Payload Fraction Example Proof

Continuing with our story from last time…

The next day, your boss pokes his head in your office and asks:

“How’s those forty trans-Mars injection stages going?”

He notices that you’re checking out scuba-dive sites in the Caribbean for your upcoming vacation with your feet up on the desk, and comes into the room with the blood rising to your face. In your defense, you blurt out that you’ve already done the analysis!

He, somewhat increduously, demands to see the results, so you show him the spreadsheet. He’s less than impressed.

“I thought you were going to design each stage! I need pictures and layouts of these things, with lists of mass and volumes and so forth…you’re showing me a little number. Furthermore, I’ve got the Ohio and Nevada congressional delegations breathing down my neck to send billions of dollars to Stan Borowski to develop an NTR that he promises will get us to Mars faster. You’ve got to show me more detail for this. And how do you even know that your equations are correct?”

Now on the defensive, you offer to try to quickly verify two of the comparison cases for your boss. He looks through your results and decides to pick a comparison point in the middle of the trade space: 4000 m/s delta-V and 0.5 initial thrust-to-weight. Your spreadsheet quickly predicts that the NTR stage will have a payload fraction of 0.3828 and that the chemical stage will have a payload fraction of 0.3955, with a ratio of the two of 1.033, but your boss wants to see proof that your equations are correct.

So how do you go about turning these expressions into masses and volume and graphics?

chemStageProperties

ntrStageProperties

chemNTRstageGraphics

This time, it took considerably longer than ten minutes, but you showed that your equations match up with reality (at least to the resolution of your initial assumptions). You can see, physically, that the NTR stage is much larger than the chemical stage, even though both carry roughly the same payload. The NTR stage needs almost 6 engines to meet the T/W requirements, while the chemical stage only needs 4. The real difference between the two stages, from a mass perspective, is in the engine weight. You can see that the total engine weight on the chem stage is 1480 lb, while on the NTR stage it is 29,333 lb. This is a staggering difference, due almost entirely to the wretched thrust-to-weight ratio of the NTR engines. This could also lead to another problem. There will be every incentive to try to remove weight from the NTR engines, and with six engines in close proximity, it is almost certain that there will be a lot of neutronic leakage from one engine to another. This means that the engines won’t be able to be controlled individually, but will have to be controlled as a group. It may not be possible to shut one of them down in flight. It also means that they might have to be tested as a group which will drive costs up like crazy.

From the propellant side of the house, the NTR LH2 tank is much larger than the chem tanks, and the mass devoted to tankage on the chem side is 1554 kg, whereas the NTR tank is 4426 kg. Those big NTR LH2 tanks might cause you to hit the volume constraint on your launch vehicle before the mass constraint is reached.

Your boss is happy, for now. He’s got the numbers to show that for 4000 m/s and 0.5 T/W, the NTR stage only has a few more percent payload, and that’s not worth paying the billions to develop it.

But unfortunately for you, now he knows you can get this kind of preliminary analysis work done much faster, so he’s loaded you down with all kinds of new analyses. And that beach vacation is looking further and further away…

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MS, nuclear engineering, University of Tennessee, 2014, Flibe Energy, president, 2011-present, Teledyne Brown Engineering, chief nuclear technologist, 2010-2011, NASA Marshall Space Flight Center, aerospace engineer, 2000-2010, MS, aerospace engineering, Georgia Tech, 1999

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62 Responses to Payload Fraction Example Proof

  1. Bob Smith says:

    Trouble is Kirk I didn’t get any facts wrong. Don’t make stuff up. The maneuver you describe from French Guiana is in fact energetically trivial. And I’m quite correct that what you are calling “LEO” is energetically very different than the circular LEO used in these analyses. And about the large error in gravity losses from assuming that low thrust/weight stages start firing in LEO. But these important issues you ignore and instead you try to pick nits. Sad.

  2. No, it’s not Bob. You’re getting tiresome. Start backing up your assertions with numbers or your posts are toast. And what I call LEO is just as much a LEO as any other. You ignored Baikonur when it didn’t suit you. That’s “sad”.

    Next comment better have numbers in it or it’s getting deleted.

  3. johnhare john hare says:

    @ Kirk in #41,

    That post was supposed to have Geatano moment in brackets except I screwed it up. A couple of concepts I posted earlier here ‘might’ offer the possibility. The odds are of course that neither will work past a preliminary review.

    The tether rocket had an onboard engine spinning contrarotating tethers with pulsed rockets at the tips. The rockets pulsed during that portion of the arc when the tether velocity and rocket exhaust both point astern. The nominal ship relative exhaust velocity could be on the order of 8,000 m/s with O2 being 85% of the exhausted mass. My assumption was that a reactor driving a spinning tether would be closer to known practice than a NTR and should require less shielding mass. This is a handwaving guess.

    The other idea was a solar thermal rocket with direct after heating of the pollutants in a hydrogen exhaust plume by very large thin mirrors. Four digit Isp seems possible.

    Neither of these ideas is actually close enough to reality to warrant serious consideration. As I said, it was a Geatano moment.

    @ Bob,
    You are way off base. You clearly don’t know orbital mechanics and are insulting someone that does.

  4. Brad says:

    JG, “Running your spreadsheet though, it still doesn’t look like a game-changer (about a 30-40% better payload fraction across the board), until you start going to higher delta V’s. Like if you also had it do the Mars capture burn or something instead of doing aerocapture.

    I’m not 100% with you yet that NTRs make absolutely no sense, but they do definitely seem to be a hard sell compared to other technologies.”

    That is solid and fair analysis. It probably wouldn’t pay to develop a new type of NTR engine like Dumbo or Timberwind. But the Rover type was developed to a very high level under NERVA before the program was dropped. Just short of flight hardware. Heck, the hydrogen turbopump was used in the J-2.

  5. Robert Clark says:

    For anyone interested I posted the calculation for how the Falcon 1 first stage can become a SSTO by switching to a more efficient Russian kerosene-fueled engine on sci.astro:

    Newsgroups: sci.space.policy, sci.astro, sci.physics, sci.space.history
    From: Robert Clark
    Date: Sun, 14 Mar 2010 18:24:37 -0700 (PDT)
    Local: Sun, Mar 14 2010 9:24 pm
    Subject: Re: A kerosene-fueled X-33 as a single stage to orbit vehicle.
    http://groups.google.com/group/sci.space.policy/msg/b2dfd3ce833c4470?hl=en

    Bob Clark

  6. Robert Clark says:

    In regards to a dens propellant vehicle requiring about 300 m/s less to orbit than for a hydrogen-fueled SSTO vehicle, this is another key advantage of dense propellant vehicles. The main idea behind this is that dense propellant vehicles burn mass so much more quickly that they achieve the speed needed to attain the right altitude for orbit more quickly. Since the gravity loss is dependent on the time spent on this vertical portion of the trip, dense propellant vehicles experience less gravity loss. Still, the explanation is probably not easy to grasp unless you do the actual numerical calculations over the trajectory of the flight. However, I can show an approximate calculation that makes the idea more understandable below.
    This Wikipedia article also mentions the fact that dense propellant
    vehicles require 300 m/s less delta-V to orbit than hydrogen vehicles:

    Single-stage-to-orbit.
    # 4 Dense versus hydrogen fuels.
    “The end result is the thrust/weight ratio of hydrogen-fueled engines is 30–50% lower than comparable engines using denser fuels.”
    “This inefficiency indirectly affects gravity losses as well; the
    vehicle has to hold itself up on rocket power until it reaches orbit. The lower excess thrust of the hydrogen engines due to the lower thrust/weight ratio means that the vehicle must ascend more steeply, and so less thrust acts horizontally.
    Less horizontal thrust results in taking longer to reach orbit, and
    gravity losses are increased by at least 300 meters per second. While not appearing large, the mass ratio to delta-v curve is very steep to reach orbit in a single stage, and this makes a 10% difference to the mass ratio on top of the tankage and pump savings.”
    http://en.wikipedia.org/wiki/Single-stage-to-orbit%23Dense_versus_hydrogen_fuels

    However, the explanation given here is not quite correct. This rather implies it is a function of greater thrust/weight ratio only. But in actual fact the lowered delta-V required for dense fuels applies *even when the hydrogen and the dense fuel vehicles have the same thrust/weight ratio*.
    For the calculation of the delta-V savings for dense fuels, suppose
    both the dense-fueled and hydrogen-fueled vehicles have a initial T/W of, say, 1.3. Let Mi be the initial gross mass of the vehicle, r the constant propellant flow rate, Ve the exhaust velocity, a(t) the acceleration, changing with time, of the vehicle due to the thrust, and g the acceleration due to gravity 9.8 m/s^2. Then the
    mass of the vehicle at time t is Mi-r*t, and the thrust force is (Mi-r*t)a(t).
    We’ll use the fact that the thrust of a rocket is (propellant flow rate)x(exhaust velocity) to get the equation (Mi-r*t)a(t) = r*Ve. We can solve this for the acceleration to get a(t) = r*Ve/(Mi-r*t).
    Now because we set the initial thrust/weight ratio as 1.3 we know
    that thrust = r*Ve = 1.3(g*Mi), so Mi = r*Ve/(1.3g). Then plug this
    into the equation for acceleration to get: a(t) = r*Ve/(r*Ve/(1.3g) –
    r*t) = Ve/(Ve/(1.3.g) – t). Quite notable here is that the propellant
    flow rate cancels out and the acceleration due to the thrust depends only on the exhaust velocity Ve, or equivalently, only on the Isp.
    Then for the vertical portion of the trip where gravity drag takes
    place, the rocket’s acceleration will be Ve/(Ve/(1.3g)-t) – g. Now it may not be apparent at first glance but this formula says the acceleration is greater for a smaller exhaust velocity Ve, so for a smaller Isp. To make it clearer multiply top and bottom of the expression for a(t) by 1.3g to bring it to 1.3g*Ve/(Ve-1.3g*t). Then if you do the division this becomes 1.3g + t*(1.3g)^2/(Ve-1.3g*t). Now you see because the Ve is in the denominator the expression is larger when Ve is *smaller*.
    So a dense propellant with a lower Isp will accelerate faster during
    this vertical portion of the trip meaning it spends less time when
    gravity drag is operating so that gravity drag is reduced. You
    couldn’t make the Isp be arbitrarily small though because that would result in huge fuel loads and tanks, and, most importantly, engines to get the vehicle off the ground.

    Bob Clark

  7. Bob Clark, I make no assertions that a kerosene/LOX rocket is inherently capable of 300 m/s less delta-V to orbit than a LH2/LOX rocket. I pulled that out of thin air just to show you that LH2 still beat kerosene. I would have to do a real integrated ascent trajectory analysis to find the difference and it would likely be subject to very specific design decisions in the vehicle, such as engine number, initial thrust-to-weight, etc.

    Why are you so zealous about this subject? For a mathematics professor you seemed to miss your calling as an aerospace engineer. You write such enormous comments…why don’t you start your own blog devoted to LOX/kerosene SSTOs?

  8. It seems the LOX/LH2 vs LOX/RP1 debate has been answered with Space-X’s design studies for the Falcon 9. The find a faster trip to orbit and a higher delta-V from using RP1.

    http://commercialspace.pbworks.com/f/Markusic%20-%20SpaceX%20Propulsion%20small.pptx
    Is the presentation of the design study of the Merlin2 engine.

  9. mmeijeri says:

    That link didn’t work for me:
    There is no file called Markusic – SpaceX Propulsion small.pptx on this workspace.

  10. Mike Atkinson says:

    My understanding of why NTR gets chosen for Mars missions goes like this:
    – You need to get back so desire a large hab and a stage to place it in Earth Transfer Orbit.

    – The hab and stage together are too large in mass and volume to perform an aero-capture into Mars orbit, so you need a propulsive breaking.

    – With the delta-v of the TMI, breaking and TEI burns are combined then NTR (especially with drop tanks) becomes the best option, mass fraction wise.

    There are two counters to that argument:
    – doing one-way-to-mars you don’t need to return.
    – using a propellant depot at Mars you can aero-capture much smaller bits.

  11. Chris (Robotbeat) says:

    Your equation images have broken links!!!

    A shame, for the analysis is great.

  12. I found the missing graphics and restored them to the post.

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