Payload fraction derivation for vehicle with split delta-V (case #1)

Posted on March 14, 2021 by Kirk Sorensen

Consider a vehicle that undertakes a first $\Delta v$, then picks up a payload and undertakes a second $\Delta v$. It carries the propellant for both maneuvers, but only on the second maneuver does it have the added mass of the payload. This situation might be representative of:

1. a one-way lunar vehicle, arriving in low Earth orbit fully fueled but with no payload. It executes a trans-lunar injection burn and then a lunar orbit insertion burn, all with no payload. It then picks up a payload in lunar orbit and descends to the surface with the payload and lands with essentially all its propellant expended.

2. a reusable lunar lander, based on the surface of the Moon, fully fueled by lunar propellant but lacking any payload, which ascends to a lunar orbit and recovers a payload in that orbit, then descends to the surface, landing with nearly all its propellant expended.

3. a space tug that departs for geosynchronous orbit and recovers a satellite, then returns with it to a low-Earth orbit for repair.

First, define the mass conditions at the beginning and end of $\Delta v_1$:

(1)   \begin{equation*} \eta_1 \equiv \exp(\Delta v_1/v_e) = \frac{m_\text{vehicle} + m_\text{prop1} + m_\text{prop2}}{m_\text{vehicle} + m_\text{prop2}} \end{equation*}

Alternatively, and just as importantly, the conditions bracketing $\Delta v_1$ can be described in terms of an initial mass:

    \begin{displaymath} \eta_1 = \frac{m_\text{initial}}{m_\text{initial} - m_\text{prop1}} \end{displaymath}

This expression can be conveniently rearranged to yield the propellant mass consumed by the vehicle in $\Delta v_1$:

    \begin{displaymath} m_\text{prop1} = m_\text{initial} \left(1 - \dfrac{1}{\eta_1}\right) \end{displaymath}

In a similar manner, we define the mass conditions at the beginning and end of $\Delta v_2$:

(2)   \begin{equation*} \eta_2 \equiv \exp(\Delta v_2/v_e) = \frac{m_\text{vehicle} + m_\text{payload} + m_\text{prop2}}{m_\text{vehicle} + m_\text{payload}} \end{equation*}

We can also express the conditions bracketing $\Delta v_2$ in another way, in terms of initial mass:

    \begin{displaymath} \eta_2 = \frac{m_\text{initial} + m_\text{payload} - m_\text{prop1}}{m_\text{initial} + m_\text{payload} - m_\text{prop}} \end{displaymath}

Yet another approach will later yield a useful relationship:

    \begin{displaymath} \eta_2 = \frac{m_\text{initial}/\eta_1 + m_\text{payload}}{m_\text{vehicle} + m_\text{payload}} \end{displaymath}

    \begin{displaymath} \eta_2\left(m_\text{vehicle} + m_\text{payload}\right) = m_\text{initial}/\eta_1 + m_\text{payload} \end{displaymath}

    \begin{displaymath} \eta_2 m_\text{vehicle} + \eta_2 m_\text{payload} = m_\text{initial}/\eta_1 + m_\text{payload} \end{displaymath}

    \begin{displaymath} \eta_2 m_\text{vehicle} + (\eta_2 - 1) m_\text{payload} = m_\text{initial}/\eta_1 \end{displaymath}

    \begin{displaymath} m_\text{initial} = \eta_1\eta_2 m_\text{vehicle} + \eta_1(\eta_2 - 1) m_\text{payload} \end{displaymath}

The propellant mass consumed by the vehicle in $\Delta v_2$ can also be expressed in a manner analogous to prop1:

    \begin{displaymath} m_\text{prop2} = \left(1 - \dfrac{1}{\eta_2}\right)\left(\dfrac{m_\text{initial}}{\eta_1} + m_\text{payload}\right) \end{displaymath}

    \begin{displaymath} m_\text{prop2} = m_\text{initial}\left(\dfrac{1}{\eta_1} - \dfrac{1}{\eta_1\eta_2}\right) + m_\text{payload}\left(1 - \dfrac{1}{\eta_2}\right) \end{displaymath}

Now we are positioned to calculate the total propellant load:

    \begin{displaymath} m_\text{prop} = m_\text{prop1} + m_\text{prop2} \end{displaymath}

substituting the definitions for prop1 and prop2

    \begin{displaymath} m_\text{prop} = m_\text{initial} \left(1 - \dfrac{1}{\eta_1}\right) + m_\text{initial}\left(\dfrac{1}{\eta_1} - \dfrac{1}{\eta_1\eta_2}\right) + m_\text{payload}\left(1 - \dfrac{1}{\eta_2}\right) \end{displaymath}

collecting terms and simplifying

    \begin{displaymath} m_\text{prop} = m_\text{initial} \left(1 - \dfrac{1}{\eta_1} + \dfrac{1}{\eta_1} - \dfrac{1}{\eta_1\eta_2}\right) + m_\text{payload}\left(1 - \dfrac{1}{\eta_2}\right) \end{displaymath}

(3)   \begin{equation*} m_\text{prop} = m_\text{initial} \left(1 - \dfrac{1}{\eta_1\eta_2}\right) + m_\text{payload}\left(1 - \dfrac{1}{\eta_2}\right) \end{equation*}

now let us define the vehicle’s “dry” mass entirely in terms of initial-mass-sensitive ($\phi$), propellant-mass-sensitive ($\lambda$), and payload-mass-sensitive ($\epsilon$) mass terms. This is a substantial simplification, but it should do for now.

    \begin{displaymath} m_\text{vehicle} = \phi m_\text{initial} + \lambda m_\text{prop} + \epsilon m_\text{payload} \end{displaymath}

    \begin{displaymath} m_\text{initial} = \eta_1(\eta_2 - 1)m_\text{payload} + \eta_1\eta_2 m_\text{vehicle} \end{displaymath}

substituting the definition of the vehicle’s mass in

    \begin{displaymath} m_\text{initial} = \eta_1(\eta_2 - 1)m_\text{payload} + \eta_1\eta_2(\phi m_\text{initial} + \lambda m_\text{prop} + \epsilon m_\text{payload}) \end{displaymath}

we collect terms related to the initial mass on the left hand side

(4)   \begin{equation*} m_\text{initial}(1 - \phi\eta_1\eta_2) = m_\text{payload}(\eta_1(\eta_2 - 1) + \epsilon\eta_1\eta_2) + \lambda\eta_1\eta_2 m_\text{prop} \end{equation*}

Take equation(3) and multiply it through by $\lambda\eta_1\eta_2$:

    \begin{displaymath} \lambda\eta_1\eta_2 m_\text{prop} = \lambda\eta_1\eta_2 m_\text{initial} \left(1 - \dfrac{1}{\eta_1\eta_2}\right) + \lambda\eta_1\eta_2 m_\text{payload}\left(1 - \dfrac{1}{\eta_2}\right) \end{displaymath}

    \begin{displaymath} \lambda\eta_1\eta_2 m_\text{prop} = m_\text{initial} \left(\lambda\eta_1\eta_2 - \dfrac{\lambda\eta_1\eta_2}{\eta_1\eta_2}\right) + \lambda\eta_1\eta_2m_\text{payload}\left(\dfrac{\eta_2 - 1}{\eta_2}\right) \end{displaymath}

(5)   \begin{equation*} \lambda\eta_1\eta_2 m_\text{prop} = m_\text{initial} \left(\lambda\eta_1\eta_2 - \lambda\right) + m_\text{payload}\lambda\eta_1(\eta_2 - 1) \end{equation*}

then substitute the RHS for equation(5) in equation(4) and collect terms

    \begin{displaymath} m_\text{initial}(1 - \phi\eta_1\eta_2 - \lambda\eta_1\eta_2 + \lambda) = m_\text{payload}(\eta_1(\eta_2 - 1) + \epsilon\eta_1\eta_2 + \lambda\eta_1(\eta_2 - 1)) \end{displaymath}

further simplifying

    \begin{displaymath} m_\text{initial}(1 - (\phi + \lambda)\eta_1\eta_2 + \lambda) = m_\text{payload}(\eta_1(\eta_2 - 1)(1 + \lambda) + \epsilon\eta_1\eta_2) \end{displaymath}

With all terms relating only to initial mass and payload mass, a general expression for payload fraction can at last be defined:

    \begin{displaymath} \dfrac{m_\text{payload}}{m_\text{initial}} = \dfrac{1 - (\phi + \lambda)\eta_1\eta_2 + \lambda}{\eta_1(\epsilon\eta_2 + (\eta_2 - 1)(1 + \lambda))} \end{displaymath}

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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

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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2 Responses to Payload fraction derivation for vehicle with split delta-V (case #1)

  1. gbaikie says:
    March 19, 2021 at 5:25 pm

    I wondering about landing a lunar lander on a lunar lander.
    First I think one should have a lunar ascent vehicle which has 2 stages.
    And first stage adds about 500 m/s, and second stage goes to lunar orbit and returns to
    the lunar surface. But what wondering about was landing the 2nd stage “back on” the first stage. Or the first stage is mobile landing/launch pad. And could use mostly before one makes “permanent” landing pads- or for the early stages of lunar {or Mars} development or exploration.

  2. TheRadicalModerate says:
    March 20, 2021 at 12:27 am

    Gotta love some gnarly algebra.

    So what are reasonable values for ϕ, λ, and ε? Three cases to think about:

    1) Standard mass estimating relationships, as sort of a baseline.

    2) An aerobraking case for cheap return to LEO. (ϕ would be larger to account for a ship-sized aeroshell or ballute-like thingy.)

    3) A case for a “flimsy” initial prop tank. The idea here is that, since most prop is burned during your departure maneuver (the first part of η₁), it could be serviced by a very low pressure tank. That likely requires just-in-time fueling on-orbit to avoid massive boiloff, but note that tank mass is roughly proportional to pressure * volume, so taking a flimsy tank from 2ish bar down to 20kPa, with no insulation, could result in a substantial reduction in λ. Then you’d have a sorta-kinda regular tank for the later burns (the remainder of η₁ and η₂). Note that a tradeoff here is that you need a bigger pump for your engine(s), which increases ϕ somewhat.

    Note that there’s some more parameterization to do to make the benefits (or lack thereof–I haven’t done more than a hand-wave at the analysis) of the flimsy tank clear. I think that v_departure / (v_arrival + v₂) will turn out to be a figure of merit, but that’s as far as I’ve gotten. (Just thought I’d dangle that in case anybody’s got a bad case of algebra fever…)

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