Boost pressures and math just for fun

[dracor85]

Ok guys I was thinking about boost pressures in relation to the compression ratio of the engine. So all of my calculations are at sea level atmospheric pressure of 14.7psi.

calculations are air pressure (PSI) first then compression ratio second

Idle no boost
14.7psi x 21.1 = 310.17psi
14.7psi x 18.0 = 264.6psi
diffrence = 45.57psi

so you can see there is a fair diffrence in the final pressure inside the combustion chamber with out any boost from the turbo

at 15psi turbo boost plus 14.7psi atmospherific
29.7psi x 21.1 = 626.67psi
29.7psi x 18.0 = 534.6psi
diffrence = 92.07psi

here you can see as the boost rises the diffrence in the overall combustion chamber pressure increses


now if you run a 18.0 to 1 engine at 20psi the combustion chamber pressure comes out about the same as a 21.1 to 1 engine at 15psi

20psi boost + 14.7psi atmospheric = 34.7psi
34.7psi x 18.0 = 624.6psi


now for a comparison, all those duramax and cummins folks running 60psi boost in a 15.5 to 1 compression ratio engine here is their final combustion chamber pressure.

60psi boost + 14.7psi atmospheric = 74.7psi
74.7psi x 15.5 = 1157.85psi



so to sum this up all you guys that are running a 18 to1 engines at 20psi are just as safe as us with standard 21.1 to 1 engines running 15psi. If Im off on any of this let me know guys.

Justin

 

[WarWagon]

You forget the difference in the outlet air temperature. PSI may be the same, but, less O2. A non intercooled turbo vs. a turbo setup with an intercooler is not a good comparison.

 

[6.2 turbo]

Ever since I lowered my compression I've been a little unsure about it, it looks like Bill Heath is right. This an area of much debate for me. Looks like I could go back to stock compression,and drop my boost from 60 to 40 and cyl pressure would still be less,than before. This calculation also holds true because I went from 18 to 1 compression to 16 to 1 and my head gasket still blew almost right away.

 

[SmithvilleD]

Regarding the head gasket failure issues, isn't peak cylinder pressure (& temp), an important variable to consider?

My point being peak pressure happens after the combustion process has begun. Certainly compression ratio & boost are factors, but other variables like fuel rate, charge air temp, inj timing/duration, etc., are in play as well.

Has anyone seen any data showing peak cyl pressure for a 6.5? Seems a fairly safe guess it would be =/> 2000 psi.

If anybody has seen any peak cyl pressure data for our 6.5, please share in this thread.

 

[625fireman]

Justin not pickin at you by any means but your math only shows compression psi as would be in a air compressor. There is a big jump in cylinder psi once the fuel gets ignited which is what drives the piston back down. You are on the right track starting though, you just need to factor in the cetane rating of the fuel along with volume of fuel and % of effciantcy of fuel ignition. I can't help any further though with numbers for this though as I dont have any of the figures for the fuel...... Great start though. If I would guess at cylinder PSI with ignition I would say its pushing 3000 with decent boost psi's.

 

[Wrecker]

I copied this because it is a reasonable description and I was too lazy to do it myself.

On Edit: I can however speak to it with a reasonable amount of clarity if it doesn't make sense to someone.

http://www.physicsforums.com/archive/index.php/t-155498.html

"You can express the mean effective pressure (in kPa) as:

mep = (P*n)/(V*N)

where:
P is the power in kW
n is the number of crank revolutions per power stroke (ie n=1 for a 2 stroke, n=2 for a 4 stroke)
V is the displaced volume per cycle in cubic meters
N is the engine speed in revs per second

Now, this is the mean effective pressure. This is a very useful parameter when dealing with engines, because comparisons of mep are not engine-specific, such that comparisons can be made between different sizes and types of engines.

It follows that the mep can be expressed in terms of torque, as follows:

mep = (6.28*n*T)/V

where T is the torque in Nm.

Now then, peak pressures are much more difficult to calculate, because of the sheer number of different factors which can affect it: Engine speed, load, throttle position, fuel energy content, knock, volumetric efficiency, atmospheric pressure, residual cylinder temperatures, valve timing and overlap, compression ratio, spark timing, swirl, squish, tumble....

If you want values for peak cylinder pressures, you really need to be measuring them."

 

[buddy]

Peak cylinder pressures are at least twice those numbers at ignition.

 

[Wrecker]

Is this based on measurement or theory?

 

[buddy]

Both, many diesels are built for peak cylinder pressures of 1800-2500psi.

And even using the equation in the post above its like 1400psi for 300hp at 3400rpm, which would be at the engine/flywheel, which seems reasonable for a modded 6.5 running 15psi boost and max fuel on reflashed PCM.

 

[Wrecker]

Cool, thanks for the clarification:thumbsup:

 

[SmithvilleD]

Has anyone heard recent information about a glowplug design that can also provide some feedback on actual cylinder pressures?

Seems like I've read new technology teaser sort of articles occasionally that hint such a product will come to market. Intriguing idea if it could provide useful information at a cost reasonable for us diesel enthusiasts.

 

[dracor85]

Thanks guys for all of the replies, I was in no way trying to calculate the cylinder pressure once fuel is introduced and combustion started because of the amount of variables involved, I was just trying to figure out how much boost would be comparable depending on what compression ratio you were running.

Thanks Wrecker for the formula for figuring out cylinder pressure, and thanks buddy for clarifying the formula Wrecker posted.

SmithvilleD I vaguely recall something about the glow plugs that read cylinder pressure, I think there was something about the new 2011 Duramax's having them I'll have to see if I can pull up the article.

Justin

 

[buddy]

And the original math isnt quite that simple, as you have to take into account temperature and not just volume change. Higher compression obviously squishes it more but also make it higher temp so even higher pressure.

PV=nRT

V is the factor you were varying.

n is the same, R is the same, but T changes significantly.

P=nRT/V

So if there is atmospheric plus 15psi boost, 29.7psi in the cylinder at intake closure shortly after Bottom Dead Center at 200F. Then V is a 1/21 at TDC so the calculations are correct up to there. Then change Temp from 366Kelvin to 550Kelvin, and the pressures are 50% more just before ignition.

at 15psi turbo boost plus 14.7psi atmospherific
29.7psi x 21.1 x 1.5 = 940psi
29.7psi x 18.0 x 1.5 = 802psi
diffrence = 138psi

20psi boost plus 14.7psi atmospheric
34.7 x 18 x 1.5 = 937psi

So the difference is even more than you thought

 

[SmithvilleD]

There was an article RJ Schoolcraft wrote in the 3rd of Jim Bigley's 6.5 manuals that discusses the theory on why lowering the compression a bit may result in some durability benefits.

Believe the short version was theoretically, at equivalent power output (lower comp ratio running a bit more boost), the lower comp engine would see ~ equivalent average cylinder pressures, but lower peak pressure & temperature.

The idea being that head gasket & piston/bore seizure failures are likely to be somewhat proportional to how high the peak cyl pressures/temps are.