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Effect of pressure ratio on jet/rocket efficiency Q&A by msat on 14 Apr, 2014 22:11
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Hello all,
I know this topic has been addressed all over the web, but as someone who doesn't really have a grasp on the concepts and terminology of thermodynamics nor calculus, seeing explanations such as "Because formula X(YZ) says so" doesn't really help me, unfortunately. I've also read some explanations on other forums but some of them leave me with the feeling that they're wrong. Since I've come across this forum numerous times while searching various topics and saw how many smart people post here, I figured this might be a good place to get some clarification on my questions. Hopefully I'll be welcomed, and you won't all be too hard on me
My question is more focused on jet engines, but I assume the concept holds true for rocket engines operating in an atmosphere. I'm going to state my understanding of the subject, and am hoping that someone will tell me where I'm wrong, and perhaps even show me how thermodynamic concepts like entropy, enthalpy, etc. relate to aspects of the topic.
So here we go.. Lets say we have two (ram or turbo)jet engines - one operates at a pressure ratio of 10 while the other operates at a PR of 20. It's generally assumed the engine with the higher PR will have better efficiency (so more thrust for a given mass flow rate, or better SFC for a given thrust level). If we keep the combustion temperature between the two engines the same, is the increased efficiency of the engine with the higher PR derived primarily from the nozzle, specifically because it allows a higher throat to nozzle exit area ratio? I understand (hopefully correctly) that mass flow rate through a nozzle throat can be determined by the upstream pressure, so for a given throat area, an engine with a higher pressure ratio will have a higher flow rate. However, it's the divergent section of a nozzle where the potential energy of the gas is converted to kinetic energy (reducing the gas temperature in the process), correct? Since a higher PR allows for a higher throat to nozzle exit area ratio (exit area being determined by ambient pressure), more of the potential energy can be extracted in the expansion phase, right? Is it safe to say that any heat not absorbed by the expansion process, and thus [presumably] above ambient temperature is a waste?
Now, when it comes to rocket engines outside the atmosphere, is it the lack of ambient pressure which allows us to utilize large throat to nozzle area ratios (ignoring practical constraints), and thus reducing the pressure ratio requirements to obtain high efficiencies (except where it significantly affects combustion temps)?
Cheers,
Mark -
#1 by Proponent on 15 Apr, 2014 10:49
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I think you've got the right idea: higher chamber pressure (and lower ambient pressure) allows for more expansion, which means conversion of a larger fraction of the enthalpy (heat- and pressure-related energy) of combustion into kinetic energy. Another contributing factor is that the equation for the thrust of an engine includes a term equal to nozzle-exit area times the amount by which exit pressure exceeds ambient.
One minor detail: even in the convergent part of the nozzle, enthalpy is being converted into kinetic energy; pressure and temperature both decrease as the throat is approached. -
#2 by msat on 15 Apr, 2014 17:48
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Thanks for the confirmation and additional details, Proponent. Good to know I wasn't way off the mark.
"Another contributing factor is that the equation for the thrust of an engine includes a term equal to nozzle-exit area times the amount by which exit pressure exceeds ambient."
I'm assuming you're talking about underexpansion? I can see how it would affect efficiency.
"One minor detail: even in the convergent part of the nozzle, enthalpy is being converted into kinetic energy; pressure and temperature both decrease as the throat is approached."
I was thinking about this last night, but I'm not sure I understand how pressure ratio affects efficiency in such cases (or if it even does). I can see how a higher PR would allow for higher exhaust velocities, which I believe I've heard but don't mentally grasp, that either it's necessary in order for an aircraft attain high velocities, or it's more efficient for such use cases. -
#3 by Proponent on 17 Apr, 2014 12:47
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"Another contributing factor is that the equation for the thrust of an engine includes a term equal to nozzle-exit area times the amount by which exit pressure exceeds ambient."
I'm assuming you're talking about underexpansion? I can see how it would affect efficiency.
In part, that's what I'm talking about. In the simple one-dimensional-flow model of a rocket engine, thrust is equal to
F = q ve + Ae(pe - pa) ,
where q is the mass flow rate (usually represented by m-dot, of course), ve is the exit velocity, Ae is the exit area, pe is the exit pressure, and pa is the ambient pressure. Right?
Now, let's do the thought experiment of starting up a rocket engine that's underexpanded at some non-zero ambient pressure. Underexpansion means pe > pa. So, the second term in the thrust equation is positive. The exhaust gas at the exit is pushing on the atmosphere, and that helps generate thrust.
Now let's lengthen the nozzle to the point that the exit pressure drops to ambient. The second term vanishes. However, the exit velocity increases. In fact, the increase in velocity more than offsets the the loss in the second term. Obviously, we need to consider equations other than just the thrust equation to determine this, but that's how it works out.
If we further extend the nozzle, the exit pressure drops below ambient, and the second term goes negative. The nozzle is now sucking on the atmosphere. Meanwhile, exhaust velocity has increased, but not enough to overcome the growth in the second term. The moral of the story is that the engine is most efficient when exit pressure matches ambient.Quote"One minor detail: even in the convergent part of the nozzle, enthalpy is being converted into kinetic energy; pressure and temperature both decrease as the throat is approached."
I was thinking about this last night, but I'm not sure I understand how pressure ratio affects efficiency in such cases (or if it even does). I can see how a higher PR would allow for higher exhaust velocities, which I believe I've heard but don't mentally grasp, that either it's necessary in order for an aircraft attain high velocities, or it's more efficient for such use cases.
I'm not sure I understand by what you mean by "in such cases." After all, every de Laval nozzle begins with a converging section. -
#4 by MP99 on 17 Apr, 2014 18:16
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One minor detail: even in the convergent part of the nozzle, enthalpy is being converted into kinetic energy; pressure and temperature both decrease as the throat is approached.
Really? I'd assume the pressure would increase towards the throat, then decrease after?
cheers, Martin -
#5 by aero on 17 Apr, 2014 18:43
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One minor detail: even in the convergent part of the nozzle, enthalpy is being converted into kinetic energy; pressure and temperature both decrease as the throat is approached.
Really? I'd assume the pressure would increase towards the throat, then decrease after?
cheers, Martin
That would be wrong. Velocity increases as the flow from the chamber approaches the throat (continuity) hence heat energy converts to kinetic energy, temperature drops and via the gas laws, pressure drops. -
#6 by msat on 18 Apr, 2014 22:18
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F = q ve + Ae(pe - pa) ,
Now, let's do the thought experiment of starting up a rocket engine that's underexpanded at some non-zero ambient pressure. Underexpansion means pe > pa. So, the second term in the thrust equation is positive. The exhaust gas at the exit is pushing on the atmosphere, and that helps generate thrust.
Now let's lengthen the nozzle to the point that the exit pressure drops to ambient. The second term vanishes. However, the exit velocity increases. In fact, the increase in velocity more than offsets the the loss in the second term. Obviously, we need to consider equations other than just the thrust equation to determine this, but that's how it works out.
Thanks for pointing this out! I admit to having been confused by this equation because it does indeed look like it's desirable to have nozzle exit pressure above ambient if you neglect (like I have done) to account for the fact that you trade off the chance for higher exhaust velocity utilizing a higher throat to nozzle exit area ratio (assuming a nozzle with a divergent section is being used).Quote"One minor detail: even in the convergent part of the nozzle, enthalpy is being converted into kinetic energy; pressure and temperature both decrease as the throat is approached."
I was thinking about this last night, but I'm not sure I understand how pressure ratio affects efficiency in such cases (or if it even does). I can see how a higher PR would allow for higher exhaust velocities, which I believe I've heard but don't mentally grasp, that either it's necessary in order for an aircraft attain high velocities, or it's more efficient for such use cases.
I'm not sure I understand by what you mean by "in such cases." After all, every de Laval nozzle begins with a converging section.
By "in such cases", I meant an engine with high PR that does not utilize a con-di nozzle. I suppose it makes sense for turbofan engines where most of energy is used up driving the turbines, so there might not be much pressure left to recover. What I'm less clear on is what's going on behind the turbine blades here: http://farm8.staticflickr.com/7223/7275290886_df36fca31a.jpg . Is the diverging section acting as a diffuser for pressure recovery which is presumably attached to a convergent nozzle?
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#7 by Proponent on 22 Apr, 2014 05:29
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One minor detail: even in the convergent part of the nozzle, enthalpy is being converted into kinetic energy; pressure and temperature both decrease as the throat is approached.
Really? I'd assume the pressure would increase towards the throat, then decrease after?
cheers, Martin
That would be wrong. Velocity increases as the flow from the chamber approaches the throat (continuity) hence heat energy converts to kinetic energy, temperature drops and via the gas laws, pressure drops.
To amplify on aero's reply, let me put it this way: if the pressure didn't decrease toward the throat, what would make gas flow into the throat? -
#8 by msat on 28 Apr, 2014 22:12
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I haven't stopped contemplating this topic, and I don't think I have a grasp on it, at least from a thermodynamics standpoint. I've been focusing too much on the nozzle, and not how PR affects SFC.
Correct me if I'm wrong, but ideally (i.e. assuming 100% efficiency of the components), the amount of energy the turbine needs to extract from the gas flow is equal to that necessary to drive the compressor. So, if we have two engines with the same mass flow rate, the amount of fuel needed to, say, double the amount of energy in the gas between them is the same, regardless of PR. Therefore, if one of those engines has a PR of 10, while the other 20, after extracting the energy needed to drive the compressor via the turbine, we still have double the amount of (potential?) energy in the higher PR engine (again, neglecting losses) left over that we can use to do work.
