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Many space exploration fans are familiar with the rocket equation [https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation] $$\Delta V = V_{e} \times ln\frac{M_0}{M_F}$$ where $V_e$ is the exhaust velocity (= $I_{sp}g$), $M_0$ is the initial mass, and $M_F$ is the final mass, with the difference between $M_0$ and $M_F$ being the mass of the propellant used, $M_P$. That equation makes it look like near-infinite ∆V is possible: just make $\frac{M_0}{M_F}$ arbitrarily large and ∆V gets arbitrarily large. You make $M_0$ large by making the propellant mass large: $M_0 = M_F + M_P$. Simple!

Except that in the real world $M_F$ cannot be divorced from $M_P$. $M_F$ must include the tank(s) to contain the propellant. Conventional wisdom has been that tank mass is proportional to the mass of propellant contained. Assume that the mass of any ancillary equipment involved in properly delivering the propellant to the feed lines, such as pressurant and pressurant tanks, is included in the effective propellant tank mass. Also assume the mass of things such as engines and turbopumps are included with the "inert mass" of the spacecraft bus and don't scale with propellant mass. In that case:

QUESTION: Is there a ∆V limit for a single-stage rocket with propellant tank mass proportional to propellant mass? If so, what is that limit?

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  • $\begingroup$ This is great! I adjusted the wording of the title though to be easier on the eyes (in my opinion). I think it's identical mathematically. If you don't like it just roll back to and earlier version. To do that you first click edited i.stack.imgur.com/6S0Tu.png and then click rollback i.stack.imgur.com/2gnuy.png to the previous version (which is currently indicated with the large "2". $\endgroup$ – uhoh Aug 28 '18 at 5:44
  • $\begingroup$ At certain point the specific impulse of the engine (and velocity of the exhaust gases) won't suffice to make them escape the rocket's own gravity well... $\endgroup$ – SF. Aug 29 '18 at 10:37
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The simple answer: yes, there is a limit. Here I derive the equations showing what that limit is.

Per the question's assumptions, I'll assume that the mass of any ancillary equipment involved in properly delivering the propellant to the feed lines, such as pressurant and pressurant tanks, is included in the effective propellant tank mass. Also I'll assume the mass of things such as engines and turbopumps are included with the "inert mass" of the spacecraft bus and don't scale with propellant mass. There are some caveats riding along with those assumptions, and I'll comment on those later. Such as: assuming that the mass of the propellant and the mass of the tank(s) are the only things that are scaling—everything else stays constant.

For now, assume that the tank mass $M_T$ is proportional to propellant mass, such that $$M_T = \xi_T M_P \tag{2}$$ Then $$M_O = M_{SC} + M_P + M_T = M_{SC} + M_P + \xi_T M_P \tag{3}$$ or $$M_O = M_{SC} + (1 + \xi_T) M_P \tag{4}$$where $M_{SC}$ is the spacecraft mass that includes everything (including engines and propellant lines) except the propellant and tanks; and $$M_F = M_{SC} + M_T = M_{SC} + \xi_T M_P \tag{5}$$ Substituting those into the Tsiolkovsky equation, $$\Delta V = V_e \times ln\frac{M_{SC} + (1 + \xi_T) M_P}{M_{SC} + \xi_T M_P} \tag{6}$$

If we're looking for absolute limits then let $M_P$ approach infinity. In that case, in Eq. 6, $M_{SC}$ becomes insignificant (say, for instance, 1000 kg on top of $10^{100}$ kg of propellant and tank), so the $M_P$'s cancel out of what remains. So in the limiting case $$\Delta V = V_e \times ln\frac{(1 + \xi_T)}{\xi_T } \tag{7}$$

This is the absolute limit to the ∆V for a single-stage rocket with propellant tank mass proportional to propellant mass. Given a $\xi_T$, this says getting a ∆V larger than that is not difficult—it's impossible!

So that's the answer. But there are some caveats I alluded to in the introduction.

For a significant range of propellant masses it appears from empirical data that tank mass is indeed roughly proportional to propellant mass, as Pietrobon found in his original 2008 publication, "Analysis of Propellant Tank Masses". But when he examined very large systems he found that they wound up lighter than predicted by that linear relation. A later (2009) version of his paper skips mention of the linear model and goes straight to a different model where tank mass is proportional to propellant mass to the 0.856 power. This would change the assumption of linearity that results in Eq. 7, but that model is fairly crude, with not a lot of data at the very large masses, so the power-law relation is far from confirmed.

Equation 7 assumes that nothing outside of the propellant mass and tank mass scales as the propellant mass changes. This is not strictly true. No real rocket engine can burn ("process") an unlimited quantity of propellant. As propellant mass grows unchecked, eventually you have to add more engines or use larger engines that can process more propellant. Either way the mass spent in engines (I include things like pumps and propellant lines in this) increases. If, like the tank mass, the mass of the engines winds up being proportional to propellant mass, then you have a new variable $M_E = \xi_E M_P$ split out of $M_{SC}$, and it gets treated much like $M_T$. $\xi_E$ and $\xi_T$ will add to produce an effective propulsion system mass fraction parameter $\xi_{PS}$ which can be substituted for $\xi_T$ in Eq. 7. Since $\xi_{PS}$ is larger than $\xi_T$, the resulting ∆V limit is smaller. In general, this approach applies to any component that scales linearly with propellant mass.

It doesn't end there. Spacecraft are notorious for the ripple effect: increase the mass of one item, and usually the masses of other things increase as well, and often those increase the masses of still other things. If you increase the number of engines you'll need more structure mass, even if they're not all running simultaneously. If they are all running simultaneously, you'll need even more structure mass, and probably will need more electric power to operate all the engines' valves and such, so more mass in the power system. If you increase the size (thrust) of the engines you'll definitely need more structure mass and more power system mass. The ripple effect says that most likely, the ∆V limit will be even smaller than indicated by the modified Eq. 7.

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  • $\begingroup$ To zeroth approximation, engine mass is proportional to tank mass; to first approximation it's slightly sublinear, much like the tank mass. Since you're parameterizing anyway, it makes sense to combine all the dry mass components into a $\xi_D$ and introduce one proportionality-exponent term to abstract all the nonlinearity. Since both $\xi_D$ and the exponent are dependent on how much engineering you're willing to invest in, though, there's not really a hard limiting value, just a gradient of difficulty. $\endgroup$ – Russell Borogove Aug 27 '18 at 19:56
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    $\begingroup$ Your equation 7 really is just the Tsiolkovsky with a different way of expressing $M_0$ and $M_f$. $\endgroup$ – Russell Borogove Aug 27 '18 at 20:00
  • $\begingroup$ @RussellBorogove The problem arises from components with different exponents of proportionality. When those exponents are different, those terms cannot be rigorously combined. The best you can hope for is a "closest approximation" that unfortunately will become increasingly inaccurate as propellant mass increases far beyond the data. Of course, in the limiting case (propellant mass growing beyond any bound) the term with the smallest exponent will wind up dominant. $\endgroup$ – Tom Spilker Aug 27 '18 at 20:04
  • $\begingroup$ @RussellBorogove Yes, but the reformulation gives you explicit insight into the effects of the tank mass that is absent in the simple Tsiolkovsky equation. Notably, that there is a finite ∆V limit. That's why I did this exercise, for the benefit of those who aren't aware of this limitation. $\endgroup$ – Tom Spilker Aug 27 '18 at 20:08
  • $\begingroup$ When a genius answers a question asked by yet another genius... great quality stuff. I didn't realize looking at the rocket equation with Mt in terms of Mp gives you this limit. Neat. $\endgroup$ – Magic Octopus Urn Aug 28 '18 at 0:14
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The beauty of @TomSpilker's tutorial here will be lost on some people because it is long but doesn't have any bling or eye-candy for those who don't think equations are bling or eye-candy, so here's a plot:

THere's certainly more subtlety to the answer than is shown here, but this still may be helpful

For those of you who haven't already read The Tyranny of the Rocket Equation by Expedition by 30/31 Flight Engineer Don Pettit, I'd recommend it. The context is single-stage to orbit, but if you look at those velocities, their ratios, and the propellant mass ratios there, then Tom's tutorial comes into even sharper focus:

enter image description here

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