Re: Correcting for 80% VE - just more questions then

[QSHIPQ@aol.com]

Based on a couple of recent posts, I will, for the B.O.D., revise my original
post numbers, only slightly, however, and hardly necessary really.  Please be
advised, my original post (and all subsequent) assumed a VE of 85%, which for
the 10v motor has been argued as prolly high (heck I even have on occasion).
 For a 20v motor about right, and my RS2 equations for 4 valve motors do use
this number> has to be less for the 10v engine, I can accept 80% easily.  I
won't change the givens, cuz I make no assumptions of them, only in the
results section of them.  The givens are as follows:

 S.O.c claims and audi facts, THESE STAY THE SAME @:
 *  26psi in 2nd thru 5th
 *  2.5 bar PT (given)
 *  16+psi in 1st gear
 *  .5 PD accross modded IC
 *  Audi --> RS2 gets 310hp @ 460CFM (given)
 *  1325 EGT
 *  Audi --> MC motor = 134ci (given)
 *  Stock WG spring
 *  "...An RSR turbo, a Ned Ritchie turbo, blows the RS2 into the weeds"
 *  "... the car now will outrun just about any other MC engine around.
Especially any of the "Rs2"
wannabe's.........  "

Correcting my post to the "homework" section, with a 80% VE the questions
become:
 
 The homework would be
 A) How does a 136ci motor flow 482CFM <was 513> in 10v trim?  Can it stock?
 B)  Why such a large PR?  I thought Density Ratio was more the goal?
 C)  Given a 2.79 PR vs a 2.50PR with the same output, which turbo should one
use?
 D)  Then, how does an S.O.c. car "blow the RS2 wannabes (me by definition I
suppose) into the weeds?"
 E)  How does a PT that can control boost to 2.50PR by definition, control
boost to 2.79PR with a stock WG spring?

Newly arrived back to the list, Bruce queries:

>On this issue of flow, and Scott's assumption that a motor is flowing 513
cfm.  >Assuming the number is correct, >again, the number has only been
assumed so far, why is it a bad thing?

Bruce, for engine physics, the formula for base CFM is:
ci motor * peak power output * .5 (filling motor only .5 times for a 4 cycle
motor) * VE (I assumed 85%, some want to see 80%, I'm game)/1728 (converts to
CFM) = Baseline CFM N/A motor

Bruce, we have a given of ci = 136 and PPO = 5500, so the ONLY assumption we
make is VE, about 1/2 the material I have says 80% the others say 85% for ohc
motors.  If we look at the 20v and the 10v with the same displacement, it
looks like the 80% applies well to the 10v and 85% well to the 20v (cross
checking it against a general "guru" accepted 1.5 correction factor for CFM
vs HP, this looks to be a really good set of numbers).  90% would be REALLY
high, 70% would be REALLY low, so this is OK with me, and won't really change
things much.  So, mathmatically:

136Cubic Inches (corrected to MC = 2.22L) * 5500rpm * .5Fill * .80VE/1728to
CFM = 173CFM is baseline 10v motor

Another piece of math from any turbocharger book shows:

Baseline CFM * Pressure Ratio = boosted engine CFM  (look hard so far,
everyone, there are NO mods here)  

So:

173 * 2.79 = 482CFM

Then a10vt motor with a 2.79PR flows the corrected 482CFM.  Some more
calculations from someone here can determine  what the runner and effective
valve size have to be (this is where I get paid, so someone else can do it).
 Can it be done stock?  IF the runner and effective valve size HAS to be
increased over stock, then 482CFM is now min, cuz changing runner and/or
effective valve size from a stock motor, increases it's baseline
efficiency>larger CFM baseline AND boosted.  Cool eh?  So really, Motor Mods
aren't going to help in the argument at all.

> I mean, my concern with non-turbo motors 
>in the old days was creating a lean condition / detonation, which in turn
led to >destruction in some form.   Does the >turbo thing create some
different problem?  
>If you have good exhaust gas temps [egt's], and, from what I see, Eric's 
>are very good, fuel certainly is not a >problem, correct?
>Where am I going wrong here?  I mean, no lean condition; no detonation; good
egt's; isn't that really more to the point >of it all.  

Interesting questions Bruce.  What you are missing is a couple of things that
might/should concern you.  2.79PR =26psi gives you a BMEP (someone else can
calculate that, I think it's too funny to put a number to) on a 7.8CR motor,
what does that mean for you with 8.4?  Now, the accepted rule of thumb is:

Turbocharged cars perform best with the lowest Pressure Ratio at the Highest
Density ratio (this assumes CFM to be a constant)

(To confirm this, see mathmatical example in IC efficiency section of my
original post, it is correct as printed.)

So, a hybrid k26/27 turbo needs 2.79PR (26psi corrected) to put out the same
flow as a RS2 at 2.50PR (21.75 corrected), which should one choose given the
above ROT?  Specifically, how does a 2.79PR turbo "blow the RS2 into the
weeds, when, in fact, the PR is higher, and the DR is lower too, by
definition.  

What does this mean to the confused?  It really doesn't change my original
conclusion that the wrong turbo is doing the wrong job on the wrong engine.
 A box stock RS2, in this case, MODS ASIDE, will perform on Eric's car better
than the one he has.  By definition:  Turbocharger Efficiency is higher, BMEP
is lower, PR is lower, and Density Ratio is higher.  These are all win/win
concepts to a turbocharged application from what I know, and most gurus will
tell you.  

Now, after reading thru the original post to the claims, my VE correction
doesn't affect much of it at all, but fair is fair.  Assuming the givens
above, the corrected (from 85% to 80% VE) airflow rates are as follows:

WAS:
"Baseline N/A motor airflow for a 134ci (MC) engine at 85% VE = 184CFM
********  CORRECTED TO:***************************
 Baseline N/A motor airflow for a 136ci (MC) motor at 80% VE = 173CFM
(Since 173 * 1.43PR (stock MC turbo motor - bentley) = 244CFM boosted/1.5
(est correction to HP) = 164.9 pretty good, thanks Orin.

So, a claimed 2.79PR, by math means:
173 * 2.79 = 482CFM @ 5500 (RSR turbo 26/27 hybrid @ ?? efficiency @ 5500

Compare say the RS2 turbo 
184 (remember I keep 85% efficiency here for the 4v head) * 2.50 = 460CFM

Ok, and the stage II mods (assuming 2.0 bar)
173 * 2.00 = 340CFM    ****  346/1.5 correction = 230hp for you HP dudes, a
little high in my estimation, your IC efficiency is probably dragging that
down some.

in the grand scheme of things, my corrected assumption for VE is pretty minor
when I review the posts.  The 1.5 rule does show that the efficiencies of the
10v vs the 20v are pretty good.  However, none of this really changes things,
in fact, the obvious questions just become more blatent, the charge air temps
(thankfully) are for PR vs DR not flow, so they don't change.  Neither does
the really big question, how can a hybrid "RSR turbo, a Ned Ritchie turbo,
blow the RS2 into the weeds -claim above" when the PR is lower on the RS2 for
the same flow?   That defies physics and logic and accepted theory and work
output by the best of the gurus (baseline: I'm full of crap:).

I wholy accept that the 1325 EGT is darn good.  And really don't argue with
the fuel part for now, minor issue in the grand scheme o' things.

What really creates a question, is how a claim of 2.79bar can be had with a
stock spring on a 2.5bar PT computer box.  That leaves only 1 of 4 answers:
 A) 26psi is guage error (but I have the same calibrated aircraft guage that
S.O.c. does, and it can predict weather fronts, it's that good, and if it
were failing, the pressure would read low, not high), discount that.  2)
 There is external boost control, not the claim, read above, or 3) The PT
voltage is divided by a % so the computer is interpreting a lower PR.  I
discount this for a variety of reasons, the most obvious being, you don't
need a 2.5PT to do that, a 2.0PT can do the same thing.  4)  There isn't
2.79PR, is the claim, read above.

By definition, it has to be one of the above.  Why?  Cuz you can't program
WGFV control over the measure limit of the PT, Period.   We also know that a
2.79PR means that there is no overboost protection using the same logic
above, which means above 2.50PR the PT goes to a constant voltage (actualy
it's usually less than that, but BOD).  So what stops the motor from going to
3bar?  We now know by the claim, it's not the Baseline Spring Pressure in the
WG.   So, what do we assume?  Eric?  Randall?  Anyone?  Mysterious, magic,
logic, claims vs math and physics now comes to a head.  It's HOLLOWEEN.
 Trick or treat?

Questions still remain, numbers (revised by request) still don't add up.  


Scott Justusson, RS2 wannabe
'87 5ktqwRS2
'87 5ktq
'86 5ktqw
'84 Urq

P.S.
From:  Mr. Science aka, Mr. Turbo
To:  Dave Lawson 1/2 of Dave, Dave et. al.
RE:  Homework

Thanks for the work, as always Dave.
I believe using the 72% IC number for now might be more correct.  I also
might remind a few (you know who you are) that a stock IC in single pass
configuration won't flow 482CFM.  That creates heat> reduces PR and DR.

Dave, please correct for 80% VE which seems to confirm itself with the 1.5 ck
benchmark.  Also, unless you live in Death Valley 14.7 is usually not used
for Po.  As a general rule, minor variations in altitude (and storm fronts)
here in the midwest might dictate that 14.5 is a more accurate and accepted
number.