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Thermal Efficiency Of Engines  
User currently offlineJetmech From Australia, joined Mar 2006, 2635 posts, RR: 53
Posted (4 years 6 months 1 week 4 hours ago) and read 7769 times:

G’day Techies,

Recently, I have been pondering about the thermal efficiency of engines, more specifically, the effect of increasing the compression ratio and heat difference across the engine. Equations and formulas will tell you that increasing the compression ratio and difference between the temperatures of heat addition and heat rejection increases the thermal efficiency of an engine, but why? What is the intuitive, non mathematical reason for this?

I think I can intuitively figure out the effect of increasing the heat difference across the engine. A given working fluid at a “high” temperature has very energetic particle motion. The collision of these particles with any sort of boundary creates an elevated pressure. In other words, a “hot” working fluid wants to expand. The hotter it is, the more pressure it subjects a boundary to, and the more it wants to expand.

However, this will only occur if it is allowed to flow to a region of “lower” pressure. Such a region of “lower” pressure exists when the working fluid is at a “lower” or ambient temperature. Thus, having a higher temperature difference across the engine causes the “hot” working fluid to expand faster. When confined to the casing of a jet engine, this results in exhaust gases passing through the turbine section with a higher momentum, and thus provides an increased opportunity for the turbines to extract work.

The effect of increasing compression ratio however, is harder for me to fathom. My initial thought was that increasing the compression on the working fluid through the compressor directly allows for greater expansion through the turbine, but this fails to make sense on closer inspection as it is tantamount to a proposition for a perpetual energy device.

My second thoughts on the issue was that compressing the working fluid to a higher degree allows one to burn more fuel which increases the temperature of the working fluid providing an increase in thermal efficiency due to greater expansion. Again, this seems to run into a wall.

The mass flow rate of working fluid through the core engine is constant at all cross sections. Thus, the rate at which one can burn fuel would also be the same at any cross section. So why does increasing compression ratio make such a difference? Compressing the working fluid increases the density only, not the mass flow rate. So what is it about a denser working fluid that increases thermal efficiency? Does fuel combusted in a denser charge of air burn hotter?

Regards, JetMech


JetMech split the back of his pants. He can feel the wind in his hair.
34 replies: All unread, showing first 25:
 
User currently offlineReheat From Germany, joined Jan 2009, 4 posts, RR: 0
Reply 1, posted (4 years 6 months 6 days 15 hours ago) and read 7687 times:

I'd say you are on the right track with your second idea.

Thermal efficiency is usually defined as net work extracted from the fluid divided by the added heat (in the case of a gas turbine engine, by combusting the fuel/air mixture). The maximum temperature after the combustion chamber (turbine inlet temperature) is limited by the turbine materials in a real engine. The temperature rise between the outside air at the engine inlet and the turbine inlet is split into two steps: heating by compression and by combustion. If this balance is shifted towards the compression, i.e. a greater part of the temperature rise happens in the compressor, less heat (read: fuel) needs to be introduced in the combustor to attain the same turbine inlet temperature. This increases thermal efficiency, while at the same time lowering the amount of work that can be extracted from the engine as the higher compression ratio requires more power.

From a thermodynamics standpoint, the ideal Joule/Brayton cycle with the highest thermal efficiency has a close to infinite compression ratio and produces almost no useful work. This is obviously not workable in a real cycle, but I hope the explanation helps.


User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 2, posted (4 years 6 months 6 days 8 hours ago) and read 7649 times:



Quoting Jetmech (Thread starter):
Equations and formulas will tell you that increasing the compression ratio and difference between the temperatures of heat addition and heat rejection increases the thermal efficiency of an engine, but why? What is the intuitive, non mathematical reason for this?

There's a couple of ways of doing it. If you're comfortable with entropy, the best explanation works from:

Q = TdS

In English, the amount of heat transferred equals temperature times entropy change. As you correctly note, the mass flow-rate is constant and (ignoring material temperature limits for the moment) that means your max fuel flow rate is also constant, so your energy input (Q) is constant. If Q is constant, and T goes up, then dS goes down.

For a given amount of energy, the entropy change is smaller if you do it at high temperature. Thus, adding your heat at a very high temperature causes the cycle to "lose" less entropy, increasing efficiency (the limiting Carnot cycle is zero entropy loss).

The other way of looking at it is that, although the way the math comes out it's easiest to express the efficiency in terms of compression ratio, that's really a side effect of the math and not so much physical...it's the temperature difference that really matters. If you assume perfect compressors and turbines, and ideal gas in the engine, then the temperature rise in the compressor must exactly equal the temperature drop in that portion of the turbine that's powering the compressor. In other words, if your compressor increases the air temp by 200 C, the air temp has to drop by 200 C in the turbine in order to provide the energy to power the compressor. Whatever's left is available to be used for something useful, like pushing the aircraft or spinning the fan. So, the higher your turbine inlet temperature, the more you have "left" after extracting the power needed to run the compressor.

Quoting Jetmech (Thread starter):
The mass flow rate of working fluid through the core engine is constant at all cross sections. Thus, the rate at which one can burn fuel would also be the same at any cross section. So why does increasing compression ratio make such a difference? Compressing the working fluid increases the density only, not the mass flow rate. So what is it about a denser working fluid that increases thermal efficiency?

It's not that the working fluid is denser, it's that it's hotter. More compression equals higher temperature entering the combustor, equals higher temperature entering the turbine (since the temperature rise from fuel is constant for a particular mass flow and fuel/air ratio). More temperature going into the turbine means more energetic fluid available to do useful work after you've powered the compressor.

This is why one of the fundamental performance parameters for a jet is tau_lamda...the ratio of maximum allowable turbine inlet temperature to the ambient air temperature.

Tom.


User currently offlineJetmech From Australia, joined Mar 2006, 2635 posts, RR: 53
Reply 3, posted (4 years 6 months 4 days 22 hours ago) and read 7522 times:

Quoting Reheat (Reply 1):
The maximum temperature after the combustion chamber (turbine inlet temperature) is limited by the turbine materials in a real engine.



Quoting Tdscanuck (Reply 2):
This is why one of the fundamental performance parameters for a jet is tau_lamda...the ratio of maximum allowable turbine inlet temperature to the ambient air temperature.

I suppose this is one of the benefits of flying at high altitude. You can reject heat to a much "colder" fluid reservoir compared with sea level conditions.

Quoting Reheat (Reply 1):
heating by compression and by combustion. If this balance is shifted towards the compression



Quoting Tdscanuck (Reply 2):
More compression equals higher temperature entering the combustor, equals higher temperature entering the turbine

Fair enough, but surely the temperature increase through the compressor is a side effect of less that 100% isentropic compressor efficiency, as opposed to something that the engineers deliberately design for? They would want 100% isentropic efficiency, but I believe that compressors hover around the 90% isentropic efficiency level. Thus, only 10% of the shaft work absorbed by the compressor turns into heat, which I suspect is a non desirable entity. What is it about the 90% of the shaft work that provides a denser working fluid that benefits thermal efficiency?

Quoting Tdscanuck (Reply 2):
Thus, adding your heat at a very high temperature causes the cycle to "lose" less entropy, increasing efficiency

I can see how that works from the Q = TdS equation. I suppose what the equation means intuitively, is that adding energy to an already energetic working fluid results in a relatively modest "calming down" (entropy loss) of the newly added energy. Adding energy to a cooler, or less energetic fluid reservoir means that much of newly added energy is "calmed down" significantly (large reduction in disorder of newly added energy).

Quoting Tdscanuck (Reply 2):
it's the temperature difference that really matters


Quoting Tdscanuck (Reply 2):
It's not that the working fluid is denser, it's that it's hotter. More compression equals higher temperature entering the combustor, equals higher temperature entering the turbine (since the temperature rise from fuel is constant for a particular mass flow and fuel/air ratio). More temperature going into the turbine means more energetic fluid available to do useful work after you've powered the compressor.

I did suspect that some associated side effect of the compression ratio benefitted thermal efficiency, as opposed to the compression ratio itself. As you point out, it is the heat difference that is of primary importance. If so, why not design a compressor with 0% isentropic efficiency, and convert all the compressor shaft work to heat? Why do current engines only convert a small amount of the compressor shaft work to heat if it is heat that benefits thermal efficiency?

Quoting Tdscanuck (Reply 2):
The other way of looking at it is that, although the way the math comes out it's easiest to express the efficiency in terms of compression ratio, that's really a side effect of the math and not so much physical...

So basically, the compression process in an engine is nothing more than a method to mechanically add heat to the working fluid?

Regards, JetMech

[Edited 2009-10-17 07:50:43 by JetMech]


JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlineJetMech From Australia, joined Mar 2006, 2635 posts, RR: 53
Reply 4, posted (4 years 6 months 4 days 22 hours ago) and read 7506 times:



Quoting Tdscanuck (Reply 2):
Q = TdS



Quoting Reheat (Reply 1):
The temperature rise between the outside air at the engine inlet and the turbine inlet is split into two steps: heating by compression and by combustion. If this balance is shifted towards the compression, i.e. a greater part of the temperature rise happens in the compressor, less heat (read: fuel) needs to be introduced in the combustor to attain the same turbine inlet temperature.

Damn it, I think I get it now! The compressor is indeed there to mechanically add heat to the working fluid. The isentropic efficiency of the compressor is therefore an indicator of what process is used to add heat, as opposed to indicating how much heat is added.

A 100% isentropic compression process means that one brings all the working fluid particles closer together without exciting or increasing the "agitation" of any of the particles. The engineers aim for a 100% isentropically efficient compressor as from the TdS equation, the temperature rise approaches infinity as the entropy change approaches zero for any given quantity of heat addition. Thus, a 100% isentropically efficient compressor provides the most amount of temperature rise for a given amount of shaft work.

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 5, posted (4 years 6 months 4 days 17 hours ago) and read 7464 times:



Quoting Jetmech (Reply 3):

Fair enough, but surely the temperature increase through the compressor is a side effect of less that 100% isentropic compressor efficiency, as opposed to something that the engineers deliberately design for?

An isentropic compressor still has a temperature rise, it just has the lowest rise you can get for an adiabatic process and a given pressure rise. The efficiency is a measure of how close you get to the ideal.

Quoting JetMech (Reply 4):
The compressor is indeed there to mechanically add heat to the working fluid. The isentropic efficiency of the compressor is therefore an indicator of what process is used to add heat, as opposed to indicating how much heat is added.

The major purpose of the compressor is to increase pressure...temperature rise is a side effect. I wasn't very good in my prior post about distinguishing about heat added by the compressor and heat added by the fuel. Heat added by the compressor isn't "free" because you had to pull that energy out of the turbine. You want to get as much pressure rise as you, with the lowest temperature rise, in the compressor. The compressor effiicency is a measure of how well you raise the pressure compared to the temperature. A good compressor approaches the isentropic limit...a bad one will increase temperature a bunch without increasing pressure much.

Quoting Jetmech (Reply 3):
What is it about the 90% of the shaft work that provides a denser working fluid that benefits thermal efficiency?

Total temperature rise is limited by what the turbine inlet can withstand. The amount of power you can get out of the turbine is a function of the temperature and pressure at the inlet. So you want the maximum pressure and temperature at the turbine inlet. All your pressure rise comes from the compressor; some of your temperature rise comes from the compressor and some from the fuel addition. Since the compressor is "robbing" energy from the turbine, you want to minimize the amount of energy used in the compressor and maximize the amount of fuel you can add. That means you want the biggest pressure increase you can in the compressor for the lowest temperature increase...that gives you the most temperature margin to burn fuel with.

Quoting Jetmech (Reply 3):
If so, why not design a compressor with 0% isentropic efficiency, and convert all the compressor shaft work to heat?

Because then you'd have no pressure rise. The turbine is driven by pressure differential, not temperature differential. Temperature differential is what allows you to have excess energy in the turbine, after you've driven the compressor, to do useful work with.

Quoting Jetmech (Reply 3):
So basically, the compression process in an engine is nothing more than a method to mechanically add heat to the working fluid?

No. I think I've accidentally lead you astray. The compressor is there to raise the pressure. The turbine needs pressure to operate. The heat addition in the combustor is there to increase the fluid energy so that, after the turbine drives the compressor, you've still got energy left over to do something with. The ideal case is that the compressor provides a very large pressure rise for a very small temperature rise...then you can add an enourmous amount of heat in the combustor and get a lot of useful work out of your turbine.

If you have a really lossy compressor, the temperature entering the combustor is already high and you can't add much fuel before you hit your temperature limit...that means that there's not much left after the turbine drives the compressor, and the useful work extraction from the overall engine is low.

Quoting JetMech (Reply 4):

A 100% isentropic compression process means that one brings all the working fluid particles closer together without exciting or increasing the "agitation" of any of the particles.

That would be an isothermal compression...which is actually better (it's what intercoolers are for), but very very difficult to implement in a jet engine.

Quoting JetMech (Reply 4):
The engineers aim for a 100% isentropically efficient compressor as from the TdS equation, the temperature rise approaches infinity as the entropy change approaches zero for any given quantity of heat addition. Thus, a 100% isentropically efficient compressor provides the most amount of temperature rise for a given amount of shaft work.

You need to differentiate between work and heat...the compressor takes in work, not heat, and turns it into both pressure and temperature rise in the working fluid. A 100% isentropically efficient compressor provides the lowest temperature rise for a particular adiabatic compression. Or, flipped around, it gives you the maximum pressure for a particular temperature rise.

That gets you to the pressure you want, where you can now add heat (fuel) up until you hit the turbine inlet temperature limit.

Did I just makes things better or worse?

Tom.


User currently offlineReheat From Germany, joined Jan 2009, 4 posts, RR: 0
Reply 6, posted (4 years 6 months 4 days 16 hours ago) and read 7457 times:

My initial explanation may have been a tad misleading, as I brought both real engines and ideal cycles into play. To rephrase:
Basically, there are two ways to look at this. Either you constrain turbine inlet temperature and compress higher in order to require less heat in the combustion chamber. Or you keep the fuel flow/heat addition constant while compressing higher, which results in a higher turbine inlet temperature and, consequently, more work extracted by the expansion. The former results in less heat input required while the latter yields more work, both act to increase thermal efficiency.

Generally, though, Tdscanuck did a better job explaining this matter  Smile


User currently offlineDocLightning From United States of America, joined Nov 2005, 18696 posts, RR: 58
Reply 7, posted (4 years 6 months 4 days 14 hours ago) and read 7440 times:



Quoting Reheat (Reply 1):
This increases thermal efficiency, while at the same time lowering the amount of work that can be extracted from the engine as the higher compression ratio requires more power.

This is exactly where I am losing you. OK, so you've decreased the amount of actual heat you need to add to the gas during combustion, but you aren't getting energy for nothing. In fact, heating by the compressor should be less efficient than heating by combustion because the energy to heat the gas by compression comes from the combusted gas turning the turbine.

So wouldn't it be better to figure out a way to compress the gas as much as possible while heating it as little as possible?


User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 8, posted (4 years 6 months 4 days 13 hours ago) and read 7436 times:



Quoting DocLightning (Reply 7):
This is exactly where I am losing you. OK, so you've decreased the amount of actual heat you need to add to the gas during combustion, but you aren't getting energy for nothing. In fact, heating by the compressor should be less efficient than heating by combustion because the energy to heat the gas by compression comes from the combusted gas turning the turbine.

It's the difference between efficiency and power output. As the compression ratio goes up, the compressor outlet temperature goes up. That means you can inject less fuel before you hit your temperature limit. Energy in to the engine goes down (less fuel flow). However, since the pressure ratio across the turbine is now larger, you get more power out of the turbine. Subtract the compressor power from the turbine power and you've got your useful energy output. The ratio of that to the energy input by the fuel is the overall efficiency.

Going to very high compression ratio increases efficiency but, for a fixed turbine inlet temperature, drops absolute power output.

Quoting DocLightning (Reply 7):

So wouldn't it be better to figure out a way to compress the gas as much as possible while heating it as little as possible?

Yes. That's always the goal. That's why, when you can, you do quasi-isothermal compression (zero heating, arbitrarily large compression). This is usually implemented as staged nearly isentropic compressors with intercoolers.

In a jet engine, the intercooler part is extremely difficult, so you go with adiabatic compression and make it as lossless (nearly isentropic) as you can. If you want a particular pressure ratio, and you take as a given that your heat in/out of the compressor is negligible (which it is in jet engines), then isentropic compression is the limiting state...minimum temperature rise for a particular pressure rise.

Tom.


User currently offlineDocLightning From United States of America, joined Nov 2005, 18696 posts, RR: 58
Reply 9, posted (4 years 6 months 4 days 13 hours ago) and read 7433 times:



Quoting Tdscanuck (Reply 8):

Going to very high compression ratio increases efficiency but, for a fixed turbine inlet temperature, drops absolute power output.

So you're saying that the loss of power output is not as great as the decrease in fuel flow (if there were a way to express them in equivalent units)?

Quoting Tdscanuck (Reply 8):

In a jet engine, the intercooler part is extremely difficult, so you go with adiabatic compression and make it as lossless (nearly isentropic) as you can.

Why adiabatic? It's been a *long* time since I took my one required semester of thermodynamics, but isn't an adiabatic expansion/contraction a way to lose/gain the most heat?


User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 10, posted (4 years 6 months 3 days 20 hours ago) and read 7359 times:



Quoting DocLightning (Reply 9):

Quoting Tdscanuck (Reply 8):
Going to very high compression ratio increases efficiency but, for a fixed turbine inlet temperature, drops absolute power output.

So you're saying that the loss of power output is not as great as the decrease in fuel flow (if there were a way to express them in equivalent units)?

Yes. If you express fuel flow in Watts (fuel mass flow x fuel heat of combustion) you can compare them directly. If you raise compression ratio and hold turbine inlet temperature so that fuel flow drops by, say, 10%, then the drop in useful turbine output will be less than 10%. As a result, the efficiency (power out / power in) will go up.

Quoting DocLightning (Reply 9):
Quoting Tdscanuck (Reply 8):
In a jet engine, the intercooler part is extremely difficult, so you go with adiabatic compression and make it as lossless (nearly isentropic) as you can.

Why adiabatic? It's been a *long* time since I took my one required semester of thermodynamics, but isn't an adiabatic expansion/contraction a way to lose/gain the most heat?

Adiabatic is, by definition, zero heat loss/gain. It's a model of a perfectly insulating container where no heat enters or leaves the fluid, just work in/out. The compressor does work on the fluid; it does not introduce heat.

The reason you can model compressors that way (to first order) is that the flow rates in a jet engine are so large, and the heat capacity of air so high, that the amount of heat conduction between the air and the engine parts is quite small compared to the overall enthalpy of the fluid.

If you have adiabatic compression (or expansion) without losses, it's isentropic (basically by definition). Isentropic is the best you can get if you don't have the ability to add/remove heat from the fluid during the cycle; this is why ideal efficiency of compressors/turbines is the isentropic limit.

The combustor *does* have the ability to add heat; as a result, there's no such thing as an isentropic combustor.

If you have the ability to remove/add heat during the compression and expansion cycle, you get a different cycle that, at the limit, will reach the Carnot efficiency, which is as good as you can get for a particular temperature difference. This is implemented with jet engines, usually, by putting intercoolers between compressor stages and reheaters between turbine stages. This works fine at power plants and other stationary applications, but is too complex and heavy to do on practical jet engines today.

If you go all the way to the limit of isothermal compression, isobaric combustion, isothermal expansion, and isobaric recovery, you get the Ericsson cycle, which has the maximum possible efficiency (equal to the Carnot efficiency).

tom.


User currently offlineNoWorries From United States of America, joined Oct 2006, 539 posts, RR: 1
Reply 11, posted (4 years 6 months 3 days 16 hours ago) and read 7328 times:



Quoting Tdscanuck (Reply 2):
For a given amount of energy, the entropy change is smaller if you do it at high temperature. Thus, adding your heat at a very high temperature causes the cycle to "lose" less entropy, increasing efficiency (the limiting Carnot cycle is zero entropy loss).

Not sure I get this. The Carnot efficiency is 1 - Tsink/Tsource and represents the theoretical maximum proportion of work that can be extract between two reservoirs, the remainder being heat. Even an ideal Carnot engine with reversible processes would still reject some heat and result in some entropy increase -- at least that's my understanding? If Tsink were zero or Tsource were infinite, then entropy increase would be zero.

On a broader note, the posts here are rather formidable in terms of understanding them all -- but I think

Quoting Jetmech (Thread starter):
increasing the compression on the working fluid through the compressor directly allows for greater expansion

is still a pretty good intuitive answer without resorting to all sorts of thermodynamic constructs -- after all, the amount of work that comes out of a system is (roughly) proportional to the amount of expansion and the amount of expansion is going to be (roughly) proportional to the amount of compression.


User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 12, posted (4 years 6 months 3 days 9 hours ago) and read 7267 times:



Quoting NoWorries (Reply 11):
Not sure I get this. The Carnot efficiency is 1 - Tsink/Tsource and represents the theoretical maximum proportion of work that can be extract between two reservoirs, the remainder being heat. Even an ideal Carnot engine with reversible processes would still reject some heat and result in some entropy increase -- at least that's my understanding? If Tsink were zero or Tsource were infinite, then entropy increase would be zero.

The ideal Carnot does have some entropy increase (unless you have an infinite temperature reservoir). *But* all the entropy change happens during the heat input/rejection portions of the cycle. Carnot cycles have isetropic compression and expansion. When I said "the limiting Carnot cycle is zero entropy loss" that was a really bad phrase on my part...I was mixing between the heat addition (i.e. combustion) and compression parts of the cycle and not being clear at all about the distinction.

The Carnot cycle gets more efficient as temperature goes up because the entropy change during the heat addition goes down (but never goes to zero unless you have infinite temperature). Part and parcel of the ideal cycle is that you have isentropic compression and expansion.

The isentropic expansion/compression is common between the Carnot and Brayton (jet engine) cycles. The difference is in the heat addition/exhaust stages...Carnot is isothermal while Brayton is isobaric.

Tom.


User currently offlineJetMech From Australia, joined Mar 2006, 2635 posts, RR: 53
Reply 13, posted (4 years 6 months 5 hours ago) and read 7092 times:



Quoting Tdscanuck (Reply 5):
Did I just makes things better or worse?

You and Reheat have made a sterling effort! I am now much clearer on the subject. I threw a big spanner in the works with my "Eureka" moment of reply four  Sad ! Never post on Tech-Ops at 2 am !

Quoting Tdscanuck (Reply 5):
No. I think I've accidentally lead you astray. The compressor is there to raise the pressure. The turbine needs pressure to operate.

So much for getting it! I just realised that if adding heat to the working fluid was all the compressor was there for, it would be better not to have the compressor at all, as more heat would be removed from the working fluid by the turbine (to drive the compressor) than could be added back by the compressor.

Quoting Tdscanuck (Reply 5):
An isentropic compressor still has a temperature rise,

Sure, and for some reason I always thought it did not. I think I assumed this erroneous train of thought whilst I was learning thermo. Less than 100% isentropic efficiency always resulted in heat, so I assumed a perfectly isentropic process did not produce heat.

Quoting Tdscanuck (Reply 5):
The compressor efficiency is a measure of how well you raise the pressure compared to the temperature. A good compressor approaches the isentropic limit...a bad one will increase temperature a bunch without increasing pressure much.

I see. A 0% isentropically efficient compressor makes lots of heat with no pressure, a 100% isnetropically efficient compressor gives the maximum pressure increase with minimum temperature rise. Either way, the temperature will increase.

Quoting Tdscanuck (Reply 5):
The turbine is driven by pressure differential, not temperature differential. Temperature differential is what allows you to have excess energy in the turbine, after you've driven the compressor, to do useful work with.



Quoting Tdscanuck (Reply 5):
That means you want the biggest pressure increase you can in the compressor for the lowest temperature increase...that gives you the most temperature margin to burn fuel with.



Quoting Tdscanuck (Reply 5):
The ideal case is that the compressor provides a very large pressure rise for a very small temperature rise...then you can add an enormous amount of heat in the combustor and get a lot of useful work out of your turbine.

So for a given fuel (and fuel flow rate?), you are going to get a given flame temperature and heat release no matter what. If the compressor discharge is too hot, you will then need to cool your turbine blades more, which directly reduces thermal efficiency? This is in addition to having less energy left for the turbine stages not driving compressors.

Quoting Tdscanuck (Reply 5):
That gets you to the pressure you want, where you can now add heat (fuel) up until you hit the turbine inlet temperature limit.



Quoting Tdscanuck (Reply 5):
The turbine is driven by pressure differential

I see. All the turbine cares about is maximum working fluid momentum. It just so happens that burning fuel directly in the engine and producing heat is the most practical way of doing this. Fluid momentum is related to velocity, which is achieved by having a pressure differential. The compressor discharge pressure needs to be higher than the turbine inlet pressure otherwise there would be reverse flow.

Quoting Tdscanuck (Reply 5):
Temperature differential is what allows you to have excess energy in the turbine, after you've driven the compressor, to do useful work with.

This makes sense, and gets rid of the "perpetual machine" proposition I mentioned in the OP. The addition of heat reduces the rate of "momentum deterioration" of the working fluid through the turbine section. The "momentum deterioration" rate of the working fluid through the turbine without adding heat would be much quicker, so much so the engine ceases to run.

So basically, compression ratio is desirable for thermal efficiency as it allows one to make full use of the heat release and resultant expansion of the working fluid when fuel is added. If the compression ratio is too low, one then needs to find a way of limiting the resultant heat release and expansion due to the combustion of a given fuel?

Quoting Tdscanuck (Reply 5):
Did I just makes things better or worse?



Quoting Tdscanuck (Reply 5):
That would be an isothermal compression

Much better! I think I now pretty much understand the desirable characteristics one wishes to have in compressors and turbines. It's simpler yet more complex at the same time. What I need to do is read up more on entropy!

Quoting NoWorries (Reply 11):
pretty good intuitive answer without resorting to all sorts of thermodynamic constructs

It always makes me feel better and more accepting of the formula if I can form some sort of intuitive understanding. Doesn't always happen unfortunately!

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlineJetMech From Australia, joined Mar 2006, 2635 posts, RR: 53
Reply 14, posted (4 years 6 months 5 hours ago) and read 7085 times:



Quoting JetMech (Reply 13):
This makes sense, and gets rid of the "perpetual machine" proposition I mentioned in the OP. The addition of heat reduces the rate of "momentum deterioration" of the working fluid through the turbine section. The "momentum deterioration" rate of the working fluid through the turbine without adding heat would be much quicker, so much so the engine ceases to run.

On second thoughts, the addition of heat increases the overall momentum of the working fluid. The rate of momentum deterioration through the turbine section should be the same regardless of the temperature and heat contained in the working fluid.

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 15, posted (4 years 5 months 4 weeks 7 hours ago) and read 6966 times:



Quoting JetMech (Reply 13):
So for a given fuel (and fuel flow rate?), you are going to get a given flame temperature and heat release no matter what.

You will get a given heat release, no matter what. The temperature *rise* will be the same for a given mass flow and fuel/air ratio...so the temperature going in to the turbine will be the fuel deltaT added to whatever came out of the compressor.

Quoting JetMech (Reply 13):
If the compressor discharge is too hot, you will then need to cool your turbine blades more, which directly reduces thermal efficiency?

That is one way to think of it, but more practically you'd fnd that you just can't dump in as much fuel. Jet engines run *really* lean...the limit on how much fuel you can dump comes primarily from the temperature limit on the turbine. If you can stand a higher turbine temperature you can get a lot more power out of the core without changing the size of anything (except maybe the shaft strength).

Quoting JetMech (Reply 13):
This is in addition to having less energy left for the turbine stages not driving compressors.

Exactly.

Quoting JetMech (Reply 13):
All the turbine cares about is maximum working fluid momentum.

It's not exactly momentum...momentum doesn't take the energy you've got as heat into account. Fundamentally, it's all momentum of molecules wizzing around, but random motion (heat energy) has zero net momentum.

This it the major reason you use enthalpy, not internal energy, when working with turbines and other continuous flow cycles. Enthalpy is internal energy plus (pressure * volume)...enthalpy is a constant most of the time, even when internal energy, pressure, or volume aren't.

Quoting JetMech (Reply 13):
The compressor discharge pressure needs to be higher than the turbine inlet pressure otherwise there would be reverse flow.

That's basically true. If you've got momentum (it's back!) you can overcome an adverse pressure gradient for a little while (this is what keeps the air on the top surface of the wing going) but you can't do it for very long. But, in a real jet engine, the peak total pressure is always at the compressor outlet.

Quoting JetMech (Reply 13):
So basically, compression ratio is desirable for thermal efficiency as it allows one to make full use of the heat release and resultant expansion of the working fluid when fuel is added.

That is the best explanation I've seen anywhere in this thread, by anybody.

Quoting JetMech (Reply 13):
If the compression ratio is too low, one then needs to find a way of limiting the resultant heat release and expansion due to the combustion of a given fuel?

I'm not quite sure about this. If the compression ratio is too low, the difference in work between the compression and expansion parts of the cycle is very small, and the overall output of the cycle is small.

Quoting JetMech (Reply 13):
It's simpler yet more complex at the same time.

Yup. This is (one of) the reasons I really hate undergraduate thermodynamics...it's just enough to make it look simple, but it glosses over all the nasty complexities lying just beneath.

Quoting JetMech (Reply 13):
What I need to do is read up more on entropy!

If you get a handle on entropy, a lot of thermodynamics makes more sense. Unfortunately, entropy is among the most difficult topics (for me, anyway) to get a really good quantitative handle on. Strictly speaking, since energy is conserved, what we really "consume" is negentropy (the opposite of entropy)...but I still haven't really figured out how to *use* that for anything.

Tom.


User currently offlineJetMech From Australia, joined Mar 2006, 2635 posts, RR: 53
Reply 16, posted (4 years 5 months 4 weeks 2 hours ago) and read 6933 times:



Quoting Tdscanuck (Reply 15):
The temperature *rise* will be the same for a given mass flow and fuel/air ratio...so the temperature going in to the turbine will be the fuel deltaT added to whatever came out of the compressor.



Quoting Tdscanuck (Reply 15):
...the limit on how much fuel you can dump comes primarily from the temperature limit on the turbine

Fair enough. Which is why you would want the minimum compressor discharge temp in order to make full use of the temperature rise provided by combusting the fuel - which you will get no matter what - whilst staying below the maximum T.I.T..

Quoting Tdscanuck (Reply 15):
It's not exactly momentum...momentum doesn't take the energy you've got as heat into account.

I see. So the heat liberated by combusting fuel manifests as an increase in internal energy ( molecular motion ), not so much an increase in bulk fluid motion ( mass* velocity ) ?

Quoting Tdscanuck (Reply 15):
That is the best explanation I've seen anywhere in this thread, by anybody.

Thanks! I was a little wary of proposing it, as apparently, piston engines also benefit in terms of thermodynamic efficiency by having a higher compression ratio. In this case however, the pressure during the combustion process far exceeds that at the end of the compression stroke. So there must also be another element to the desirability of having a high compression ratio  scratchchin  .

Quoting Tdscanuck (Reply 15):
If the compression ratio is too low, the difference in work between the compression and expansion parts of the cycle is very small, and the overall output of the cycle is small.

True. What I was thinking however was what would happen if the compressor discharge pressure dropped suddenly. You would then need to quickly cut your fuel flow to minimise any additional damage from flow reversal through the engine.

Quoting Tdscanuck (Reply 15):
Unfortunately, entropy is among the most difficult topics

I don't really get it either. However, I have heard it been said that the concept of entropy means that the universe will eventually be homogenously filled with diffuse energy / radiation / heat.

The homogenous part strikes a chord with me. Entropy predicts the natural direction of processes, such as the cooling of a hot cup of coffee in a cooler surrounding. Whatever process we carry out results in a net increase of entropy, or homogeneity. In other words, the temperature of the coffee and surroundings become homogenous.

This ties in with engines. When we burn fuel and pass it through a turbine, we get mechanical work, but also a lot of waste heat. When rejected to the cooler heat reservoir (atmosphere) this waste heat acts like a cup of coffee.

We could also state this in a more mechanical fashion. A volume of liquid kerosene is very distinct and non-homogenous from the atmosphere, but through the process of combustion, we tend to bring these two entities closer in form, and make them more alike (homogenous).

That’s my intuitive explanation anyway  blush  !

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlineNoWorries From United States of America, joined Oct 2006, 539 posts, RR: 1
Reply 17, posted (4 years 5 months 3 weeks 6 days 23 hours ago) and read 6913 times:

The other thing worth noting here is that entropy is strictly a statistical measure. At the particle level, there is no entropy (or temperature for that matter). Entropy and temperature are convenient ways of dealing with large populations for which individual information (momentum, position, etc) is impossible to obtain. Down at the particle level, every interaction is completely reversible. I don't know if it's strictly correct to think of it this way, but I think of entropy as a consequence of the expanding universe.

User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 18, posted (4 years 5 months 3 weeks 6 days 20 hours ago) and read 6894 times:



Quoting JetMech (Reply 16):
I see. So the heat liberated by combusting fuel manifests as an increase in internal energy ( molecular motion ), not so much an increase in bulk fluid motion ( mass* velocity ) ?

Yes. Since the mass of fuel added is usually very small compared to the air mass, the change in bulk fluid motion through the combustor is very small...first order jet engine analysis usually assumes it's zero. What changes a bunch is the internal energy (random molcular motion). That energy is then extracted (with some necessarily loses) by the turbine (as shaft power) and later by the nozzle (as fluid momentum increase). It's vaguelly similar to a rocket nozzle...inside a rocket combustion chamber, the momentum is very low (relative to the exhaust plume), but the internal energy is *really* high. The rocket nozzle converts that thermal energy to directed momentum.

Quoting JetMech (Reply 16):
I was a little wary of proposing it, as apparently, piston engines also benefit in terms of thermodynamic efficiency by having a higher compression ratio. In this case however, the pressure during the combustion process far exceeds that at the end of the compression stroke. So there must also be another element to the desirability of having a high compression ratio

The same underlying principles are at work, but the details are different because it's a different cycle.

Quoting JetMech (Reply 16):
True. What I was thinking however was what would happen if the compressor discharge pressure dropped suddenly. You would then need to quickly cut your fuel flow to minimise any additional damage from flow reversal through the engine.

That's exctly what happens in a compressor stall...compressor discharge pressure drops, combustor flow reversers, and hot gas burps out the inlet.

Quoting NoWorries (Reply 17):
I don't know if it's strictly correct to think of it this way, but I think of entropy as a consequence of the expanding universe.

I think the expanding universe aggravates it, but I think it would happen anyway. The most general explanation I've heard is that "entropy is the tendendy of the universe to smooth out gradients." Big temperature differences even out, mountains & valleys erode to flat plains, etc.


Tom.


User currently offlineJetmech From Australia, joined Mar 2006, 2635 posts, RR: 53
Reply 19, posted (4 years 5 months 3 weeks 6 days 6 hours ago) and read 6835 times:



Quoting NoWorries (Reply 17):
I don't know if it's strictly correct to think of it this way, but I think of entropy as a consequence of the expanding universe.

To me, this explanation certainly ties in with things becoming more diffuse.

Quoting Tdscanuck (Reply 18):
What changes a bunch is the internal energy (random molcular motion). That energy is then extracted (with some necessarily loses) by the turbine (as shaft power) and later by the nozzle (as fluid momentum increase).

I see. It definitely makes sense theoretically but I'll have to ponder for a while to get an intuitive understanding  scratchchin 

Quoting Tdscanuck (Reply 18):
The most general explanation I've heard is that "entropy is the tendendy of the universe to smooth out gradients." Big temperature differences even out, mountains & valleys erode to flat plains, etc.

Homogeneity  Smile ! Thanks for all you input BTW. Most informative and helpful  bigthumbsup  !

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlineThegeek From Australia, joined Nov 2007, 2638 posts, RR: 0
Reply 20, posted (4 years 5 months 3 weeks 6 days 6 hours ago) and read 6832 times:



Quoting Tdscanuck (Reply 2):
In other words, if your compressor increases the air temp by 200 C, the air temp has to drop by 200 C in the turbine in order to provide the energy to power the compressor. Whatever's left is available to be used for something useful, like pushing the aircraft or spinning the fan. So, the higher your turbine inlet temperature, the more you have "left" after extracting the power needed to run the compressor.

Never thought of this before! But Q = W + delta U (or something like that). If Q = 0, i.e. adiabatic then shaft power = change in enthalpy. You're quite correct for a theoretical engine.

Quoting Tdscanuck (Reply 10):
If you have the ability to remove/add heat during the compression and expansion cycle, you get a different cycle that, at the limit, will reach the Carnot efficiency, which is as good as you can get for a particular temperature difference. This is implemented with jet engines, usually, by putting intercoolers between compressor stages and reheaters between turbine stages. This works fine at power plants and other stationary applications, but is too complex and heavy to do on practical jet engines today.

Why would reheaters increase efficiency? I'd think that the best option would be intercoolers + regenerators.


User currently offlineSNAFlyboy From United States of America, joined Oct 2007, 86 posts, RR: 0
Reply 21, posted (4 years 5 months 3 weeks 6 days 5 hours ago) and read 6827 times:



Quoting Tdscanuck (Reply 15):
Yup. This is (one of) the reasons I really hate undergraduate thermodynamics...it's just enough to make it look simple, but it glosses over all the nasty complexities lying just beneath.

I always had this nagging feeling that my Thermo professor wasn't telling me the whole story...  worried 

~SNAFlyboy


User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 22, posted (4 years 5 months 3 weeks 5 days 21 hours ago) and read 6782 times:



Quoting Thegeek (Reply 20):

Why would reheaters increase efficiency? I'd think that the best option would be intercoolers + regenerators.

When I say "reheater" I don't mean an afterburner (I forgot about the terminology change when you cross the pond), I mean adding heat between turbine stages. The heat source could be more fuel, or some other part of the engine (which is what, I think, you're calling a regenerator).

However, many jet engines have temperatures in the turbine so far above temperatures in the compressor that you can't use the heat from the intercooler to run the regenerator, or at least not do it through the whole turbine.

Either way, you're doing an isothermal expansion, the difference is where you're getting the heat from.

Tom.


User currently offlineThegeek From Australia, joined Nov 2007, 2638 posts, RR: 0
Reply 23, posted (4 years 5 months 3 weeks 5 days 13 hours ago) and read 6747 times:



Quoting Tdscanuck (Reply 22):
(which is what, I think, you're calling a regenerator).

FWIW, I was thinking of waste heat from the exhaust being applied to the compressor output. I suppose even stationary engines still suffer from practicalities in that you have to trade off efficiency versus power output, and that's where adding fuel after the first turbine stage can be beneficial. Although I'd think that cooling of the turbine blades would then be exceedingly troublesome.


User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 24, posted (4 years 5 months 3 weeks 5 days 9 hours ago) and read 6728 times:



Quoting Thegeek (Reply 23):
FWIW, I was thinking of waste heat from the exhaust being applied to the compressor output.

Ah, gotcha. There's one biz jet engine with a really goofy flow path that sort of does that, but I can't remember what it's called right now. Yes, that would be more efficient from a strictly thermodynamic point of view, although it might not be practical in an aircraft due to the heat exchanger intricacies.

Quoting Thegeek (Reply 23):
I suppose even stationary engines still suffer from practicalities in that you have to trade off efficiency versus power output, and that's where adding fuel after the first turbine stage can be beneficial.

Very true...the only applications where I've seen close-to-ideal performance are those places where weight and size are mostly irrelevant and the utilization is so high that small efficiency increases pay off: gas turbine power plants and cargo ships.

Quoting Thegeek (Reply 23):
Although I'd think that cooling of the turbine blades would then be exceedingly troublesome.

Unless you were asking for trouble, you wouldn't want to heat it any hotter than the turbine inlet, so the cooling isn't worse than what you have now, there's just more of it.

Tom.


25 Thegeek : Even getting to that hot would be hard though, wouldn't it? Doesn't the first stage of a jet engine turbine use some funny concept where the hottest
26 Tdscanuck : The inlet guide vanes and first turbine stage are usually hollow with a lot of holes in the surface that they blow "cold" compressor air through (it'
27 Thegeek : You could. It's just that you've already mixed your compressor air and your combustor output. If you split the compressor air, you are wasting compre
28 Dakota123 : "Seems like a lot of hassle", but film or impingement cooling of downstream stages (blades, nozzles or both) is incorporated into many designs of aer
29 Thegeek : Is it used on second and later turbine stages in "many designs of aero engines"? That's what we were discussing.
30 Dakota123 : As was I. Yes, 2nd stage nozzle and/or blade active cooling is common, at least as far back as the CF6-50. Mike
31 Post contains links JetMech : Sorry to revive such an old thread, but I found some new info with respect to compression ratio and reciprocating engines. According to Wikipedia, hig
32 Tdscanuck : There is, but it only shows up when you go outside the underlying assumptions of the thermodynamics...if you get compression high enough that the gas
33 Thegeek : I've read that the effect of increasing compression improving efficiency reaches a limit at about 16:1 (by volume). So that means many diesels are bey
34 Post contains links and images JetMech : Fair enough. I erroneously thought that thermal efficiency did mathematically top out at about 60 %, but this is apparently not so. Plotting thermal
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