Rocket exhaust and atmosphere

[DaveC426913]

This is perhaps a somewhat naive question about rocket exhaust.

If I understand correctly, one of the challenges in designing a rocket nozzle - say, the main thrusters for the Saturn V - is that it needs to operate efficiently at sea level as well as at altitude. The difference in air pressure affects expansion of the gases, which require a different shape to the bell.

Exhaust nozzles must be designed as a compromise to operate reasonably well - though not optimally - at the range of altitudes required. (This is one of the challenges that the aerospike engine is designed to overcome, by essentially having only one side of the bell and letting the atmo do the rest.)

Do I have that right?

What I don't understand is how the thrust coming out of a Saturn V nozzle - all 1.5 million pounds of it - cares at all about 1 atmosphere of pressure. Surely, it's essentially vacuum compared to the thrust. What am I missing?

 

[phinds]

Dave, I don't have any answer to your question but one thing that does occur to me is that those thrusters are really big, so it's not like you are comparing 1,500,000 psi to 14 psi, it's more like 14 psi against 1,500,000 p / a whole bunch of si. I agree that it still seems like the atm would be negligible but you might see if you can find the total size of the exit area of the thrust.

 

[Nidum]

Basic explanation here :

https://en.wikipedia.org/wiki/Rocket_engine_nozzle

 

[boneh3ad]

You can't compare thrust directly to atmospheric pressure. You have to compare the pressure in the exhaust gas to the atmospheric pressure. If the two are equal at the exit, the engine is operating at optimal efficiency. However, for a given pressure in the combustion chamber and a given nozzle shape, there is only one possible exit pressure, so as the rocket ascends, there is only one point where the exhaust pressure and ambient pressure are equal.

 

[cjl]

Boneh3ad has it correct here - you need to compare the pressure of the gas at the exit plane of the nozzle to atmospheric, and interestingly (and perhaps nonintuitively), that will almost always be below atmospheric. The thrust comes from the acceleration of the gas to a very high exit speed (highest possible momentum flux out of the rocket per unit fuel), and the acceleration occurs because of the expansion of the gas. Therefore, you want to expand it as much as possible to accelerate it to the highest possible speed, but if you expand it too much, it can separate from the nozzle and lose efficiency (and possibly cause damage to the nozzle, depending on the details of the separation). In extreme cases, a shock can even form inside an overexpanded nozzle, slowing the flow to subsonic before it ever leaves the nozzle.

For a concrete example, look at the space shuttle. The space shuttle main engines have an exhaust pressure at the exit plane of below 5 PSI, if I remember right. This is about a third of the ambient pressure, and very close to causing problems with the flow separating from the nozzle. You can see this if you look at a picture of the engine in operation at sea level - the first thing the flow does after leaving the nozzle is to start contracting (see here or https://californiasciencecenter.org/sites/default/files/styles/full-breakpoints_theme_labrat_l_1x/public/media/image/ssme-test-fire-marshall-sfc.jpg?itok=gHLlUnSD&timestamp=1405633558). This contraction is because the flow is well below ambient pressure, and is trying to equalize with the environment around it.

 

[DaveC426913]

You can see this if you look at a picture of the engine in operation at sea level - the first thing the flow does after leaving the nozzle is to start contracting (see here or https://californiasciencecenter.org/sites/default/files/styles/full-breakpoints_theme_labrat_l_1x/public/media/image/ssme-test-fire-marshall-sfc.jpg?itok=gHLlUnSD&timestamp=1405633558). This contraction is because the flow is well below ambient pressure, and is trying to equalize with the environment around it.

I've seen this on the shuttle exhaust before, yes. A pulsed jet, well back of the nozzle.

But back to the main question. I'm afraid I still don't understand how such a great volume of gas belching from the nozzle can result in a what is essentially a partial vacuum.

If I stood next to it, would I get sucked toward the nozzle (before blowing my eardrums, and burning to a cinder)?

 

[boneh3ad]

I've seen this on the shuttle exhaust before, yes. A pulsed jet, well back of the nozzle.

That's not really a pulsed jet you are seeing. It is an alternating pattern of shocks and expansions due to the difference in pressure between the exhaust and the atmosphere. They are typically called Mach diamonds or shock diamonds. It is actually an entirely steady phenomenon.

But back to the main question. I'm afraid I still don't understand how such a great volume of gas belching from the nozzle can result in a what is essentially a partial vacuum.

As the flow expands, its velocity increases greatly, so its pressure drops accordingly. See: de Laval nozzle.

If I stood next to it, would I get sucked toward the nozzle (before blowing my eardrums, and burning to a cinder)?

If you were standing to the side of it, then yes. If you were standing in line with it, then no, since the velocity is so high that the stagnation pressure you would feel when it hit you would be extraordinarily high, similar to that in the combustion chamber.

 

[DaveC426913]

As the flow expands, its velocity increases greatly, so its pressure drops accordingly. See: de Laval nozzle.

Like Bernoulli's Law or a Venturi Tube?

[EDIT] I see by the de Laval nozzle diagram that it's not really the same.

 

[boneh3ad]

Like Bernoulli's Law or a Venturi Tube?

[EDIT] I see by the de Laval nozzle diagram that it's not really the same.

Bernoulli's equation and the most common concept of a Venturi tube (which is still largely Bernoulli) are only valid in incompressible flows. The idea of a Venturi apparatus can be extended into compressible and supersonic regimes but the equations change.

However, the inverse relationship between velocity and pressure persists. For an adiabatic flow of a perfect gas, for example, conservation of energy implies
$$c_p T + \dfrac{v^2}{2} = h_0 = \mathrm{constant},$$
and
$$p=\rho R T.$$
So,
$$\dfrac{c_p p}{\rho R} + \dfrac{v^2}{2} = h_0.$$
Rocket exhaust is very unlikely to be a perfect gas, but I'm just using that for illustration. Either way it shows an inverse relationship between $p$ and $v$ (though keep in mind that density is not constant there).

Anyway, in a rocket nozzle with sufficient chamber pressure, the flow accelerates through the converging section up to the throat according to the rules of subsonic flow where the pressure falls along with the rising velocity and velocity increases as the flow constricts. At the throat it reaches the speed of sound, at which point the relationship between the duct area and velocity reverses, as expressed by
$$\dfrac{du}{u} = \dfrac{1}{M^2-1}\dfrac{dA}{A}.$$

So after passing the throat, it continues its acceleration and expansion. The larger the eventual exit area, the faster the exhaust velocity and the lower the static pressure in the exhaust.