our intuition.

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13 Responses to Perfectly energy-efficient rocket vehicle

  1. Isaac Kuo says:

    If you are limited to a particular finite amount of propellant (obviously you do not have an infinite amount of propellant), then the ideal exhaust velocity to optimize energy efficiency is that which has a constant – but non-zero – exhaust velocity relative to the launch frame of reference.

    In other words, you don’t truncate things at a particular minimum exhaust velocity. Instead, you add that minimum exhaust velocity to your current velocity. The end result is that all of the bits of exhaust you trail behind will end up with zero velocity with respect to each other – but they are all moving a bit rearward compared to the original launch rest frame.

    This is a bit of a navel gazing intellectual exercise, of course, because the real world involves more significant compromises (such as propellant tank mass/volume, gravity drag, air resistance, logistics costs, and so on). The point of the original version is to provide some insight as to why it’s better to start with low Isp and end with high Isp. This slightly more optimal wrinkle doesn’t really add anything to that insight.

  2. Chris Stelter says:

    Thanks, Isaac. You did a better job explaining the point of my post than I did. 🙂

  3. Paul D. says:

    Another interesting example that conservation of energy can illuminate:

    Suppose you have a car equipped with a pump and a tank of propellant. The pump is connected to the wheels. When the car is moving forward, the wheels drive the pump which ejects propellant backwards. Assume perfect efficiency, and assume the propellant is ejected so that its motionless in the rest frame of the road.

    It’s easy to see that the vehicle accelerates forward, because the initial kinetic energy of the ejected propellant (which, after ejection, has zero kinetic energy) is concentrated in the vehicle. The mass ratio of this system goes as the square of the ratio of the final and initial speeds.

  4. Paul D. says:

    “Overall, I appreciate the change in focus to energy.”

    It’s ironic then that NTRs are being talked about, since these aren’t energy limited in the same sense chemical rockets are. The reactor can produce essentially unlimited amounts of energy.

    The limit on NTRs is not energy, it’s entropy. All the entropy produced in the system has to be carried away in the propellant stream, and it has a limited capacity to do that. When designing a nuclear rocket, then, a key is to minimize thermodynamic irreversibility, especially in the components that are temperature limited. There are various ways to do this that will boost the exhaust velocity (for a given reactor temperature) over and above what an ordinary NTR will achieve, without the need for another entropy dump (radiators).

  5. Jim Davis says:

    “Anyway, there is also an energy-ideal exhaust velocity if you can’t adjust the Isp. I believe it’s about 5/8th the final delta-v, if I remember geoffrey’s talk. I may try to recreate this calculation.”

    The calculation is fairly straightforward. Assume the use of consistent units throughout.

    Efficiency is defined as the final kinetic energy of the vehicle divided by the chemical energy consumed:

    eta = (mf)(dv)^2/2/(mp)/q Eq.1

    where

    eta = efficiency
    mf = final mass
    dv = change in velocity
    mp = propellant consumed
    q = chemical energy per unit mass of propellant

    The rocket equation is:

    dv = Isp * ln (1+ (mp)/(mf)) Eq.2

    where

    Isp = specific impulse

    Assuming perfect thermodynamic efficiency:

    Isp^2 = 2*q Eq.3

    Substituting Eq. 2 and Eq.3 into Eq.1 gives:

    eta = (dv/Isp)^2/(exp(dv/Isp) – 1) Eq.4

    It can easily be shown than Eq.4 is maximized when dv/Isp = 1.5936 or Isp = 0.6275 dv which is just about the 5/8ths Landis mentioned. The corresponding maximum eta is 0.6476.

  6. Chris Stelter says:

    Thanks, Jim Davis, for doing the math for me!

    I was going to do that for another post. I may do it anyway, typesetting it nicely in Latex.

  7. Robert Clark says:

    Thanks for the article. I had been thinking about the how would going for energy efficiency effect the rocket equation for orbital rockets, specifically using the fact the exhaust velocity should match the craft velocity to maximize efficiency. BTW, what would be the propellant requirements here compared to the usual case of the rocket equation where the exhaust velocity is kept constant? That is, how much propellant would be saved?

    About the problem you mentioned of needing high exhaust velocity ca. 7,000 m/s to 9,000 m/s on reaching orbit there were a couple of ideas I was thinking of.

    One, use air as the reaction mass, the propellants being used just to generate sufficient energy to propel the air backwards. But the problem here is if the air is propelled *exactly* at the craft speed, then you aren’t adding to the air’s relative speed it already has with respect to the craft, so you get *zero* thrust.

    So in actuality you need to have some *additional* speed added to the incoming air. And that’s where I get stuck. How do you optimize the added speed at each point in the flight to minimize the total energy required? Should you keep the added speed constant? Should it vary with the craft speed? Or is that you can actually make the added additional speed as small as you want and the smaller you make the added speed, so the exhaust velocity is as close as possible to the craft speed, the smaller would be the energy required.

    This last might be true but it seemed inconsistent to me since the added speed and thrust might be so small that it could take days to get to orbit. Another complication is that the energy applied at each point depends also on the amount of air you are taking in via the intakes. So though it could be the added speed to the air is large at high craft speed, the energy would be small by reducing the air taken in.

    The second method where you can get high exhaust speed does use the propellant as the reaction mass. So how can you get it at a higher exhaust velocity than the maximum chemical exhaust velocity? This is done by drawing off some of the energy produced by the combustion and storing it for later, then applying that energy to later combustion to increase the energy applied to that later combustion.

    Not a trivial task however. First, this will decrease the energy and speed of the early combustion. But actually that’s OK because you want the speed of the exhaust to be low to match the craft speed then anyway.

    One big problem is storing it in a lightweight way, important for an aircraft or rocket. There are lots of ways of storing energy, such as batteries. However they usually are rather massive. This is a problem for example for power companies because you want to save energy produced during off-peak hours to be used during peak hours. I am looking at such methods.

    Then you have to reapply this stored energy to the later combustion. A problem here is you could not apply it as heat. Here’s the reason: by one of the versions of the second law of thermodynamics you could not use this stored energy as heat to raise the temperature of the later combustion. The temperature of the early combustion will be at a certain set temperature in accordance with the propellant combination. And the temperature of the later combustion will be at the same point. Then by the 2nd Law you can’t raise the temperature by combining the two together.

    No, what you would have to do is first convert this stored heat energy to another form, say electrical, as batteries, or mechanical, as flywheels. Then you can apply this energy to the later combustion to increase the temperature and therefore the exhaust speed of these combustion products.

    That conversion would cost some energy and add weight, so you have to make a careful consideration of the trade offs.

    Bob Clark

  8. Bob Steinke says:

    Bob,

    “How much propellant would be saved?”

    Depends on what you mean by propellant. Rocket propellant serves two functions. It is both an energy source and reaction mass. Using this method would require less energy, but more reaction mass. So if you are stuck with getting both energy and reaction mass out of a single propellant it doesn’t take any less propellant. In fact it takes more to get enough reaction mass.

    This method applies more to the case when your energy and reaction mass come from different sources such as a nuclear or beamed power rocket, but then are you really energy limited or just power limited? An air breather is another beast again because, as you point out, you can collect as much reaction mass as you want, but doing so creates inlet drag so your exhaust has to be faster than your vehicle speed to generate any thrust at all.

    Your final idea is interesting of stealing energy from early propellant and adding it to later propellant, but there’s one snag. If you have some kind of energy storage system why steal energy from early propellant to charge it? Why not just lift off with it fully charged to start with? And what is the most convenient form of extra energy storage you can add to your rocket? Make the tanks bigger and hold more propellant!

  9. Robert Clark says:

    Bigger tanks also mean bigger engines to lift them. The shuttle and SLS had huge tanks and huge costs. The idea is to increase efficiency so that you can loft the same payload with a smaller rocket.

    Bob Clark

  10. Jim Davis says:

    “But the problem here is if the air is propelled *exactly* at the craft speed, then you aren’t adding to the air’s relative speed it already has with respect to the craft, so you get *zero* thrust.”

    Not necessarily. For an airbreather thrust is:

    T = (mdotair + mdotprop) ve – (mdotair) va + (pe – pa)Ae

    where

    T = net thrust of engine
    mdotair = mass flow of air through engine
    mdotprop = mass flow of propellant into engine
    ve = engine exit velocity
    va = flight velocity
    pe = engine exit pressure
    pa = ambient pressure
    Ae = engine exit area

    so even in the case where ve = va you still have the mdotprop*ve and (pe – pa)Ae terms contributing thrust. Arranging matters thus would require that mdotair >>> mdotprop so this isn’t really practical but it is a theoretical possibility.

  11. Bob Steinke says:

    “Bigger tanks also mean bigger engines to lift them.”

    Yes, that’s true. I was commenting on this proposal:

    “convert this stored heat energy to another form, say electrical, as batteries, or mechanical, as flywheels. Then you can apply this energy to the later combustion to increase the temperature and therefore the exhaust speed of these combustion products.

    That conversion would cost some energy and add weight, so you have to make a careful consideration of the trade offs.”

    Any energy storage system will also need bigger engines to lift it. Unless the energy storage system has higher specific energy than your propellants the best trade-off is just to carry more propellant.

  12. Michael Egorov says:

    Actually, apart from nuclear engines, there is also laser propulsion where you can change Isp quite easily in very broad range (2 s to 10k s was shown).

    Not sure if “energy-efficient rockets” will be needed for anything practical, but the ability to have this degree of freedom of changing Isp is interesting.

  13. MBMelcon says:

    Jim Davis, your equation 1 has a typo. Easier to follow is

    eta = [(1/2) (mf)(dv)^2 ] / [(mp)*q)]

    -MBM

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