# In search of an MC baseline

```Clean copy of SAE 860103 in hand. (This is a paper written by
D. Stock of Audi, Neckarsulm, to do a little engineering
bragging about the MC engine.) Here is a second take on the
numbers looking at only a single operating point. That is WOT,
5500 rpm, 5th gear. I gleaned the following info.

Just to verify which engine we are discussing
Displacement      =  2.226 L
CR                     =  7.8:1
bhp @ 5500 rpm  = 162
IC = two pass
turbo = K26-2664-GA 6.95

Performance data provided*

Pio = 1.42 bar  (IC outlet)
Pco - Pio  = deltaPi = 0.14 bar   (Figure 6)
P0 = 0.981 bar  (*estimate only; a typical KKK map P0)

Tio = 38C = 311 K  (Figure 6)
Tco - Tio = deltaTi = 50C = 50 K (Figure 7
T0 = 20C = 293 K

Vehicle velocity approximately 135 mph (217 Km/h)

Calculations:

Tco = Tio + deltaTi = 38C + 50C = 88C = 361 K

IC efficiency:   Eic =  deltaTi / (Tco-T0) = 50 / (361 - 311)

Eic  =  50 / 68 = 0.735

73.5% doesn't seem like a bad IC efficiency but bear in mind
the vehicle velocity of 135 mph and the excellent location of
the IC.  It doesn't take much sniffing around the Spearco
catalog to realize that compared to this two pass, a decent IC
in the same environment would get you into the 90s and
shave quite a few mbar off the deltaPi.

Pco = Pio + deltaPi = 1.42 + 0.14 = 1.56 bar

Pressure ratio:     PR  =  Pco / P0 = 1.56 / 0.981 = 1.59

Turbo efficiency:    Et = [T0 (PR^0.28  -1)] / (Tco - T0)

Et = [293(1.59^0.28 - 1)] / (361 - 293)

Et = 293(0.139) / 68  =  0.60

This is the part that strikes me as a bit odd.  Looking at a
K26 map, it's hard to see how the efficiency could be less
than 70% at this PR and flow level (somewhere around 0.1 m^3/s).
One possibility is that unaccounted for pressure drops
significantly increase the operating PR.  Working backward
from Et = 70%,

PR  =  [1 + {Et(Tco - T0) / T0}]^(1/.28)   =  [1 + 0.1625]^3.57

PR  =  1.71   This is no problem for the K26 which
appears to be pretty much loafing in this system.

But to get this PR, we would need a smaller P0 and/or a higher
Pco.  If we assigned most of the pressure loss to the German
carburetor CIS intake garbage and intake filter, instead of the throttle
body and the other usual suspects, the effective P0, P0e, would be

P0e  =  Pco / PR  =  (1.42 + 0.14) / 1.71  =  0.912  bar

The pressure loss = P0 - P0e = 0.981 - 0.912   =  0.069 bar
or about 1.0 psi.  I think this is reasonable considering the
2.0 psi drop across the IC.  If the turbo was actually even
more efficient than 70% (could go to 75% here), the
pressure losses could be even greater than 3 psi total.

It would be a fun exercise to see what the system would look
like after you eliminate 2 of the 3 or so psi in losses and up
the IC efficiency to 95% @ 135 mph.  (Note: the air
velocity through the IC is _much_ lower than vehicle air
speed and can only be determined by diect measurements.)
For example, assume Et = 70% and compute the new PR
that would get you the same 38C intake manifold air temp.

DeWitt Harrison    de@aztek-eng.com
Boulder, CO
88 5kcstq

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