Why do turbine engines require a compressor section [Archive]

Slippery_Pete

11th Nov 2011, 11:09

Hi QJB.

Sorry no-one has answered the question you originally answered. I'm sure you realise the answers given about volumetric efficiency have nothing to do with your question about thermodynamic efficiency of an engine.

The best way to answer you question about the thermodynamic efficiency of an engine increasing with a higher compression ratio, is to consider losses due to heat.

If you consider two cups of water - 1x 50 degrees celsius, 1x 100 degrees celsius... if you put them over a flame of 200 degrees for exactly one second, the cooler cup of water will absorb more heat (because the temperature split between the two is larger).

The same applies in an engine cylinder. When the ignition occurs, a lower compression ratio engine will have a cooler air/fuel charge in the cylinder - and so it will absorb more energy (which is wasted as exhaust gas heat).

A high compression ratio engine will ignite a hotter air/fuel charge which will absorb less heat. Less energy wasted as heat = more energy transferred to the crank.

Of course, it follows then that if you were to have two almost identical piston engines (one low/one high compression) burning exactly the same amount of fuel, the exhaust gases from the higher compression engine would be slightly cooler than the low compression engine.

Make sense?

oggers

11th Nov 2011, 11:33

A high compression ratio engine will ignite a hotter air/fuel charge which will absorb less heat. Less energy wasted as heat = more energy transferred to the crank....Make sense?

No not really. It was the crank that turned the piston that compressed the air in the first place, so no gain there only a net loss. :rolleyes:

barit1

11th Nov 2011, 12:35

One might approach this from the ramjet end of the spectrum. What makes a ramjet work, and why won't it create thrust while standing still? Once you internalize this problem, you begin to understand what makes a gas turbine work.

lomapaseo

11th Nov 2011, 12:51

the compressor ensures that the air will stay in the combustor long enough to burn and move rearward out through the turbine.

Take away the compressor and all you have is a bonfire

oggers

11th Nov 2011, 14:14

Take away the compressor and all you have is a bonfire

Best Answer (rated by viewers)

Chu Chu

11th Nov 2011, 16:43

For the turbine to work, the pressure in the combustion chamber needs to be higher than ambient pressure (otherwise there would be no flow across the turbine). But combustion requires air, and that air needs to be forced in against the combustion chamber pressure.

I think that's the simplest answer, though it ignores a lot of stuff.

balsa model

11th Nov 2011, 17:40

And let's not forget the other essential function of the compressor which is to finely chop up any meandering birds so that they pass through the combustion chamber without clogging it up. Inferior designs end up falling into water and generating lot's of Pprune traffic ;)
(PS: It's Friday.)

Slippery_Pete

11th Nov 2011, 22:11

No not really. It was the crank that turned the piston that compressed the air in the first place, so no gain there only a net loss. :rolleyes:
Mate, are you serious? :ugh: The losses incurred by compressing the air have nothing to do with it, because that energy is regained... - ie, if you ignore any fuel going into the engine, if more energy is used to compress the air, at the top of the cycle there is more energy to push the piston down. Compressing an air charge with a piston is like a squashing a spring - you get back what you put in (ignoring friction losses).

I think you are just trying to shoot me down because I said you comments about volumetric efficiency didn't answer the original post. :D You can argue all you like, but I have a physics degree and the principles of thermodynamics have been unchallenged for a few hundred years.

A certain amount of available energy enters the engine in the form of fuel. A heat engine converts this into three types of energy - sound, heat and useable work.

The sound is such a tiny percentage it can be ignored. As a result, the efficiency of an engine is essentially a function of how much waste heat it DOESN'T produce. The difference in heat between the air entering the front of the engine compared to the heat of the exhaust air is proportional to the thermodynamic efficiency of the engine.

QJB, if my previous posts didn't make sense... you might as well start here.

Thermal efficiency - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Thermal_efficiency)

grounded27

12th Nov 2011, 00:16

Fact is that does not need one but (if working) would be considered more of a ramjet with a turbine disk to slow it down inside. The fan is a compression stage...

oggers

15th Nov 2011, 20:24

Mate, are you serious?

This is my serious face.

You can argue all you like, but I have a physics degree and the principles of thermodynamics have been unchallenged for a few hundred years.

I'm not challenging the laws of thermodyamics. What I'm disagreeing with is how you've explained them as they apply to a real engine. You may have a degree in physics but your post will have to stand on its own merit.

"Sorry no-one has answered the question you originally answered. I'm sure you realise the answers given about volumetric efficiency have nothing to do with your question about thermodynamic efficiency of an engine."

His original question was 'why does a turbine engine need a compressor' and that was answered in a number of ways including one direct reference to the Brayton cycle, which is correct.

Volumetric efficiency impacts on thermal efficiency though. If the VE is poor then you are suffering pumping losses and using more energy just to keep the engine running. Idealised heat engine models are all well and good up to a point but you have to move beyond a PV diagram to understand the constraints on the efficiency of a real engine.

"When the ignition occurs, a lower compression ratio engine will have a cooler air/fuel charge in the cylinder - and so it will absorb more energy (which is wasted as exhaust gas heat)."

I don't think so. You want the working fluid to absorb all available heat from combustion so as to raise the pressure. That's where the umph comes from. Waste heat in the exhaust is simply that which you are unable to turn into work before the power stroke ends.

"A high compression ratio engine will ignite a hotter air/fuel charge which will absorb less heat. Less energy wasted as heat = more energy transferred to the crank."

Again, the method of transferring energy to the crank is to heat the working fluid so it expands against the piston. The fluid needs the heat if it is to do any work. For this reason modern engines are run a little hotter than they used to be so that less of the combustion heat is taken out of the working fluid.

You are barking up the wrong tree. The significance of running the compression ratio high is to get more heat into the charge (without going too far) in order to promote better combustion. It is NOT to get more heat into the charge so the charge then absorbs less heat from combustion because that is not going to work:

The losses incurred by compressing the air have nothing to do with it, because that energy is regained...Compressing an air charge with a piston is like a squashing a spring - you get back what you put in (ignoring friction losses)

As well as friction you have blow-by and heat loss to the coolant (which both increase if CR increases I believe). Therefore, you need to get more out of the power stroke than you took from the crank to achieve the extra compression. That 'more' comes from the improved combustion, not from the heat of compression itself. It is to do with better mixing of the fuel and increased flame speed.

Slippery_Pete

16th Nov 2011, 02:24

Hi Oggers.

Yes, you are correct that there are other factors at play in a real engine.

Volumetric efficiency impacts on thermal efficiency though. If the VE is poor then you are suffering pumping losses and using more energy just to keep the engine running. Idealised heat engine models are all well and good up to a point but you have to move beyond a PV diagram to understand the constraints on the efficiency of a real engine.

The loss of efficiency due to pumping losses (ie the engine having to suck air in and blow it out) 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.

Waste heat in the exhaust is simply that which you are unable to turn into work before the power stroke ends.

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).

You are barking up the wrong tree. The significance of running the compression ratio high is to get more heat into the charge (without going too far) in order to promote better combustion. It is NOT to get more heat into the charge so the charge then absorbs less heat from combustion because that is not going to work:

Sorry, disagree. I do agree higher compression engines have better mixing and better burn, but this is not the main reason they are more efficient.
You need to think again about the scenario I posed earlier. Consider two engines, one high and one low compression ratio, burning 1L of fuel per minute. I guarantee the higher compression ratio will have a lower exhaust gas temperature once back at atmospheric pressure.

Your argument is that higher compression engine will have hotter exhaust gas and that better and more complete combustion is the only reason it will be more efficient. You can't get more "umph", plus hotter exhaust gases too - where is all this extra energy coming from? 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.

aerobat77

16th Nov 2011, 06:31

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.

blackhand

16th Nov 2011, 07:11

i think slippery pete is pretty right with the higher compression- lower exhaust temperatures at pistons
Not understanding why higher compression ratio leads to lower exhaust temps.
One would have thought that exhaust temp was a function of fuel air ratio.

BH

Saint Jack

16th Nov 2011, 07:35

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.

MathFox

16th Nov 2011, 08:52

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 (http://en.wikipedia.org/wiki/Brayton_cycle)

Slippery_Pete

16th Nov 2011, 10:48

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.

oggers

16th Nov 2011, 14:31

Slippery:

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.

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:

the answers given about volumetric efficiency have nothing to do with your question about thermodynamic efficiency of an engine.

However, with your new post you have now introduced some more errors:

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

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.

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).

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 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.

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.

You can't get more "umph", plus hotter exhaust gases too - where is all this extra energy coming from?

Strawman. I have never said that.

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"?

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.

More energy is transferred to useful work because less heat is wasted heating the fluid. It's that simple

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?

oggers

16th Nov 2011, 19:19

aerobat:

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.

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.

Luc Lion

16th Nov 2011, 19:40

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

MathFox

16th Nov 2011, 22:02

QJB had another question:
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)?

The mathematics behind all this is quite complex, but it all hinges around Carnot's theorem Carnot's theorem (thermodynamics) - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Carnot%27s_theorem_%28thermodynamics%29)

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

Ref: Thermal efficiency - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Thermal_efficiency)
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.

Slippery_Pete

16th Nov 2011, 22:56

Hi Oggers.

I'll say it one more time, and if you can't understand it after that, then I will consider :E 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.

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.

Meanwhile, the "comparatively small" pumping losses you mention are big enough to explain the efficiency difference between diesel and petrol engines.

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.

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.

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.

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?

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.

Chu Chu

17th Nov 2011, 00:35

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.

blackhand

17th Nov 2011, 03:52

@chu chu
Think BMEP and rework you supposition.

Cheers

oggers

17th Nov 2011, 09:04

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 have absolutely nothing to do with it. You simply CAN NOT burn the fuel "faster" to get more energy out of it.

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. :rolleyes: 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. :ok: 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.

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...

to suck the air in and blow it out? Hahaha, you've got to be joking

...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. :8

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.

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". :rolleyes: I would say you've had an epiphany but you're actually just swapping between two positions.

Slippery_Pete

17th Nov 2011, 10:06

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 :ok:

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'm listening.

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.

And then you accused me of not understanding timing with this...

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. http://images.ibsrv.net/ibsrv/res/src:www.pprune.org/get/images/smilies/thumbs.gif I guess they missed that bit out when you were doing physics at uni

Oops, guess you didn't read (or understand) what I'd already written.

19% is your figure from the back of a fag packet.

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).

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?

:D

oggers

17th Nov 2011, 12:57

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

Don't worry. there's so much wrong that I'm still considering where to begin!

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.

Yes I did miss that, thanks for pointing it out 'cos that's badly wrong as well! All you'll do is punch a hole in the piston. :ok: 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"

...and then with a straight face you pretend you know something about timing!

Luc Lion

17th Nov 2011, 14:54

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

http://www.grc.nasa.gov/WWW/k-12/airplane/Images/braytonts.gif

The image is from the www.grc.nasa.gov (http://www.grc.nasa.gov) web page : Turbine Engine Thermodynamic Cycle - Brayton Cycle (http://www.grc.nasa.gov/WWW/k-12/airplane/brayton.html)

oggers

17th Nov 2011, 16:12

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 ?

Luc, both myself and slippery gave our direct answers to QJB at the top. With all due respect you aren't a moderator. := 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.

Slippery_Pete

17th Nov 2011, 21:25

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.

Don't worry. there's so much wrong that I'm still considering where to begin!

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.

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.

I'm STILL listening.

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?

It's because the fluid absorbs less heat :ugh:

Anyway, back to Google for you :D

blackhand

17th Nov 2011, 21:50

@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

aerobat77

17th Nov 2011, 23:57

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/papers/RDH_Engine_Performance.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.

Slippery_Pete

18th Nov 2011, 00:29

Glad you've come over to the dark side, aerobat! :ok:

All the arguments Oggers posted relating to small throttle settings (VE), or flame front speed, or burning fuel faster, or ignition timing blah blah blah - don't relate to the original post about thermodynamic cycle efficiency of low vs. high compression. :ugh:

In fact, these problems disappear in a constant pressure heat engine like a turbine, where the concept is the same:

Less heat transferred to the fluid = more useable energy available
More useable energy available from same energy input = higher thermodynamic efficiency :ok:

I'd really like to hear Oggers explain how ignition timing, or valve timing, or mixtures, or flame fronts, or pumping losses, or any other of his false arguments explain why a high compression turbine engine is more efficient than a low compression turbine. :D

blackhand

18th Nov 2011, 00:55

I'd really like to hear Oggers explain how ignition timing, or valve timing, or mixtures, or flame fronts, or pumping losses, or any other of his false arguments explain why a high compression turbine engine is more efficient than a low compression turbine. :D
Now that is very clever, change the parameters of your argument.
You were talking about the Otto cycle, not the Brayton cycle:D

aerobat77

18th Nov 2011, 06:16

I'd really like to hear Oggers explain how ignition timing, or valve timing, or mixtures, or flame fronts, or pumping losses, or any other of his false arguments explain why a high compression turbine engine is more efficient than a low compression turbine.

gents, are you talking turbines or pistons ? valve timing in a turbine ? when it comes to turbine engines we also discuss pressure ratio, not compression. maybe its just a misunderstanding since different things are mixed up.

the original question is well answered but of course a little bit like asking why a piston engine needs pistons . also the ram-jet explanation directs a wrong way since a ram jet is not a turbine engine. the job of a turbine disc is ( beyond driving the propeller via a gearbox at a turboprop) to drive the compressors , and when there are no compressors before the combustion chamber , for what does the turbine disc spin ?

maybe the thread beginner meant a jet engine, since the question for what a compressor in a turbine engine is self answering.

Slippery_Pete

18th Nov 2011, 09:59

Now that is very clever, change the parameters of your argument.
You were talking about the Otto cycle, not the Brayton cycle:D
Sorry Blackhand, you really have missed the boat here.

I originally chose to talk about the Otto cycle because I thought it was easier to explain why thermodynamic efficiency goes up with higher compression of the fluid. The fact is that in both the Otto and Brayton cycles, more compression before the combustion results in less heat exchange to the fluid - and a thermodynamically more efficient engine.

I obviously made a mistake, because I didn't count on people like yourself and Oggers being unable to understand such a simple concept. Oggers then went off on several tangents, and used a myriad of false arguments about ignition timing, and valve timing, and mixing, and flame fronts, and all sorts of other things to try and squeeze his way out of the fact that nothing of those things have to do with the original question about the thermodynamic efficiency of higher compression.

gents, are you talking turbines or pistons ? valve timing in a turbine ? when it comes to turbine engines we also discuss pressure ratio, not compression. maybe its just a misunderstanding since different things are mixed up.

Hi Aerobat. Yes, these things don't occur in a turbine - this was the point I was trying to make. Oggers used these arguments (valve timing, fuel mixing, incomplete combustion, VE at low throttle) as his reasoning for the difference between the efficiencies of high and low compression engines.

My switch over to turbine was to show that his arguments don't apply on a turbine, where in fact the same thermodynamic principle does (higher compression before combustion results in less energy wasted heating the fluid = more available work).

Luc Lion

18th Nov 2011, 12:38

Slippery_Pete,

you stated several time the idea
Less heat transferred to the fluid = more useable energy available
I may be misunderstanding the point you intend to make, but I think that the statement is either wrong or more or less correct but very misleading.

Your sentence suggests that we add energy to the fluid and that this energy is split between useful work and heat. And so, the less heat the more work.
That's misleading.
Actually, the sequence of events is:
- we take a gas that has an internal energy U1
- optionally, we increase its energy to U2
(this is the compression phase. Mandatory for turbines. For steam machines, just assume U2=U1)
- we pour in some heat Q and get the internal energy U3 = U2 + Q
- we extract a percentage of the internal energy U3 into work through an expansion of the gas (usually adiabatic expansion).
- the resulting energy U4 = U3 - W
The percentage of extracted energy depends on the pressure ratio.
Actually it depends on the temperature ratio T4/T3 and is 1 - T4/T3 .
For adiabatic expansions, it can be shown that 1 - T4/T3 = 1 - (p4/p3)^((γ-1)/γ)

You can see that, for instance if the extraction efficiency is 40%, we extract 40% of U3 which is 40% of U2 plus 40% of Q. The more heat is added to the fluid, the more work is extracted.

This picture is, of course, a simplification. I looked at the efficiency of just one phase of the cycle rather than looking at the efficiency of the whole cycle. However it shows that:
- heat is the object of the energy extraction efficiency and is not an adjustement variable of the efficiency
- theoritical efficiency is just linked to the temperature ratio during expansion phases or to the pressure ratio which can be considered as a surrogate for temperature ratio

Note that the 40% U2 extracted is not magic energy ; it is only if the gas has been compressed initially from U1 to U2 with U1 being 60% of U2 that the final pressure ratio allows a 40% reduction. So the 40% extracted of U2 is just some invested energy that is reclaimed.
If there is no initial compression (steam engine) the increase of pressure obtained by adding Q in a fixed volume results in a pressure increase that allows a very poor extraction efficiency, much less than Q/U1.

Note also that the final state U4 is equivalent to the initial state U1 except that we extracted only a part of Q. So grossly, U4 = U1 + 60%.Q
And the final temperature of the gas is not a variable that can be acted upon ; it is dictated by be heat that could not be extracted.

Luc

oggers

18th Nov 2011, 16:02

I'd really like to hear Oggers explain how ignition timing, or valve timing, or mixtures, or flame fronts, or pumping losses, or any other of his false arguments explain why a high compression turbine engine is more efficient than a low compression turbine.

Slippery_pete: please don't do this to yourself. You know full well that all of those things were being discussed in the context of the piston engine. Anyone who reviewed our exchange could see that - the clue is in the liberal use of the terms 'crank', 'piston', and 'cylinder', including in your first post where this started :rolleyes:

aerobat77: thanks for the link to that paper. There won't be any surprises in there for those who've been schooled in piston engine theory and it's a useful case study for those that might be interested. :8

Specifically on the point of exhaust temperature there is no surprise that it dropped as compression rose: this is not in dispute.

lomapaseo

18th Nov 2011, 17:11

Folks when you get all done with this piston engine stuff (no compressors or turbines), could you do a google-idiot style translation into turbine engines for the casual thread reader.

Maybe a distinction could then be made between TIT (Turbine Inlet Temperature), EGT (Turbine Exhaust Gas temperature) out the tailpipe and pressure out of the compressor,

I doubt that timing enters into this for a gas turbine since the flowrate is constant and so is the fuel and ignition.

The casual reader would appreciate simple comparisons with everyday effects e.g. welding torches, bunsen burners, baloons, home oil heaters etc.

blackhand

18th Nov 2011, 18:58

Turbine inlet temperature - Not indicated on turbine engines I work on.
Usually ITT, interstage turbine temperature between compressor turbine and power turbine sections is indicated
EGT is idicated on piston engines as an aid to leaning the mixture, not so much on turbine engines.
In any case will be lower than ITT and TIT.
Pressure out of the compressor is a function of the compressor stages in axial flow and compressor size/effenciency in centrifugal compressor; and is expressed as a ration over inlet pressure.
Which leads us to bleed valves and interstage bleeds to decrease stall margins etc.
BH

Slippery_Pete

18th Nov 2011, 21:46

Hi Oggers.

I've already explained why I used the words crank, piston and cylinder in my first post - because I thought it would be easier for people like you to understand the thermodynamic concept ask by the OP if I talked about the Otto Cycle. I used those words to try and explain the concept to most readers on this forum, but I never said anything other than the fact that the main reason for thermodynamic efficiency increase in high compression is reduced heat waste to the fluid. All of the false arguments were introduced by you.

I'd really like to hear Oggers explain how ignition timing, or valve timing, or mixtures, or flame fronts, or pumping losses, or any other of his false arguments explain why a high compression turbine engine is more efficient than a low compression turbine. :D

I'm listening.

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.

FOR THE FOURTH TIME, I'm still listening. You really are choosing to avoid this.

Luc,

Why don't you read all my posts, rather than glaze over the last one?

The point regarding adding heat to the fluid is the total heat added to the fluid over the entire cycle (ie, once the fluid is returned to atmospheric pressure).

I'll say it again, for either a piston or turbine pressure, then higher the compression ratio or pressure ratio, the less overall energy will have been absorbed by the fluid by the time it has returned to atmospheric pressure.

Higher compression/pressure ratio means for a given fuel flow, less heat exchange has occurred to the exhaust gas by the end of the cycle. It allows more energy to be extracted as useful work.

aerobat77

18th Nov 2011, 23:52

ok, when we come to turbines i will give here some impressions on the turboprop GA aircraft i currently fly- a cheyenne III with the pt6a-41 engines. the pt6a-41 is a turboprop two shaft engine which incorporates a three stage axial compressor followed by a 1 stage centrifugal compressor driven by a 1 stage compressor turbine. after this we have a 2 stage power turbine which drives the prop via a gearbox. the both shafts are not mechanically coupled and the low pressure turbine drives only the prop - so we call the pt6a is a free turbine.

the engine is like common nowaday design flat rated - so the rated power output is a mechanical limit and the engine is able to keep rated power above ISA or keep rated power in thinner air when you climb until its thermodynamical limit ( ITT or compressor speed) is reached. in other words- the engine could develop on ground more power that the gearbox is approved for.

basicly on this engine you give with the power levers an input to the FCU ( fuel control unit ) to set a target compressor speed. in regard to air density and outside temperature a given amount of fuel is needed to keep this speed. this will result in a given force to the power turbine and a given torque - so power output.

at take off you are mostly torque ( so power output) limited and the turbine is at its mechanical limit . the ITT and compressor speed are below its limit. when you climb out and do not touch the power levers the compressor speed stays the same. the ITT also but torque and fuel flow decreases. this is due the fact the FCU ( fuel control unit) keeps like said a target compressor speed . in a climb out the air gets thinner and the "resistance" on the compressor stages also. so the compressors try to spin faster and the FCU has to decrease fuel flow to keep the same speed. due to less fuel and gas driving the power turbine the torque ans power output also decreases.

when you want to keep the same power output in a thinner air in a climb you will have to push the power levers more and more forward. this will result in a faster and faster spinning compressors and a higher and higher ITT until at a given altitude you match the maximum ITT or compressor speed. here the turbine reaches its thermodynamical limit.

sooo... when the air gets thinner and the compressors deal with a lower pressure ratio ( in pistons compression) the ITT rises .thats a fact. i found and attached a pic at our top of climb in FL 280 with a cheyenne III with pt6a-41 engines so you can have a look what the torque, ITT, compressor speed amd fuel flow is here. at this altitude the engine is at its thermodynamical limit - so the compressor speed is at company limits resulting in a given ITT and torque far below its redline ( mechanical limit)

now we can talk why it is so a a turbine engine.

oggers

19th Nov 2011, 11:10

Aerobat77:

now we can talk why it is so a a turbine engine.

Turbine engines utilise air for cooling of the combustion chamber and the turbines themselves. ITT will rise due to the reduced mass flow rate of air going to the combustion chamber as you climb.

Less mass of air going into combustion chamber, but same mass required for combustion itself = less left over for cooling. I'm not saying this is the only cause (I have a mere pilot's understanding of turbines) but it is the only cause I was taught at the Royal Naval Flying Training School. It is important to helicopter pilots because you can become power limited in the hover due to turbine temp before you reach the service ceiliing - a factor in so called 'hot and high' operations.

Mr Optimistic

20th Nov 2011, 18:54

...if you didn't have a compressor it would be a ram jet. Anyone mentioned stoichiometric yet ? As someone said, need to have the pressure gradient to avoid a bonfire - imagine starting a ram jet at zero speed.

chris weston

20th Nov 2011, 20:14

Gentlemen, as a mere physical chemist and at the risk of causing more confusion, may I suggest we try to stick to what is to me the fundamental process here?

The more we compress the air the more oxygen molecules we supply per unit time to the combustion chambers so we can burn more mostly hydrocarbon fuel per unit time and release more energy to do useful work per unit time.

That's the simple central point here.

Let's dig a little deeper.

The Second Law of Thermodynamics can be expressed as

ΔG = ΔH -TΔS

The combustion products are hotter than the fuel or the air reactants so the ΔS term becomes more positive as their entropy is now larger, the negative sign above then makes the TΔS term negative.

T is large and positive again making the TΔS term negative

The combustion process (yes yes ideally stoichiometrically , we'll have that discussion if you wish but not here for the moment please) is an exothermic reaction that mostly releases energy to the surroundings in the form of heat so ΔH has a negative sign on this side of the pond.

So ΔG (the free energy) is large and negative and we have lots of free energy available to do useful work - such as provide thrust.

Finally, to go back to the original question of "why compress?", we can now say "so as to release more energy per unit time from the fuel"

CW

barit1

20th Nov 2011, 21:52

oggers:ITT will rise due to the reduced mass flow rate of air going to the combustion chamber as you climb.

Well, this may or may not be true for a given engine type, depending on the characteristics of the control system (whether governing to a constant core speed, constant corrected speed, constant torque, constant CDP, etc etc.)

If an electronic control, it's all done with ones and zeroes; if hydromechanical, then it's bellows and cams and flyweights.

Dan Winterland

21st Nov 2011, 02:33

Blimey - PPRuNe at it's best. A simple question asked by someone seeking enlightenment and they get a thesis and a sh!t fight!

I would like to answer the original queries by by answering the second question first. In simple terms, the power output of a reciprocating engine can be expressed as bore x stroke x rpm x compression ratio. Based on this, if you force more of a charge into the engine by using a turbo or superchager, the compression ratio will increase and the power output will go up.

As for the first part, obviously you understand the OTTO cycle an the principle of induction, compression, conmbustion and exhaust, or suck, squeeze bang and blow. The turbojet engine does the same. The suck is the flow into the engine, the squeeze comes from the compressor, the bang occurs in the combustion chamber and the blow is the thurst. If you look at it like this, if you didn't have the compressor, the engine wouldn't work and as someone else said, all you would have is a bonfire!

oggers

21st Nov 2011, 09:54

Barit:

Well, this may or may not be true for a given engine type [that ITT rises in the climb due to reduced mass flow rate of air], depending on the characteristics of the control system (whether governing to a constant core speed, constant corrected speed, constant torque, constant CDP, etc etc.)

No doubt you're quite right. However, I was responding to aerobat's specific scenario regarding his specific type in which he observes an ITT rise.

If you follow the thread back you will find this comment by aerobat which gives context:

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

My point is that if you observe such a rise in those circumstances there is a good alternative explanation; namely less cooling air available.

TURIN

21st Nov 2011, 11:08

Pedant head on.

Internal combustion engine (Otto) cycle - suck, Squeeze, bang, fart.

Gas turbine (I refuse to call it Brayton as Mr Whittle formulated it years before) cycle - Suck, squeeze, burn, blow.

Just sayin'. :\

Anyway, to answer the original question...

If the turbine engine didn't have a compressor, then the turbine is just a spinning disc getting in the way. It has to drive something.

I suppose in theory you could have a turbo-prop with no compressor and rely on a very efficient ram effect for compression but as no-one has done it yet I suspect it won't work, for all the very clever and frankly complex answers given on the last 3 pages. :ok:

lomapaseo

21st Nov 2011, 12:24

If the turbine engine didn't have a compressor, then the turbine is just a spinning disc getting in the way. It has to drive something

actually it won't spin unless there is a pressure drop across it. The turbine needs a forced movement of air through it.

TURIN

21st Nov 2011, 20:03

Quote:
If the turbine engine didn't have a compressor, then the turbine is just a spinning disc getting in the way. It has to drive something

actually it won't spin unless there is a pressure drop across it. The turbine needs a forced movement of air through it.

Erm, I don't quite understand your point. :confused:

EG. A ram jet with a turbine stuck up it's arris. :\ Pointless but you get my drift?

lomapaseo

21st Nov 2011, 20:36

EG. A ram jet with a turbine stuck up it's arris. http://images.ibsrv.net/ibsrv/res/src:www.pprune.org/get/images/smilies/wibble.gif Pointless but you get my drift

yea, but how do you get it up to speed to ram it? Difficult to start one on the ground.

TURIN

21st Nov 2011, 20:59

Yes, I know but that wasn't the point I was trying to make.

chris weston

21st Nov 2011, 21:47

A ram jet has sufficient compression without the need for a fan assembly.

Go sub routine velocity dependence!

Why do we compress - to release more energy to do useful work per unit time within the engine (be it piston or gas turbine).

CW

Slippery_Pete

21st Nov 2011, 23:29

Why do we compress - to release more energy to do useful work per unit time within the engine (be it piston or gas turbine).

It's so that less heat is added to the air over the entire cycle.

Burning a certain amount of fuel produces a certain amount of energy, let's call it X. You can't change the amount of energy released by burning a fuel by burning it faster or slower or under more pressure. Burn it as fast or slow or under as much pressure as you like, the amount of energy stored in that fuel is constant.

The energy produced by X is divided into waste heat, useful energy and sound by an engine. The sound is so negligible against the power output of the engine, it can be ignored.

So energy from the fuel = useful work + waste heat

The ONLY way to get more energy out of the same amount of fuel is to make less waste heat. Higher compression results in less waste heat transferred to the air by the end of the cycle.

barit1

22nd Nov 2011, 03:20

Any aircraft engine supplies a propulsive force by capturing a mass of air and then accelerating that air. It may be a propeller doing the job, or it may be a gas turbine (we call it a jet engine). In any case, a pressure increase is needed to force (accelerate) the air mass aft. Thus, the need for a compressor. The prop is a compressor; the fan in a turbofan is a compressor; and so is the compressor (doh!) in a straight jet or a turboprop. :)

oggers

22nd Nov 2011, 10:48

slippery....

The ONLY way to get more energy out of the same amount of fuel is to make less waste heat. Higher compression results in less waste heat transferred to the air by the end of the cycle.

If you confine yourself to framing everything in terms of thermal efficiency you will never be able to explain how or why an engine works.

The topic of this thread is simply "why do turbine engines require a compressor section". That is the question that was being answered by CW. His answer emphasised the requirement for power.

Nobody wants an aircraft that can taxi out on a thimble of fuel before failing miserably to accelerate to flying speed.

chris weston

22nd Nov 2011, 11:17

Slippery Pete, sir,

Saying that "the main reason for thermodynamic efficiency increase in high compression is reduced heat waste to the fluid" is, in my view, wrong. It is the "main reason" aspect of the statement that I quarrel with.

The primary reason for compressing the air is the enhanced power per unit time the engine can then produce; heat transference losses are also reduced it is true for the reasons you give but I submit that that it is a secondary (albeit welcome) effect of the compression.

With compressed air you have more oxygen molecules in the combustion chamber and you can burn more fuel per unit time. You release more energy to do useful work per unit time. The time dependence is critical.

The main reason is power but you are right in that, under high compression, the loss of heat this way is a smaller % of the process i.e. it helps but it's not the key driver.

If we really want to get obscure the parallel with Le Chateliers Principle when justifying running big pressures on systems that are making fewer moles of gas in production processes is a good one - you "notice" the drop in entropy less than at low pressures.

Yes yes there's a rate of reaction factor at elevated pressures too but let's not go there either!

I suspect we shall have to agree to differ short of doing a lot of sums in public.

CW

barit1

22nd Nov 2011, 12:51

Methinks you are all missing the point. Suppose I build an aircraft that is electrically powered (motor driving a prop), or hydraulically powered, or compressed air, or human powered. In all cases the prop is a compressor - a device which captures a mass of air and accelerates it aft.

Now if we put a simple turbojet in the plane, it still has to - ahem - capture a mass of air and accelerate it aft. But we have to somehow drive the compressor, and so we have a burner and turbine to recover some of the energy in the airflow to create a continuous Brayton cycle (named after American engineer George Brayton (1830–1892)).

Lonewolf_50

22nd Nov 2011, 14:18

Luc/MathFox/Dan and many others -- thank you for a fun discussion.

Note as to reasons why we use a compressor, per the original question.
Chris: Why do we compress - to release more energy to do useful work per unit time within the engine (be it piston or gas turbine).

Yes.
Slippery_Pete :It's so that less heat is added to the air over the entire cycle.
Not quite.

That may be an outcome of the process of burning oxygen to release energy in the fluid to get something to turn (shaft or turbine) but it isn't the reason to do it.

Other point made earlier: without a delta P, (Pressure differential between inside the turbine engine and the outside air) you get no work out of the turbine, be it in turbo shaft, turbo prop, turbofan, or turbojet.

You build turbine engines to get work out of them.

My degree isn't important, but I did take some thermogoddamics and a few courses on compressible flow and turbomachinery. Otto, Brayton, Carnot ... all familiar names.

Put energy in, get directed energy out to make something turn.

That is why you have an engine.

Slippery_Pete

23rd Nov 2011, 16:18

Hi Chris Weston, respectfully I disagree.

With compressed air you have more oxygen molecules in the combustion chamber

More oxygen molecules in the combustion chamber? Only in a turbine. In a piston engine, an 8:1 engine has the same volume of air in the combustion chamber as a 12:1 compression ratio engine. If your theory of more air was important, it would not effect a piston engine (which it does).

and you can burn more fuel per unit time

Burning more fuel per unit time has nothing to do with it. The entire CONCEPT of the original post is about thermodynamic efficiency (ie being able to get more useable work out of the same amount of fuel). Anyone can just increase the fuel flow and get more power out of an engine ... has nothing to do with thermodynamic efficiency.

Let's make it absolutely clear... we are talking about two identical engines, same size, same fuel flow, same RPM, same EVERYTHING - except one has a higher compression ratio - why is the one with higher compression getting more energy out of the same fuel? That's what we are talking about here.

Slippery_Pete

23rd Nov 2011, 20:35

Hi Blackhand.

Are you sure about this?

Look, I'm sure we all know that a piston engine with the same physicals can't have a different compression ratio, and in order to change it without changing the shape of the cylinder you would end up changing the volumetric capacity very slightly.
So, for the purposes of answering the original post, yes, if the capacity of the engines is the same, the compression ratio will not change the amount of air molecules. Lets say 2x 350 cubic inch engines - you will theoretically have the same amount of air in the cylinder at TOPD if it has a 8:1 or 15:1 compression ratio.

From work on performance engines, it is the compression pressures and not really the comp ratio that is important. Compression pressures are related to the comp ratio but are more influenced by valve and ignition timing. Of course the dynamics of the intake and combustion chambers has a direct affect.

I've talked about this before, and the fact that it relates to limitations of how a piston engine is operated (ie at inefficient high RPM or across a wide RPM band, low throttle settings etc). Not related to the OP, which was about how compression, and compression alone makes a difference in thermodynamic efficiency.

I work on the practical side, so is interesting talking to a theorist - unless of course you are just a sciolist.

Nope, no sciolist. Just someone with a basic knowledge of physics. As an engine tuner, I'm sure you know more about me on how to tune an engine on a dyno for best power and driveability over a large RPM range... interesting, but not related to the OP, nor the thermodynamic efficiency of an engine. It really is simple physics.

blackhand

24th Nov 2011, 03:09

In a piston engine, an 8:1 engine has the same volume of air in the combustion chamber as a 12:1 compression ratio engine.Are you sure about this?
we are talking about two identical engines, same size, same fuel flow, same RPM, same EVERYTHING - except one has a higher compression ratio -
I can see that you are talking theoretical engine, as some parameter of the engine has to be changed to alter compression ratio; combustion bowl size, head height, or piston stroke.
From work on performance engines, it is the compression pressures and not really the comp ratio that is important. Compression pressures are related to the comp ratio but are more influenced by valve and ignition timing. Of course the dynamics of the intake and combustion chambers has a direct affect.
I work on the practical side, so is interesting talking to a theorist - unless of course you are just a sciolist.

Gulfstreamaviator

24th Nov 2011, 03:23

So typical of PPrune...

I love the technical questions, and really would like to learn the answer....but with so many answers I often end up more confused..
and it is almost Friday...

What about a mod filter, after say 1 week to concatinate all the answers, and make 1 complete response......

I must go to Friday Jokes Forum....

glf

blackhand

24th Nov 2011, 04:19

Taking the brayton cycle to the next level
Supercritical carbon dioxide Brayton Cycle turbines promise giant leap (http://www.eurekalert.org/pub_releases/2011-03/dnl-scd030311.php)

chris weston

24th Nov 2011, 14:53

Slippery

I suspect trolling here so more fool me but I will have one final attempt and share some simple physics with you.

pV = nRT

so

p = n/V x RT

Putting that V R and T are constant we have that

p is directly proportionate to n

The greater pressure you have the more moles of gas you will have in a fixed volume.

CW

Slippery_Pete

24th Nov 2011, 20:47

Chris,

I'll say it again REALLY SLOWLY so you can understand.

Your theory of more compression = more air molecules only applies in a turbine with a fixed volume combustion chamber. It doesn't apply in a fixed volume piston engine (because the swept volume determines the number of molecules you can get in there). And yet a higher compression piston engine is also more efficienct. If it was (as you argue) all about the number of oxygen molecules, a high compression piston engine would be no more thermodynamically efficienct than a low compression one.

As they say on mythbusters, myth BUSTED.

In fact, thermodynamic efficiency has nothing to do with how many oxygen molecules are there - the only time that becomes important is if there are not enough oxygen molecules to combust all the fuel.

When will all the lies and myths end? Every new page on this thread brings more people with more false arguments. So far we've had volumetric efficiency (consequence of operating a piston engine inefficiently), faster burning fuel, incomplete combustion, valve timing, ignition timing... it goes on and on and on.

These are all important things for tuning or selecting and engine for an application, because they make it practical and useable over a large RPM range (rather than super-efficienct at one particular RPM), and because we have to manage inadequacies of the fuel, and the temperature of the components.

But the fact remains, the only thermodynamic reason a higher compression or pressure ratio engine is more efficient (THE QUESTION IN THE OP) - is because by the end of the cycle, less waste heat has been transferred to the air. IT'S SIMLPE FN PHYSICS.

You put in an amount of energy (X) into an engine in the form of fuel, you want to get some of that energy out in the form of useful work (Y), but unfortunately you get a lot of waste heat (W). Maximum thermodynamic efficiency is the concept of ... for a given amount of fuel energy (X), you get the maximum useful work (Y). High compression or pressure ratio engines do this better because less of the original fuel energy (X) is converted to waste heat (W).

Conservation of energy - I would expect most high school students to know this.

Anyway, let the next round of myths begin on thermodynamic efficiency at higher compression. I wonder what we'll get next...

MathFox

24th Nov 2011, 22:30

Taking the brayton cycle to the next level
Supercritical carbon dioxide Brayton Cycle turbines promise giant leap

That seems to apply to stationary power generation, as a replacement for steam turbines. I don't see it replace turbo jet engines as these CO₂ turbines require an external heat source.

aerobat77

25th Nov 2011, 07:30

the propeller on a turboprop is in such a point of view also a "compressor" but its main work is to create thrust, it has very little effect on pre compressing the air which enters the inlet for combustion, regardless if its a multishaft free turbine ( eg.pt6a) or a singleshaft fixed turbine ( rr dart, tpe331 e.g ) .

when it comes again to ( 4 stroke) pistons the amount of air in the cylinder is determined on the suction cycle and the suction, as we all know it ends with the piston at bottom dead centre. the amount of air being able to suck in is describes in volumentric efficiency. - at natural aspirated engines typically below 1 , at turbocharged engines above 1 - here the engine not just sucks the air, it is forced into the cylinder. the the piston moves up again and compresses the amount that was sucked in.

so when the amount of sucked air in suction cycle stays the same a low compression engine will have the same amount of air as the high compression engine at end of the compression cycle. at the higher compression engine this amount of air is just squeezed more .

its the same when you take an airballon , put an amount of air in it and then squeeze it. sqeezing it more will rise the compression inside the ballon, not the amount of air in it ( since no further air is forced inside when you squeeze it)

cheers !

TURIN

25th Nov 2011, 08:57

The prop is a compressor; the fan in a turbofan is a compressor; and so is the compressor (doh!) in a straight jet or a turboprop

Erm, I have to query that.
Does a prop actually compress? If so what is it's actual practical effect on the overall compression cycle?

The gas turbine section of a turboprop can be (but isn't always) de-linked from the prop, a free power turbine drives the prop. I would have to see the numbers but I'm not sure what contribution to the overall compression cycle the prop will actually give. Especially in some of the more complex reverse flow turbine engines.
In a directly linked turboprop such as the RR Dart, would there be any compression effect from the prop? I can't see how there would be much flow increase from the root of the blade into the intake. Discuss.

The fan in a turbofan can be referred to as the LP compressor (EG RR RB211) and I can see how that would work.

In an aft-fan config of course it would have no effect on the compression at all. Maybe that's one of the reasons the idea was only used briefly. Convair 880 is the only on I can think of, no idea what make/model the engine is.)

No axe to grind just trying to understand what the OP is after. :ok:

oggers

25th Nov 2011, 12:21

Slippery_pete:

It doesn't apply in a fixed volume piston engine (because the unswept volume determines the number of molecules you can get in there [there being the combustion chamber])

Wrong again. The "unswept volume" is the combustion chamber (or at least part of it depending how semantic you want to be). It is the swept volume x the volumetric efficiency that determines the number of molecules you get in - not the "unswept volume".

barit1

25th Nov 2011, 12:55

TURIN:Erm, I have to query that.
Does a prop actually compress? If so what is it's actual practical effect on the overall compression cycle?

Well, a prop blade is an airfoil, and Bernoulli tells us all about that. Static pressure is increased on the aft face of the blade. This may have little or no effect on the ENGINE cycle, but it's this pressure rise that integrates to propeller thrust i.e. the motive force for the airplane.

Granted the pressure rise is not huge, but the prop disc area is relatively large, and F=p x A. The prop moves a large mass of air at relatively low pressure.

Now if the prop has many blades and is enclosed in a shroud (i.e. a ducted fan), then the pressure rise is greater although the area may be less. This higher pressure acts to supercharge the downstream core engine.

Slippery_Pete

26th Nov 2011, 01:58

Hi Oggers.

Wrong again. The "unswept volume" is the combustion chamber (or at least part of it depending how semantic you want to be). It is the swept volume x the volumetric efficiency that determines the number of molecules you get in - not the "unswept volume".

Sorry mate, that was a typo, you are correct - I'll fix it up.

I'd really like to hear Oggers explain how ignition timing, or valve timing, or mixtures, or flame fronts, or pumping losses, or any other of his false arguments explain why a high compression turbine engine is more efficient than a low compression turbine. :D

I'm still listening for a response to this, and have been listening for the last 3 pages of this post, and to the PM I sent and which you ignored.

Are you still debunking the simple concept I explained on page 1? Here it is for you to ignore 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.

Mr Optimistic

26th Nov 2011, 02:13

I am surprised this is still going. Without the turbines the engine would just be a venturi, symmetrical (as far as the air can tell) front and back. Light it up on the ground and which way will the hot exhaust go ? Both ways ie bonfire with no thrust. Add an 'exhaust' turbine and the obstruction will encourage the exhaust to go the other way, ie out of the front. Need to create a preferred route, preferably backwards.

grounded27

26th Nov 2011, 03:15

Suck squeeze bang boom. How many times does this need to be said to kill this thread.

twochai

27th Nov 2011, 02:33

Isn't it amazing that such a simple question can generate so much heat???

Old Fella

27th Nov 2011, 03:18

"Why do turbine engines require a compressor section"? Simply to make the engine work efficiently. Adding to the pressure ratio results in better specific fuel consumption for a given thrust. Higher pressure ratio = improved specific fuel consumption, one of the reasons axial flow compressors are favoured in high thrust engines.

aerobat77

27th Nov 2011, 07:33

well, the original question is really self explanatory i think. like written above- when no compressor , for what a turbine disk in a TURBINE engine ???

for what do you need a piston in a piston engine would be a similar question .

oggers

28th Nov 2011, 08:43

chris weston

28th Nov 2011, 11:11

Barit, sir.
81
Nicely put.
Agree.
CW

Slippery_Pete

28th Nov 2011, 11:57

Multiple posters have disagreed with you along the way, and your latest "concept" bears no resemblance to your original
My explanation from page 1 until page 5 has remained unchanged, and until the laws of physics change - will remain so.

the answers at #43 and #47 are still there.

Posts 43 and 47 do not answer my question. In fact, they (like your most recent post) deliberately avoid it. AGAIN. :D

Since you are still unable to explain why a high pressure ratio turbine is more thermodynamically efficient, I'm going to assume you are admitting you really don't know. :ok:

Flight Safety

28th Nov 2011, 23:11

All, a fun thread, but all of you are missing an important piece of information, the direct relationship between thermal energy and mechanical kinetic energy.

I'll skip the reason for a compressor, and go straight to why higher compression yields higher efficiency. The Brayton and Otto cycles are similar, if you look at the volume, pressure and heat curves. The important question is HOW does heat energy get converted into mechanical energy inside either a piston or a turbine engine? Remember that thermal energy in a gas is directly related to the kinetic energy of the individual molecules within that gas. The temperature of a gas is the average temperature of all the molecules within the volume being measured, as some will be hotter and some cooler. The temperature of an individual molecule is the amount of kinetic motion that molecule is experiencing. The hotter it is, the faster that molecule is vibrating or moving.

The bottom of a piston and the end of the exhaust nozzle of a turbine engine are both exposed to the ambient outside pressure. When an air/fuel mixture is compressed and ignited, the resulting heat energy is directly transferred to mechanical motion, when individual molecules transfer their heat (kinetic energy) to the metal molecules of the piston or turbine blade (think Newton's cradle). In other words, the transfer of many gas molecule's kinetic energy to the metal of the piston or turbine blade, causes the metal parts to move. This happens because the heat and kinetic energy of the gas molecules are used to create pressure within the engine by design, which causes movement of the piston or turbine blade. When that movement occurs, the gas molecules give up some of their kinetic heat energy to the moving parts, and exchange it for mechanical motion or work, which is just another form of kinetic energy (again think Newton's cradle). This lowers the temperature of the gas, since it has now given up some of its heat energy in exchange for mechanical motion.

When higher compression (piston) is used, the pressure difference after ignition between the top of the piston and the ambient outside pressure at the bottom of the piston is greater, and therefore more kinetic energy is transferred to the piston, exchanging more heat from the gas to the piston's movement. If you look at the Otto cycle curves, most of the work and most of the energy transfer takes place in the top half of the power stroke, and this is why ignition timing is so important for efficiency. Bad ignition timing means heat is released by the fuel at the wrong time, and doesn't transfer as much heat to mechanical work, thus the EGT is higher when the timing is off.

For the turbine, a greater pressure ratio means a higher pressure is created in the combustion chamber, thus the pressure differential between the combustion chamber and the exhaust nozzle (across the turbines) is greater, thus more exchange of the heat (kinetic) energy of the gas into mechanical motion. In a turbofan, this is really helpful as turning the fan produces much more thrust then the residual pressure out of the exhaust nozzle (which you still use).

To summarize, the whole point of greater compression or pressure ratios, is to create greater pressure differentials across the moving parts, so more heat within the gas can be exchanged for mechanical work. This is what creates greater efficiencies.

QJB

29th Nov 2011, 05:05

Hi Everyone,

First of all thanks for taking the time to reply to my original post. As is often the case with PPRUNE I do not really feel that I have a clear concise answer to take away from the discussion. However I believe compromise one of the universal tenets of science and indeed aviation and I certainly appreciate that there is no single factor that controls the efficiency of an engine nor the need for/benefit of a compressor section.

HOWEVER... I was really looking for a simple thermodynamic explanation that would provide the basic model on which I could add any number of exceptions and limitations. In this sense I think that slippery pete's explanation helped a lot. I have been trying to find an answer independent of this forum and have enjoyed mixed success.

From what I can summarise, and I'm happy to be corrected:

1/ Why do turbine engines require a compressor section?

There are obvious and practical reasons for having a compressor section such as were mentioned in the early replies such as the need to avoid a "bonfire" etc. but fundamentally the reason seems to be to increase the overall efficiency of the engine.

2/ Why does increasing the compression ratio (piston) or pressure ratio (turbine) improve efficiency?

I think that the general thermodynamic reasoning is that, compression when viewed as an ideal process is reversible. Meaning that if I expend a certain amount of work compressing air inside a cylinder then the same amount of work can be expelled by the piston in returning to it's original state. As it is compressed air (whether inside a piston or compressor) will increase in temperature.

The efficiency of any system is:

what you get out / what you put in

Carnot's Theorem tells us that in reality even with an idealised engine (no friction losses etc) efficiency is determined by the difference between the temperature at which heat enters the engine and the ambient temperature of the surrounding environment. Since we cannot really hope to change the ambient temperature of the environment the best we can do is improve the temperature at which the heat enters the engine. In aircraft this is done by increasing the temperature at which the fuel air mixture is ignited.

One must also consider of course the effect of practical real world factors such as material temperature limitations, volumetric efficiency effects, flame propagation, points of maximum pressure and so on so forth.

Any thoughts,

J

Luc Lion

29th Nov 2011, 09:48

but fundamentally the reason seems to be to increase the overall efficiency of the engine.
No.
Fundamentally, the reason is that you need a pressure differential for extracting work from heat. And since there is no pressure increase in the combustion chamber of a turbine (turbine = open flow engine), the pressure increase must be created by a compressor.

Without a compressor, there no work that can be extracted !

Luc

Golf-Sierra

29th Nov 2011, 10:12

Without a compressor, there no work that can be extracted !

Or, in other words, the efficiency of the engine drops to 0.

Golf-Sierra

Flight Safety

29th Nov 2011, 10:31

Other related tidbits:

Volumetric Efficiency - Primarily related to power production. The amount of air and oxygen pumped through a given engine's physically limited size, determines the amount of fuel that can be burned, thus peak power production. However any engine design, piston or turbine, has a sweet spot (or range) of power production that has the highest thermal efficiency (converting heat to work), thus the lowest specific fuel consumption for work produced in that power range.

Turbochargers or Superchargers - Increases the amount of air and oxygen pumped through an engine, and the amount of fuel that can be burned. Superchargers NEVER add thermal efficiency only power, however a turbocharger can add thermal efficiency in 2 ways. One, it allows a smaller engine to produce the same power as a larger engine, making it possible to operate a smaller engine closer to its optimum thermal efficiency power range in a given application. Two, the recovered waste heat from the exhaust is used by the turbo to pump air through the engine, thus taking over the engine's air pumping duties and reducing or eliminating the engine's normal pumping losses.

Atkinson Cycle - A piston engine design where the power stroke is longer than the compression stroke. The purpose is to increase thermal efficiency by extracting more work from the heat of the burned air/fuel charge, at the expense of power density. True Atkinson engines used various mechanical linkages to create the dissimilar length of compression and power strokes, whereas modern Atkinson cycle engines (such as those used in hybrid autos) use clever valve timing to achieve the same result.

Lean Burning - Can increase thermal efficiency. Stoichiometric fuel burning is designed to burn just enough fuel to consume all the oxygen in an air/fuel charge, and still produce complete combustion of the hydrogen and carbon in the HC fuel, with no oxygen left over. However lean burning burns less fuel than the available oxygen can support. Less heat is added relative to the amount of air compressed in either the Otto or Brayton cycle, making it possible to convert more of a smaller heat product into mechanical work with less residual wasted heat. It can also be argued that a more open throttle for a given power output while lean burning, reduces pumping losses. Like the Atkinson cycle, lean burning produces more thermal efficiency at the expense of overall power. However clever engine design allows lean burn to be a selectable mode rather than a constant condition, to preserve an engine's high power production when needed.

pattern_is_full

29th Nov 2011, 17:06

It seems to me that this is all getting overcomplicated - AND back-to-front, as it were.

The simplest answer to "Why a compressor?" is - to provide "artificial airspeed" through the engine so that it will run even when sitting still. Actually, of course, it is REAL airspeed - but localized to within the confines of the engine.

Without that flow - you get, as previously mentioned - a bonfire. Google "turbine hot start."

Pure internal-combustion jets (e.g. ramjet) require substantial forward speed (= airflow through the engine) to function at all. At least 100 mph (160 kph) for minimal function. "Best" operational speed is at Mach 0.5 or higher.

Very hard to taxi (or pull up to the gate) while maintaining Mach .5 :}

So - there needs to be a way to keep the engine up to minimal self-sustaining spin, even with little or no external inflow of air from the speed of the aircraft/vehicle itself. Thus the compressor.

A ramjet works with neither compressor NOR power turbine! The main function of the power section in a turbojet, at least at lower aircraft speeds, is - to drive the compressor to sustain ignition! Ideally subtracting as little power from the exhaust thrust as possible.

In a turboprop/turboshaft engine, that is the main function of the power turbine - period. A turbine helicopter can hover while the compressor keeps the airflow/speed through the engine up in the mach numbers.

If an aircraft reaches a high enough speed, in theory one could fold the compressor blades out of the way, and just let the ram air do its thing unimpeded. Hard to build a hinge strong enough to take the stresses, though, and in fact, development went the other direction (fanjets and turboprops) where the compressor (or at least a part of it) becomes the primary source of thrust.

Now, there is lots of room for theory and engineering to make the compressor (and the turbine driving it) work as effectively and efficiently as possible, which is the reason for all the theory and measurements and charts.

But that answers not "Why a compressor" - but "HOW a compressor?"

@ FS (below) - yes, OK, I was referring to compressor as the physical bit of machinery. A ramjet does need compression - from ram airspeed, augmented by venturi compression (usually). Just not the spinning fan.

Flight Safety

29th Nov 2011, 17:13

Actually, both ramjets and scramjets still require the "compressor" function, as they are both still heat engines using external oxygen for combustion. It's just that the "compressor" is external to the engine, using a ram effect from high speed airflow, instead of rotating blades.

barit1

29th Nov 2011, 18:02

A ramjet works with neither compressor NOR power turbine!

Well, almost. The ramjet needs a high-pressure fuel pump, and it's generally driven by an air turbine, by tapping some inlet air from the diffuser.

barit1

29th Nov 2011, 18:08

If an aircraft reaches a high enough speed, in theory one could fold the compressor blades out of the way, and just let the ram air do its thing unimpeded.

This effect can be achieved by valving, gradually opening a bypass duct while closing down airflow to the gas turbine. Several schemes have been tested, going back to P&W's entry in the SST race of the late 60s.

Flight Safety

29th Nov 2011, 18:13

Barit1, I believe the J58 of the Blackbird worked the way you described. Accord to Ben Rich, 80% of the power produced by the J58 at Mach 3.2 came from the resulting ram effect.

Lonewolf_50

29th Nov 2011, 19:11

The Blackbird used a hybrid engine before hybrids were cool. :ok: (IIRC, the engine begins as a turbojet and as it increases in speed, the flow more closely approximates a ramjet by use of valves and moving the nose cone backwards. This picture from a wiki article seems to tell that tale as well).

http://upload.wikimedia.org/wikipedia/commons/thumb/8/89/SR71_J58_Engine_Airflow_Patterns.svg/220px-SR71_J58_Engine_Airflow_Patterns.svg.png

Mr Optimistic

29th Nov 2011, 20:39

Without the pressure differential the heated air would still do work, it's just that it wouldn't be much use. Gas flows from a high pressure region to a low pressure region. No compressor and the heated gas, which is now at higher pressure (pv=nkt) sees lower and equal pressure at either end of the tube. Compressor (or ram effect, or reflected shock as in those exhaust pipe trick engines) gives the right gradient to send the gas to the designated exhaust where it expands, cools, and does work. The higher the initial pressure of the incoming gas, the higher its density and so it can support a richer mix, which increases temperature,etc.

TTex600

29th Nov 2011, 22:19

Why do turbine engines require a compressor section?

Hi guys,

Can anyone give me an answer on this? 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?

Cheers,

J

First off, I'm just a pilot. My bachelors is in professional aviation. I normally don't do math in public.

I think the answer to the "why" question is simple. It's because they are intended to produce thrust in a specific speed range. (my remarks apply to airplanes only, but could likely be used in other discussions). What is thrust? "Thrust is a reaction force described quantitatively by Newton's second and third laws. When a system expels or accelerates mass in one direction the accelerated mass will cause a force of equal magnitude but opposite direction on that system."

You can't have an equal and opposite reaction without something to push against. When the exhaust gasses move, they have to push against something in order for there to be an opposite reaction. In a piston type reciprocating engine, the expanding gases push against the piston, which is connected to a crankshaft which converts the linear force to rotating force. In a turbine engine, the expanding gases ultimately have to have something to push against and absorb the energy produced. The compressor blades absorb that push, transfer it to the bearings on their shaft which are connected to the engine cases which are connected to the engine pylon which is connected to the airframe and the entire airframe moves opposite of the exhaust.

I realize that ram/scram jets fly, but they are highly specialized and also work in a specific velocity range. I'm out of my league here, but I think that ramjets also have "something to push against" but that "thing" isn't a structure, it is a pressure wave inside the combustion chamber. Hitler's "buzz bomb" used a pulse jet which was a ramjet with a flapper door at the entrance to the combustion chamber. That flapper provided the expanding exhaust gas with something to push against.

Luc Lion

30th Nov 2011, 15:07

No compressor and the heated gas, which is now at higher pressure (pv=nkt) sees lower and equal pressure at either end of the tube
No.
This is a very common misunderstanding of turbine engine.
The gas heating is done at constant pressure not at constant volume !
The temperature increase is exactly compensated by a density decrease.
Without compressor, the heated air would just remain at atmospheric pressure and expand in both forward and backward direction and wouldn't produce any work (no pressure differential).

Luc

Yankee Whisky

30th Nov 2011, 22:53

Mc2

Increase the mass (using compression i.e.take a mass of

air and compress into a small space) to increase power output.

In a jet engine airmass accelleration is caused by heating the air in

combustion chambers.This action (accelleration) causes an equal reaction

(thrust).

To use only ram air,which I think the questioner is thinking of, there'd be

a relatively low mass, hence low thrust. A hypersonic engine, on the other

hand, relies on ram air, but requires very high velocities to compress

before heating.

This is the way I remember and understand how it works.:):)

barit1

30th Nov 2011, 23:34

TTex600:You can't have an equal and opposite reaction without something to push against.

Absolutely NOT true, and the obvious example is a rocket engine in a vacuum. The thrust is the result of the mass flow of gas and its exit velocity, and does not require "something to push against".

So the example of the piston in a cylinder has no relevance to the propulsive utility of a jet or rocket. The piston does not propel the aircraft; its mechanical linkage (connecting rod, crankshaft, perhaps a gearset) is to a propeller, which does.

barit1

30th Nov 2011, 23:40

Does anyone remember the pulsejet (http://en.wikipedia.org/wiki/Pulse_jet_engine) engine? The German V-1 used it.

Lyman

1st Dec 2011, 00:11

barit1

Add a second nozzle pointing in the opposite direction from the one you started with, on the other side of the reaction chamber.

The Third law is honored, and movement (velocity) is nil.

Slippery_Pete

1st Dec 2011, 01:16

Hi QJB.

Glad you came back and posted again.

The efficiency of any system is:

what you get out / what you put in

Carnot's Theorem tells us that in reality even with an idealised engine (no friction losses etc) efficiency is determined by the difference between the temperature at which heat enters the engine and the ambient temperature of the surrounding environment. Since we cannot really hope to change the ambient temperature of the environment the best we can do is improve the temperature at which the heat enters the engine. In aircraft this is done by increasing the temperature at which the fuel air mixture is ignited.

Yup, exactly what I've been saying from the start - this is the thermodynamic reason why higher compression engines are more efficient. Glad it helped you answer the question from your OP.

Obviously how the engine is designed, the operating RPM range, type of fuel etc. etc. are all things which are important for designing an engine for a certain application.

TTex600

1st Dec 2011, 01:31

barit1

In your rocket in a vacuum, is the rocket motor open on both ends?

Does anyone remember the pulsejet engine? The German V-1 used it.

You bust on me, and you didn't even read my post. It doesn't lend you much credibility.

oggers

19th Dec 2011, 14:19

2/ Why does increasing the compression ratio (piston) or pressure ratio (turbine) improve efficiency?

...Carnot's Theorem tells us that in reality even with an idealised engine (no friction losses etc) efficiency is determined by the difference between the temperature at which heat enters the engine and the ambient temperature of the surrounding environment. Since we cannot really hope to change the ambient temperature of the environment the best we can do is improve the temperature at which the heat enters the engine. In aircraft this is done by increasing the temperature at which the fuel air mixture is ignited.

You will not find the answer to your question in the Carnot cycle where, for two given heat reservoirs, you can achieve the same efficiency regardless of compression ratio used. With your statement above you are effectively hoping to use heat from the hot reservoir to increase the temperature difference between the two reservoirs. A hopeless endeavour.

If you want to consider the effect of increased CR in the Otto cycle with a notional perfect constant volume combustion process occuring just after TDC then you are simply left with a longer expansion path, a correspondingly longer compression path and therefore the difference between the two is also maintained over a greater path ie its more; more net work.

But you cannot increase the theoretical efficiency of the Otto cycle by taking energy out the crank and feeding it back to increase the temperature of the fuel/air charge prior to combustion. This is where slippery_pete has given you a bum steer. If you look beyond the cycle itself to the combustion process then the picture changes because you can increase the speed of combustion by increasing the temperature and pressure in the combustion chamber.

chris weston

19th Dec 2011, 18:29

Spot on Oggers

At high compression the rate of the combustion reaction increases as the effective concentration of the fuel/air mix increases.

This happens because there is now a greater statistical probability of collisions at above or equal to the Energy of Activation for the reaction

Crucially we're back to time dependence. A more rapid reaction means more bangs per unit time and more power per unit time.

Why do we compress? To make the engine more powerful.

Carnot is predicated on the most efficient cycle for converting a given amount of thermal energy into work

Carnot cycle - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Carnot_cycle)

CW

Jason Burry

19th Dec 2011, 19:29

To further Chris's post, compressorless atmospheric jet engines are actually quite common. There's one in most DIY's tool kit, it's his blow torch.

The blow torch makes pretty minimal thrust for the fuel used. The combustion chamber pressure is limited to about atmospheric, and air is induced only by venturi effect from the high speed fuel jet supplying the burner. Without combustion chamber pressure, there is no exhaust jet thrust. I've measured the thrust from a small torch (bored at work one day). Used a torch lighter. It managed to make about 1/10gram of thrust. Nothing.

Compressorless jet engines also exist in the non-air-breathing set. Oxy-fuel torches all the way to liquid fuel rocket engines. These ARE compressorless jet engines. In liquid fuelled rockets, the combustion chamber pressures are limited only by material limitations and the pressure your fuel/oxy pumps can deliver. Thus, the chamber pressure can be high, and so can the thrust.

Without a method to adequately concentrate the oxidizer, and supply it to the burner at a pressure above chamber pressures, jet engines cannot produce meaningful thrust on a continuous basis. It is possible intermittently, as in the Argus pulsejet (or indeed any pulsejet engine) on the V1 (pulsejets aren't truely compressorless either, actually, as they use tuned pipe resonance to achieve compression acoustically).

Food for thought.

J

barit1

19th Dec 2011, 19:51

In liquid fuelled rockets, the combustion chamber pressures are limited only by material limitations and the pressure your fuel/oxy pumps can deliver. Thus, the chamber pressure can be high, and so can the thrust.

Exactly, and the oxidizer pump performs exactly the same function as a gas turbine's compressor. Without pressure in the rocket's chamber, there is no thrust. And without a fuel pump supplying a like pressure, fuel would never flow into the pressurized rocket chamber.

barit1

19th Dec 2011, 20:00

TTex600:Add a second nozzle pointing in the opposite direction from the one you started with, on the other side of the reaction chamber.

The Third law is honored, and movement (velocity) is nil.

Sorry, I had missed your point. You are of course correct.

But I was addressing the common misconception of the escaping exhaust of a jet or rocket "reacting" against the atmosphere behind the vehicle. I used to have to shoot down that "logic" in the classes I taught.

You are exactly right, the gases in the rocket chamber force against the forward wall, and that is where the reaction is felt. :)

Slippery_Pete

24th Dec 2011, 06:28

Hello all, and Season's Greetings to you.

Why do we compress? To make the engine more powerful.

I'd agree with that in principle. It allows useful work to be extracted over a wide range of conditions.

If you look beyond the cycle itself to the combustion process then the picture changes because you can increase the speed of combustion by increasing the temperature and pressure in the combustion chamber.

This is misleading. You can't increase the speed of the combustion reaction to extract more energy - we've been over this before. Please see post #70. I'll repost the question I posed for you below so you don't ignore it again, and you can consider them turbine engines (this will prevent you bringing in your multiple false arguments again.

Consider two identical TURBINE engines, same size, same fuel flow, same RPM, same EVERYTHING - except one has a higher compression ratio - why is the one with higher compression getting more energy out of the same fuel?

lomapaseo

24th Dec 2011, 14:01

To tell a man something requires belief and acceptance

For a man to learn something you must teach him not tell him

oggers

27th Dec 2011, 12:47

Slippery

"My explanation from page 1 until page 5 has remained unchanged, and until the laws of physics change - will remain so."

It certainly isn't the laws of physics at issue, so let's have another look at your first post:

"If you consider two cups of water - 1x 50 degrees celsius, 1x 100 degrees celsius... if you put them over a flame of 200 degrees for exactly one second, the cooler cup of water will absorb more heat (because the temperature split between the two is larger)... The same applies in an engine cylinder."

No. In a cylinder, when you burn a given quantity of fuel 'Q', you get:

ΔU = Q - W

The idealised cycle has a constant volume combustion process and looks like this:

http://upload.wikimedia.org/wikipedia/commons/c/cf/P-V_otto.png

The time may vary by a millisecond or two but ALL the heat of combustion goes into the working fluid because W = 0 during this idealised process [#2 to #3 on the diagram].

Bearing that in mind it is clear that your first post is ill-founded:

"When the ignition occurs, a lower compression ratio engine will have a cooler air/fuel charge in the cylinder - and so it will absorb more energy (which is wasted as exhaust gas heat).

A high compression ratio engine will ignite a hotter air/fuel charge which will absorb less heat. Less energy wasted as heat = more energy transferred to the crank."

To be clear, the energy absorbed during this idealised heat addition process will NOT vary with CR in the way you suggest. No useful work can be done - in theory or in practice - with any of the heat until it HAS been absorbed. The clue is in the phrase 'heat addition process'.

"Of course, it follows then that if you were to have two almost identical piston engines (one low/one high compression) burning exactly the same amount of fuel, the exhaust gases from the higher compression engine would be slightly cooler than the low compression engine."

The exhaust gas temperature would vary with the change in PdV work during the power stroke. Not because "A high compression ratio engine will ignite a hotter air/fuel charge which will absorb less heat".

Mr Optimistic

27th Dec 2011, 12:58

http://web.me.unr.edu/me372/Spring2001/Brayton%20Cycle.pdf

twistedenginestarter

28th Dec 2011, 22:37

The problem here is a lot of what is being said is true. There is not one answer to the original question, which in turn is rather artificial.

It is true you need a compressor otherwise you haven't got a jet engine; all you've got is an oil fire. Also higher compression tends to lead to more fuel being consumed. However the point of compressors is to raise pressure. The thermodynamic efficiency of internal combustion engines increases with pressure and temperature. The explanations given above may be true; it doesn't really matter. All you need to know is you want to get the pressure as high as you can without suffering from side effects like spontaneous combustion/explosion.

Note that pressure is not the same as compression ratio. That's why superchargers and turbochargers raise pressure - they help stop brake mean effective pressure falling as rotational speed increases or air thins.

oggers

29th Dec 2011, 12:03

Slippery

The edit to your last post is welcome. Crucially you now agree with what CW has been saying in principle, so well done, we are halfway there.

As for the other change:

This is misleading. You can't increase the speed of the combustion reaction to extract more energy - we've been over this before.

I'm not going to rehash the whole thread to prove that you are taking my comments out of context. I will just paste your reply from the time:

"the ignition timing is simply adjusted to ensure the maximum pressure in the cylinder is occuring at TDC."

Which isn't a good concept to be promoting.

Flight Safety

29th Dec 2011, 12:52

Oggers, I'm reposting your Otto cycle diagram from your interesting post #118.

http://upload.wikimedia.org/wikipedia/commons/c/cf/P-V_otto.png

Work accomplished can be seen in the diagram. Energy added by burning the air/fuel charge and the rapid pressure rise is represented by line 2-3. Residual heat remaining when the exhaust valve opens at point 4, is represented by line 4-1. Notice carefully that line 2-3 is longer than line 4-1. The relationship between heat and pressure in a gas is well known, therefore heat has been converted into mechanical work between points 3 and 4, to yield a shorter pressure line 4-1 than pressure line 2-3.

Efficiency could therefore be expressed as the ratio between the heat added to the process to raise the pressure from point 2 to point 3 (a larger pressure change), and the residual heating remaining in the pressure drop from point 4 to point 1.

Why then does higher compression yield more efficiency? Work is done by the pressure difference between the top and bottom of the piston. Higher compression increases the pressure difference, and therefore converts more heat in the pressurized gas into mechanical work, yielding a residual heat value that is lower, and an efficiency ratio that is higher. Notice too that most of the work is done in the top half of the expansion (or power) stroke, where most of the benefit from higher compression is located and is most useful.

In the turbine and rocket engine, higher pressures yield higher velocities imparted to the reaction mass, and thrust as we know is a product of the velocity and mass. The higher the velocity imparted to the same mass, the higher the thrust. However is a turbine, lower velocities imparted to a larger air mass (the turbofan) is more efficient that adding higher velocities to a lower air mass (pure turbojet). With either turbine type, higher compression (or pressure ratios) yields higher velocities.

Slippery_Pete

31st Dec 2011, 09:53

Oggers...

Crucially you now agree with what CW has been saying in principle, so well done, we are halfway there.

Regarding my agreement with Chris Weston, my posts have ALWAYS related to the thermodynamic efficiency of higher combustion ratios, and in fact not the question which was in the subject line of the OP (about the fundamental purpose of a compressor) until now.

I will happily admit that my explanation in my first post was not ideal, but then it wasn't meant to be. It was, infact, meant to be an easy way for someone with little or no knowledge of basic physics and conservation of energy to grasp the concept of minimising waste heat. It certainly wasn't supposed to be taken so literally, and after you chipped me on it, I clarified what I meant in subsequent posts (temperature change over the entire cycle).

the ignition timing is simply adjusted to ensure the maximum pressure in the cylinder is occuring at TDC

See above. It's obvious why this is not going to work (or will work very poorly), but allows people to conceptualise what is going on and what is trying to be achieved.

The fact remains that EVERY FN TIME you refuse to answer my question about two different compression turbines with the same fuel flow and the difference in their thermodynamic efficiencies. How many times have you avoided this now? I've asked AT LEAST 5 TIMES now how your "better mixing, flame front speeds" BS applies in this situation, but you simply refuse to answer.

YOU JUST DON'T KNOW seems to be the only explanation.

I'm still waiting.

oggers

5th Jan 2012, 13:36

Slippers...

"Regarding my agreement with Chris Weston, my posts have ALWAYS related to the thermodynamic efficiency of higher combustion ratios, and not the question which was in the subject line of the OP until now."

:suspect: In post #63 CW answered that very question: "why compress - to release more energy to do useful work per unit time within the engine (be it piston or turbine)." But in the very next post you quoted him and replied: "It's so that less heat is added to the air over the entire cycle" and then went on about efficiency again. And so on, throughout the thread. It's clear to me you have shifted position, however, I'm not interested in a subjective exchange so I'll move along.

"I will happily admit that my explanation in my first post was not ideal, but then it wasn't meant to be."

I wouldn't say "not ideal". I would say wrong.

"It certainly wasn't supposed to be taken so literally, and after you chipped me on it, I clarified what I meant in subsequent posts (temperature change over the entire cycle)."

[BTW, small point but temp change over the entire cycle is zero in both cases. Has to be if the engine is in a steady state because internal energy is a state variable. I only mention that because you said "argue all you like, but I have a physics degree and the principles of thermodynamics have been unchallenged for a few hundred years" ;)]

I assume this was your clarification from the post after I first "chipped you" on it:

"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)."

Yes, we know the exhaust is cooler. That's a given if we consider increased efficiency in an idealised cycle. The question is why is it cooler? So back to what you wrote originally:

"A high compression ratio engine will ignite a hotter air/fuel charge which will absorb less heat. Less energy wasted as heat = more energy transferred to the crank. When the ignition occurs, a lower compression ratio engine will have a cooler air/fuel charge in the cylinder - and so it will absorb more energy (which is wasted as exhaust gas heat)."

Three questions for you:

1. If that wasn't an attempt to explain why the exhaust ended up cooler, where in that first post is it?

2. Where in any of your posts do you clarify that what you wrote above isn't meant to suggest that 'a hot charge absorbs less heat during the combustion process and vice versa'?

3. If that's not what you were getting at, why write this:

"consider two cups of water - 1x 50 degrees celsius, 1x 100 degrees celsius... if you put them over a flame of 200 degrees for exactly one second, the cooler cup of water will absorb more heat (because the temperature split between the two is larger)."

...and by post #31 you were still reiterating the point:

"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."

?

What I'm getting at is - despite what you may say now - you were definitely writing that by increasing CR, the working fluid would absorb less heat during the heat addition phase.

Finally:

"The fact remains that EVERY FN TIME you refuse to answer my question about two different compression turbines with the same fuel flow and the difference in their thermodynamic efficiencies. How many times have you avoided this now? I've asked AT LEAST 5 TIMES now how your "better mixing, flame front speeds" BS applies in this situation, but you simply refuse to answer."

I'll leave the finer points of combustion in a turbine for someone else. I'll just say this: you chose the piston engine as your original example and no matter how these factors relate to a turbine it doesn't change the fact they are critical in a piston engine, unless you limit your knowledge to an idealised approximation of the cycle. In the real world no such ideal engine exists.

ShyTorque

5th Jan 2012, 15:41

Just speed read this... strewth, talk about handbags at dawn!!

I'm surprised no-one has yet mentioned expansion ratio, rather than compression ratio (at least I don't think it was mentioned). The work output comes from the expansion of the gases, not the compression. A higher CR provides more expansion of the gases after combustion, which is an explanation for the lower EGT of a high CR engine.

The disadvantage of the use of a very high CR is the instability of the fuel/air mix as the peak cylinder temperature increases beyond a critical level, depending on which fuel is being burned. Detonation can occur if this is too high, which results in a "kick" to the piston, rather than a controlled push.

Talking of which, Slippery Pete wrote: the ignition timing is simply adjusted to ensure the maximum pressure in the cylinder is occuring at TDC.

Yes, this is incorrect. It would have been more accurate to say that peak cylinder pressure should occur just after TDC, about 17 degrees ATDC in fact. This is for geometric reasons of the conrod pushing round the crank in the most efficient way.

lomapaseo

5th Jan 2012, 18:55

I'm surprised no-one has yet mentioned expansion ratio, rather than compression ratio (at least I don't think it was mentioned). The work output comes from the expansion of the gases, not the compression

Isn't the expansion limited by the atmosphereric pressure and thus has only small variations?

It would seem that the imput pressure is what varries the most, so in what way does the term of using the expansion ratio differ from the compression ratio in a gas turbine?

We really do need to corral this discusion around gas velocities in a turbine, other wise we wouldn't have a compressor and a means of expansion across a turbine stage or an exhaust jet pipe

Chu Chu

6th Jan 2012, 01:00

I think the point is that the more compression, the higher the pressure in the combustion chamber, and the more the gas expands in the turbine section. To me the real question is why is the energy gained from the extra expansion is greater than the energy used to compress the gas further in the first place.

The only thing I can figure out is that there's effectively less gas in the compressor section (because it hasn't yet been heated in the combustion chamber), and compression therefore take less energy than you get back from expansion.

Imagine climbing a mountain with a 10 pound weight, then letting it go to roll back down. If you had a magic weight that increased to 20 pounds when you let it go, the weight would release more energy rolling down the mountain than you put into it climbing up. And the higher you climbed before you released it, the greater the energy difference would be.

Heating the fuel-air mixture in the combustion chamber is a little like increasing the weight from 10 pounds to 20, except it's due to the heat energy from combustion, not magic. But it still means you get back more energy from the turbine than you put in with the compressor.

At least that's my (doubtless somewhat confused) story.

chris weston

6th Jan 2012, 20:28

Chu Chu,

Under high compression you can burn more fuel in a given amount of time as high compression crams in more oxygen molecules et al and you simply release more energy per unit time from the increased levels of combustion.

Specifically you release enough extra energy available to do useful work than you need to use in the compression step. The relationship between compression and power out is exponential and not linear. I will play hunt the graphs.

turbo graphs - Google Search (http://www.google.co.uk/search?q=turbo+graphs&hl=en&prmd=imvns&tbm=isch&tbo=u&source=univ&sa=X&ei=Z9oKT7XeDZHbsgaf_OWCDw&sqi=2&ved=0CFAQsAQ&biw=1440&bih=785)

Increased expansion comes from now having generated more combustion product molecules CO2/H2O/NOx and friends in a given space or volume (can/cylinder etc.)

Gas turbine and piston engines are both fully open systems thermodynamically, exchanging energy and matter with their surroundings.

CW

Slippery_Pete

7th Jan 2012, 06:46

Oggers,

YOU JUST DON'T KNOW seems to be the only explanation.

Thought so.

rudderrudderrat

7th Jan 2012, 11:48

rudderrudderrat

7th Jan 2012, 12:01

In a piston engine, higher compression ration permits faster combustion (especially useful in high revving engines), thus it is possible to get the maximum mean pressure to do useful work (hence more efficient) .

A gas turbine needs high compression to force the air into the combustion chamber against the combustion pressure.
It is introduced through a much smaller surface area than the exhaust gas exit area.

Slippery_Pete

7th Jan 2012, 12:01

Dear oggers & Slippery_Pete,

You have both been slinging handbags since Nov 11, and have even developed affectionate names for one another e.g. "Slippers.."

Please could you exchange your love letters via PMs?

My wife is starting to become suspicious :ok:, so that's a great idea!

I actually went down that path about a month ago, but Oggers refused to answer my PM :D... for probably the same reason he won't answer my question in the public forum either.

barit1

7th Jan 2012, 12:13

A gas turbine needs high compression to force the air into the combustion chamber against the combustion pressure.

Well, not really. The combustor does not create pressure, unlike a piston engine. In fact there's a small pressure drop as air flows through the burner. The main restriction to flow is the turbine nozzle vanes (guide vanes) just downstream of the burner; this restriction, coupled with the pumping flow rate of the compressor, determines the pressure ratio of the machine.

It's very much like a garden hose; the spigot controls the flow into the hose, but if the nozzle is wide open, there's not much pressure in the hose. But close down the nozzle, and the pressure in the hose increases.

:)

rudderrudderrat

7th Jan 2012, 12:40

Hi Bart1,
The combustor does not create pressure,
Well if doesn't create dynamic pressure - what accelerates the gas?

The static pressure may drop through any venturi - but when you measure EPR surely the exhaust pressure exceeds inlet pressure by the indicated ratio.

Turbine D

7th Jan 2012, 17:45

The combustion system in a turbine engine receives engine airflow that is highly compressed from the compressor, adds heat energy to this airflow and delivers the hot gases to the turbine. The combustor must deliver uniformly mixed hot gases to the turbine just slightly below stoichiometric fuel-air mixture combustion temperatures. At stoichiometric conditions, the maximum amount of heat is released and all the available fuel and oxygen is consumed. Over stoichiometric, excess fuel acts as a heat sink reducing the heat released.

There are generally two regions in a combustor system that are divided into about equal volumes but perform different functions. The upstream region is the primary combustion zone where nearly stoichiometric burning takes place with the correct fraction of air flow. The downstream region is the secondary or dilution zone where the excess air is mixed with the hot combustion products to provide the desired turbine inlet temperature. The average flow velocities in typical combustors range from 60 to 100 Ft/sec. All in all, it is a very complex system.

Combustor efficiency is a measure of the ratio of actual to theoretical heat release and must be as high as possible over the entire operating range of the engine.

The total pressure loss of the combustion system is defined as the difference between the averaged stream total pressure at the compressor exit station and the turbine inlet station. In general, higher pressure losses result in better combustor performance, but of course the engine cycle performance is reduced. A balance must be found between these opposing factors.

TD

barit1

8th Jan 2012, 01:37

Well if doesn't create dynamic pressure - what accelerates the gas?

The temperature rise in the burner expands the gas - thus its velocity increases smartly. But it's still subsonic flow, in an environment where the velocity of M1.0 is greatly increased compared to ambient. The flow becomes choked at the first stage turbine nozzle, and this sonic flow is what drives the turbine rotor.

Crabman

8th Jan 2012, 14:00

My vote is with SP. Assuming that the OP's question could be rephrased as "why is compression desirable" or more accurately "why does engine efficiency increase with compression ratio (or pressure ratio for turbines)". The answer is that compression causes an increase in the thermodynamic efficiency of the engine. Compression causes a temperature increase of the working fluid (gas). This results in the heat charge (combustion) being introduced at a higher temperature. The result is an increase in thermodynamic efficiency.

So the answer to the OPs question is that compression results in higher thermodynamic efficiency. Everything else is fluff...

oggers

8th Jan 2012, 15:15

Crabman: No. You have not answered the question, merely repeated it. We know there is an increase in efficiency - half the question was predicated upon that very observation.

So the answer to the OPs question is that compression results in higher thermodynamic efficiency. Everything else is fluff...

The question is why? If you read through the thread properly you will find it has been answered along with the other question posed in the OP. However, the answer slippery gave on page 1 is incorrect.

chris weston

8th Jan 2012, 17:30

Crabman 137

Crabman;

I think Oggers and Turbine D et al are right here, I suggest that Slippery P is wrong.

It's true that compression will elevate the temperature of gas phase molecules and that that takes us in the right direction, but it's a fairly trivial contribution relative to the enhanced combustion processes so nicely described by Turbine D in post 135.

Let's stay with the notion of temperature change in gas turbines.

Do the Maths; remember these are fully open systems as thermodynamically they exchange matter and energy with the surroundings.

ΔT1 and the energy from the compression of the gases (courtesy of Van der Waals + Dipole-Dipole et al bond formation) will be, at very best, 100s of degrees C or kJ if you want to calculate the energy released and (unless we're in ram jet territory which I guess we're not) is far less than we require for the energy needed for the compression process.

ΔT2 and the energy released to the surroundings from the enhanced ΔHc (combustion processes see post 128) will be orders of magnitude greater than that needed for the compression and therefore available to do useful work - such as provide thrust.

CW

Crabman

8th Jan 2012, 18:17

oggers: I think that I did answer where the increase in efficiency comes from (I certainly was trying to):

"Compression causes a temperature increase of the working fluid (gas).This results in the heat charge (combustion) being introduced at a higher temperature. The result is an increase in thermodynamic efficiency."

CW: I'm not saying that the delta T (not good at typing Greek on my ancient keyboard) of compression is somehow solely the cause of the efficiency increase. Without combustion, it would all be for naught. What I am saying is that delta T of compression results in the heat of combustion (delta Q - from which the work is derived) occurring at a higher temperature. The maths are fairly straightforward and have everything to do with the pressure ratio of the turbine (and its effect on the delta T of compression and the resulting delta T of exhaust, after combustion).

Maybe I'm not expressing it clearly (and perhaps we are even in violent agreement). I will say it simply one more time. Thermodynamic efficiency is a function of Compression Ratio (or a slightly different different function of Pressure Ratio in a turbine). Why? Because of its effect on the temperature changes of the working fluid. What is this effect? The heat of combustion is introduced into the system at a higher temperature.

Crabman

9th Jan 2012, 12:21

CW: Your reference to Mr. Van der Waals and the compression of gases left me somewhat puzzled (No, don't bother explaining it - I'm not a chemist - I only work with ideal gases), but started me thinking ... Wouldn't it be interesting if, sometime, we could all have a discussion about something everyone surely agrees upon, such as what causes lift. (and drop names like Newton and Bernoull - and maybe even Navier and Stokes).

Or, perhaps, planes on conveyor belts...

chris weston

9th Jan 2012, 21:22

Crabman,

Thank you, nice idea.

But .....don't forget the importance of the little guys in chemical bonding. The 787 is in trouble if we do.

CW

Slippery_Pete

12th Jan 2012, 01:08

Hi Crabman.

You are on the right track.

What you have to understand here, is that the concepts which others have discussed (flame front speeds, timing and better burning) relate to design parameters of a particular engine using the Otto cycle - not the theoretical thermodynamics of the engine.

Oggers and CW say that faster “flame front speeds” make engine X more efficient at higher compression. I say you can make engine X just as efficient by operating it at low RPM. In fact, this is what the world's most efficient piston engines do - they operate at extremely low RPM.

Oggers and CW say that “better burning” make engine Y more efficient. If this were true, consider what would happen if you increased the compression ratio of an engine at constant fuel flow. Oggers and CW say that you will more completely combust much more of the fuel. This should mean that the exhaust gas temperature should increase. We saw from the linked document in post 41 that in fact, completely the opposite occurs – EGT reduces. Ask them to explain why “better burning” (which obviously they are implying means less unburnt fuel being wasted) causes the EGT to reduce.

You can easisly take their piston engine arguments out of the equation by running a large diesel engine at extremely low RPM (where flame front speeds mean jack sh*t and where the fuel is essentially completely combusted). But why does compression ratio still affect the thermodynamic efficiency?

I can tell you why - because you are adding less energy to the fluid.

As for the practicalities in a car or aircraft of a large, extremely low RPM diesel - I obviously understand this and a multitude of other reasons why for practical purposes there other considerations in making piston engines adaptable and useable over a large range for a particular application. But these are inconsequential to the FUNDAMENTAL concept of the OP.

Not only will Oggers and CWs arguments not apply to a large, efficiently designed and operated piston engine, they also fail to apply to a turbine. This is why they have both failed to answer my question about this so many times – because their reasoning simply can't explain it.

rudderrudderrat

12th Jan 2012, 07:52

Hi Slippery Pete,
Ask them to explain why “better burning” (which obviously they are implying means less unburnt fuel being wasted) causes the EGT to reduce.Er... I'd say that if the fuel was burned completely by the correct moment during the cycle, then more useful work can be extracted from the energy supplied, therefore you would need less fuel compared to the inefficient cycle, therefore the EGT would reduce.

Your engine X may be very efficient at low RPM, because the combustion is completed by the time the piston is at position Z. However it is not producing as much power as a faster revving engine.

As the revs increase, we need faster burning (hence higher compression ratio) so that combustion is complete by the time the piston is at position Z.

chris weston

15th Jan 2012, 18:38

Slippery Pete,

At no stage have I referred to flame front speed or indeed to "better" burning in this thread, I have never referred solely to piston engines.

I have carefully not considered the kinetics of the hydrocarbon combustion processes as they follow and do not lead in this discussion on "why do we compress?" We can go into the kinetics if you wish presumably to bring some order into this discussion if you wish, the pun is intended.

I do however agree that rate (or flame front speed) will show a form of direct proportionality to gas pressure during combustion. It follows from simple rearrangement of the ideal Gas Equation pV = nRT. Pressure can be equated as concentration on rearrangement etc.

What I have stressed is that compression allows more combustion per unit time thereby releasing more energy to do useful work per unit time.

We compress primarily to make the engine more powerful. You can be as efficient as you like in your isolated system engine but if you lack the power to reach Vr what's the point! I believe Oggers said something similar some time ago albeit in a different way.

Your efficiency argument is predicated on thermodynamically closed systems.

Remember that piston or gas turbine engines are more or less fully open systems thermodynamically.

The heating effect due to increased combustion per unit time is orders of magnitude greater than that relating to heating from compression alone.

Beyond a defined threshold, the need to add less combustion energy to an already compressed fluid is relatively trivial in an open system.

Hope this helps!

CW

thermostat

14th Feb 2012, 15:31

This is a little off topic but valid.. 3 'planes take off to the north but the route is to the south. (A) takes off, leaves the flaps down, turns south then cleans up and accelerates. (B) takes off, cleans up the flap, remains at 0 flap speed, turns south then accelerates. (C) cleans up, accelerates to 250 Kts IAS, then turns south. Which has the best performance and burns the least fuel?
It's "C". Why? At 250 Kts IAS there is more ram air and therefore more power. "C" was further south and higher on less fuel than "A" & "B". More compression (ram) means more power. This was done in the late '70s.

barit1

14th Feb 2012, 21:58

It's "C". Why? At 250 Kts IAS there is more ram air and therefore more power. "C" was further south and higher on less fuel than "A" & "B". More compression (ram) means more power.

No doubt you climb faster, but since TAS & GS are greater, your rate of turn (assuming same bank angle) is less, time & distance traveled is greater, and total trip distance is greater.

So the "more power" aspect is fighting the physical facts of more time and distance. And you don't get "more power" except by burning more fuel. Perhaps "C" is correct, but not by much.

(Although - I certainly can sympathize with a Canadian hurrying to fly south...) :p

jh5speed

20th Feb 2012, 18:12

I read all the above and reckon there’s room for another angle on this so I’m going to resurrect this thread and wade in here – this is one of those questions which keeps me awake at night – something that’s generally accepted (e.g. because a computer model or a graph tells you) but there are rarely any good/satisfying answers to be found …

The question was put in terms of gas turbines and also reciprocating engines. I’ll answer in terms of the former because that’s what I know about. I’ll assume that the same principle can be read across to reciprocating engines. That exercise is left to the student...

So why do we compress? I distinctly remember answering this with ‘because it packs more wallop’ at a university interview nearly 30 years ago - not a bad answer really, I got a degree in the end….

Let’s put a few things to bed:

1. It is nothing to do with rate of combustion.

I can show (below) that you can get more thrust (and/or better thermal efficiency) from a turbojet with a 14:1 OPR (overall pressure ratio) than from a turbojet with a 12:1 OPR both designed for the same inlet airflow and operating at the the same fuel flow.

2. It is nothing to do with keeping the air going in the right direction.

Mr Whittle figured he’d need a compressor for cycle efficiency reasons and was probably jolly glad that this also kept things going aftwards – so that was something he didn’t have to worry about. I can also show that an engine with hardly any OPR (1.1:1) will develop some thrust but will have a shocking thermal efficiency. You don’t need much pressure ratio to keep things going rearwards – a slightly inclined tube with a fire in it does the job pretty well - think of the Mt Blanc tunnel fire of 1999…

Essentially, the answer is that for the higher OPR engine, the nozzle pressure ratio (NPR) is higher so you get more pressure thrust. Momentum thrust is the same (in our example).

For a convergent nozzle: Gross Thrust = W8.v8 + (Ps8-Pambient).A8

... where 8 is the nozzle throat plane.
W is flow
Ps is static pressure
v is velocity
A is area

Momentum thrust (W8v8) is the same because W8 is the same (we fixed that at the start in the way we set up the comparison for both engines). Exhaust jet velocity v8 is the same because T8 is the same and the nozzle is choked (Mach 1).

But why is T8 the same for the second engine, and indeed, and why is NPR higher i.e. why is nozzle entry pressure (P7) higher?

Furthermore why, for the higher compressor PR, doesn’t the turbine operate at a similarly higher pressure ratio to make the NPR the same as before?

Burning questions indeed ....

Let’s look at two design points as described above. I ran these on a proprietary Gas Turbine simulation program (GasTurb – you can download a free reduced functionality version from the website). This saved me doing the calcs myself. The first engine had 12:1 OPR and the second 14:1 at the same airflow and fuel flow (for this we need a smaller turbine throat area and final nozzle area – these areas fall out of the thermo calcs).

Sure enough – we get a higher NPR in the 14:1 case at much the the same T8 because the extra T we’ve added at the compressor is taken off by the turbine, with about the same combustor heat addition in each case. Thrust is higher (due to increased NPR) and so is thermal efficiency (obviously, because we are getting more thrust at the same fuel flow).

Furthermore, Gasturb allows you to throttle back the second engine (i.e. run slightly slower at fixed geometry using nominal component maps) to get the same thrust as the first engine. Lo and behold, the thermal effy was slightly higher in the second engine case (throttled back – so no longer operating at 14:1).

The 1.1 OPR example gave quite a bit of thrust by virtue of the momentum term (v was significantly increased due to the temp rise) but the cycle was at ~1% thermal effy!
Compare this with Engine 1 at 40% and Engine 2 at 42%.

I haven’t answered those questions yet though - all the GasTurb run showed me is what we know happens, though it does illustrate very clearly that it is not a rate-of-combustion thing...

Essentially the answer is in the relationship:

Pressure Ratio = Temp Ratio ^ (gamma/gamma-1)

Because the turbine has to develop a deltaT (starting at a higher T due to the combustion process) to drive the compressor, the turbine temp ratio is smaller than the compressor temp ratio and consequently the pressure ratio is smaller. For an increase in OPR, the turbine therefore satisfies the extra power requirement with some pressure ‘to spare’ thus NPR increases.

You can do these calcs by hand, just using the equation above – making them very simple by assuming 100% component efficiency and assuming an identical combustor temp rise for each engine (not far from the truth in these two examples – remember we set the airflow and fuel flow to be the same). You can also assume constant gas properties (gamma and Cp) – realities such as varying gamma and Cp just make the sums trickier but don’t change the basic principle.

I realise that I haven’t really explained things all that satisfactorily as my ‘explanation’ still relies on an unchallenged thermodynamic equation (I’m happy to accept it!). But it does illustrate where the benefit comes from, and possibly more clearly than ‘explaining’ things in terms of Temperature-Entropy charts where the pressure lines ‘conveniently’ diverge! (for the same reason of course).

So, a slightly different, and perhaps long-winded - angle on the subject to what’s gone before. And hopefully useful !

lomapaseo

20th Feb 2012, 19:45

2. It is nothing to do with keeping the air going in the right direction

Really :confused:

But the subject did mention Gas Turbine

From a standing start how do you expect a rotational compressor to turn if the gas doesn't move in the direction of a turbine somewhere downstream of the combustor or are we getting into a related subject like a ram-jet ?

Lyman

20th Feb 2012, 22:31

I like Chris Weston's comment about compression. Isn't it the reason computers, geting smaller, also get faster? The Physics is similar? Faster is better. And faster implies proximity, yes?

jh5speed

21st Feb 2012, 07:03

Starting is an entirely separate issue. Claarly, the stable flow/burning process needs to be initiated by persuading everything to flow in the right dierction. However what this thread is all about (I believe - correct me if I'm wrong), is why an increase in compression ratio results in an increase in thermal efficiency.

lomapaseo

21st Feb 2012, 20:42

However what this thread is all about (I believe - correct me if I'm wrong), is why an increase in compression ratio results in an increase in thermal efficiency.

The thread is whatever the responders make it

This is so often true in the technical section where the thread starter drops a one-liner subject header and disappears into the nether leaving us to expound all sorts of intelligence that falls on deaf ears.

but it's fun anyway :)

Lyman

22nd Feb 2012, 00:41

Keeping the gas going in the "right direction" is everything to do with propulsion.

The starting cycle is a simple way of understanding the need for compression. Without compression, there is no work. How can compression become irrelevant to the process at any time?

Starting is compression, it contains gases in a dynamic flow by directing its passage from a large to small annulus. Without the pipe, there is no power, and without the fan, there is only multidirectional expansion.

Water injection, Open Iris Afterburner, Hypergolic starting, Start carts, Compression is our friend. The difference between early centrifugal compressors and axial flow instructs as to the OP post. PUSH, PACK, POP, off you go.

jh5speed

22nd Feb 2012, 07:14

Lomapaseo – quite right, we do (and should) pick up stuff around the edge of the discussion, however I was addressing what I considered to be the main question. And yes – it is fun.

Lyman, you said “Keeping the gas going in the "right direction" is everything to do with propulsion”.

I’m not disagreeing with you – I was suggesting that the reason the compressor is there is for fundamental cycle effy reasons and not primarily to set the flow direction (it’s perhaps more of a philosophical point I am making here).

You also said : The starting cycle is a simple way of understanding the need for compression. Without compression, there is no work. How can compression become irrelevant to the process at any time?

I’m certainly not saying compression is irrelevant to the process. I’ve established that it is fundamental to getting work out of a machine at a good thermal efficiency. Work is required in the start phase just as in other operational phases (steady-state, accels and decels alike)

But starting is a means to an end, it is all about getting the engine to a stable, self sustaining operation to deliver the required thrust. Starting needs compression, sure, and it needs external cranking power (or windmilling). And yes, compression sets the flow direction, though you have to be especially careful to control the fuel flow (and variable geometry if you have any) otherwise the compressor might get upset (stall). Same comment applies for ‘normal’ (engine-started) operation where reverse flow may occur under some circumstances.

‘Compression is our friend’ – certainly, it keeps me in a job. Compressors however are a nuisance!

rudderrudderrat

22nd Feb 2012, 11:16

Hi jh5speed,

Our local National Trust house has a very old gas turbine installed in the kitchen.

It consists of an open coal fire (at ambient atmospheric pressure) and a very long vertical jet pipe called a chimney. A turbine (set of fan blades) is placed above the fire and its rotation is converted through a suitable system of bell cranks and levers to turn the spit and roast the meat.

It doesn't seem to be very efficient - would a compressor help?

jh5speed

22nd Feb 2012, 11:50

Without a doubt, though probably the extra cost of control-system improvements would outweigh the efficiency gain (however calculated - I guess it would need a labour cost element as well as fuel cost to account for those poor servants who, at present, have to waste valuable time checking that the meat is browning evenly, and fetching in the extra odd bucket of coal). But then, if they weren't doing that, what would they be getting up to?

Lyman will advise you on getting the fire lit without smoking the place out.

Once lit, I reckon if smoke isn't coming back into the room and the meat is turning, you'll probably be better off leaving well alone.

Oh - and it might be a good idea to send a child up the chimney with a tub of goose fat to grease all those bellcranks and pulleys (thus increasing mechanical efficiency) ... While he (or she) is up there a decoke of the turbine blades might be a good idea.

rudderrudderrat

22nd Feb 2012, 12:45

Hi jh5speed,

Thanks for the tips and advice which I've passed on to the N.T. kitchen turbine department.
Apparently they have tried using a compressor (mechanical bellows of soft leather nailed between flat wooden plates) which increases the kinetics and raised the combustion temperature. The more energetic exhaust flue gas turned the turbine more quickly.

It seems to be a win win situation, but the bellows operator is demanding extra pay because of the increase in thermal efficiency. They are now working on a mechanically driven set of bellows from the extra power from their turbine.
They want to know if it is worth adding a second turbine, downstream of the first to work the bellows?

They are grateful for the advise on the turbine wash and suitable gearbox lubrication.

Lyman

22nd Feb 2012, 14:39

jh5speed, rudderrat

Hmmm. Save the coal for your children at Christmas, seal the bellows at the top of the flue and have your valet pump the bellows into the chimney in reverse flow. Change the pitch on your fan, and use the gear to drive your abacus.

Do your part for the Planet :ok:

Basil

22nd Feb 2012, 15:54

Apologies if anyone has already mentioned the real reason . . .

...............................http://upload.wikimedia.org/wikipedia/en/thumb/e/e2/AleksandrOrlovmeerkat.jpg/200px-AleksandrOrlovmeerkat.jpg

. . which is, of course, to absorb the power developed by the turbine and prevent it from overspeeding - simples!

peter kent

28th Feb 2012, 16:18

I think this is a new angle and hope it is not in the mind-numbing category (not derogatory but complimentary in this context) of most of the previous posts. It is not in answer to any outstanding question, be it either still to be answered or of great merit.

As already stated the RJ needs (and has) a compressor just as does a jet engine. I suspect the reason for the initial observation that it doesn't have one is the belief that they are fundamentally different machines. The RJ and TF are just different breeds of the same with very similar cycle requirements but obtained from very different looking parts. I don't think there's any great stumbling block to be made of the fact that a conventional RJ doesn't work at zero speed but the TF can.

At zero airspeed the TF does no more than an RJ. It can sit on a test bed running at TO from Monday to Sunday (with slave oil supply) and it has actually done less than nothing. It has moved nothing, except the thrust cradle a few thou, but has cost a weeks worth of fuel/oil/cell occupancy/creep life/etc. Once it starts down the runway though, it has taken the first step to being a ramjet and the turbom/c compressor has taken the first step to being redundant. eg you can't use the SLS 43:1 PR of a big fan in an F-22.

The B777 at cruise has the same subsonic piece of compressor hardware as the pre-SCRJ, a piece of ducting. the only difference is that the rest of the compressor is downstream on the TF but upstream on the RJ.

I find it satisfying to look for underlying similarities rather than thinking there are fundamental differences.

To get the job done both the RJ compressor, with its attendant shock compression, and the TFC, with its supersonic regimes, treat the air with the utmost violence on the one hand, and then gently on the other, with the touchy subsonic diffusion in the duct or rotating blade rows and fixed stator passages.

The degree of brutality which the RJC metes out has always been foisted upon it by the missile cruise requirements. The turbom/c, on the other hand, has increased in brutality from the gentle subsonic compressor of a J79, for example, where the inevitable low stage PR required 17 stages to get about 13:1. You have to thrash a lot more energy into the air if you want a compact HPC where 10 stages give about 20:1. The road to this level began with turbocompressors entering the RJ compressor regime by using supersonic blade relative MN.

Therein lies a similarity. It's high relative MN between air and pieces of metal that give you the makings of a compressor. You don't necessarily need relative motion between the metal of the compressor and its 'mother'.

But the conventional RJ still can't get off the ground!
Bear in mind that the TF only exists to cruise just like the RJ. It needs to be a 'different' machine to get there, in as much as, in its money making regime, it has different ratings, ECS bleeds, turbine clearance bleeds, etc. compared to TO and CL. A B777 won't economically get to 35000ft in its cruise 'config'. The TF is a hybrid.

So what if an RJ also needs a bit of hybridization to get to cruise.