Turbo-Diesel Thermodynamics 101... 3

[ernesteugene]

I'm a retired EE, a fulltime RVer, I participate in several RV and diesel truck forums, and I'd very much appreciate a "critical review" of everything I post on this thread along with comments and suggestions for how to explain "Thermodynamics 101" to forum members who mostly have nontechnical backgrounds. As an example I need to be able to explain this equation I came up with for turbo performance and later I'll discuss several different ways to derive it using some basic thermodynamic concepts.

EGT=[AIT]x[{(0.9)(TPR^0.257)}/{(TE)(CE)(1+1/AFR)}]x[{CPR^0.283-1}/{TPR^0.257-1}] *R

The EGT=Exhaust Gas Temperature *R at the turbine inlet, AIT=Air Inlet Temperature *R at the compressor inlet, TPR=Turbine Pressure Ratio, TE=Turbine Efficiency, CE=Compressor Efficiency, AFR=Air Fuel Ratio, and CPR=Compressor Pressure Ratio. Just to make sure I typed this equation correctly I'll copy and paste it below, plug in some numbers, and do a sample calculation using TE=CE=0.7, AFR=20, CPR=3 , and TPR=2.54.

If the AP=Atmospheric Pressure is AP=14.7 psia, the AFPD=Air Filter Pressure Drop is AFPD=0.7 psi, and the ICPD=Intercooler Pressure Drop is ICPD=1.3 psi, then the BP=Boost Pressure is BP=26 psig in order to satisfy the constraint that CPR=(AP+BP+ICPD)/(AP-AFPD)=(14.7+26+1.3)/(14.7-0.7)=(42)/(14)=3.0. Assume the absolute EBP=Exhaust Back Pressure psia at the turbine inlet is equal to the absolute MAP=Manifold Air Pressure psia at the engine inlet so that EBP=MAP=AP+BP, and that the PTPD =Post Turbine Pressure Drop PTPD to the tailpipe outlet is PTPD=1.3 psi. This gives a TPR=(EBP)/(AP+PTPD)=(AP+BP)/(AP+PTPD)=(14.7+26)/(14.7+1.3)=(40.7)/(16)=2.54.

Using these values gives... EGT={AIT}x[{(0.9)(2.54^0.257)}/{(0.7)(0.7)(1+1/20)}]x[{3^0.283-1}/{2.54^0.257-1}]={AIT}x[{(0.9)(1.27)}/{(0.49)(1.05)}]x[{0.365}/{0.27}]={AIT}{(1.14)/(0.515)}{1.35}={AIT}{(2.21)(1.35)}={AIT}(2.98) ...so EGT=(2.98)(AIT) *R and converting from *R to *F you get... EGT=(2.98)(AIT+460)=(2.98)(AIT)+911 *F.

I haven't yet built a spreadsheet with these equations to check and double check them but without going into mass production it's easy to see how higher AIT effects EGT everything else being equal.

AIT*F....EGT*F
70.........1,120
90.........1,179
110.......1,239
130.......1,298

Each 20*F increase in AIT requires a 60*F increase in EGT and this means you've got to push harder on the fuel pedal to release enough additional heat energy into the exhaust flow to increase the EGT by the 60*F that's needed to maintain the assumed compressor pressure ratio of CPR=3.

You might think this table alone would be enough to convince any reasonable person that installing an "open element" air filter directly onto their compressor inlet is a bad idea because under the hood air temperature is significantly higher than ambient. However the "diesel heads" on the sites I visit claim that their upgraded intercoolers compensate for any increases in AIT and/or that the air under their hood isn't any hotter than ambient air anyway!

Well according to the laws of Thermodynamics as the AIT increases the compressor has to work harder to achieve a given CPR at the compressor outlet and this means the turbine has to work harder to power the compressor and since the turbine is powered by the heat energy in the exhaust flow this means the EGT must increase as is predicted by my equation. Since the intercooler is downstream from the compressor it can't possibly compensate for this "increase in required EGT penalty" due to the increased workload on the compressor in maintaining a given CPR for higher AIT! The effect an intercooler has on engine inlet air density will be addressed in a future post.

Another common "myth" is that adding an intercooler will reduce the EGT but if the compressor pressure ratio is to be maintained at a given value like the CPR=3 in this example then using an intercooler actually increases the required EGT! To show this is true let's remove the intercooler and redo the above example. If you keep the compressor pressure ratio at it's original value of CPR=3 and have a ICPD=0 psi then the boost pressure increases from BP=26 psig to BP=27.3 psig as is required to keep the CPR=(14.7+27.3+0)/(14.7-0.7)=(42)/(14)=3.0 as before. Now the new TPR=(EBP)/(AP+PTPD)=(AP+BP)/(AP+PTPD)=(14.7+27.3)/(14.7+1.3)=(42)/(16)=2.63 without an intercooler versus the pervious TPR=2.54 which was the case with an intercooler.

Now... EGT={AIT}x[{(0.9)(2.63^0.257)}/{(0.7)(0.7)(1+1/20)}]x[{3^0.283-1}/{2.63^0.257-1}]={AIT}x[{(0.9)(1.28)}/{(0.49)(1.05)}]x[{0.365}/{0.28}]={AIT}{(1.15)/(0.515)}{1.35}={AIT}{(2.21)(1.30)}={AIT}(2.87) ...so EGT=(2.87)(AIT) *R and converting from *R to *F you get... EGT=(2.87)(AIT+460)=(2.87)(AIT)+860 *F, and the revised EGT versus AIT table without an intercooler is...

AIT*F....EGT*F
70.........1,061
90.........1,118
110.......1,176
130.......1,233

Without an intercooler each 20*F increase in AIT requires a 57*F increase in EGT compared to the 60*F increase in EGT that's required with an intercooler. In order to maintain a CPR=3 with an intercooler in the loop you've got to push harder on the fuel pedal to release enough additional heat energy into the exhaust flow to increase the EGT by an amount given by... EGT=(0.11)(AIT)+51 *F which varies from an EGT increase of 59*F at an AIT=70*F to an EGT increase of 65*F at an AIT=130*F.

Since adding an intercooler decreases the boost from 27.3 psig to 26 psig and increases the EGT requirement for achieving a CPR=3 by 59*F to 65*F depending on the AIT you might think that comparing this EGT table without an intercooler to the previous EGT table which includes an intercooler would be enough to convince people to stop posting misguided claims about the effect an intercooler has on EGT. Even though the laws of Thermodynamics say that installing an intercooler will increase the EGT requirement for maintaining a given compressor pressure ratio this concept will be even harder to sell than convincing people that their flywheel torque is not what gets them up hills faster which of course is exactly what the motorhome salesman claimed when he did his sales pitch for a diesel versus a gas powered RV!

It's been posted so often on RV and diesel truck forums that increasing your MAF=Mass Air Flow lbm/min by whatever means possible will lower your EGT and make your turbo spool better and produce higher boost that this "myth" has become "common knowledge"! Well let me try to bust the "increased MAF lowers your EGT myth" and then post what I've got for now to test what the word limit is on this forum! I'll eventually give all the details of the summary below.

If you assume the parameter values used in the first example apply and consider a MAF=1 lbm/min airflow with an AIT=70*F then the FEHP=Flow Energy HP that's available at the compressor inlet due to the atmospheric pressure pushing the airflow into the inlet is FEHP=0.87 hp and at the compressor outlet a FEHP=1.30 hp is required to push the compressed airflow toward the engine. This leaves a net deficit of 1.30-0.87=0.43 hp that must be supplied by the CSHP=Compressor Shaft HP driving the compressor in order to maintain the airflow.

To compress a MAF=1 lbm/min airflow with an AIT=70*F using a CPR=3 requires that an IEHP=Internal Energy HP of IEHP=1.13 hp be added to the airflow as stored internal energy and this IEHP=1.13 hp must also be supplied by the CSHP. Thus the total required CSHP=1.13+0.43=1.56 hp and this is the exact TSHP=Turbine Shaft HP that must be supplied by the "turbine heat engine" in order to maintain a steady state compression of a MAF=1 lbm/min airflow with a CPR=3 and a compressor inlet temperature of AIT=70*F.

At the turbine inlet the FEHP that's available due to the hot exhaust gas pressure pushing the airflow into the inlet is FEHP=2.68 hp and at the turbine outlet the FEHP=2.28 hp for a net gain of 2.68-2.28=0.40 hp and this surplus FEHP=0.40 hp is supplied to the turbine output shaft as a TSHP=0.40 hp. As the hot exhaust gas cools by 236*F from a pre-turbine EGT=1,120*F to a post-turbine EGT=884*F it releases internal Heat Energy at a ft-lbf/min rate which is sufficient to supply a HEHP=1.16 hp to the turbine output shaft as an additional TSHP=1.16 hp which gives a total TSHP=0.40+1.16=1.56 hp and that's the exact amount required to power the compressor!

In the above calculations for the turbine the MGF=Mass Gas Flow lbm/min in the exhaust is given by MGF=(1+1/AFR)(MAF) and the reason that MAF doesn't appear in my EGT equation is that the required CSHP and the available TSHP are both proportional to MAF so when you set TSHP=CSHP to determine the required EGT to achieve an assumed operating condition the MAF term cancels and only the term (1+1/AFR) remains. An increase in MAF increases the compressor workload and the turbine work output by equal amounts and the EGT remains unchanged.

If you now push harder on the fuel pedal and have the reserve fuel flow to release enough additional heat energy into the exhaust flow to increase the EGT the TSHP will increase above TSHP=1.56 hp and this will drive the compressor harder and produce an increase in CPR until a new equilibrium is achieved where the required CSHP and the available TSHP are once again exactly the same value! So it's Heat Energy that spools the turbo to a higher CPR to produce a higher BP and not an increase in MAF!

 

[MikeHalloran]

Are you trying to pick a fight?

There are lots of serious motorheads here, but this is not an RV forum, and it's not like the other forums you may have found in your wanderings in the land of HTML.

You'll probably get the critique you asked for. I hope you have a thick skin...



Mike Halloran
Pembroke Pines, FL, USA

 

[Lou Scannon]

All I can say is that thermodynamics in general and thermodynamics for turbochargers in particular are not subjects for debate, they are established physics.
Trying to teach thermo to a non-technical audience is probably going to be akin to trying to herd cats.
I didn't take the time to read through your rather long post, as I don't have the time to parse through your information to check it against the established physics.
I suggest that instead of trying to derive and explain the physics of turbocharging, that you refer yourself and if necessary your non-technical audience to appropriate reference materials. If any questions remain after that, feel free to post them here.
One readily accessible publication I can recommend that is a good bridge between a lay and a scientific treatment of turbocharging is the book by Hugh MacInnis titled "Turbochargers".

I don't mean to brush you off entirely. I'd suggest you post your topics individually, rather than as a smorgasbord. I suspect others besides myself would find it more attractive to respond to a concise topic/question versus trying to digest & respond helpfully to a post like your OP above.

 

[JSteve2]

Just as a general note - when you are dealing with a non-technical audience (which is not this audience) it is better to focus on the consequences of the thermodynamics, not the equations. Even most technical people don't enjoy crunching through equations if it's not for a problem they care about at the moment.

 

[ernesteugene]

My retirement hobby is analyzing topics of interest I come across in the mentioned forums and in this instance the conclusions I've drawn from my EGT equation are at odds with the claims by "experts" on these forums. The point-counter points I mentioned were to highlight these areas of disagreement not to pick a fight!

Before going forward with my results I'd like to know if any of my basic conclusions are at odds with the "expert knowledge" on this forum especially my calculated hp # for running the compressor? If so I'll do some more homework and thanks to the suggestion by "hemi" I'll shortly be prepared to do so because I just ordered these three references as Amazon had a package deal!

"Turbochargers HP49 (HP Books): Turbo Design, Sizing & Matching, Spark-Ignition & Diesel Engine Applications, Water Injection, Controls, Carburetion, Intercooling, ... Street & Race Cars, Boats, Motorc"
Hugh MacInnes; Paperback; $14.93

"Turbo: Real-World High-Performance Turbocharger Systems (S-A Design)"
Jay K. Miller; Paperback; $16.47

"Maximum Boost: Designing, Testing, and Installing Turbocharger Systems (Engineering and Performance)"
Corky Bell; Paperback; $23.07

I won't bother anyone by posting more details of how I derived my EGT equation but if I ever get around to doing a Thermodynamics 101 post on another forum I thought I might illustrate how heat energy (which purists claim is a misnomer) has the potential for doing work by starting off with...

If you lift a full 16 oz can of beer to your mouth your arm exerts 1 ft-lbf of work energy and if the temperature of the beer increases by only 1*F the beer absorbs 778 ft-lbf of heat energy from your hand. If you could then "somehow" get the temperature of the beer to decrease by 1*F and capture the entire 778 ft-lbf of heat energy that's released with a "device" that converts it into kinetic energy then that 1 lbm can of beer would be traveling at exactly 152.52 mph! If this conversion of 778 ft-lbf of heat energy into 778 ft-lbf of kinetic energy occurred in a time interval of 1 sec then the "device" produced 1.42 hp! Thermodynamics is the study of the "somehow" and the "device" and Thermodynamic principles dictate what's possible and what's not and that 152.52 mph number isn't even close to being a possibility!

Of course the working fluid in a turbo diesel is air and not beer but the same Thermodynamic principles apply. The two basic differences between a beer powered turbo and one powered by the heat energy in the exhaust flow is that exhaust gas is compressible and for a given temperature change exhaust gas absorbs and releases only 17% as much heat energy as beer does! Well if there's a mistake in this analysis or with my EGT equation I'll blame it on the beer!

 

[patprimmer]

ernesteugene

Have you read the forum rules about work related and professional.

While by your own admission you seem to breach this, most here will not red flag so long as you remain polite, intelligent and don't ask to much.

Your OP is asking to much.

Do your own homework, then as others have suggested, comeback with bite sized chunks. It will also help if the chunks are relevant to real world and not academic pie in the sky nor theoretical details of no practical concequence.



Regards
Pat
See FAQ731-376 for tips on use of eng-tips by professional engineers &

http://eng-tips.com/market.cfm

for site rules

 

[Lou Scannon]

"Maximum Boost: Designing, Testing, and Installing Turbocharger Systems (Engineering and Performance)"
Corky Bell

I caution that, having had a quick read through this book, that Corky Bell evidently does not have university level training in thermodynamics and fluid dynamics.

 

[JSteve2]

ernest,
A couple of comments on your OP, just as suggestions for a better look.

First, AIT is not going up to 130 F for a vehicle that is traveling, no matter where your air is coming from. There just isn't enough time for heat transfer to that level.

Second, increases in AIT lead to about 0.8x increase in EGT, not 5x. I have literally thousands of data sets that indicate that this is pretty close in a wide variety of conditions, so while it can vary if you're not in that ballpark something is missing.

Third, you cannot extrapolate an empirical correlation too far, and you certainly can't call it a thermodynamic law. For example, you wouldn't want to take my AIT number above and call that a thermodynamic law, it's simply a correlation. It wouldn't surprise me, for example, in extreme arctic conditions if there were not some kind of fall-off that changed the typically observed ratio.

Fourth, using an intercooler is about power density, not reducing EGT. It will reduce EGT (again, I think you are abusing the empirical correlation in the steps where you remove it above, but I didn't work it all out to find the error) because where you end up post-combustion is closely related to where you started pre-combustion. However, operation with an intercooler is necessarily less fuel efficient than without and so you pay for that somewhere - it would not surprise me if EGT ended up slightly higher than otherwise predicted, but it will still be lower than with no intercooler at the same power output. If we didn't need the power density and had no emissions issues, we wouldn't include the expense and inefficiency introduced by an intercooler.

Fifth - I couldn't resist - in your beer example the kinetic energy of your armthe beer example. The kinetic energy of your arm is not what warms the beer. The beer doesn't warm 1 degree in a second in any event (can you warm a 35 degree beer to room temperature by picking it up 40 times? of course not). A beer sitting on a table will take maybe 45 minutes to warm to room temperature, and if you had a machine that would do it in 40 seconds it would in fact have to be a 1-hp machine. We call that a microwave, and they run about 1,500 watts (2 hp) in many iterations.

Personally, I have no issues with people who question the status quo - in fact we wouldn't learn anything if no one ever did. However, the number of times something that truly goes against the status quo is turned up in a well vetted area like engines is very small. Almost always, there is a flaw in the would-be-revolutionary's analysis. I don't think that means the exercise is not a good one.

 

[JSteve2]

My 5x should be 3x above.

 

[ernesteugene]

Well as a retired Ph D engineer I no longer request nor receive $ type reimbursements for my technical contributions so I suppose in that regard I'm no longer a "professional engineer" and henceforth I'll retire from this forum as well! However since I already wrote the following "non-beer related technical reply" before logging on and learning that I shouldn't be here to begin with I'll go ahead and make that my final post on this forum!

However I can't restrain myself from commenting on this one beer related point so I'll be a "nonprofessional engineer" by responding in kind to this silly statement I noticed which has been attributed to me..."The kinetic energy of your arm is not what warms the beer." ... which I take as an example that engineers on this forum can't or won't take the time to read and understand a carefully worded post any better than most members of diesel truck forums! Of course my actual statement was ..."If you lift a full 16 oz can of beer to your mouth your arm exerts 1 ft-lbf of work energy and if the temperature of the beer increases by only 1*F the beer absorbs 778 ft-lbf of heat energy from your hand."

So where does this "778 ft-lbf of heat energy from your hand" come from and assuming an initial proper beer drinking temperature of 40*F how long do you need to grasp that can of beer in your "hot hand" without imbibing for the 10 lbm blood mass in your body pulsating at 1 gal/min to circulate through your hand long enough to transfer the 778 ft-lbf of heat energy to the beer assuming your body acts as a heat reservoir with a temperature of 98.6*F? Well I'll leave this as a homework assignment but if anyone has trouble solving it just look me up on any of the popular RV or diesel truck forums and I'll give you a hand free of charge of course because after all I'm just an amateur anyway!

I've checked my derivation again and I I'm pretty sure my equation for EGTr="required EGT" is correct! However when using my equation to compare different operating conditions like I did in example #1 where TPR=2.54 with an intercooler versus example #2 where TPR=2.63 without an intercooler you need to define the independent versus dependent variables and in my conclusions I didn't mention the individual values of MAF and MFF only their ratio AFR=MAF/MFF and I didn't said anything about comparing with constant engine RPM or constant FWHP or whatever.

In the table below I use the EGTa="actual EGT" from data I collected during 10 years of towing my 5er with my old early 99 7.3L power stroke and EGTa=AIT+21,000/AFR *F is a reasonable fit to that data. For AIT=70*F my equation for EGTr predicts EGTr1={(1658.9)/(1+1/AFR)}-460 *F for example #1 and EGTr2={(1697.1)/(1+1/AFR)}-460 *F for example #2 and these equations were used to get the result given below for AIT=70*F.

AFR....EGTa....EGTr1.....EGTr2
18......1,237....1,112.....1,053
19......1,175....1,116.....1,057
20......1,120....1,120.....1,061
21......1,070....1,123.....1,065
22......1,025....1,127.....1,068

For an AFR=20 the EGTa=EGTr1 and I chose the specific parameter values in example #1 the way I did because they approximate the values from my old truck where I'd measured AFPD, removed the CAT and installed a straight-through muffler to reduce PTPD, etc.. so when my equation for CPR=3 predicted the same EGT as I'd been seeing when towing grades at near maximum boost with a CPR of about 3 I was fairly confident my equation was correct.

So if I'd actually removed the intercooler from my old truck my required EGTr would've decreased from EGTr1=1,120 *F to EGTr2=1,061*F at an AFR=20 but as can be seen from the table to actually operate without the intercooler my actual AFR would have to increase to AFR=21.06 where once again the actual EGTa=1,067 *F equals the required EGTr2=1,067 *F! Now since removing the intercooler increases the MAT and reduces the MAF I need to take that into account. Even if could retain my original MAF by say increasing RPM I'd still have to reduce MFF some to operate at the new higher AFR=21.06 and that means less FWHP and trying to increase RPM and find a self-consistent operating point where CPR=3 at the same is probably not possible so the CPR would also have to change!

However the main point I was trying to make is that adding an intercooler is not a slam dunk win-win proposition because in addition to reducing MAT an intercooler also reduces MAP and if EBP=MAP the TPR decreases and the turbine then requires higher EGT to produce the same drive hp to the compressor as before! Also with an intercooler the power band RPM lines plotted on a compressor map are typically to the right of the peak ridge of the efficiency island toward higher MAF so removing the intercooler moves these lines to the left toward the surge line where the compressor efficiency is higher and this reduces the compressor outlet temperature thus mitigating some of the increase in MAT due to removing the intercooler so when calculating the net change in MAF with and without an intercooler you need to consider all of these effects.

I was hoping someone here might possibly be able to reference a sophisticated computer model for matching points on compressor maps with corresponding points on turbine maps and automatically identify a self-consistent operating point so I could get a self-consistent set of parameters for my EGTr equation? Well if anyone's bothered to read this far I'd sure appreciate a short reply just to let me know if I've stated things in a way that might possibly be understood by members of a typical diesel truck forum for the purpose of beginning a discussion there. Of course if you've found any mistakes I'd also like to know about those as well.

For what it's worth in the 1960's I was building my own race engines and a divisional champion driver in SCCA road racing so I'm not a newcomer to engines as some have suggested! However back then a bamboo slide rule and a CRC Handbook were my only tools to aid my brainpower for doing engine analyses! I only hope you guys realize just how easy you have it today!

 

[Lou Scannon]

 

[patprimmer]

|-I

Regards
Pat
See FAQ731-376 for tips on use of eng-tips by professional engineers &

http://eng-tips.com/market.cfm

for site rules

 

[tbuelna]

ernesteugene,

Don't let these guys get you down. You're obviously a well educated person with a solid technical background, and you have posed some well considered technical questions on an engineering forum.

The problem with the equation you postulate for evaluating EGT's in a supercharged diesel engine, is not so much that it's incorrect, but that it's an over generalization of the situation. A free-wheel exhaust turbocharger is by definition a device that always operates at equilibrium between the compressor and turbine flows. And being a dynamic compression/expansion device working with a compressible fluid, the compressor and turbine performances can vary greatly with regards to variables like flow velocities, pressure ratios, flow turbulence, and densities.

Besides the variables that you have considered, EGT's (at the turbine inlet) can be affected by injection timing, intake air swirl and compression turbulence, injector nozzle spray geometry, heat transfer in the exhaust ports and manifolds, effective expansion ratio due to EVOP timing, increased trapped intake charge mass (above what would normally be produced by the nominal MAP) due to intake inertias and acoustic wave effects, combustion cycle frequencies, etc. So, as you can imagine, calculating EGT's based on some fixed set of engine operating variables is probably not going to provide you an accurate result.

As for the basic issue of using charge air coolers on an automotive diesel piston engine, they have both benefits and drawbacks. Charge air coolers are used for two basic reasons on most automotive diesel engines. 1) To reduce the peak combustion temps to minimize NOx emissions. 2) To reduce the thermal loads on the piston crown, rings and cylinder head with high performance upgrades.

But since a diesel engine is not knock limited like a gasoline engine, a charge air cooler is not necessarily required or beneficial. While the charge air cooler provides a cooler, denser intake charge at the cylinder inlet, it also produces an intake manifold air pressure drop, produces intake flow losses, increases the engine package size, weight, cost and aero drag, and rejects heat energy that adversely impacts the Brake Thermal Efficiency of the engine.

Many current production diesels use multiple turbos in series to produce very high levels of boost. Having a charge air cooler between the stages is almost always beneficial, since the denser charge at the second stage compressor inlet greatly helps the compressor's efficiency.

If you go to Garrett's website, you can get some compressor and turbine performance maps for most of their products.

http://www.turbobygarrett.com/turbobygarrett/products/turbochargers.html

Good luck and have fun in your retirement.
Terry

 

[Lou Scannon]

Terry, I agree with most of your post, however, regarding charge coolers, I would add that another significant benefit is increased power density, through increased charge-air density. I would also counter that, when proper advantage is taken (via AFR, CR, & injection timing) of a well engineered charge cooling system, thermal efficiency will be improved.

ernesteugene, I'm glad you're keeping your sense of humour. Your posts do tend to be a little long and rambling. As JSteve2 and Terry have indicated, there is a lot more to consider than EGT. The fact is, it probably isn't practical to instrument your vehicle engine sufficiently to collect the data needed to properly analyze engine and turbocharger performance, unless you're going to get really serious about it.
What is your basis for this statement: "Also with an intercooler the power band RPM lines plotted on a compressor map are typically to the right of the peak ridge of the efficiency island ". We all understand that intercooling increases mass flow vs no intercooling; do you think the engine manufacturer's ignore this when doing the sizing the compressor for an intercooled application?

 

[BrianPetersen]

I'm also not convinced that a charge air cooler would reduce the thermal efficiency, provided that the whole system is considered.

From the idealized point of view, you want to reject heat at the lowest possible temperature (think about the Carnot cycle). Rejecting heat at the intercooler involves a lower temperature than rejecting it via the exhaust pipe.

In a more practical situation, rejecting heat at the intercooler allows the part of the cycle inside the piston-and-cylinder part of the engine to happen at a slightly lower temperature, which probably reduces heat losses from the cylinder and combustion chamber to the coolant by a little bit.

In automotive applications it isn't likely to have much effect on the efficiency during "cruise" and mild acceleration at all, because the engine is only running at part load. In my VW "pumpe-duse" TDI, it operates at very little boost pressure when rolling down the road at constant speed. This means the turbo isn't doing much (this engine uses a variable-geometry exhaust turbine, whose vanes would be largely open - minimal restriction - under these conditions) and it also means the intercooler isn't doing much.

The presence of the intercooler allows a lower boost pressure to be used for a given "target" engine power output level, which means the turbo doesn't have to work as hard (i.e. it can be sized less aggressively and can operate at both lower boost pressure and lower exhaust back-pressure), and by lowering the overall cycle temperature and pressure for a given target power level, it reduces stress (both thermal and mechanical) on the engine internals. It should also reduce NOx by reducing the peak cycle temperature a little.

All production VW TDI engines have an intercooler. The older indirect-injection engines only had it on very rare versions (none of which were sold in North America). I know folks who have retrofitted them, and they make a large difference in how the engine runs and greatly reduce exhaust smoke during acceleration ... unless, of course, you adjust the injection pump to squirt in more fuel, which you can do and make more power for the same "target" smoke level. It's possible to modify an IDI to achieve about the performance of a stock first-generation TDI by doing this.

I won't comment on the equations ... I'm more of a nuts-and-bolts, hands-on, this-is-the-real-world type.

 

[JSteve2]

brianpeterson,
Because of other constraints, especially peak temperature and pressure limits, you may be able to achieve actual efficiencies in an engine as-applied. Also, the ability to change injection timing, etc. may allow you to achieve something.

I'm pretty sure improved Carnot efficiency won't be the thing that gets you there, though. First, the Carnot efficiency would be measured against exhuast temp which inter-cooling doesn't affect 1:1 as indicated above. Second, I'm pretty sure that if I can get back more energy from the combustion by inter-cooling than it costs me to cool the intake air, that I can build a perpetual motion machine on that basis (I haven't thought through all of the consequences of this result, perhaps I'm missing something).

If power density were not such a constraint on high boost systems, I would take my chances getting better efficiency without an inter-cooler. Inter-coolers induce pumping losses in the intake and coolant system, and requires more parts and a larger coolant system. Of course, if displacement has to be driven up to the point where engine size has to be increased, and you have to move that engine then maybe any efficiencies are lost again. But I guess that's just another way of saying power density is a constraint.

 

[ivymike]

if I can get back more energy from the combustion by inter-cooling than it costs me to cool the intake air that I can build a perpetual motion machine on that basis ... perhaps I'm missing something

sooo... would you say the same thing about turbocharging?

 

[JSteve2]

No, I'm recovering exhaust energy that is otherwise lost during turbocharging. If I draw my control volume around the entire engine system, where is the energy coming from for the inter-cooling? If I inter-cool 50 degrees, I get maybe a 40 degree cooler exhaust, which I haven't done the math but I'll eat my hat (although I don't wear one) if the energy recovered from a 40 degree exhaust stream could cool an intake stream 50 degrees.

 

[ivymike]

where is the energy coming from for the inter-cooling?
If I've got this straight, you're saying that cooling the intake charge to increase its density can't increase net power output of the engine, because the energy released by the additional fuel burned (maintaining constant AFR) is necessarily less than the energy removed when cooling the air?

 

[ivymike]

here, a quick test of that theory:
mass air-fuel ratio 20 kg/kg
intake air temp before cooler 50C
intake air temperature after cooler 30C
change in intake air temperature 20C
change in air density 6.6%
intake air flow as cooled 1kg/s
power removed from intake air 20kJ/s
fuel flow rate as cooled 0.050kg/s
intake air flow if not for cooler 0.938kg/s
fuel flow rate if not cooled 0.047kg/s
delta fuel flow rate 0.003kg/s
LHV of fuel 42800kJ/kg
delta heat release 132.5kJ/s

133 > 20...