# Propellant mass calculation

I have been given the following problem:

To solve it I used the formulas:

$v_e = I_{sp}.g_0$

$M_{propellant} = (e^{\Delta_v/v_e} - 1).M_{final}$

And I assumed that the apogee kick manoeuvre was executed first. So I started by calculating the propellant required for the second manoeuvre:

$v_e = 200\times9.81 = 1962m/s$

$M_{p2} = (e^{1050/1962} - 1).3000 = 2123.21Kg$

And then for the first manoeuvre I got:

$v_e = 320\times9.81 = 3139.21m/s$

$M_{p2} = (e^{2300/3139.2} - 1).(3000+2123.21) = 5536.29Kg$

Giving a total propellant mass of:

$M_p = 7659.5Kg$

However, this answer differs from the correct answer. Furthermore, if I consider the manoeuvres being performed in the reverse order, I get different masses for each propellant, but somehow, the total mass is the same, which seems strange, since the hint suggests that the order matters in the total amount of fuel required.

Could you please point out where did I make a mistake? Thank you.

• I get the same results as you if I work the equation the other direction (going from propellant amounts to ∆v, using trial and error to find the propellant amount). Where does the problem come from? It seems strange that it gives a "+/-2%" when there are no uncertainty figures given in the problem statement. Commented Sep 5, 2015 at 19:52
• This problem comes from a practice test for a course on thermal rocket propulsion. I think the +-2% is to say that the software would tolerate an answer within those margins. Probably to account for rounding errors in intermediate calculations. I'm beginning to think it's an issue with the software. I know that each student gets a slightly different version of the problem - it's possible that the solutions for the different versions got mixed up or something like that... Commented Sep 5, 2015 at 20:00
• Your calculations are correct. The only way they could differ due to the order, and the only way the mass could be less than your calculation, is if you threw off the empty apogee kick motor before using the hydrazine. However the problem does not provide that mass. You can calculate that mass from the answer though, which turns out to be 183 kg. So if 183 kg of the 3000 kg dry mass is discarded after the kick and before the auxiliary maneuvers, then you can get the 7390 kg of total propellant. Commented Sep 5, 2015 at 20:33
• Though that would be a 3.4% dry mass fraction, which is unreasonably low. Unless you're only throwing off part of the motor. Commented Sep 5, 2015 at 20:39
• Seeing as there is no information provided on this test about the weight of the apogee motor, I am more and more inclined to believe that this is an error within the test itself. I will contact my professor about this. Thank you to everyone who helped me with this problem! I was seriously thinking that I had failed to understand something. Commented Sep 5, 2015 at 20:51

## 1 Answer

As discussed in the comments, your calculated value is correct and the test solution is wrong.

I will however make a note about this part:

Furthermore, if I consider the manoeuvres being performed in the reverse order, I get different masses for each propellant, but somehow, the total mass is the same, which seems strange, since the hint suggests that the order matters in the total amount of fuel required.

This is expected!

let's rearrange your formulas slightly:

$$\frac{M_{start}}{M_{final}} = e^{\Delta v / v_e}$$

That is, each burn just describes a mass ratio of before and after the burn. Now consider the total mass of the rocket:

$$M_{total} = M_{dry} \cdot e^{\Delta v_1 / v_{e1}} \cdot e^{\Delta v_2 / v_{e2}}$$

Which is completely equal to doing the burns in the opposite order:

$$M_{total} = M_{dry} \cdot e^{\Delta v_2 / v_{e2}} \cdot e^{\Delta v_1 / v_{e1}}$$