Derivation of Propellant-Mass Sensitive Term

In a real rocket, the influence of propellant on the structures of the rocket is very significant. Tankage to hold propellant is typically the dominant structure of any rocket. The mass of that tankage is a complicated function of the loads on the tank as well as the density (and to some degree, temperature) of the propellants inside the tanks. Often, there is a very substantial difference between the density of the fuel used in the rocket and the density of the oxidizer. In some rockets, there is no oxidizer at all. It would be very helpful to have a way to “boil down” many of these factors into a single term that relates the masses of all of these tanks and structures to the overall mass of the propellant. Let us make that attempt.

Let us define a term that is dependent only on the mass of the propellant in the rocket:

     \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}

In preliminary rocket design, it in not uncommon to encounter various mass-estimating relationships (MERs) for propellant tanks and they are usually based on volume. They are simplifications but end up being very useful simplifications. For instance, a mass-estimating relationship might 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, let us rewrite the expression in terms of a factor (f) for the fuel and oxidizer tanks and their volumes. We will assume that we have obtained this factor from someone else’s work.

     \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 ($f_{FT}$) and oxidizer tank ($f_{OT}$) 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.

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.