[Author Prev][Author Next][Thread Prev][Thread Next][Author Index][Thread Index]

*To*: quattro@coimbra.ans.net*Subject*: In search of an MC baseline*From*: DeWitt Harrison <de@aztek-eng.com>*Date*: Fri, 21 Nov 97 12:09:33 MST*Cc*: scottmo@teleport.com*Sender*: owner-quattro@coimbra.ans.net

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

- Prev by Author:
**Re: Anybody ever hit a sheep?** - Next by Author:
**Fuel pump problems - continued** - Prev by thread:
**Re: Franco Sprocket - eprom burners?** - Next by thread:
**Slick 50 & friends** - Index(es):