Calculating Propellant-Mass-Sensitive Term

In my last two posts I’ve been talking about calculating payload fraction of a rocket using the mass ratio from the rocket equation and some vehicle parameters that have been sensitive to propellant mass and gross mass. To use these parameters successfully, it would be helpful to have some idea what they should be for different designs. For instance, we all know that hydrocarbon fuels are more dense than hydrogen fuel, so how would that affect that parameter? To try to answer these questions better, I’ve taken the previously-defined lambda term and created a derivation of what its value ought to be in different circumstances.

     \begin{displaymath} \lambda = \frac{m_{propellant-mass-dependent}}{m_{propellant}} \end{displaymath}

Let’s assume that the propellant-sensitive mass consists of only two things: the fuel tank and the oxidizer tank. In the expressions I abbreviate fuel tank as FT and oxidizer tank as OT.

     \begin{displaymath} \lambda = \frac{m_{FT} + m_{OT}}{m_{propellant}} \end{displaymath}

When I was at Georgia Tech, my professor gave us some mass-estimating relationships (MERs) for propellant tanks, but they were all based on volume. For instance, they would tell you to estimate the mass of a hydrogen tank as some number of kilograms per cubic meter of tank volume. So to use these mass-estimating relationships, I rewrite the expression in terms of a factor (f) for the fuel and oxidizer tanks and their volumes.

     \begin{displaymath} \lambda = \frac{f_{FT} V_{FT} + f_{OT} V_{OT}}{m_{propellant}} \end{displaymath}

The volume of the fuel and oxidizer tanks will simply be the mass of the fuel or oxidizer they contain divided by the density of the fuel or oxidizer. To account for the role of ullage (extra volume) in the tank, we throw an ullage factor in the denominator, to make the tank have a little more volume than it would otherwise have if it was 100% full of fuel or oxidizer.

     \begin{displaymath} V_{FT} = \frac{m_{fuel}}{\rho_{fuel} (1 - f_{ullage})} \end{displaymath}

     \begin{displaymath} V_{OT} = \frac{m_{ox}}{\rho_{ox} (1 - f_{ullage})} \end{displaymath}

With expressions for fuel and oxidizer volume computed, we substitute these expressions back into the overall expression for lambda.

     \begin{displaymath} \lambda = \dfrac{f_{FT} \left(\dfrac{m_{fuel}}{\rho_{fuel} (1 - f_{ullage})}\right) + f_{OT} \left(\dfrac{m_{ox}}{\rho_{ox} (1 - f_{ullage})}\right)}{m_{propellant}} \end{displaymath}

The ullage term is collected and moved to the denominator.

     \begin{displaymath} \lambda = \dfrac{f_{FT} \left(\dfrac{m_{fuel}}{\rho_{fuel}}\right) + f_{OT} \left(\dfrac{m_{ox}}{\rho_{ox}}\right)}{m_{propellant} (1 - f_{ullage})} \end{displaymath}

Now the expression has fuel mass, oxidizer mass, and overall propellant mass terms in it. But we know that these terms aren’t actually independent. The fuel mass plus the oxidizer mass equals the propellant mass. And the oxidizer mass divided by the fuel mass gives us the mixture ratio (MXR). I use MXR in the expression because I already used MR for the mass ratio, and I want to limit confusion as much as possible.

     \begin{displaymath} MXR = \frac{m_{ox}}{m_{fuel}} \end{displaymath}

     \begin{displaymath} m_{propellant} = m_{fuel} + m_{ox} \end{displaymath}

     \begin{displaymath} m_{propellant} = m_{fuel} (1 + MXR) \end{displaymath}

Thanks to mixture ratio and the propellant summation, we can calculate fuel and oxidizer mass entirely in terms of propellant mass and mixture ratio.

     \begin{displaymath} m_{fuel} = \frac{m_{propellant}}{1 + MXR} \end{displaymath}

     \begin{displaymath} m_{ox} = \frac{(MXR) m_{propellant}}{1 + MXR} \end{displaymath}

Substituting these definitions for fuel mass and oxidizer mass back into the lambda expression, we get something that looks complicated but is on its way to being simplified.

     \begin{displaymath} \lambda = \dfrac{f_{FT} \left(\dfrac{m_{propellant}}{\rho_{fuel} (1 + MXR)}\right) + f_{OT} \left(\dfrac{(MXR) m_{propellant}}{\rho_{ox} (1 + MXR)}\right)}{m_{propellant} (1 - f_{ullage})} \end{displaymath}

With propellant mass showing up in the numerator and in the denominator, it cancels out nicely. The mixture ratio terms moves to the denominator, and we get a nice compact expression.

     \begin{displaymath} \lambda = \dfrac{(MXR) \left(\dfrac{f_{OT}}{\rho_{ox}}\right) + \left(\dfrac{f_{FT}}{\rho_{fuel}}\right)}{(1 + MXR) (1 - f_{ullage})} \end{displaymath}

There’s some really nice aspects to this simple expression for lambda. Propellant mass has been completely removed from the equation, which means you don’t need to know how big or small your rocket is to calculate lambda. You need to know the mixture ratio (MXR), which is determined by whatever rocket engine you choose. Choosing the rocket engine also chooses the fuel and oxidizer, which lets you plug in their densities. Assuming you have an idea what ullage would be and what the factors for the fuel tank (fFT) and oxidizer tank (fOT) would be, you can calculate lambda pretty quickly.

Let’s use this equation in an example. Suppose we had an LH2/LOX rocket engine that had a mixture ratio of 5.8 to 1 (5.8 kg of LOX for every kg of LH2). LOX has a density of 1142 kg/m3 and LH2 has a density of 71 kg/m3. Let us assume that we have a ullage of 3% and that the tank factor for the hydrogen tank is 9 kg/m3 and the factor for the oxygen tank is 12 kg/m3.

     \begin{displaymath} \lambda = \frac{(5.8) \left(\dfrac{12 \text{kg/m}^3}{1142 \text{kg/m}^3}\right) + \left(\dfrac{9 \text{kg/m}^3}{71 \text{kg/m}^3}\right)}{(1 + 5.8) (1 - 0.03)} \end{displaymath}

We can evaluate some of the math but stop short of the answer to examine some of the implications. We can see that the effect on the overall factor due to the LH2 tank is significantly greater than the LOX tank. That’s because LOX is 16 times more dense than LH2. Even multiplying by the mixture ratio of 5.8 still doesn’t bring the effect of the LOX tank up to parity with the LH2 tank.

     \begin{displaymath} \lambda = \frac{(0.061) + (0.127)}{6.596} = 0.028 \end{displaymath}

So the masses of the LH2 and LOX tanks will together be about 2.8% of the mass of the propellant they carry within them, and the LOX tank will account for (0.061/(0.061 + 0.127)) = 32% of this effect and the LH2 tank will account for (0.127/(0.061 + 0.127)) = 68% of this effect.

Now let’s try the calculation with a strange case–a nuclear thermal rocket using LH2 that has no oxidizer (MXR = 0).

     \begin{displaymath} \lambda = \frac{(0) \left(\dfrac{12 \text{kg/m}^3}{1142 \text{kg/m}^3}\right) + \left(\dfrac{9 \text{kg/m}^3}{71 \text{kg/m}^3}\right)}{(1 + 0) (1 - 0.03)} \end{displaymath}

     \begin{displaymath} \lambda = \frac{0.127}{0.97} = 0.131 \end{displaymath}

The factor is far, far worse in the case of a nuclear thermal rocket, at over 13% compared to less than 3% for the LH2/LOX case. This is because the only propellant is hydrogen and it has a very low density. So this is a major factor is performance difference between an NTR stage and a chemically-fueled stage–the tankage for the chemically-fueled stage will be far less.

To help get things started, here’s a table of tank factors based on both real and conceptual tank designs, from Langley Research Center. Remember to make sure that you’re using the same units for tank factor and propellant density (kg/m3 or lbm/ft3).

And here’s some different calculations I did based on a couple of different engines (including a nuclear thermal engine, which has no oxidizer at all and a mixture ratio of zero) showing the effects of propellant selection and mixture ratio on lambda.

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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

About Kirk Sorensen

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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5 Responses to Calculating Propellant-Mass-Sensitive Term

  1. Jonathan Goff Jonathan Goff says:


    Just a few points that might be worth adding. You didn’t mention it in the post, but the MER’s chart does mention tank ullage pressure (which will be a factor if you’re using a different feed pressure like say for a pressure fed tank). Also, you have to factor in what FOS you’re using on the tank structure.

    Also the “scales linearly with volume and pressure” assumption only works really well for larger tanks (ie bigger than the ones we’re working on at MSS), because you have certain structures like the skirts, flanges, sumps, slosh baffles, etc that are more nonlinear. Past a certain size they’re small enough that they’re in the noise, but below a certain size and they start becoming a bigger and bigger fraction of the tank weight. And that’s actually long before you hit actual minimum gage issues.


  2. Garland says:

    This is the reason I read Marvelous post.

  3. Pingback: Selenian Boondocks » Blog Archive » Using Payload Fraction Expressions in an Example

  4. Tretos says:

    Hi, Very interesting but how would you calculate Propellant Mass from LEO to the moon (including landing) ?

  5. Greg says:

    Why did you choose an ullage of 3%. Do you know where I could find a source for that?

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