Competition Timing and cylinder pressures

[NoSeeUm]

Consider:

• it takes a finite amount of time for the initial fuel to begin evaporating, thus producing vapors that will burn
• as the fuel evaporates, it cools, thus reducing available heat, to some degree

The injection of fuel has to be timed so that a portion of the fuel has evaporated and started burning before TDC is reached. As long as the fuel-air mix is swirling around in the cylinder, the fuel will continue to evaporate and burn after TDC, thus maintaining high cylinder pressure even as the piston is dropping.

Yeah OK this makes sense now. In a gas engine the fuel is almost completely vaporized at ignition and the vapor / air is almost fully mixed at the right ratio for combustion. So nearly all the fuel combusts almost instantly at ignition. Hence the dramatic pressure increase at ignition / combustion.



While in the the diesel engine the fuel is only atomized, not vaporized. Vaporization and combustion occur during the same time period. Hence the constant pressure at combustion, as it takes more time for process to occur. And in that regard the actual pressure in the cylinder probably controls the rate of vaporization / combustion. Then throw boost into the mixture with varying charge temperatures I can see how it gets complex to hit the timing exactly.

A 12V's standard injection timing is around 12 degrees BTC, which probably produces optimal operating conditions around 2000 RPM (and, of course, sub-optimal conditions around low idle and governed RPM). Consider if you have your fuel injected at 6 degrees BTC. The engine would run much quieter because ignition likely isn't happening until after TDC, and peak cylinder pressure is considerably lower. Remember, the usual way to reduce a 12V's NOx emissions is to retard the timing, to reduce peak cylinder pressure, which prevents NOx from forming. Of course, power is reduced as well.

#ad


I am thinking this is closer to OEM under NOx restrictions. The higher compression is at, at ignition then the higher the combustion temperature will be and the higher the NOx partition of exhaust gases would be.



#ad


And this is closer to a "electronically" tuned engine or a 12v at its "tuned" RPM. I could see how a CR could really hit it dead on, just like a electronically tuned fuel injected gas engine. And a VP44 engine could get pretty darn close, but not quite like a CR could over the whole RPM range.



#ad


On an electronically tuned engine the combustion pressure spike as illustrated above could be more or less an internet myth I think. I am talking about the over the counter style tuners for liability reasons on the part of the tuner manufacturer.

As I said, things to ponder.

Thanks for the wonderful insight;

Jim

 

[Gary - K7GLD]

Yes I know that. The P-V does not describe the whole process. The idea is that it describes (or tries to) the "engine" heat cycle. Another example would be the Rankine cycle for steam turbines or the Otto cycle for gas engines.



FWIW for the life of me I can not find what a diesel cycle is called other than a Deisel cycle.



Jim

This may be too basic to be of help - but check it out:



The Diesel Engine

 

[NoSeeUm]

This diagram not draw to scale.

I got lazy, but it works for illustration only

#ad


Here is one showing different boosts. Notice the Intake Exhaust Lines are moved upwards. The Intake due to boost, the Exhaust due to drive pressure. Notice there is also a much larger area inside the curves for more Hp output.



The big difference is the compression ratio pressure which is about 729 PSIA vice 240 PSIA. The difference in cylinder pressure is dramatic. It rises 240 psi for every 14. 7 psi increase in boost and is also a killer of head gaskets.



#ad


This one shows the effect of drive pressure or how I think it would look. The boost pressure is 0 and drive pressure is 10. Notice the shortened and raised Power Stroke line and the portion cut off at the bottom right of the curves. I would think that indicates less Hp higher EGT.



Jim

 

[NoSeeUm]

This may be too basic to be of help - but check it out:



The Diesel Engine

LOL



Gary,



Go back to my first post in this thread and follow the links. I have been staring at this very webpage off and on for a couple of months now. You should also click on the ones for the Otto (Gas engine) cycle I posted if you are interested enough.



Jim

 

[Gary - K7GLD]

LOL



Gary,



Go back to my first post in this thread and follow the links. I have been staring at this very webpage off and on for a couple of months now. You should also click on the ones for the Otto (Gas engine) cycle I posted if you are interested enough.



Jim

I was afraid it wouldn't help - but worth a try...

 

[GHolcomb]

Good post by fest3er! Alot going on in getting the most out of our engines!



Jim... Please clarify... boost pressure vs. drive pressure? Also... Fuel=50 vs Fuel=100



Thanks!

 

[NoSeeUm]

Jim... Please clarify... boost pressure vs. drive pressure?

Thanks!

Not sure what you mean, but for all of my drawings boost pressure = drive pressure except for one. This is means the same pressure on the pistons for the intake and exhaust strokes. So the net pressure across the engine is zero. When drive is greater than boost, the engine most force the gases from the engine. This requires Hp to do that.



#ad


In this drawing notice how the lines for the Intake Stroke and the Exhaust Stroke are at the same pressure. They are superimposed over each other.



#ad


In this drawing notice how the Intake Stroke and the Exhaust Stroke are not at the same pressure. The Exhaust Stroke is at a higher pressure. My conception of the effects of a drive pressure ratio greater than one. It breaks the whole curve up.

Also... Fuel=50 vs Fuel=100

Thanks!

#ad


This one is 50% fuel no timing. The 50% is not a real number, just relative.



#ad


This one is 100% fuel no timing. The 100% is not a real number, just relative. Notice the lengths of the Heat Input (RED) and the Heat Output (Blue) lines are much longer. In this case the more fuel was added and the duration for combustion was longer. Combustion is occuring almost through out the entire Power Stroke. As you can see, and ends fairly close to the . 98 liter displacement vertical line (BDC).



#ad


Here is a drawing on how Hp is utilized in this type of an engine. But I quick note, I made a mistake. I believe the very bottom horizontal line should be even with the 14. 7 psia line vice the 0 psia line, but it works for illustration.



Top Left

The diesel engine curve for reference



Bottom Left Orange Shaded Area

The gross Hp of the engine. It is basically porportional to how much energy from fuel combustion that gets added to make the whole thing work.



Bottom Right Blue Shaded Area

The loss Hp of the engine. Basically, it is proportional to how much energy from fuel combustion that does nothing but make the engine work.



Top Left Green Shaded Area

The shaft Hp of the engine. The result is proportional to the total area of Orange minus the total area of Blue or just area inside the curves indicated by the total Green area.



You will read often that what powers an internal combustion engine is the difference in pressure between the Compression Stroke and the Power Stroke. The difference is created, by of course combustion. The Green area is this difference.



Jim

 

[fest3er]

NoSeeUm,

What the graphs don't seem to show is the high pressure spikes that should occur just before TDC compression. If you are working with a N/A engine and assume 14. 7 PSI absolute, 16:1 compression would yield about the 240 PSI you indicate. But once ignition commences, the pressure should rise (or even spike) up to, potentially, 8000 PSI or so; this comes from a calculation I once made while attempting to determine how much cylinder pressure is needed to lift the head off the block (stretch the head bolts). Given the strength of the steel bolts and how many of them hold the head down around a cylinder, 8K-10K PSI cylinder pressure was my rough guess. If a N2O-injected engine is too aggressive on injection timing, the fuel could be burning well before TDC of compression; inertia combined with another cylinder still pushing down in its power stroke could well drive the 'compressing' piston on upward, causing a tremendous spike in pressure: enough to stretch the head bolts and blow out the gasket.

If you think about it, the 8K-10K pressure handily explains why diesel injection pressure is so high; it has to overcome the rising cylinder pressure to be able to continue atomizing the fuel. If it didn't, the fuel would come out in a solid stream, which won't evaporate as well. Think of a full hypo, and what happens when you squirt it out slowly and when you squirt it out as fast as possible.

Also bear in mind the total mechanical disadvantage the piston/conn-rod have at TDC; specifically, all the pressure in the universe will not make the piston turn the crank at TDC. Now, *past* TDC, the mechanical advantage rises until the conn rod is perpendicular to the moment arm of the crank, when it begins to fall again.

If I guess correctly, optimal bore and stroke must differ between spark and compression ignition in an otherwise identical engine. A spark engine needs as much mechanical advantage it can get near the top of the stroke. A diesel engine applies power over a longer portion of the stroke.

Maximizing cylinder pressure is 'generally' important in a spark engine. But a diesel is all about controlling fuel injection timing to smoothly ramp up to peak pressure, and maintaining combustion during a portion of the power stroke.

Time to make supper. The ancient ones are getting restless.

 

[nwpadmax]

Yeah, maybe I'm reiterating the obvious, but I agree with Fester. The diesel cycle plots are just idealized cycles (which explain a lot) but don't in fact mirror reality all that well.

If you go and do a Google search on "diesel cylinder pressure" or "IMEP", you'll turn up a lot of interesting research. Most of it is focused on emissions, but don't let that get in the way. You'll invariably find some nice real-time cylinder pressure plots that show far from constant-pressure burns. The way the fuel in injected (one shot vs. pilot + main for CR guys), what the compression ratio is, and many other things have an effect on the real-time cylinder pressure.

For sure, there are spikes in cylinder pressures for advanced-timing injection.

I for one would sure like to understand the cylinder pressure traces for a CR engine with stock injectors and wide pulse width (and advanced timing is required for big pulse widths) versus a big injector that can be started much later, with a much narrower pulse width. At the same fuel delivery, such an experiment would tell you a lot about atomization and "the burn. "

 

[NoSeeUm]

Keep in mind these P-V curves are not actual, they are Ideal.



Fes3ter;



AFAIK the maximum pressure of a VP44 is 300-330 bar or 4351-4786 psi. I am not sure of the pop pressure of the injectors, but it seems that it could only be less. The purpose of the high injection pressure is for more complete atmization vice over coming cylinder pressures?



Given a constant pressure combustion as I show, this leaves injection pressure 4100 psi or so above cylinder pressure at zero boost and about 3000 or so psi over at 60 psi boost. Given your 8000-10000 psi cylinder pressure number no fuel could be injected past the initial onset of the injection event by a VP44 truck. How would it fuel?



A CR injects at 1600 bar or about 23206 psi up 1800 bar or 26107 psi. Really optimizing the atomization I would think, but given that the compression ratio is close to the same as a VP44 as would be the cylinder pressure.



Are you referring to the clamp pressure of the head studs due to the tensional force placed on them due to torqueing? Or the yield strength?



NWPADMAX;



I can't find any links depicting a good P-V curve. I am a boob sorry.



I did find one that cylinder pressure was in the neighborhood of 8000-9000 Kpa or 1160-1305 psi for a big displacement low RPM marine diesel. There was no great spike. It just what looked like a normal adiabatic compression curve with a hump in the middle. The hump was smoothed out a bit over the top due to the effects of combustion. Then the curved followed what looked like a normal adiabatic decompression curve. It was not a typical P-V diagram, it was a Pressure-Crank Position diagram.



Care to post a link or two to illustrate your point?



FWIW this P-V Curve shows a spike. Its not a diesel.



Thanks guys, now we are getting somewhere with this discussion;

Jim

 

[NoSeeUm]

I did a couple more drawings.



#ad


This is a standard P-V which I have drawn on a different scale. It is 0 timing and 50% fuel.



#ad


This P-V shows a rapid pressure increase at injection. It is -30 timing and 50% fuel .



I don't think I got the Heat Input - Heat Output part correct after looking at it some more. IE: In this drawing Q1 >>> Q2. But it works somewhat for illustration. If I made Q1 = Q2 then it would make the right upper point of the curve way higher, because the Q2 line would extend to be about 5 times longer. It should be the equivalent length of about 35 degrees of duration. And to that end make this particular heat engine even more optimistic than it already is.



Notice how vastly greater the area inside the 2nd curve is compared to the 1st curve. This would correspond to a very large Hp increase, just eyeballing it, of at least over 10 times. I am not sure any known fuel could actully release that much energy that fast or be so efficient in doing it. But I am not sure at all.



I would think that this curve is nearly impossible based on my experience in the real world.



LOL Yeah I do live in the real world occasionally



Opinions?



Jim

 

[nwpadmax]

Jim, the only curves I have seen that I can grasp are pressure vs. time.

Lemme see if I can dig up a link or two.

 

[NoSeeUm]

Well... . I got to re-thinking my heat balance and advanced timing. Now I really don't know what to think exactly, but take a look at the curves.



#ad


This one is for reference. It has 0 timing, 50 fuel, and 15 duration. I changed the scale once again.



#ad


This one has -30 timing, 50 fuel and 15 duration. I am starting to understand maybe a little bit of more. Notice the increase cylinder pressure. IMO the area inside the curve is just too much of and improvement over the reference drawing. I must be missing some component that has to do with negative thrust during the compression stroke due to BTDC combustion. So for now consider the drawing for illustration only.



#ad


This one has -30 timing, 50 fuel and 15 duration. It also has two injection events just for fun. Ah the versatility of a CR.



But.....



By far in the curves of simply advancing the timing has a very great effect. This is even over a standard diesel heat cycle, I feel I must be missing a critical ingredient. Or the timing would be set at -170 for an even greater effect.



How close am I getting?



Jim

 

[fest3er]

Keep in mind these P-V curves are not actual, they are Ideal.



Fes3ter;



AFAIK the maximum pressure of a VP44 is 300-330 bar or 4351-4786 psi. I am not sure of the pop pressure of the injectors, but it seems that it could only be less. The purpose of the high injection pressure is for more complete atmization vice over coming cylinder pressures?

The pop-off pressure is in the range of 5000 PSI, IIRC. I thought the P7100 puts out around 12K-15K PSI, and the VP44 perhaps a little higher. CR pressure, of course, commonly ranges up to 26K PSI these days, or higher.

Are you referring to the clamp pressure of the head studs due to the tensional force placed on them due to torqueing? Or the yield strength?

More accurately, I believe I am referring to the elastic tensile strength of the bolts. I suppose 'clamp pressure' is close enough; but clamp pressure also includes how much torque is applied to the bolt.



I had all kinds of other stuff typed, but deleted it in favor of the following SWAG. As a SWAG, it might not be in the ballpark, but I hope it's at least a sign pointing the way to the ballpark.



The table below represents the cylinder pressure, in PSI, required to produce 500 lb-ft of torque during the first 80 degrees ATC of crank rotation, based on the 6B's bore/stroke (4. 02/4. 72). I also assume that the maximum advantage of the con-rod happens at about 79 degrees ATC (when the con-rod is perpendicular to the crank's arm). This is 'simple' math. I didn't fully take into account the complex interaction of the piston, con-rod and crank positions and angles. The equation I used is:

TQ / AREA / ARM / SIN(ANG * CRC), where

• TQ is the desired torque output
• AREA is the area of the piston ((4. 02/2)^2)*PI)
• ARM is the moment arm of the crank, in feet (4. 72/12/2)
• ANG is degrees ATC in the power stroke
• CRC is a guess at a con-rod constant (90/79) that accounts for the con-rod being perpendicular to the crank at (a guessed) 79 degrees

 Degrees   CP (PSI)

   ATC

 2 5038. 37

   11  923. 17

   21  493. 95

   31  346. 5

   41  275. 19

   51  235. 94

   61  213. 86

   71  202. 87

   81  200. 47

In theory, these pressures should produce smooth output torque. At least between 2 and 79 degrees ATC. Before 2 degrees, the con-rod doesn't have enough mechanical advantage to generate much torque. After 79 degrees or so, cylinder pressure would have to start rising to maintain 500 lb-ft torque; instead, torque drops off until the next cylinder fires.



According to this 'theory', there is about a 40 degree gap where torque falls off before the next cylinder starts making power. So to maintain an *average* of 500 lb-ft of torque, greater torque must be produced earlier to take up the slack in that 20 degree gap. As an aside, this 40 degree torque gap might explain why the Cummins breaks stuff so easily.



So, you should now have calculated cylinder pressures that are in the ballpark of what's needed to produce 500 lb-ft of torque. In a one-lunger. An exercise for the student is to determine why these pressure values are too low for a multi-cylinder engine.

 

[NoSeeUm]

.....

I had all kinds of other stuff typed, but deleted it in favor of the following SWAG. As a SWAG, it might not be in the ballpark, but I hope it's at least a sign pointing the way to the ballpark.

.....

Thanks fest3er;



A SWAG is only as far as I have gotten to date. I am still trying to build mental conceptions. I have not even tried to put in real numbers yet.



It will take me a bit of time to digest your cylinder pressure curve.



Jim

 

[NoSeeUm]

Cycle ProcessCompression Heat Addition Expansion Heat Rejection



Otto (Petrol) adiabatic isometric adiabatic isometric

Diesel adiabatic isobaric adiabatic isometric

Brayton (Jet) adiabatic isobaric adiabatic isobaric



An adiabatic process is a process in which no heat is transferred to or from working fluid. Got this one



An isobaric process is a thermodynamic process in which the pressure stays constant: Δp = 0. Confusion



An isometric process is a thermodynamic process in which the volume stays constant; ΔV = 0. Confusion



More confused than ever;

Jim

 

[fest3er]

Cycle ProcessCompression Heat Addition Expansion Heat Rejection



Otto (Petrol) adiabatic isometric adiabatic isometric

Diesel adiabatic isobaric adiabatic isometric

Brayton (Jet) adiabatic isobaric adiabatic isobaric



An adiabatic process is a process in which no heat is transferred to or from working fluid. Got this one



An isobaric process is a thermodynamic process in which the pressure stays constant: Δp = 0. Confusion



An isometric process is a thermodynamic process in which the volume stays constant; ΔV = 0. Confusion



More confused than ever;

Jim

Consider an internal combustion piston to be an imperfect isobaric process. Ideally, the cylinder pressure would be at maximum at TDC and stay there until BDC; reality rears its ugle head here, in that fuel burns too fast. Or too slow, depending on your perspective. Almost ideally, ignition would occur such that pressure reaches a maximum just after TDC and stays there until just before BDC; again reality kicks us in the gluteous maximus in that fuel oxidation is not that controllable nor is the probability of fuel molecules being near free oxygen. The reality is that cylinder pressure peaks rapidly at or just after TDC, and then falls arithmetically (or exponentially?) as the piston accelerates downward.



A rocket engine might be closer to being an ideal isometric process, because the combustion chamber is fixed in volume. A missile (HAWK, Stinger, SAM, etc. ) is driven by, essentially, a controlled explosion; the fuel doesn't last long, but it is fed very rapidly to maximize combustion pressure.



I suppose a hot air balloon could be considered an example of and isometric process. The bag's volume is constant; what changes inside is the pressure (well, OK, what *really* changes inside is the density). Contrast this with a toy balloon, which is an imperfect isobaric process; the pressure is relatively constant, whereas the volume changes. Of course, if you release the balloon before tying it shut, it'll take off; the pressure is relatively constant whilst the volume shrinks.



So, how does this relate to a gasoline engine?

• Compression is generally adiabatic; P and V are both changing.
• Ignition is generally isometric; combustion takes place while the volume is nearly constant.
• Power is generally adiabatic; heat is converted to mechanical energy.
• And exhaust/intake is generally isobaric.

In a diesel engine, the ignition process starts out isometric (same as a gasser) but seems to be followed by an isobaric process (more fuel is burned, while volume increases and pressure is nearly constant). This isobaric process shortens the subsequent adiabatic power process. The extra thermodynamic process in the diesel's reciprocation probably accounts for the extry work the diesel produces.



As always, treat this as the ramblings of a madman. It's probably not very accurate, scientifically, but I hope illustrates the ideas well enough to jump-start your neural processes.

 

[GOT-Torque]

Nice thread, I finally understand all this stuff:-laf



Thinking outside the box for a moment, could someone drill a hole in the head (say 1/16" diameter) to one of the combustion chambers and then mount a pressure sensor of some sort on the head to monitor cylinder pressures? I wonder how high they get?



I'm sure its possibly to calculate how much cylinder pressure it would take to lift the head (stretch the ARP studs) and blow the HG. Any ideas?

 

[fest3er]

... I'm sure its possibly to calculate how much cylinder pressure it would take to lift the head (stretch the ARP studs) and blow the HG. Any ideas?

There are 6 bolts around each cylinder. Multiply that by the tensile strength of each bolt (or stud). That should give you an idea of the maximum clamping force per cylinder. Divide by 50. 776 (area of the bore). That should give you an idea of the cylinder pressure (PSI) needed to float the head. You might guess that cylinders 1 and 6 are more apt to pop first, 2 and 5 next, and 3 and 4 last.



For kicks, find the shear strength of both the bolts and the cast iron used in the block. Get the length of the bolt threads that are in the block when torqued. Get the inner (minimum) diameter of the bolt threads and the inner (maximum) diameter of the block threads. Compute the surface area of each thread's cylinder (Pi*diameter*length). Divide the bolt's shear strength by the bolt's thread area to find how much it'll take to strip the bolt threads. Divide the block's shear strength by the block's thread area to find how much it'll take to strip the block's threads. You should find that both are far greater than the tensile cohesion of the bolt, even if you derate the computed strengths by a significant amount. (In other words, you'll stretch the bolt/stud like putty before you strip the threads. )



These are, of course, ball-park computations. You should consult a mechanical engineer if you want or need actual values.

 

[miteshshetty]

Intake and exhaust pressure

How do I calculate the intake and the exhaust pressures of the diesel engine considering ideal diesel cycle?