mrx

Hi,

i have a question concerning engine stress and power in combination with compression and boost. As you know i want to run moderate boost levels on a high compression b16 engine. So in read very much (here in the forum and a lot of websites) but a littlebit is anyway not clear to me...

So, fact is:

1) you can make more power with bosst then with compression, thats fact but i don't want exorbitant high power outputs, i want a street driveable car so thats not my goal.
2) another fact is that power destroyes an engine, not boost alone.
3) more boost=> more fuel=> more power
4) boost is limites by compression because higher compression=>higher pressures and temps so you get knocking.

Whats about the following example:
Get the same engine, turbo and so on... Original compression of the engine was: 10:1

1st CR=7.7:1 / Boost=16psi / Effective compression=13.08:1
2nd CR=10:1 / Boost=7psi / Effective compression=13.04:1

So if you know that setup 1 is running without knocking and the engine don't blow up... isn't it so that you have:

1) Less Power on setup 2 because you have lower boost (that menas less fuel) against setup no. 1
2) better off boost performance because of higher static compression on setup no. 2

And now the question i don't know the answer:

Whats about the stress of the engine and the Temperatures on both of the setups?

What setup is "hotter" i.e. exhaust gas temp. and why?
what setup is more stress to parts like pistons, rods and so on and why?
can i take the effective compression to see how much stress the engine has? or isn't it compareable?

Thanks
Malte.

JonnyCoupe

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by mrx &raquo;</TD></TR><TR><TD CLASS="quote">

So, fact is:

1st CR=7.7:1 / Boost=16psi / Effective compression=13.08:1
2nd CR=10:1 / Boost=7psi / Effective compression=13.04:1
</TD></TR></TABLE>
WTf you on, boost is the pressure ontop of atmospheric pressure, its not the absolute pressure of the air in teh manifold..

7psi of boost is air at 22.7psi.

All the boost guage shows its 'dial pressure' ie a pressure differential.
An actuator/wastegate however reads absloute pressure so some days you get the same pressure in your manifold but your boost guage varies.

Ie up a mountain where pressure is lower the pressure in engine will be the same, but the boost guage will read higher because its the difference from atmospheric (now lower) and MAP.

mrx

I don't see where your problem is on my calculation?!

i know that boost pressure is the pressure above athmosphere... but what did this have to do with the effective compression and the engine stress?

i found another formula for effective compression here in the forum:
E = C((B / 14.7) + 1)
Where E= Effective Compression, B= boost psi, and C= Static compression. Also remember that 14.7 is equal to 1 bar of boost.

with this formula i get allmost higher effective compressions with my example engine then with my formula... but i even don't see what this has to do with my question...

maybe you exmplain what do you want to say... maybe you show us your calculation so we can see what you did...

Bye
Malte.

JonnyCoupe

Surly you should look at cylinder static compression figures?

Ie 10:1 at 7 psi boost = 230psi
7.7:1 at 16 psi = 238psi

Of course this doesn't account for valve effects though and volumetric efficient is never 100% so you can never totally fill the cylinder os a induction stroke.

mrx

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JonnyCoupe &raquo;</TD></TR><TR><TD CLASS="quote">Surly you should look at cylinder static compression figures?

Ie 10:1 at 7 psi boost = 230psi
7.7:1 at 16 psi = 238psi

Of course this doesn't account for valve effects though and volumetric efficient is never 100% so you can never totally fill the cylinder os a induction stroke. </TD></TR></TABLE>

Ah ok, i see... but isn't it the same? i mean if i understand the stuff right effective compression ratio is only another form of display the cylinder pressure or am i wrong?

can you please tell me the formula to calculate the cyl. pressure?

if i take your example then the both cyl. pressures are nearly the same (only 8psi difference) so can i expect that the stress for the engine is nearly the same?! (mainly this was my initial question )

Bye
Malte.

PH_B16-VTi

E = C((B / 14.7) + 1)

1:
C= 10:1
B= 7psi = 0,48bar

10*1,48 = 14,8 E.CR

2:
C= 7.7:1
B = 16psi = 1,09bar

7,7*2,09 = 16,09 E.CR

There you go .....

ZoRG

the stress on components will not be only due to boost, for instance rods have a higher stress on higher rpm's, pistons stress comes mainly from increased heat, which your higher boost engine will create simply due to its higher power output. Crank have both high rpm and high boost stress... Why are you so concerned though? Are you actually building a engine, or do you just want to play with the numbers, if you want to build a strong engine, then just overengineer everything .

At only 8psi the stress in the high boost engine will be much more since it will create more heat and power.

mrx

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by PH_B16-VTi &raquo;</TD></TR><TR><TD CLASS="quote">E = C((B / 14.7) + 1)</TD></TR></TABLE>

Ok, thats the exact formula i found here in the forum also (my second post)
in my book about engines i found this formula to calculate the effective compression ratio: Eeff = Estatic * (PL/P0)^(1/1.4)

The P's are absolute pressures.
with this formula i get a bit other values...

does anyone have the formula to calculate the cylinder pressure?

Thanks
Malte.

mrx

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ZoRG &raquo;</TD></TR><TR><TD CLASS="quote">the stress on components will not be only due to boost, for instance rods have a higher stress on higher rpm's, pistons stress comes mainly from increased heat, which your higher boost engine will create simply due to its higher power output. Crank have both high rpm and high boost stress... Why are you so concerned though? Are you actually building a engine, or do you just want to play with the numbers, if you want to build a strong engine, then just overengineer everything .

At only 8psi the stress in the high boost engine will be much more since it will create more heat and power.</TD></TR></TABLE>

Hi,

you are right, i'am building an engine at the moment... and thats my problem... i don't know what compression i want to go...
What do you mean with your last sentence? "the high boost engine" do you mean this on with the high compression? why does he produce more heat and power then the one with the lower compression?

i'am a littlebit confused

ZoRG

no no, I meant the high boost low compression engine. by the 8 psi i meant the difference between 230 and 238...

Johnyquest

higher static compression will always yeild faster spool, more power for a given boost level, and better off boost response. What's the question again? Alright, just kidding. But honestly, if you want a daily driver with some power, 10:1 is a great compression if you know how to tune well or you know someone who is good. Low compression just isn't necessary if you have 93/94 octane gas regularly available and a good tuner. Low compression cars (especially with HUGE turbos) are almost always dogs off of boost, and suck to drive.

mrx

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ZoRG &raquo;</TD></TR><TR><TD CLASS="quote">no no, I meant the high boost low compression engine. by the 8 psi i meant the difference between 230 and 238...</TD></TR></TABLE>

hi... oh ok... i see... Thanks very much guyes...

i think that answers my question.

JonnyCoupe

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JonnyCoupe &raquo;</TD></TR><TR><TD CLASS="quote">

Ie 10:1 at 7 psi boost = 230psi
7.7:1 at 16 psi = 238psi
</TD></TR></TABLE>

Made a error there as was doing in head with a heangover

10:1 CR is ............... 10 x (7+14.7) = 217psi
7.7:1 CRis ................7.7 x (16+14.7) = 236psi

With the 10:1 however you get this compression for much more of the rev range, negating turbo lag. The only trouble is flame propogation and timing becomes more critical with less area for error with the higher engine compression ratios.

Bailhatch

Do static compression #s really matter as far as how mechanicly stressed the motor is? For example, I have read (on here, naturally) that if you are running a really agressive cam set in an NA application then you would need to have higher static compression because once the motor is running the overlap tends to let some of the pressure out the exhaust.

I realize that with a turbo motor the 'momentum' of the charge for cylinder filling is not really important but woulden't you lose some anyway especially with a stock NA cam? If not please fill me in...

I like the idea of chosing what works by other experiance rather than the numbers anyway, some of the turbos people pick based on the flow maps just don't perform as predicted, and some that shoulden't work on paper are great on the street, maybe this is the same kind of thing...$.02