Why do turbine engines require a compressor section
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i think slippery pete is pretty right with the higher compression- lower exhaust temperatures at pistons . you can see this effect also in tubine engines where maintaining the same power output / fuel flow in a climb will result in exhaust temperature RISING due to thinner air and worsening compression of the compressor stages.
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i think slippery pete is pretty right with the higher compression- lower exhaust temperatures at pistons
One would have thought that exhaust temp was a function of fuel air ratio.
BH
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The original question was "Why do turbine engines require a compressor section", how did we get into diesel engines, pistons, turbo- and super-chargers?
The simple answer to this very straitforward queation is that a compressor, delivering high pressure air, is required to ensure self-sustaining operation. The compressor delivery pressure (and the velocity) at the inlet of the combustion chamber must be high enough to ensure that the pressure in the combustion chamber results in the combustion gasses flowing through the turbine section and and not 'backwards' into the compressor.
The simple answer to this very straitforward queation is that a compressor, delivering high pressure air, is required to ensure self-sustaining operation. The compressor delivery pressure (and the velocity) at the inlet of the combustion chamber must be high enough to ensure that the pressure in the combustion chamber results in the combustion gasses flowing through the turbine section and and not 'backwards' into the compressor.
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Another reason to use a compressor is that the efficiency of a gas turbine increases with the pressure ratio (p_combustion_chamber/p_environment). Thanks to the compressor it is possible to run a turbofan or turboprop efficiently at low airspeeds.
See: Brayton cycle - Wikipedia, the free encyclopedia
See: Brayton cycle - Wikipedia, the free encyclopedia
Hi Oggers.
I thought of an even better example to explain it. Compare a gasoline to a diesel (low compression to high compression).
Diesel fuel has 11% more energy by volume than gasoline.
A diesel engine is approximately 30-35% more efficient at the same RPM and fuel flow. Let's assume 30%.
So that's 30% more power from 11% better fuel. If you take the 11% off to even the playing field of fuel energy density, that's 19% more efficient.
Are you trying to tell me that 19% effciency gain comes about solely because of more complete combustion and "mixing"?
Going on your theory, that means 19% of the fuel in a gasoline engine must not be combusting.
I guarantee if your gasoline engine was putting 19% of it's gasoline down the exhaust, your typical cat converter wouldn't last more than a week.
Unfortunately, the fallacy that higher compression engines gain their efficiency from more complete combustion will continue for years to come.
More energy is transferred to useful work because less heat is wasted heating the fluid. It's that simple.
I thought of an even better example to explain it. Compare a gasoline to a diesel (low compression to high compression).
Diesel fuel has 11% more energy by volume than gasoline.
A diesel engine is approximately 30-35% more efficient at the same RPM and fuel flow. Let's assume 30%.
So that's 30% more power from 11% better fuel. If you take the 11% off to even the playing field of fuel energy density, that's 19% more efficient.
Are you trying to tell me that 19% effciency gain comes about solely because of more complete combustion and "mixing"?
Going on your theory, that means 19% of the fuel in a gasoline engine must not be combusting.
I guarantee if your gasoline engine was putting 19% of it's gasoline down the exhaust, your typical cat converter wouldn't last more than a week.
Unfortunately, the fallacy that higher compression engines gain their efficiency from more complete combustion will continue for years to come.
More energy is transferred to useful work because less heat is wasted heating the fluid. It's that simple.
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Slippery:
You are indeed slippery! That sounds suspiciously like what I said to begin with: 'The goal is to get maximum charge into a given combustion chamber. That is, basically, volumetric efficiency'. You're meant to be arguing against it because you said:
However, with your new post you have now introduced some more errors:
You have to remember that the fuel metering system will have reduced the fuel so as to maintain the desired mixture. Therefore what you have described is a loss of torque or power, not a loss of efficiency per se. Meanwhile, the "comparatively small" pumping losses you mention are big enough to explain the efficiency difference between diesel and petrol engines.
I don't really understand where you are coming from there. If the working fluid absorbs less heat its pressure will rise less. Less work will be done on the piston.
It's not about unburnt fuel so much as flame front speeds and a few other things that a chemical engineer might discuss. Essentially the flame propagates faster at high temp which has benefits under the heading 'timing' I can't be bothered to explain, but which the mechanical engineer and engine tuner inside me finds useful.
Strawman. I have never said that.
No I'm not! The difference is mainly to do with pumping losses. I did mention them in my previous post but you decided they are "comparatively small". The diesel doesn't have a throttle it just fills the cylinder with maximum air each time and varies the power by injecting less fuel. No throttle restriction = less pumping losses. Which renders the rest of your latest post redundant.
This is just wrong. If you don't put the heat in the fluid you can't expand the fluid against the piston. Perhaps you could just explain how you're going to expand the fluid without heating it?
The loss of efficiency due to pumping losses are comparatively small. The biggest loss of efficiency in an engine with poor VE is because of a drop in effective compression ratio. For example, if an engine can only suck in half a cylinder full of air because of intake/exhaust restrictions or throttle position - a 10:1 compression ratio engine is actually only effectively producing a 5:1 compression against atmospheric pressure. I still don't think VE is important in the question of the OP.
the answers given about volumetric efficiency have nothing to do with your question about thermodynamic efficiency of an engine.
if an engine can only suck in half a cylinder full of air because of intake/exhaust restrictions or throttle position - a 10:1 compression ratio engine is actually only effectively producing a 5:1 compression against atmospheric pressure
Exactly. This is exactly what I said. A higher compression ratio adds the heat to a hotter air charge, so once the engine reaches BDC the higher compression engine "fluid" will be cooler. By "absorb less heat", I meant at the end of the cycle the fluid has absorbed less total heat during the cycle (not saying it's cooler at the point of ignition - it is, in fact, hotter as you said).
The difference in percentage of unburnt fuel between a low and high compression engine is quite small - trust me - I've done it on a lab dyno at university.
You can't get more "umph", plus hotter exhaust gases too - where is all this extra energy coming from?
A diesel engine is approximately 30-35% more efficient at the same RPM and fuel flow. Let's assume 30%.
So that's 30% more power from 11% better fuel. If you take the 11% off to even the playing field of fuel energy density, that's 19% more efficient.
Are you trying to tell me that 19% effciency gain comes about solely because of more complete combustion and "mixing"?
So that's 30% more power from 11% better fuel. If you take the 11% off to even the playing field of fuel energy density, that's 19% more efficient.
Are you trying to tell me that 19% effciency gain comes about solely because of more complete combustion and "mixing"?
More energy is transferred to useful work because less heat is wasted heating the fluid. It's that simple
Last edited by oggers; 16th Nov 2011 at 20:19.
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aerobat:
The old correlation is not causation phrase springs to mind. There is another simpler explanation though: in a turbine, air is used to cool the combustion chamber but seeing as you are still using the same amount for combustion you have less left over for cooling as you go higher. Less cooling air passing through combustion chamber means less in the zorst as well.
At least that's what I was taught at groundschool and have always attributed the rise in PTIT to. If there's another explanation I'm all ears.
i think slippery pete is pretty right with the higher compression- lower exhaust temperatures at pistons . you can see this effect also in tubine engines where maintaining the same power output / fuel flow in a climb will result in exhaust temperature RISING due to thinner air and worsening compression of the compressor stages.
At least that's what I was taught at groundschool and have always attributed the rise in PTIT to. If there's another explanation I'm all ears.
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QJB,
For your first question "Why do turbine engines require a compressor section ?", many people have shown the right direction for the answer and many others have indicated parts of the reason.
However, I haven't seen a comprehensive answer.
Here it is.
In short, a turbine engine requires a compressor by design and thermodynamic constraints.
The constraints stem from linking the following facts:
1. A thermodynamic engine requires that at least one phase of its cycle increases the gas pressure.
A thermodynamic machine produces mechanical work when the gas pass through an expansion and it consumes work when the gas is compressed.
This positive or negative work is visible in a pV (pressure/volume) diagram as the integral beneath the line depicting the state change ( dW= p.dV ).
The machine produce a net work if the expansion is done at a higher pressure than the compression. Thus a thermodynamic machine must have at least one phase where the pressure increases.
2. There are only 2 ways of increasing the gas pressure :
a. mechanically (ie, with a compressor)
or b. through putting some heat in the gas whilst keeping it in a limited volume.
3. A turbine is an open flow thermodynamic engine. This implies that the pressure in the combustion chamber keeps a constant value.
An open flow thermodynamic engine means that, by progressing from one place to the next, the state variables of the gas are evolving and complete the transformations of the thermodynamic cycle. However, in any fixed place, these state variables stay constant though time. Thus the combustion chamber pressure is constant.
4. As no pressure increase occurs in the combustion chamber and as it is the only place where heat is added to the gas, then the pressure increase must happen through a mechanical compression.
Hence the requirement of a compressor.
Luc
For your first question "Why do turbine engines require a compressor section ?", many people have shown the right direction for the answer and many others have indicated parts of the reason.
However, I haven't seen a comprehensive answer.
Here it is.
In short, a turbine engine requires a compressor by design and thermodynamic constraints.
The constraints stem from linking the following facts:
1. A thermodynamic engine requires that at least one phase of its cycle increases the gas pressure.
A thermodynamic machine produces mechanical work when the gas pass through an expansion and it consumes work when the gas is compressed.
This positive or negative work is visible in a pV (pressure/volume) diagram as the integral beneath the line depicting the state change ( dW= p.dV ).
The machine produce a net work if the expansion is done at a higher pressure than the compression. Thus a thermodynamic machine must have at least one phase where the pressure increases.
2. There are only 2 ways of increasing the gas pressure :
a. mechanically (ie, with a compressor)
or b. through putting some heat in the gas whilst keeping it in a limited volume.
3. A turbine is an open flow thermodynamic engine. This implies that the pressure in the combustion chamber keeps a constant value.
An open flow thermodynamic engine means that, by progressing from one place to the next, the state variables of the gas are evolving and complete the transformations of the thermodynamic cycle. However, in any fixed place, these state variables stay constant though time. Thus the combustion chamber pressure is constant.
4. As no pressure increase occurs in the combustion chamber and as it is the only place where heat is added to the gas, then the pressure increase must happen through a mechanical compression.
Hence the requirement of a compressor.
Luc
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QJB had another question:
The mathematics behind all this is quite complex, but it all hinges around Carnot's theorem Carnot's theorem (thermodynamics) - Wikipedia, the free encyclopedia
Ref: Thermal efficiency - Wikipedia, the free encyclopedia
For a gas turbine this translates to 1-(P2/p1)^(1-y/y) with y the adiabatic index (~7/5 for air)
A turbo charger can add to the efficiency of a Diesel engine, but not to that of a petrol (Otto) engine, as the compression is limited by self combustion of the mixture. A turbo charger allows to obtain more power from a smaller engine too.
why is it that both the piston engine and the turbine engine can have their efficiencies increased by increasing the pressure ratio (compression ratio for piston)?
No engine operating between two heat reservoirs can be more efficient than a Carnot engine operating between the same reservoirs.
The maximum efficiency is 1-T_H/T_C
The maximum efficiency is 1-T_H/T_C
For a gas turbine this translates to 1-(P2/p1)^(1-y/y) with y the adiabatic index (~7/5 for air)
A turbo charger can add to the efficiency of a Diesel engine, but not to that of a petrol (Otto) engine, as the compression is limited by self combustion of the mixture. A turbo charger allows to obtain more power from a smaller engine too.
Hi Oggers.
I'll say it one more time, and if you can't understand it after that, then I will consider
to stop wasting my breath.
Flame front speeds have absolutely nothing to do with it. You simply CAN NOT burn the fuel "faster" to get more energy out of it. You seem to have a fundamental problem understanding this. You can burn the fuel as quickly or as slowly as you like, and it produces the same amount of energy. Even if gasoline fuel burns "slowly" or "quickly", the ignition timing is simply adjusted to ensure the maximum pressure in the cylinder is occuring at TDC. This has been happening for the past 20 years with electronic ignition control, and for the 30 years before that, with centrifugal spark advance. Welcome to the 1960s.
This is exactly right - YOU DON'T UNDERSTAND. The comment about the fluid absorbing less heat is once the fluid is returned to atmospheric pressure. I'll now write it very, very simply for you:
The change in heat between the air which enters the engine (at standard atmospheric pressure) and the air which leaves the engine (one it is returned back to atmospheric pressure) is what I am talking about here. If you fail to understand that, and keep incorrectly assuming I'm saying cylinder pressures and temperatures at TDC are lower, you will never understand this most basic of principles.
The fluid pressure and temperature are both HIGHER at ignition than in a low compression engine. The DIFFERENCE between the fluid temperature and the burning temperature of the fuel at the point of ignition is LOWER in a high compression engine.
It's really not that hard.
Completely disagree. Are you telling me a gasoline engine uses 19% of it's energy (say, 30hp in a standard car) to suck the air in and blow it out? Hahaha, you've got to be joking.
So what you are trying to tell me here that an engine which is effectively only compressing the air to 5x the atmospheric pressure rather than 10x, is still able to extract the same amount of energy per unit of fuel? Give me a break! If this were true, then high compression and low compression engines would have identical thermodynamic efficiency, and we wouldn't be discussing the differences between the two.
See above. It's about the total fluid heat change from start to finish. It is lower in a high compression engine. You need to stop assuming this means less absolute pressure/temperature at TDC. It doesn't.
I'll say it one more time, and if you can't understand it after that, then I will consider
to stop wasting my breath.Flame front speeds have absolutely nothing to do with it. You simply CAN NOT burn the fuel "faster" to get more energy out of it. You seem to have a fundamental problem understanding this. You can burn the fuel as quickly or as slowly as you like, and it produces the same amount of energy. Even if gasoline fuel burns "slowly" or "quickly", the ignition timing is simply adjusted to ensure the maximum pressure in the cylinder is occuring at TDC. This has been happening for the past 20 years with electronic ignition control, and for the 30 years before that, with centrifugal spark advance. Welcome to the 1960s.
I don't really understand where you are coming from there. If the working fluid absorbs less heat its pressure will rise less. Less work will be done on the piston.
The change in heat between the air which enters the engine (at standard atmospheric pressure) and the air which leaves the engine (one it is returned back to atmospheric pressure) is what I am talking about here. If you fail to understand that, and keep incorrectly assuming I'm saying cylinder pressures and temperatures at TDC are lower, you will never understand this most basic of principles.
The fluid pressure and temperature are both HIGHER at ignition than in a low compression engine. The DIFFERENCE between the fluid temperature and the burning temperature of the fuel at the point of ignition is LOWER in a high compression engine.
It's really not that hard.
Meanwhile, the "comparatively small" pumping losses you mention are big enough to explain the efficiency difference between diesel and petrol engines.
You have to remember that the fuel metering system will have reduced the fuel so as to maintain the desired mixture. Therefore what you have described is a loss of torque or power, not a loss of efficiency per se.
If you don't put the heat in the fluid you can't expand the fluid against the piston. Perhaps you could just explain how you're going to expand the fluid without heating it?
Last edited by Slippery_Pete; 16th Nov 2011 at 23:17.
I'm probably under thinking this, but a higher compression ratio means higher peak cylinder pressure and therefore more force against the piston. Of course, a higher compression ratio also means more energy to compress the charge. But the compression takes less energy than the piston absorbs on the power stroke, so, say, doubling both should create a net gain.
To use a very simple example, if the piston expended 1 unit of energy on the compression stroke and absorbed 2 units on the power stroke, the net gain would be 1. If you doubled the pressure so that compression took 2 units and combustion gave 4, the net gain would be 2.
Of course, reality isn't nearly that simple (for example, there would be a real pressure difference only near TDC), but I don't see why the principle wouldn't hold. Or at least I don't now -- it will probably take 5 minutes for someone to point out the fatal flaw.
To use a very simple example, if the piston expended 1 unit of energy on the compression stroke and absorbed 2 units on the power stroke, the net gain would be 1. If you doubled the pressure so that compression took 2 units and combustion gave 4, the net gain would be 2.
Of course, reality isn't nearly that simple (for example, there would be a real pressure difference only near TDC), but I don't see why the principle wouldn't hold. Or at least I don't now -- it will probably take 5 minutes for someone to point out the fatal flaw.
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Slippery:
That is essentially one big mess of obfuscation and strawmen. It is pretty much a gish gallop so I'll just pick on a few points for now.
Flame front speeds are very important for timing reasons. It's a basic. Google it and see what you get or check out a good textbook.
The second sentence is the strawman here. We all know you don't get more enrgy from the fuel if you burn it faster.
But you have to realise that in an engine, having got the heat into the fuel you then have convert it to work on the piston during the limited time available on the power stroke. This is where timing comes in. 
I guess they missed that bit out when you were doing physics at uni 
No. 19% is your figure from the back of a fag packet. I'm telling you that pumping losses are a big part of the efficiency difference. Again google it and see what you get. And right in there is another error...
...the efficiency gain on the diesel is on the 'sucking in' side only, where the throttle is on a petrol engine. So laugh away - the joke's on you. 
That part is pure obfuscation and you have gone full circle. At the very beginning:
1) I explained the importance of VE and 2. I explained why increased compression positively impacted on efficiency.
...that is where you came in with "volumetric efficiency has nothing to do with thermal efficiency". I would say you've had an epiphany but you're actually just swapping between two positions.
That is essentially one big mess of obfuscation and strawmen. It is pretty much a gish gallop so I'll just pick on a few points for now.
Flame front speeds have absolutely nothing to do with it. You simply CAN NOT burn the fuel "faster" to get more energy out of it.
The second sentence is the strawman here. We all know you don't get more enrgy from the fuel if you burn it faster.
But you have to realise that in an engine, having got the heat into the fuel you then have convert it to work on the piston during the limited time available on the power stroke. This is where timing comes in. I guess they missed that bit out when you were doing physics at uni
Completely disagree. [that pumping losses account for the difference between petrol and diesel efficiency] Are you telling me a gasoline engine uses 19% of it's energy (say, 30hp in a standard car) to suck the air in and blow it out? Hahaha, you've got to be joking.
to suck the air in and blow it out? Hahaha, you've got to be joking
So what you are trying to tell me here that an engine which is effectively only compressing the air to 5x the atmospheric pressure rather than 10x, is still able to extract the same amount of energy per unit of fuel? Give me a break! If this were true, then high compression and low compression engines would have identical thermodynamic efficiency, and we wouldn't be discussing the differences between the two.
1) I explained the importance of VE and 2. I explained why increased compression positively impacted on efficiency.
...that is where you came in with "volumetric efficiency has nothing to do with thermal efficiency".
I would say you've had an epiphany but you're actually just swapping between two positions.
My dear friend Oggers,
Your attempt at ignoring the main point of my last post (which I repeated a myriad of times) is noble, but not unnoticed. I will AGAIN post it here for you to consider
I'm listening.
I also wrote this
And then you accused me of not understanding timing with this...
Oops, guess you didn't read (or understand) what I'd already written.
Nope, that's calculated from the chemical energy density of the fuels (MJ/L gasoline vs. diesel) and the proven efficiency of diesel engines. The 30% I assumed as the difference between gasoline/diesel engines is actually 30-35%, so it could have actually been as high as 24%.
Poor volumetric effficiency is a consequence of not operating at it's most efficient (small throttle settings), not the thermodynamic efficiency of the cycle which was the question in the OP (posted here for your benefit).
Your attempt at ignoring the main point of my last post (which I repeated a myriad of times) is noble, but not unnoticed. I will AGAIN post it here for you to consider
It's about the total fluid heat change from start to finish (once the fluid is returned to atmospheric pressure). It is lower in a high compression engine. You need to stop assuming this means less absolute pressure/temperature at TDC. It doesn't.
I also wrote this
Even if gasoline fuel burns "slowly" or "quickly", the ignition timing is simply adjusted to ensure the maximum pressure in the cylinder is occuring at TDC. This has been happening for the past 20 years with electronic ignition control, and for the 30 years before that, with centrifugal spark advance. Welcome to the 1960s.
But you have to realise that in an engine, having got the heat into the fuel you then have convert it to work on the piston during the limited time available on the power stroke. This is where timing comes in. 
I guess they missed that bit out when you were doing physics at uni
I guess they missed that bit out when you were doing physics at uni
19% is your figure from the back of a fag packet.
Poor volumetric effficiency is a consequence of not operating at it's most efficient (small throttle settings), not the thermodynamic efficiency of the cycle which was the question in the OP (posted here for your benefit).
Also why is it that both the piston engine and the turbine engine can have their efficiencies increased by increasing the pressure ratio (compression ratio for piston)? Is there some sort of simple thermodynamic explanation for this?
Last edited by Slippery_Pete; 17th Nov 2011 at 10:47.
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Slippers...
Your attempt at ignoring the main point of my last post (which I repeated a myriad of times) is noble, but not unnoticed. I will AGAIN post it here for you to consider
Even if gasoline fuel burns "slowly" or "quickly", the ignition timing is simply adjusted to ensure the maximum pressure in the cylinder is occuring at TDC.
You really don't seem to understand timing. I'll try to sum up the key points:You don't want maximum pressure at TDC because that is very ineffective. Google ineffective crank angle. But the later you time the ignition the less time you'll get to extract work from the fluid before the exhaust valve opens.
And a lot more besides - none of which is about getting maximum pressure at TDC!
You said:
"Flame front speeds have absolutely nothing to do with it"
Last edited by oggers; 17th Nov 2011 at 18:34.
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oggers, Slippery_Pete,
do you really think that QJB (or anyone else having the same question) will have the patience of reading all your duet and will try to find an answer in it ?
@QJB,
MathFox has actually written a complete and concise answer in post #30.
If you want to go further and understand or visualize the formula, you'll have to get somewhat familiar with T-s diagrams (temperature-entropy).
The theoritical diagram for the Brayton cycle (turbines) is underneath.
These diagrams are easy to read because they hold together the heat that enters the engine, the (lost) heat that exits the device, and the work produced.
- The heat that enters the device is depicted by all the area that is underneath the upper line(s) of the cycle, descending down up to the axis. Hereunder, it is all the area that is beneath the curved line 3-4.
- The lost heat that exits the device is depicted by all the area that is underneath the lower line(s) of the cycle, descending down up to the axis. Hereunder, it is all the area that is beneath the curved line 8-0.
- The work that can be produced (theoritically) is the difference of the two, that is, the area within the cycle 0-3-4-8-0.
- The efficiency of the cycle (theoritical) is the ratio of the inner surface divided by the whole surface underneath the line 3-4.
The points of the upper curved line 3-4 are at a constant pressure ; the pressure of the combustion chamber (p3).
The points of the lower curved line are at a constant pressure ; the external atmospheric pressure (p0).
The point 0 is at the entry of the inlet and the point 8 is somewhere in the exhaust flow behind the nozzle. The linking from 8 to 0 represents an hypothetical closure of the cycle by dragging the exhaust air from the nozzle to the inlet whilst letting it cool down at constant pressure up to the ambiant temperature.
It is clear that the lower curve is constrained by the value of the atmospheric pressure (p0).
It is also clear that increasing the combustion chamber pressure (p3) will push the line 3-4 higher on a parallel curve line and will increase the ratio of the inner surface (work) to the area underneath 3-4 (input heat), and thus will increase the efficiency of the cycle.
Luc
The image is from the www.grc.nasa.gov web page : Turbine Engine Thermodynamic Cycle - Brayton Cycle
do you really think that QJB (or anyone else having the same question) will have the patience of reading all your duet and will try to find an answer in it ?
@QJB,
MathFox has actually written a complete and concise answer in post #30.
If you want to go further and understand or visualize the formula, you'll have to get somewhat familiar with T-s diagrams (temperature-entropy).
The theoritical diagram for the Brayton cycle (turbines) is underneath.
These diagrams are easy to read because they hold together the heat that enters the engine, the (lost) heat that exits the device, and the work produced.
- The heat that enters the device is depicted by all the area that is underneath the upper line(s) of the cycle, descending down up to the axis. Hereunder, it is all the area that is beneath the curved line 3-4.
- The lost heat that exits the device is depicted by all the area that is underneath the lower line(s) of the cycle, descending down up to the axis. Hereunder, it is all the area that is beneath the curved line 8-0.
- The work that can be produced (theoritically) is the difference of the two, that is, the area within the cycle 0-3-4-8-0.
- The efficiency of the cycle (theoritical) is the ratio of the inner surface divided by the whole surface underneath the line 3-4.
The points of the upper curved line 3-4 are at a constant pressure ; the pressure of the combustion chamber (p3).
The points of the lower curved line are at a constant pressure ; the external atmospheric pressure (p0).
The point 0 is at the entry of the inlet and the point 8 is somewhere in the exhaust flow behind the nozzle. The linking from 8 to 0 represents an hypothetical closure of the cycle by dragging the exhaust air from the nozzle to the inlet whilst letting it cool down at constant pressure up to the ambiant temperature.
It is clear that the lower curve is constrained by the value of the atmospheric pressure (p0).
It is also clear that increasing the combustion chamber pressure (p3) will push the line 3-4 higher on a parallel curve line and will increase the ratio of the inner surface (work) to the area underneath 3-4 (input heat), and thus will increase the efficiency of the cycle.
Luc
The image is from the www.grc.nasa.gov web page : Turbine Engine Thermodynamic Cycle - Brayton Cycle
Last edited by Luc Lion; 17th Nov 2011 at 15:17. Reason: quoting sources
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oggers, Slippery_Pete,
do you really think that QJB (or anyone else having the same question) will have the patience of reading all your duet and will try to find an answer in it ?
do you really think that QJB (or anyone else having the same question) will have the patience of reading all your duet and will try to find an answer in it ?
This is a forum and it is in the nature of these things for a thread to develop. It is not necessary that the thread continues to address the original question. We are still talking about related issues.
Oggers, oggers, oggers...
You can rely on the first pages a quick google search brings you as much as you like, I'll rely on physics, thermodynamics and conservation of energy.
The problem is you just keep arguing about inefficiencies caused by operating the engine outside its most efficient range and somehow try to relate this to the thermodynamic efficiency of the cycle. In the same way that small throttle settings relate to inefficient operation, so does your high rpm "not much time for combustion, inefficient valve/igntion timing, or poor mixing/incomplete combustion" arguments.
They are consequences of the way the engine is operated, not the thermodynamic efficiencies of the cycle. To make it absolutely clear and to prevent you muddying the relationship between thermodynamics/compression ratio and operating an engine inefficiently, let's consider a very low RPM, full throttle engine.
This ACTUALLY means you just don't know. This is now the third time you've denied answering this. Instead, each time you seem to cherry pick my posts and then wander off talking about consequences of how the engine is operated. Don't worry, physicists have been trying to find away around the conservation of energy for hundreds of years - they too have been unable to.
Here it is, AGAIN.
I'm STILL listening.
It's because the fluid absorbs less heat 
Anyway, back to Google for you
You can rely on the first pages a quick google search brings you as much as you like, I'll rely on physics, thermodynamics and conservation of energy.
The problem is you just keep arguing about inefficiencies caused by operating the engine outside its most efficient range and somehow try to relate this to the thermodynamic efficiency of the cycle. In the same way that small throttle settings relate to inefficient operation, so does your high rpm "not much time for combustion, inefficient valve/igntion timing, or poor mixing/incomplete combustion" arguments.
They are consequences of the way the engine is operated, not the thermodynamic efficiencies of the cycle. To make it absolutely clear and to prevent you muddying the relationship between thermodynamics/compression ratio and operating an engine inefficiently, let's consider a very low RPM, full throttle engine.
Don't worry. there's so much wrong that I'm still considering where to begin!
Here it is, AGAIN.
It's about the total fluid heat change from start to finish (once the fluid is returned to atmospheric pressure). It is lower in a high compression engine.
Also why is it that both the piston engine and the turbine engine can have their efficiencies increased by increasing the pressure ratio (compression ratio for piston)? Is there some sort of simple thermodynamic explanation for this?
Anyway, back to Google for you
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@slippery pete
I must say I find it amusing that you, as a nonmechanic, with no understanding of how the Otto cycle(modified) works is espousing inaccurate information on ignition timing, flame front propagation, VE(volumetric efficency), conversion of fuel energy into mechanical power and so forth.
Your theory of the infernal combustion engine is enlightening.
Cheers
BH
I must say I find it amusing that you, as a nonmechanic, with no understanding of how the Otto cycle(modified) works is espousing inaccurate information on ignition timing, flame front propagation, VE(volumetric efficency), conversion of fuel energy into mechanical power and so forth.
Your theory of the infernal combustion engine is enlightening.
Cheers
BH
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calm down gents, its an interesting discussion, no need to rush, and i must say when it comes to
( automotive?) pistons i thought about it and must admit i am not pretty sure here.
due to my knowledge hot exhaust gases are a "waste" not being able to convert into useful output. it is also a common knowledge that rising the compression rises thermal efficiency- so a higher compressed piston engine , a more efficient engine will give less "waste" hot exhaust gases and so the exhaust temperature will be lower. but i googled it and found this pdf.
http://rescomp.stanford.edu/~efroeh/...erformance.pdf
when you search for the "exhaust" or "temperature" you will find statements and graphs showing that indeed the EGT is lower at a higher compression and as well it is lower when spark timing is advanced.
so slippery truly seems to be right here when we discuss the theoretical coupling of EGT and compression.
but ots not thats not that simple- fuel mixture, piston design, cylinderhead design and much more seem to have an influence on it. and oggers gave also a lot of useful information. and all this seem to be valid on piston engines.
cheers.
( automotive?) pistons i thought about it and must admit i am not pretty sure here.
due to my knowledge hot exhaust gases are a "waste" not being able to convert into useful output. it is also a common knowledge that rising the compression rises thermal efficiency- so a higher compressed piston engine , a more efficient engine will give less "waste" hot exhaust gases and so the exhaust temperature will be lower. but i googled it and found this pdf.
http://rescomp.stanford.edu/~efroeh/...erformance.pdf
when you search for the "exhaust" or "temperature" you will find statements and graphs showing that indeed the EGT is lower at a higher compression and as well it is lower when spark timing is advanced.
so slippery truly seems to be right here when we discuss the theoretical coupling of EGT and compression.
but ots not thats not that simple- fuel mixture, piston design, cylinderhead design and much more seem to have an influence on it. and oggers gave also a lot of useful information. and all this seem to be valid on piston engines.
cheers.