5
$\begingroup$

enter image description here

First of all this code is not mine I am trying to understand how the equation for m_u works. This comes from a master thesis called Trajectory design for a Titan mission using the Direct Fusion Drive as shown it is a different variant of the Tsiolkovsky rocket equation I believe for the propellant weight on the rocket in this case it is being used to plot the change in delta V relevant to mission length so Delta-t is total number of seconds in 0 to 15 years. Can anyone link to elsewhere this equation can be found?

To make it clearer the equation is enter image description here

Where Ve is (Isp * Gravity constant).

Link to thesis is https://webthesis.biblio.polito.it/15184/1/tesi.pdf

$\endgroup$
3
  • $\begingroup$ Do you have a link to the thesis? $\endgroup$ Commented Jul 2, 2022 at 14:00
  • 1
    $\begingroup$ @OrganicMarble Here is the link webthesis.biblio.polito.it/15184/1/tesi.pdf $\endgroup$ Commented Jul 2, 2022 at 14:21
  • $\begingroup$ i know specific power/ total efficiency gets you the beam efficiency in Ion thrusters so i assume its similar in the fusion drive discussed in the thesis i just can find a reference to how that effects the fuel amount $\endgroup$ Commented Jul 2, 2022 at 14:23

1 Answer 1

4
$\begingroup$

This is my first answer, if this won't be clear tell me and I'll write it better. I've seen that equation when treating electric propulsion, I'll write the derivation here. You can write your payload mass as the initial mass of your spacecraft, minus the mass of the electric power generator $m_s$ and the propellant mass $m_p$, so you can write the payload fraction as $$\frac{m_u}{m_0}=1-\frac{m_s}{m_0}-\frac{m_p}{m_0}$$

The propellant mass is the only one ejected from the spacecraft, so it's the difference between initial and final mission's mass: $m_p=m_0-m_f$. Put that in the payload fraction and you get $$\frac{m_u}{m_0}=\frac{m_f}{m_0}-\frac{m_s}{m_0}$$

You can think of the electrical generator mass being proportional to the generated power $P_E$ through a constant specific power $\alpha$, so $m_s=\alpha P_E$. The electrical power dictates a variation in kinetic energy of the spacecraft through the global efficiency of the propulsor $\eta$, so $\eta P_E = \frac{1}{2}\dot{m}_pc$, where $c$ is the exhaust speed. Since the mass flow is related to the thrust $T$ by $\dot{m}_pc=T$ you can write $m_s$ as: $$m_s = \frac{\alpha}{\eta}\frac{Tc}{2}$$

We can now write the "generator mass fraction" $\frac{m_s}{m_0}$ by using the definition of mass flow $\dot{m}_p = \frac{m_p}{\Delta t}$, where $\Delta t$ is the time span of thrust and the rocket equation $\frac{m_p}{m_0}=(1-e^{-\frac{\Delta V}{c}})$: $$\frac{m_s}{m_0}=\frac{\alpha}{\eta}\frac{T}{m_0}\frac{c}{2}$$ $$\frac{m_s}{m_0}=(1-e^{\frac{\Delta V}{c}})\frac{\alpha}{\eta}\frac{c^2}{2\Delta t}$$

Looking again at the payload fraction $\frac{m_u}{m_0}$ you can write the final to initial mass fraction with the rocket equation as $\frac{m_f}{m_0}=e^{-\frac{\Delta V}{c}}$, substitute the $\frac{m_s}{m_0}$ calculated before, and multiply and divide everything by $\Delta V^2$ to get: $$\frac{m_u}{m_0}=e^{-\frac{\Delta V}{c}}-\frac{\alpha}{\eta}(\frac{c}{\Delta V})^2\frac{(\Delta V)^2}{2\Delta t}(1-e^{-\frac{\Delta V}{c}})$$

$\endgroup$
6
  • 2
    $\begingroup$ Great answer, tiny mistake: you switched around mf and m0 for the mp equation $\endgroup$
    – Ruben
    Commented Jul 4, 2022 at 12:13
  • 1
    $\begingroup$ Oh you're right, I edited it $\endgroup$
    – JuliusC
    Commented Jul 4, 2022 at 12:21
  • 1
    $\begingroup$ Outstanding answer, welcome to space stack exchange. $\endgroup$ Commented Jul 4, 2022 at 12:26
  • 1
    $\begingroup$ you sir a legend thank you so much $\endgroup$ Commented Jul 4, 2022 at 16:39
  • $\begingroup$ Do you have a link to a text book or reference i can learn more from and where i can reference the equation came from $\endgroup$ Commented Jul 4, 2022 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.