# Overcoming pressure differential to feed fuel into rocket engines

When pumping fuel and oxidizer into a rocket engine, I presume the pump must create more pressure than the inside of the combustion chamber, otherwise the liquid would back-flow. Is this right? Then a rocket that uses a pressurized fuel tank instead of pumps would have to have more pressure in the whole fuel tank than in the engine! So why have the engine at all? the pumps have already achieved the pressure! The full flow Raptor engine really baffles me in this regard. There must be more to it. Can someone explain how this works?

• – Erik Jan 3 '17 at 20:44

You are correct, the injection pressure is larger than the chamber pressure of the engine. According to "Modern Engineering Design of Liquid Propellant Rocket Engines", the Space Shuttle Main Engine featured turbopump characteristics as follows:

1.) High-Pressure Oxidizer Turbopump (HPOTP) boost discharge pressure is 6,952.2 psia, which flows into the oxidizer preburner.
2.) High-Pressure Fuel Turbopump (HPFTP) discharge pressure is 6,024.8 psia, which flows into the fuel preburner.

All the while, the SSME had a nominal chamber pressure of approximately, 3,000 psia.

As for your question, why even have the engines, if the turbopumps can supply propellant at such high pressures? Well, the answer to your question is the role of combustion. First off, we must look at what is our equation for thrust. From simple control volume analysis, we obtain a thrust equation of the form, $$F = \dot{m}v_2 + (p_2-p_3)A_2$$ where $F$ is the thrust force, $\dot{m}$ is the engine flow rate, $v_2$ is the gas velocity at the nozzle exit, $p_2$ is the gas pressure at the nozzle exit, $A_2$ is the cross sectional area at the nozzle exit, and $p_3$ is the ambient pressure outside the engine. Using quasi one-dimensional gas dynamics and isentropic relations, we can write the nozzle exit velocity in the following form,

$$v_2 = \sqrt[]{\frac{2\gamma}{\gamma-1} R T_1 \left[1-\left(\frac{p_2}{p_1}\right)^{(\gamma-1)/\gamma}\right]}$$

where $\gamma$ is the ratio of specific heats, $R$ is the specific gas constant, and $T_1$ is the chamber temperature (very important for combustion). As shown here, $v_2 \sim \sqrt[]{T_1}$. Similarly, the mass flow rate through the engine can be obtained from the mass flow parameter, which we will assume choked flow at the throat, $$\dot{m} = \frac{A_t p_1}{\sqrt[]{T_1}} \ \sqrt[]{\frac{\gamma}{R}} \left(\frac{\gamma+1}{2}\right)^{-(\gamma+1)/[2(\gamma-1)]}$$

where $p_1$ is the chamber pressure. Once again, we can see that mass flow rate of the engine is influence by the chamber temperature as $\dot{m} \sim 1/\sqrt[]{T_1}$. Combining these two expressions into the equation of thrust yields,

$$F = A_t p_1 \sqrt[]{\frac{2\gamma^2}{\gamma-1}\left(\frac{2}{\gamma+1}\right)^{(\gamma+1)/(\gamma-1)}\left[1-\left(\frac{p_2}{p_1}\right)^{(\gamma-1)/\gamma}\right]} + (p_2-p_3)A_2$$

Now we will consider some hypothetical scenarios in regard to your question. First we will look at the conventional rocket engine. Suppose we have a chamber pressure similar to the SSME at approximately, $p_1$ = 3,000 psia. Also, for simplifying analysis, suppose we have a nozzle that perfectly expands the gas to atmospheric pressure at sea level, which would yield a nozzle pressure ratio of $p_2/p_1$ = 14.7/3000 = 0.0049. Additionally, we will assume the throat area is 0.5 ft$^2$. Since we expand into the atmospheric condition, $p_2-p_3 = 0$ and we have no pressure area force in our equation and the thrust is purely momentum flux through the nozzle. The final thing to determine is the ratio of specific heat $\gamma$ for the engine. We will assume we are using LOX/LH2 propellant, which features a burned product ratio of specific heat of approximately $\gamma \approx$ 1.2 at 6,000 F. Substituting all of this into our expression for thrust we obtain for a rocket (very similar to the SSME, as expected): $$\boxed{F = 372,064 \ \ \text{lbf} \\ \dot{m} = 1,030 \ \text{lbm/s} \\ v_2 = 11,614 \ \text{ft/s} \\ I_{sp} = 361 \ \text{sec} }$$

Now let us consider your case, what if we only used the turbopump injection pressure as our means of thrust generation. For this case, we will assume the injection pressure is the discharge pressure from the SSME and nominally equal to $p_1$ = 6,500 psia. In this case, as we are abandoning the engine concept, hence, we do not require combustion and can focus on simply blasting O$_2$ at 6,500 psia through a nozzle that perfectly expands to atmospheric sea level condtitions. Now visual inspection of the SSME plumbing system indicates that the 6,500 psia is going into a plumbing network that is roughly $\approx$1/5th the diameter of the primary injection lines at a temperature of -258 F (above boiling point for LOX). Moreover, the primary thrust chamber feed lines are operating nominally at roughly 3,100 psia and 728 F, so we can consider 2 cases for your proposed question. The first, is considering an injection pressure and temperature of $p_1$ = 6,500 psia and $T_1$ = -258 F, but with a throat area of 1/5th our rocket example above for choked flow. Now in these cases we do not have combustion and our ratio of specific heats will stay at roughly $\gamma$ = 1.4. The second case we will consider is maintaining the same throat area from the previous rocket example, but with an injection pressure and temperature of 3,100 psia and 728 F. The results for your case go as follows:

Your case 1: $p_1$ = 6,500 psi, $T_1$ = -258 F, and $A_t$ = (1/25)(0.5 ft$^2$) = 0.02 ft$^2$

$$\boxed{F = 30,796 \ \text{lbf} \\ \dot{m} = 215 \ \text{lbm/s} \\ v_2 = 4,613 \ \text{ft/s} \\ I_{sp} = 143 \ \text{sec} }$$

Your case 2: $p_1$ = 3,100 psi, $T_1$ = 758 F, and $A_t$ = 0.5 ft$^2$

$$\boxed{F = 357,865 \ \text{lbf} \\ \dot{m} = 6,212 \ \text{lbm/s} \\ v_2 = 1,854 \ \text{ft/s} \\ I_{sp} = 58 \ \text{sec} }$$

In both of your cases, we come up short as far as the thrust produced and specific impulse without combustion. In fact, the thrust produced in the second scenario is comparable to the thrust produced for a engine similar to the SSME, however, the specific impulse is horrendous. Operating the rocket in this fashion would be light years beyond inefficient. The inefficiency comes from the low exhaust velocity as a consequence of $v_2 \sim \sqrt[]{T_1}$ and the large mass flow rate as a consequence of $\dot{m} \sim 1/\sqrt[]{T_1}$. Hence, a way to obtain high $v_2$ and low $\dot{m}$ for the same throat area, is to obtain a large $T_1$, which requires combustion. Even if we considered using the H$_2$ line instead of the O$_2$, we would only obtain an $I_{sp}$ of 167 for the latter scenario, which is comparable to a solid rocket booster, while using a liquid propellant, i.e. awful performance.

• Not that it changes your excellent argument, but the SSME throat area was about 0.65 ft^2 (93 in^2). nasa.gov/sites/default/files/files/3PFD.pdf – Organic Marble Jan 3 '17 at 12:44
• I've always looked for the SSME nozzle throat dimensions, but could never come across it! Thanks! – TRF Jan 3 '17 at 15:33
• This is not an answer, it's a guy showing off his math. It is not required for an explanation, there is no hard-science tag. This answer could easily be improved by removing the unnecessary math and offering a simple explanation instead. KISS. – Innovine Jan 20 '17 at 18:16
• If only rocket engine design question didn't involve math... – TRF Jan 20 '17 at 20:39
• @karthikeyan Note, the ratio of $p_2/p_1$ can be replaced with $T_2/T_1$ using isentropic flow relations. Additionally, the mass flow rate through the engine is proportional to $1/\sqrt[]{T_1}$, while the exhaust velocity is proportional to $\sqrt[]{T_1}$. Hope this helps. – TRF Jul 25 '18 at 22:46

Let me try to bite, putting it in easy terms.

The purpose of engine is to provide thrust - by ejecting propellant (combustion products) at as high velocity as possible.

The fuels store a lot of chemical energy - more than any closed chamber could contain. If they were allowed to convert all that energy to pressure in one place, the chamber would explode, period. The pressure would be many orders of magnitude of what the pumps can provide.

That's why the combustion chamber has a throat and a nozzle. As the fuels burn, they would increase pressure, but they are simultaneously vented through the nozzle - they expand, pressure drops, their velocity increases. The turbopump doesn't have to overcome the pressure that would result from combustion in enclosed space - it just needs to overcome the equilibrium pressure between what builds up from combustion and what is lost due to venting that through the nozzle. Pressure is kept in balance - volume grows. This growth of volume directly converts into movement - gas ejected at high speed through the nozzle - and propulsion.

Plus the pumps need to pump liquid fuel instead of combustion gasses - pump's work is proportional to pressure and volume pumped, not mass. Pumping 100l of water at 10 bar is the same work as pumping 100l of steam at 10 bar, created through boiling a liter of water. Same here - most of work is done when burning the fuel, creating a lot of fast-moving gas. A pump to move the same mass of gas at the same pressure would need to provide far, far more work - because the volume of that gas would be vastly bigger.

• upvoted for being much clearer than the other answers. – Innovine Jan 20 '17 at 18:18
• In addition to increasing volume (which increases the velocity of the gas escaping through the throat), combustion increases temperature, i.e. random movement of molecules. Nozzle then converts most of this random movement into ordered movement in the direction of the exhaust, contributing to thrust (when sideways-moving molecules hit the nozzle they push the rocket forward because of the slope). – al13n Jan 21 '17 at 0:35
• This is an EXCELLENT answer. Wow. I couldn't figure it out, and now I understand. I think I could even explain it to someone else. Thank you so much. From my point of view the key to conceptualizing this is the mass. As you said, the pump is pumping more the massive liquid. And the exhaust, in expelling gasses is creating the high exhaust velocity. Because of course, a given amount energy will propel lighter medium faster than a heavier one. Thank you. – Johnny Robinson Jan 21 '17 at 14:38
• @JohnnyRobinson: Given amount will propel lighter medium faster (very true; $H_2$ is an excellent propellant as exhaust gas, e.g. for NTR engine) but also great most of that energy comes from combustion. The exhaust gasses are often tapped into to propel the turbopump - as only a bit larger volume of exhaust is needed to pump a volume of liquid to the same pressure. But you need only a drop of that liquid to produce that larger volume of exhaust. – SF. Jan 21 '17 at 20:08
• @MikeW: It's an approximation, which is not valid for high throughput low pressure differential (lots of mass moved, not pressurized) but definitely works for high pressure differential. The amount of work needed to pressurize 1m^3 of any medium from 1 to 10 bar totally overshadows amount of work needed to move that medium the distance which corresponds to the resultant volume change. Though yes, pumping 100l of air from 1 bar tank to 1.01 bar tank will take less work than doing the same to 100l of hot tar. – SF. May 19 '17 at 9:55

You are correct that for a pressure fed engine, the combustion chamber pressure must be lower than the tank pressure. But as for "why have the engine at all".... an important thing happens in the combustion chamber. Namely, combustion! That process uses the chemical energy of the propellants to accelerate the resultant gases through the nozzle. And acceleration of gas to high velocity is what it's all about. Thrust is exhaust velocity times mass flow (plus a pressure term related to the exit plane pressure...typically a much smaller contributor to the thrust [sometimes a detractor]).

I am not sure what your question about the Raptor engine is. It has been discussed in this question and full flow staged combustion has been discussed in this question.

• Isn't the point of combustion to create engine pressure? Isn't exhaust velocity created by engine pressure pushing the reaction mass out the constricted nozzle opening? See my point? If the tank pressure is higher than the engine pressure, then you could just put a nozzle on the tank and get the same exhaust velocity (or rather a higher one). There must be more to this. – Johnny Robinson Jan 3 '17 at 5:47
• The combustion process is called deflagration and is actually constant pressure combustion. So no, the combustion does not raise the pressure of the gas. – TRF Jan 3 '17 at 6:30
• "That process adds energy to the system" - not really. It only turns potential chemical energy into kinetic energy, aka "releases" energy from fuel. Small thing, but it's noce to be precise when it does not require much more words, right? – Mołot Jan 3 '17 at 11:44
• @JohnnyRobinson The point of combustion is to raise the temperature. See TRF's answer if you want the math. – Organic Marble Jan 3 '17 at 12:29
• @TRF interesting to hear deflagration inside engine! Would the pressure of combustion be same as injection pressure or lesser than that? – karthikeyan Jul 19 '18 at 3:23