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I was playing around with the math for getting a pressure-fed rocket to orbit and came across something that I haven't seen addressed anywhere.

In a pressure-fed rocket, the chamber pressure of the engine is related to the pressure of the propellant tanks. It is equal to the pressure in the propellant tanks minus the pressure drops in the feed system and across the injector.

The ISP of a rocket engine increases with chamber pressure. A higher ISP means more delta-V with a given mass fraction. However, getting a higher chamber pressure requires more tank pressure, which requires stronger tanks, which requires more wall thickness, which will increase tank mass. As far as I understand, this is a ROUGHLY linear relationship (e.g. double the tank pressure, double the wall thickness, double the mass).

The weird part is that (after playing around in RPA) ISP doesn't seem to decrease linearly with chamber pressure. That means that decreasing tank pressure (and thus chamber pressure) decreases ISP but actually INCREASES delta-V because the mass fraction improves at a faster rate than ISP falls off.

My first question: Is it correct that ISP does not fall off linearly with chamber pressure?

My second question: Are there other inherent factors that stop delta-V from increasing with decreasing ISP?

My third question: What factors (if any) prevent a rocket from taking advantage of this and running at extremely low chamber pressures (like 5bar)?

Follow-up question: I assume combustion instability is the limiting factor, correct? Can that be mitigated by using more, smaller nozzles?

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    $\begingroup$ And that’s why orbital rockets are fed by a turbo pump ;) $\endgroup$
    – Antzi
    Sep 16 '20 at 16:58
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    $\begingroup$ Don't forget that a rocket needs to provide a certain thrust - at half the pressure all the plumbing needs to be twice as large. $\endgroup$
    – asdfex
    Sep 16 '20 at 17:09
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    $\begingroup$ @asdfex For sure. One thing I learned while trying to find answers to my question was that the Sea Dragon concept rocket was designed with a pressure-fed engine. According to Wikipedia the engine would operate at <300psi (~20bar). I guess that's why the nozzle needed to be larger in diameter than the rocket itself (to achieve sufficient mass flow rates at low pressures). $\endgroup$ Sep 16 '20 at 17:17
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    $\begingroup$ I suspect there's a rather large book or two which deal with all the conflicting parameters involved here. $\endgroup$ Sep 17 '20 at 12:41
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    $\begingroup$ @CallMeTom What are you talking about here "same amount of fuel mass but half the pressure means double the tank size" ? You can't be talking about liquid propellants can you? $\endgroup$ Sep 18 '20 at 21:36
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First of all: great observation! This is indeed the reason why pressure fed rocket engines are limited in possible chamber pressure, the added weight from the tanks isn't worth it at a certain point. Which is why we have pump fed rocket engines.

Question 1:

Some equations from Ideal Rocket Theory:

Specific Impulse is the equivalent velocity divided by standard gravity $$I_{sp} = \frac{v_{eq}}{g_0}$$

Standard gravity is a constant so $I_{sp}$ scales linearly with the equivalent velocity. For simplicity's sake we assume the equivalent velocity to be equal to the exhaust velocity, which is given by the following equation:

$$v_e=\sqrt{\frac{2\gamma}{\gamma-1}\cdot R\cdot T_C \bigg[1-\Big(\frac{p_e}{p_C}\Big)^\frac{\gamma-1}{\gamma}\bigg]}$$

Now we aren't really interested in any of the other parameters, we just want to know what happens if we change the chamber pressure, $p_C$. So we make all the stuff before the brackets 1 constant, let's call it $a$. (Important to note is that the chamber temperature $T_C$ is thus assumed constant, while it actually depends on $p_C$, but let's ignore that for simplicity as well. $\gamma$ and $R$ aren't dependent on $p_C$ for an ideal gas)

Furthermore we fill in some typical values for the ratio of specific heats $\gamma$ and the exit pressure, $p_e$.Let's say $1.2$ and $10^5\ Pa$ (approximately 1 atmosphere) respectively. You then get:

$$v_e=a\sqrt{1-\Big(\frac{10^5}{p_C}\Big)^\frac{.2}{1.2}}\approx a\sqrt{1-6.8\Big(\frac{1}{p_C}\Big)^{0.17}}$$

Obviously, that isn't linear, but just to give you an idea of what it looks like if you plot this with $\frac{v_e}{a} = y$ and $p_C = x$:

Rough indication of relation chamber pressure and exhaust velocity(divided by a constant)

Question 2:

As you can see from the graph above, the exhaust velocity drops faster and faster if you decrease the chamber pressure, so at a certain point the opposite of your observation will be true: loss of pressure will not be worth the decrease in $I_{sp}$ relative to the savings in weight.

Question 3:

I think should be answered by the previous two: There is an optimal pressure for the design of a pressure fed engine. A lower pressure will mean a bigger decrease in $\Delta V$ due to less specific impulse than an increase in $\Delta V$ due to weight savings. A higher pressure than this optimum will be the opposite: less gain in DV due to increased specific impulse than loss due to increased weight.


Just as an extra: a graph that shows some rockets plotted by their tank volume and tank pressure. It shows that choosing between pressure or pump fed is actually largely determined by size (volume of propellant). It also shows that the lowest pressure designed pressure-fed engine had a tank pressure of about 2-3 MPa, so 20-30 bar. Even accounting for feed line and injector losses, the chamber pressure would be way higher than 5 bar. (It uses data from 1975 though, so there might be newer engines with lower operating pressures)

tank volume vs pressure graph Graph taken from Humble's Space Propulsion Analysis and Design (1995). The 1975 NASA report it references only has the data, not the graph.

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  • $\begingroup$ Could you possibly resize that first graph so we can see the numbers/read the axes? Thanks $\endgroup$
    – Ingolifs
    Oct 5 '20 at 1:47
  • $\begingroup$ No, I don't have the original image anymore. But it is just meant as a vague indication anyway, because we made some over simplifying assumptions and the y-axis values aren't useful ($\frac{v_e}{a}$ isn't a very useful parameter except to illustrate the point made in this answer). Also the image is pretty decent quality, if you click on it you should be able to read it, no? $\endgroup$
    – Ruben
    Oct 5 '20 at 7:50

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