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Formulation:
$$e^{\Delta v/v_{exhaust}} - 1=\frac{mass_{fuel}}{mass_{structure}}$$ $$\text{let } k=e^{\Delta v/v_{exhaust}} - 1 \implies mass_{fuel}= k \times mass_{structure} $$ According to SMAD (Space Mission Analysis and Design book), overall tank weight is $1.25\times (0.1\times mass_{fuel})$ (the meaning behind those values is 10% of propellant and 25% extra of the weight for PMDs 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!

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  • $\begingroup$ 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, 2018 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, 2018 at 2:52

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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

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