2
$\begingroup$

Formulation - $$e^{\Delta v/v_{exhaust}} - 1=mass_{fuel}/mass_{structure}$$ $\text{let } k=e^{\Delta v/v_{exhaust}} - 1 $ $$\implies mass_{fuel}= k \times mass_{structure} $$ According to SMAD (Space Mission Analysis and Design book), overall tank weight is $1.25 (0.1\times mass_{fuel})$ (10% of propellant and 25% extra of the weight for PMD's and hardware) According to this assumption, $$ mass_{fuel}= k \times (mass_{structure-tank}+mass_{tank})$$ $$ mass_{fuel}= k \times (mass_{structure-tank}+0.125 \times mass_{fuel})$$ $$ mass_{fuel}\times (1-0.125k)=k \times mass_{structure-tank}$$ As k can reach 50, $mass_{fuel } $ would become -ve which should not be so. I would be grateful if you could tell me what exactly I have missed. Thanks!

$\endgroup$
  • $\begingroup$ If I'm understanding correctly, the 1.25(0.1 x mf) formulation determines what k is; you can't vary k while keeping the 1.25 and 0.1 constant. (Not sure why tank weight isn't just 0.125mf; is there something else being let out?) $\endgroup$ – Russell Borogove Mar 14 '18 at 18:17
  • $\begingroup$ @RussellBorogove $k=e^{\Delta v/v_{exhaust}}-1$, so k is a constant. Tank weight is 0.125 mf. $\endgroup$ – Infi Mar 15 '18 at 2:52
0
$\begingroup$

K value is too large. In most cases major $\Delta V$ is provided by gravity assists. So if you consider $\Delta V$ performed just by the propulsion systems, $k$ value would decrease drastically. An excerpt from SMAD $-$ $$$$ enter image description here

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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