2
$\begingroup$

My idea: If the propellent mass to tank mass ratio remains constant when increasing/decreasing the amount of propellant then you could increase $\sum \Delta v$, by splitting the propellent tank and leaving it behind once empty.

My professor disagrees that the ratio stays constant. His counter argument was that as the number of tanks tend to infinity, you would have more tank mass than propellent mass because of increased propellent surface area and in turn would have a lower $\sum \Delta v$.

2nd idea: The thickness of the propellent tanks would decrease as you increase the number of tanks keeping the propellent mass to tank mass ratio constant. I am not sure if this is right.

In any case, how can I find the optimal number of propellent tanks?

This is for a project to find the fastest time we can reach Alpha Centauri in the near future.

Improve this question
$\endgroup$
1
  • $\begingroup$ If you scale all measures of a tank by the same factor, surface scales by the square of that factor and volume and weight of the tank scale by the cube. Tank wall thickness scales linearily. But when you keep wall thickness constant, surface and tank structure weight scales by the square. When tank pressure is constant, wall thickness should scale linearily to keep material stress constant. So constant wall thickness is too optimistic. $\endgroup$
    – Uwe
    Apr 4, 2019 at 11:06

1 Answer 1

Reset to default
2
$\begingroup$

You are presumably going to make the tanks as thin-walled (more generally low-mass) as you can. To understand this you need to know what stresses the tank walls are there to resist. To a first approximation, there are two: the pressure of the fuel inside (if the fuel is a gas to a liquid with significant vapour pressure) and the forces due to the acceleration of the rocket. If you are going to alpha centauri you are almost certainly limited by the maount of propellant you have, rather than the acceleration, so we can assume you accelerate very gently and ignore those stresses.

The forces due to fuel pressure can be modelled easily enough. We can assume the tank is spherical (the maximum volume for a given surface area, and a strong shape for resisting pressure). Suppose the radius is $R$ and the pressure $P$ and the tank thickness $t$. The tension in the skin of the tank is fairly easily calculated to be $PR/t$. For a given tank material this is limited to some yield strength $Y$ so we get $t = PR/Y$.

So now the mass of the tank is $4\pi t R^2\rho$ ($\rho$ is the density of the material) which is $4\pi PR^3 \rho/Y$.

For gaseous fuel, the mass of fuel is proportional to the volume times the pressure, so to $PR^3$ so the mass of the tank is actually a fixed proportion of the mass of the fuel, depending only on properties of the tank material. Given that, we may as well use small tanks and drop them as we go along.

For liquid fuel, $P$ is a constant, depending on temperature and fuel but teh above calculation still works, although if $P$ is small, the tension in the tanks will be small and it will be hard to keep other forces low enough.

For solid fuel (well below its sublimation point) no tanks at all are needed.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I don't think t may be constant if P is constant. $\endgroup$
    – Uwe
    Apr 4, 2019 at 11:14
  • $\begingroup$ The mass of a spherical fuel tank would be $4 \pi tR^2 \rho$ only if $t \ll R$ right? If so, avoiding approximations, there should be an optimal number of fuel tanks. $\endgroup$
    – user29999
    Apr 11, 2019 at 0:17

Your Answer

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