Fuel Calculations

[RedDevilEA]

I'm trying to write a formula to determine which fuel tank to use for a rocket, based on the mass of the tank. I'm working backwards from the rocket equation:

ÃŽâ€-ν = ISP â— ln(MFull / MEmpty) â— 9.81

If I know the mass of the fuel tank I'm using, I can add it's empty mass to MEmpty and calculate the amount of fuel burned with this:

MFuel = MFull - MFull/{e^[ÃŽâ€-ν/(ISPâ—Â9.81)] }

My question is how do I determine the weight of the fuel tank I would need for a given ÃŽâ€-ν and ISP and mass of everything but the tank?

The assumption is that for whatever mass of the tank, 10% is the empty tank and 90% is the fuel.

Oh, man. I hope that made sense.

Edited March 5, 2014 by RedDevilEA
Answered

[Dave Kerbin]

You sort of go around a bit, but it sounds like you know your payload (engines, capsule, basically everything but the fuel tanks) and you know how much delta-v you need, you just need to know how many fuel tanks you need to bring along.

With the exception of the smallest fuel tanks (the Round8 and Oscar) all the stock fuel tanks have a ratio of 8:1 for fuel to tank. If we assume that we will only use those tanks we can figure it out without too much trouble, though because of the way the fuel ratios work (and the fact that I don't know how to format equations on these forums) I'll break the problem down into 2 parts.

The first part is figuring out what value is going into ln which is the wet to dry mass ratio. Let's calculate a value for MassRatio as follows,

MassRatio = EXP(DeltaV / 9.81 / ISP)

Using that ratio lets figure out the total weight of the fuel tanks (including the fuel in them)

WetTankMass = (9* Payload * (MassRatio - 1)) / ( 9 - MassRatio)

You now have the mass of your fuel tanks and fuel. So if WetTankMass = 4.5t that means you need exactly one T800 tanks (360 units of liquid fuel plus 440 units of oxidizer). If you want you can check that I haven't made a mistake (and that you got it correctly) by calculating the delta-v of a particular combination of engine and fuel tank, then reverse it by using my equestion to figure out the tank using the weight of the engine and the delta-v.

[capi3101]

Oh....Dave, that's wrong, the mass ratio is 9:1, not 8:1.

The rest of that is right, though - the mass of everything but your tanks can be broken down into individual components, the tank mass and everything else. That "everything else" is your dead mass, which you can represent by X, thus the rocket equation becomes:

dV = ln(M+x / Md + x) * Isp * Go

And now you put in the relationship M = 9Md:

dV = ln(9Md+x / Md + x) * Isp * Go

And solve the equation for Md:

e^(dV/(Isp*Go)) = y = 9Md+x / Md + x

9Md+x = yMd+yx

(9-y)Md = (y-1)x

Md = (y-1)x/(9-y)

Solve for that, then feed the result back into M = 9Md, and you have the fuel mass you need.

Then take that result and divide it by .05625, the full mass of an FL-T100 fuel tank, and round up; the result is the number of FL-T100 equivalents you'll need to generate roughly the amount of delta-V you want. By rounding up, you don't short yourself. An FL-T200 is the same as 2 FL-T100s, an FL-T400 is the same as 4 FL-T100s, an FL-T800/X200-32 is the same as 8 FL-T100s, an X200-16 is the same as 16 FL-T100s, an X200-32 is the same as 32 FL-T100s, and a Jumbo64 (the orange tank) is the same as 64 FL-T100s.

Equation in action: I want 4550 m/s of delta V with a 300 Isp Engine, with 1 tonne of deadmass. Plug everything in:

y = e^(dV/(Isp*Go)) = e^(4550/(300*9.81)) = 4.693

Md = (y-1)x/(9-y) = ((4.693-1)*1)/(9-4.683) = 3.693 / 4.317 = .855

M = 9Md = 7.699, 7.699 / 0.5625 = 13.68, round up to 14 FL-T100s (A combination FL-T800, FL-T400 and FL-T200 stack).

Hope that helps.

Edited March 5, 2014 by capi3101

[RedDevilEA]

Wow, that's awesome. Thank you so much!

[Dave Kerbin]

Oh....Dave, that's wrong, the mass ratio is 9:1, not 8:1.

The ratio between a full and empty tank is 9:1 and it's the easier number to use in most equations. That's why it's easy to overlook that the ratio between the fuel and the tank itself is 8:1. For example a full Rockomax X200-16 is composed 1 of ton of kaluminum tank, 3.6 tons of liquid fuel and 4.4 tons of oxidizer. 9:1 is easiest to use for math problems while 8:1 is easiest to use when explaining how fuel storage works: You need 1 ton of tank to hold 8 tons of fuel.

[capi3101]

Ah. Okay. In any event, it looks like we both wound up giving the exact same advice, just with different terms.

Let me ask you this: have you ever come across a situation where the MassRatio is greater than nine (i.e. a negative result)? That's happened to me a few times before and I'm not sure what exactly that signifies.

[Dave Kerbin]

Ah. Okay. In any event, it looks like we both wound up giving the exact same advice, just with different terms.

Let me ask you this: have you ever come across a situation where the MassRatio is greater than nine (i.e. a negative result)? That's happened to me a few times before and I'm not sure what exactly that signifies.

It means that what you are asking is impossible. This generally happens when you ask for the fuel requirements to give a very large payload a very large delta-v. As you know there are diminishing returns as you add more fuel because some of that new fuel will be burned to push all the other fuel and tanks. It reaches a tipping point where the fuel mass you add (say 8 tons) is not enough to even move the fuel tank (1 ton) that it came in. At that point the MassRatio formula is basically going to come back and say you need some kind of negative-mass fuel to get that much delta-v for the given ISP constant. If you follow through the formula I think one of the terms ends up being infinity.

[RedDevilEA]

I used 10:1 because I'm using KW tanks. 20t tank = 2 tons empty. At any rate, thanks for the replies, guys. I was completely lost.

[capi3101]

It means that what you are asking is impossible. This generally happens when you ask for the fuel requirements to give a very large payload a very large delta-v. As you know there are diminishing returns as you add more fuel because some of that new fuel will be burned to push all the other fuel and tanks. It reaches a tipping point where the fuel mass you add (say 8 tons) is not enough to even move the fuel tank (1 ton) that it came in. At that point the MassRatio formula is basically going to come back and say you need some kind of negative-mass fuel to get that much delta-v for the given ISP constant. If you follow through the formula I think one of the terms ends up being infinity.

I was afraid you were going to tell me that...I have a challenge going on and was having issues applying the equation for my VAB entry. Tiny one too, though apparently not tiny enough...

[RedDevilEA]

Out of curiosity, how did you get from 9Md+x = yMd+yx

to

(9-y)Md = (y-1)x ?

[mhoram]

I was afraid you were going to tell me that...I have a challenge going on and was having issues applying the equation for my VAB entry. Tiny one too, though apparently not tiny enough...

If you need more dV in a single stage, you could combine a LV-N with multiple lower-ISP-engines and let the fuel to them run out. Tavert managed in my Delta-V Maximization Challenge to pack 9124 m/s dV into a single stage.

[capi3101]

Out of curiosity, how did you get from 9Md+x = yMd+yx

to

(9-y)Md = (y-1)x ?

Start: 9Md+x = yMd+yx

Subtract yMd from both sides: 9Md + x - yMd = yx

Subtract x from both sides: 9Md - yMd = yx - x

Take out the common factors: (9-y)Md = (y-1)x

And there you go.

[RedDevilEA]

Start: 9Md+x = yMd+yx

Subtract yMd from both sides: 9Md + x - yMd = yx

Subtract x from both sides: 9Md - yMd = yx - x

Take out the common factors: (9-y)Md = (y-1)x

And there you go.

Thanks. I am not a clever man.